KNL3353_Control_System_Engineering_Lectu.ppt

P R E P A R E D B Y
H A Z R U L B I N M O H A M E D B A S R I
D E P A R T M E N T O F E L E C T R O N I C S E N G I N E E R I N G
F A C U L T Y O F E N G I N E E R I N G
U N I V E R S I T I M A L A Y S I A S A R A W A K
( U N I M A S )
KNL 3353
Control System Engineering

Lecture Note Outlines
2
 Introduction to Control Systems
 Modeling in Frequency Domain
 Modeling in Time Domain
 Time Response
 Multiple Subsystems
 Stability
 Steady State Errors
 Root Locus Techniques
 Frequency Response Techniques

LEARNING UNIT 1
INTRODUCTION TO CONTROL SYSTEM
KNL 3353 – CONTROL
SYSTEM ENGINEERING

Introduction
4
Synopsis
This course introduces the basic concepts of
control system. It includes modeling of control
systems in frequency and time domain,
introduction of transfer function and state-space
representation, study of time response, stability
and steady-state errors, application of root locus
techniques, design via Bode Plots and root locus.

Introduction
5
Week Topics CO
1
(12/9/11 – 18/9/11)
Introduction to Control Systems.
Open-Loop Systems.
Closed-Loop Systems.
CO1
2
(19/9/11 – 25/9/11)
Modelling in the Frequency Domain
Laplace Transform
The Transfer Function - Electrical Network
CO1
3
(26/9/11 – 2/10/11)
Modelling in the Frequency Domain
(Continued)
The Transfer Function - Mechanical Systems
CO1
4
(3/10/11 – 9/10/11)
Modelling in the Time Domain
The State-Space Representation
Converting a Transfer Function to State Space
Converting a State Space to a Transfer Function
CO1
5
(10/10/11 – 16/10/11)
Time Response
Poles, Zeroes and System Response
First-Order Systems
Second-Order Systems
CO2
6
(17/10/11 – 23/10/11)
Multiple Subsystem
Block Diagrams
Signal-Flow Graphs CO2
7
(24/10/11 – 30/10/11)
Multiple Subsystem (Continued)
Mason's Rule
Alternative Representations In State Space
CO2

Introduction
6
8
(31/10/11 – 6/11/11)
Stability
Routh-Hurwitz Criterion
Stability in State-Space
9
(14/11/11 – 20/11/11)
Steady-State Errors
Steady-State Error for Unity Feedback Systems
Static Error Constants and System Type
10
(21/11/11 – 27/11/11)
Steady-State Errors (Continued)
Steady-State Error for Non-Unity Feedback
Systems
11
(28/3/11 – 4/12/11)
Root Locus Techniques
Complex Numbers
Root Locus - Define, Properties, Sketch
12
(5/12/11 – 11/12/11)
Root Locus Techniques (continued)
Breakaway and break-in points
13
(12/12/11 – 18/12/11)
Frequency Response Techniques
Bode Plots
14
(19/12/11 – 25/12/11)
Frequency Response Techniques (continued)
Stability, Gain Margin and Phase Margin

Introduction
8
Assessment Marks
Coursework
Quiz (s) / Tutorial(s)/
Assignment(s)/ Peer
Evaluation/ PBL(s)/
Project(s)/Attendance/
Report(s)/ Oral
Presentation(s) etc
30%
Test(s) 20%
Final Exam 50%

Introduction to Control System
9
Automatic control is
essential in any field of
engineering and
considered as integral part
of robotic systems, space
vehicle systems, modern
manufacturing systems
etc.

Introduction to Control System
10

Open Loop and Closed Loop Systems
11
Open-Loop Control Systems.
Those systems in which the output has no
effect on the control action are called open-
loop control systems

12
Closed-Loop Control Systems. Feedback control
systems are often referred to as closed-loop control
systems. In a closed-loop control system the
actuating error signal, which is the difference
between the input signal and the feedback signal
(which may be the output signal itself or a function of
the output signal and its derivatives and/or
integrals), is fed to the controller so as to reduce the
error and bring the output of the system to a desired
value.

13
Block diagram of room temperature control system

14
• Simple construction and
ease of maintenance.
• Less expensive than a
corresponding closed-loop
system.
• There is no stability
problem.
• Convenient when output
is hard to measure or
measuring the output
precisely is economically
not feasible.
Advantages of
Open Loop
• Disturbances and changes
in calibration cause errors,
and the output may be
different from what is
desired.
• To maintain the required
quality in the output,
recalibration is necessary
from time to time.
• They are less accurate.
• If external disturbances
are present, output differs
significantly from the
desired value.
Disadvantages
of Open Loop

15
• They are more accurate.
• The effect of external
disturbance signals can be
made very small.
• The variations in
parameters of the system
do not affect the output of
the system i.e. the output
may be made less sensitive
to variation is parameters.
Hence forward path
components can be of less
precision. This reduces the
cost of the system.
Advantages of
Closed Loop
• They are more complex
and expensive
• They require higher
forward path gains.
• The systems are prone to
instability. Oscillations in
the output many occur.
• Cost of maintenance is
high.
Disadvantages
of Closed
Loop

Classification of Control Systems
16
 Continuous Time Feedback Control Systems
If the signals in all parts of a control system are continuous
functions of time
 Discrete Data Feedback Control Systems
Discrete data control systems are those systems in which at one
or more pans of the feedback control system, the signal is in the
form of pulses. In most sampled data control systems, the signal
is reconstructed as a continuous signal, using a device called
'hold device'.

17
Discreet data control systems, in which a digital
computer is used as one of the elements, are known
as digital control systems. The input and output to
the digital computer must be binary numbers and
hence these systems require the use of digital to
analog and analog to digital converters.

18
 Linear control systems
If a system obeys superposition principle, the system is said to
be a linear system. Let
and be two inputs to a system and and
Implies
 Non-Linear Control Systems
Any system which does not obey superposition principle is said
to be a non-linear system. Physical systems are in general non-
linear and analysis of such systems is very complicated. Hence
these systems are usually linearlised and well known linear
techniques are used to analyse them.
)
(
1 t
x )
(
2 t
x )
(
1 t
y )
(
2 t
y
)
(
1 t
x )
(
1 t
y
)
(
2 t
x )
(
2 t
y
)
(
.
)
(
. 2
2
1
1 t
x
K
t
x
K  )
(
.
)
(
. 2
2
1
1 t
y
K
t
y
K 

Definitions and terminologies
19
 Controlled Variable and Manipulated Variable.
The controlled variable is
the quantity or condition that is measured and
controlled. The manipulated variable is the quantity or
condition that is varied by the controller so as to affect
the value of the controlled variable. Normally, the
controlled variable is the output of the system. Control
means measuring the value of the controlled variable of
the system and applying the manipulated variable to the
system to correct or limit deviation of the measured
value from a desired value.

20
 A plant may be a piece of equipment, perhaps just a
set of machine parts functioning together, the
purpose of which is to perform a particular
operation. In this
 A system is a combination of components that act
together and perform a certain objective.

21
 A disturbance is a signal that tends to adversely
affect the value of the output of a system. If a
disturbance is generated within the system, it is
called internal, while an external disturbance is
generated outside the system and is an input
 Feedback control refers to an operation that, in
the presence of disturbances, tends to reduce the
difference between the output of a system and some
reference input and does so on the basis of this
difference

22
End of Introduction to Control System

LEARNING UNIT 2
MODELING IN THE FREQUENCY
DOMAIN
KNL3353
23

Outlines
 Introduction
 Laplace transform
 Transfer function – Electrical network
 Transfer function – Mechanical systems
24

Introduction
 Mathematical models of physical systems are key
elements in the design and analysis of control
systems
 Desired outcomes :
- Mathematical models of physical systems are key
elements in the design and analysis of control
systems
- Understand the application of Laplace transforms
and their role in obtaining transfer functions.
25

Laplace Transform – A review
26
Let f(t) be a function of time t such that f(t) = 0 for t < 0
and s is a complex variable. We use L as a symbolic
operator to stand for Laplace integral on the quantity that
is prefixed by it.
The Laplace transform of f(t) is defined as
L [f(t)]=F(s)= 0
∞
𝑓(𝑡)𝑒−𝑠𝑡
𝑑𝑡
F(s) denotes the Laplace transform of f(t).
The function f(t) in time domain can be found from the
Laplace transform F(s) by the reverse process known as the
inverse Laplace transformation

27
Laplace transform of a common function.
i) Exponential function 𝑓 𝑡 = 𝑎𝑒−𝛼𝑡 for 𝑡 ≥ 0, 0
otherwise.
F(s)=L [𝑎𝑒−𝛼𝑡
]= 0
∞
𝑎𝑒−𝛼𝑡
𝑒−𝑠𝑡
𝑑𝑡
]= 0
∞
𝑎𝑒−(𝛼+𝑠)𝑡
𝑑𝑡
]=𝑎 0
∞
𝑒−(𝛼+𝑠)𝑡
𝑑𝑡
]=
𝑎
𝑠+𝛼
The exponential function in time is found to produce a pole
for the complex function

28
ii) Step function
𝑓 𝑡 = 𝐴𝑢 𝑡 for 𝑡 ≥ 0
𝑓 𝑡 = 0 for 𝑡 < 0
-where R is a constant and us(t) is an unit step function.
-The unit-step function has a height of unity
F(s)=L [𝑅𝑢 𝑡 ]=A 0
∞
1. 𝑒−𝑠𝑡𝑑𝑡
F(s)=L [𝑅𝑢 𝑡 ]=A
𝑒−𝑠𝑡
−𝑠 0
∞
F(s)=L [𝑅𝑢 𝑡 ]=
𝐴
𝑠

29
iii) Ramp function
𝑓 𝑡 = 𝑅𝑡𝑢 𝑡 for 𝑡 ≥ 0
𝑓 𝑡 = 0 for 𝑡 < 0
-where R is the slope of a linear function and us(t) is an unit
step function.
-The unit-step function has a height of unity
F(s)=L [𝑅𝑡𝑢 𝑡 ]=R 0
∞
𝑡. 𝑒−𝑠𝑡
𝑑𝑡
F(s)=L [𝑅𝑢 𝑡 ]=𝑅 𝑡
𝑒−𝑠𝑡
−𝑠 0
∞
− R 0
∞ 𝑒−𝑠𝑡
−𝑠
𝑑𝑡
F(s)=L [𝑅𝑢 𝑡 ]=
𝑅
𝑠2

30
iv) Sinusoidal function
𝑓 𝑡 = 𝐴𝑠𝑖𝑛(𝜔𝑡) for 𝑡 ≥ 0
𝑓 𝑡 = 0 for 𝑡 < 0
Where
sin 𝜔𝑡 =
1
2𝑗
𝑒𝑗𝜔𝑡 − 𝑒−𝑗𝜔𝑡
F(s)=L [𝐴 sin 𝜔𝑡 ]=
𝐴
2𝑗 0
∞
𝑒𝑗𝜔𝑡 − 𝑒−𝑗𝜔𝑡 . 𝑒−𝑠𝑡𝑑𝑡
𝐴
2𝑗
[ 0
∞
𝑒𝑗𝜔𝑡−𝑠𝑡𝑑𝑡 + 0
∞
𝑒−𝑗𝜔𝑡−𝑠𝑡𝑑𝑡]

Laplace transform – A review
31
𝐴
2𝑗
[ 0
∞
∞
𝐴
2𝑗
[ 0
∞
∞
𝐴
2𝑗
𝑒𝑗𝜔𝑡−𝑠𝑡
𝑗𝜔−𝑠
−
𝑒−𝑗𝜔𝑡−𝑠𝑡
−𝑗𝜔−𝑠
𝐴𝜔
𝑠2+𝜔2
Performing the same calculation, one can obtain
F(s)=L [𝐴𝑐𝑜𝑠(𝜔𝑡)]=
𝐴𝑠
𝑠2+𝜔2

Laplace transform – A review
32
v) Pulse function
𝑓 𝑡 =
𝐴
𝑇
for 0 < 𝑡 ≤ 𝑇
𝑓 𝑡 = 0 for 𝑡 > T, 𝑡 < 0
𝑓 𝑡 =
𝐴
𝑇
𝑢 𝑡 −
𝐴
𝑇
𝑢 𝑡 − 𝑇
F(s)=L [𝑓 𝑡 ] = L
𝐴
𝑇
𝑢 𝑡 −
𝐴
𝑇
𝑢 𝑡 − 𝑇
F(s)=L [𝑓 𝑡 ] = L
𝐴
𝑇
𝑢 𝑡 −
𝐴
𝑇
𝑢 𝑡 − 𝑇
F(s)=L [𝑓 𝑡 ] =
𝐴
𝑇𝑠
−
𝐴
𝑇𝑠
𝑒−𝑇𝑠
F(s)=L [𝑓 𝑡 ] =
𝐴
𝑇𝑠
(1 − 𝑒−𝑇𝑠
)

Laplace Transform – A Review
33
vi) Impulse function
𝑓 𝑡 = lim
𝑇→0
𝐴
𝑇
for 0 < 𝑡 ≤ 𝑇
𝑓 𝑡 = 0 for 𝑡 > T, 𝑡 < 0
F(s)=L [𝑓 𝑡 ] = L lim
𝑇→0
𝐴
𝑇
F(s)=L [𝑓 𝑡 ] = lim
𝑇→0
𝐴
𝑇𝑠
(1 − 𝑒−𝑇𝑠)
F(s)=L [𝑓 𝑡 ] = A

34
Properties of Laplace Transform
1- Multiplication by a constant
L [𝑎𝑓 𝑡 ] = 𝑎L [ 𝑓 𝑡 ] = 𝑎𝐹 𝑠
L [𝑓1(𝑡) + 𝑓2(𝑡)] = L [ 𝑓1 𝑡 ] + L [ 𝑓2 𝑡 ] = 𝐹1(𝑠) + 𝐹2(𝑠)
2- Multiplication by 𝑒−𝛼𝑡
L [𝑒−𝛼𝑡𝑓 𝑡 ] = 0
∞
𝑒−𝛼𝑡𝑓 𝑡 𝑒−𝑠𝑡 = 𝐹 𝑠 + 𝛼
3-Translation in time , 𝛼 > 0
L [𝑓 𝑡 − 𝛼 ] = 𝑒𝛼𝑠𝐹(𝑠)

35
4- Change in time scale, 𝛼 ≠ 0
L [𝑓
𝑡
𝛼
] = 0
∞
𝑓
𝑡
𝛼
𝑑𝑡 = 𝛼F(𝛼s)
5- Real differentiation
L
𝑑
𝑑𝑡
𝑓(𝑡) = 𝑠𝐹 𝑠 − 𝑓 0
L
𝑑𝑛
𝑑𝑡𝑛 𝑓(𝑡) = 𝑠𝑛𝐹 𝑠 − 𝑠𝑛−1𝑓 0 − 𝑠𝑛−2𝑓′ 0 −
𝑠𝑛−3
𝑓" 0 − ⋯ − 𝑠𝑓(𝑛−2)
(0) − 𝑓(𝑛−1)
(0)

36
6-Initial value theorem
𝑓 0 = lim
𝑠→∞
𝑠𝐹(𝑠)
7-Final value theorem
𝑓 ∞ = lim
𝑠→0
𝑠𝐹(𝑠)
8-Real integration
L 𝑓 𝑡 𝑑𝑡 =
𝐹(𝑠)
𝑠
+
𝑓 𝑡 𝑑𝑡|𝑡=0
𝑠

37
9-Complex differentiation
L 𝑡𝑛
𝑓(𝑡) = −1 𝑛 𝑑𝑛
𝑑𝑡𝑛 𝐹(𝑠)
10- Convolution integral
L 0
𝑡
𝑓1 𝑡 − 𝜏 𝑓2 𝑡 𝑑𝜏 = L 𝑓1 ∗ 𝑓2 = 𝐹1(𝑠)𝐹2(𝑠)

Laplace Transforms– A Review
40
 Partial Fraction Expansion (Example)
 It can be expressed as the summation of partial
fraction.
 Where k1 and k2 are the coefficient need to be
determined

41
Partial-Fraction Expansion Method for Finding
Inverse Laplace Transforms
Partial-Fraction Expansion when
F(s) Involves Distinct Poles Only.
Partial-Fraction Expansion when
F(s) Involves Multiple Poles
)
)...(
)(
(
)
)...(
)(
(
)
(
)
(
)
(
2
1
2
1
n
m
p
s
p
s
p
s
z
s
z
s
z
s
K
s
A
s
B
s
F








where p1, p2, . . . , pn and z1, z2,. . . , zm are either real or complex quantities
for each complex pi or zi, there will occur the complex conjugate of pi or zi,
respectively

42
)
)...(
)(
(
)
)...(
)(
(
)
(
)
(
)
(
2
1
2
1
n
m
p
s
p
s
p
s
z
s
z
s
z
s
K
s
A
s
B
s
F








Partial-Fraction Expansion when F(s) Involves Distinct Poles Only.
)
(
...
)
(
...
)
(
)
(
)
(
)
(
)
(
2
2
1
1
n
n
k
k
p
s
a
p
s
a
p
s
a
p
s
a
s
A
s
B
s
F










The coefficient ak is called the residue at the pole at s = -pk. The value of ak can be
found by multiplying both sides by ( s + pk) and letting s = -pk, which give

43
Partial-Fraction Expansion when F(s) Involves Multiple Poles
Example :
     k
k
k
k
k
p
s
c
p
s
c
p
s
c
p
s
c
p
s
s
A
s
F










 

1
1
2
2
1
...
)
(
)
(
)
(
   
   
      k
k
k
k
k
k
k
k
c
p
s
c
p
s
c
p
s
c
p
s
c
p
s
c
p
s
c
p
s
s
F
p
s

























1
2
2
1
1
2
2
1
...
...
)
(
 
       
 
      1
2
3
2
2
1
1
2
2
1
1
2
...
)
2
(
)
1
(
...
)
(

























k
k
k
k
k
k
k
k
k
c
p
s
c
p
s
c
k
p
s
c
k
c
p
s
c
p
s
c
p
s
c
ds
d
s
F
p
s
ds
d
 
      2
4
2
3
1
3
2
2
2
...
)
2
)(
3
(
)
1
)(
2
(
)
( 












 k
k
k
c
p
s
c
k
k
p
s
c
k
k
s
F
p
s
ds
d
 
   
  3
5
2
4
1
3
3
3
2
3
...
)
2
)(
3
)(
4
(
)
1
)(
2
)(
3
(
)
(

















k
k
k
c
p
s
c
k
k
k
p
s
c
k
k
k
s
F
p
s
ds
d

Laplace transform– A Review
44
  k
p
s
k
c
s
F
s 



)
(
1
 
  1
)
( 



 k
p
s
k
c
s
F
p
s
ds
d
 
   
  2
2
2
2
2
2
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3
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)
(
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3
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)
( 
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Transfer Function
45
 The transfer function of a linear systems
 Let us begin by writing a general nth-order, linear, time-
invariant differential equation,
 where c(t) is the output, r(t) is the input,

Transfer Function
46
 ipdjvaj
Electrical networks

Transfer function – Electrical networks
47

Transfer function – Mechanical Systems
48
Transfer function – Electrical networks

49
End of Modeling in Frequency
Domain

LEARNING UNIT 3
MODELING IN TIME DOMAIN
KNL3353
50

Outlines
 Objective
 Introduction
 State space representation
 Electrical network model in state space representation
 Conversion of transfer function to state space and vice-
versa
 Time response analysis
51

Objectives
After completing this chapter, the student will be able
to :
 Find a mathematical model, called a state-space
representation, for a linear, time invariant system
 Model electrical systems in state space
 Convert a transfer function to state space
 Convert a state-space representation to a transfer
function
 Analyze time response of a system
52

Introduction
Two approaches
Frequency domain
Based on converting a system's
differential equation to a transfer
function, thus generating a
mathematical model of the
system that algebraically relates
a representation of the output to
a representation of the input
Tim e domain
Unified method for modeling,
analyzing, and designing a wide
range of systems.
53

State space representation
 The state and output equations can be written in vector-
matrix form if the system is linear
𝑥 = 𝐴𝑥 + 𝐵𝑢
𝑦 = 𝐶𝑥 + 𝐷𝑢
Where
𝑥= state vector (contains the state variables)
x = derivative of the state vector with respect to time
y = output vector
u = input or control vector
A = system matrix
B = input matrix
C = output matrix
D = feedforward matrix
54

State vector must be chosen according to the following considerations :
A minimum number of state
variables must be selected as
components of the state
vector. This minimum
number of state variables is
sufficient to describe
completely the state of the
system.
Typically, the minimum
number required equals the
order of the differential
equation describing the
system.
The components of the state
vector (that is, this minimum
number of state variables)
must be linearly independent
55

 Example :
56

 Example : Find the state space representation
59

Conversion of transfer function to state space and
vice-versa
Consider the differential equation
𝑑𝑛
𝑦
𝑑𝑡𝑛
+ 𝑎𝑛−1
𝑑𝑛−1
𝑦
𝑑𝑡𝑛−1
+ 𝑎𝑛−2
𝑑𝑛−2
𝑦
𝑑𝑡𝑛−2
+ ⋯ + 𝑎1
𝑑𝑦
𝑑𝑡
+ 𝑎0 = 𝑏0𝑢
62

 Finally
vice-versa
63

Example : Find the state-space representation in
phase-variable form for the transfer function
𝐶(𝑠)
𝑅(𝑠)
=
24
𝑠3 + 9𝑠2 + 26𝑠 + 24
This yields 𝐶(𝑠)(𝑠3 + 9𝑠2 + 26𝑠 + 24) = 24𝑅(𝑠)
𝑐 + 9𝑐 + 26𝑐 + 24𝑐 = 24𝑟
vice-versa
64

vice-versa
65

Example : Find the state-space representation of the
transfer function
vice-versa
66

vice-versa
67

68
 Given the state and output equations
𝑥 = 𝐴𝑥 + 𝐵𝑢
𝑦 = 𝐶𝑥 + 𝐷𝑢
take the Laplace transform assuming zero initial
conditions
𝑠𝑋 𝑠 = 𝐴𝑋 𝑠 + 𝐵𝑈 𝑠
𝑌(𝑠) = 𝐶𝑋(𝑠) + 𝐷𝑈(𝑠)
Solving X(s) : (𝑠𝐼 − 𝐴)𝑋 𝑠 = 𝐵𝑈 𝑠
Or, 𝑋 𝑠 = (𝑠𝐼 − 𝐴)−1𝐵𝑈 𝑠
Conversion of state space to transfer function

69
Considering the output :
𝑌 𝑠 = 𝐶. (𝑠𝐼 − 𝐴)−1
𝐵𝑈 𝑠 + 𝐷𝑈(𝑠)
Put into common factor
𝑌 𝑠 = 𝐶. (𝑠𝐼 − 𝐴)−1𝐵 + 𝐷 𝑈(𝑠)
Finally

70
 Example: Given the system, find the transfer
function, T(s) = Y(s)/U(s), where U(s) is the input
and Y(s) is the output.

71

72

THE END
73
Questions and Answers

LEARNING UNIT 4
TIME RESPONSE
KNL3353
74

Outlines
75
 Introduction
 Poles, Zeros and Systems Response
 First order system
 Second order system

Introduction
76
The output response of a system is the sum of two responses:
the forced response and the natural response.
The use of poles and zeros and their relationship to the time
response of a system is such a technique
The concept of poles and zeros, fundamental to the analysis
and design of control systems, simplifies the evaluation of a
system's response

Poles, Zeros and Systems Response
77
The poles of a transfer function are
 the values of the Laplace transform variable s, that cause
the transfer function to become infinite
 any roots of the denominator of the transfer function that
are common to roots of the numerator.
However, if a factor of the denominator can be canceled by
the same factor in the numerator, the root of this factor no
longer causes the transfer function to become infinite

78
The zeros of a transfer function are
 the values of the Laplace transform variable s, that cause
the transfer function to become zero
 any roots of the numerator of the transfer function that
are common to roots of the denominator.
The roots of the numerator are values of s that make the
transfer function zero and are thus zeros.
However, if a factor of the numerator can be canceled by
the same factor in the denominator, the root of this factor
no longer causes the transfer function to become zero

79
Poles and Zeros of a first order system: An example
Given the transfer function G(s), a pole exists at s =- 5 , and a zero
exists at -2. These values are plotted on the complex s-plane using
an ‘x’ for the pole and a ‘o’ for the zero. Let us find the unit step
response of the system.
𝐶 𝑠 =
𝑠 + 2
𝑠(𝑠 + 5)
=
𝐴
𝑠
+
𝐵
𝑠 + 5
=
2
5
𝑠
+
3
5
𝑠 + 5
Thus
𝑐 𝑡 =
2
5
+
3
5
𝑒−5𝑡

80
 a. System showing
input and output;
 b. pole-zero plot of
the system;
 c. evolution of a
system response.

81
 A pole of the input
function generates the
form of the forced
response (that is, the
pole at the origin
generated a step
function at the output).
 A pole of the transfer
function generates the
form of the natural
response (that is, the
pole at - 5 generated
𝑒−5𝑡
.

82
 A pole on the real axis
generates an exponential
response of the form 𝑒−𝛼𝑡
where −𝛼 is the pole
location on the real axis.
Thus, the farther to the left
a pole is on the negative
real axis, the faster the
exponential transient
response will decay to zero
 The zeros and poles
generate the amplitudes for
both the forced and natural
Responses

First order system
83
Consider a unity feedback system with
𝐺 𝑠 =
1
𝜏𝑠
The closed loop transfer function of the system is given by
𝑇 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
1
1 + 𝜏𝑠

First order system
84
For a unit step input 𝑅 𝑠 =
1
𝑠
and the output is given by
C 𝑠 =
1
𝑠(1+𝜏𝑠)
=
1
𝑠
+
1
1+𝜏𝑠
Inverse Laplace transformation yields 𝑐 𝑡 = 1 − 𝑒−
𝑡
𝜏
The response is an exponentially
increasing function and it
approaches a value of unity as 𝑡 → ∞
At t = 𝜏, the response reaches a value
𝑐 𝑡 = 1 − 𝑒−1 ≈ 0.632 which is 63.2
percent of the steady value. This
time, 𝜏 , is known as the time
constant of the system.

85
Response to a Unit Ramp Input or Unit Velocity Input
𝐶 𝑠 =
1
𝑠2(1 + 𝜏𝑠)
The time response is obtained by taking inverse
Laplace transform
𝑐 𝑡 = 𝑡 − 𝜏(1 − 𝑒−
𝑡
𝜏)

Second order system
86
 We shall undertake here the analysis of a
second-order system that will help to form a
basis for the understanding of analysis and
design of higher-order systems.
 The open-loop transfer function of the
system is
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
𝑠(𝑠 + 2𝜉𝜔𝑛)
Where
𝜔𝑛=undamped natural frequency
𝜉=damping ratio
 The closed-loop transfer function of the
system with H(s) = 1 is
𝐶(𝑠)
𝑅 𝑠
=
𝜔𝑛
2
𝑠2 + 2𝜉𝜔𝑛𝑠 + 𝜔𝑛
2

Second order system
87
The characteristic equation of the closed-loop system
is obtained by setting the denominator zero.
Δ 𝑠 = 𝑠2 + 2𝜉𝜔𝑛𝑠 + 𝜔𝑛
2 = 0
The roots of the characteristic equation are given by
𝑠1,2 = −𝜉𝜔𝑛 ± 𝜔𝑛 𝜉2 − 1
The dynamic behavior of the second-order system can,
therefore, be described in terms of two parameters of
the roots of the characteristic equation, 𝜁 and 𝜔𝑛,
respectively.

Second order system
88
From the point of view of transient response, four distinct
cases are of interest to us when 𝜔𝑛is held constant and the
damping ratio 𝜉 takes on different values.
Underdamped
case : 0 < 𝜉 < 1
Critically damped
case : 𝜉=1
Overdamped case
𝜉>1
Undamped case
𝜉=0

Second order system
89
Underdamped case: 𝟎 < 𝝃 < 𝟏
𝑠2 + 2𝜉𝜔𝑛𝑠 + 𝜔𝑛
2 = 0
𝑎 = 1, 𝑏 = 2𝜉𝜔𝑛, 𝑐 = 𝜔𝑛
2
Δ = 𝑏2 − 4𝑎𝑐 = 4𝜔𝑛
2(𝜉2 − 1)
If 0 < 𝜉 < 1, then 𝜉2 − 1 < 0 →two complex conjugate
roots
𝑠1,2 = −𝜉𝜔𝑛 ± 𝑗𝜔𝑛 1 − 𝜉2
𝑠1,2 = −𝜉𝜔𝑛 ± 𝑗𝜔𝑑
which are complex conjugates and lie on the semicircle in
the left-half s-plane

Second order system
90
Step input response for under-damped case
In this case, Y(s)/R(s) can be written as
𝑌(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
(𝑠+𝜉𝜔𝑛+𝑗𝜔𝑑)(𝑠+𝜉𝜔𝑛−𝑗𝜔𝑑)
Or
𝑌 𝑠 =
𝜔𝑛
2
𝑠 𝑠+𝜉𝜔𝑛+𝑗𝜔𝑑 𝑠+𝜉𝜔𝑛−𝑗𝜔𝑑
=
1
𝑠
−
𝑠+2𝜉𝜔𝑛
𝑠2+2𝜉𝜔𝑛𝑠+𝜔𝑛
2
𝑌 𝑠 =
1
𝑠
−
𝑠+𝜉𝜔𝑛
𝑠+2𝜉𝜔𝑛
2+𝜔𝑑
2 −
𝜉𝜔𝑛
𝑠+2𝜉𝜔𝑛
2+𝜔𝑑
2
𝑌 𝑠 =
1
𝑠
−
𝑠+𝜉𝜔𝑛
𝑠+2𝜉𝜔𝑛
2+𝜔𝑑
2 −
𝜉
1−𝜉2
𝜔𝑑
𝑠+2𝜉𝜔𝑛
2+𝜔𝑑
2

Second order system
91
The inverse Laplace
𝐶 𝑠 =
1
𝑠
−
𝑠+𝜉𝜔𝑛
𝑠+𝜉𝜔𝑛
2+𝜔𝑑
2 −
𝜉
1−𝜉2
𝜔𝑑
𝑠+𝜉𝜔𝑛
2+𝜔𝑑
2
L−1
𝑠 + 𝜉𝜔𝑛
𝑠 + 𝜉𝜔𝑛
2 + 𝜔𝑑
2
= 𝑒−𝜉𝜔𝑛𝑡cos(𝜔𝑑𝑡)
L−1
𝜔𝑑
𝑠 + 𝜉𝜔𝑛
2 + 𝜔𝑑
2
= 𝑒−𝜉𝜔𝑛𝑡
sin(𝜔𝑑𝑡)
L−1 𝐶(𝑠) = 1 − 𝑒−𝜉𝜔𝑛𝑡 cos 𝜔𝑑𝑡 −
𝜉
1 − 𝜉2
𝑒−𝜉𝜔𝑛𝑡sin(𝜔𝑑𝑡)
𝑐(𝑡) = 1 − 𝑒−𝜉𝜔𝑛𝑡 cos 𝜔𝑑𝑡 +
𝜉
1 − 𝜉2
sin(𝜔𝑑𝑡)
Or
𝑐 𝑡 = 1 −
𝑒−𝜉𝜔𝑛𝑡
1 − 𝜉2
sin 𝜔𝑑𝑡 + 𝜃 , 𝜃 = 𝑡𝑎𝑛−1
1 − 𝜉2
𝜉
= 𝑐𝑜𝑠−1(𝜉)

Second order system
92
Critically damped case: 𝝃 = 𝟏
𝑠2
+ 2𝜉𝜔𝑛𝑠 + 𝜔𝑛
2
= 0
𝑎 = 1, 𝑏 = 2𝜉𝜔𝑛, 𝑐 = 𝜔𝑛
2
Δ = 𝑏2 − 4𝑎𝑐 = 4𝜔𝑛
2(𝜉2 − 1)
If 𝜉 = 1, then 𝜉2 − 1 = 0 →two same roots
𝑠1,2 = −𝜔𝑛
The step response is monotonic without any
overshoots and undershoots.

Second order system
93
For a unit step input for critically damped case
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
(𝑠+𝜔𝑛)2
Or
𝐶(𝑠) =
𝜔𝑛
2
𝑠(𝑠+𝜔𝑛)2 =
𝐴
𝑠
+
𝐵
(𝑠+𝜔𝑛)2 +
𝐶
𝑠+𝜔𝑛
𝐴 = lim
𝑠→0
𝜔𝑛
2
𝑠 + 𝜔𝑛
2
= 1, 𝐵 = lim
𝑠→−𝜔𝑛
𝜔𝑛
2
𝑠
= −𝜔𝑛, 𝐶 = lim
𝑠→−𝜔𝑛
𝑑(
𝜔𝑛
2
𝑠
)
𝑑𝑡
= −1
𝑐 𝑡 = L−1 1
𝑠
−
1
𝑠 + 𝜔𝑛
2
−
1
𝑠 + 𝜔𝑛
𝑐 𝑡 = 1 − 𝑒−𝜔𝑛𝑡(𝜔𝑛𝑡 + 1)

Second order system
94
Over damped case: 𝝃 > 𝟏
𝑠2
2
= 0
𝑎 = 1, 𝑏 = 2𝜉𝜔𝑛, 𝑐 = 𝜔𝑛
2
Δ = 𝑏2 − 4𝑎𝑐 = 4𝜔𝑛
2(𝜉2 − 1)
If 𝜉 > 1, then 𝜉2 − 1 > 0 →two real and distinct roots
𝑠1,2 = −𝜉𝜔𝑛 ± 𝜔𝑛 𝜉2 − 1
𝑠1 = −𝜉𝜔𝑛 + 𝜔𝑛 𝜉2 − 1 , 𝑠2 = −𝜉𝜔𝑛 − 𝜔𝑛 𝜉2 − 1
both the roots are lying on real axis

Second order system
95
For a unit step input for critically damped case
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
(𝑠+𝜔𝑛)2
Or
𝐶 𝑠 =
𝜔𝑛
2
𝑠 𝑠 − 𝑠1 𝑠 − 𝑠2
=
𝐴
𝑠
+
𝐵
𝑠 − 𝑠1
+
𝐶
𝑠 − 𝑠2
𝑠1 = −𝜉𝜔𝑛 + 𝜔𝑛 𝜉2 − 1 , 𝑠2 = −𝜉𝜔𝑛 − 𝜔𝑛 𝜉2 − 1
𝑐 𝑡 = 1 −
𝑒
− 𝜉− 𝜉2−1 𝜔𝑛𝑡
2 −𝜉2 + 𝜉 𝜉2 − 1 + 1
−
𝑒
− 𝜉+ 𝜉2−1 𝜔𝑛𝑡
2 −𝜉2 − 𝜉 𝜉2 − 1 + 1

Second order system
96
Undamped case: 𝝃 = 𝟎
𝑠2
2
= 0
𝑎 = 1, 𝑏 = 2𝜉𝜔𝑛, 𝑐 = 𝜔𝑛
2
Δ = 𝑏2 − 4𝑎𝑐 = 4𝜔𝑛
2(𝜉2 − 1)
If 𝜉 = 0, then 𝜉2 − 1 < 0 →two pure imaginary roots
𝑠1,2 = ±𝑗𝜔𝑛
𝑠1 = +𝑗𝜔𝑛,𝑠2 = −𝑗𝜔𝑛
both the roots are lying on imaginary axis.
The transient response is oscillatory in nature and
does not die out

Second order system
97
For a unit step input for undamped case. In this case, Y(s)/R(s) can be
written as
𝐶(𝑠)
𝑅(𝑠)
=
𝜔𝑛
2
𝑠2 + 𝜔𝑛
2
Or
𝑐 𝑠 =
𝜔𝑛
2
𝑠(𝑠 + 𝑗𝜔𝑛)(𝑠 − 𝑗𝜔𝑛)
=
𝐴
𝑠
+
𝐵
𝑠 + 𝑗𝜔𝑛
+
𝐶
𝑠 − 𝑗𝜔𝑛
𝐴 = lim
𝑠⟶0
𝜔𝑛
2
(𝑠 + 𝑗𝜔𝑛)(𝑠 − 𝑗𝜔𝑛)
= 1, 𝐵 = lim
𝑠⟶𝑠+𝑗𝜔𝑛
𝜔𝑛
2
𝑠(𝑠 − 𝑗𝜔𝑛)
= −
1
2
, 𝐶 = lim
𝑠⟶𝑠−𝑗𝜔𝑛
𝜔𝑛
2
𝑠(𝑠 + 𝑗𝜔𝑛)
= −
1
2
𝑐 𝑠 =
𝜔𝑛
2
𝑠(𝑠 + 𝑗𝜔𝑛)(𝑠 − 𝑗𝜔𝑛)
=
1
𝑠
−
𝑠
𝑠2 + 𝜔𝑛
2
L−1
1
𝑠
−
𝑠
𝑠2 + 𝜔𝑛
2
= 1 − cos(𝜔𝑛𝑡)

Second order system
100
Peak time : the peak time 𝑡𝑝 by differentiating y(t) with respect to time and letting
this derivative equal to zero
𝑑
𝑑𝑡
𝑐 𝑡
𝑡=𝑡𝑝
=
𝑑
𝑑𝑡
1 −
𝑒−𝜉𝜔𝑛𝑡
1 − 𝜉2
sin 𝜔𝑑𝑡 + 𝜃 = 0
𝜔𝑛𝑒−𝜉𝜔𝑛𝑡𝑝
1 − 𝜉2
𝜉 sin 𝜔𝑑𝑡𝑝 + 𝜃 − 1 − 𝜉2cos(𝜔𝑑𝑡𝑝 + 𝜃) = 0
tan 𝜔𝑑𝑡𝑝 + 𝜃 =
1 − 𝜉2
𝜉
= tan 𝜃
𝜔𝑑𝑡𝑝 = 0 + 𝑘𝜋, 𝑘 ∈ ℤ
For 𝑘 = 1,
𝑡𝑝 =
𝜋
𝜔𝑑
=
𝜋
𝜔𝑛 1 − 𝜉2
The peak time 𝑡𝑝 corresponds to one-half cycle of the frequency of damped
oscillation

Second order system
101
Maximum overshoot 𝑴𝒑 : The maximum overshoot occurs at the peak
time at 𝑡 = 𝑡𝑝 =
𝜋
𝜔𝑑
.
𝑀𝑝 = 𝑦 𝑡𝑝 − 𝑦𝑠𝑠 = 𝑦 𝑡𝑝 − 1
𝑀𝑝 = −
𝑒
−
𝜉𝜋
1−𝜉2
1 − 𝜉2
sin 𝜋 + 𝜃 , 𝜃 = 𝑡𝑎𝑛−1
1 − 𝜉2
𝜉
= 𝑐𝑜𝑠−1
(𝜉)
𝑀𝑝 = −
𝑒
−
𝜉𝜋
1−𝜉2
1 − 𝜉2
sin 𝜋 cos(𝜃) + cos 𝜋 sin(𝜃)
𝑀𝑝 = −𝑒
−
𝜉𝜋
1−𝜉2 𝜉
1 − 𝜉2
sin 𝜋 + cos 𝜋
𝑀𝑝 = 𝑒
−
𝜉𝜋
1−𝜉2

Second order system
102
Settling time 𝒕𝒔: The exact analytical expression for the
settling time even for a simple second order system cannot
be derived. However, it can be approximated for damping
ratios in the range 0 < 𝜉 < 0.7 by the envelope of the
damped sinusoidal output 𝑦 𝑡 .
1 ±
𝑒−𝜉𝜔𝑛𝒕𝒔
1 − 𝜉2
= 1 ± 0.02
𝑡𝑠 = −
1
𝜉𝜔𝑛
𝑙𝑛 0.02 1 − 𝜉2 , 0.01 < 𝜉 < 0.7
𝑡𝑠 ≈
4
𝜉𝜔𝑛
, ±2%, 𝑡𝑠 ≈
3.12
𝜉𝜔𝑛
, ±5%

LEARNING UNIT 5
MULTIPLE SUBSYSTEMS
KNL3353

Outlines
 Introduction
 Block Diagrams
 Signal Flow Graph
 Mason’s Rule

Introduction
 More complicated systems, however, are represented
by the interconnection of many subsystems.
 In this chapter, multiple subsystems are represented
in two ways: as block diagrams and as signal-flow
graphs.
 Signal-flow graphs represent transfer functions as
lines, and signals as small circular nodes. Summing
is implicit.

Objective
 Reduce a block diagram of multiple subsystems to a
single block representing the transfer function from
input to output
 Convert block diagrams to signal-flow diagrams
 Find the transfer function of multiple subsystems
using Mason's rule
 Represent state equations as signal-flow graphs

Block Diagrams
 Cascade form

Block Diagrams
 Parallel form

Block Diagrams
 Feed back form 𝐶(𝑠) = 𝐺(𝑠)𝐸(𝑠)
𝐸(𝑠) = 𝑅(𝑠) – 𝐵(𝑠) = 𝑅(𝑠) – 𝐻 𝑠 𝐶(𝑠)
𝐶(𝑠) = 𝐺(𝑠)[𝑅(𝑠) – 𝐻 𝑠 𝐶(𝑠)]
𝐶 𝑠 = 𝐺 𝑠 𝑅 𝑠 − 𝐺(𝑠) 𝐻 𝑠 𝐶(𝑠)
𝐶 𝑠 (1 + 𝐺 𝑠 𝐻 𝑠 ) = 𝐺 𝑠 𝑅 𝑠
Finally
𝐶(𝑠)
𝑅(𝑠)
=
𝐺(𝑠)
1 + 𝐺 𝑠 𝐻 𝑠

Block Diagrams
Moving Blocks to Create Familiar Forms –
Example :
Block diagram algebra for
summing junctions—
equivalent forms for moving a
block a. to the left past a
summing junction; b. to the
right past a summing junction

Block Diagrams
 Example: Reduce the block diagram to a single
transfer function.

Block Diagrams
Example : Reduce the system to a single transfer
function

Signal Flow Graphs
 Signal-flow graphs are an alternative to block diagrams. Unlike block
diagrams, which consist of blocks, signals, summing junctions, and
pickoff points, a signal-flow graph consists only of branches, which
represent systems, and nodes, which represent signals.
Nodes
Nodes
Nodes

Signal Flow Graphs
 Signal-flow graphs are an alternative to block diagrams. Unlike block
diagrams, which consist of blocks, signals, summing junctions, and
pickoff points, a signal-flow graph consists only of branches, which
represent systems, and nodes, which represent signals.
Systems Systems

Signal Flow Graphs
V(s) = R1(s)G1(s)-R2(s)G2(s) +R3(s)G3(s).
C1(s)=V(s)G4(s)=(R1(s)G1(s)-R2(s)G2(s) +R3(s)G3(s))G4(s)
C2(s)=V(s)G5(s)=(R1(s)G1(s)-R2(s)G2(s) +R3(s)G3(s))G5(s)
C3(s)=-V(s)G6(s)=-(R1(s)G1(s)-R2(s)G2(s) +R3(s)G3(s))G6(s)
Notice that in summing negative signals
we associate the negative sign with the
system and not with a summing
junction, as in the case of block
diagrams.

Signal Flow Graphs
124
Converting a Block Diagram to Signal Flow Graph

Converting a Block Diagram to Signal Flow Graph
125
Notice that the negative signs at the
summing junctions of the block
diagram are represented by the
negative transfer functions of the
signal-flow graph

Mason’s Rule
126
 Technique for reducing signal-flow graphs to single
transfer functions that relate the output of a system
to its input.
 Mason's rule for reducing a signal-flow graph to a
single transfer function requires the application of
one formula.
 Mason's rule easier to use than block diagram
reduction.
 Mason's formula has several components that must
be evaluated.

Mason’s Rule
127
Loop gain :
The product of branch gains
found by traversing a path that
starts at a node and ends at the
same node, following the
direction of the signal flow,
without passing through any
other node more than once.
1) 𝐺2 𝑠 𝐻1 𝑠
2) 𝐺4 𝑠 𝐻2 𝑠
3) 𝐺4 𝑠 𝐺5 𝑠 𝐻3 𝑠
4) 𝐺4 𝑠 𝐺6 𝑠 𝐻3 𝑠

Mason’s Rule
128
Forward path gain :
The product of gains found by
traversing a path from the input
node to the output node of the
signal-flow graph in the
direction of signal flow
1) 𝐺1 𝑠 𝐺2 𝑠 𝐺3 𝑠 𝐺4 𝑠 𝐺5 𝑠 𝐺7 𝑠
2) 𝐺1 𝑠 𝐺2 𝑠 𝐺3 𝑠 𝐺4 𝑠 𝐺6 𝑠 𝐺7 𝑠

Mason’s Rule
129
Non touching loops:
Loops that do not have any
nodes in common
loop G2(s)H1(s) does not touch loops
G4(s)H2(s), G4(s)G5(s)H3(s), and
G4(s)G6(s)H3(s).

Mason’s Rule
130
Non touching loops gain:
The product of loop gains from non
touching loops taken two, three, four, or
more at a time.
Product of loop gain G2(s)H1(s) and loop
gain G4(s)H2(s) is a non touching-loop
gain taken two at a time.
In our example there are no non touching-
loop gains taken three at a time since three
non touching loops do not exist in the
example.
1) 𝐺2(𝑠)𝐻1(𝑠) 𝐺4(𝑠)𝐻2(𝑠)
2) 𝐺2(𝑠)𝐻1(𝑠) 𝐺4(𝑠)𝐺5(𝑠)𝐻3(𝑠)
3) 𝐺2(𝑠)𝐻1(𝑠) 𝐺4(𝑠)𝐺6(𝑠)𝐻3(𝑠)

Mason’s rule
131
The transfer function :
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
𝑘 𝑇𝑘∆𝑘
∆
𝑘=number of forward path
𝑇𝑘=the 𝑘𝑡ℎ
forward path gain
∆= 1 − loop gain + non touching loop gains taken two at
a time- non touching loop gains taken three at a time+….
∆𝑘= ∆- loop gain terms in ∆ that touch the 𝑘𝑡ℎ
forward
path. In other words, ∆𝑘 is formed by eliminating from ∆
those loop gains that touch the 𝑘𝑡ℎ
forward path.

Mason’s Rule
132
Example : Find the transfer function, C(s)/R(s) for the
following signal-flow graph

Mason’s Rule
133
1𝑠𝑡
• Identify the forward-path gains
• 𝐺1(𝑠)𝐺2(𝑠)𝐺3(𝑠)𝐺4(𝑠)𝐺5(𝑠)
2𝑛𝑑
• Identify the loop gains.
• 𝐺2(𝑠)𝐻1(𝑠)
• 𝐺4(𝑠)𝐻2(𝑠)
• 𝐺7(𝑠)𝐻4(𝑠)
• 𝐺2(𝑠)𝐺3(𝑠)𝐺4(𝑠)𝐺5(𝑠)𝐺6(𝑠)𝐺7(𝑠)

Mason’s Rule
134
3𝑟𝑑
• Identify the non touching loops taken two at a time
• loop 1 does not touch loop 2
• Notice that loops 1, 2, and 3 all touch loop 4
Loop 1 and loop 2 : G2(s)H1(s)G4(s)H2(s)
Loop 1 and loop 3 : G2(5)H1(s)G7(5)H4(5)
Loop 2 and loop 3 : G4(s)H2(s)G7(s)H4(s)

Mason’s Rule
135
4𝑡ℎ
• Identify the non touching loops taken three at a time
• Loops 1, 2, and 3 : G2(s)H1(s)G4(s)H2(s)G7(s)H4(s)
5𝑡ℎ
• Identify the non touching loops taken four, five …. at a
time
• -None in this example-

Mason’s Rule
136
6𝑡ℎ
• Calculate △
• △= 1 − [𝐺2 𝑠 𝐻1 𝑠 + 𝐺4 𝑠 𝐻2 𝑠 + 𝐺7 𝑠 𝐻4 𝑠 +

Mason’s Rule
137
7𝑡ℎ
• Calculate △𝑘
• Only one forward path, △1
• △1= 1 − 𝐺7(𝑠)𝐻4(𝑠)
8𝑡ℎ
• Apply the Mason’s formula
• 𝐺 𝑠 =
𝑇1△1
△
=
𝐺1(𝑠)𝐺2(𝑠)𝐺3(𝑠)𝐺4(𝑠)𝐺5(𝑠) 1−𝐺7(𝑠)𝐻4(𝑠)
△

Alternative Representations in State Space
138
Cascade form
𝐶(𝑠)
𝑅(𝑠)
=
24
(𝑠 + 2)(𝑠 + 3)(𝑠 + 4)
First order
system

139
𝑥1 = −4𝑥1 + 𝑥2
𝑥2 = −3𝑥2 + 𝑥3
𝑥3 = −2𝑥3 + 24𝑟
𝑦 = 𝑐 𝑡 = 𝑥1
The state representation is completed by re-writting in vector-matrix form
𝑥 =
−4 2 0
0 −3 1
0 0 −2
𝑥 +
0
0
24
𝑟
𝑦 = 1 0 0 𝑥

140
Parallel form
𝐶(𝑠)
𝑅(𝑠)
=
24
(𝑠+2)(𝑠+3)(𝑠+4)
=
12
(𝑠+2)
−
24
(𝑠+3)
+
12
(𝑠+4)
𝑥1 = −4𝑥1 + 𝑥2
𝑥2 = −3𝑥2 + 𝑥3
𝑥3 = −2𝑥3 + 24𝑟
𝑦 = 𝑐 𝑡 = 𝑥1
Yields
𝑥 =
−2 0 0
0 −3 0
0 0 −4
𝑥 +
12
−24
12
𝑟
𝑦 = 1 1 1 𝑥

141
Example : If the denominator of the transfer function has repeated real
roots, the parallel form can still be derived from a partial-fraction
expansion. 𝐶(𝑠)
𝑅(𝑠)
=
𝑠 + 3
(𝑠 + 1)2 (𝑠 + 2)
=
2
𝑠 + 1 2 −
1
𝑠 + 1
+
1
𝑠 + 2
𝑥1 = −𝑥1 + 𝑥2
𝑥2 = −𝑥2 + 2𝑟
𝑥3 = −2𝑥3 + 𝑟
𝑦 = 𝑐 𝑡 = 𝑥1 −
1
2
𝑥2 + 𝑥3
𝑥 =
−1 1 0
0 −1 0
0 0 −2
𝑥 +
0
2
1
𝑟
𝑦 = 1 −
1
2
1 𝑥

142
Controller Canonical Form
𝐺 𝑠 =
𝐶(𝑠)
𝑅(𝑠)
=
𝑠2
+ 7𝑠 + 2
𝑠3 + 9𝑠2 + 26𝑠 + 24

LEARNING UNIT 6
STABILITY
KNL 3354 – Control System Engineering
144
KNL3354

Outlines
145
 Introduction
 Routh-Hurwitz Criterion
 Stability in State-Space

Introduction
146
Terminologies
and definition A linear, time-invariant system is stable if the natural
response approaches zero as time approaches infinity.
A linear, time-invariant system is unstable if the natural
response grows without bound as time approaches infinity.
A linear, time-invariant system is marginally stable if the
natural response neither decays nor grows but remains
constant or oscillates as time approaches infinity.
A system is stable if every bounded input yields a bounded
output (BIBO).

Introduction
147
BIBO
A system is stable if every
bounded input yields a bounded
output.
A system is unstable if any
bounded input yields an
unbounded output.

Introduction
148
Poles in the left
half-plane (lhp)
yield either pure
exponential
decay or damped
sinusoidal
natural
responses.
That is, stable
systems have
closed-loop
transfer
functions with
poles only in the
left half-plane.

Introduction
149
All the poles
of the closed
loop system
are located
at lhs of the
s-plane
Decaying
exponential
)
047
.
1
164
.
0
)(
047
.
1
164
.
0
)(
672
.
2
(
3
2
3
3
)
2
(
)
1
(
)
(
3
)
2
(
)
1
(
3
)
(
)
(
)
2
(
)
1
(
3
)
(
2
3
j
s
j
s
s
s
s
s
s
s
s
s
T
s
s
s
s
R
s
C
s
s
s
s
G






















Figure 1

Introduction
150
)
505
.
1
434
.
0
)(
505
.
1
434
.
0
)(
087
.
3
(
7
2
3
7
)
2
(
)
1
(
)
(
7
)
2
(
)
1
(
7
)
(
)
(
)
2
(
)
1
(
7
)
(
2
3
j
s
j
s
s
s
s
s
s
s
s
s
T
s
s
s
s
R
s
C
s
s
s
s
G






















Some of the
poles of the
closed loop
system are
located at rhs
of the s-
plane.
Increasing
exponential
Figure 2

Routh Hurwitz Criterion
151
Generate a data table called a
Routh table
Interpret the Routh table to tell
how many closed-loop system
poles are in the left half-plane,
the right half-plane, and on the
j𝜔-axis.
Routh-Hurwitz
Criterion
Routh-Hurwitz criterion declares that the number of roots of
the polynomial that are in the right half-plane is equal to the
number of sign changes in the first column.

152
n
s 2

n
s 4

n
s .......
Figure 4 : Equivalent closed-loop transfer
function
Figure 5 : Initial layout for Routh table

153
Figure 5 : Completed Routh table

154

155

156

157

158

159

160

161

162
The Routh-Hurwitz criterion will be
applied to this denominator. First label
the rows with powers of s from 𝑠3
down to 𝑠0 in a vertical column

163
Now form the second row with the
coefficients of the denominator
skipped in the previous step.
Subsequent rows are formed with
determinants,

164
For convenience, any row of the Routh
table can be multiplied by a positive
constant without changing the values
of the rows below. Care must be taken
not to multiply the row by a negative
constant.

165
 Stability via Epsilon Method (Zero only at the first
column)
Determine the stability of the closed-loop transfer
function Two sign
changes in the
first column of
the Routh-
Hurwitz table
indicates that
the
characteristic
equation have
two poles at
right half plane
and three poles
at left half plane.

166
 Routh-Hurwitz with Row of Zeros
Find the number of poles in the left half-plane, the
right half-plane, and on the jw-axis for the system
below

167
A row of zeros appears in the 𝑠5
row. Return to the 𝑠6
row and form the
even polynomial

168
From row 𝑠5
to row 𝑠0
, There are two sign changes
2 rhp poles
Even polynomial must have equal number of rhp as lhp. (2
lhp)
The remaining: 6 poles-2 rhp poles- 2 lhp=2 𝑗𝜔 poles
From row 𝑠8 to row 𝑠6, There are no sign changes
2 lhp
 4 lhp, 2 rhp and 2 𝑗𝜔 poles

169
If row of zeros
occurs, there are
2 possibilities
Possibility 1:
The roots
are
symmetrical
and
imaginary
Possibility 2 :
The roots are
quadrantal
and located
at rhp

Stability in state space
170
Stability
in State
Space
The values of the system's poles are equal to
the eigenvalues of the system matrix, A
The eigenvalues of the matrix A were solutions
of the equation det (sI - A) = 0, which also
yielded the poles of the transfer function.
The characteristic equation of a state space is
given by det(sI - A) = 0.

Stability in state space
171
 Example: find out how many poles are in the left
half-plane, in the right half-plane, and on the 𝑗𝜔-
axis.

LEARNING UNIT 7
STEADY STATE ERRORS
KNL3353

Introduction
173
Steady-state
error is the
difference
between the input
and the output for
a prescribed test
input as t tend to
infinity.
Test waveforms
for evaluating
steady-state
errors of position
control systems

Introduction
174
 Example
Output 1 has zero steady-state error, and output 2
has a finite steady-state error, 𝑒2(∞).

Steady-State Error of a unity
feedback system
175
R(s) is the input, C(s) is the
output, and E(s) = R(s)-
C(s) is the error
Apply final value
theorem to
calculate the
steady state error
The combination
of the first two
equation give

feedback system
176
 Step Input : R(s) = 1/s,
The term is the dc gain of the forward transfer
function
Discussion : In order to have zero steady-state error,
must tend to infinity. The transfer
function G(s) must contain at least 1 pure integrator.

feedback system
177
 Ramp Input : R(s) = 1/𝑠2
, we obtain
The transfer function G(s) must contain at least 2 pure
integrators.

feedback system
178
 Parabolic Input: R(s) = 1/𝑠3
, we obtain
The transfer function G(s) must contain at least 3 pure
integrators.

feedback system
179
 Example: Find the steady-state errors for inputs of
5𝑢(𝑡), 5𝑡𝑢(𝑡), and 5𝑡2𝑢(𝑡) of the system below. The
function u(t) is the unit step.
inputs of 𝟓𝒖(𝒕)
inputs of 𝟓𝒕𝒖(𝒕)
inputs of 𝟓𝒕𝟐
𝒖(𝒕)

feedback system
180
 Example: Find the steady-state errors for inputs of
5𝑢(𝑡), 5𝑡𝑢(𝑡), and 5𝑡2𝑢(𝑡) of the system below. The
function u(t) is the unit step.
inputs of 𝟓𝒖(𝒕)
inputs of 𝟓𝒕𝒖(𝒕)
inputs of 𝟓𝒕𝟐
𝒖(𝒕)

181
 focus on unity negative feedback systems and define
parameters that we can use as steady-state error
performance specifications, just as we defined
damping ratio, natural frequency, settling time,
percent overshoot, and so on as performance
specifications for the transient response.
 These steady-state error performance specifications
are called static error constants.

182
For a step input, 𝑢(𝑡),
For a ramp input, t𝑢(𝑡),
For a ramp input,
1
2
𝑡2
𝑢(𝑡),
position constant, Kp, velocity constant, Kv, acceleration constant, Ka
static error constants

183
For a step input, 𝑢(𝑡),
For a ramp input, t𝑢(𝑡),
For a ramp input,
1
2
𝑡2
𝑢(𝑡),
position constant, Kp, velocity constant, Kv, acceleration constant, Ka
static error constants

184
 Example : evaluate the static error constants and
find the expected error for the standard step, ramp,
and parabolic inputs.

185

186
 Example 2 : evaluate the static error constants and
find the expected error for the standard step, ramp,
and parabolic inputs.

187

188
 System Type
We define system type to be the value of n in
the denominator or, equivalently, the number of
pure integrations in the forward path.

189
 Steady-State Error for Disturbances
Final value theorem

190
 Steady-State Error for Disturbances
steady-state error due to R(s), steady-state error due to D(s),

191
Example :Find the steady-state error component due
to a step disturbance for the system

Steady-State Error for Nonunity Feedback
Systems
192
Steady-State Error for Nonunity Feedback Systems
𝐺(𝑠) = 𝐺1(𝑠)𝐺2(𝑠)
H(𝑠) = 𝐻1(𝑠)/𝐺1(𝑠)

Systems
193
Systems
Unity
Feedback
Systems
𝒆 ∞ = lim
𝒔→𝟎
𝒔𝑹(𝒔)
𝟏 + 𝑮𝒆(𝒔)

Systems
194
 Example : find the system type, the appropriate error
constant associated with the system type, and the
steady-state error for a unit step input.

Systems
195
Example (Cont.)
The appropriate static error
constant is
The steady-state error is

Systems
196
Nonunity feedback control system with disturbance
Superposition theorem + 𝑒 ∞ = 𝑟 ∞ − 𝑐(∞) give

-The End-
197

LEARNING UNIT 8
ROOT LOCUS TECHNIQUES
198
KNL3354

Outlines
199
 Define a root locus
 State the properties of a root locus
 Sketch a root locus
 Find the coordinates of points on the root locus and
their associated gains

Learning Unit 8 – Root locus techniques
200
 Root locus, a graphical presentation of the closed-
loop poles as a system parameter is varied, is a
powerful method of analysis and design for stability
and transient response (Evans, 1948; 1950)
 The root locus is a graphical technique that gives us
the qualitative description of a control system's
performance that we are looking for and also serves
as a powerful quantitative tool that yields more
information than the methods already discussed.

201
 The Control System Problem
poles of the open-loop transfer function are
easily found (typically, they are known by
inspection and do not change with changes
in system gain), the poles of the closed-loop
transfer function are more difficult to find
(typically, they cannot be found without
factoring the closed-loop system's
characteristic polynomial, the denominator
of the closed-loop
transfer function), and further, the closed-
loop poles change with changes in system
gain
Equivalent closed loop transfer
function

202
 Equivalent closed loop transfer
function

203
 Assume a function
 The function defines the complex arithmetic to be
performed in order to evaluate F(s) at any point, s.
 Magnitude, M, of F(s) at any point, s, is

204
 (𝑠 − 𝑧𝑖) , is the magnitude of the vector drawn from
the zero of F(s) at - 𝑧𝑖 to the point s, and a pole
length, (𝑠 − 𝑝𝑖) is the magnitude of the vector drawn
from the pole of F(s) at −𝑝𝑖 to the point s.

205
 Example : Find F(s) at the point s =3+j4.

206

207

208

209
 Sketching the Root Locus
Number of branches.
• Each closed-loop pole moves as the gain is varied. If we define a
branch as the path that one pole traverses, then there will be one
branch for each closed-loop pole. Our first rule, then, defines the
number of branches of the root locus:
• The number of branches of the root locus equals the number of
closed-loop poles
Symmetry.
• If complex closed-loop poles do not exist in conjugate pairs, the
resulting polynomial, formed by multiplying the factors containing
the closed-loop poles, would have complex coefficients. Physically
realizable systems cannot have complex coefficients in their transfer
functions. Thus, we conclude:
• The root locus is symmetrical about the real axis.

210
Real-axis segments
• On the real axis, for K > 0 the root locus exists to the left of an
odd number of real-axis, finite open-loop poles and/or finite
open-loop zeros.
Starting and ending points.
• The root locus begins at the finite and infinite poles of G(s)H(s)
and ends at the finite and infinite zeros of G(s)H(s).
Behavior at infinity.
• The root locus approaches straight lines as asymptotes as the
locus approaches infinity. Further, the equation of the asymptotes
is given by the real-axis intercept, 𝜎𝑎 and angle, 𝜃𝑎as follows

211
 Example: Sketching a Root Locus with
Asymptotes
Asymptotes : 𝜎𝑎 =
( − 1 − 2 − 4 ) − ( − 3 )
4 − 1
= −
4
3
The angles of the lines that intersect at -4/3 :
𝜃𝑎 =
(2k+1)π
#finite poles − #finite zeros
𝜃𝑎 =
π
3
if k=0
𝜃𝑎 = π if k=1
𝜃𝑎 =
5π
3
if k=2

212
 Refining the Sketch
Real-Axis
Breakaway
and Break-In
Points
Numerous root loci
appear to break away
from the real axis as the
system poles move from
the real axis to the
complex plane. At other
times the loci appear to
return to the real axis as a
pair of complex poles
becomes real.
The point where the
locus leaves the real axis,
−𝜎1 , is called the
breakaway point, and
the point where the locus
returns to the real axis,
𝜎2, is called the break-in
point.

213
 Real-Axis Breakaway and Break-In Points
Breakaway and Break-in Points via
Differentiation
-Maximize and minimize the gain, K,
using differential calculus
Variation on the differential calculus
method
-Also called the transition method, it
eliminates the step of differentiation

214
 Breakaway and Break-in Points via Differentiation
For all points on the root locus:
𝐾 = −
1
𝐺 𝑠 𝐻(𝑠)
 For points along the real-axis segment of the root locus where
breakaway and breaking points could exist, 𝑠 = 𝜎
𝐾 𝜎 = −
1
𝐺 𝜎 𝐻 𝜎
 Points of maximum and minimum gain and hence the breakaway
and break-in points:
𝑑𝐾(𝜎)
𝑑𝜎
= 0

215
 Example : Find the breakaway and break-in points
for the root locus of Figure , using differential
calculus.
𝐾𝐺 𝑠 𝐻 𝑠 =
𝐾(𝑠 − 3)(𝑠 − 5)
(𝑠 + 1)(𝑠 + 2)
=
𝐾(𝑠2
− 8𝑠 + 15)
𝑠2 + 3𝑠 + 2
𝐾𝐺 𝜎 𝐻 𝜎 =
𝐾(𝜎2
− 8𝜎 + 15)
𝜎2 + 3𝜎 + 2
𝐾(𝜎2
− 8𝜎 + 15)
𝜎2 + 3𝜎 + 2
= −1
𝐾 = −
𝜎2 + 3𝜎 + 2
𝜎2 − 8𝜎 + 15
𝑑𝐾(𝜎)
𝑑𝜎
=
11𝜎2 − 26 − 61
𝜎2 − 8𝜎 + 15 2
= 0
𝜎 = -1.45 and 3.82

216
 Method no 2 : Variation on the differential calculus
method
Breakaway and break-in points satisfy the relationship
𝑖=1
𝑚
1
𝜎 + 𝑧𝑖
=
𝑖=1
𝑛
1
𝜎 + 𝑝𝑖
Where zi and pi are the negative of the zero and pole
values, respectively, of G(s)H(s)

217
Example :
1
𝜎 − 3
+
1
𝜎 − 5
=
1
𝜎 + 1
+
1
𝜎 + 2
Simplifying
11𝜎2 − 26 − 61 = 0
𝜎 = -1.45 and 3.82

218
The 𝒋𝝎-Axis Crossings
The value of 𝜔 at the axis crossing
yields the frequency of oscillation
To find the; 𝑗𝜔-axis crossing, we can
use the Routh-Hurwitz criterion.
The 𝑗𝜔-axis crossing is a point on the
root locus that separates the stable
operation of the system from the
unstable operation.

219
 Example : find the frequency and gain, K, for which
the root locus crosses the imaginary axis. For what
range of K is the system stable?

220
 Solution :
The closed-loop transfer function for the system
𝑇 𝑠 =
𝐾(𝑠+3)
𝑠4+7𝑠3+14𝑠2+ 8+𝐾 𝑠+3𝐾
.

221
 Solution (Cont)
A complete row of zeros yields the possibility for
imaginary axis roots. For positive values of gain, those
for which the root locus is plotted, only the 𝑠1
row can
yield a row of zeros. Thus,
−𝐾2−65K+720=0
From this equation K is evaluated as K=9.65
Forming the even polynomial by using the 𝑠2
row with
K=9.65, we obtain
(90 − 𝐾)𝑠2
+ 21𝐾 = 80.35𝑠2
+ 202.7 = 0
and s is found to be equal to ±j1.59. Thus the root
locus crosses the 𝑗𝜔-axis at ±j1.59 at a gain of 9.65.
We conclude that the system is stable for 0 < K <
9.65.

222
Angles of Departure and Arrival
 Recall :
- The root locus starts at the open-loop poles and ends
at the open-loop zeros.
- the sum of angles drawn from all finite poles and
zeros to this point is an odd multiple of 180°.

223
 Angle of departure

224
 Angle of arrival

225
 Example : Given the unity feedback system, find the
angle of departure from the complex poles and
sketch the root locus.
Calculate the sum of angles drawn to a point 𝜀 close to
the
complex pole, -1 +j1, in the second quadrant. Thus
−𝜃1 − 𝜃2 + 𝜃3 − 𝜃4 = −𝜃1 −90° −𝑡𝑎𝑛−1 1
1
− 𝑡𝑎𝑛−1 1
2
−𝜃1 − 𝜃2 + 𝜃3 − 𝜃4 = 180°
For which 𝜃1 = −251.6° = 108.4°

226
Example (Cont)

227
END OF LEARNING UNIT 8

LEARNING UNIT 9
FREQUENCY RESPONSE
TECHNIQUES
KNL3353

Learning Unit 9 – Frequency Response
Techniques
229
 Define and plot the frequency response of a system
 Plot asymptotic approximations to the frequency
response of a system
 Find stability and gain and phase margins
 Find the closed-loop time response parameters of peak
time, settling time, and percent overshoot given the
open-loop frequency response

230
 The Concept of Frequency Response
Techniques
In the steady state, sinusoidal inputs to a
linear system generate sinusoidal responses of
the same frequency.
These responses are of the same frequency as
the input, they differ in amplitude and phase
angle from the input.
Sinusoids can be represented as complex
numbers called phasors. 𝑀1 cos 𝜔𝑡 + 𝜙1 = 𝑀1∠𝜙1

Techniques
231
a. system;
b. transfer function;
c. input and output
waveforms

Techniques
232
 Steady-state output sinusoid is
 The system function is given by
 We call M(𝜔) the magnitude frequency response
and ϕ(𝜔) the phase frequency response.
 The combination of the magnitude and phase
frequency responses is called the frequency
response and is M(𝜔)∠ϕ(𝜔)

Techniques
233
 Analytical Expressions for Frequency
Response
𝑟 𝑡 = 𝐴 cos 𝜔𝑡 + 𝐵 sin 𝜔𝑡
𝑟 𝑡 = 𝑀𝑖 cos 𝜔𝑡 + 𝜙𝑖
𝑟 𝑡 = 𝑀𝑖 𝑒𝑗𝜙𝑖
𝜙𝑖 = 𝑡𝑎𝑛−1(𝐵/𝐴), 𝑀𝑖 = 𝐴2 + 𝐵2

Techniques
234
 Recall on complex number
Let 𝑧 = 𝑎1 + 𝑗𝑏1 𝑎2 + 𝑗𝑏2 … 𝑎𝑛 + 𝑗𝑏𝑛
Angle(z)=𝑡𝑎𝑛−1
(
𝑏1
𝑎1
)+𝑡𝑎𝑛−1
(
𝑏2
𝑎2
)+…+𝑡𝑎𝑛−1
(
𝑏𝑛
𝑎𝑛
)
Mag (z)= 𝑎1
2 + 𝑏1
2
𝑎2
2 + 𝑏2
2
… 𝑎𝑛
2 + 𝑏𝑛
2
Mag (z) dB= 20 log( 𝑎1
2 + 𝑏1
2
) +20 log ( 𝑎2
2 + 𝑏2
2
)
+20 log 𝑎2
2 + 𝑏2
2
+…+20 log ( 𝑎𝑛
2 + 𝑏𝑛
2
)

Techniques
235
Asymptotic Approximations: Bode Plots
 The log-magnitude and phase frequency response curves
as functions of log 𝜔 are called Bode plots or Bode
diagrams.
 Sketching Bode plots can be simplified because they can
be approximated as a sequence of straight lines.
 Straight-line approximations simplify the evaluation of
the magnitude and phase frequency response.

Techniques
236

Techniques
237
 Consider the following transfer function
𝐺 𝑠 =
𝐾 𝑠 + 𝑧1 𝑠 + 𝑧2 … (𝑠 + 𝑧𝑘)
𝑠𝑚 𝑠 + 𝑝1 𝑠 + 𝑝2 … (𝑠 + 𝑝𝑛)
 The magnitude frequency response is the product of
the magnitude frequency responses of each term, or
𝐺 𝑗𝜔 =
𝐾 𝑠 + 𝑧1 𝑠 + 𝑧2 … 𝑠 + 𝑧𝑘
𝑠𝑚 𝑠 + 𝑝1 𝑠 + 𝑝2 … 𝑠 + 𝑝𝑘 𝑠=𝑗𝜔
𝐺 𝑗𝜔 =
𝐾 𝑗𝜔 + 𝑧1 𝑗𝜔 + 𝑧2 … 𝑗𝜔 + 𝑧𝑘
(𝑗𝜔)𝑚 𝑗𝜔 + 𝑝1 𝑗𝜔 + 𝑝2 … 𝑗𝜔 + 𝑝𝑘

Techniques
238
Converting the magnitude response into dB, we
obtain
20 log 𝐺 𝑗𝜔 =
20 log 𝐾 + 20 log 𝑗𝜔 + 𝑧1 + 20 log 𝑗𝜔 + 𝑧2 + ⋯
+ 20 log 𝑗𝜔 + 𝑧𝑘 − 20 log 𝑗𝜔 𝑚 − 20 log 𝑗𝜔 + 𝑝1
− 20 log 𝑗𝜔 + 𝑝2 − ⋯ − 20 log 𝑗𝜔 + 𝑝𝑘
Phase frequency response is the sum of the phase frequency
response curves of the zero terms minus the sum of the phase
frequency response curves of the pole terms.

Techniques
239
𝐺(𝑠) = 𝑠 + 𝑎
𝐺(𝑗𝜔) = 𝑗𝜔 + 𝑎
𝐺 𝑗𝜔 = 𝑗𝜔 + 𝑎 = 𝑎(𝑗
𝜔
𝑎
+ 1)
Low frequency 𝜔 ⟶ 0
𝐺 𝑗𝜔 ≈ 𝑎
-The magnitude response in
dB is 20 log 𝑀 = 20 log 𝑎
- The angle approaches zero
High frequency 𝜔 ≫ 𝑎
𝑗𝜔
𝑎
= 𝜔∠90°
The magnitude response in
dB is 20 log 𝑀 = 20 log 𝑎 +
20 log
𝜔
𝑎
=20 log 𝜔

Techniques
240
If we plot dB, 20 𝑙𝑜𝑔(𝑀) , against 𝑙𝑜𝑔(𝜔) ,
20 𝑙𝑜𝑔(𝜔) becomes a straight line: y=20x
where y = 20 𝑙𝑜𝑔 (𝑀), and 𝑥 = 𝑙𝑜𝑔 (𝜔). The line has a
slope of 20 when plotted as dB vs. 𝑙𝑜𝑔(𝜔)
We call the straight-line approximations asymptotes.
The low-frequency approximation is called the low-
frequency asymptote, and the high-frequency
approximation is called the high-frequency
asymptote. The frequency, «, is called the break
frequency because it is the break between the low-
and the high-frequency asymptotes.

Techniques
241
-The magnitude response in
dB is 20 log 𝑀 = 20 log 𝑎
- The angle approaches zero
The magnitude response in
dB is 20 log 𝑀 = 20 log 𝑎 +
20 log
𝜔
𝑎
=20 log 𝜔
Low frequency 𝜔 ⟶ 0 High frequency 𝜔 ≫ 𝑎

Techniques
242
Low frequency 𝜔 ⟶ 0
The angle approaches zero
High frequency 𝜔 ≫ 𝑎
𝑗𝜔
𝑎
= 𝜔∠90°

Techniques
243
Bode Plots for 𝐺 𝑠 =
1
𝑠+𝑎
, 𝐺 𝑗𝜔 =
1
𝑗𝜔+𝑎
=
1
𝑎(
𝑗𝜔
𝑎
+1)
Magnitude M (dB) :
20log 𝐺(𝑗𝜔) = 20log 1 − 20log 𝑎 − 20log (
𝜔
𝑎
)2+1
20log 𝐺(𝑗𝜔) = −20log 𝑎 − 20log (
𝜔
𝑎
)2+1
Angle :
𝜑 = 𝑡𝑎𝑛−1
0
1
− 𝑡𝑎𝑛−1
0
𝑎
𝜔
𝑎
𝜑 = −𝑡𝑎𝑛−1
𝜔
𝑎

Techniques
244
𝐺 𝑠 =
1
𝑠 + 𝑎
Magnitude M (dB)
: 20log 𝐺(𝑗𝜔) =
− 20log 𝑎 −
20log (
𝜔
𝑎
)2+1
Angle :
𝜑 = −𝑡𝑎𝑛−1
𝜔
𝑎
Low frequency
𝜔 → 0
Magnitude:−20log 𝑎
Angle:𝜑 = −𝑡𝑎𝑛−1 0
𝑎
=0
High frequency
𝜔 ≫ 𝑎,
𝜔
𝑎
≫1, (
𝜔
𝑎
)2
+1 ≈ (
𝜔
𝑎
)2
Magnitude =−20log(𝑎) −
20log
𝜔
𝑎
=−20log 𝜔
Angle=−𝑡𝑎𝑛−1 ∞ = −
𝜋
2

Techniques
245

246
Bode Plots for 𝐺 𝑠 = 𝑠 , 𝐺 𝑗𝜔 = 𝑗𝜔
Magnitude M (dB) :
20log 𝐺(𝑗𝜔) = 20log 𝜔
Angle :
𝜑 = 𝑡𝑎𝑛−1 𝜔
0
=𝑡𝑎𝑛−1
∞ =
𝜋
2
Techniques

Techniques
247
Bode Plots for 𝐺 𝑠 = 𝑠 , 𝐺 𝑗𝜔 = 𝑗𝜔
Magnitude M (dB) : 20log 𝐺(𝑗𝜔) = 20log 𝜔
Angle : 𝜑 = 𝑡𝑎𝑛−1 𝜔
0
=𝑡𝑎𝑛−1
∞ =
𝜋
2

Techniques
248
Bode Plots for 𝐺 𝑠 = 1/𝑠 , 𝐺 𝑗𝜔 = 1/𝑗𝜔
Magnitude M (dB) :
20log 𝐺(𝑗𝜔) = 20log 1/𝜔 = −20log 𝜔
Angle :
0
1
𝜔
0
= −𝑡𝑎𝑛−1 ∞ = −
𝜋
2

Techniques
249
Bode Plots for 𝐺 𝑠 = 1/𝑠 , 𝐺 𝑗𝜔 = 1/𝑗𝜔
Magnitude M (dB) : 20log 𝐺(𝑗𝜔) = −20log 𝜔 , Angle : 𝜑 = −
𝜋
2

Techniques
250
 Bode Plots for 𝑮 𝒔 = 𝒔𝟐 + 𝟐𝜻𝝎𝒏𝒔 + 𝝎𝒏
𝟐
𝐺 𝑗𝜔 = (𝑗𝜔)2
+2𝑗𝜁𝜔𝑛𝜔 + 𝜔𝑛
2
𝐺 𝑗𝜔 = 𝜔𝑛
2
−
𝜔
𝜔𝑛
2
+ 2𝑗𝜁
𝜔
𝜔𝑛
+ 1
Magnitude M (dB) : 20 log 𝐺(𝑠) = 20 log 𝜔𝑛
2
+
20 log 1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
Angle : 𝜑 = 𝑡𝑎𝑛−1
(
2𝜁
𝜔
𝜔𝑛
1−
𝜔
𝜔𝑛
2)

Techniques
251
Low frequency : 𝝎 ≪ 𝝎𝒏, 𝝎 → 𝟎.
Magnitude (dB)=20 log 𝜔𝑛
2=40 log 𝜔𝑛
0 = 0°
High Frequency : : 𝝎 ≫ 𝝎𝒏,
𝝎
𝝎𝒏
≫ 𝟏, 𝝎 → ∞,
M (dB)=20 log 𝜔𝑛
2
+ 20 log 1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
≈
Magnit≈ 20 log 𝜔𝑛
2
+20 log
𝜔
𝜔𝑛
2
=40 log 𝝎
𝑮 𝒋𝝎 ≈ −𝝎𝟐
=𝝎𝟐
∠𝟏𝟖𝟎°
M (dB) =20 log 𝜔𝑛
2
+ 20 log 1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
, 𝜑 = 𝑡𝑎𝑛−1
(
2𝜁
𝜔
𝜔𝑛
1−
𝜔
𝜔𝑛
2)

Techniques
252

Techniques
253
Normalized and scaled log-magnitude response for 𝑮 𝒔 = 𝒔𝟐
+
𝟐𝜻𝝎𝒏𝒔 + 𝝎𝒏
𝟐

Techniques
254
Scaled phase response for 𝑮 𝒔 = 𝒔𝟐
+ 𝟐𝜻𝝎𝒏𝒔 + 𝝎𝒏
𝟐

Techniques
255
 𝝎𝒏 is the break frequency for the second order polynomial.

Techniques
256
 Bode Plots for 𝑮 𝒔 =
𝟏
𝒔𝟐+𝟐𝜻𝝎𝒏𝒔+𝝎𝒏
𝟐
𝐺 𝑗𝜔 =
1
(𝑗𝜔)2+2𝑗𝜁𝜔𝑛𝜔 + 𝜔𝑛
2
𝐺 𝑗𝜔 =
1
𝜔𝑛
2 −
𝜔
𝜔𝑛
2
+ 2𝑗𝜁
𝜔
𝜔𝑛
+ 1
Magnitude M (dB) : 20 log 𝐺(𝑠) = −20 log 𝜔𝑛
2
−
20 log 1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
Angle : 𝜑 = −𝑡𝑎𝑛−1
(
2𝜁
𝜔
𝜔𝑛
1−
𝜔
𝜔𝑛
2)

Techniques
257
Low frequency : 𝝎 ≪ 𝝎𝒏, 𝝎 → 𝟎.
Magnitude (dB)=−20 log 𝜔𝑛
2 = −40 log 𝜔𝑛
0 = 0°
High Frequency : : 𝝎 ≫ 𝝎𝒏,
𝝎
𝝎𝒏
≫ 𝟏, 𝝎 → ∞,
M (dB)=−20 log 𝜔𝑛
2
− 20 log 1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
≈
Magnit≈ −20 log 𝜔𝑛
2 − 20𝑙𝑜𝑔
𝜔
𝜔𝑛
2
= −40 log 𝝎
𝑮 𝒋𝝎 ≈ −
𝟏
𝝎𝟐=
𝟏
𝝎𝟐 ∠ − 𝟏𝟖𝟎°
M (dB) =−20 log 𝜔𝑛
2
− 20 log 1 −
𝜔
𝜔𝑛
2 2
+ 2𝜁
𝜔
𝜔𝑛
2
, 𝜑 = −𝑡𝑎𝑛−1
(
2𝜁
𝜔
𝜔𝑛
1−
𝜔
𝜔𝑛
2)

Techniques
258
Normalized and scaled log-magnitude response for 𝑮 𝒔 = 𝟏/(𝒔𝟐
+
𝟐𝜻𝝎 𝒔 + 𝝎 𝟐
)

Techniques
259
Scaled phase response for 𝑮 𝒔 = 𝟏/(𝒔𝟐
+ 𝟐𝜻𝝎𝒏𝒔 + 𝝎𝒏
𝟐
)

Techniques
260
 Example: Bode Plots for Ratio of First-Order
Factors
𝐺 𝑠 =
(𝑠 + 3)
𝑠(𝑠 + 1)(𝑠 + 2)

Techniques
261

Techniques
262
𝐺 𝑠 =
(𝑠 + 3)
𝑠(𝑠 + 1)(𝑠 + 2)

Techniques
263

Techniques
264
 Example: Bode Plots for Ratio of First- and Second-Order
Factors
𝐺 𝑠 =
𝑠 + 3
(𝑠 + 2)(𝑠2 + 2𝑠 + 25)
Magnitude diagram slopes

Techniques
265

Techniques
266

Techniques
267

Techniques
268
THE END OF LEARNING UNIT

References
269
[1] S. Yaduvir, S. Janardhanan, Modern Control
Engineering, Cengage Learning Asia Pte Ltd, 2011
[2] N. S. Nise, Control System Engineering, Sixth
Edition, Wiley, 2009
[3]Katsuhiko Ogata, Modern Control Engineering,
International Edition, Pearson, 2010.

KNL3353_Control_System_Engineering_Lectu.ppt

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