1. R.C. Dorf and R.H. Bishop, Modern Control Systems, 11th Edition, Prentice Hall, 2008.2. J.J. DiStefano, A. R. Stubberud, I. J. Williams, Feeedback and Control Systems, Schaum's Outline

Series, McGraw-Hill, Inc., 19903. N.S. Nise, Control Systems Engineering, 3rd Edition, John Wiley & Sons, Inc., NY, 2002.4. R.T. Stefani, B. Shahian, C.J. Savant, Jr., G.H. Hostetter, Design of Feedback Control Systems, Oxford

University Press, NY, Oxford, 2002.5. K. Ogata, Modern Control Engineering, Prentice-Hall, 4th Edition, NJ, 2002.6. Chi-Tsong Chen, Analog and Digital Control System Design, Saunders College Publishing, 2000.

References:

Digital Control 4th Class, Production Engineering

Department of Production Engineering and Metallurgy,University of Technology, Baghdad

Lecturer: Dr. Laith Abdullah Mohammed

►Requirements and Grading:First term Exam: 10%, Second term Exam: 10%Final Exam: 60%Homworks, Quizess and Class attendance: 10%Laboratory (CNC Lab): 10%

►Software:

Maple MATLAB & Simulink

Email: dr.laith@uotechnology.edu.iq

Control System:What is a control system?

To answer this, we need to understand the “objectives” of the system to be controlled.In Industry: the objectives of manufacturing processes are satisfy the precision and cost requirements of the products.In Transportation: control automobile and airplane to satisfy accurate and safe trans.Then [The means of achieving these objectives involve the use of control systems that implement certain control strategies].Control systems found in all sectors of industry such as:[Quality Control of manufactured products, automatic assembly line, machine tool control, space technology, weapon systems, computer control, robotics,………].

Basic Components of a Control System:

Control SystemObjectives(Inputs, actuating signals)

Results(Outputs,

controlled variables)

Basic Definitions:

Control System is a combination of components or subsystems connected in such a way to give a desired/specified system performance.

Plant/System is the physical object that needs to be controlled, examples include power system, chemical process, spacecraft, economic system etc.

Disturbance is an unwanted signal that affects the output of a system adversely.

Feedback Control System is a system where in the actual input to the system being controlled is the difference between the reference input and the actual output.

Automatic Regulating System is a feedback system in which the reference input (or desired output) is a constant value. Example: electric power system

Process Control System is an automatic regulating system in which the output variable or variables include temperature, pressure, flow, liquid level, etc.

Open-loop Control System is a system in which the output has no effect on the input to the system. Examples: washing machine, electric switch, Automatic Door Opening & Closing system.

Closed-loop Control System is a feedback control system.Examples: Human being, Home Heating system, Ship Stabilization system.

Mathematical Modeling: Finding a mathematical model of the system to be controlled is the first step in the design of control systems. There are two types of mathematical models we study in control. They are: Transfer Functions and State-Space Equations.

Brain Hands

Eyes

Desired position of the hands

Reference positionof Book

Input Error signal

Closed Loop Sytem for A person wants to reach for a book on the table

Types of Control Systems:1- Open Loop Control System:

Controlled ProcessController

ReferenceInput (r)

ActuatingSignal (u)

ControlledVariable (y)

Features: Less accurate, Simple, cheap.Controller can be an amplifier, mechanical linkage, or other control element.Examples: Electric Washing Machine, Spraying, Drilling, On/Off processes.

2- Closed Loop Control System (Feedback Control System)

Controller Controlled Process

Sensor

Error Detector Controlled

Variable (y)ReferenceInput (r)

Difference Variable

Controller Engine

Speed Transducer

Error Detector Controlled

Variable (w)wewr

+-

TL

+

+

Block Diagram of closed loop speed control system.To obtain more accurate control, the controlled signal (y) should be feedback and compared with the reference input (r), and an actuating signal proportional to the difference of the input and the output must be sent through the system to correct the error.Reference input (wr) is desired speed, Engine speed (w) should equal (wr) and any difference such as load torque (TL) is sensed by the speed transducer and the error detector. The controller will operate on the difference and provide a signal to correct error.

Application of Load Torque (TL)

Application of Load Torque (TL)

TimeTime

Desired Speed (wr)

Desired Speed (wr)

Typical Response of the Open Loop Speed Control System

Typical Response of the Closed Loop Speed Control System

Regulator System

The effect of feedback on a control system:

1- reduction of system error.2- stability.3- sensitivity.

Dynamic systems:Studying the time behavior of the physical system under investigation. Most physical processes can be described by balance equations of the form,

Continuous systems: the rate of change component gives rise to a first-order differential equation.Discrete systems: events occur at discrete time intervals and the balance equation is written in terms of discrete difference equations.

Single input-single output (SISO) systems : a single differential equation is used to describe the relationship between the input quantity, u(t), and the output quantity, y(t). For Example: a position control system has only one input (desired position) & one output (actual output position).Multiple input and multiple outputs (MIMO) systems : described by a set of differential equations. For such systems, the input and output quantities are represented as vector quantities, and , respectively. State Variable Approach : The formal matrix methods treatment of dynamic system analysis.

Continuous Versus Discrete ModelingFirst Order Continuous Time SystemFor continuous time systems that display exponential growth or decay, the basic growth/decay law can be stated as "The rate of change is proportional to the amount present." This can be written mathematically as

The solution for this system is given as,

All system variables are the functions of a continuous time variable (t).Example: the speed control of a D.C. Motor using tachogenerator feedback.

First Order Discrete Time SystemFor discrete time systems that display geometric growth or decay, the basic growth/decay law can be stated as "The value at interval k+1 is proportional to the value at k." Mathematically, this can be written as

and k is the discrete time index. This discrete time variable is sometimes written as , where T is the sampling period. The solution to this difference equation can be obtained by assuming a solution of the form , where c is some constant to be determined. In this case, it is easy to show that and the solution becomes a simple geometric series,

In discrete time system one or more system varibles are known only at certain discrete intervals of time.Example: Microprocessor or computer based systems.

Digital & Sampled Data Control System:

In recent years, Microprocessors & microcomputers are used in the control systems to obtain necessary controlling action. Such controllers use digital signals which exists only at finite instants, in the form of short pulses (Digital controllers). Thus the digital control system is hybrid system using the combination of continuous time signals & digital signals. To obtain analog signal from digital, Digital to Analog (DAC) converters are used while to obtain digital signals from analog, Analog to Digital (ADC) converters are used. The input & output of a digital controller are both in digital form. The digital signals exist in the form of coded digital data at discrete intervals of time. Such signals are obtained from computers, microprocessors, ADC and digital sensing elements.Applications: radar tracking systems, Industrial robots, Modern Industrial control system,aircraft control systems.

Digital Control System :

Digital InputSignal Controlled

output

Fig. Digital Control Systemusing Digital Controller

Sampled Data System :Many times digital signals are obtained by sampling the continuous time signals at regular intervals. The switch can be used as a sampler. When switch is closed for short duration of time, signal is available at the output and otherwise it is zero. Such signal is called sampled signal which exists in a digital form.The digital controllers accept such sampled error signals to produce controlled variable in digital form. This is converted to analog signal using DAC and hold circuits. The hold circuits convert sampled signal back to analog signal. This signal is used to control the process. The system using such sampler and hold circuits is called sampled data control system. The input & feedback signals both are continuous in nature. The accuracy of sampled signals is less than the digital signals. Hence digital control systems are more accurate than sampled data systems.

Analog continuous time signal Controlled Output

Continuous time feedback signal

Switch is closed for short duration after regular time interval T

Use of sampler to obtain sampled signal

Sampled Data control system

Transform and Frequency Domain Methods Laplace Transforms

Definition: The Laplace transform is defined by the linear transformation

where is an arbitrary complex number and f(t) is function of time (t)

In practice, one usually does not need to perform a contour integration in the complex plane. Instead, a "dictionary" of Laplace transform pairs is generated and some simple rules allow one convert between the time domain solution, f(t), and the frequency or s-plane solution, F(s).

Some Laplace Transform Pairs1- Unit Step Signal:

where:

2- Exponential Signal:

where:

وهي دالة (الى صيغة المعادلة الجبرية ) tوهي دالة للزمن (تستخدم صيغة تحويل البالس لتحويل المعادلة التفاضلية .وذلك للتحويل من الحقل الزمني الى الحقل الترددي) sللمتغير المركب

The Laplace Transform is one of the mathematical tools used to solve linear ordinary differential equations.

s=σ+jω σ: Real part of complex variable (s).ω: Imaginary part of complex variable (s).

Pierre-Simon Laplace (1749–1827).

Lecturer: Dr. Laith Abdullah Mohammed

For the 2nd derivative:

In general, the nth derivative is:

Differentiation:

Superposition (Sum and Difference):

--

Some Laplace Transforms

Note that Matlab has built-in capability within the Symbolic Toolbox to generate Laplace transforms and inverse Laplace transforms (see commands LAPLACE and ILAPLACE, respectively).

Time Function f(t) Laplace Transform F(s) Function Plot

Step function f(t)=h

where: h=constant

Pulse functionf(t)=h when 0<t<t0f(t)=0 when t>t0

Exponential functionf(t)=e-at

f(t)=eat

Sine wave functionf(t)=sin wt

f(t)

t, time

h

)1()( 0steshsF −−=

shsF =)(

f(t)

t, time

h

t0

assF

assF

−=

+=

1)(

1)(f(t)

t, time

22)(ws

wsF+

=f(t)

t

Time Function f(t) Laplace Transform F(s)

Polynomial function f(t)=tn

(n=positive integer)

f(t)=tne-at

f(t)=tneat

Multiplication by a Constant Kf(t)(K= constant)

K F(s)

1

!)( += nsnsF

1)(!)( −−

= nasnsF

1)(!)( ++

= nasnsF

Z=X+jY

X, σ

Y, ω

Real Part

Imag

inar

y P

art

Complex Plane (Z plane)

s=σ+jω

Laplace Transform Calculation using Matlab

Inverse Laplace Transform Calculation using Matlab

13175)(

175]13)[(0)(15)(32)5()(

]5)([3)]0()([33

2)5()()0()0()(

2

2

2

222

2

+++

=∴

+=++

=+−+−−∴

−==− →

−−==

−=− →

ssssy

ssssysyssyssys

ssytyssydtdy

ssysdttdytsysys

dtyd

Laplace

Laplace

Example [1]: Given initial condition (y=5), (dy/dt=2) at t=0. Write Laplace Transform of differential equation:

Solution:

032

2

=++ ydtdy

dtyd

Example [2]: Given initial condition (y=3), (dy/dt=0) at (t=0). Write Laplace Transform of differential equation, then find the solution of y as a function of time:

Solution:

05 =+ ydtdy

tTable

at

Laplace

Laplace

etyetfs

sy

ssysyssysyy

ssytyssydtdy

53)()(

53)(

3]5)[(0)(53)(

)(55

]3)([)]0()([

−

−

= →∴

=+

=∴

=+=+−∴

→

−==− →

Laplace Transform of differential equation calculation using Matlab

Laplace transform solutions to nth-order parameter linear time-invariant systems are typically of the form:

where the are referred to as the zeros of F(s) and the are the poles of F(s).

The main principal method for finding when F(s) is a ratio of polynomials is Partial Fraction Expansion

Inverse Transforms:

Non-Repeated Linear Factors

Lecture.3.

Lecturer: Dr. Laith Abdullah Mohammed

3( )( 1)( 2)

sF ss s

+=

+ +

( )1 2

F ss

A Bs

= ++ +

2( ) 1( )t tf t e eA B t− − = + 3( 1) ( 1)

( 1)( 2) 1 2s A Bs s

s s s s+ + = + + + + + +

3 ( 1)( 2) 2s BA ss s+ = + + + +

2

3( 1) s

sBs =−

+= +

2A =

A:

3( 2) ( 2)( 1)( 2) 1 2

s A Bs ss s s s

+ + = + + + + + +

3 ( 2)( 1) 1s As Bs s+ = + + + +

1

3( 2) s

sAs =−

+= +

1B = −

B:

2( ) ( ) 1(12 )t tf t e e t− − −= +

Example: Solve the following using Partial Fraction Expansion:

Solution:

Block Diagram Representations Large dynamic systems can usually be broken into several blocks that have relatively simple SISO relationships. These blocks can be manipulated to obtain the overall transfer function.Transfer Function TF is defined to be the relation of the Laplace transform of the input to the Laplace transform of the output, With all initial conditions assumed to be Zero, The TF represents the dynamics of the system. The TF allows to determine the response of the system to various input signals.Block diagram algebra is identical for the scalar and matrix cases except one must always be careful to maintain the order of matrix operation. The goal is to generate an equivalent overall system transfer function matrix for various common block diagram configurations.

Process G(s)Input Signal U(s) Output Signal Y(s)

Overall System Transfer Function = Y(s)/U(s) = G(s)

. Block Diagramتمثيل نظم السيطرة يتم من خالل التعبير عن العالقات الرياضية بشكل مخططات كتلية :عناصر هذه المخططات

1- Summing Point: تستخدم ألضافة أو طرح األشارات بحيث يمكن ان تدخل أكثر من اشارة لها.كما هو موضح في ادناه. وتخرج اشارة واحدة فقط

+

+

+

-Z(s)=X(s)+W(s)-Y(s)Z=X+Y X(s)X

Y(s)Y

+W(s)

2- Take off Point: تسمح باستخدام المتغير في أكثر من مكان حيث التتغير قيمة المتغير عند مروره في)نقطة التفرع(هذه النقطة

X X

X

3- Function Box : رمز الضرب عند بناء المعادالت الخاصة بنظام السيطرة حيث نحصل على األشارة. (G(s))مع الدالة الموجودة قي الصندوق (X)عن طريق ضرب األشارة الداخلة (Y)الخارجة

G(s)X Y=G(s)*X

Basic Operations in Block Diagram Algebra

1- Blocks in Series: TF connected in series are combined by multiplication.

2- Blocks in Parallel: TF connected in Parallel are combined by addition.

G H G*H=

= G+HG

H

+

+

3- Closed Loop Function Formula:

G

H

+

±

x1 e x2

x3

e=x1 ± x3G=x2/e , H=x3/x2x2=G*e=G*(x1 ± x3)=G*(x1 ± x2*H)x2=G*x1 ± G*H*x2 ÷x1x2/x1 = G ± G*(x2/x1)*H(x2/x1) ± (x2/x1)*G*H=G ⇒ (x2/x1)*(1 ± GH)=G

GHG

xx

112=∴

=GHG1

x2x1

Equivalent Transfer Function

Definitions:G(s): Forward Path Transfer Function.H(s): Feedback Path Transfer Function.G(s)H(s): Open-Loop Transfer Function or “loop gain”.

Summary of Block Diagram Reduction Rules:

Transformation Block Diagram Equivalent Block Diagram

Transformation Block Diagram Equivalent Block Diagram

Procedure to Solve Block Diagram reduction Problems:

Step 1 : Reduce the blocks connected in series.Step 2 : Reduce the blocks connected in parallel.Step 3 : Reduce the minor internal feedback loops.Step 4 : As far as poosible try to shift take off points towards right and summing points to the left.Step 5 : Repeat steps 1 to 4 till simple form is obtained.Step 6 : Using standard Transfer Function of simple closed loop system, obtain the closed loop Transfer Function C(s)/R(s) of the overall system.

Example: Block Diagram Reduction:

The Goal is to reduce to a block diagram with fewer blocks.

Step 1

Moving a Takeoff Point beyond a Block

Cascaded Blocks

Eliminating a Feedback Loop

Step 2

Step 3

Step 4Final Block Diagram

Eliminating a Feedback Loop

Home Work:1- Rewrite the following equation using Partial-Friction Expansion expanded form:

)3)(2)(1(35)(

++++

=sss

ssG

Answer:

36

27

11)(

+−

++

+−

=sss

sG

2- Consider the differential equation, where u(t) is the unit-step function. The initial conditions are y(0)=-1, dy/dt =2 at t=0. solve the differential equation using Laplace transform, then find the complete solution as y(t).

)(5)(2)(3)(2

2

tutydt

tdydt

tyd=++

Answer:tt eety 2

235

25)( −− +−=

)2)(1(5)(

2

+++−−

=ssssssy

The differential Equation of a Linear System is shown below, find Transfer Function Y(s)/R(s) , assume all initial

conditions are zero.

)()(10)(6)(5)(2

2

3

3

trtydt

tdydt

tyddt

tyd=+++

10651

)()(

23 +++=

ssssRsY

Answer

(MASON) تحليل المنظومات المتداخلة باستخدام طريقة

∆+∆+∆

=........2211 TT

lInputSignaalOutputSign

∆=1-(L1+L2+L3+…………) +(Sum of Product of any two nontouching loops) -(Sum of Product of any three nontouching loops)+……..

where:

L = Loops.

T1,T2 = Forward Transmission from Input Signal to Output Signal.

∆1= ∆+L1 =value of ∆ after removing the loops that touch (T1).

∆2= ∆+L2 =value of ∆ after removing the loops that touch (T2).

In block diagram represenattion, we have to apply reduction rules, one after the other to obtain simple of the system and hence overall transfer function. We have to draw the reduction block diagram after every step. This is time consuming. In signal flow graph (SFG) approach, once SFG is obtained, direct use of one formula leads to the overall system transfer function C(s)/R(s). This formula is stated by S.J.Mason (1953) and hence referred as Mason‘s Gain formula.

Lecturer: Dr. Laith Abdullah Mohammed

Example: Find the Response of the following control system at: D2=1, A=1, K1=1, K2=2, KH=0.5, B=-5, D1=1/6.

when:1- v is step function with constant value. (i.e. u=0)2- u is step function with constant value. (i.e. v=0)

AsD

K

1

1

1+ sDK

2

2

1+

B

KH

v c

u

+

-

e

G1

++ G2

H

Input Signal

Output Signal

Solution:

1- v is step function with constant value. (i.e. u=0).

Using MASON method:

L1= -G1G2HT1=G1G2A∆=1-L1=1-(-G1G2H)=1+G1G2H∆1=∆+L1=1+G1G2H-G1G2H=1

tt eetcssss

CsC

sCsc

sRootssss

sc

sonStepFunctivwhen

ssssHGGAGGT

vc

lInputSignaalOutputSign

43

321

221

2111

341)()4(

3)3(

41)4()3(

)(

4,3,0:)4)(3(

12)(

1:

)4)(3(12

12712

1))(1(

−− +−=∴

++

+−=

++

++=

−−=∴++

=∴

==

++=

++=

+=

∆∆

==

v A G1 G2 cL1

T1B

u=0

-H

e

2- when (u) is step function with constant value. (i.e. v=0).

Using MASON method:

L1= -G1G2HT1=BG2∆=1-L1=1-(-G1G2H)=1+G1G2H∆1=∆+L1=1+G1G2H-G1G2H=1

tt eetcssss

CsC

sCsc

sRootssss

ssc

sonStepFunctiuwhen

sss

HGGBGT

uc

lInputSignaalOutputSign

43

321

21

211

5105)()4(

5)3(

105)4()3(

)(

4,3,0:)4)(3(

)6(10)(

1:

)4)(3()6(10

1)1)((

−− −+−=∴

+−

++

−=

++

++=

−−=∴++

+−=∴

==

+++−

=+

=∆∆

==

v=0 A G1 G2 cL1

T1B

u

-H

e

Example: Convert the block diagram representation of a system as shown below into a signal flow graph. Hence find the transfer function C(s)/R(s) of the closed loop system from the signal flow graph.

Example: Find the overall Transfer Function by using Mason‘s gain formula for the single flow graph in the figure below.

Home Work: Find the Response for the speed control system at: D=0.25, K1=1, K2=0.75, u=Load Torque, no=Output Speed, ni =Input Speed, and all initial conditions are zero.

when:1- ni is step function with constant value. (i.e. u=0)2- u is step function with constant value. (i.e. ni =0)

sK1

DsK+1

2nino

u

+

-

e

G1

+- G2

Input Speed

Output Speed

1. no(t)=1+0.5e-3t-1.5e-tAnswer:

2. no(t)=-1.5e-t+1.5e-3t

Mason's Rule MATLAB function

DescriptionMason.m uses mason's rule to simplify signal flow graphs. It takes a file describing the network and produces a symbolic equation relating a dependent output node to an independent input node. The routine requires the user to create a ".txt" file describing a network's signal paths.

s11

s21

s22

s12

R2

This program generates these equations for a given network and pair of nodes.

Using the ProgramIt is important that the lines in the net file be ordered so that the coefficient numbers count from 1 up. Don't use 0 to number the coefficients or nodes! Once you have made the net file, run 'mason.m' from Matlab, as described below:

USAGE:[Numerator,Denominator] = mason(Netfile,StartNode,StopNode)

Netfile - is a string with the name of the netfile to loadStartNode - is the integer number describing the independent input nodeStopNode - is the integer number describing the dependent output nodeNumerator - is a string containing the equation for the NumeratorDenominator - is a string containing the equation for the Denominator

Try out the example network! To recreate the above examples use:

[Numerator,Denominator] = mason('example.net',1,3)[Numerator,Denominator] = mason('example.net',1,2)

Presentation by studentsEach student prepares a presentation on one of the following topics.No. of Slides: Less than 10 (using Microsoft PowerPoint), Duration: 2 weeks. [Draw the block diagram (The Control System) of the following systems showing the input variables,

the output variables, and inside the block (gain)]1 Potentiometer 11 Automatic

Elevator21 Suspension

system in the car31 Self-Guided

Vehicle41 Active Vibration

Absorber

2 Winders 12 Toaster 22 Actuator 32 Thermostat 42 Tachometer

3 Nuclear reactor 13 DVD Player 23 Hydraulic pump 33 Floppy Disk Drive 43 Antenna Azimuth

4 Control the fluid level in a home tank

14 Remote controlled robot arm

24 Automobile Guidance System

34 Electric Ventricular Assist Device (EVAD)

44 Position control system in NC machine

5 Dynamometer 15 Steam Boiler 25 Furnace 35 Wind turbines 45 Voltage stabilizer

6 Grinder system 16 Automatic ship steering system

26 Walking Robots (Hannibal)

36 Arc Welding Robot

46 High speed rail pantograph

7 High speed proprtional solenoid valve

17 Navigation system of missiles

27 Magnetic Levitation Transportation System

37 Heat Exchanger Process

47 Sunseeker solar system

8 The pupil of human eye

18 Guidance system of Space shuttle

28 Cutting forces during machining operation

38 Coordinate Measuring Machine (CMM)

48 Automatic controlled Load tester

9 A Segway Human Transporter

19 Steel Plate Finishing mill

29 Dynamic Voltage Restorer (DVR)

39 Charge-Coupled Device (CCD)

49 CameraMan (Automatic Presenter Camera system)

10 Automatic Door operating system

20 Continuous Casting machine

30 3D Full body Scanning

40 Anti-lock braking (ABS)

50 conveyor system

Determine the Transfer Function C(s)/R(s) of the system shown in the Fig. below.

Modeling of Physical Systems:

Modeling of Mechanical Systems Elements:

The motion of mechanical elements can be described as translational, rotational, or combination of both.

Translational Motion:The motion takes place along a straight or curved path. The variables that are used to describe this motion are acceleration, velocity, displacement.Newton’s Law of Motion states: Σ forces = Mawhere: M: mass (kg), a: acceleration (m/sec2), force: Newton (N)

1. Mass: it is a property of an element that stores the kinetic energy of translation motion. It is analogous to the inductance of electric networks.M= W/gwhere: W: weight of a body,g: acceleration of the body due to gravity (g=9.8066 m/sec2)For Force-mass system, the force equation is:

where: v(t) : linear velocity (m/s).

dttdvM

dttydMtMatf )()()()( 2

2

===M

y(t)

f(t)Displacement

Force

)()( 2 syMssf =Taking laplace and neglecting initial conditions:

Lecturer: Dr. Laith Abdullah Mohammed

2- Linear Spring: it is an element that stores potential energy. It is analogous to a capacitor in electric networks. f(t) = K y(t)where: K: spring constant (stiffness) (N/m)

y(t)f(t)

K

Force-Spring System

If the spring is preloaded with a preload tension of T, then:f(t) – T = K y(t)

KM2 M1

f(t)

y2(t)

y1(t)

Consider the spring connected between the two moving elements having masses M1 and M2 where force is applied to mass M1.Now mass M1 will get displaced by y1(t) but mass M2 will get displaced by y2(t) as spring of constant (K) will store some potential energy and will be the cause for change in dispalcement. Consider free body diagram of spring as shown in figure beside. Net dispacement in the spring is y1(t)-y2(t) and opposing force by the spring is proportional to the net displacement i.e. y1(t)-y2(t) K

y2(t)

y1(t)(y1-y2)

Fspring = K [ y1(t) – y2 (t) ] Fspring = K [ y1(s) – y2 (s) ]Taking Laplace

dttdyBtf )()( =

y(t)

f(t)

Bdy/dtf

B=slope

3- Friction for Translation Motion: three different types of friction are used in practical systems, viscous friction, static friction, Coulomb friction.A- Viscous Friction (Dashpot/Damper): retarding force that is a linear relationship between the applied force and velocity. B: viscous frictional coefficient (N/m/sec)

Frictional Force

Taking laplace and neglecting initial conditions, Ffrictional(s)= B s y(s)

y1By2

M2 M1

Friction between two moving surfaces causes change in displacement.

In such case, opposing force is given by,

−=

dttdy

dttdyBtFfrictional

)()()( 21 Taking Laplace, [ ])()()( 21 sysyBssFfrictional −=

Rotational Motion:Motion of a body about a fixed axis. The force gets replaced by a moment about fixed axis (force * distance from fixed axis) which is called Torque. Using Newton’s Law of Motion:Σ Torques = J αwhere: J: inertia (kg . m2) , α: angular acceleration.

1.Inertia (J): property of an element that stores the kinetic energy of rotational motion. (J) depends on the geometric composition about the axis of rotation and its density.Example:Inertia of a circular disk or shaft about its axis = J = (M r2)/2For Torque-inertia system:

2

2 )()()()(dt

tdJdt

tdJtJtT θωα ===

T(t) Ө(t)J

Torque-inertia systemwhere: T: torque (N.m)Ө(t) : angular displacement (radian).ω(t) : angular velocity (rad/sec).α(t) : angular acceleration.M: mass (kg)

1 rad = 180/π = 57.3 deg

1 rpm = 2 π/60 = 0.1047 rad/sec = 6 deg/sec

)()( 2 sJssT θ=Taking Laplace,

2- Torsional Spring:

For Torque torsional spring system: T(t) = K Ө(t)where: K: torsional spring constant (N.m/rad) T(t)

Ө(t)K

If the torsional spring is preloaded by a preload torque of TP , the equation:

T(s) – TP = K Ө(s) 3- Friction for Rotational Motion:The three types of friction (viscous, static, coulomb) carried over to the motion of rotation.For Viscous friction:

dttdBtT )()( θ

=

)()( sKsT θ=Taking Laplace,

Taking Laplace:

)()( sBssT θ=

B T(t)

Ө(t)

Example 1: Find y1/y2 for the mechanical system shown.

Solution: Fspring= K (y1-y2)

FDamper= B D y2

Fspring= FDamper

K (y1-y2) = B D y2Ky1 = Ky2 + B D y2

KBDK

yy +

=21

K

B

y1

y2

Example 2: Find the Transfer Function for the mechanical system shown.

K B

y

B

F

M

Solution:

KBsMssFsXsG

yieldsfunctionstransfertheforSolvingsFsYKBsMsorsFsKYsBsYsYMs

ConditionsInitialZeroAssumeTransformLaplaceTaking

tKydt

tdyBdt

tydMtF

NewtonLawFFFF SpringDamperMass

++==

=++

=++

++=

++=

2

2

2

2

1)()()(

:)()()()()()()(

)()()()(

F

Spring

Damper

Example 3: Find for the mechanical system shown.

Solution:

K

B

y1

y2

)()(1

)()(2

SFSYand

SFSY

F

M

)()()(1

)()()(1

))(

1(1

)(1

)(1

21.13.

)(1

)()(2

)3..(........................................)(

1)()(2

)(2

2)(2.12.

).1.........(..............................21:)1(

)2.....(..............................)(21

2221)1....(....................22)21(

2

2

2

2

2

2

2

2

2

2

2

2

2

2

BSMSKKBSMS

SFSY

BDMDKKBDMD

tFty

BDMDKFKy

BDMDKFFKy

BDMDFKKyF

KyKyFainSub

BSMSSFSY

BDMDtFty

BDMDyF

KyK

KBDMDyKF

ainSubaKyKyFFrom

KKBDMDyy

yMDBDyKyKyBDyyMDyyKF

FFFF DMs

+++

=

+++

=

++=

++=

+−=

−=

+=∴

+=∴

+=

−++

=

−=

++=∴

++=

+=−=

=+=

Mass

Damper

Spring

Mass-Dampening-Stiffness

Test m File:

The Results:

Mass Matrix

Dampening Matrix

Stiffness Matrix

Conversion between Translation and Rotational Motions:

W X(t)

RackPinion

T(t)Drive Motor

r

θ(t)

Rack and Pinion

W

T(t)Drive Motor

r r

PulleyBelt

X(t)

θ(t)

Belt and Pulley

Rotary – to – Linear motion control system:

r= radius of pulley (Belt and Pulley)r= radius of pinion (Rack and Pinion)M= mass (kg)W= weight of the body (N).g= acceleration of the body due to gravity (g=9.8066 m/sec2)

22 rg

WMriaMotorInertJ ===

Lecture [6]

Lecturer: Dr. Laith Abdullah Mohammed

WMotor

T(t), θ(t)X(t)

Lead Screw

Rotary – to – Linear motion control system (Lead screw):

L: screw lead, distance that the mass travels per revolution of the screw (mm).W: weight of the body (N).g: acceleration of the body due to gravity (g=9.8066 m/sec2)

L

2

2

==πL

gWInertiaJ

Electrical Control Components Representation:

+ -VR

R

+ -VL

L

+ -VC

C

1/R 1/LD CDVR I VL I VC I

IVR R= dtdI

VL L

/=

dtdVIC

c /=

)اوم(المقاومة )هنري(المحاثة )فاراد(السعة Let: D= d /dt

Example: Find the transfer function for the following circuit.

+VR

R

VL

L C

-VC

I

V

Similar to the mechanical systems, very commonly used systems are of electrical type. The behavior of such systems is governed by Ohm’s Law. The dominate elements of an electrical system are, Resistor, Inductor, Capacitor.

++=∴

==

==

=

=

==

++=∴

++=++=

++=

CRDLDQV

CQ

CDDQ

CDI

RDQRIQLDLDI

eqinDQIsubbydtdD

DQdtdQI

CDRLDIV

CDIRILDI

CDIRI

dtdILV

VVVV CRL

1

)1(................

)1...().........1(

2

2

CRDLD 11

2 ++V Q

Transfer function of the RC network

)/1/(1

)1)((1

)()()(

)()()(

)())(

1()(

1

2

2

1

TsTsRCsvsvsG

sCsisv

sisC

Rsv

+=

+==

=

+=

Analogous Systems:

In between electrical & mechanical systems there exists a fixed analogy and their exists a similarity between their equilibrium equations. Due to this, it is possible to draw an electrical system which will behave exactly similar to the given mechanical system, this is called electrical analogous of given mechanical system and vice versa.

Consider simple mechanical system, shown in Fig. Due to the applied force, mass M will displace by an amount x(t) in the direction of force f(t). According to Newton’s law of motion, applied force will cause displacement x(t) in spring, acceleration to mass M against frictional force having constant B.f(t)=Ma+Bv+Kx(t)Where, a=acceleration, v=velocity

Tabular Form of Force-Voltage Analogy

Translational Rotational ElectricalForce F Torque T Voltage VMass M Inertia J Inductance L

Friction constant B Torsional friction constant B

Resistance R

Spring constant K (N/m)

Torsional spring constant K (Nm/rad)

Reciprocal of capacitor 1/C

Displacement x θ Charge qVelocity x=dx/dt θ=dθ/dt = ω Current I=dq/dt

Comparing equations for F(s) and V(s) it is clear that,

Inductance L is analogous to mass M.Resistance R is analogous to friction B.Reciprocal of capacitor i.e. 1/C is analogous to spring of constant K.

Thermal Control Component Representation:

RT , T

Q

T1

Q: Heat flow rate.h: Heat transfer factor.A: Surface area of body.T: Body Temperature.T1: Outer Temperature.RT: Thermal Resistance.C: Specific heat of body.M: Mass of body.CT: Thermal Capacitor

Body

[ ]{ }

[ ]DTCQMCCalso

MCDTQdtdDlet

dtdTMCQalso

RTTQ

hAR

TThAQ

T

T

T

T

=∴=

=∴

=

=

−=∴

=

−=

:

:

:

)(

1)(

1

1

ElectricalCDEIREIalso

ThermalDTCQR

TTQ TT

==

=−

=∴

,

,)( 1

For Electrical Circuit:

+=⇒

=−

=

==

RCDEE

CDER

EEI

CDEREI

11

1

For Thermal Circuit:

+

=⇒

=−

=

DCRTT

DTCR

TTQ

TT

TT

1

)(

1

1

This is known as direct analog between thermal and electrical circuits.

i.e. T α E , C α CT , RT α R , Q α I

Models of Thermal Systems1. Heat Transfer System

Fig. shown a Thermal System

Electric heating element is provided in the tank to heat the water. The tank is insulated to reduce heat to the surroundings.

The necessary simplifying assumptions are:

1. There is no heat storage in the insulations.2. All the water in the tank is perfectly mixed and hence at a uniform temperature.

θI = Inlet water temperature in °Cθo= Outlet water temperature in °C.θ = surrounding temperature.q = Rate of heat flow from heating element in J/secqi = Rate of heat flow to the water.qt = Rate of heat flow through tank insulation.

Transfer function is:

θo(s)/Q(s)=R/(1+sCR)

The time constant of the system is RC

2. Thermometer:Consider a thermometer placed in a water bath having temperature θi, as shown.θo is the temperature indicated by the thermometer. The rate of heat flow into the thermometer through its wall is,dq/dt = (θi,- θo)/RWhere: R = Thermal resistance of the thermometer wallThe indicated temperature, rises at rate of

– State: تحليل مجال الحالة Space Analysis

هذا . تزودنا هذه الطريقة باسلوب عام لتفسير نظم السيطرة بحيث يالئم تحليل نظم السيطرة المعقدة وغير الخطيةحيث يكون (Transfer Function)األسلوب يمكن استخدامه في معظم األنظمة الحركية بدال من استخدام

هذه الصيغة تسهل بناء . بداللة سلسلة من المعادالت التفاضلية من الدرجة األولى والتي تكون بصيغة مصفوفة.برامج كومبيوتر قادرة على تحليل وتصميم األنظمة المعقدة

البرمجة –البرمجة المباشرة : الطرق الشائعة للحصول على تمثيل مجال الحالة لنظام السيطرة هي :تمثيل النظامواختيار تمثيل مجال الحالة يعتمد على طبيعة الحالة المراد . والبرمجة العامة–البرمجة المتوالية –المتوازية

.حلها

Direct Programming:Example: Write the state model for the system described by transfer function.

)()2)(1(

3)( tfDD

Dtc++

+=

:الى الصيغ التالية) تعني مشتقة D(نحول المقام والبسط للمعادلة التفاضلية -1 :الحل

)()2)(1(

1)3()(

1

1

tfDD

x

xDtc

++=

+=

State-space models are models that use state variables to describe a system by a set of first-order differential or difference equations, rather than by one or more nth-order differential or difference equations. State variables x(t) can be reconstructed from the measured input-output data, but are not themselves measured during an experiment.

Lecturer: Dr. Laith Abdullah Mohammed

)3.....(..........

)2.......().........(32)(23

:)(23

)(23

1)()2)(1(

1)1......(..........3)(

:

3)(

21

212

122

21

111

21

12

21

11

xx

tfxxxtfxxx

xxlettfxxx

tfDD

tfDD

x

xxtcxxlet

xxtc

=

+−−==++∴

==++∴

++=

++=

+==

+=∴

مع مالحظة ان عدد متغيرات الحالة يساوي درجة المعادالت ). برمز الحالة(نرمز لكل مشتقة برمز يعرف -2.التفاضلية للنظام المطلوب تحليله

(state variables)بصيغة المصفوفات مع ترتيب المعامالت لمتغيرات الحالة 1،2،3نعيد كتابة المعادالت -3(x1,x2,x3,……) للحصول على معادلة الحالة ومعادلة األخراج.

)(10

3210

2

1

2

1 tfxx

xx

+

−−

=

System Matrix State Vector

Control Vector

x1 x2

State Equation

[ ]

=

2

113xx

c

Output Equation

Example: Write the state model for the system described by transfer function.

)(396

5)( 23 sfsss

sc+++

=

}3)..{(5693)(5693)(5396

}2......{..............................}1....{........................................

:Assume)(5396

)(396

5)(:Let

:Solution

32131

1111

1111

321

21

1111

231

1

sfxxxxxsfxxxxsfxxxx

xxxxx

sfxxxx

sfsss

x

xsc

+−−−==+−−−=+−−−=

===

=++++++

=

=

[ ] [ ] 1

3

2

1

3

2

1

3

2

1

3

2

1

3213

3212

3211

001001)(

:EquationOutput

)(500

693100010

,

:Equation State:formMatrix toequations above all Transform

)(5693)(0100)(0010

:produce }3,2,1{ Equations ofent Rearrangem

xxxx

xsc

sfxxx

xxx

xxxx

x

sfxxxxsfxxxxsfxxxx

=

==

+

−−−=

=

=

+−−−=+++=+++=

System Matrix Control Vector

State Vector

tf2ss:

Convert transfer function filter parameters to state-space form.Syntax:[A,B,C,D] = tf2ss(b,a)Description:tf2ss converts the parameters of a transfer function representation of a given system to those of an equivalent state-space representation.[A,B,C,D] = tf2ss(b,a) returns the A, B, C, and D matrices of a state space representation for the single-input transfer function

The input vector a contains the denominator coefficients in descending powers of s. The rows of the matrix b contain the vectors of numerator coefficients (each row corresponds to an output).

)(23

3)( 2 tfDD

Dtc++

+=

Example: Consider the following system

Using Matlab, convert this system to state-space, type,

Example:

Stability of Linear Control Systems: Routh’s Criterion:For Design purposes, there will be unknown or variable parameters imbedded in the characteristic equation, so it will not be feasible to use the root finding programs. The Routh method is use for determining the stability of linear continuous data systems without involving root solving. The Routh criterion is best limited to equations with at least one unknown parameter. It gives the necessary and sufficient condition for all roots of the characteristic equation of a linear time-variant SISO system to lie in the left of the s-plane.

0........)( 011

1 =++++= −− asasasasF n

nn

n

This method determine the location of zeros (roots) of a polynomial with constant real coefficients. The form of linear time SISO system is:

Unstable Region

Unstable Region

Stable Region

Stable Region

jw

σ0

S-plane

Lecturer: Dr. Laith Abdullah Mohammed

Step-2- Routh’s Array.

Step-3- Investigate the signs of the coefficients in the first column of the tabulation, which contains information on the roots of the equation. The number of changes of the signs in the elements of the first column equal to the number of roots.

0........)( 015

56

6 =++++= asasasasForder equation.-the sixthExample:

............

7531

642

−−−−

−−−

nnnn

nnnn

aaaaaaaa

Step-1- Arrange the coefficients of the above equation into two rows:

EXAMPLE 1: Consider the characteristic equation of a linear system, check the stability of the equation using Routh stability test.[2s4+s3+3s2+5s+10=0]

Solution:

s4 2 3 10s3 1 5 0

s2 [(1)(3)-(2)(5)]/1 = -7 10 0s1 [(-7)(5)-(1)(10)]/-7=6.4 0 0s0 10 0 0

System is unstable because of changes (two changes) in first column sign.

Sign changeSign change

Example 2:

Homework 1: Considering the characteristic equation of a closed loop control system.

s3+3Ks2+(K+2)s+4=0

Find the range of K so that the system is stable.

Answer:

Homework

Homework 2: Consider the characteristic equation of a linear system, check the stability of the equation using Routh stability test.[s4+s3+2s2+2s+3=0]

s4 1 2 3s3 1 2 0s2 0.001 3 0s1 -2998 0s0 3

Answer:

System is unstable because of changes (two changes) in first column sign.

Routh-Hurwitz stability criterion Calculation using Matlab

Example:

The Nyquist Stability Criterion قاعدة األستقرار لنايكوست

. هي طريقة لقياس أستقرارية النظام من خالل الربط بين شكل الرسم التخطيطي القطبي والسلوك الحركي للنظام.(Phase Angle)وزاوية الطور |Amplitude|هو رسم العالقة بين مطلق القيمة : (Polar Plot)الرسم القطبي

:خطوات قاعدة األستقرار

وذلك بأن (Sinsoidal Form)الى الصيغة الجبرية (Laplace Transform)تحويل صيغة النظام من -1(Sinsoidal Input Signal)اي نفرض ان أشارة األدخال هي أشارة جيبية (w:rad/sec) (s=wj)نفرض

−

==∠

+

+==

∠==

−− ∑∑ ReImtan

ReImtanAngle Phase)(

ReImReImAmplitude)(

:)()()()(

:FunctionTransfer Sinsoidal

11

22

22

ο

ο

wjG

wjG

wherewjGwjGwjGsG

مجموع زوايا البسط مجموع زوايا المقام

للبسط

للمقام

dr.laith@uotechnology.edu.iq

Lec.9

: (Polar Plot)رسم النقاط على المخطط القطبي -2

Re(w)

Im(j)

):يجب تحقق الشرطين معا ليكون النظام مستقرا( تحقيق شروط األستقرارية -31- Phase Margin = 180°-Phase Angle = βif β=+ , Phase Angle < 180° → Stable Systemif β=- , Phase Angle > 180° → Unstable System

2- Gain Margin = L = 1-Amplitudeif L<-1 → Stable Systemif L>-1 → Unstable System

Re(w)

Im(j)

Unit Circle

Nyquist Path

Polar Plot[Stable System]

PhaseAngle

β-1

L Amplitude

Example 1: Draw the Nyquist plot for the following system:

121)(+

=s

sG

)2(tan1

2tan10tan

ReImtan

ReImtanAngle Phase)(

141

1)2(10

ReImReIm)(

)()(12jw

1G(jw) jws

:

11111

222

22

22

22

wwjwG

wwAmplitudejwG

jwGjwGlet

Solution

−−−−−°

°

−=

−

=

−

==∠

+=

+

+=

+

+==

∠=+

=⇒=

w (rad/sec) Amplitude Phase Angle

0 1 0

1 0.447 -63.4

2 0.242 -75.9

3 0.164 -80.5

4 0.124 -82.8

1-1 Re

Im

w

w=0

w=1w=2

Polar Plot (Stable System)

Example 2: Draw the Nyquist Diagram and check if the system is stable for (w = 0,1,2,3,4,10):

)11.0)(11.0)(1(5)(

+++=

ssssG

)1.0(tan)1.0(tan)(tan

11.0tan

11.0tan

1tan

50tan)(Angle Phase

1)1.0(1)1.0(150)(

)11.0)(11.0)(1(5)()(

111

1111

222222

22

www

wwwjwG

wwwjwGAmplitude

jwjwjwjwGsG

−−−

−−−−°

−−−=

+

+

−

=∠=

+++

+==

+++==

w(rad/sec)

Amplitude Phase Angle

0 5 01 3.501 -56.4212 2.15 -86.0553 1.451 -104.9644 1.04 -119.567

10 0.249 -174.289∞ 0 -270

5-1

w=2

w=1

w=0

w=∞

Amplitude=0.206w=10.95Phase Angle=-180°

L

Phase Angle

Polar Plot[Stable System]

β

Unit Circle

w

Im

Re

Homework:Draw the Nyquist Diagram for the following control systems and check the stability.

)105.0)(12.0(10)(

)2)(1(5)(

++=

++=

ssssG

ssssF

dr.laith@uotechnology.edu.iq

dr.laith@uotechnology.edu.iq