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ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _________________ SR. NO. PRACTICAL EXPERIMENTS DATE SIGN. 1 Install process and control instrumentation Simulating and Analyzing softwares and familiarize with the system requirements and essential features/specifications of the software in use. 2 Develop mathematical model for a single process parameter of a tank system. 3 Simulate a simple feedback loop for process parameter and obtain output varying set point. 4 Determine poles and zeros of given first order transfer function. 5 Obtain poles and zeros of given second order transfer function. 6 Simulate P -control action on a given system for given step/ramp input and set point. Obtain the effect on output varying Kp of the system. 7 Simulate I -control action on a given system for given step/ramp input and set point. Obtain the effect on output varying Ki of the system. 8 Simulate P+I -control action on a given system for given step/ramp input and set point. Obtain the effect on output varying Kp and Ki of the system. 9 Simulate P+D -control action on a given system for given step/ramp input and set point. Obtain the effect on output varying Kp and Kd of the system. 10 Simulate P+I+D -control action on a given system for given step/ramp input and set point. Obtain the effect on output varying Kp, Kd and Ki of the system. 11 Simulate a simple feed forward loop for process parameter and obtain output varying set point. 12 Simulate a simple cascade loop for process parameter 13 Simulate a simple ratio control for process parameter and obtain output varying set point. 14 Simulate non interacting level system using process control simulator. 15 Simulate interacting level system using process control simulator. 16 Simulate On-Off LEVEL control system using process control simulator.

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Page 1: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

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INDEX

Subject Code ISP (3341705) Enrol No _________________

SR NO

PRACTICAL EXPERIMENTS

DATE

SIGN

1 Install process and control instrumentation Simulating and

Analyzing softwares and familiarize with the system

requirements and essential featuresspecifications of the

software in use

2 Develop mathematical model for a single process parameter of a tank system

3 Simulate a simple feedback loop for process parameter and obtain output varying set point

4 Determine poles and zeros of given first order transfer function

5 Obtain poles and zeros of given second order transfer function

6 Simulate P -control action on a given system for given stepramp input and set point Obtain the effect on output varying Kp of the system

7 Simulate I -control action on a given system for given stepramp input and set point Obtain the effect on output varying Ki of the system

8 Simulate P+I -control action on a given system for given stepramp input and set point Obtain the effect on output varying Kp and Ki of the system

9 Simulate P+D -control action on a given system for given stepramp input and set point Obtain the effect on output varying Kp and Kd of the system

10 Simulate P+I+D -control action on a given system for given stepramp input and set point Obtain the effect on output varying Kp Kd and Ki of the system

11 Simulate a simple feed forward loop for process parameter and obtain output varying set point

12 Simulate a simple cascade loop for process parameter

13 Simulate a simple ratio control for process parameter and obtain output varying set point

14 Simulate non interacting level system using process control simulator

15 Simulate interacting level system using process control simulator

16 Simulate On-Off LEVEL control system using process control simulator

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17 To convert 10-degree celsius into fahrenheit using LABVIEW

18 Simulate three element boiler control system

19 Practice exercise-1

20 Practice exercise-2

21 Practice exercise-3

22 Practice exercise-4

23 Practice exercise-5

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PRACTICAL-1 Aim Install process and control instrumentation Simulating and Analysing software and

familiarize with the system requirements and essential featuresspecifications of the

software in use

SOFTWARE REQUIRED

bull MATLAB R2012a

INSTALLING PROCEDURE

Step 1 Start the Installer Step 2 Choose Whether to Install Using the Internet Step 3 Review the Software License Agreement Step 4 Log In to Your MathWorks Account Step 5 Select the License You Want to Install Step 6 Choose the Installation Type Step 7 Specify the Installation Folder Step 8 Specify Products to Install (Custom Only) Step 9 Specify Installation Options (Custom Only) Step 10 Confirm Your Choices Step 11 Complete the Installation

INTRODUCTION MATLAB which stands for MATrix LABoratory is a technical computing environment for high-performance numeric computation and visualization SIMULINK is a part of MATLAB that can be used to simulate dynamic systems Starting and Quitting MATLAB To start MATLAB click on the MATLAB icon or type in MATLAB followed by pressing the enter The screen will produce the MATLAB prompt gtgt which indicates that MATLAB is waiting for a command to be entered In order to quit MATLAB type quit or exit after the prompt followed by pressing the enter or return key Display Windows MATLAB has three display windows They are 1 A Command Window which is used to enter commands 2 A Graphics Window which is used to display plots and graphs 3 An Edit Window which is used to create and modify M-files M-files are files that contain a program or script of MATLAB commands Entering Commands MATLAB commands are case sensitive and lower case letters are used throughout To execute an M-file (such as Project_1m) simply enter the name of the file without its extension (as in Project_1)

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The Semicolon () If a semicolon () is typed at the end of a command the output of the command is not displayed Typing When percent symbol () is typed in the beginning of a line the line is designated as a comment When the enter key is pressed the line is not executed The clc Command Typing clc command and pressing enter cleans the command window Once the clc command is executed a clear window is displayed Help MATLAB has a host of built-in functions For a complete list refer to MATLAB userrsquos guide or refer to the on line Help Statements and Variables Statements have the form gtgt variable = expression The equals (ldquo=rdquo) sign implies the assignment of the expression to the variable

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MATLAB BASICS 1 Start Matlab and then the Simulink environment by typing simulink to the matlab prompter

2 Open a new Simulink model window from File 1048774 New 1048774 Model

3 You can construct your block diagram by drag-and-dropping the appropriate blocks from the main Simulink windows

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4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

Faculty Sign

Grade

Date

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PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

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PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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Grade

Date

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

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Date

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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Conclusion

Question bank

1)

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Grade

Date

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PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 2: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 2

17 To convert 10-degree celsius into fahrenheit using LABVIEW

18 Simulate three element boiler control system

19 Practice exercise-1

20 Practice exercise-2

21 Practice exercise-3

22 Practice exercise-4

23 Practice exercise-5

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PRACTICAL-1 Aim Install process and control instrumentation Simulating and Analysing software and

familiarize with the system requirements and essential featuresspecifications of the

software in use

SOFTWARE REQUIRED

bull MATLAB R2012a

INSTALLING PROCEDURE

Step 1 Start the Installer Step 2 Choose Whether to Install Using the Internet Step 3 Review the Software License Agreement Step 4 Log In to Your MathWorks Account Step 5 Select the License You Want to Install Step 6 Choose the Installation Type Step 7 Specify the Installation Folder Step 8 Specify Products to Install (Custom Only) Step 9 Specify Installation Options (Custom Only) Step 10 Confirm Your Choices Step 11 Complete the Installation

INTRODUCTION MATLAB which stands for MATrix LABoratory is a technical computing environment for high-performance numeric computation and visualization SIMULINK is a part of MATLAB that can be used to simulate dynamic systems Starting and Quitting MATLAB To start MATLAB click on the MATLAB icon or type in MATLAB followed by pressing the enter The screen will produce the MATLAB prompt gtgt which indicates that MATLAB is waiting for a command to be entered In order to quit MATLAB type quit or exit after the prompt followed by pressing the enter or return key Display Windows MATLAB has three display windows They are 1 A Command Window which is used to enter commands 2 A Graphics Window which is used to display plots and graphs 3 An Edit Window which is used to create and modify M-files M-files are files that contain a program or script of MATLAB commands Entering Commands MATLAB commands are case sensitive and lower case letters are used throughout To execute an M-file (such as Project_1m) simply enter the name of the file without its extension (as in Project_1)

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The Semicolon () If a semicolon () is typed at the end of a command the output of the command is not displayed Typing When percent symbol () is typed in the beginning of a line the line is designated as a comment When the enter key is pressed the line is not executed The clc Command Typing clc command and pressing enter cleans the command window Once the clc command is executed a clear window is displayed Help MATLAB has a host of built-in functions For a complete list refer to MATLAB userrsquos guide or refer to the on line Help Statements and Variables Statements have the form gtgt variable = expression The equals (ldquo=rdquo) sign implies the assignment of the expression to the variable

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MATLAB BASICS 1 Start Matlab and then the Simulink environment by typing simulink to the matlab prompter

2 Open a new Simulink model window from File 1048774 New 1048774 Model

3 You can construct your block diagram by drag-and-dropping the appropriate blocks from the main Simulink windows

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4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 8

PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

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PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

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Date

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

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Date

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PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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Conclusion

Question bank

1)

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Date

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PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 3: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 3

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PRACTICAL-1 Aim Install process and control instrumentation Simulating and Analysing software and

familiarize with the system requirements and essential featuresspecifications of the

software in use

SOFTWARE REQUIRED

bull MATLAB R2012a

INSTALLING PROCEDURE

Step 1 Start the Installer Step 2 Choose Whether to Install Using the Internet Step 3 Review the Software License Agreement Step 4 Log In to Your MathWorks Account Step 5 Select the License You Want to Install Step 6 Choose the Installation Type Step 7 Specify the Installation Folder Step 8 Specify Products to Install (Custom Only) Step 9 Specify Installation Options (Custom Only) Step 10 Confirm Your Choices Step 11 Complete the Installation

INTRODUCTION MATLAB which stands for MATrix LABoratory is a technical computing environment for high-performance numeric computation and visualization SIMULINK is a part of MATLAB that can be used to simulate dynamic systems Starting and Quitting MATLAB To start MATLAB click on the MATLAB icon or type in MATLAB followed by pressing the enter The screen will produce the MATLAB prompt gtgt which indicates that MATLAB is waiting for a command to be entered In order to quit MATLAB type quit or exit after the prompt followed by pressing the enter or return key Display Windows MATLAB has three display windows They are 1 A Command Window which is used to enter commands 2 A Graphics Window which is used to display plots and graphs 3 An Edit Window which is used to create and modify M-files M-files are files that contain a program or script of MATLAB commands Entering Commands MATLAB commands are case sensitive and lower case letters are used throughout To execute an M-file (such as Project_1m) simply enter the name of the file without its extension (as in Project_1)

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The Semicolon () If a semicolon () is typed at the end of a command the output of the command is not displayed Typing When percent symbol () is typed in the beginning of a line the line is designated as a comment When the enter key is pressed the line is not executed The clc Command Typing clc command and pressing enter cleans the command window Once the clc command is executed a clear window is displayed Help MATLAB has a host of built-in functions For a complete list refer to MATLAB userrsquos guide or refer to the on line Help Statements and Variables Statements have the form gtgt variable = expression The equals (ldquo=rdquo) sign implies the assignment of the expression to the variable

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MATLAB BASICS 1 Start Matlab and then the Simulink environment by typing simulink to the matlab prompter

2 Open a new Simulink model window from File 1048774 New 1048774 Model

3 You can construct your block diagram by drag-and-dropping the appropriate blocks from the main Simulink windows

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4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 8

PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

ISP(3341705) IC DEPT

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

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GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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Grade

Date

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GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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Grade

Date

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GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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Grade

Date

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GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

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Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

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PRACTICAL-1 Aim Install process and control instrumentation Simulating and Analysing software and

familiarize with the system requirements and essential featuresspecifications of the

software in use

SOFTWARE REQUIRED

bull MATLAB R2012a

INSTALLING PROCEDURE

Step 1 Start the Installer Step 2 Choose Whether to Install Using the Internet Step 3 Review the Software License Agreement Step 4 Log In to Your MathWorks Account Step 5 Select the License You Want to Install Step 6 Choose the Installation Type Step 7 Specify the Installation Folder Step 8 Specify Products to Install (Custom Only) Step 9 Specify Installation Options (Custom Only) Step 10 Confirm Your Choices Step 11 Complete the Installation

INTRODUCTION MATLAB which stands for MATrix LABoratory is a technical computing environment for high-performance numeric computation and visualization SIMULINK is a part of MATLAB that can be used to simulate dynamic systems Starting and Quitting MATLAB To start MATLAB click on the MATLAB icon or type in MATLAB followed by pressing the enter The screen will produce the MATLAB prompt gtgt which indicates that MATLAB is waiting for a command to be entered In order to quit MATLAB type quit or exit after the prompt followed by pressing the enter or return key Display Windows MATLAB has three display windows They are 1 A Command Window which is used to enter commands 2 A Graphics Window which is used to display plots and graphs 3 An Edit Window which is used to create and modify M-files M-files are files that contain a program or script of MATLAB commands Entering Commands MATLAB commands are case sensitive and lower case letters are used throughout To execute an M-file (such as Project_1m) simply enter the name of the file without its extension (as in Project_1)

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The Semicolon () If a semicolon () is typed at the end of a command the output of the command is not displayed Typing When percent symbol () is typed in the beginning of a line the line is designated as a comment When the enter key is pressed the line is not executed The clc Command Typing clc command and pressing enter cleans the command window Once the clc command is executed a clear window is displayed Help MATLAB has a host of built-in functions For a complete list refer to MATLAB userrsquos guide or refer to the on line Help Statements and Variables Statements have the form gtgt variable = expression The equals (ldquo=rdquo) sign implies the assignment of the expression to the variable

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MATLAB BASICS 1 Start Matlab and then the Simulink environment by typing simulink to the matlab prompter

2 Open a new Simulink model window from File 1048774 New 1048774 Model

3 You can construct your block diagram by drag-and-dropping the appropriate blocks from the main Simulink windows

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4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

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PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

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PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

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Grade

Date

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 5: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 5

The Semicolon () If a semicolon () is typed at the end of a command the output of the command is not displayed Typing When percent symbol () is typed in the beginning of a line the line is designated as a comment When the enter key is pressed the line is not executed The clc Command Typing clc command and pressing enter cleans the command window Once the clc command is executed a clear window is displayed Help MATLAB has a host of built-in functions For a complete list refer to MATLAB userrsquos guide or refer to the on line Help Statements and Variables Statements have the form gtgt variable = expression The equals (ldquo=rdquo) sign implies the assignment of the expression to the variable

ISP(3341705) IC DEPT

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MATLAB BASICS 1 Start Matlab and then the Simulink environment by typing simulink to the matlab prompter

2 Open a new Simulink model window from File 1048774 New 1048774 Model

3 You can construct your block diagram by drag-and-dropping the appropriate blocks from the main Simulink windows

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GPGANDHINAGAR Page 7

4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 8

PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

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PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

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Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 6: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 6

MATLAB BASICS 1 Start Matlab and then the Simulink environment by typing simulink to the matlab prompter

2 Open a new Simulink model window from File 1048774 New 1048774 Model

3 You can construct your block diagram by drag-and-dropping the appropriate blocks from the main Simulink windows

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GPGANDHINAGAR Page 7

4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

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PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

ISP(3341705) IC DEPT

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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GPGANDHINAGAR Page 12

models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

ISP(3341705) IC DEPT

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 15

Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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GPGANDHINAGAR Page 16

Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 23

OUTPUT

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

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4 Start the simulation You can see the graph on scope screen

ESSENTIAL FEATURESSPECIFICATIONS OF THE SOFTWARE

bull High-level language for numerical computation visualization and application development

bull Interactive environment for iterative exploration design and problem solving

bull Mathematical functions for linear algebra statistics Fourier analysis filtering optimization numerical integration and solving ordinary differential equations

bull Built-in graphics for visualizing data and tools for creating custom plots

bull Development tools for improving code quality and maintainability and maximizing performance

bull Tools for building applications with custom graphical interfaces

Conclusion From this practical we can learn installing procedure of process and control instrumentation Simulating and Analyzing software and its features

Question bank

1) Give the full form of Matlab

2) Give the name of Simulating and Analysing softwares

3) which indicates that MATLAB is waiting for a command to be entered a) gtgt b) clc c) d) ^ 4) Which command is used for clear screen 5) Matlab has Built-in graphics for visualizing data and tools for creating custom plots True or False

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PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

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PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

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Grade

Date

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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Date

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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Date

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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Date

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 8: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 8

PRACTICAL 2

Aim Develop mathematical model for a single process parameter of a tank system

Tank system Parameters

Height of tank

024m Maximum

fill level 02m

Radius of

tank0105m

Tank capacity volume 00069237m3

Figure 1 Fig shows process model Figure 2 Fig shows level sensor probe

Liquid

inflow qi(t)

Liquid

outflow

qo(t) Height

outflow h

Area A

Rate of change of volume = inflow -outflow

119889119907

119889119905= 119902119894(119905) minus 1199020(119905)

Now volume = area x height

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GPGANDHINAGAR Page 9

119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

ISP(3341705) IC DEPT

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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GPGANDHINAGAR Page 11

The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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GPGANDHINAGAR Page 12

models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 13

Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 15

Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 16

Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 9: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 9

119889119860ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

119860119889ℎ

119889119905= 119902119894(119905) minus 1199020(119905)

Two assumptions are made here

1 Cross sectional area is constant 2 Output flow is proportional to height

1199020 =ℎ

119877

119860119889ℎ

119889119905= 119902119894(119905) minus

119877

Where R=restriction

Taking Laplace transform on both the side of equation we get

119860119904ℎ(119904) = 119902119894(119904) minusℎ(119904)

119877

119860119877119904ℎ(119904) = 119877119902119894(119904)ℎ(119904)

Now taking h(s) as common we get

ℎ(119904)(119860119877119904 + 1) = 119877119902119894(119904)ℎ(119904)

ℎ(119904)

119902119894(119904)=

119877

119860119877119904 + 1

Now k = R t = A R

R = k = 1000secm2

t = A R = 4 140 = 335195 min

Now transfer function become

119877

119860119877119904 + 1=

1000

335195119904 + 1

Conclusion From this practical we can learn how to find the process model for the given process control system

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PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 23

OUTPUT

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 10: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 10

PRACTICAL-3 Aim Simulate a simple feedback loop for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

To derive a first-order-plus-deadtime model of the heat exchanger characteristics inject a step disturbance in valve voltage V and record the effect on the tank temperature T over time The measured response in normalized units is shown below

heatex_plotdata

title(Measured response to step change in steam valve voltage)

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GPGANDHINAGAR Page 11

The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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GPGANDHINAGAR Page 12

models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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GPGANDHINAGAR Page 13

Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

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GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 11: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 11

The values t1 and t2 are the times where the response attains 283 and 632 of its final value You can use these values

to estimate the time constant tau and dead time theta for the heat exchanger

t1 = 218 t2 = 360

tau = 32 ( t2 - t1 )

theta = t2 - tau

tau = 213000

theta = 147000

Verify these calculations by comparing the first-order-plus-deadtime response with the measured response

s = tf(s)

Gp = exp(-thetas)(1+taus)

Gp =

1

exp(-147s) ----------

213 s + 1

Continuous-time transfer function

hold on step(Gp) hold off

title(Experimental vs simulated response to step change)

Feedback Control

The transfer function

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models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 23

OUTPUT

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 12: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 12

models how a change in the voltage V driving the steam valve opening affects the tank temperature T while the transfer

function

models how a change d in inflow temperature affects T To regulate the tank temperature T around a given setpoint Tsp

we can use the following feedback architecture to control the valve opening (voltage V)

Figure Feedback Control

In this configuration the proportional-integral (PI) controller

Kc = 0859 (theta tau)^(-0977)

tauc = ( tau 0674 ) ( theta tau )^0680

C = Kc (1 + 1(taucs))

Kc = 12341

tauc = 245582

the feedback loop and simulate the response to a set point change

Tfb = feedback((GpC)1)

step(Tfb) grid on

title(Response to step change in temperature setpoint T_sp)

ylabel(Tank temperature)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 13

Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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GPGANDHINAGAR Page 15

Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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GPGANDHINAGAR Page 16

Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

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PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

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Question bank

1) Draw a simple close loop and open loop

2) Give the function of hold on and hold off instruction

3) Write instruction for feedback loop 4) Which controller is used for above feedback loop 5) Write instruction for axis label

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PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 14: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 14

PRACTICAL-4 Aim Determine poles and zeros of given first order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Response of First Order Transfer Functions

Laplace Domain

gtgtnum=[0 1]

gtgt den=[10 1]

gtgt s=tf(numden)

s = 1

--------

10 s + 1

Continuous-time transfer function

gtgt pzmap(s)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

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Unit step

num=[1 2]

den=[5 2 0]

s=tf(numden)

s = s + 2

-----------

5 s^2 + 2 s

Continuous-time transfer function

gtgt pzmap(s)

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Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

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4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 16: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 16

Unit ramp

num=[1 2]

den=[5 5 0 0]

s=tf(numden)

s = s + 2

-------------

5 s^3 + 5 s^2

Continuous-time transfer function

gtgt pzmap(s)

Conclusion From this practical we can determine the pole and zero on pole zero map in

matlab

Question bank

1) What is TF

2) Give the equation of First Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 17: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 17

4) Which instruction is used to denote poles and zeros in MATLAB

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 18

PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

ISP(3341705) IC DEPT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

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GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 18: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

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PRACTICAL-5 Aim Obtain poles and zeros of given second order transfer function

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Transfer function of second order

Where

k the gain of the system

ζ the damping ratio of the system

ωn the (undamped) natural frequency of the system

Three cases

ζ = 1 critically damped case

ζ gt 1 overdamped case

0 lt ζ lt 1 underdamped case

Transfer function=1( s^2 + 12 s + 1)

gtgtnum=[0 1]

gtgtden=[1 12 1]

gtgts=tf(numden)

s = 1

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--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

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Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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Date

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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Grade

Date

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

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Date

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

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Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 19: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 19

--------------

s^2 + 12 s + 1

Continuous-time transfer function

gtgt pzmap(s)

Transfer function=1( s^2 + 08 s + 1)

gtgtnum=[1]

gtgtden=[1 08 1]

gtgt g=tf(numden)

g = 1

---------------

s^2 + 08 s + 1

gtgtpzmap(g)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 20: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 20

Conclusion From this practical we can determine the pole and zero of second order transfer

function on pole zero map in matlab

Question bank

1) What is TF

2) Give the equation of second Order Transfer Function

3) Which instruction is used to denote Transfer Function in Matlab

4) Which instruction is used to denote poles and zeros in Matlab

5) Poles and zeros are indicated with which sign in map

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 21

PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 21: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

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PRACTICAL-6 Aim Simulate P -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

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Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 22: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 22

Proportional Control

From the table shown above we see that the proportional controller (Kp) reduces the rise time increases the overshoot

and reduces the steady-state error

The closed-loop transfer function of the above system with a proportional controller is

Let the proportional gain ( ) equal 300 and change the m-file to the following

INPUT

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OUTPUT

INPUT

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OUTPUT

For ramp signalkp=300

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 23: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 23

OUTPUT

INPUT

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 24: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

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GPGANDHINAGAR Page 24

OUTPUT

For ramp signalkp=300

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GPGANDHINAGAR Page 25

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

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PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

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PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

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PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 25: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

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bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

Kp=200

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Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

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PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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For step signalki=70

INPUT

OUTPUT

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For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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For ramp signalki=100

INPUT

OUTPUT

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Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

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PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

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PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

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PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

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PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

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PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

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PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

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PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

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PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 26: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 26

Question bank

1) Give the name of continuous controller

2) what is value of step function if tgt0

3) what is kp

4) from above step outputs derive steady state values

5) which instruction is used for p-controller

Faculty sign

Date

Grade

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 27: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 27

PRACTICAL-7 Aim Simulate I -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open Matlab

bull Open Matlab simulink library

bull Add blocks from simulink library to untitled window

bull Press double click to change the block parameter

bull Start simulation and observe graph

system

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 28: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 28

For step signalki=70

INPUT

OUTPUT

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GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 29: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 29

For step signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

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Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 30: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 30

For ramp signalki=70

INPUT

bull

OUTPUT

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GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

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GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 31: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 31

For ramp signalki=100

INPUT

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 32: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 32

Question bank

1) Which system is used in above practical

2) Which to signals are used in above practical

3) What is ki

4) How can we change the value of block

5) Write the TFof above system

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 33: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 33

PRACTICAL-8 Aim Simulate P+I -control action on a given system for given stepramp input and set point Obtain

the effect on output varying Kp and Ki of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

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GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

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GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

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GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

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GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

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GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

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Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 34: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 34

Proportional-Integral Control

Before going into a PID control lets take a look at a PI control From the table we see that an

integral controller (Ki) decreases the rise time increases both the overshoot and the settling

time and eliminates the steady-state error For the given system the closed-loop transfer

function with a PI control is

Lets reduce the to 30 and let equal 70 Create an new m-file and enter the following

commands

Kp = 30

Ki = 70

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

with Kp = 30 Ki = 70

Continuous-time PI controller in parallel form

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 35: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 35

Continuous-time transfer function

Run this m-file in the MATLAB command window and you should get the following plot

We have reduced the proportional gain (Kp) because the integral controller also reduces the

rise time and increases the overshoot as the proportional controller does (double effect) The

above response shows that the integral controller eliminated the steady-state error

For kp=40ki=80

Kp = 40

Ki = 80

C = pid(KpKi)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki ---

s

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

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GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 36: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 36

with Kp = 40 Ki = 80

Continuous-time PI controller in parallel form

T =

40 s + 80

------------------------

s^3 + 10 s^2 + 60 s + 80

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 37: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 37

For ramp signalKp=30 Ki=70

INPUT

OUTPUT

For ramp signalKp=40 Ki=80

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GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 38: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 38

INPUT

OUTPUT

Question bank

1) Integral controller (Ki) eliminates the steady-state error TRUE or FALSE

2) what is value of unit step function if tlt0

3) What is effect on peak values if we are changing kpki

4) Integral controller (Ki) decreases the rise timeTRUE or FALSE

5) Which instruction is used for p-i controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 39: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 39

PRACTICAL-9 Aim Simulate P+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp and Kd of the system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

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Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 40: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 40

Proportional-Derivative Control

Now lets take a look at a PD control From the table shown above we see that the derivative controller (Kd) reduces

both the overshoot and the settling time The closed-loop transfer function of the given system with a PD controller is

(8)

Let equal 300 as before and let equal 10 Enter the following commands into an m-file and run it in the MATLAB

command window

Kp = 300

Kd = 10

C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

with Kp = 300 Kd = 10

Continuous-time PD controller in parallel form

T = 10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function

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This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 41: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 41

This plot shows that the derivative controller reduced both the overshoot and the settling time and had a small effect on

the rise time and the steady-state error

with Kp = 320 Kd = 20

Kp = 320

gtgt Kd =20

gtgt C = pid(Kp0Kd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

Kp + Kd s

Continuous-time PD controller in parallel form

T = 20 s + 320

----------------

s^2 + 30 s + 340

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Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 42: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 42

Continuous-time transfer function

Question bank

1) Give the type of pd controller

2) what is value of step function if tlt0

3) What are peak values if we are changing kpkd

4) Obtain the output of pd when we are using ramp as input signal

5) which instruction is used to draw p-d controller

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

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Date

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PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

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GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

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GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

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Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

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Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

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PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

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Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

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PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

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2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 43: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 43

PRACTICAL-10 Aim Simulate P+I+D -control action on a given system for given stepramp input and set point

Obtain the effect on output varying Kp Kd and Ki of the system

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

PROGRAM

Suppose we have a simple mass spring and damper problem

The modeling equation of this system is

Taking the Laplace transform of the modeling equation we get

The transfer function between the displacement and the input then becomes

Let

M = 1 kg b = 10 N sm k = 20 Nm F = 1 N

Plug these values into the above transfer function

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Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

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General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 44: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 44

Proportional-Integral-Derivative Control

Now lets take a look at a PID controller The closed-loop transfer function of the given system

with a PID controller is

After several trial and error runs the gains = 350 = 300 and = 50 provided the

desired response To confirm enter the following commands to an m-file and run it in the

command window You should get the following step response

Kp = 350

Ki = 300

Kd = 50

C = pid(KpKiKd)

T = feedback(CP1)

t = 00012

step(Tt)

C =

1

Kp + Ki --- + Kd s

s

with Kp = 350 Ki = 300 Kd = 50

Continuous-time PID controller in parallel form

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 45: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 45

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system follow the steps shown below to

obtain a desired response

1 Obtain an open-loop response and determine what needs to be improved

2 Add a proportional control to improve the rise time

3 Add a derivative control to improve the overshoot

4 Add an integral control to eliminate the steady-state error

5 Adjust each of Kp Ki and Kd until you obtain a desired overall response

Question bank

1) Give the equation of pid controller from above process

2) Obtain the output of pid when kp=200kd=100ki=25

3) Which instruction is used for pid controller

4) Obtain the output of pid when we are using ramp as input signal

5) Which type of pid controller is used

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 46: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 46

PRACTICAL-11 Aim Simulate a simple feed forward loop for process parameter and obtain output varying

set point

SOFTWARE REQUIRED

bull MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

This example shows how to design feedforward compensators to regulate the temperature

of a chemical reactor through a heat exchanger

Heat Exchanger Process

A chemical reactor called stirring tank is depicted below The top inlet delivers liquid to

be mixed in the tank The tank liquid must be maintained at a constant temperature by

varying the amount of steam supplied to the heat exchanger (bottom pipe) via its control

valve Variations in the temperature of the inlet flow are the main source of disturbances

in this process

Figure 1 Stirring Reactor with Heat Exchanger

Interactive Simulation

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

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GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

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GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 47: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 47

To gain additional insight and interactively tune the feedforward use the companion GUI and

Simulink model

gtgtheatex

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 48: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 48

Conclusion From the above practical we can Simulate a simple feed forward loop for

process parameter and obtain output varying set point

Question bank

1) What do you mean by feed forward control

2) What indicates dTv

3) Which instruction is used for feed forward process

4) What indicates C Gd Gp

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 49: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 49

PRACTICAL-12 Aim Simulate a simple cascade loop for process parameter

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

Cascade control is mainly used to achieve fast rejection of disturbance before it

propagates to the other parts of the plant The simplest cascade control system involves

two control loops (inner and outer) as shown in the block diagram below

Controller C1 in the outer loop is the primary controller that regulates the primary

controlled variable y1 by setting the set-point of the inner loop Controller C2 in the inner

loop is the secondary controller that rejects disturbance d2 locally before it propagates

to P1 For a cascade control system to function properly the inner loop must respond

much faster than the outer loop

In this example you will design a single loop control system with a PI controller and a

cascade control system with two PI controllers The responses of the two control systems

are compared for both reference tracking and disturbance rejection

Plant

In this example the inner loop plant P2 is

The outer loop plant P1 is

P2 = zpk([]-23)

P1 = zpk([][-1 -1 -1]10)

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GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

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GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

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GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

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GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

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Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

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GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 50: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 50

Designing a Single Loop Control System with a PI Controller

Use pidtune command to design a PI controller in standard form for the whole plant

model P = P1 P2

The desired open loop bandwidth is 02 rads which roughly corresponds to the response

time of 10 seconds

The plant model is P = P1P2

P = P1P2

Use a PID or PIDSTD object to define the desired controller structure

C = pidstd(11)

Tune PI controller for target bandwidth is 02 rads

C = pidtune(PC02)

C

C =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 00119 Ti = 0849

Continuous-time PI controller in standard form

Designing a Cascade Control System with Two PI Controllers

The best practice is to design the inner loop controller C2 first and then design the outer

loop controller C1 with the inner loop closed In this example the inner loop bandwidth

is selected as 2 rads which is ten times higher than the desired outer loop bandwidth In

order to have an effective cascade control system it is essential that the inner loop

responds much faster than the outer loop

Tune inner-loop controller C2 with open-loop bandwidth at 2 rads

C2 = pidtune(P2pidstd(11)2)

C2

C2 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0244 Ti = 0134

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GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

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GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 51: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 51

Continuous-time PI controller in standard form

Tune outer-loop controller C1 with the same bandwidth as the single loop system

Inner loop system when the control loop is closed first

clsys = feedback(P2C21)

Plant seen by the outer loop controller C1 is clsysP1

C1 = pidtune(clsysP1pidstd(11)02)

C1

C1 =

1 1

Kp (1 + ---- ---)

Ti s

with Kp = 0015 Ti = 0716

Continuous-time PI controller in standard form

Performance Comparison

First plot the step reference tracking responses for both control systems

single loop system for reference tracking

sys1 = feedback(PC1)

set(sys1NameSingle Loop)

cascade system for reference tracking

sys2 = feedback(clsysP1C11)

set(sys2NameCascade)

plot step response

figurestep(sys1rsys2b)

legend(showlocationsoutheast)

title(Reference Tracking)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 52: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 52

Secondly plot the step disturbance rejection responses of d2 for both control systems

single loop system for rejecting d2

sysd1 = feedback(P1P2C)

set(sysd1NameSingle Loop)

cascade system for rejecting d2

sysd2 = P1(1+P2C2+P2P1C1C2)

set(sysd2NameCascade)

plot step response

figurestep(sysd1rsysd2b)

legend(show)

title(Disturbance Rejection)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 53: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 53

From the two response plots you can conclude that the cascade control system performs

much better in rejecting disturbance d2 while the set-point tracking performances are

almost identical

Conclusion From this practical we can simulate cascade loop using matlab

Question bank

1) how many loops are there in cascade control

2) Give the value of P2if P2= P2 = zpk([]-23)

3) Which response is ploted in graph

4) Which controller is used in cascade process

5) Give Continuous-time PI controller in standard form

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

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GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

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Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

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GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

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Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

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Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 54: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 54

PRACTICAL-13 Aim Simulate a simple ratio control for process parameter and obtain output varying set

point

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Procedure

The two variables are usually flow rates a manipulated variable u and a

disturbance variable d Thus the ratio R=du is controlled rather than the

individual variables

A ratio control scheme is to be used to maintain a ratio of H2 and N2 as the feed

to an ammonia synthesis reactor Individual flow controllers will be used for both

the H2 and N2 streams Draw a schematic diagram for the ratio control scheme

and specify the appropriate gain for the ratio station KR

Solution

The equation for the ammonia synthesis reaction is

3H2+N2=2NH3

In order to introduce the feed mixture in proportions the ratio of the molar flow

rates (H2N2) should be 31 For the sake of simplicity we assume that the ratio

of the molar flow rates is equal to the ratio of the volumetric flow rates But in

general the volumetric flow rates also depend on the temperature and pressure of

each stream (the ideal gas law)

The schematic diagram for the ammonia synthesis reaction is shown in Fig The

H2 flow rate is considered to be the disturbance variable although this choice is

because both the H2 and N2 flow rates are controlled Note that the ratio station

is merely a device with an adjustable gain The input signal to the ratio station is

dm the measured H2 flow rate Its output signal usp serves as the set point for

the N2 flow control loop It is calculated as

usp = KR dm

From the equation it follows that the desired ratio is Rd = ud = 13

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 55: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 55

Fig

Block representation in Matlab

OUTPUT

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 56: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 56

Conclusion

Question bank

1) Give the equation of ratio controller

2) Give ammonia synthesis reaction equation

3) Describe the function of mux block

4) Prepare ratio control for water

5) What is dm

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 57: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 57

PRACTICAL-14 Aim Simulate non interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Non-Interacting Liquid Level System

The two tank non-interacting liquid level system is shown in Fig 1 It consists of two

tanks namely tank 1 and tank 2 Various parameters and their values are mentioned in

Table I The outlet flow from the tank 1 discharges directly into tank 2 Moreover liquid

flow through valve R1 depends only on h1 The variation in h2 in tank 2 does not affect

the transient response occurring in tank 1 Hence this type of a system is known as a

non-interacting system By simple mathematical calculation one can get transfer function

of system Let us apply mass balance equation to tank 1 as

helliphelliphelliphelliphelliphellip(1)

Similarly applying mass balance to tank 2 gives

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 58: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 58

helliphelliphelliphelliphelliphellip (2)

The flow-head relationships for the two linear resistances are given by

helliphelliphelliphelliphelliphelliphelliphellip (3)

helliphelliphelliphelliphelliphelliphelliphelliphelliphellip(4)

Combining (1)-(4) transfer functions for tank 1 and tank 2 are obtained as

helliphelliphelliphelliphelliphellip (5)

helliphelliphelliphelliphelliphelliphellip(6)

where τ1= R1 A1 and τ 2= R2 A2 Therefore the overall transfer function is determined

as

helliphelliphelliphelliphellip (7)

Substituting the values from Table I the complete transfer function is calculated as

helliphelliphelliphelliphelliphelliphellip (8)

It should be noted that the variable to be controlled is h2 ie height of tank 2 by

manipulating the flow rate in tank 1 ie q The roots of denominator polynomial are found

to be (-1 -2) Hence the open loop response is observed to be overdamped

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=1(05s^2+15s+1)

Transfer function

1

-------------------

05 s^2 + 15 s + 1

gtgt step(G)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 59: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 59

Perform PID control using matlab Simulink with trial error method

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 60: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 60

PRACTICAL-15 Aim Simulate interacting level system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Introduction to Cascade Control

The controls of liquid level in multiple tanks and flow between the tanks are basic

problems in the process industries The process industries require liquid to be pumped and

stored in the tanks and then pump it to another tank Often the tanks are so coupled

together that the levels interact and these must also be controlled

About 95 of process control loops are of PID or PI type Since its inception over eighty

years ago the proportional-integral-derivative (PID) control structure has remained the

most commonly used single-input single-output (SISO) technique in industry primarily

because it is one of the simplest

Modeling Of Two Tank Interacting Liquid Level System

Consider the process consisting of two interacting liquid tanks in the Figure 1 The

volumetric flow into tank1 is qin (cm3min) the volumetric flow rate from tank1 to tank2

is q1(cm3min) and the volumetric flow rate from tank2 is qo(cm3min) The height of

the liquid level is h1 (cm) in tank1 and h2 in tank2 (cm)

Both tanks have the same cross sectional area denotes the area of tank1 is A1 (cm2) and

area of tank2 is A2(cm2)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 61: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 61

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 62: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 62

Matlab Command

gtgt s=tf(s)

Transfer function

s

gtgt G=001(025s^2+75s+1)

Transfer function

001

--------------------

025 s^2 + 75 s + 1

gtgt step(G)

gtgt

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 63: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 63

Perform PID control using matlab Simulink with following parameters

1 P=20

2 P=20 I=12

3 P=20 D=10

4 P=12 I=4 D=10

Comment on each output above

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 64: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 64

PRACTICAL-16 Aim Simulate On-Off LEVEL control system using process control simulator

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

Output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 65: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 65

Conclusion

Question bank

1)

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 66: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 66

PRACTICAL-17 Aim LabVIEW Program for degree Celsius to Fahrenheit Conversion

Step 1 Open the LabVIEW and click on Blank VI

Step 2 Now you will see two windows on your computer screen One is lsquofront panelrsquo and other is

lsquoblock diagramrsquo To properly adjust both of them on your screen press ldquoCtrl+trdquo from your

keyboard

Step 3 Now you will see both the windows are adjusted on a single screen Move your cursor on

block diagram window and right click on empty space The available function will appear Click

on structures and then click on while loop Now draw the while loop (box) by dragging your

mouse on block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 67: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 67

Step 4 In this first LabVIEW program we are converting temperature readings from degree

Celsius to degree Fahrenheit So in short we are implementing the formula to convert CgtF as

shown below

Step 5 First of all we have to take a slider to vary the Celsius readings To bring a slider right

click on the front panel Available controls will appear Move cursor on ldquoNumericrdquo and click on

ldquovertical fill sliderdquo You can give a name to the slider Here we are using this slider for degree

Celsius readings therefore I have named as ldquoDeg Crdquo In this way we have completed ldquoCrdquo Now

we have to multiply this ldquoCrdquo with ldquo95rdquo as shown in the formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 68: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 68

Step 6 To multiply ldquoCrdquo with ldquo95rdquo we will first multiply it with 9 and then divide it by 5 To

multiply ldquoCrdquo with 9 we have to take ldquomultiplierrdquo Right-click on block diagram gt numeric gt

multiply Same steps will be needed for taking divider and for addition with 32 as from formula

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 69: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 69

Step 7 for taking ldquo9rdquo ie constant right click on block diagram gt numeric gt numeric constant

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 70: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 70

Step 8 Similarly take divider and divide by 5 Take addition and add 32 Now your block

diagram will look like this

Step 9 Now we have shown this converted temperature readings (which are now in Fahrenheit)

on the numeric indicator Right click on front panel gt numeric gt numeric indicator I have given

the name as ldquoDeg Frdquo to this numeric indicator

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 71: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 71

Step 10 Finally we have to create a control for the while loop to stop the program at any time To

create a control right click on red button ie ldquoloop conditionrdquo and select ldquocreate controlrdquo

Step 11 Save the program by pressing ldquoCtrl+Srdquo

Step 12 Click on run button available in the block diagram window

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 72: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 72

Step 13 Now vary the slider you will see a corresponding change in degree Fahrenheit output

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 73: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 73

PRACTICAL-18 Aim Simulate three element boiler control system

SOFTWARE REQUIRED

MATLAB R2012a

PROCEDURE

bull Open MATLAB

bull Open new M-file

bull Type the program

bull Save in current directory

bull Compile and Run the program

bull For the output see command window Figure window

Three element drum level control Three-element level control as shown in Fig1 is the most

common boiler drum level control strategy A feed water flow loop slave is added to the two-

element strategy Three-element level control linearizes the feed water flow with respect to the

steam flow and the level controller output The control loop now requests volumetric flow

change not just a change in the valve position This strategy attempts to compensate for

changes or disturbances in steam flow and feed water flow based on the principle that flow in

equals flow out The installed characteristics of the feed water valve are no longer an issue

because the flow controller can compensate Using this approach the steam feed forward

element can be a simple gain without requiring characterization

The relationship between the feed water flow rate and drum level for the boiler process are

expressed by the following equations The process function valve function and disturbance

function is shown below

Gp(s)= [025(-s+1)] [s(2s+1)]

Gv(s) = 1[015s+1]

Gd(s)=[ -025(-s+1)] [s(s+1)(2s+1)]

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 74: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 74

Following are the process used to determine the PID gain parameter

1) ZieglerndashNichols Method

This method is introduced by John G Ziegler and Nathaniel B Nichols In this method the

Ki and Kd gains are first set to zero The Kp gain is increased until it reaches the ultimate

gain Ku at which the output of the loop starts to oscillate Ku is found to be 351 Pu is

98Ku and the oscillation period Pu are used to set the gains as shown in Table 1

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 75: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 75

Conclusion

Question bank

1) Which measurement is available in three element boiler control system

2) Give the Transfer Function

3) Which controller is used in above process

4) Which method is used to determine the PID gain parameter 5) What is Ku and Pu

Faculty Sign

Grade

Date

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 76: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 76

PRACTICAL 19 1 Obtain pole zero amp gain values of a transfer function

a 119918(119956) =119956120784+120786119956+120785

(119956+120787)(119956120784+120786119956+120789)

b 119918(119956) =120783

(119956120784+119956+120786)

c 119918(119956) =120787

(119956120784+120791)

d 119918(119956) =120783

(119956120784+120785119956+120787)(119956+120785)(119956+120787)

2 Write matlab code to obtain transfer function of a system from its pole zero gain values

a Poles= -2 1 Zero=-1 and gain=7

b Poles = -1+i -1-i -4 Zeros = -2 -5 gain = 1

c Poles = -1+4i -1-4i -5 Zeros = -8 -5 gain = 75

3 Obtain step response of a unity feedback system having forward path transfer function of

a 119918(119956) =120783

(119956+120786)

b 119918(119956) =120785

(119956120784+120786119956+120791)

c 119918(119956) =120785

(119956+120783) (119956120784+120786119956+120791)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 77: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 77

PRACTICAL 20 1 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783120782) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784+120783

119956120784+120786119956+120786) 119918120786(119956) =

119956+120783

119956+120788

119919120783(119956) =119956+120783

119956+120784 119919120784(119956) = 120784 119919120785(119956) = 120783

MATLAB solution

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 78: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 78

2 For the following multi-loop feedback system get closed loop transfer function and the

corresponding pole-zero map of the system

119918120783(119956) =120783

(119956+120783) 119918120784(119956) =

120783

(119956+120783) 119918120785(119956) =

119956120784

119956120784+120786119956+120786 119918120786(119956) =

120783

119956+120788

119919120783(119956) = 120783 119919120784(119956) = 120784 119919120785(119956) = 120783

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 79: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 79

PRACTICAL 21

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PD controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1= [1 10 20]

g2=tf (num1 den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Kd=10

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Kd=5

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t4=feedback(g31)

step(t4y)

hold on

Kp=500

Kd=01

numc=[Kd Kp]

numo=conv(numcnum)

deno=den

g3=tf(numodeno)

t5=feedback(g31)

step(t5r)

hold on

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 80: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 80

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PD controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 81: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 81

PRACTICAL 22

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

(119956120784+120783120782119956+120784120782)

Show the effect of addition of a PI controller on the system performance

Matlab Code

num=1

den=[1 10 20]

g1=tf (numden)

t1=feedback(g11)

step(t1g)

hold on

num1=10

den1=[1 10 20]

g2=tf (num1den1)

t2=feedback(g21)

step(t2m)

hold on

Kp=500

Ki = 1

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t3=feedback(g31)

step(t3b)

hold on

Kp=500

Ki = 100

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

t4=feedback(g31)

step(t4r)

hold on

Kp=500

Ki = 500

numc=[Kp Ki]

denc= [1 0]

numo=conv(numcnum)

deno=conv(dendenc)

g3=tf(numodeno)

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 82: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 82

t5=feedback(g31)

step(t5g)

hold on

Output

2 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PI controller on the system performance

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)

Page 83: INDEX [mgdic.files.wordpress.com] · 2019-12-16 · ISP(3341705) IC DEPT GP,GANDHINAGAR Page 1 INDEX Subject Code: ISP (3341705) Enrol. No.: _____ SR. NO. PRACTICAL EXPERIMENTS DATE

ISP(3341705) IC DEPT

GPGANDHINAGAR Page 83

PRACTICAL 23

1 Consider a unity feedback system with forward path transfer function 119918(119956) =120783

119956(119956+120783120788) Show

the effect of addition of a PID controller on the system performance

2 Show how to transfer step input plant input and plant output data to MATLAB workspace

for a unity feedback system with 119918(119956) =120783

119956(119956+120783120788)