control3.pptx

ADIGRAT UNIVERSITY
COLLAGE OF ENGINEERING &
TECHNOLOGY
DEPARTMENT OF ELECTRICAL AND
COMPUTER ENGINEERING
Course name: Introduction to control
systems
Course code: ECEg3153
Course instructor: Getnet Z.
Contact information:
Email: abaye.get@gmail.com
Consultation hours:
Wednesday and Friday : from 8:00-
11/12/2023
ADU,
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ECEg3153
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OUTLINE
 Introduction
 Time domain analysis of
 First order systems
 Second order systems
 Steady state error
 Stability analysis's
 Routh’s stability criteria
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INTRODUCTION
Time Domain Analysis of Control Systems
 In time-domain analysis the response of a dynamic
system to an input is expressed as 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.
 After the engineer obtains a mathematical
representation of a system, the system is analyzed for
its transient and steady-state responses to see if these
characteristics yield the desired behavior
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 The concept of poles and zeros are fundamental to the
analysis and design of control systems which simplifies
the evaluation of a system's response.
Pole, Zeros and System Response
 Pole of a Transfer Function: The values of the Laplace
transform variable, s, that cause the transfer function to
become infinite
 Zeros of a Transfer Function: The values of the Laplace
transform variable, s, that cause the transfer function to
become zero
INTRODUCTION
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 Usually, the input signals to control systems are not
known fully ahead of time.
 For example, in a radar tracking system, the position
and the speed of the target to be tracked may vary in a
random fashion.
 It is therefore difficult to express the actual input signals
mathematically by simple equations.
 So in order to get System Response:
 The commonly used test input signals are step
functions, ramp functions, acceleration function, impulse
functions and sinusoidal functions.
INTRODUCTION
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Standard Test Signals
 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.
 Another standard signal of great importance is a
sinusoidal signal.
INTRODUCTION
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 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.
INTRODUCTION
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INTRODUCTION
 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 t
u(t)
A
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INTRODUCTION
 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 t
r(t)
r(t)
unit ramp signal
r(t)
ramp signal with slope A
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 Standard test signal
INTRODUCTION
 Parabolic signal
 The parabolic signal imitate
the constant acceleration
characteristic of actual input
signal.
 If A=1, the parabolic signal is
called unit parabolic signal.
0 t
p(t)
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INTRODUCTION
 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
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 Laplace Transform of Test Signals
INTRODUCTION
Impulse
Step






0
0
0
t
t
A
t)
(

A
s
t
L 
 )
(
)}
(
{ 







0
0
0
t
t
A
t
u )
(
S
A
s
U
t
u
L 
 )
(
)}
(
{
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 Laplace transform of test signals
INTRODUCTION
Ramp
Parabolic






0
0
0
t
t
At
t
r )
(
2
s
A
s
R
t
r
L 
 )
(
)}
(
{








0
0
0
2
2
t
t
At
t
p )
(
3
2
S
A
s
P
t
p
L 
 )
(
)}
(
{
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 Time Response of Control Systems
 Time response of a dynamic system is response to an
input expressed as a function of time.
 The time response of any system has two components
 Transient response
 Steady-state response.
INTRODUCTION
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TIME DOMAIN ANALYSIS OF 1ST ORDER
SYSTEMS
Transient Response Analysis
 First Order System:
 First order system without zero
 The first order system has only one pole considering with
gain K.
 Where K is the D.C gain and T or 1/a 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
)
(
)
(
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 As an example For the first order system given below
 D.C gain is 10 and time constant is 3 seconds.And for
following system
 D.C Gain of the system is 3/5 and time constant is 1/5
seconds.
 In general having
 E.g: If unit step input is applied to this first order system
1
3
10


s
s
G )
(
5
3


s
s
G )
(
1
5
1
5
3


s
/
/
SYSTEMS
Cn(t): (natural),transited
response
Cf(t): forced (ss) response
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Testing 1st order systems
 Impulse Response of 1st Order System
 Consider the following 1st order system
 Re-arrange above equation as
 In order to represent the response of the system in time
domain we need to compute inverse Laplace transform of
the above equation.
SYSTEMS
1

Ts
K
)
(s
C
)
(s
R
0
t
δ(t)
1
1
)
(
)
( 
 s
s
R 
1
)
(


Ts
K
s
C
T
s
T
K
s
C
/
/
)
(
1


T
t
e
T
K
t
c /
)
( 

at
Ae
a
s
A
L 









1
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Testing 1st order systems
 Step Response of 1st Order System
 Consider the following 1st order system
 In order to find out the inverse Laplace of the above equation,
we need to break it into partial fraction expansion
 Taking inverse Laplace
 Having u(t)=1 and t=T(at which the response reach's 63%
of final value)
SYSTEMS
1

Ts
K
)
(s
C
)
(s
R
s
s
U
s
R
1

 )
(
)
(  
1


Ts
s
K
s
C )
(
1



Ts
KT
s
K
s
C )
(
Forced Response Natural Response









1
1
Ts
T
s
K
s
C )
(
 
T
t
e
t
u
K
t
c /
)
(
)
( 


 
T
t
e
K
t
c /
)
( 

 1   K
e
K
t
c 632
0
1 1
.
)
( 

 
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 Step Response of 1st Order System
 E.g: If K=10 and T=1.5s then
 Exercise: fined 1st order response for parabolic and
ramp signal input.
SYSTEMS
 
T
t
e
K
t
c /
1
)
( 


0 1 2 3 4 5 6 7 8 9 10
0
1
2
3
4
5
6
7
8
9
10
11
Time
c(t)
K*(1-exp(-t/T))
Unit Step Input
Step Response
1
10



Input
output
state
steady
K
Gain
C
D.
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 Relation Between Step and impulse response
 The step response of the first order system is
 Differentiating c(t) with respect to t yields
SYSTEMS
  T
t
T
t
Ke
K
e
K
t
c /
/
)
( 




 1
 
T
t
Ke
K
dt
d
dt
t
dc /
)
( 


T
t
e
T
K
dt
t
dc /
)
( 
 (impulse response)
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 Example: Impulse response of a 1st order system is
given below.
 Find out
 Time constant T
 D.C Gain K
 Transfer Function
 Step Response
t
e
t
c 5
0
3 .
)
( 

solution
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.
t
e
t
c 5
0
3 .
)
( 

)
(
.
.
)
( s
S
S
s
C 






5
0
3
1
5
0
3
5
0
3
.
)
(
)
(
)
(
)
(



S
s
R
s
C
s
s
C

1
2
6


S
s
R
s
C
)
(
)
(
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Transient Response Performance Specification:
 Time Constant: T or a is the only parameter needed
to describe the transient response We call 1/a or T the
time constant of the response.
 The time constant can be described as the time for
exp(-at) to decay to 37% of its initial value. Alternately,
the time constant is the time it takes for the step
response to rise to 63% of its final value
 It is related to the speed at which the system responds
to a step input.
 Since the pole of the transfer function is at -a, we can
say the pole is located at the reciprocal of the time
constant, and
 The farther the pole from the imaginary axis, the faster
the transient response.
SYSTEMS
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 Rise Time, Tr: Rise time is defined as the time for the
waveform to go from 0% or 10% to 90% or 100% of its
final value.
 Rise time calculated by solving for the difference in time
at c(t) = 0.9 and c(t) = 0.1. Hence,
 Settling Time, Ts: Settling time is defined as the time
for the response to reach, and stay within, 2% of its
final value.
 Letting c(t) = 0.98 and solving for time, t, we find the
 settling time to be
SYSTEMS
Or Ts=4T
Or Tr=2.2T
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 Could be summarized as:
SYSTEMS
Or Slope= 1/T
Ts=4T
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 Example: a system has a transfer function
 Find the time constant, Tc, settling time, Ts, and rise
time, Tr ?
 Solution:
 a=50
 Tc= 1/a=0.02sec
 Tr = 2.2/a = 0.044sec
 Ts= 4/a = 0.08sec
 What will be the total time step response and show the
natural and forced responses ?
 1-exp(-50t)-Total response
 1- forced response
 exp(-50t)- Natural response
SYSTEMS
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 1st Order System with a Zero
 Zero of the system lie at -1/α and pole at -1/T.
 Step response of the system would be:
1
)
1
(
)
(
)
(



Ts
s
K
s
R
s
C 
 
1
1



Ts
s
s
K
s
C
)
(
)
(

 
1




Ts
T
K
s
K
s
C
)
(
)
(

T
t
e
T
T
K
K
t
c /
)
(
)
( 


 
Partial Fractions
Inverse Laplace
SYSTEMS
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 1st Order System with & W/O Zero (Comparison)
 If T>α the shape of the step response is approximately
same (with offset added by zero)
1
1



Ts
s
K
s
R
s
C )
(
)
(
)
( 
T
t
e
T
T
K
K
t
c /
)
(
)
( 


 
 
T
t
e
K
t
c /
)
( 

 1
1


Ts
K
s
R
s
C
)
(
)
(
T
t
e
n
T
K
K
t
c /
)
(
)
( 










  T
t
e
T
n
K
t
c /
1
)
(
0 5 10 15
6.5
7
7.5
8
8.5
9
9.5
10
Time
c(t)
Unit Step Response
1
3
2
1
10



s
s
s
R
s
C )
(
)
(
)
(
3
3
2
3
10
10 /
)
(
)
( t
e
t
c 



offset
𝑜𝑓𝑓𝑠𝑒𝑡 = 𝐾 +
𝐾
𝑇
(𝛼 − 𝑇)
SYSTEMS
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 If T<α the response of the system would look like
1
5
1
2
1
10



s
s
s
R
s
C
.
)
(
)
(
)
(
5
1
1
2
5
1
10
10 .
/
)
(
.
)
( t
e
t
c 



0 5 10 15
9
10
11
12
13
14
Time
Unit
Step
Response
Unit Step Response of 1st Order Systems with Zeros
0 5 10 15
6
7
8
9
10
11
12
13
14
Time
Unit
Step
Response
Unit Step Response of 1st Order Systems with Zeros


T


T
Comparing the 2-cases
SYSTEMS
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 Examples of First Order Systems
 Armature Controlled D.C Motor (La=0)
 Electrical System
 Mechanical System
 
 
a
b
t
a
t
R
K
K
B
Js
R
K
U(s)
Ω(s)



u
ia
T
Ra La
J

B
eb
1
1


RCs
s
E
s
E
i
o
)
(
)
(
1
1


s
k
b
s
X
s
X
i
o
)
(
)
(
SYSTEMS
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2nd order systems
 We have already discussed the affect of location of
pole and zero on the transient response of 1st order
systems.
 Compared to the simplicity of a first-order system, a
second-order system exhibits a wide range of
responses that must be analyzed and described.
 Varying a first-order system's parameters (T, K) simply
changes the speed and offset of the response
 Whereas, changes in the parameters of a second-order
system can change the form of the response.
 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.
TIME DOMAIN ANALYSIS OF 2ND ORDER
SYSTEMS
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2nd order systems
 A general second-order system (without zeros) is
characterized by the following transfer function.
SYSTEMS
2
2
2
2 n
n
n
s
s
s
R
s
C






)
(
)
(
)
2
(
)
(
2
n
n
s
s
s
G



 Open-Loop Transfer
Function
Closed-Loop
Transfer Function
un-damped natural frequency of the second
order system, which is the frequency of
oscillation of the system without damping.
n

damping ratio of the second order system, which is a
measure of the degree of resistance to change in the
system output.

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 Lets study the effect of pole location before finding
response of the system.
 lets have the characteristic equation or denominator
 The following conclusion could be drawn:
SYSTEMS
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Second Order Systems:
 Let’s take numerical examples of the second-order
system responses which has two finite poles and no
zeros.
SYSTEMS
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SYSTEMS
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 NB: Varying a first-order system's parameter simply
changes the speed of the response, changes in the
parameters of a second-order system can change the
form of the response.
 Therefore; dep. On damping ratio ( ) the response
 Over damped responses( )
 Poles: Two real at (i.e )
 Natural response: Two exponentials with time constants
equal to the reciprocal of the pole locations,
SYSTEMS

1


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 Under damped responses ( )
 Poles: Two complex at (i.e )
 Natural response: Damped sinusoid with an exponential
envelope whose time constant is equal to the reciprocal of
the pole's real part. The radian frequency of the sinusoid,
the damped frequency of oscillation, is equal to the
imaginary part of the poles,
 Un damped responses ( )
 Poles: Two imaginary at (i.e )
 Natural response: Un damped sinusoid with radian
 frequency equal to the imaginary part of the poles,
SYSTEMS
1
0 
 
0


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 Critically damped responses( )
 Poles: Two real at (i.e )
 Natural response: One term is an exponential whose
time constant is equal to the reciprocal of the pole
location. Another term is the product of time, t and an
exponential with time constant equal to the reciprocal of
the pole location,
SYSTEMS
1


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General Second Order System
 Two physically meaningful specifications for second-
order systems transient response like that of time
constant for first order response.
 Natural Frequency, ωn: The natural frequency of a
second-order system is the frequency of oscillation of
the system without damping.
 For example, the frequency of oscillation of a series
RLC circuit with the resistance shorted would be the
natural frequency
 Damping Ratio, It is the ratio of exponential decay
frequency of the envelope to the natural frequency. This
ratio is constant regardless of the time scale of the
response.
SYSTEMS
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 Mathematically;
Expressing second order system in terms of natural
frequency and damping ratio
Let,
 Without damping, the poles would be on the jω-axis, and the
response would be an un damped sinusoid. For the poles to
be purely imaginary, a = 0. Hence,
 By definition, the natural frequency, ω, is the frequency of
oscillation of this system. Since the poles of this system are
on the jω-axis at ,
Then,
SYSTEMS
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 For the above general second order system transfer
function, assuming under damped system, the complex
poles have a real part = –a/2 and the magnitude of this
value is the exponential decay frequency.
 Having this general 2nd order unit FB transfer function:
the general 2nd order time response of sinusoidal and
exponential parts.
SYSTEMS
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 Cont…
SYSTEMS











 
t
t
e
t
c d
d
t
n





sin
cos
)
(
2
1
1
1
2


 


 n
n
Real Part Imaginary Part
11/12/2023
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 Example1: For a system with a transfer function
given below calculate the damping ratio and the natural
frequency of the system.
 Solution:
 From the TF ,
 substituting the value of damping ratio
 NB: The damping ratio and the natural frequency of
a system can be
 related with the pole location of the system : which
means if the information about these two parameter is
known we can calculate the poles of the system.
SYSTEMS
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• Example 2; The natural frequency of closed loop poles
of 2nd order system is 2 rad/sec and damping ratio is
0.5. Determine the location of closed loop poles so that
the damping ratio remains same but the natural
undamped frequency is doubled.
• Solution
SYSTEMS
4
2
4
2 2
2
2
2






s
s
s
s
s
R
s
C
n
n
n



)
(
)
(
-2 -1.5 -1 -0.5 0 0.5
-3
-2
-1
0
1
2
3
0.28
0.38
0.5
0.64
0.8
0.94
0.5
1
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
0.08
0.17
0.28
0.38
0.5
0.64
0.8
0.94
0.08
0.17
Pole-Zero Map
Real Axis
Imaginary
Axis
11/12/2023
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SYSTEMS
11/12/2023
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Under damped Second Order System
 The under damped second-order system is a
common model for physical problems which displays
unique behavior that have to be itemized;
 Hence, a detailed description of the underdamped
response is necessary for both analysis and design.
 The step response for the general second-order
system is:
SYSTEMS
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 Step Response of underdamped System
 The partial fraction expansion of above equation is
given as
 Above equation can be written as
SYSTEMS
2
2
2
2
2
2
2
2
1
n
n
n
n
n
s
s
s
s
s
C














)
(
2
2
2
2
1
n
n
n
s
s
s
s
s
C








)
(
 2
2 n
s 

 
2
2
1 
 
n
   
2
2
2
1
2
1










n
n
n
s
s
s
s
C )
(
2
2
2
2 n
n
n
s
s
s
R
s
C






)
(
)
(
 
2
2
2
2
)
(
n
n
n
s
s
s
s
C






Step Response
  2
2
2
1
d
n
n
s
s
s
s
C








)
(
11/12/2023
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 Where , is the frequency of transient oscillations and
is called damped natural frequency.
 The inverse Laplace transform of above equation can be obtained
easily if C(s) is written in the following form:
    2
2
2
2
1
d
n
n
d
n
n
s
s
s
s
s
C














)
(
    2
2
2
2
2
2
1
1
1
d
n
n
d
n
n
s
s
s
s
s
C



















)
(
t
e
t
e
t
c d
t
d
t n
n



 

sin
cos
)
( 





2
1
1
    2
2
2
2
2
1
1
d
n
d
d
n
n
s
s
s
s
s
C

















)
(
SYSTEMS
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 Cont…
 When
 Example: Consider Effect of varying damping ratio
with constant natural frequency for under damped
systems.
 I,
 II,
 III,











 
t
t
e
t
c d
d
t
n





sin
cos
)
(
2
1
1
n
n
d






 2
1
0


t
t
c n

cos
)
( 
 1
sec
/
. rad
n
and
if 3
1
0 
 

sec
/
. rad
n
and
if 3
5
0 
 

sec
/
. rad
n
and
if 3
9
0 
 

SYSTEMS
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 The response looks:
 I II
 III
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 2 4 6 8 10
0
0.2
0.4
0.6
0.8
1
1.2
1.4
SYSTEMS
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 Home work: Step Response of overdamped and
critically damped Systems ?
SYSTEMS
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Time Domain Specifications of Underdamped system
For 0< ζ <1 and ωn > 0, the 2nd order system’s response
due to a unit step
SYSTEMS
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 Rise time, Tr: The time required for the waveform to go
from 0% or 10% of the final value to 90% or 100% of
the final value.
 Peak time, Tp: The time required to reach the first the
maximum, peak.
 Percent overshoot, %OS: The amount that the
waveform overshoots the steady-state, or final value at
the peak time, expressed as a percentage of the
steady-state value.
 Settling time, Ts: The time required for the transient's
damped oscillations to reach and stay with in ―2% ( •
}
5%) of the steady-state value.
 The delay (td) time: is the time required for the
response to reach half the final value the very first time.
SYSTEMS
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 NB: All definitions are also valid for systems of order
higher than 2, however analytical expressions for these
parameters cannot be found unless the response of the
higher-order system can be approximated as a second-
order system,
 Rise time, peak time, and settling time yield information
about the speed of the transient response.
 This information can help a designer to determine
whether the speed and the nature of the response do or
do not degrade the performance of the system.
SYSTEMS
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 Evaluating the performance parameters
 Time Domain Specifications (Rise Time)











 
t
t
e
t
c d
d
t
n





sin
cos
)
(
2
1
1
equation
above
in
Put r
t
t 











 
r
d
r
d
t
r t
t
e
t
c r
n





sin
cos
)
(
2
1
1
1

)
c(tr
Where











 
r
d
r
d
t
t
t
e r
n





sin
cos
2
1
0
0

  r
nt
e 










 r
d
r
d t
t 


 sin
cos
2
1
0
SYSTEMS
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 Time Domain Specifications (Rise Time)
as
writen
-
re
be
can
equation
above
0
1 2










 r
d
r
d t
t 


 sin
cos
r
d
r
d t
t 


 cos
sin
2
1





2
1


r
d t
tan









 

 



2
1 1
tan
r
dt







 

 




 n
n
d
r
t
2
1 1
tan
1
d
r
t


 


b
a
1

 tan

SYSTEMS
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 Time Domain Specifications (Peak Time)











 
t
t
e
t
c d
d
t
n





sin
cos
)
(
2
1
1
• In order to find peak time let us differentiate above
equation w.r.t t.






















 

t
t
e
t
t
e
dt
t
dc
d
d
d
d
t
d
d
t
n
n
n









 

cos
sin
sin
cos
)
(
2
2
1
1













 
t
t
t
t
e d
d
d
d
d
n
d
n
t
n












cos
sin
sin
cos
2
2
2
1
1
0














 
t
t
t
t
e d
n
d
d
d
n
d
n
t
n













cos
sin
sin
cos
2
2
2
2
1
1
1
0
SYSTEMS
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 Time Domain Specifications (Peak Time)
 Time Domain Specifications (Maximum Overshoot)
0
1 2
2











d
n
dt 




sin 0
1 2
2











d
n




0

t
d

sin
0
1

 sin
t
d

d
t


 
,
,
, 2
0

• Since for underdamped stable systems first peak is
maximum peak therefore,
d
p
t



SYSTEMS
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












p
d
p
d
t
p t
t
e
t
c p
n





sin
1
cos
1
)
(
2
1

)
(
c
100
1
1
1
2

























p
d
p
d
t
p t
t
e
M p
n





sin
cos
equation
above
in
Put
d
p
t



100
1 2
























d
d
d
d
p
d
n
e
M











sin
cos
SYSTEMS
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100
1 2
1 2



































sin
cos
n
n
e
M p
  100
0
1
2
1



















e
M p
100
2
1






e
M p
equation
above
in
Put 2
1-ζ
ω
ω n
d 
SYSTEMS
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 Time Domain Specifications (Settling Time)
• Settling time (2%) criterion
• Time consumed in exponential decay up to 98% of
the input.
n
s T
t

4
4 

• Settling time (5%) criterion
• Time consumed in exponential decay up to 95%
of the input.
n
s T
t

3
3 

n
T

1

SYSTEMS
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 Example1:Consider the system shown in following
figure, where damping ratio is 0.6 and natural
undamped frequency is 5 rad/sec. Obtain the rise time
tr, peak time tp, maximum overshoot Mp, and settling
time 2% and 5% criterion ts when the system is
subjected to a unit-step input.

SYSTEMS
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 Solution
d
r
t


 

Rise Time

2
1
141
.
3






n
r
t
rad
93
0
1 2
1
.
)
(
tan 

 
n
n




s
tr 55
0
6
0
1
5
93
0
141
3
2
.
.
.
.




d
p
t



Peak Time
s
tp 785
0
4
141
3
.
.


SYSTEMS
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 Cont…
 Example: calculate Tp, Ts, %OS and Tr for a
system with a transfer function;
n
s
t

4

Settling Time (2%)
n
s
t

3

Settling Time (4%)
s
ts 33
1
5
6
0
4
.
.


 s
ts 1
5
6
0
3



.
100
2
1






e
M p
Maximum Overshoot
100
2
6
0
1
6
0
141
3

 


.
.
.
e
M p
SYSTEMS
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Relationship b/n %OS, Tp and Ts to the location of
the poles of the system
 The pole plot for a general, underdamped second-order
system, From the Pythagorean theorem that the radial
distance from the origin to the pole is the natural
frequency, ωn, and the cosine of the angle formed b/n
this line & the real axis be the damping ratio.
 Pole plot for second order underdamped system
SYSTEMS
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 From the previous peak time and settling time equation
for second order step response. Comparing the values
to the pole locations;
 Where; ωd is the imaginary part of the pole and is called
the damped frequency of oscillation, σd is the
magnitude of the real part of the pole and is the
exponential damping frequency.
SYSTEMS
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 From the equations, Tp is inversely proportional to the
imaginary part of the pole. Since horizontal lines on the
s-plane are lines of constant imaginary value, they are
also lines of constant peak time.
 The settling time is inversely proportional to the real
part of the pole.
 Since vertical lines on the s-plane are lines of constant
real value, they are also lines of constant settling time.
 Finally, since ϛ = cosѳ, radial lines are lines of constant
ϛ
 Since percent overshoot is only a function of ϛ, radial
lines are thus lines of constant percent overshoot,
%OS.
SYSTEMS
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 Lines of constant peak time, Tp, settling time, Ts, and
percent overshoot, %OS.
SYSTEMS
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SYSTEMS
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 Example 1: For the system shown in Figure-(a),
determine the values of gain K and velocity-feedback
constant Kh so that the maximum overshoot in the unit-
step response is 0.2 and the peak time is 1 sec. With
these values of K and Kh, obtain the rise time and
settling time. Assume that J=1 kg-m2 and B=1 N-
m/rad/sec.
SYSTEMS
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 Solution: reduce the block in to
Nm/rad/sec
and
Since 1
1 2

 B
kgm
J
K
s
KK
s
K
s
R
s
C
h 



)
(
)
(
)
(
1
2
2
2
2
2 n
n
n
s
s
s
R
s
C






)
(
)
(
• Comparing above T.F with general 2nd order T.F
K
n 

K
KKh
2
1 )
( 


SYSTEMS
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 Cont…
• Maximum overshoot is 0.2.
 
2
0
2
1
.
ln
)
ln( 




e
• The peak time is 1 sec
d
p
t



2
456
0
1
141
3
.
.


n

2
1
141
3
1

 

n
.
53
3.

n

SYSTEMS
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 Cont…
K
n 
 K
KKh
2
1 )
( 


K

53
3.
5
12
53
3 2
.
.


K
K
)
.
(
.
. h
K
5
12
1
5
12
2
456
0 


178
0.

h
K n
s
t

4

n
s
t

3

2
1 






n
r
t
s
tr 65
0.

s
ts 48
2.

s
ts 86
1.

96
3.

n

Then the rise and settling
time can be calculated as:
SYSTEMS
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 Example2: Given the system shown in following figure,
find J and D to yield 20% overshoot and a settling time
of 2 seconds for a step input of torque T(t).
 Solution : model T.F
SYSTEMS
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 Cont…
SYSTEMS
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 Exersice1:Figure (a) shows a mechanical vibratory
system. When 2 lb of force (step input) is applied to the
system, the mass oscillates, as shown in Figure (b).
Determine m, b, and k of the system from this response
curve.
SYSTEMS
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 Exersice2: When the system shown in Figure(a) is
subjected to a unit-step input, the system output
responds as shown in Figure(b). Determine the values
of a and c from the response curve.
)
( 1

cs
s
a
SYSTEMS
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Effect of Adding Pole and Zero on the Transient
Response of Second Order System
 Adding pole for second order system:
 If a system has more than two poles or has zeros, we
cannot use the formulas to calculate the performance
specifications that we derived.
 However, under certain conditions, a system with more
than two poles or with zeros can be approximated as a
second-order system that has just two complex
dominant poles.
 Then, the formulas for percent overshoot, settling time,
and peak time can be applied to these higher-order
systems by using the location of the dominant poles.
SYSTEMS
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 Considering a three-pole system with complex poles
and a third pole on the real axis.
 Assuming that the complex poles are at
and the real pole is at -αr, the step response of the
system can be determined from a partial-fraction
expansion.
 And the system response become;
SYSTEMS
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SYSTEMS
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 If (Case II), the pure exponential will die out much
more rapidly than the second-order underdamped step
response.
 If the pure exponential term decays to an insignificant
value at the time of the first overshoot, such parameters
as percent overshoot, settling time, and peak time will
be generated by the second-order underdamped step
response component.
 Thus, the total response will approach that of a pure
second order system (Case III).
SYSTEMS
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 NB: If αr is not much greater than (Case I), the real
pole's transient response will not decay to
insignificance at the peak time or settling time
generated by the second order pair.
 In this case, the exponential decay is significant, and
the system cannot be represented as a second-order
system.
 How much should further the additional real pole from
the dominant pole depends on the accuracy for which
we are looking. But most of the time five time constants.
 Hence, the real pole should five times farther to the left
than the dominant poles, to represent the system by its
dominant second-order pair of poles.
SYSTEMS
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 We have to also consider also the effect of the residue
on the magnitude of the exponential magnitude.
 Example: Comparing the effect of adding real pole on
step response for the following three system
SYSTEMS
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 Adding Zeros on the system:
 The zeros of a response affect the residue, or
amplitude, of a response component but do not affect
the nature of the response, exponential, damped
sinusoid, and so on.
 Let’s add a real-axis zero to a two-pole system. We can
add a zero either in the left half plane or in the right half
plane.
SYSTEMS
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 The closer the zero is to the dominant poles, the
greater its effect on the transient response. As the zero
moves away from the dominant poles, the response
approaches that of the two-pole system.
SYSTEMS
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Second order systems
 Exercise: 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 own
revision
SYSTEMS
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STEADY STATE ERROR
Steady State Response:
 Control systems analysis and design focus on three
specifications:
 (1) Transient response
 (2) Steady-state errors
 (3) Stability
 Steady State Error: Steady-state error is the
difference between the input and the output for a
prescribed test input as t ∞.
E(s) = C(s) – R(s)
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Steady state error for unity feedback system
 It can be calculated from a system's closed-loop
transfer function, T(s), or the open-loop transfer
function, G(s), for unity feedback systems.
 Steady-State Error in Terms of T(s)
STEADY STATE ERROR
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STEADY STATE ERROR
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 Considering G(s)
STEADY STATE ERROR
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 Any physical control system inherently suffers steady-state
error in response to certain types of inputs.
 A system may have no steady-state error to a step input, but
the same system may exhibit nonzero steady-state error to a
ramp input.
 Whether a given system will exhibit steady-state error for a
given type of input depends on the type of open-loop
transfer function of the system.
Classification of Control Systems
 Control systems may be classified according to their ability
to follow step inputs, ramp inputs, parabolic inputs, and so
on.
 The magnitudes of the steady-state errors due to these
individual inputs are indicative of the goodness of the
system.
STEADY STATE ERROR
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 Consider the unity-feedback control system with the
following open-loop transfer function
• It involves the term sN in the denominator,
representing N poles at the origin.
• A system is called type 0, type 1, type 2, ... , if
N=0, N=1, N=2, ... , respectively.
• As the type number is increased, accuracy is
improved.
STEADY STATE ERROR
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 However, increasing the type number aggravates the
stability problem.
 A compromise between steady-state accuracy and
relative stability is always necessary.
Steady State Error of Unity Feedback Systems
 Consider the system shown in following figure.
 The closed-loop transfer function is
STEADY STATE ERROR
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 Steady State Error of Unity Feedback Systems
 The transfer function between the error signal E(s)
and the input signal R(s) is
)
(
)
(
)
(
s
G
s
R
s
E


1
1
• The final-value theorem provides a convenient way to find
the steady-state performance of a stable system.
• Since E(s) is
• The steady state error is
STEADY STATE ERROR
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Static Error Constants
 The static error constants are figures of merit of control
systems. The higher the constants, the smaller the
steady-state error.
 In a given system, the output may be the position,
velocity, pressure, temperature, or the like.
 Therefore, in what follows, we shall call the output
“position,” the rate of change of the output “velocity,”
and so on.
 This means that in a temperature control system
“position” represents the output temperature, “velocity”
represents the rate of change of the output
temperature, and so on.
STEADY STATE ERROR
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 Static Position Error Constant (Kp)
 The steady-state error of the system for a unit-step
input is
• The static position error constant Kp is defined by
• Thus, the steady-state error in terms of the static
position error constant Kp is given by
STEADY STATE ERROR
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Static Position Error Constant (Kp)
 For a Type 0 system
 For Type 1 or higher systems
 For a unit step input the steady state error ess is
STEADY STATE ERROR
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Static Velocity Error Constant (Kv)
 The steady-state error of the system for a unit-ramp
input is
 The static position error constant Kv is defined by
 Thus, the steady-state error in terms of the static
velocity error constant Kv is given by
STEADY STATE ERROR
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 For Type 1 systems
 For type 2 or higher systems
STEADY STATE ERROR
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 For a ramp input the steady state error ess is
Static Acceleration Error Constant (Ka)
 The steady-state error of the system for parabolic input
is
STEADY STATE ERROR
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 The static acceleration error constant Ka is
defined by
 Thus, the steady-state error in terms of the static
acceleration error constant Ka is given by
 For Type 1 systems
STEADY STATE ERROR
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 For type 2 systems
 For type 3 or higher systems
 For a parabolic input the steady state error ess is
STEADY STATE ERROR
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 Summery
 E.g.1:For the system shown in figure below evaluate
the static error constants and find the expected
steady state errors for the standard step, ramp and
parabolic inputs.
C(S)
R(s)
)
)(
(
)
)(
(
12
8
5
2
100
2




s
s
s
s
s
STEADY STATE ERROR
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 Solution: (evaluation of Static Error Constants)
)
)(
(
)
)(
(
)
(
12
8
5
2
100
2





s
s
s
s
s
s
G
)
(
lim s
G
K
s
p
0















 )
)(
(
)
)(
(
lim
12
8
5
2
100
2
0 s
s
s
s
s
K
s
p


p
K
)
(
lim s
sG
K
s
v
0















 )
)(
(
)
)(
(
lim
12
8
5
2
100
2
0 s
s
s
s
s
s
K
s
v


v
K
)
(
lim s
G
s
K
s
a
2
0















 )
)(
(
)
)(
(
lim
12
8
5
2
100
2
2
0 s
s
s
s
s
s
K
s
a
4
10
12
0
8
0
5
0
2
0
100
.
)
)(
(
)
)(
(














a
K
0
 0
 09
0.

STEADY STATE ERROR
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 Ex: Find the steady-state errors for inputs of 5u(t),
5tu(t), and 5t²u(t) to the system shown in below
STEADY STATE ERROR
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STABILITY ANALYSIS
 Stability: is the most important system specification.
 If a system is unstable, transient response and steady
state response (i.e. steady state error) are nothing.
 There are d/t definition of stability for a certain system
In general:
 Stable: if natural response approaches zero as time
approaches infinity.
 Unstable: if the natural response grow without bound
as time approaches infinity.
 Marginal Stability: neither grows nor decay but remain
constant or oscillate.
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 An Other Definition:
 A system is stable if every bounded input yields a
bounded output.
 A system is unstable if any bounded input yields an
unbounded output. (we call it BIBO stability definition)
 Stability and Roots of Characteristics Equation
 Characteristics Equation: is the denominator of our
system transfer function. For instance for a second
order system,
 Then, the roots will be:
STABILITY ANALYSIS
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 Based on the system pole location definitions of
stability:
 Stable systems have closed-loop transfer functions
with poles only on the left half-plane.
 If the closed-loop system poles are on the right half of
the s-plane i.e. having a positive real part, the system is
unstable.
 Thus the roots of the characteristics equation
determine the transient response of the system.
 a second order system could be:
 Over damped:
 Critically damped:
 Under damped:
STABILITY ANALYSIS
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 If the coefficient of b in the above equation be negative,
the roots will be:
 In general if any of the roots of the characteristics
equation has positive real parts, the system will be
unstable.
STABILITY ANALYSIS
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 Cont…
STABILITY ANALYSIS
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 Cont…
STABILITY ANALYSIS
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Routh Hurwitz Stability Criterion
 Routh and Hurwitz give a method of indicating the
presence and number of unstable roots, but not their
value, using Routh table.
 The Routh-Hurwitz criterion states that “the number of
roots of the characteristic equation with positive real
parts is equal to the number of changes in sign of the
first column of the Routh array”.
 This method yields stability information without the
need to solve for the closed-loop system poles.
 Using this method, we can tell how many closed-loop
system poles are in the left half-plane, in the right half-
plane, and on the jω-axis. (Note: we are saying how
many, not where.)
STABILITY ANALYSIS
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The method requires two steps:
1. Generate a data table called a Routh table.
2. interpret the Routh table to tell how many closed-loop
system poles are in the LHP, the RHP, and on the jw-axis.
 The characteristic equation of the n -order continuous
system can be written as:
 The stability criterion is applied using a Routh table
which is defined as:
STABILITY ANALYSIS
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 Cont…
 How to generate Routh Table:
 First label the rows with powers of s from highest power
of s down to lowest power of s in a vertical column.
 Next form the first row of the Routh table, using the
coefficients of the denominator of the closed-loop
transfer function (characteristic equation).
STABILITY ANALYSIS
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 Start with the coefficient of the highest power and skip
every other power of s.
 Now form the second row with the coefficients of the
denominator skipped in the previous step.
 The table is continued horizontally and vertically until
zeros are obtained.
 For convenience, any row can be multiplied or divide
by a positive constant before the next row is computed
without changing the values of the rows below and
disturbing the properties of the Routh table.
STABILITY ANALYSIS
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Routh Table Interpretation
 If the closed-loop transfer function has all poles in the
left half of the s-plane, the system is stable. Thus, a
system is stable if there are no sign changes in the first
column of the Routh table.
 The Routh-Hurwitz criterion declares that the number
of roots of the polynomial that are lies in the right half-
plane is equal to the number of sign changes in the first
column.
 Hence the system is unstable if the poles lies on the
right hand side of the s-plane.
STABILITY ANALYSIS
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 For there to be no roots with positive real parts then
there is necessary, but not sufficient, condition that all
coefficients in the characteristics equation have the
same sign and that none are zero.
 If the above is satisfied, the necessary and sufficient
condition for stability is all coefficients of the first column
of Routh array have the same sign. The number of sign
changes indicate the number of unstable roots.
STABILITY ANALYSIS
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 Only the first 2 rows of the array are obtained from the
characteristic equation the remaining are calculated as
follows;
STABILITY ANALYSIS
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 Example-2: Find the stability of the system shown
below using Routh criterion.
 The Routh table of the system is:
 System is unstable because there are two sign
changes in the first column of the Routh’s table. Hence
the equation has two roots on the right half of the
splane.
STABILITY ANALYSIS
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 Exercise : Find the stability of the system using Routh
Hurwitz Criterion.
Special Cases:
 Case 1: A Zero Only in the First Column, substitute
with a very small number ε 0 and proceed
 Example: determine the stability of the system
 Stability via Epsilon Method Routh table
STABILITY ANALYSIS
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 Cont…
 If ϵ is chosen positive, the table shows a sign change
from the s³ row to the s² row, and there will be another
sign change from the s² row to the s row. Hence, the
system is unstable and has two poles in the right half
plane.
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 Case-II: Entire Row is Zero
 Sometimes while making a Routh table, we find that an
entire row consists of zeros.
 This happen because there is an even polynomial that
is a factor of the original polynomial.
 This case must be handled differently from the case of
a zero in only the first column of a row.
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 Cont…
STABILITY ANALYSIS
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 Application of routh’s stability criteria to control system
analysis
 This criteria has limited use in LTI systems because it
does not suggest how to improve relative stability or
how to stabilize unstable systems.
 However we can consider the system in which the range
of parameters value of which determines stability of the
system(like the gain or DC gain od system)
 Example: Determining acceptable gain values for a
system For the system being stable: Consider a system
whose closed-loop transfer function is
STABILITY ANALYSIS
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 Cont…
STABILITY ANALYSIS
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 Cont…
STABILITY ANALYSIS
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Thank you dear students
Any questions?
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