Lecture 13 14-time_domain_analysis_of_1st_order_systems

Feedback Control Systems (FCS)

Dr. Imtiaz Hussainemail: [email protected]

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

Lecture-13-14 Time Domain Analysis of 1st Order Systems

mailto:[email protected]

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.

1

Ts

K

sR

sC

)()(

Introduction• For the first order system given below

13

10

ssG )(

5

3

ssG )(

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

1Ts

K)(sC)(sR

0 t

δ(t)

1

1 )()( ssR

1

Ts

KsC )(

Impulse Response of 1st Order System

• Re-arrange above equation as1

Ts

KsC )(

Ts

TKsC

//

)(1

TteT

Ktc /)(

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

atAeas

AL

1

Impulse Response of 1st Order SystemTte

T

Ktc /)( • 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

1Ts

K)(sC)(sR

ssUsR

1 )()(

1

Tss

KsC )(

1

Ts

KT

s

KsC )(

• 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

1

1

Ts

T

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

1

10

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

TtKeKdt

d

dt

tdc /)(

TteT

K

dt

tdc /)( (impulse response)

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

31

50

3

50

3

.)()(

)()(

SsR

sC

s

sC

12

6

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 .)(

12

6

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

12

6

SsR

sC

)()(

12

6

SssC )(

1212

6

s

B

s

A

Ss

ssRsR

1)(,)( input step a is since

50

66

12

6

.

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)

kps

r

ps

r

ps

r

sx

sy

n

n

2

2

1

1

)()(

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

86

842

ss

s

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

84

ps

r

ps

r

ss

s

2

8

4

12

86

842

ssss

s

Partial Fraction Expansion in Matlab

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

12

6

SssC )(

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

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

ssss

6

50

6

12

6

.

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

1Ts

K)(sC)(sR

2

1

ssR )(

12

Tss

KsC )(

• The ramp response is given as

TtTeTtKtc /)(

0 5 10 150

2

4

6

8

10

Time

c(t)

Unit Ramp Response

Unit Ramp

Ramp 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 Ramp

Ramp 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

1Ts

K)(sC)(sR

3

1

ssR )(

13

Tss

KsC )(

• 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.


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

1

Ts

K

sR

sC

)()(


• 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:

77

55

1130

720

..

..

)()(

sssR

sC

7

5

ssR

sC

)()(

1st Order System with a Zero

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

1

1

Ts

sK

sR

sC )()()(

11

Tss

sKsC

)()(

• Step response of the system would be:

1

Ts

TK

s

KsC

)()(

TteTT

KKtc /)()(

Partial Fractions

Inverse Laplace

1st Order System with & W/O Zero (Comparison)

1

1

Ts

sK

sR

sC )()()(

TteTT

KKtc /)()( TteKtc /)( 1

1

Ts

K

sR

sC

)()(

• If T>α the shape of the step response is approximately same (with offset added by zero)

TtenT

KKtc /)()(

Tte

T

nKtc /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

13

2110

s

s

sR

sC )()()(

3323

1010 /)()( tetc offset

𝑜𝑓𝑓𝑠𝑒𝑡=𝐾 +𝐾𝑇

(𝛼−𝑇 )

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

151

2110

s

s

sR

sC

.)(

)()(

511251

1010 ./)(

.)( tetc

0 5 10 159

10

11

12

13

14

Time

Uni

t S

tep

Res

pons

e

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 S

tep

Res

pons

eUnit Step Response of 1st Order Systems with Zeros

T

T

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

1st Order System with & W/O Zero

T

T

1st Order System Without Zero

Home Work

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

1

1

Ts

sK

sR

sC )()()(

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°C/min, how much error does the thermometer show?

PZ-map and Step Response

1

Ts

K

sR

sC

)()(

-1-2-3δ

jω

1

10

ssR

sC

)()( sT 1


1

Ts

K

sR

sC

)()(

-1-2-3δ

jω

2

10

ssR

sC

)()( sT 50.

150

5

ssR

sC

.)()(


1

Ts

K

sR

sC

)()(

-1-2-3δ

jω

3

10

ssR

sC

)()( sT 330.

1330

33

ssR

sC

..

)()(

Comparison

1

1

ssR

sC

)()(

10

1

ssR

sC

)()(

Step Response

Time (sec)

Am

plitu

de

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)

Am

plitu

de

First Order System With Delays

• Following transfer function represents the 1st order system with time lag.

• Where td is the delay time.

dsteTs

K

sR

sC

1)()(


dsteTs

K

sR

sC

1)()(

1

Unit StepStep Response

ttd


sessR

sC 2

13

10

)()(

0 5 10 15

0

2

4

6

8

10

Step Response

Time (sec)

Am

plit

ude

st d 2sT 3

10K

Examples of First Order Systems

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

abt

at

RKKBJs

RK

U(s)

Ω(s)

uia T

Ra La

J

B

eb

V f=constant


• Electrical System

1

1

RCssE

sE

i

o

)()(


• Mechanical System

1

1

sk

bsX

sX

i

o

)()(


• Cruise Control of vehicle

bmssU

sV

1

)()(

END OF LECTURES-13-14

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Lecture 13 14-time_domain_analysis_of_1st_order_systems

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