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Time response analysis - First Order Systems - Impulse and Step
Response analysis of second order systems - Steady state errors – P,
PI, PD and PID Compensation, Analysis using MATLAB (9)
15UEC904 – LINEAR CONTROL ENGINEERING
REGULATION - 2015
CO.2
Analyze the time response of first and Second order systems.
(K4 - Analyze)
UNIT II - TIME RESPONSE ANALYSIS
Topics to be covered
 Time response
 Types of test input
 I order/II order system responses
 Time domain specification
 Error Coefficients
 Steady state errors
 P. PI, PD, PID controllers
Introduction
 In time-domain analysis the response of a dynamic system to an input is
expressed as a function of time.
 Time response of the system is defined as the output of a system when
subjected to an input which is a function of time.
 It is possible to compute the time response of a system if the nature of
input and the mathematical model of the system are known.
 A control system generates an output or response for given input.
 The input represents the desired response while the output is actual
response of system.
r(t) c(t)
system
Time Response of Control Systems
• When the response of the system is changed from equilibrium it takes
some time to settle down.
• This is called transient response.
0 2 4 6 8 10 12 14 16 18 20
0
1
2
3
4
5
6
x 10
-3
Step Response
Time (sec)
Amplitude
Response
Step Input
Steady
State
Response
• The response of the system after
the transient response is called
steady state response.
Time Response of Control Systems
• Transient response depend upon the system poles only and not on the
type of input.
• It is therefore sufficient to analyze the transient response using a step input.
• The steady-state response depends on system dynamics and the input
quantity.
• It is then examined by using different test signals.
Standard Test Signals
 For time response analysis of control systems, we need to subject the system to
various test inputs.
 Test input signals are used for testing how well a system responds to known input.
 The characteristics of actual input signals are a sudden shock, a sudden change, a
constant velocity, and constant acceleration.
 The dynamic behavior of a system is therefore judged and compared under
application of standard test signals – an impulse, a step, a constant velocity, and
constant acceleration.
 The other standard signal of great importance is a sinusoidal signal.
Standard Test Signals
 Impulse signal
 The impulse signal imitate the sudden shock
characteristic of actual input signal.
 If A=1, the impulse signal is called unit impulse
signal.
0 t
δ(t)
A






0
0
0
t
t
A
t)
(

Standard Test Signals
 Step signal
 The step signal imitate the sudden change
characteristic of actual input signal.
 If A=1, the step signal is called unit step signal






0
0
0
t
t
A
t
u )
(
0
t
u(t)
A
Standard Test Signals
 Ramp signal
 The ramp signal imitate the constant
velocity characteristic of actual input
signal.
 If A=1, the ramp signal is called unit
ramp signal






0
0
0
t
t
At
t
r )
(
0
t
r(t)
Standard Test Signals
 Parabolic signal
 The parabolic signal imitate the constant acceleration
characteristicof actual input signal.
 If A=1, the parabolic signal is called unit parabolic signal.








0
0
0
2
2
t
t
At
t
p )
(
0
t
p(t)
Relation between standard Test Signals
Impulse
Step
Ramp
Parabolic






0
0
0
t
t
A
t)
(







0
0
0
t
t
A
t
u )
(






0
0
0
t
t
At
t
r )
(








0
0
0
2
2
t
t
At
t
p )
(


 dt
d
dt
d
dt
d
ORDER OF THE SYSTEMS
The value of n gives the no. of poles in the transfer
function.
Hence the order is also given by the no. of poles of the
transfer function.
The response of the system when the input signal is an Impulse signal.
TIME RESPONSE OF FIRST ORDER SYSTEMS
For Impulse Input:
1

 )
(
)
( s
s
R 
 The first order system has only one pole.
 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.
1


Ts
K
s
R
s
C
)
(
)
(
Impulse Response of 1st Order System
 Re-arrange following equation as
1


Ts
K
s
C )
(
T
s
T
K
s
C
/
/
)
(
1

 T
t
e
T
K
t
c /
)
( 

• In order to compute the response of the system in time domain we need to
compute inverse Laplace transform of the above equation.
at
Ce
a
s
C
L 









1
 The first order system given below.
1
3
10


s
s
G )
(
5
3


s
s
G )
(
1
5
1
5
3


s
/
/
• D.C gain is 10 and time constant is 3 seconds.
• For the following system
• D.C Gain of the system is 3/5 and time constant is 1/5 seconds.
Impulse Response of 1st Order System
T
t
e
T
K
t
c /
)
( 

• If K=3 and T=2s then
0 2 4 6 8 10
0
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 with step input
1

Ts
K
)
(s
C
)
(s
R
s
s
U
s
R
t
t
u
t
r
input
step
For
1
)
(
)
(
0
;
1
)
(
)
(
;





   
1
1
1
/
1
1
)
(
)
(






Ts
s
Ts
s
Ts
s
R
s
C
1
;
1
1
1
)
(
)
(




 K
Let
Ts
Ts
K
s
R
s
C
Step Response of 1st Order System
   
T
s
s
T
Ts
s
Ts
s
R
s
C
/
1
/
1
1
/
1
1
)
(
)
(






Bs
T
s
A
T
T
s
B
s
A
T
s
s
T
s
C














)
1
(
/
1
1
1
/
1
)
(
1
;
/
1
1
;
0






B
T
s
put
A
s
put
T
s
s
s
C
1
1
1
)
(




Step Response of 1st Order System
 Taking Inverse Laplace of this equation
 
T
t
e
t
u
t
c /
)
(
)
( 


• Where u(t)=1  
T
t
e
t
c /
1
)
( 


  632
.
0
1
)
( 1


 
e
t
c
• When t=T (time constant)
T
s
s
s
C
1
1
1
)
(




Step Response of 1st Order System
• T=1, 3, 5, 7 seconds  
T
t
e
t
c /
1
)
( 


0 5 10 15
0
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 2nd Order System
Damping - A reduction in the amplitude of an oscillation as a result of energy
being drained from the system to overcome frictional or other resistive forces.
The response has no oscillations, but it require longer time to reach its
steady state value
Cs
Bs
Aw
As
w
s
C
Bs
w
s
A
w
w
s
C
Bs
s
A
w
s
s
w
n
n
n
n
n
n
n













2
2
2
2
2
2
2
2
2
2
2
2
)
(
)
(
)
(
1
1
:
tan
0
0
:
0
:
2
2
2












B
and
A
Aw
w
t
cons
of
ts
coefficien
the
Compare
C
C
s
of
ts
coefficien
the
Compare
B
A
B
A
s
of
ts
coefficien
the
Compare
n
n
2
2
2
2
2
2
1
)
(
0
1
n
n
n
w
s
s
s
s
C
w
s
s
s
w
s
C
Bs
s
A











wt
w
s
s
L
t
w
t
u
t
c
t
u
s
L
T
L
inverse
Ontaking
that
know
we
w
s
s
s
s
C
n
n
cos
)
(
cos
)
(
)
(
)
(
)
1
(
:
.
.
:
1
)
(
2
2
1
1
2
2










t
w
t
c
input
step
ForUnit
system
order
II
of
response
step
Unit
n
cos
1
)
(
:


n
n
n
n
n
n
n
n
n
n
n
n
n
w
B
w
B
w
w
s
Put
A
Aw
w
s
Put
w
s
Cs
Bs
w
s
A
w
w
s
C
w
s
B
s
A
w
s
s
w






















)
(
1
0
)
(
)
(
)
(
)
(
2
2
2
2
2
2
2
2
1
2
2
2
3
1
]
2
3
[
2
3
2
4
2
4
2
2
2
2
2
2
2
2
2
2
2
2

















C
C
C
C
w
w
w
C
w
w
w
C
w
w
w
w
C
Bw
w
A
w
w
s
Put
n
n
n
n
n
n
n
n
n
n
n
n
n
n
The response has no oscillations
We know that;
The response oscillates before setting to a final value. The oscillations
depend on the damping ratio.
The response has no oscillations but it takes longer time to reach steady
state value.
TIME DOMAIN SPECIFICATIONS
 The transient response of a practical control system often exhibits damped
oscillations before reaching steady state.
 Actual output behavior according to the various time response specifications
TIME DOMAIN SPECIFICATIONS
 The typical damped oscillatory response of a system is shown
1) Delay Time, Td
It is the time taken for response to reach 50 % of the final value for the very
first time
2) Rise Time, Tr
 It is the time taken for response to raise from 0 % to 100 % of the final
value for the very first time. It is for undamped system.
 For over damped system. It is the time taken for response to raise from
10 % to 90 % .
 For critically damped system. It is the time taken for response to raise
from 5 % to 95 %
3) Peak Time, Tp
It is the time required for the response to reach its peak value for the very
first time.
4) Peak Overshoot, Mp
It is defined as the ratio of the maximum peak value to the final value
5) Settling Time, Ts
Time required for the response to decrease and stay within specified
percentage of its final value (Usual tolerable error : 2 % or 5 % of final
value)
Example #1
1
4
4
0
)
1
(
)
4
(
)
4
)(
1
(
4
4
1
)
4
)(
1
(
4


















A
A
s
Put
s
s
C
s
s
B
s
s
A
s
C
s
B
s
A
s
s
s
3
/
1
)
3
)(
4
(
4
4
3
/
4
3
4
1














C
C
s
Put
B
B
s
Put
Example #2
Example #3
B
A
B
A
s
of
coeff
the
Compare
Cs
Bs
A
As
As
s
C
s
B
s
s
A
s
s
C
Bs
s
A
s
s
s






















0
:
16
4
16
)
(
)
16
4
(
16
16
4
)
16
4
(
16
2
2
2
2
2
2
4
4
0
1
.
1
&
1
16
16
:
tan
)
1
(
4
0
:













C
C
Eqn
in
values
above
the
Substitute
B
A
A
coeff
t
cons
the
Compare
C
A
s
of
coeff
the
Compare
sec
2
4
5
.
0
4
4
)
%
2
(
sec
5
.
1
4
5
.
0
3
3
)
%
5
(








n
s
n
s
error
t
time
Settling
error
t
time
Settling


Exercise Problem #1
A unity feedback control system has an open loop transfer function G(s)=6/s(s+5). Obtain
the response of the system and find the rise time, peak time, delay time, maximum peak
overshoot and settling time for a unitstep input.
Solution:
Steady State Error
Steady-state error is defined as the difference between the input and the output of a
system in the limit as time goes to infinity (i.e. when the response has reached steady
state).
The parameter that is important in this is the steadystate error (Ess).
Steady stateerror is error at t→∞.
)
4
)(
1
(
6
)
(
)
(
)
4
)(
1
(
7
)
(
)
(
2







s
s
s
s
H
s
G
s
s
s
s
s
H
s
G
The steady state error associated with the any one of the
following constants:
Exercise problem:
Innovative Method: Flipped class room – P, PI, PID controllers
1.Controllers and types of controllers
2.Advantages and disadvantages of controllers
3.Real time application of controllers
Innovative Method: Flipped class room – P, PI, PID controllers
https://www.youtube.com/watch?v=9H3-xM70j5c
https://www.youtube.com/watch?v=7oZgcAtgG90
https://www.youtube.com/watch?v=OYPXiC7QjUM
https://www.youtube.com/watch?v=UR0hOmjaHp0
https://www.youtube.com/watch?v=CQA8miYlY1Q
https://www.youtube.com/watch?v=sFqFrmMJ-sg
https://www.youtube.com/watch?v=JFTJ2SS4xyA
https://www.youtube.com/watch?v=d4sCLmTJbvM
Thank You

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