Lecture 13 14-time_domain_analysis_of_1st_order_systems

Feedback Control Systems (FCS)

Lecture-13-14
Time Domain Analysis of 1st Order Systems
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/

Introduction
• The first order system has only one pole.
C(s)
R( s )

K
Ts

1

• 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.

Introduction
• For the first order system given below
10

G(s)

3s

1

• D.C gain is 10 and time constant is 3 seconds.
• And for following system
3/5

3

G(s)
s

5

1 / 5s

1

• 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
δ(t)

1

K

R( s )
0

C(s)

1

Ts
t

R( s )

(s)

K

C(s)
Ts

1

1

K

C(s)

1

Ts

• Re-arrange above equation as
K /T

C(s)
s

1/T

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

A

1

s

Ae
a

at

c( t )

K
T

e

t /T

K

• If K=3 and T=2s then c ( t )

e

t /T

T
K/T*exp(-t/T)
1.5

c(t)

1

0.5

0

0

2

4

6
Time

8

10

Step Response of 1st Order System
K

R( s )

Ts
R( s )

C(s)

1
1

U (s)

s
K

C(s)

s Ts

1

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

C(s)

K
s

Natural Response

KT
Ts

1

C(s)

K

1

T

s

1

Ts

• Taking Inverse Laplace of above equation

c( t )

K u(t )

e

• Where u(t)=1
c( t )

K 1

e

t /T

t /T

• When t=T
c( t )

K 1

e

1

0 . 632 K

• If K=10 and T=1.5s then

c( t )

K 1

e

t /T

K*(1-exp(-t/T))
11
10
Step Response

9
8

K

10

Input

D .C Gain

steady state output

1

c(t)

7

63 %

6
5
4
3
2

Unit Step Input
1
0

0

1

2

3

4

5
Time

6

7

8

9

10

• If K=10 and T=1, 3, 5, 7

c( t )

K 1

e

t /T

K*(1-exp(-t/T))
11
10

T=1s

9
8

T=3s

c(t)

7

T=5s

6
T=7s

5
4
3
2
1
0

0

5

10
Time

15

Step Response of 1st order System
• System takes five time constants to reach its
final value.

• If K=1, 3, 5, 10 and T=1

c( t )

K 1

e

t /T

K*(1-exp(-t/T))
11
10
K=10

9
8

c(t)

7
6

K=5

5
4
K=3

3
2

K=1

1
0

0

5

10
Time

15

Relation Between Step and impulse
response
• The step response of the first order system is
c( t )

K 1

t /T

e

K

Ke

t /T

• Differentiating c(t) with respect to t yields
dc ( t )

d

dt

dt

K

Ke

dc ( t )

K

dt

T

e

t /T

t /T

(impulse response)

Example#1
• Impulse response of a 1st order system is given below.
c(t )

• Find out
–
–
–
–

Time constant T
D.C Gain K
Transfer Function
Step Response

3e

0 .5 t

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.
c(t )

3e

3

C(s)

3

1

0 .5

S

0 .5 t

0 .5

S

C(s)

C(s)

(s)

R( s )

C(s)
R( s )

3
S
6

2S

1

0 .5

(s)

Example#1
• Impulse response of a 1st order system is given below.
c(t )

3e

0 .5 t

• Find out
–
–
–
–
–

Time constant T=2
D.C Gain K=6
6
Transfer Function C ( s )
R( s )
2S 1
Step Response
Also Draw the Step response on your notebook

Example#1
• For step response integrate impulse response
c(t )

0 .5 t

3e

c ( t )dt

0 .5 t

3 e

c s (t )

dt

0 .5 t

6e

C

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

6e
C

c s (t )

0 .5 0

C

6

6

6e

0 .5 t

Example#1

• If initial Conditions are not known then partial fraction
expansion is a better choice
C(s)
R( s )

6
2S

since R ( s ) is a step input , R ( s )

1

1
s
6

C(s)

s 2S

6
s 2S

A
1

6
s 2S

1

s

B
2s

6
1

c( t )

s

6

1
6

s

0 .5

6e

0 .5 t

Partial Fraction Expansion in Matlab
• If you want to expand a polynomial into partial fractions use
residue command.
y( s )
x( s )

r1
s

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

r2
p1

s

rn


p2

s

k
pn

• If we want to expand following polynomial into partial fractions
4s

Y=[-4
8];
X=[1
6 8];

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

8]
-2]

s

2

8

6s

8

4s
s

2

8

r1

6s
4s
s

2

8
8

6s

s

r2
p1

s

12
8

s

4

p2
8

s

2

• If you want to expand a polynomial into partial fractions use
residue command.
C(s)

Y=6;
X=[2 1 0];
r =[ -6
p =[-0.5
k = []

6]
0]

6
s 2S

1

6
s 2s

6
1

s

6

0 .5

s

Ramp Response of 1st Order System
K

R( s )

C(s)

1

Ts

1

R( s )

s

2

K

C(s)
s

2

1

Ts

• The ramp response is given as

c( t )

K t

T

Te

t /T

• If K=1 and T=1

c( t )

K t

T

Unit Ramp Response
10

Unit Ramp
Ramp Response

c(t)

8

6

4

error

2

0

0

5

10

Time

15

Te

t /T

• If K=1 and T=3

c( t )

K t

T

Unit Ramp Response

10

Unit Ramp
Ramp Response

c(t)

8

6

4

error

2

0

0

5

10
Time

15

Te

t /T

Parabolic Response of 1st Order System
K

R( s )

R(s)

Ts
1
s

• Do it yourself

3

Therefore,

C(s)

1

C(s)

K
3

s Ts

1

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.

C(s)
R( s )

K
Ts

1

Practical Determination of Transfer Function
of 1st Order Systems
• For example, assume the unit
step response given in figure.

K=0.72

• 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.
• K is simply steady state value.

C(s)

5

R( s )

s

7

T=0.13s

• Thus transfer function is
obtained as:
C(s)
R( s )

0 . 72
0 . 13 s

5 .5
1

s

7 .7

1st Order System with a Zero
C(s)

K (1

s)

R( s )

Ts

1

• Zero of the system lie at -1/α and pole at -1/T.
• Step response of the system would be:
K (1

C(s)

s)
1

s Ts
K(

K

C(s)

s

c( t )

K

Ts

K
T

(

T)

Partial Fractions

1

T )e

t /T

Inverse Laplace

1st Order System with & W/O Zero (Comparison)
C(s)
R( s )
c( t )

K 1

C(s)

Ts
e

1
t /T

K (1

R( s )

K

Ts

c( t )

K

K

(

s)
1
T )e

t /T

T

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

K

K

( n )e

t /T

T

c (t )

K 1

n
T

e

t /T

1st Order System with & W/O Zero
• If T>α the response of the system would look like
Unit Step Response
10

10 (1

R( s )

3s

2s)

9

1
c(t)

C(s)

9.5

8.5
8
7.5

c( t )

10

10
3

(2

3 )e

t/3

7
6.5

offset
0

5

10
Time

15

• If T<α the response of the system would look like
Unit Step Response of 1st Order Systems with Zeros
14

10 (1

R( s )

c(t )

1 .5 s

10

10
1 .5

2s)

13

1

(2

Unit Step Response

C(s)

1)e

t / 1 .5

12

11

10

9

0

5

10
Time

15

1st Order System with a Zero
Unit Step Response of 1st Order Systems with Zeros
14

Unit Step Response

13
12
11

T
10

T

9
8
7
6

0

5

10
Time

15

Unit Step Response of 1st Order Systems
14

Unit Step Response

12

T

10

T
8
6
1st Order System Without Zero

4
2
0

0

2

4

6
Time

8

10

Home Work
• Find out the impulse, ramp and parabolic
response of the system given below.
C(s)

K (1

R( s )

Ts

s)
1

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
jω

C(s)
R( s )
C(s)
R( s )

K
Ts

1

10
s

1s

T

1

-3

-2

-1

δ

jω

C(s)
R( s )

K

C(s)
R( s )

1

Ts

T

10
s

2

C(s)

5

R( s )

0 .5 s

-3

1

0 .5 s

-2

-1

δ

jω

C(s)
R( s )

K

C(s)
R( s )

1

Ts

T

10
s

3

C(s)

0 . 33 s

-3

3 .3

R( s )

0 . 33 s

1

-2

-1

δ

Comparison
C(s)

C(s)

1

R( s )

R( s )

1

s

Step Response

s

10

Step Response

1

0.1

0.08

0.6

0.06

Amplitude

0.8

Amplitude

1

0.4

0.2

0

0.04

0.02

0

1

2

3
Time (sec)

4

5

6

0

0

0.1

0.2

0.3
Time (sec)

0.4

0.5

0.6

First Order System With Delays
• Following transfer function represents the 1st
order system with time lag.
C(s)
R( s )

K
Ts

1

• Where td is the delay time.

e

st d

C(s)
R( s )

K
Ts

1

st d

e

1

Unit Step
Step Response

td

t

Step Response
10
K

10

C(s)
R( s )

10
3s

1

e

2s

Amplitude

8

6

4

2

td
0
0

2s

T

3s

5

10
Time (sec)

15

Examples of First Order Systems
• Armature Controlled D.C Motor (La=0)
Ra

La
B

u

ia

Ω(s)
U(s)

eb

T

K t Ra
Js

B

K t K b Ra

J

• Electrical System

Eo (s)

1

Ei (s)

RCs

1

• Mechanical System

X o(s)
X i(s)

1
b
k

s

1

• Cruise Control of vehicle

V (s)
U (s)

1
ms

b

To download this lecture visit
http://imtiazhussainkalwar.weebly.com/

END OF LECTURES-13-14

Lecture 13 14-time_domain_analysis_of_1st_order_systems

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