Second Order System Time Response

ELECTRICAL ELECTRONICS COMMUNICATION INSTRUMENTATION
CONTROL SYSTEMS
B.Madhuri

Time response of a second order control
system?
Second Order System
System with two poles

Second order system
• Two independent energy storage elements
– Exchange of energy through mechanical mass and
fluid capacitance (tank) through piston
– Exchange between electrical inductance and
mechanical inertia in electric motor

system
Over-damped Time Response
a
acbb
2
42
2,1


146.1
854.7
2
459
)1(2
)9)(1(499
2
1
2
2,1









tt
eKeKKtc 21
321)(  

General solution
2
3
1
21
21 ))((
9
)(
 





s
K
s
K
s
K
sss
sC

system?
Under-damped Time Response
a
acbb
2
42
2,1


81
81
2
322
)1(2
)9)(1(422
2
1
2
2,1
j
j
j









)cos()( 
 
d
td
Aetc
General solution
nd j
j




2,1
2,1 81

system
Under-damped Time Response
)cos()( 
 
d
td
Aetc

system
Un-damped Time Response
a
acbb
2
42
2,1


3
3
2
36
)1(2
)9)(1(400
2
1
2
2,1
j
j
j









)cos()(   Atc
General solution

Time response of a second order
control system
Critically-damped Time Response
a
acbb
2
42
2,1


3
2
6
)1(2
)9)(1(466
2,1
2
2,1







tt
teKeKtc 11
21)(  

General solution

Time response of second order control
system - Summary
Over-damped
Under-damped
Un-damped
Critically-damped
Poles are real
and distinct
Poles are real
and imaginary
Poles are
imaginary
Poles are real
and repeated

system - Summary

system - Example
a
acbb
2
42
2,1


23.1352,
2
)175(410
)1(2
)200)(1(41010
1
2,1
2
2,1
j
j








The poles are real and imaginary.
Therefore UNDER-DAMPED.
)cos()( 
 
d
td
AetcGeneral
solution of
under-damped
)23.13cos()( 5
  t
Aetc

system - Example
2
2
)(


s
sG
What are poles and zeroes?
Plot on s-plane
Write the general expression of the step response?
State the nature of response

system - Example
)6)(3(
5
)(


ss
sG
Plot on s-plane
State the nature of response

system? - Example
)20)(10(
)7(10
)(



ss
s
sG
Plot on s-plane
State the nature of response.

system
So far, we based on the poles of the
second order system
a
acbb
2
42
2,1



system
We are going to learn two
quantities that will help us analyze
second order system
Natural
Frequency
n
Damping
Ratio


system- Natural Frequency
Natural frequency of a second
order system is the frequency of
oscillation of the system without
damping
Natural
Frequency
n

system- Natural frequency
bass
b
sG

 2
)( Natural frequency of a
second order system is the
frequency of oscillation of the
system without damping
bass
b
sG

 2
)(
bs
b
sG

 2
)( Poles:
njs
bjs
bs
bs



 02
Therefore,
2
n
n
b
b





system- Damping Ratio
Damping
Ratio
 (rad/sec)FrequencyNatural
FrequencyDecaylExponentia


system- Damping Ratio
Damping
Ratio
n
a


2

bass
b
sG

 2
)(

system
bass
b
sG

 2
)(
22
2
2
)(
nn
n
ss
sG




General Form

system
362.4
36
)( 2


ss
sG Find damping ratio and
the natural frequency?
6
362


n
n


Natural Frequency
35.0
)6(2
2.4
2.42



n
Damping Ratio

system
22
2
2
)(
nn
n
ss
sG




a
acbb
s
2
42
2,1


Find poles in
general form.
 
1
2
422
2
2,1
22
2,1





nn
nnn
s
s

system.
12
2,1   nns
Case 1: 0 njs 2,1

system
12
2,1   nns
Case 2: 10 
2
2,1 1   nn js

system
12
2,1   nns
Case 3: 1 ns 2,1

system
12
2,1   nns
Case 4:
1
12
2,1   nns

system

1
1
10 
0
OVERDAMPED
CRITICALLY DAMPED
UNDER DAMPED
UNDAMPED

system- Example
40012
400
)( 2


ss
sG
90090
900
)( 2


ss
sG
22530
225
)( 2


ss
sG
625
625
)( 2


s
sG
3.0
)20(2
12
20400



n
UNDERDAMPED
5.1
)30(2
90
30900



n
OVER-DAMPED
1
)15(2
30
15225



n
CRITICALLY-DAMPED
0
)25(2
0
25625



n
UN-DAMPED

Time response of second order control system. Special
Case: Performance of under-damped System
Under-damped:
10 
Four performance
parameters:
1.Rise time
2.Peak time
3.Percent overshoot
4.Settling time

Rise Time
General
definition: The
time required for
the waveform to
go from 0.1 of the
final value to 0.9
of the final value
rnT

Underdamped case: The time taken
to reach the final value the first time.
𝑡 𝑟 =
𝜋 − 𝑡𝑎𝑛−1 (1 − 𝜉2)
𝜉
𝜔 𝑛 (1 − 𝜉2)

Peak Time
The time
required to
reach the first
maximum
peak
2
1 



n
pT

Percent Overshoot
The amount that
the waveform
overshoots the
steady-state











2
1
100%


eOS

Settling Time
The time required
for the transient
damped
oscillations to
reach and stay
within 2% of the
steady-state
value
n
sT

4


Case: Performance of under-damped System – Example
36116
361
)( 2


ss
sG Find: OSTTT rpsn ,%,,,,
19361 n
421.0
)19(2
16

5.0
)19(421.0
44

n
sT

182.0
421.01191 22








n
pT
%3.23100100%
22
421.01
421.0
1




















 


eeOS

Case: Performance of under-damped System – Example
36116
361
)( 2


ss
sG
Find:
rT
19n
421.0
079.0
19
501.1
501.1


r
nr
T
T 
Or interpolate

system? –Lessons
22 February 2015 37
1
1
10 
0
OVERDAMPED
CRITICALLY DAMPED
UNDER DAMPED
UNDAMPED
WHERE DO WE GET 
22
2
2
)(
nn
n
ss
sG





system? -
22 February 2015 38
What Interpretations?
Performances of second order under-damped
control system are determined by two quantities
called natural frequency and damping ratio
Four performance parameters:
1.Rise time
2.Peak time
3.Percent overshoot
4.Settling time
2
1 



n
pT
n
sT

4
2
1
100%




 eOS
n
r
f
T

)(


system? -
22 February 2015 39
a
acbb
2
42
2,1


22
2
2
)(
nn
n
ss
sG




2
2,1 1   nn js
dd js  2,1
For under-damped system
)cos( 

system? -
22 February 2015 40
n
sT

4

Line of constant settling time

system?
22 February 2015 41
2
1 



n
pT
Line of constant peak time

system?
22 February 2015 42
2
1
100%




 eOS)cos( 
Line of constant %OS

system?
22 February 2015 43
Constant Settling Time

system?
22 February 2015 44
Constant Peak Time

system?
22 February 2015 45
Constant Percent Overshoot

Second Order System Time Response

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