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 )
K

R( s ) 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.
10
G( s ) 
3s  1
• D.C gain is 10 and time constant is 3 seconds.
• And for following system
3
3/ 5
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)
K
Ts  1
R(s )
1
0
t
R( s )   ( s )  1
K
C( s ) 
Ts  1
C(s )
Impulse Response of 1st Order System
K
C( s ) 
Ts  1
• Re-arrange following equation as
K /T
C( s ) 
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 
C 
L 
  Ce  at
sa
K t / T
c(t )  e
T
Impulse Response of 1st Order System
K t / T
• If K=3 and T=2s then c(t ) 
e
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
Ts  1
C(s )
1
R( s )  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
K
KT
C( s ) 

s 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 t / T
• Where u(t)=1


c(t )  K 1  e
t / T
• When t=T


c(t )  K 1  e 1  0.632K
Step Response of 1st Order System

c(t )  K 1  e t / T
• If K=10 and T=1.5s then

K*(1-exp(-t/T))
11
10
Step Response
9
8
D.C Gain  K 
c(t)
7
63%
6
steady state output 10

Input
1
5
4
3
2
Unit Step Input
1
0
0
1
2
3
4
5
Time
6
7
8
9
10
Step Response of 1st Order System

c(t )  K 1  e t / T
• If K=10 and T=1, 3, 5, 7

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.
Step Response of 1st Order System
• 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  e t / T  K  Ket / T
• Differentiating c(t) with respect to t yields

dc(t ) d

K  Ke t / T
dt
dt
dc(t ) K t / T
 e
dt
T

Example#1
• Impulse response of a 1st order system is given below.
c(t )  3e 0.5t
• Find out
–
–
–
–
Time constant T
D.C Gain K
Transfer Function
Step Response
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 0.5t
3
3
C( s ) 
1 
  (s)
S  0.5
S  0.5
C( s ) C( s )
3


 ( s ) R( s ) S  0.5
C( s )
6

R( s ) 2 S  1
Example#1
• Impulse response of a 1st order system is given below.
c(t )  3e 0.5t
• Find out
–
–
–
–
–
Time constant T=2
D.C Gain K=6
Transfer Function C ( s )  6
R( s ) 2 S  1
Step Response
Also Draw the Step response on your notebook
Example#1
• For step response integrate impulse response
c(t )  3e 0.5t
0.5t
c
(
t
)
dt

3
e
dt


cs (t )  6e 0.5t  C
• We can find out C if initial condition is known e.g. cs(0)=0
0  6e 0.50  C
C6
cs (t )  6  6e 0.5t
Example#1
• If initial Conditions are not known then partial fraction
expansion is a better choice
C( s )
6

R( s ) 2 S  1
1
since R( s ) is a step input , R( s ) 
s
C( s ) 
6
s2S  1
6
A
B
 
s2S  1 s 2s  1
6
6
6
 
s2S  1 s s  0.5
c(t )  6  6e 0.5t
Partial Fraction Expansion in Matlab
• If you want to expand a polynomial into partial fractions use
residue command.
rn
r1
r2
y( s )



k
x( s ) s  p1 s  p2
s  pn
Y=[y1 y2 .... yn];
X=[x1 x2 .... xn];
[r p k]=residue(Y, X)
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
4s  8
r1
r2


2
s  6s  8 s  p1 s  p2
4s  8
12
8


2
s  6s  8 s  4 s  2
Partial Fraction Expansion in Matlab
• If you want to expand a polynomial into partial fractions use
residue command.
6
C( s ) 
s2S  1
Y=6;
X=[2 1 0];
[r p k]=residue(Y, X)
r =[ -6
p =[-0.5
k = []
6]
0]
6
6
6


s2s  1 s  0.5 s
Ramp Response of 1st Order System
• Consider the following 1st order system
K
Ts  1
R(s )
R( s ) 
C( s ) 
C(s )
1
s2
K
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t / T
Unit Ramp Response
10
Unit Ramp
Ramp Response
c(t)
8
6
4
error
2
0
0
5
10
Time
15

Ramp Response of 1st Order System

• If K=1 and T=3
c(t )  K t  T  Tet / T
Unit Ramp Response
10
Unit Ramp
Ramp Response
c(t)
8
6
4
error
2
0
0
5
10
Time
15

Parabolic Response of 1st Order System
• Consider the following 1st order system
K
Ts  1
R(s )
R( s ) 
• Do it yourself
1
s
3
Therefore,
C( s ) 
C(s )
K
s 3 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 )
K

R( s ) 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 )
0.72
5.5


R( s ) 0.13 s  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  s )
C( s ) 
sTs  1
K K (  T )
C( s )  
Ts  1
s

c(t )  K 1  e
t / T

K
c(t )  K  (  T )e t / T
T
1st Order System with & W/O Zero
C ( s ) K (1  s )

R( s )
Ts  1
C( s )
K

R( s ) Ts  1

c(t )  K 1  e
t / T

K
c(t )  K  (  T )e t / T
T
• If T>α the response will be same
K
c(t )  K  ( n )e t / T
T
Kn t / T 

c(t )  K 1 
e

T


1st Order System with & W/O Zero
• If T>α the response of the system would look like
Unit Step Response
10
9.5
C ( s ) 10(1  2s )

R( s )
3s  1
c(t)
9
8.5
8
7.5
10
c(t )  10  ( 2  3)e t / 3
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
10
c(t )  10 
( 2  1)e t / 1.5
1.5
13
Unit Step Response
C ( s ) 10(1  2s )

R( s )
1.5s  1
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 ) K (1  s )

R( 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
jω
C( s )
K

R( s ) Ts  1
C( s )
10

R( s ) s  1
T  1s
-3
-2
-1
δ
PZ-map and Step Response
jω
C( s )
K

R( s ) Ts  1
C( s )
10

R( s ) s  2
C( s )
5

R( s ) 0.5s  1
T  0.5s
-3
-2
-1
δ
PZ-map and Step Response
jω
C( s )
K

R( s ) Ts  1
C( s )
10

R( s ) s  3
C( s )
3. 3

R( s ) 0.33 s  1
T  0.33s
-3
-2
-1
δ
Comparison
C( s )
1

R( s ) s  1
C( s )
1

R( s ) s  10
Step Response
Step Response
0.1
0.8
0.08
0.6
0.06
Amplitude
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 is the generic
representation of 1st order system with time
lag.
C( s )
K

e  std
R( s ) Ts  1
• Where td is the delay time.
First Order System With Delays
C( s )
K
 std

e
R( s ) Ts  1
1
Unit Step
Step Response
td
t
First Order System With Delays
Step Response
10
K  10
C( s )
10  2 s

e
R( s ) 3 s  1
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
u
ia
eb
T
J


K t Ra 
Ω(s)

U(s) Js  B  K t K b Ra 
Examples of First Order Systems
• Liquid Level System
H (s)
R

Qi ( s ) ( RCs  1)
Examples of First Order Systems
• Electrical System
Eo ( s )
1

Ei ( s ) RCs  1
Examples of First Order Systems
• Mechanical System
X o (s)
1

X i (s) b
s 1
k
Examples of First Order Systems
• Cruise Control of vehicle
V (s)
1

U ( s ) ms  b
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END OF LECTURES-20-21