Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: imtiaz.hussain@faculty.muet.edu.pk

URL :http://imtiazhussainkalwar.weebly.com/

Lecture-20-21 Time Domain Analysis of 1st Order Systems

Introduction• The first order system has only one pole.

• Where K is the D.C gain and T is the time constant of the system.

• Time constant is a measure of how quickly a 1st order system responds to a unit step input.

• D.C Gain of the system is ratio between the input signal and the steady state value of output.

1TsK

sRsC)()(

Introduction• The first order system given below.

1310

s

sG )(

53

s

sG )(151

53

s//

• D.C gain is 10 and time constant is 3 seconds.

• And for following system

• D.C Gain of the system is 3/5 and time constant is 1/5 seconds.

Impulse Response of 1st Order System

• Consider the following 1st order system

1TsK )(sC)(sR

0 t

δ(t)

1

1 )()( ssR

1TsKsC )(

Impulse Response of 1st Order System

• Re-arrange following equation as1

TsKsC )(

TsTKsC /

/)(1

TteTKtc /)(

• In order represent the response of the system in time domain we need to compute inverse Laplace transform of the above equation.

atCeas

CL

1

Impulse Response of 1st Order SystemTte

TKtc /)( • If K=3 and T=2s then

0 2 4 6 8 100

0.5

1

1.5

Time

c(t)

K/T*exp(-t/T)

Step Response of 1st Order System• Consider the following 1st order system

1TsK )(sC)(sR

ssUsR 1

)()(

1

TssKsC )(

1TsKT

sKsC )(

• In order to find out the inverse Laplace of the above equation, we need to break it into partial fraction expansion

Forced Response Natural Response

Step Response of 1st Order System

• Taking Inverse Laplace of above equation

11

TsT

sKsC )(

TtetuKtc /)()(

• Where u(t)=1 TteKtc /)( 1

KeKtc 63201 1 .)(

• When t=T

Step Response of 1st Order System• If K=10 and T=1.5s then TteKtc /)( 1

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

11

Time

c(t)

K*(1-exp(-t/T))

Unit Step Input

Step Response

110

Input

outputstatesteadyKGainCD .%63

Step Response of 1st Order System• If K=10 and T=1, 3, 5, 7 TteKtc /)( 1

0 5 10 150

1

2

3

4

5

6

7

8

9

10

11

Time

c(t)

K*(1-exp(-t/T))

T=3s

T=5s

T=7s

T=1s

Step Response of 1st order System

• System takes five time constants to reach its final value.

Step Response of 1st Order System• If K=1, 3, 5, 10 and T=1 TteKtc /)( 1

0 5 10 150

1

2

3

4

5

6

7

8

9

10

11

Time

c(t)

K*(1-exp(-t/T))

K=1

K=3

K=5

K=10

Relation Between Step and impulse response

• The step response of the first order system is

• Differentiating c(t) with respect to t yields

TtTt KeKeKtc //)( 1

TtKeKdtd

dttdc /)(

TteTK

dttdc /)(

Example#1• Impulse response of a 1st order system is given below.

• Find out– Time constant T– D.C Gain K– Transfer Function – Step Response

tetc 503 .)(

Example#1• The Laplace Transform of Impulse response of a

system is actually the transfer function of the system.• Therefore taking Laplace Transform of the impulse

response given by following equation.tetc 503 .)(

)(..)( sSS

sC

50

3150

3

503

.)()(

)()(

SsR

sCssC

126

SsR

sC)()(

Example#1• Impulse response of a 1st order system is given below.

• Find out– Time constant T=2– D.C Gain K=6– Transfer Function – Step Response– Also Draw the Step response on your notebook

tetc 503 .)(

126

SsR

sC)()(

Example#1• For step response integrate impulse response

tetc 503 .)(

dtedttc t 503 .)(

Cetc ts 506 .)(

• We can find out C if initial condition is known e.g. cs(0)=0

Ce 05060 .

6Ct

s etc 5066 .)(

Example#1• If initial Conditions are not known then partial fraction

expansion is a better choice

126

SsR

sC)()(

126

Ss

sC )(

12126

sB

sA

Ss

ssRsR 1

)(,)( input step a is since

5066

126

.

ssSs

tetc 5066 .)(

Partial Fraction Expansion in Matlab• If you want to expand a polynomial into partial fractions use

residue command.

Y=[y1 y2 .... yn];X=[x1 x2 .... xn];[r p k]=residue(Y, X)

kpsr

psr

psr

sxsy

n

n

2

2

1

1

)()(

Partial Fraction Expansion in Matlab• If we want to expand following polynomial into partial fractions

8684

2 sss

Y=[-4 8];X=[1 6 8];[r p k]=residue(Y, X)

r =[-12 8] p =[-4 -2] k = []

2

2

1

12 86

84psr

psr

sss

28

412

8684

2

ssss

s

Partial Fraction Expansion in Matlab

• If you want to expand a polynomial into partial fractions use residue command.

126

Ss

sC )(

Y=6;X=[2 1 0];[r p k]=residue(Y, X)

r =[ -6 6]p =[-0.5 0]k = []

ssss6

506

126

.

Ramp Response of 1st Order System• Consider the following 1st order system

1TsK )(sC)(sR

21s

sR )(

12

TssKsC )(

• The ramp response is given as

TtTeTtKtc /)(

0 5 10 150

2

4

6

8

10

Time

c(t)

Unit Ramp Response

Unit RampRamp Response

Ramp Response of 1st Order System• If K=1 and T=1 TtTeTtKtc /)(

error

0 5 10 150

2

4

6

8

10

Time

c(t)

Unit Ramp Response

Unit RampRamp Response

Ramp Response of 1st Order System• If K=1 and T=3 TtTeTtKtc /)(

error

Parabolic Response of 1st Order System• Consider the following 1st order system

1TsK )(sC)(sR

31s

sR )( 13

TssKsC )(

• Do it yourself

Therefore,

Practical Determination of Transfer Function of 1st Order Systems

• Often it is not possible or practical to obtain a system's transfer function analytically.

• Perhaps the system is closed, and the component parts are not easily identifiable.

• The system's step response can lead to a representation even though the inner construction is not known.

• With a step input, we can measure the time constant and the steady-state value, from which the transfer function can be calculated.

Practical Determination of Transfer Function of 1st Order Systems

• If we can identify T and K from laboratory testing we can obtain the transfer function of the system.

1TsK

sRsC)()(

Practical Determination of Transfer Function of 1st Order Systems

• For example, assume the unit step response given in figure.

• From the response, we can measure the time constant, that is, the time for the amplitude to reach 63% of its final value.

• Since the final value is about 0.72 the time constant is evaluated where the curve reaches 0.63 x 0.72 = 0.45, or about 0.13 second.

T=0.13s

K=0.72

• K is simply steady state value.

• Thus transfer function is obtained as:

7755

1130720

..

..

)()(

sssRsC

75

ssR

sC)()(

1st Order System with a Zero

• Zero of the system lie at -1/α and pole at -1/T.

11

Ts

sKsRsC )()()(

11

Tss

sKsC )()(

• Step response of the system would be:

1

Ts

TKsKsC )()(

TteTTKKtc /)()( TteKtc /)( 1

1st Order System with & W/O Zero

11

Ts

sKsRsC )()()(

TteTTKKtc /)()( TteKtc /)( 1

1TsK

sRsC)()(

• If T>α the response will be same

TtenTKKtc /)()(

Tte

TKnKtc /)( 1

1st Order System with & W/O Zero• If T>α the response of the system would look like

0 5 10 156.5

7

7.5

8

8.5

9

9.5

10

Time

c(t)

Unit Step Response

132110

s

ssRsC )()()(

33231010 /)()( tetc

1st Order System with & W/O Zero• If T<α the response of the system would look like

1512110

ss

sRsC

.)(

)()(

5112511010 ./)(.)( tetc

0 5 10 159

10

11

12

13

14

Time

Uni

t Ste

p R

espo

nse

Unit Step Response of 1st Order Systems with Zeros

1st Order System with a Zero

0 5 10 156

7

8

9

10

11

12

13

14

Time

Uni

t Ste

p R

espo

nse

Unit Step Response of 1st Order Systems with Zeros

T

T

1st Order System with & W/O Zero

0 2 4 6 8 100

2

4

6

8

10

12

14

Time

Uni

t Ste

p R

espo

nse

Unit Step Response of 1st Order Systems with Zeros

T

T

1st Order System Without Zero

Home Work

• Find out the impulse, ramp and parabolic response of the system given below.

11

Ts

sKsRsC )()()(

Example#2

• A thermometer requires 1 min to indicate 98% of the response to a step input. Assuming the thermometer to be a first-order system, find the time constant.

• If the thermometer is placed in a bath, the temperature of which is changing linearly at a rate of 10°min, how much error does the thermometer show?

PZ-map and Step Response

1TsK

sRsC)()(

-1-2-3δ

jω

110

ssR

sC)()( sT 1

PZ-map and Step Response

1TsK

sRsC)()(

-1-2-3δ

jω

210

ssR

sC)()( sT 50.

1505

ssR

sC.)(

)(

PZ-map and Step Response

1TsK

sRsC)()(

-1-2-3δ

jω

310

ssR

sC)()( sT 330.

133033

ssR

sC.

.)()(

Comparison

11

ssR

sC)()(

101

ssR

sC)()(

Step Response

Time (sec)

Ampl

itude

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

0 0.1 0.2 0.3 0.4 0.5 0.60

0.02

0.04

0.06

0.08

0.1Step Response

Time (sec)

Ampl

itude

First Order System With Delays

• Following transfer function is the generic representation of 1st order system with time lag.

• Where td is the delay time.

dsteTsK

sRsC

1)()(

First Order System With Delays

dsteTsK

sRsC

1)()(

1

Unit StepStep Response

ttd

First Order System With Delays

sessR

sC 2

1310

)(

)(

0 5 10 15

0

2

4

6

8

10

Step Response

Time (sec)

Ampl

itude

st d 2 sT 3

10K

Examples of First Order Systems

• Armature Controlled D.C Motor (La=0)

abt

at

RKKBJsRK

U(s)Ω(s)

uia T

Ra La

J

B

eb

V f=constant

Examples of First Order Systems

• Liquid Level System

)()()(

1RCsR

sQsH

i

Examples of First Order Systems

• Electrical System

11

RCssE

sE

i

o

)()(

Examples of First Order Systems

• Mechanical System

1

1

skbsX

sX

i

o

)()(

Examples of First Order Systems

• Cruise Control of vehicle

bmssUsV

1)()(

END OF LECTURES-20-21

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