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
Impulse Response of 1st Order System
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
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
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
Step Response of 1st Order System
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
Step Response of 1st Order System
• 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
Step Response of 1st Order System
• 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.
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
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
Partial Fraction Expansion in Matlab
• If we want to expand following polynomial into partial fractions
4s
Y=[-4
8];
X=[1
6 8];
[r p k]=residue(Y, X)
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
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 2S
1
6
s 2s
6
1
s
6
0 .5
s
Ramp Response of 1st Order System
• Consider the following 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
Ramp Response of 1st Order System
• 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
Ramp Response of 1st Order System
• 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
• Consider the following 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
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 (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
1st Order System with & W/O Zero
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
PZ-map and Step Response
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
PZ-map and Step Response
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
First Order System With Delays
C(s)
R( s )
K
Ts
1
st d
e
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
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
Examples of First Order Systems
• Electrical System
Eo (s)
1
Ei (s)
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
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END OF LECTURES-13-14
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