TABLE OF CONTENTS

CONTENTS Pages

1.0 ABSTRACT 2

2.0 INTRODUCTION 3

3.0 LITERATURE REVIEW 5

4.0 EXPERIMENT OBJECTIVE 7

5.0 METHODOLOGY 8

6.0 RESULT AND DISCUSSION 10

6.1 Result 10

6.2 Discussion 19

7.0 CONCLUSION 21

7.1 Conclusion 21

7.2 Recommendations 22

8.0 REFERENCES 23

9.0 APPENDICES 24

1

1.0 ABSTRACT

Processes respond to the typical changes in their environment especially changes to their

inputs. In this experiment, the properties of first order system and dynamic properties of second

order system was demonstrated. The dynamic response of first order and second order system

was also illustrated. The demonstration of properties and responses for both systems were

important for understanding the behavior of the process due to its wide applications in process

dynamic and control in the industry. To be able to demonstrate and illustrate the process system,

computer software MATLAB precisely, was used with pre-constructed mathematical model

provided by instructor. The inputs were manipulated so that the output can be observed. Two

problems were given where they test the understanding of both systems. The problems required

students to derive the transfers function for the both unknown system. The finding for the first

problem is the magnitude versus time graph is critically damped and fits into the following

function:-

Y (s )= 0.05143.2107 s+1

The finding for the second problem is the magnitude versus time graph is under-damped and fits

into the following function:-

G (s )= 15

137.382 s2+5.743 s+1

In a nutshell, this experiment provides the basic understanding of behaviors of first order

and second order systems in demonstration using MATLAB. Besides, derivation of function

from unknown systems further reinforces the understanding of both systems. It is recommended

that students prepared themselves in basic usage of MATLAB before starting this experiment for

the purpose of easier navigation in MATLAB to increase efficiency.

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2.0 INTRODUCTION

1.1 Background

Industrial processes respond to changes in their environment, that is, their inputs. Process

inputs fall into two category,input that can be manipulated to control the process and inputs that

are not manipulated which classified as disturbance variables. Therefore, dynamic for typical

processes were derived and put into standard transfer function form such as first order and

second order system. The transfer function representation makes it easy to compare the effects of

different inputs. Besides, the transfer functions help engineer to easily generalize the dynamic

behaviour of a given process. With the transfer function, ordinary differential equation weren’t

necessarily to resolve the function when process gain (K), time constant (τ), deriving expressions

Y(s) or U(s) changes. We use the general first-order transfer function;

Y (s )U ( s)

= Kτs+1

1.1 Theory

Time behaviour of a system is important. When design a system, the time behaviour may

well be the most important aspect of it’s' behaviour. These examples are intended to show that

the ability to predict details of how a system responds is important when design systems. How

quickly a system responds is important. If have a control system that's controlling a temperature,

how long it takes the temperature to reach a new steady state is important. Then trying to control

a temperature and want the temperature to be 300°C. If the temperature goes to 350°C before it

settles out, we want to know that. Control systems designers worry about overshoot and how

close a system comes to instability.

If try to control speed of an automobile at 55mph and the speed keeps varying between 70mph

and 90mph, the design isn't very good. Oscillations in a system are not usually desirable. If

trying to control any variable and want to control it accurately, so will need to be able to predict

the steady state in a system.

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Speed of response, relative stability of the system and stability is the aspects of the

system’s time behaviour need to be concerns when design systems or circuit. Before build the

system, must know how it will perform and need to make predictions.

This dynamics systems will involves a differential equation. The differential equation is

an equation that involves the derivatives of a function as well as the function itself. If partial

derivatives are involved, the equation is called a partial differential equation; if only ordinary

derivatives are present, the equation is called an ordinary differential equation. Differential

equations play an extremely important and useful role in applied math, engineering, and physics,

and much mathematical and numerical machinery has been developed for the solution of

differential equations.

Overshoot is when a signal or function exceeds its target. It arises especially in the step

response of band limited systems such as low-pass filters. It is often followed by ringing, and at

times conflated with this latter. In control theory, overshoot refers to an output exceeding its

final, steady-state value. For a step input, the percentage overshoot is the maximum value minus

the step value divided by the step value. In the case of the unit step, the overshoot is just the

maximum value of the step response minus one. In electronics, when describing a voltage or

current step function, rise time refers to the time required for a signal to change from a specified

low value to a specified high value.

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3.0 LITERATURE REVIEW

There are measurements of the step response of a linear system such as be able to

recognize the responses of first order systems and to determine the time constant and system

gain. The recognition of the response of second order systems with complex poles or with two

real poles.

According to Seborg et all , 2011, process input refer to any variable that influences the

process output Standard process input divided into 5 types that are step input, ramp input,

rectangular pulse, sinusoidal input and impulse input. Step input is a sudden change in a process

variable can be approximated by a step change of magnitude. Industrial processes often

experience “drifting disturbances”, that is, relatively slow changes up or down for some period of

time. The rate of change is approximately constant. It is called ramp input. Rectangular pulse

represents a brief, sudden step change that then returns to its original value. Sinusoidal input is

the processes subject to periodic, or cyclic, disturbances. They can be approximated by a

sinusoidal disturbance. Sinusoidal inputs are particularly important because they play a central

role in frequency response analysis. Impulse input represents a short, transient disturbance. To

obtain an impulse input, it is necessary to inject a finite amount of energy or material into a

process in an infinitesimal length of time, which is not possible.

First order system has three types of respond that are step response, ramp response and

sinusoidal response. Same with first order, second order system also has three types of response

that are over-damped, critically damped and under-damped. The fastest response without

overshoot is obtained for the critically damped case. The under-damped response faster than

critically damped and over-damped but, oscillating with progressively decreasing amplitude.

Based on the dynamic of the first order system, it should be able to determine system gain,

natural frequency and damping ratio for systems with complex poles. Other is to determine

system gain, and values for the two poles for systems with two real poles.

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Basically the system gain is a proportional value that shows the relationship between the

magnitudes of the input to the magnitude of the output signal at steady state. Many systems

contain a method by which the gain can be altered, providing more or less power to the system.

However, increasing gain or decreasing gain beyond a particular safety zone can cause the

system to become unstable.

“Second order systems are important for a number of reasons. They are the simplest

systems that exhibit oscillations and overshoot. Many important systems exhibit second order

system behaviour” claimed by James (2006).

Second order behaviour is part of the behaviour of higher order systems and understanding

second order systems helps to understand higher order systems. Thus, the properties and

characteristic of the system depend on the order of the system.

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4.0 OBJECTIVES

The first objective of this experiment is to demonstrate the properties of a first order system for

various values of the system gain and time constant and also to demonstrate the dynamic

properties of a second order system for various values of the system gain, natural period and

damping coefficient. The second objective is to illustrate the dynamic response of a first order

and second order system to different input signals.

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5.0 METHODOLOGY

4.1 PART A: First Order System

Click once on the First and Second Order Systems button from the Main Menu then select the

First Order System button to start the first order system. The two windows will be display which

are system window and the input/output window.First,the system gain Kp (numerator

coefficient) and the system time constant τP (denominator coefficient) was set to 10.0 by

clicking once on the first order system block. After that, the step time was set to 10.0, initial

value of the step function to 0.0 and the final value of the step function to 1.0 by clicking once on

the step function block. Click OK after done with the setting. From the Simulation menu,select

Start to run the stimulation.For the output to reach the new steady state (sec), record the new

steady state value and the length of time.To responds the curve,used the Pointer button to take

several points along. Repeat step 3 to increase the value of Kp to 40.0.Set Kp back to 10.0 and

increasing the value τP of to 20.0.Besides, decrease the value of the Kp to 5.0 and decrease the

value of τP to 5.0.The data wass record in Table 1.

4.2 PART B: Second Order System

Click once on the First and Second Order Systems button from the Main Menu then select the

Second Order System button to start the first order system. The two windows will be display

which are system window and the input/output window. The following transfer function is be

used for second order:

y ( s )=K P

As2+Bs+1

By clicking Second Order System block set the system gain (Kp) to 10.0 (numerator coefficient),

the value of A to 40.0 and the value of B to 14.0 (denominator coefficient). By clicking once on

the Step Function block, the initial value of the Step Function was set to 0.0 and the final value

of the Step Function to 1.0.To start the simulation, select Start from the Simulation menu.

Recognizes either the system are over-damped, under-damped or critically damped and calculate

their characteristic likes overshoot, decay ratio, rise time, settling time and the period of

oscillation. Change the value of A to 18 and the value of B to 2. Repeat the simulation note down

their characteristic. Change the value of A to 42.25 and the value of B to 13. Repeat the

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simulation. Is the system over-damped, under-damped or critically damped. If the system is

under-damped, what is the overshoot, decay ratio, rise time, settling time and the period of

oscillation.

4.3 SYSTEM IDENTIFICATION PROBLEM

Click the left mouse button and select CLOSE on the upper left hand box of both windows to close the two windows. Select the System Identification Problem 2, from the First and Second Order Systems Menu. Run the simulation by using a step input to generate the data that can be used to detrmine the system gain, time constant and the damping cefficient.

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6.0 RESULT AND DISCUSSION

6.1 Results

A) First Order SystemTable 1

Kp τP Time Output

10 10 62 10

40 10 62 40

10 20 118 10

5 5 35 5

i) Questions

1. What is the slope of initial response?

Slope = 4−0.41

91.46−0=0.039 ≈ 0.04

2) Calculate the final output value minus the initial output value.

Δ y = 3.24 – 0

= 3.24

3) Fill in the following table with the parameter values you calculated:

Kp 3.24

τP 0.04

10

4) Give the first order transfer function of this unknown system.

Y (s )= K pτ P+1

= 3.240.04+1

ii) Exercise

1) What effect does increasing the gain have on the system output?

When the gain is increased, the system output is also increased because it is directly proportional

to the system gain Y (s ) α K p.

2) What is meant physically by a system with a large gain?

Since the equation for the output signal is ( s )= K pτ P+1

. Obviously the output is directly

proportional to the system gain kp. The larger gain will produce larger output.

3) What effect does decreasing the time constant have on the system output?

It does not have effect on the system output. It will effect on the time required for the output to

reach its first maximum value. If the time constant is decreasing means the time taken to reach

the first maximum value is shorter.

4) What is meant physically by a system with a small time constant?

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The time constant tp is the time required for the output to reach its first maximum value. If the

time constant is small mean the time required for the system to reach the first maximum value is

fast.

5) Is it possible for a system to have a negative gain? What is the expected behavior?

It is possible that for a system to have a negative gain. The system output is expected to start at

negative magnitude which will then increasing by time until certain point and reach steady state

at the input given. A negative feedback amplifier (or more commonly simply a feedback

amplifier) is an amplifier which combines a fraction of the output with the input so that

a negative feedback opposes the original signal. The applied negative feedback improves

performance (gain stability, linearity, frequency response, step response) and reduces sensitivity

to parameter variations due to manufacturing or environment. Because of these advantages,

negative feedback is used in this way in many amplifiers and control systems.

6) Is it possible for a system to have a negative time constant? What is the expected

behavior?

No, the time constant is the time required to reach the first maximum point. The time cannot be

negative. If the time is negative, it means the process reach the first maximum point before the

process started.

G (s )=O(s)I (s )

= K pτ P+1

7) Determine the response of the output if the input is a sinusoidal:

I(t) = Asin (ωt ¿

Which may transformed to:

I(s) = Aω

s2+ω2

The system response, O(s), is then:

12

O(s) = A

ωT

(s+1T )(s2+ω2)

=a

s+1T

+b

s+iω+

cs−iω

Solving then gor a,b,and c:

To solve for a, multiply by s + 1/ τ and ;et s = - 1/ τ

Aωτ

1

τ2 +ω2=a=

A τ

1+τ2 ω2

To solve for b, multiply s+ Iω, let s = -iω ,

Aωτ

(−iω+1τ )(−iω−iω)

=A

ωτ

2 i2 ω2−2iω

τ

=A

i2 2 τ ω−2 i

b=−A /2τ ω+ i ( τ ω−i

τ ω−i )=−A

2(τ ω−i)

τ2ω2+1

c will be the complex conjugate of b is:

c=

−A2

(τ ω+i)

τ2 ω2+1

Using the solution in the S&C for a sinusoidal function, the solution becomes:

L−1[ B+iCs−r−iω

B−iCs−r+iω ]=2√B2+C2ert sin (τ ω+ϕ)

Where : ϕ=tan−1(C

B )C=( A

τ2ω2+1 )( 12 )B=( −A

τ2 ω2+1 )( τ ω2 ) , r=0

2( A

τ2 ω2+1 )(12)√τ

2ω

2+1 (e0 t ¿ sin (τ ω+ϕ)

ϕ= tan−1(−τ ω)

B) Second Order SystemTable 3

Kp A B Type Overshoo Decay Rise Settling period

13

t ratio time time

10 40 14 Critically

damp

- - - - -

10 18 2 Under

damp

0.45 0.22 18.37 85 27

10 42.25 13 Critically

damp

- - - - -

For second order Kp 10 a18 b2 from the graph,

Length of a = 1.8 cm

Length of b = 4.0 cm

Length of c = 0.4 cm

Length of τr from 0 s = 1.8 cm

Length of τs from 50 s = 3.4 cm

Length of τP from 0s = 2.4cm

Length of c from 0s = 5cm

Therefore,

Overshoot, os=ab=1.8

4.0=0.45

Decay RatioDR= ca=0.4

1.8=0.22

Rise time, t r=1.2 cm3.4 cm

× 50 s=17.65 s

Settling time, t s=50 s+( 3.4 cm4.9 cm

×50 s)=84.69≈ 85 s

Period, P=(50+ 0.15.0

× 50 s)−( 2.45.0

×50 s )=51−24

= 27s

τP = 2.45

x 50=24 s

i) Questions

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1) What is the overshoot in the response?

Cmax = 4.5 Cfinal = 10.0

Overshoot = 4.510

=0.45

2) What is the period of the oscillatory response?

Period = 51 – 24 = 27s

3) Calculate the final output value minus the initial output value.

Kp = final output – initial input = 15.0 – 0.0 = 15.0

4) Fill in the following table with the parameter values you calculated:

Kp 15

τp 24

ξ 0.245

Overshoot: OS=exp ( −πζ

√1−ζ 2)

Overshoot that calculate previously = 0.4683

ξ=√ [ ln(OS )]2

π 2+[ ln(OS )]2

¿√¿¿¿

=√¿¿¿

= 0.245 (ζ <1, underdamped)

15

Period: P= 2πτ

√1−ζ 2

Period, P that calculated previously = 30.8085 s

τ=√1−ξ2

2πP

¿√1−¿¿¿ = 4.166 s

5. Derive the second order transfer function for this unknown system.

From equation 5.40, the general transfer function is shown as below

G (s )= K

τ2 s2+2 ζτs+1

¿ 100¿¿

¿ 100

17.35 s2+2.04 s+1

ii) Exercise

Consider the following values for the damping coefficient for a second order dynamic system.

Region **

Expression is faulty

**

Region **

Expression is faulty

**

Region **

Expression is faulty

**

ζ < 1 ζ = 1 ζ > 1

1. What types of poles does this system have? What types of response would be expected

for a system with a damping coefficient in Region I, II and III?

Types Region **

Expression is

faulty **

Region ** Expression

is faulty **

Region **

Expression is

faulty **

Types of response Underdamped Critically damped Overdamped

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2. Sketch the corresponding response of the output variable to a step input in Region I, II

and III.

3. How does a decrease in the damping coefficient affect the speed of response?

The damping coefficient decreases as the speed of response will decrease.

4. Which of the three responses would be expected to have a shorter response time and

sluggish?

Region I.

5. What is the trade-offs from a control perspective of the different responses?

The parameters represented in the previous equations, Kp, Tn, ζ , characterize a second

order system and determine the shape of the system’s response.

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Parameter Description

Kp Steady-state process gain.

Tn

Natural period. A measure of the process’s dynamic

response speed which is equal to the inverse of the

frequency of the oscillations (radians/time).

ζDamping factor. Determines the general shape of the

dynamic response.

Along with these variables, the poles of a function are helpful in determining the behavior

of a system. The poles of a function are found by setting the denominator of a transfer

functions equal to zero and solving for the values of s (shown in the equations below).

Based on the poles of the function, the system falls into 1 of 5 categories. If the poles

produce two negative real distinct roots, over-damped behavior is observed. Here, the

damping coefficient is greater than 1 (ζ>1). When the system produces repeated negative

roots, a damping coefficient equal to one (ζ=1) is evident and critically damped behavior

is seen. An under-damped system is experienced when the roots produce a complex

conjugate pair with a negative real part which occurs if the damping coefficient is

between 0 and 1(0<ζ<1). If ζ<0, an unstable system with exponentially growing

oscillations will occur.

τp2 s2+2 ζ τp s+1=0

18

Damping

CoefficientSystem Response

ζ>1 Over-damped behavior

ζ=1 Critically damped behavior

0<ζ<1 Under-damped behavior

ζ=0 Undamped behavior

ζ<0Unstable with exponentially

growing oscillations

6.2 Discussions

Graph 1 response is based on the first order equation which is:

Y (s )= 1010 s+1

From the equation above, we can describe that the system gain and the time constant is 10,

respectively. The output response for the steady state of the system occurred at the magnitude of

10 and 62 seconds.

Graph 2 response is based on the first order equation as below:

Y (s )= 4010 s+1

From the equation above, the value of the gain is 40 and the time constant is 10. The graph

shown a little bit different compared to graph 1 due to the value of input for gain has been

changed from 10 to 40. The steady state for the input of gain, 40 and time constant, 10 occurred

at magnitude 40 and 62 seconds.

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Graph 3 response is based on the first order equation as below:

Y (s )= 1020 s+1

From the equation above, we can describe that the gain is 10 and the time constant is 20. The

graph shown a little bit different compared to the graph 6 due to the value of input for time

constant has been changed from 10 to 20. The different can be seen through the time that has

taken to become steady even though the magnitude was the same for graph 1, the time taken is

only 62 seconds but for graph 2, the time taken was 124 seconds.

Graph 4 response is based on the first order equation as below:

Y (s )= 55 s+1

From the equation above, the value of the gain and time constant is 5 and 5, respectively. The

graph shown a little bit different compared to graph 1 due to the change of the value of input and

time constant from 10 to 5. The steady state for the input of gain, 5 and time constant, 5 occurred

at magnitude 5 and 35 seconds.

As for the conclusion, the increasing value of gain will increase the value of output of the

system.

For the second order system, graph 6 and graph 8 are critical damped. But graph 7 is under

damped. It is because the ζ value is in between 0 to 1. The second system is different compare

with first order where the first order system does not have an over shoot. The second order

system has the overshoot value. The first order system is always for the slow changes process

while the second order is usually for the fast response process.

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7.0 CONCLUSION AND RECOMMENDATIONS

7.1 Conclusions

As a conclusion, in demonstrating the dynamic properties of a first and second order

system for various values of the system gain and time constant, it is shown that the ideal

performance for applications that cannot sustain oscillatory behavior is a critical damping. In

order to achieve critically damped step response, the controller must react to an error as fast as

possible without causing the process variable to overshoot the set point. The system is said to be

overdamped when the response is slower than this meanwhile the system is said to be

underdamped when the response is faster than this. In the dynamic properties, there are two

dynamic behaviors which is first order process and second order process. For the first order

system, the value of Kp is directly proportional to the value of output which gives higher value

of Kp it will give the same goes to the value of output and the value of output approach to the

value of Kp. For the second order process, the underdamped response showing that it has shorter

response time. From the theory, the best controller for the best system would have to be

reasonably fast but less oscillatory happened.

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7.2 Recommendations

The first order system, it is aimed to find the fastest response to reach the steady state

meanwhile second order system, it is aimed to get the most ideal performance for the process.

There are some recommendations in obtaining the better result:

a) For the first order system, the result shown that the less value of time constant, τp, the

faster the reaction times taken to reach the steady state. So, we need to decrease the value

of τpin order to get the fastest results.

b) For the 2nd order process, the most ideal process is under – damped process because this

process is the fastest response although the process is fluctuates (oscillate waves).

Compared to the over – damped and critical damped process is more stable, the time

taken for the response is slower than the under – damped process. Therefore, the under –

damped process is chosen as the ideal process although it have an unstable process

(fluctuate) and the oscillate waves could be minimize by gaining the most correct

damping coefficient ξ. From the results, it’s shown that the ξ>1 is the most ideal but the

value should be near to 1 in order to minimize the oscillate waves amplitude.

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8.0 REFERENCES

1) Seborg, D. E., Edgar, T. F., Mellichamp, D. A., & Doyle III, F. J. (2011). Process

dynamics and control.NJ: John Wiley & Sons, Inc.

2) RDM, (2000). Retrieved on January 21st 2011 from

http:// www .engin.umich.edu/group/ctm/working/mac/second_order/system_id.htm

3) Mastascusa E.J (2003). Retrieved on January 21st 2011 from

http://www.facstaff.bucknell.edu/mastascu/econtrolhtml/SysDyn/SysDyn1.html

4) www.wikipedia.com

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9.0 APPENDICES

24