Note extracted from

Control Systems Engineering (CSE), 5th Edition

Norman S. Nise,

Chapter 4: Transient Response

Recap

• What is control system• Element an components of control system• System representation, process and subsystem

of control system• Mathematical modeling of control system• La place function approximation of control

system • Transfer function determination• Types and classification of control system• Terminologies of control system

Objective

• Control system process, subsystem

• System stability for time and transient response of control system

• For system operation and maintenance

• For system design

IQRA: Students are advised to use part of their study time read and study this chapter

A. automatic control process

Control Loops

• An open loop system has no feedback and controller action is not related to final result. Consider a domestic central heating system as an example, in which fuel supply is varied manually or automatically by external ambient temperature. Room temperature will be maintained at a reasonable value related to outside conditions. However room temperature does not control fuel supply so that this is open loop. The word loop is really a misnomer.

•

Control Loops

to the open loop shown, add human operator, so closing the loop (dotted lines on sketch). This is a manually controlled closed loop system.

Control Loops(Source: Instrumentation and Control Systems by Leslie Jackson)

Control system and subsystem

Automatic control process• It refers to automatically controlled systems. The control action is

dependent on the output. A detecting or measuring element will obtain a signal related to this output which is fed to the transmitter.

• From the transmitter the signal is then passed to a comparator.• The comparator will contain some set or desired value of the

controlled condition which is compared to the measured value signal.

• Any deviation or difference between the two values will result in an output signal to the controller. The controller will then take action in a manner related to the deviation and provide a signal to a correcting unit.

• The correcting unit will then increases or decrease its effect on the system to achieve the desired value of the system variable. e.g. automatic control of the temperature of engine cooling water.

• Note: The comparator is usually built in to the controller unit. The transmitter, controller and regulating unit are supplied with an operating medium e.g. compressed air, hydraulic oil or electricity, in order to function.

Automatic control process

Automatic control process

• The basic theory involved in the maintenance of an engineering system, such as boiler water level or jacket cooling water temperature of a diesel engine, at a required operating condition without human intervention.

• The majority of such systems employ automatic closed loop control which may be defined as a system in which, without humans intervention, the actual value of a controlled condition, such as level, flow, temperature, viscosity or pressure is compared with a desired (or set) value representing the required operating condition, with corrective action being taken should a deviation or difference occur between these two values.

Automatic control process

• To fully understand the theory involved it is necessary to use terminology peculiar to this subject, such terminology will, however, be kept to a minimum and is based on the British Standards Publication BS 1523.

• Some of the more common terms are listed below and their application to a typical system is shown above. The operation of a plant under automatic control in order to control such variables as temperature, level, flow, viscosity etc., is know as Process Control. Here the process is the cooling of the lubricating oil in the heat exchanger.

Automatic control process

• Working round the system from the Process, the Controlled Condition is the temperature of the lubricating oil, and this is monitored by a sensor or Detecting Element, which could be a filled systems thermometer connected to a Bourdon tube.

• This operates a nozzle/flapper device which produces a pneumatic signal, known as the Measured Value, which is directly related to the temperature of the lubricating oil. This nozzle/flapper amplifier is known in control engineering terms as a value signal is taken to a Comparing Element or Comparator forming part of the Automatic Controller or Controlling Unit.

Automatic control process

• Here it is compared (one method is shown) with a signal representing the required lubricating oil operating temperature or the Set Value (Set Point) or Desired Value. (There could be a difference between these terms which will be explained later in the text).

• If the Set Value and Measured Value are the same, the beam will not move, but if there is a difference between these signals, known as the Deviation or Error it means that the lubricating oil temperature at the outlet from the cooler is not at the required operating temperature (Set Point, etc). Action has to be taken to restore it.

• The difference in signal pressures on the diaphragms will rotate the beam about the pivot, the movement being the Error Signal and this will operate the Controlling Element.

Automatic control process

B. System , subsystem stability analysis

Classification of control system

• First order control system

• Second order control system

Control system

First order control system (Refer to pg 154,CSE)

i. Time response from the transfer function

ii. Poles and zeros to determine the response of a control system

iii. Transient response of systems

iv. Approximate higher-order system as first or second order

• The output response of a system = the forced response (steady-state response) + natural response (zero input response)

• The poles of a transfer function are:(1) the values of the Laplace transform variable, s , that

cause the transfer function to become infinite, or (2) any roots of the denominator of the transfer function that are common to roots of the numerator.

• The zeros of a transfer function are: (1) the values of the Laplace transform variable, s , that

cause the transfer function to become zero, or (2) any roots of the numerator of the transfer function that are common to roots of the denominator.

Output response, Poles, and Zeros (Pg 154)

Figure 4.2 (pg.156)

Control system

a. System showing input and output;

b. pole-zero plot of the system;

c. evolution of a system response.

i. A pole of the input function generates the form of the forced response.

ii. A pole of the transfer function generates the form of the natural response.

iii. A pole on the real axis generates an exponential response.

iv. The zeros and poles generate the amplitudes for both the forced and natural responses.

Poles

fig_04_02

Effect of real axis on transient response

)5)(4)(2(

3

sss

sssR

1)( )(sC

542)( 4321

s

k

s

k

s

k

s

ksC

Natural response

Forced response

ttt ekekekktc 54

43

221)(

Poles Example 4.1

Given the system in fig 4.3 , write output, c(t) in general terms, specify the forced and natural parts of the solution

Do exercise 4.1!!

First-Order Systems (without zero)

)()()()(

ass

asGsRsC

atnf etctctc 1)()()(

Figure 4.4a. First-order system

b. pole plot

Input -> unit step , R (s)= 1/s)

Step response

Where:Forced response cf(t)=1, system pole is at –a, generated response is ate

Time constant

37.01

1

eeat

at

• The time constant can be described as the time for to decay to 37% of its initial value. Alternately, the time constant is the time it takes for the step response to rise to of its final value.

Figure 4.5 First-order system

response to a unit step 63.037.011)(1

1

at

at

atetx

The reciprocal of the time constant has the units (1/seconds), or frequency. Thus, we call the parameter the exponential frequency.

Time response terms

In first response change of parameter result to change of speed of response

• Rise time: Rise time is defined as the time for the waveform to go from 0.1 to 0.9 of its final value.

• Settling time: Settling time is defined as the time for the response to reach, and stay within, 2% (or 5%) of its final value.

First order transfer function via testing (Refer to pg 159)

• Difficult analytical method• System boxed up• Access difficulty• Identify k and from lab testing

• Time constant (TC)=63% of t• a=1/TC • T=K/a• Substitute values back to the equation

Do exercise 4.2!!

Second-Order Systems (pg 160)

• Involve wide range of response• Change of parameter lead to change of form of

response• Mathematical analysis involve: unit step input

definition->partial fraction expansion->inverse la place transform

• Solution through pole and zero facilitate determination of response that bypass mathematical analysis

Second-Order Systems

Figure 4.7Second-ordersystems, pole plots,and step responses

Changing and give different types of response

Figure 4.10Step responses for second-order system damping cases

Second-Order Systems

Solve exercise 4.3

• 1. Overdamped response:

Poles: Two real at

Natural response: Two exponentials + time constants equal to the reciprocal of the pole location

21,

tt ekektc 2121)(

Natural response of second-Order Systems

• 3. Undamped response:Poles: Two imaginary at Natural response: Undamped sinusoid with radian frequency = to the imaginary part of the poles

• 4. Critically damped responses:Poles: Two real at Natural response: One term ->is an exponential whose time constant is equal to the reciprocal of the pole location.

Another term ->is the product of time and an exponential with time constant equal to the reciprocal of the pole location

1j

)cos()( 1 tAtc

1

tt tekektc 1121)(

Natural response of second-Order Systems

Figure 4.8Second-order step response componentsgenerated by complex poles

dd j

• 2. Underdamped responses:

Poles: Two complex at

Natural response: Damped sinusoid + exponential envelope whose time constant is equal to the reciprocal of the pole’s real part.

The radian frequency of the sinusoid+ (the damped frequency of oscillation=is equal to the imaginary part of the poles)

)cos()( tAetc dtd

fig_04_09

Example (Refer to Pg 162)

• Natural Frequency: The natural frequency of a second-order system is the frequency of oscillation of the system without damping.

• Damping Ratio: The damping ratio is defined as the ratio of exponential decay frequency to natural frequency.

Consider the general system:

Without damping, bass

bsG

2)(

bbs

bsG n

2)(

b

a

radfrequencyNatural

frequencydecaylExponentia

n

2/||

sec)/(

2

4

2,

2

4

2

2

2

2

1

abj

as

abj

as

na 2 22

2

2)(

nn

n

sssG

General second – order system

Figure 4.11Second-order response as a

function of damping ratio

fig_04_12

Example 4.4 (Pg 167)

For each of the following system, find the damping ratio, and report the kind of response

Transform eqution to required formFind a and Wnsubstitute a and e from the equation@ >1overdamped, =1critically damped<1underdamped=0undamped

Underdamped Second-Order Systems

)1()(

1)(

)2()(

222

2321

22

2

nn

nn

nn

n

s

KsK

s

K

ssssC

)1()(

11

)(11

222

2

2

nn

nn

s

s

s

ttetc nn

tn 2

2

2 1sin1

1cos1)(

Step response

Taking the inverse Laplace transform

te ntn 2

21cos

1

11

where

2

1

1tan

Figure 4.13Second-order underdamped

responses for damping ratio values

Underdamped damping ration variation

Figure 4.14Second-order underdamped

response specifications

%2

• Peak time: The time required to reach the first, or maximum, peak.

• Percent overshoot: The amount that the waveform overshoots the steady-state, or final, value at the peak time, expressed as a percentage of the steady-state value.

• Settling time: The time required for the transient’s damped oscillations to reach and stay within (or ) of the steady-state value.

• Rise time: The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value.

%5

22

2

2)()]([

nn

n

ssssCtcL

)1()(

11

222

2

2

nn

nn

s

tetc ntn n 2

21sin

1)(

Setting the derivative equal to zero yields

Peak time: 2

2

11

n

n

ntornt

21

n

pt

Analysis of underdamped second order

Evaluation of peak time(Tp):

Completing the square we have

Evaluation of percent overshoot ( ):

Evaluation of settling time:

The settling time is the time it takes for the amplitude of the decaying

sinusoid to reach o.o2, or

,

where is the imaginary part of the pole and is called the damped

frequency of oscillation, and is the magnitude of the real part of the

pole and is the exponential damping frequency.

100% max

final

final

c

ccOS

)1/(max

2

1)( etcc p 100% )1/( 2

eOS

02.01

12

tne

nnst

4102.0ln( 2

dn

pt

21 dn

st 44

d

d

fig_04_15

fig_04_16

Example 4.5

Figure 4.17Pole plot for an underdamped

second-order system

)1)(1(2)(

22

2

22

2

nnnn

n

nn

n

jsjssssG

Figure 4.18Lines of constant peak timeTp , settlingtimeTs , and percent overshoot, %OS

Note: Ts2 < Ts1 ;Tp2 < Tp1; %OS1 <%OS2

Figure 4.19Step responses of

second-orderunderdamped systems

as poles move:

a. with constant real part;

b. with constant imaginary part;

c. with constant damping ratio

dn

pt

21 100% )1/( 2

eOS

2173 nndd jjj

dnst

44

Find peak time, percent overshoot, and settling time from pole location.

, ,

Figure 4.21Rotational mechanical system

Design: Given the rotational mechanical system, find J and D to yield 20% overshoot and a settling time of 2 seconds for a step input of torque T(t).

Under certain conditions, a system with more than two poles or with zeros can be approximated as a second-order system that has two complex dominant poles. Once we justify this approximation, the formulae for percent overshoot, settling time, and peak time can be applied to these higher-order systems using the location of the dominant poles.

rdn

dn

s

D

s

CsB

s

AsC

22)(

)()(

Figure 4.23Component responses of a three-pole system:

a. pole plot;b. component responses: nondominant pole is near dominant second-order pair (Case I),

far from the pair (Case II), and at infinity (Case III)

Remark:If the real pole is five timesfarther to the left than the dominant poles, we assumethat the system is represented by its dominant second-order pair of poles.

fig_04_24

Example 4.8: Control system engineering

cs

bcac

bs

cbab

cs

B

bs

A

csbs

assT

)/()()/()(

))(()(

System response with zeros (Pg 182, control system engineering)

If the zero is far from the poles, then is large comared to and . ca b

))((

)/(1)/(1)(

csbs

a

cs

bc

bs

cbasT

Figure 4.25Effect of adding

a zero to a two-pole system

)828.21( j

Figure 4.26Step response of a nonminimum-phase system

)()()()( saCssCsCas

fig_04_27

Example 4.8: Control system engineering (Pg. 184)

fig_04_29a

Effect of nonlinearity on system response

Time Domain Solution of State Equations

)()()( tButAxtx

t tAAt dBuexetx0

)( )()0()(

t

dButxt0

)()()0()(

where is called the state-transition matrix.Atet )(

)0()()]0()([)]([ 1xAsIxttx LL

)(])[( 1 tAsI -1L

fig_04_28

fig_04_29b

fig_04_30a

fig_04_30b

fig_04_31a

fig_04_31b

fig_04_32

fig_04_33

fig_04_34

fig_04_35

fig_04_36

Home Work

• 1,2,4,6, 10, 12, 13,15,16,, 17,18,21,22, 23, 24, 25,23, 26,27, 53,58,62

fig_04_21

fig_04_22

figu

n_04_01

figun_04_02

figun_04_03

figun_04_05

figun_04_06

figun_04_09a

figun_04_09b

figun_04_09c

figun_04_10

figun_04_11

figun_04_13

figun_04_14

figun_04_15

figun_04_16

figun_04_21

figun_04_24

Summary

• Automatic control process

• System, subsystem and stability analysis

Test 1 Material to Study for test Problem set and short question from Control system engineering book

chapter 1 and 2 and 5 PPT-L1; PPT-L2; PPT-L3 Lecture note is a gift The basics of control system Understand analysis and design objective of control and automation system the basics of control system components Study example of control system Study system representation of and automation system Types and classification of control system Mathematical modeling of mechanical and electrical control system Partial fraction, La place transform and transfer function of control system Study the examples and solutions Block manipulation