cupdf.com_1-chapter-4-transient-steady-state-response-analysis.ppt

1
CHAPTER 4
Transient & Steady State Response
Analysis

2
Previous Class
 In Chapter 3:
 Block diagram reduction
 Signal flow graphs (SFGs)
Transfer Function
Transfer Function

Today’s class
 Chapter 4 – Transient & Steady
State Response Analysis
 First Order System
 Second Order System
3

Learning Outcomes
 At the end of this lecture, students
should be able to:
 Identify between transient response &
steady state response
 Obtain the transfer function for a first
order system based on graph analysis
 To become familiar with the wide range
of response for second order systems
4

5
In Chapter 4
What you are expected to learn :
What you are expected to learn :
First order system
First order system
Second order system
Second order system
Routh-Hurwitz Criterion
Routh-Hurwitz Criterion
Steady-state error
Steady-state error
Chapter 4

6
Introduction
The time response of a control system
The time response of a control system
consists of two parts:
consists of two parts:
1. Transient response
- from initial state to the final
state – purpose of control
systems is to provide a desired
response.
2. Steady-state response
- the manner in which the
system output behaves as t
approaches infinity – the error
after the transient response has
decayed, leaving only the
continuous response.

7
Introduction
Transient Steady-state

8
Performances of Control Systems
 Specifications (time domain)
 Max OS, settling time, rise time, peak time,
 Standard input signals used in design
 actual signals unknown
 standard test signals:
 step, ramp, parabola, impulse, etc. sinusoid
 (study freq. response later)
 Transient response
 Steady-state response
 Relate to locations of poles and zeros

9
First Order System
s

1
R(s) C(s)
E(s)
Test signal is step function, R(s)=1/s

10
First – order system
1
1
)
(
)
(


s
s
R
s
C

A first-order system without zeros can be
represented by the following transfer function
• Given a step input, i.e., R(s) = 1/s , then
the system output (called step response in
this case) is


 1
1
1
)
1
(
1
)
(
1
1
)
(







s
s
s
s
s
R
s
s
C

11

t
e
t
c


1
)
(
Taking inverse Laplace transform, we have the step response
Time Constant: If t= , So the step response is
C( ) = (1− 0.37) = 0.63

is referred to as the time constant of the response.
In other words, the time constant is the time it takes
for the step response to rise to 63% of its final value.
Because of this, the time constant is used to measure
how fast a system can respond. The time constant has
a unit of seconds.






12
Plot c(t) versus time:

13
The following figure gives the measurements of the step
response of a first-order system, find the transfer function
of the system.
Example 1

14
Transient Response Analysis
Rise Time Tr:
The rise-time (symbol Tr units s) is defined as the time
taken for the step response to go from 10% to 90%
of the final value.
Settling Time Ts:
Defined the settling-time (symbol Ts units s) to be the
time taken for the step response to come to within
2% of the final value of the step response.


 2
.
2
11
.
0
31
.
2 


r
T

4

s
T

15
a
1



16
Second – Order System
 Second-order systems exhibit a wide range of
responses which must be analyzed and described.
• Whereas for a first-order system, varying a
single parameter changes the speed of response,
changes in the parameters of a second order
system can change the form of the response.
 For example: a second-order system can display
characteristics much like a first-order system or,
depending on component values, display damped
or pure oscillations for its transient response.

17
- A general second-order system is characterized by
the following transfer function:
- We can re-write the above transfer function in the
following form (closed loop transfer function):

18
- referred to as the un-damped natural
frequency of the second order system, which
is the frequency of oscillation of the system
without damping.
- referred to as the damping ratio of the
second order system, which is a measure of
the degree of resistance to change in the
system output.
Poles;
Poles are complex if ζ< 1!

19
- According the value of ζ, a second-order system
can be set into one of the four categories:
1. Overdamped - when the system has two real
distinct poles (ζ >1).
2. Underdamped - when the system has two
complex conjugate poles (0 <ζ <1)
3. Undamped - when the system has two
imaginary poles (ζ = 0).
4. Critically damped - when the system has two
real but equal poles (ζ = 1).

Time-Domain Specification
20
2
2
2
2
)
(
)
(
)
(
n
n
n
s
s
s
R
s
C
s
T







Given that the closed loop TF
The system (2nd
order system) is parameterized by ς and ωn
For 0< ς <1 and ωn > 0, we like to investigate its response
due to a unit step input
Transient Steady State
Two types of responses that
are of interest:
(A)Transient response
(B)Steady state response

(A) For transient response, we
have 4 specifications:
21
(a) Tr – rise time =
(b) Tp – peak time =
(c) %MP – percentage maximum overshoot =
(d) Ts – settling time (2% error) =
2
1 





n
2
1 



n
%
100
2
1
x
e 



n

4
(B) Steady State Response
(a) Steady State error

22
Question : How are the performance
related to ς and ωn ?
- Given a step input, i.e., R(s) =1/s, then the system output
(or step response) is;
- Taking inverse Laplace transform, we have the step
response;
Where; or )
(
cos 1

 


23
Mapping the poles into s-plane
2
2
2
2
)
(
)
(
)
(
n
n
n
s
s
s
R
s
C
s
T








24
Lets re-write the equation for c(t):
Let: 2
1 
 

2
1 

 
 n
d
and
Damped natural frequency
d
n 
 
Thus:
 






 
t
e
t
c d
t
n
sin
1
1
)
(
)
(
cos 1

 

where

25
1) Rise time, Tr. Time the response takes to rise from
0 to 100%
2
1 






n
r
T
  1
sin
1
1
)
( 


 





t
e
t
c d
t
r
T
t
n
0
 0












)
0
(
sin
0
)
sin(
1
r
d
r
d
T
T

26
2) Peak time, Tp - The peak time is the time required for
the response to reach the first peak, which is given by;
0
)
( 


p
T
t
t
c
  0
1
)
cos(
)
sin(
)
(
1
)
( 2
1







 














n
d
t
d
t
n
p
T
t
t
e
t
e
t
c n
n
)
cos(
)
sin(
2
1





 













 

p
d
T
p
d
T
n
T
e
T
e p
n
n
p
n




2
1
)
tan(



p
dT 




1
tan

27
We know that )
tan(
)
tan( 

 

So, )
tan(
)
tan( 


 


p
dT
From this expression:










p
d
p
d
T
T
2
1 







n
d
p
T

3) Percent overshoot, %OS - The percent overshoot is
defined as the amount that the waveform at the peak time
overshoots the steady-state value, which is expressed as a
percentage of the steady-state value.
%
100
)
(
)
(
)
(
% x
C
C
T
C
MP
p




100
max
% x
Cfinal
Cfinal
C
OS


OR

29
29
 
%
100
%
100
)
sin(
%
100
sin
1
%
100
sin
1
2
2
2
2
1
1
1
1
x
e
x
e
x
e
x
e
d
d
n
n




























































  %
100
sin
1
%
100
1
1
)
(
x
t
e
x
T
C
d
t
p n







 
2
1
sin 
 

2
1 
 

From slide 24

30
- For given %OS, the damping ratio can
be solved from the above equation;
Therefore,
%
100
%
2
1
x
e
MP 




 
 
100
/
%
ln
100
/
%
ln
2
2
MP
MP






31
4) Setting time, Ts - The settling time is the time
required for the amplitude of the sinusoid to decay
to 2% of the steady-state value.
To find Ts, we must find the time for which c(t) reaches & stays
within +2% of the steady state value, cfinal. The settling time is the
time it takes for the amplitude of the decaying sinusoid in c(t) to
reach 0.02, or
02
.
0
1
1
2




 s
nT
e
Thus,
n
s
T

4


32
UNDERDAMPED
Example 2: Find the natural frequency
natural frequency and damping
damping
ratio
ratio for the system with transfer function
Solution: 36
2
.
4
36
)
( 2



s
s
s
G
Compare with general TF
•ωn= 6
•ξ =0.35

33
Example 3
Example 3:
: Given the transfer function
UNDERDAMPED
s
T
OS
s
T p
s 475
.
0
%,
838
.
2
%
,
533
.
0 


p
s T
OS
T
find ,
%
,
Solution:
75
.
0
10 
 
n

35
a = 9
s= 0; s = -7.854; s = -1.146
s= 0; s = -7.854; s = -1.146 ( two real poles)
( two real poles)
)
146
.
1
)(
854
.
7
(
9
)
9
9
(
9
)
( 2






s
s
s
s
s
s
s
C
1


Overdamped Response
Overdamped Response

36
t
t
e
K
e
K
K
t
c 146
.
1
3
854
.
7
2
1
)
( 




OVERDAMPED
OVERDAMPED RESPONSE !!!
RESPONSE !!!

37
Underdamped Response
Underdamped Response
)
598
.
2
sin
598
.
2
cos
(
)
( 3
2
5
.
1
1 t
K
t
K
e
K
t
c t


 
s = 0; s = -1.5 ± j2.598
s = 0; s = -1.5 ± j2.598 ( two complex poles)
( two complex poles)
a = 3
1
0 


38
UNDERDAMPED
UNDERDAMPED RESPONSE !!!
RESPONSE !!!

39
Undamped Response
Undamped Response
a = 0
t
K
K
t
c 3
cos
)
( 2
1 

s = 0; s = ± j3
s = 0; s = ± j3 ( two imaginary poles)
( two imaginary poles)
0



40
UNDAMPED
UNDAMPED RESPONSE !!!
RESPONSE !!!

41
a = 6
Critically Damped System
Critically Damped System
t
t
te
K
e
K
K
t
c 3
3
3
2
1
)
( 




S = 0; s = -3,-3
S = 0; s = -3,-3 ( two real and equal poles)
( two real and equal poles)
1



42
CRITICALLY DAMPED
CRITICALLY DAMPED RESPONSE !!!
RESPONSE !!!

45
Effect of different damping ratio, ξ
Increasing ξ

46
Example 4: Describe the nature of the second-order
system response via the value of the damping ratio for
the systems with transfer function
12
8
12
)
(
.
1 2



s
s
s
G
16
8
16
)
(
.
2 2



s
s
s
G
20
8
20
)
(
.
3 2



s
s
s
G
Do them as your
Do them as your
own revision
own revision

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