Transcript Lecture-20-21: Time Domain Analysis of 1st

Feedback Control Systems (FCS)
Lecture-20-21
Time Domain Analysis of 1st Order Systems
Dr. Imtiaz Hussain
email: [email protected]
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
• The first order system given below.
G(s) 
10
3s  1
• D.C gain is 10 and time constant is 3 seconds.
• And for following system
G(s) 
3
s5

3/5
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
Ts  1
t
R(s)   (s)  1
C(s) 
K
Ts  1
C(s)
Impulse Response of 1st Order System
C(s) 
K
Ts  1
• Re-arrange following equation as
C(s) 
K /T
s  1/T
• In order represent the response of the system in time domain
we need to compute inverse Laplace transform of the above
equation.
1 

 at
L 

Ce

sa
C
c(t ) 
K
T
e
t / T
Impulse Response of 1st Order System
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
• Consider the following 1st order system
R(s)
K
C(s)
Ts  1
R( s )  U ( s ) 
1
s
C(s) 
K
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

KT
Ts  1
Natural Response
Step Response of 1st Order System
T 
1
C(s)  K  

 s Ts  1 
• 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
t / T
K*(1-exp(-t/T))
11
10
Step Response
9
8
D .C Gain  K 
7
c(t)
•
Step Response of 1st Order System

c ( t )  K 1  e
If K=10 and T=1.5s then
steady state output
Input
63 %
6

10
1
5
4
3
2
Unit Step Input
1
0
0
1
2
3
4
5
Time
6
7
8
9
10
t / T
K*(1-exp(-t/T))
11
10
T=1s
9
8
T=3s
7
c(t)
•
Step Response of 1st Order System

c ( t )  K 1  e
If K=10 and T=1, 3, 5, 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.
t / T
K*(1-exp(-t/T))
11
10
K=10
9
8
7
c(t)
•
Step Response of 1st Order System

c ( t )  K 1  e
If K=1, 3, 5, 10 and T=1
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  e
t / T
  K  Ke
t / T
• Differentiating c(t) with respect to t yields
dc ( t )
dt
d

dt
dc ( t )
dt
K  Ke

K
T
e
t / T
t / T

Example#1
• Impulse response of a 1st order system is given below.
c ( t )  3e
• Find out
–
–
–
–
Time constant T
D.C Gain K
Transfer Function
Step Response
 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
C(s) 
3
1 
S  0 .5
C(s)
 (s)

 0 .5 t
C(s)
R( s )
C(s)
R( s )

3
S  0 .5

  (s)
3
S  0 .5
6
2S  1
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
Transfer Function C ( s )  6
R( s )
2S  1
Step Response
Also Draw the Step response on your notebook
Example#1
• For step response integrate impulse response
c ( t )  3e
 0 .5 t
 c ( t )dt  3  e
c s (t )  6e
 0 .5 t
 0 .5 t
dt
C
• We can find out C if initial condition is known e.g. cs(0)=0
 0 .5 0
0  6e
C  6
C
c s (t )  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  1
since R ( s ) is a step input , R ( s ) 
1
s
C(s) 
6
s 2 S  1 
6
s 2 S  1 
6
s 2 S  1 

A

s

6
s

B
2s  1
6
s  0 .5
c( t )  6  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  p1
Y=[y1 y2 .... yn];
X=[x1 x2 .... xn];
[r p k]=residue(Y, X)

r2
s  p2

rn
s  pn
k
Partial Fraction Expansion in Matlab
• If we want to expand following polynomial into partial fractions
4s  8
Y=[-4
8];
X=[1
6 8];
[r p k]=residue(Y, X)
r =[-12
p =[-4
k = []
8]
-2]
s
2
 6s  8
4 s  8
2
s  6s  8

4s  8
s
2
 6s  8
r1
s  p1

 12
s4


r2
s  p2
8
s2
Partial Fraction Expansion in Matlab
• If you want to expand a polynomial into partial fractions use
residue command.
C(s) 
Y=6;
X=[2 1 0];
[r p k]=residue(Y, X)
r =[ -6
p =[-0.5
k = []
6]
0]
6
s 2 S  1 
6
s 2 s  1 

6
s  0 .5

6
s
Ramp Response of 1st Order System
• Consider the following 1st order system
K
R(s)
C(s)
Ts  1
R( s ) 
1
s
2
K
C(s) 
s
2
Ts
 1
• The ramp response is given as

c ( t )  K t  T  Te
t / T

Ramp Response of 1st Order System
• If K=1 and T=1

c ( t )  K t  T  Te
Unit Ramp Response
10
Unit Ramp
Ramp Response
c(t)
8
6
4
error
2
0
0
5
10
Time
15
t / T

Ramp Response of 1st Order System
• If K=1 and T=3

c ( t )  K t  T  Te
Unit Ramp Response
10
Unit Ramp
Ramp Response
c(t)
8
6
4
error
2
0
0
5
10
Time
15
t / T

Parabolic Response of 1st Order System
• Consider the following 1st order system
K
R(s)
R(s) 
Ts  1
1
s
• Do it yourself
C(s)
3
Therefore,
C(s) 
K
s Ts  1
3
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)
R( s )

5
s7
T=0.13s
• Thus transfer function is
obtained as:
C(s)
R( s )

0 . 72
0 . 13 s  1

5 .5
s  7 .7
1st Order System with a Zero
C(s)
R( s )

K (1   s )
Ts  1
• Zero of the system lie at -1/α and pole at -1/T.
• Step response of the system would be:
C(s) 
C(s) 
K (1   s )
s Ts  1 
K

s

c(t )  K 1  e
t / T

c( t )  K 
K
T
K (  T )
Ts
 1
(  T )e
t / T
1st Order System with & W/O Zero
C(s)
R( s )


K
C(s)
Ts  1
R( s )
c(t )  K 1  e
t / T


c( t )  K 
K (1   s )
Ts  1
K
T
• If T>α the response will be same
c( t )  K 
K
(  n )e
t / T
T
Kn  t / T 

c(t )  K 1 
e

T


(  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
C(s)
10 (1  2 s )
9
3s  1
c(t)
R( s )

9.5
8.5
8
7.5
c ( t )  10 
10
3
( 2  3 )e
t / 3
7
6.5
0
5
10
Time
15
1st Order System with & W/O Zero
• If T<α the response of the system would look like
Unit Step Response of 1st Order Systems with Zeros
14
R( s )

10 (1  2 s )
13
1 .5 s  1
c ( t )  10 
10
1 .5
( 2  1)e
Unit Step Response
C(s)
 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
1st Order System with & W/O Zero
Unit Step Response of 1st Order Systems with Zeros
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)
R( s )

K (1   s )
Ts  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°min, how much error does the thermometer
show?
PZ-map and Step Response
C(s)
R( s )
C(s)
R( s )
jω


K
Ts  1
T  1s
10
s 1
-3
-2
-1
δ
PZ-map and Step Response
C(s)
R( s )
C(s)
R( s )
C(s)
R( s )
jω



K
Ts  1
T  0. 5 s
10
s2
5
0 .5 s  1
-3
-2
-1
δ
PZ-map and Step Response
C(s)
R( s )
C(s)
R( s )
C(s)
R( s )
jω



K
Ts  1
10
s3
3 .3
0 . 33 s  1
T  0. 33 s
-3
-2
-1
δ
Comparison
C(s)
R( s )

1
C(s)
s 1
R( s )
Step Response
0.1
0.8
0.08
0.6
0.06
Amplitude
Amplitude
s  10
Step Response
1
0.4
0.2
0

1
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 is the generic
representation of 1st order system with time
lag.
C(s)
R( s )

K
Ts  1
• Where td is the delay time.
e
 st d
First Order System With Delays
C(s)
R( s )

K
Ts  1
e
 st d
1
Unit Step
Step Response
td
t
First Order System With Delays
Step Response
10
K  10
C(s)
R( s )

10
3s  1
e
2s
Amplitude
8
6
4
2
t d  2s
0
0
T  3s
5
10
Time (sec)
15
Examples of First Order Systems
• Armature Controlled D.C Motor (La=0)
Ra
La
B
ia
u
eb
T
J

Ω(s)
U(s)

K t
Ra 
Js   B  K t K b R a 
Examples of First Order Systems
• Liquid Level System
H (s)
Qi ( s )

R
( RCs  1)
Examples of First Order Systems
• Electrical System
Eo (s)
Ei (s)

1
RCs  1
Examples of First Order Systems
• Mechanical System
X o (s)
X i(s)

1
b
k
s 1
Examples of First Order Systems
• Cruise Control of vehicle
V (s)
U (s)

1
ms  b
To download this lecture visit
http://imtiazhussainkalwar.weebly.com/
END OF LECTURES-20-21