Control Systems (CS)

Engr. Mehr GulLecturer Electrical Engg. Deptt.

LectureTime 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 constantof 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 inputsignal and the steady state value of output.

1Ts

K

sR

sC

)(

)(

Introduction The first order system given below.

13

10

ssG )(

5

3

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 1/5seconds.

Impulse Response of 1st Order System

Consider the following 1st order system

1Ts

K)(sC)(sR

0t

(t)

1

1 )()( ssR

1Ts

KsC )(

Impulse Response of 1st Order System

Re-arrange following equation as

1Ts

KsC )(

Ts

TKsC

/

/)(

1

TteT

Ktc /)(

In order represent the response of the system in time domainwe need to compute inverse Laplace transform of the aboveequation.

atCeas

CL

1

Impulse Response of 1st Order System

TteT

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 )(

1Ts

KT

s

KsC )(

In order to find out the inverse Laplace of the above equation, weneed 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 itsfinal 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 /)(

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 impulseresponse 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

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 .

6C

ts etc

5066 .)(

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

expansion is a better choice

12

6

)(

)(

SsR

sC

126

SssC )(

12126

s

B

s

A

Ss

ssRsR

1)(,)( input step a is since

5066

12

6

.

ssSs

tetc 5066 .)(

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'stransfer function analytically.

Perhaps the system is closed, and the component parts arenot easily identifiable.

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

With a step input, we can measure the time constant and thesteady-state value, from which the transfer function can becalculated.

Practical Determination of Transfer Function of 1st Order Systems

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

1Ts

K

sR

sC

)(

)(

Practical Determination of Transfer Function of 1st Order Systems

For example, assume the unitstep response given in figure.

From the response, we canmeasure the time constant, thatis, the time for the amplitude toreach 63% of its final value.

Since the final value is about0.72 the time constant isevaluated where the curvereaches 0.63 x 0.72 = 0.45, orabout 0.13 second.

T=0.13s

K=0.72

K is simply steady state value.

Thus transfer function isobtained as:

77

55

1130

720

.

.

.

.

)(

)(

sssR

sC

7

5

ssR

sC

)(

)(

END.....