36336583 control-systems-lecture-notes

Control
Systems
Lecture Notes for
KJM597
Faculty of Mechanical
Engineering, UiTM Shah Alam

2
KJM597 Control Systems
Faculty of Mechanical Engineering
UiTM Shah Alam
CHAPTER 1.0 MATHEMATICAL REVIEW
1.1 COMPLEX VARIABLES AND COMPLEX FUNCTION
This chapter will outline an overview of basic mathematical formulation in solving control
systems problem that you will encounter throughout the course.
1.1.1 Complex Variables Concept
A complex variable denoted by s consists of two components: a real component x and an
imaginary axis component y. Graphically, the real component of s is represented by a x-axis in
the horizontal direction, and the imaginary component is measured along the vertical jy-axis.
Figure 1.0 illustrates the complex s-plane.
Figure 1.0: Complex s-plane
(source: http://mathworld.wolfram.com)
Using notation 𝑗 = −1, all numbers in engineering calculations can be re-written as
𝑧 = 𝑥 + 𝑗𝑦
Where z is called a complex number. Note that j is the only imaginary quantity in the expression.
The magnitude, |z| and angle, 𝜃 of z can be obtained mathematically,
Magnitude of z=|z|= 𝑥2 + 𝑦2, angle of z=𝜃 = tan−1 𝑦
𝑥

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A complex number can be written in rectangular form or in polar form as follows:
i. Rectangular forms
𝑧 = 𝑥 + 𝑗𝑦
𝑧 = 𝑧 (cos 𝜃 + 𝑗 sin 𝜃)
ii. Polar forms
𝑧 = |𝑧|∠𝜃
𝑧 = |𝑧|𝑒𝑗𝜃
In converting complex numbers to polar form from rectangular, we use
𝑧 = 𝑥2 + 𝑦2, 𝜃 = tan−1 𝑦
𝑥
To convert complex number to rectangular form from polar, we employ
𝑥 = |𝑧| cos 𝜃, 𝑦 = |𝑧| sin𝜃
(source: Ogata)
Notes:

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Notes:

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1.1.2 Complex Function Concept
a) Complex Function
A complex function F(s,) a function of s, has a real component and imaginary component, or
𝐹 𝑠 = 𝐹𝑥 + 𝐹𝑦
where 𝐹𝑥and 𝐹𝑦 are real quantities. The magnitude of 𝐹(𝑠)is 𝐹𝑥
2 + 𝐹𝑦
2, and the angle 𝜃 of 𝐹(𝑠)
is tan−1 𝐹𝑦
𝐹𝑥
. The angle is measured counterclockwise from the positive real axis.
b) Single-valued Function
In complex function analysis, we are interested in Single-Valued Function that can uniquely
determine the value of s. For instance, given the function
𝐹 𝑠 =
1
𝑠(𝑠 + 1)
𝐹 𝑠 = ∞ is mapped onto two points, s=0 and s=-1, in the s-plane
c) Poles and zeros of a Function
Poles are the value of s that will make the function F(s) become infinity. In other words, poles
are the roots of the denominator of F(s). If the denominator of F(s) involves k-multiple factors
(𝑠 + 𝑝)𝑘
, then 𝑠 = −𝑝 is called a multiple poles and of order 𝑘or repeated pole of order 𝑘. If
𝑘 = 1, the pole is called a simple pole.
Zeros are the value of s that will make the function F(s) become zero. In other words, zeros are
the numerator of F(s).
As an illustrative example, consider the following complex function
𝐺 𝑠 =
𝑠 + 2 (𝑠 + 10)
𝑠 𝑠 + 1 𝑠 + 5 (𝑠 + 15)2
G(s) has zeros at 𝑠 = −2 and 𝑠 = −10, simple poles at 𝑠 = 0, 𝑠 = −1 and 𝑠 = −5, and a
double pole (multiple pole of order 2) at 𝑠 = −15. Note that G(s) becomes zero at 𝑠 = ∞.
G(s) is therefore has 2 zeros and 5 poles.
d) Singularities of a Function
The singularities of a function are the points in the s-plane at which the function or its
derivatives do not exist. A pole is the most common of singularities and plays a very important
role in studies of classical control theory. (source: ogata)

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1.2 REVIEW OF DIFFERENTIAL EQUATIONS, LINEAR SYSTEMS, IMPULSE RESPONSE AND LAPLACE
TRANSFORMATIONS. DEFINITION OF STABILITY. INTRODUCTION TO STATE EQUATIONS AND
TRANSFER FUNCTIONS.
1.2.1 Review of Differential Equations
Differential equations generally involve derivatives and integrals of the dependant variables
with respect to the independent variable. For instance, a shock absorber system of a car as in
figure 1.2 can be represented by the differential equation,
 

 )
(
)
(
1
)
(
)
( t
v
dt
t
i
C
dt
t
di
L
t
Ri
Figure 1.2: RLC Circuit
where R is the resistance, L the inductance, C the capacitance, i(t) the current and v(t) the
applied voltage. The dependent variable i(t) is determined by solving the equation.
In general, a differential equation of nth-order is written as
)
(
)
(
)
(
)
(
)
(
0
1
1
1
1 t
f
t
y
a
dt
t
dy
a
dt
t
y
d
a
dt
t
y
d
n
n
n
n
n




 

 
Which is also known as a linear ordinary differential equation if the coefficients a0
, a1
, … ,an-1
are
not a function of y(t).

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1.2.2 Laplace Transforms
a) Laplace Transform
Laplace transform is used to convert from time domain to s-domain. Working with differential
equation is rather complicated. In analyzing and designing a control system it is easier to work in
s-domain. Laplace transform is defined as;
ℒ 𝑓 𝑡 = 𝐹 𝑠 = 𝑓 𝑡 𝑒−𝑠𝑡
𝑑𝑡
∞
0
Where 𝑠 = 𝑥 + 𝑗𝑦, a complex variable.
Example 2.1: Let f(t) be a unit-step function that is defined as






0
,
0
0
,
1
)
(
t
t
t
u
The Laplace transform of f(t) is obtained as
s
e
s
dt
e
t
u
s
F st
st 1
1
)
(
)
(
0 0



 
 


Example 2.2: Consider the exponential function
0
,
)
( 
 
t
e
t
f t

where α is real constant. The Laplace transform of f(t) is written as

 









0 0
)
(
1
)
(




s
s
e
dt
e
e
s
F
t
s
st
t

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Table 2.1: Laplace Transform table for input responses
Notes:

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b) Laplace Transform Theorems
The laplace transform has a set of theorems to solve a complex mathematical equations.
Table 2.2 summarizes the Laplace Transform theorems
Table 2.2: Laplace Transform Theorems

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c) Inverse Laplace Transformation Using Partial Fraction Method
Given the Laplace transform F(s), the operation of obtaining f(t) is termed the inverse
Laplace Transformation and is denoted by:
𝑓 𝑡 = ℒ−1
[𝐹 𝑠 ]
Inverse Laplace Transform is used when we want to convert from s-domain to time domain.
The inverse Laplace transform of rational functions are normally carried out using partial-
fraction expansion and the Laplace transform table.
Consider a rational function
𝐺 𝑠 =
𝑄(𝑠)
𝑃(𝑠)
where Q(s) and P(s) are polynomials of s. It is assume that the order of P(s) in s is greater
than of Q(s). The polynomial P(s) may be written as
0
1
1
1
)
( a
s
a
s
a
s
s
P n
n
n




 
 
where a0
, a1
, … ,an-1
are real coefficients. This method will be emphasized for the cases of
simple poles, multiple-order poles and complex poles.
Case 1: Simple poles
If all the poles of G(s) are simple and real, then G(s) can be written as
𝐺 𝑠 =
𝑄(𝑠)
𝑃(𝑠)
=
𝑄(𝑠)
𝑠 + 𝑠1 𝑠 + 𝑠2 … (𝑠 + 𝑠𝑛 )
, 𝑤𝑕𝑒𝑟𝑒 𝑠1 ≠ 𝑠2 ≠ ⋯ 𝑠𝑛
Applying partial-fraction expansion, the equation can be written as
𝐺 𝑠 =
𝐾−𝑠1
𝑠 + 𝑠1
+
𝐾−𝑠2
𝑠 + 𝑠2
+ ⋯ +
𝐾−𝑠𝑛
𝑠 + 𝑠𝑛
Where
i
s
s
s
P
s
Q
s
s
K i
si











)
(
)
(
)
(
The numerator of each fraction is called the residue. 𝐾−𝑠𝑖 is called the residue of G(s)
for the pole 𝑠 = −𝑠𝑖.
The inverse transform is the written as
t
s
s
t
s
s
t
s
s
n
n
e
K
e
K
e
K
t
g 




 


 
2
2
1
1
)
(

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Example 2.3:
Consider the function
)
3
)(
2
)(
1
(
3
5
)
(





s
s
s
s
s
G
which is written in the partial-fraction expanded form:
3
2
1
)
( 3
2
1





 


s
K
s
K
s
K
s
G
The coefficients 𝐾−1 , 𝐾−2 , 𝐾−3 are determined as follows:
 
 
  6
)
2
3
)(
1
3
(
3
)
3
(
5
3
)
(
)
3
(
7
)
3
2
)(
1
2
(
3
)
2
(
5
2
)
(
)
2
(
1
)
3
1
)(
2
1
(
3
)
1
(
5
1
)
(
)
1
(
3
2
1









































s
s
G
s
K
s
s
G
s
K
s
s
G
s
K
Thus,
3
6
2
7
1
1
)
(








s
s
s
s
G
The inverse transform or time function is
t
t
t
e
e
t
g 3
2
6
7
)
( 






Case 2: Multiple-order poles
If r of the n poles is identical, G(s) is written as
𝐺 𝑠 =
𝑄(𝑠)
𝑃(𝑠)
=
𝑄(𝑠)
𝑠 + 𝑠1 𝑠 + 𝑠2 … 𝑠 + 𝑠𝑛−𝑟 (𝑠 + 𝑠𝑖)𝑟

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Then G(s) can be expanded as




 




 






 




 


poles
repeated
of
r terms
)
(
)
(
poles
simple
of
r terms
-
n
)
( 2
2
1
2
1
)
(
2
1
r
i
r
i
i
r
n
s
s
s
s
s
A
s
s
A
s
s
A
s
s
K
s
s
K
s
s
K
s
G r
n

















 
The (n-r) coefficients K-s1, K-s2, … , K-s(n−r) which correspond to simple poles may be
evaluated as explained before. The coefficients A1 … Ar
are evaluated as follows:
 
 
 
 
i
s
s
s
G
s
s
ds
d
r
A
i
s
s
s
G
s
s
ds
d
A
i
s
s
s
G
s
s
ds
d
A
i
s
s
s
G
s
s
A
r
i
r
r
r
i
r
r
i
r
r
i
r





















)
(
)
(
)!
1
(
1
)
(
)
(
!
2
1
)
(
)
(
)
(
)
(
1
1
1
2
2
2
1

Example 2.4:
Consider the function
2
)
2
)(
1
(
2
)
(



s
s
s
G
G(s) can be written as
2
2
1
1
)
2
(
)
2
(
)
1
(
)
(





 
s
A
s
A
s
K
s
G
The coefficient corresponding to the simple pole is
2
1
)
2
(
2
2
1 











s
s
K

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and those of second order-pole are
2
2
)
1
(
2
2
)
1
(
2
2
2
)
1
(
2
2
1
2



































s
s
s
s
ds
d
A
s
s
A
The completed partial-fraction expansion is
2
)
2
(
2
2
2
1
2
)
(






s
s
s
s
G
The time function is
t
t
t
te
e
e
t
g 2
2
2
2
2
)
( 





Case 3: Simple complex-conjugate poles
Suppose that G(s) contains a pair of complex poles:



 j
-
-
s
and 


 j
s
The corresponding coefficients of these poles are






j
s
s
G
j
s
K j







 )
(
)
(






j
s
s
G
j
s
K j







 )
(
)
(
Example 2.5:
Considering transfer function G(s)
2
1
2
1
)
2
1
)(
2
1
(
3
)
5
2
(
3
)
(
2
1
2
1
0
2
j
s
K
j
s
K
s
K
j
s
j
s
s
s
s
s
s
G
j
j




















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5
3
0
5
2
3
2
0 










s
s
s
K



















































2
1
1
2
2
1
1
2
20
3
5
3
)
(
)
1
2
(
20
3
2
1
)
2
1
(
3
)
1
2
(
20
3
2
1
)
2
1
(
3
2
1
2
1
j
s
j
j
s
j
s
s
G
j
j
s
j
s
s
K
j
j
s
j
s
s
K
j
j
and the time function is given as
 
 














 

























t
t
e
j
e
e
e
e
e
e
e
j
e
e
e
e
j
e
j
t
g
t
t
j
t
j
t
j
t
j
t
t
j
t
j
t
j
t
j
t
t
j
t
j
2
sin
2
1
2
cos
5
3
5
3
2
2
2
4
20
3
5
3
)
(
)
2
2
(
20
3
5
3
)
1
2
(
)
1
2
(
20
3
5
3
)
(
2
2
2
2
2
2
2
2
)
2
1
(
)
2
1
(

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TUTORIAL 1: MATHEMATICAL REVIEW
1. Derive equations for a unit step, ramp, impulse and sinusoidal response in time domain.
2. In a unit step response graph, what is the relationship between final value theorem and steady
state error?
3. Find the Laplace transform of time function 𝑓 𝑡 = 5 + 3𝑒−2𝑡
.
4. Verify question (3) above by using MATLAB application.
MATLAB hint
>>syms s t; % Command to run MATLAB in s and t domains
>>f=5+3*exp(-2*t) % Entering the function
>>F=laplace(f,t,s) % Executing Laplace Transform command
5. Find the inverse Laplace Transform of a rational function and
𝐹 𝑠 =
5
𝑠2 + 3𝑠 + 2
6. Find the inverse Laplace transform of a rational function
𝐹 𝑠 =
2
𝑠 + 1 (𝑠 + 2)2
7. Verify the result in question (6) above using MATLAB application.
MATLAB hint
>> syms st;
>>F=2/((s+1)*(s+2)^2)
>>f=ilaplace(F,s,t)

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CHAPTER 2.0 INTRODUCTION TO CONTROL SYSTEMS
Control systems can be placed into three broad functional groups:
 Monitoring systems, such as Supervisory Control and Data Acquisition (SCADA) systems, which
provide information about the process state to the operator;
 Sequencing systems, used where some process must follow a pre-defined sequence of discrete
events;
 Closed-loop systems, which is widely taught in engineering course, are typically implemented to
give some process a set of desired performance characteristics
The history of feedback control system begun as early as in 1769 when James Watt’s steam engine and
governor are developed. The Watt stem engine often used to mark the beginning of the Industrial
Revolution in England. The revolution of automatic control system continues in which the first ever
autonomous rover vehicle, known as Sojourner was invented in 1997.
In summary below is the history of feedback control system
1769 - James Watt’s flyball governer
Figure 2.0: James Watt’s flyball governer
1868 - J. C. Maxwell’s model of governer
1927 - H. W. Bode’s feedback amplifiers
1932 - H. Nyquist’s stability theory

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UiTM Shah Alam
1954 - George Devol’s robot design
1970 - State-variable models and optimal control theory
1980 - Robust control system design
1997 - First ever autonomous rover vehicle “Sojourner”
Figure 2.1: Sojourner
But before we go into further details, we have to know control systems’ terms and concepts. The
frequently used terms and concepts are as follow:
Automation - The control of a process by automatic means
Control system - An interconnection of components forming a system
configuration that will provide a desired response
Controlled
variable
- Quantity or condition that is measured and controller.
Normally it is the output of the system
Manipulated
variable
- Quantity or condition that is varied by the controller so
as to affect the value of the controlled variable
Plant - A plant is a piece of equipment, perhaps just a set of
machine parts functioning together, the purpose of
which to perform a particular operation. Any physical
object to be controller (such as heating furnace, a
chemical reactor etc) is called a plant
Processes - A process can be defined as a natural, progressively
continuing operation or development marked by a
Info: The mobile Sojourner had a mass
of 10.5kg and 0.25 square meter solar
array

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UiTM Shah Alam
series of gradual changes that succeed one another in a
relatively fixed way and lead towards a particular result
or end
Disturbances - A disturbance is a signal which tends to adversely affect
the value of the output of the system. If a disturbance is
generated within the system, it is called internal; which
an external disturbance is generated outside the
system.
Feedback
control
- Feedback control is an operation which in the presence
of disturbances, tends to reduce the difference
between the output of a system and the reference
input and which does so on the basis of the difference.
Feedforward - Feedforward has a reference signal which is act as an
additional input.
Source: AAMI, Fac of Mech Eng., UiTM
Figure 2.2: Input-output configuration of control system (souce: AAMI)

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UiTM Shah Alam
Figure 2.3: Input-output configuration of a closed-loop control system (source: AAMI)
2.1 OPEN LOOP AND CLOSED-LOOP SYSTEMS
2.1.1 Open Loop Control System
A system is said to be an open loop system when the system’s output has no effect on the control
action. In open loop system, the output is neither measured nor fed back for comparison with the input.
Figure 2.4: Open loop control system
An open loop control system utilizes an actuating device (or controller) to control the process directly
without using feedback as shown in Figure 2.4.
The advantages and the disadvantages of an open-loop control system is tabulated in table 2.1 below
ADVANTAGES DISADVANTAGES
Simple and ease of maintenance Disturbances and changes in calibration
cause errors
Less expensive
Stability is not a problem Output may be different from what is
desired
Convenient when output is hard to
measure

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UiTM Shah Alam
2.1.2 Closed-loop control system
A system that maintains a prescribed relationship between the output and the reference input is called a
closed-loop system or a feedback control system. The system uses a measurement of the output and
feedback of the signal to compare it with the desired output.
Figure 2.5: Closed loop control system
In a closed-loop control system, the actuating error signal, which is the difference between the input
signal and the feedback signal, is fed to the controller so as to reduce the error and bring the output of
the system to a desired value.
2.1.3 Comparison between open loop and closed-loop control system.
The table below shows the comparison between the two systems:
OPEN LOOP CLOSED LOOP
System stability is not a major problem,
therefore easier to build
The use of feedback makes the system
response relatively insensitive to external
disturbances and internal variations in
system parameters
Use open loop only when the inputs are
known ahead of time and there is no
disturbances
System stability is a major problem
because the system tends to overcorrect
errors that can cause oscillations or
changing amplitude.
2.2 TRANSFER FUNCTION
The transfer function of a linear system is defined as the ratio of the Laplace transform of the output
variable to the Laplace transform of the input variable, with all initial conditions assumed to be zero. The
Transfer function of a system (or element) represents the relationship describing the dynamics of the
system under consideration. A transfer function may be defined only for a linear, stationary (constant
parameter) system. A non-stationary system often called a time-varying system, has one or more time-
varying parameters, and the Laplace transformation may not be utilized. Furthermore, a transfer
function is an input-output description of the behavior of a system. Thus the transfer function
description does not include any information concerning the internal structure of the system and its
behavior.

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UiTM Shah Alam
2.2.1 The Transfer function of linear systems
The transfer function of a LTI system is defined as the Laplace transform of the impulse response, with
all the initial conditions set to zero.
)]
(
[
)
( t
g
L
s
G 
The transfer function is related to the Laplace transform of the input and the output through the
following relation:
)
(
)
(
)
(
s
R
s
Y
s
G 
where all the initial conditions set to zero, and )
(s
Y and )
(s
R are the Laplace transform of )
(t
y and
)
(t
r respectively.
Although the transfer function of a linear system is defined in terms of the impulse response, in practice,
the input-output relation of a linear time-invariant system with continuous–data input is often described
by the differential equation, so it is more convenient to derive the transfer function directly from the
differential equation.
Let us consider that the input-output relation of a linear time-invariant system is described by the
following nth-order differential equation with constant real coefficients:
)
(
)
(
.....
)
(
)
(
)
(
)
(
......
)
(
)
(
0
1
1
1
1
0
1
1
1
1 t
r
b
dt
t
dr
b
dt
t
r
d
b
dt
t
r
d
b
t
y
a
dt
t
dy
a
dt
t
y
d
a
dt
t
y
d
m
m
m
m
m
m
n
n
n
n
n








 





To obtain the transfer function of the linear system that is represented by Eq. (2.3), we simply take the
Laplace transform on both sides of the equation and assume zero initial conditions. The result is
   R(s)
b
s
b
s
b
s
b
Y(s)
a
s
a
s
a
s m
m
m
m
n
n
n
0
1
1
1
0
1
1
1 







 


 

The transfer function between )
(t
r and )
(t
y is given by:
0
1
1
1
0
1
......
....
..........
)
(
)
(
)
(
a
s
a
s
a
s
b
s
b
s
b
s
R
s
Y
s
G n
n
n
m
m








 


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UiTM Shah Alam
The transfer function is said to be strictly proper if n
m  . If n
m  then the transfer function is proper.
It is improper if n
m  .
 Characteristic Equation: The characteristic equation of a LTI system is defined as the equation
obtained by setting the denominator polynomial of the transfer function to zero. Thus, the
characteristic equation of the system described by the Eq. (2.4) is
0
a 0
1
1
1 



 
 a
s
s
a
s n
n
n

Later, we shall show that the stability of a linear single-input single-output system is governed
completely by the roots of the characteristic equation.
2.2.2 Transfer function of multivariable system
The definition of a transfer function is easily extended to a system with multiple inputs and outputs. A
system of this type is often referred to as a multivariable system. Figure 2.6 shows a control system with
two inputs and two outputs.
Figure 2.6: General block representation of a two-input, two-output system
Since the principle of superposition is valid for linear systems, the total effect on any output due to all
the inputs acting simultaneously is obtained by adding up the outputs due to each input acting alone.
Thus, using transfer function relations we can write the simultaneous equations for the output variables
as
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
2
22
1
21
2
2
12
1
11
1
s
R
s
G
s
R
s
G
s
Y
s
R
s
G
s
R
s
G
s
Y





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UiTM Shah Alam
where )
(s
Gij is the transfer function relating the ith
output to the jth
input variable. Thus
)
(
)
(
s
R
s
Y
G
j
i
ij 
In general, for j inputs and i outputs, we can write the simultaneous equations for the output variables
as









































)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
2
1
2
1
2
22
21
1
12
11
2
1
s
R
s
R
s
R
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
Y
s
Y
s
Y
j
ij
i
i
j
j
i









It is convenient to express Eq. (2.7) in a matrix-vector form
G(s)R(s)
Y(s) 
where













)
(
)
(
)
(
)
( 2
1
s
Y
s
Y
s
Y
s
Y
i

is the i  1 transformed output vector; whereas















)
(
)
(
)
(
)
(
2
1
s
R
s
R
s
R
s
R
j

is the j  1 transformed input vector; and















)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
2
1
2
22
21
1
12
11
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
G
s
G
ij
i
i
j
j







is the i  j transfer-function matrix.

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UiTM Shah Alam
2.3 DEFINITION OF STABILTY
A stable system is defined as a system which gives a bounded output in response to a bounded input.
The concept of stability can be illustrated by considering a circular cone placed on a horizontal surface,
as shown in Fig. 2.7 and Fig. 2.8.
Figure 2.7: The stability of a cone.
----------------------------------------------------------------------------------------------------
Figure 2.8: Stability in the s-plane.
The stability of a dynamic system is defined in a similar manner. Let u(t), y(t), and g(t) be the input,
output, and impulse response of a linear time-invariant system, respectively. The output of the system is
given by the convolution between the input and the system's impulse response. Then




0
)
(
)
(
)
( 

 d
g
t
u
t
y

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UiTM Shah Alam
This response is bounded (stable system) if and only if the absolute value of the impulse response, g(t),
integrated over an infinite range, is finite. That is




0
)
( 
 d
g
Mathematically, Eq. (4.24) is satisfied when the roots of the characteristic equation, or the poles of G(s),
are all located in the left-half of the s-plane.
A system is said to be unstable if any of the characteristic equation roots is located in the right-half of
the s-plane. When the characteristic equation has simple roots on the j-axis and none in the right-half
plane, we refer to the system as marginally stable.
The following table illustrates the stability conditions of a linear continuous system with reference to the
locations of the roots of the characteristic equation.
STABILITY CONDITION LOCATION OF THE ROOTS
Stable All the roots are in the left-half s-plane
Marginally stable of marginally unstable At least one simple root and no multiple
roots on the j-axis; and no roots in the
right-half s-plane.
Unstable At least one simple root in the right-half s-
plane or at least one multiple-order root
on the j-axis.
The following examples illustrate the stability conditions of systems with reference to the poles of the
closed-loop transfer function M(s).
   
3
2
1
20
)
(




s
s
s
s
M
Stable
)
2
2
)(
1
(
)
1
(
20
)
( 2





s
s
s
s
s
M
Unstable due to the pole at s = 1
)
4
)(
2
(
)
1
(
20
)
( 2




s
s
s
s
M
Marginally stable or marginally unstable due to s =
j2.

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UiTM Shah Alam
)
10
(
)
4
(
10
)
( 2
2



s
s
s
M
Unstable due to the multiple-order pole at s = j2.
2.3.1 Open loop and Closed loop stability
A system is open-loop stable if the poles of the loop transfer function G(s)H(s) are all in the left hand
side of s-plane.
Figure 2.9: A typical closed-loop system
A system is closed0loop stable (or simply stable) if the poles of the closed-loop transfer function (or
zeros of 1+G(s)H(s) are all in the left hand side of s-plane
2.4 BASIC CONTROL ACTIONS
The following six basic control actions are very common among industrial automatic controllers:
1. Two-position or on-off controller
2. Proportional controller
3. Integral controller
4. Proportional-plus-integral controller
5. Proportional-plus-derivative controller
6. Proportional-plus-derivative-plus-integral controller
2.4.1 Two-position of on-off control action
In a two-position control system, the actuating element has only two fixed positions which are, in many
cases, simply on and off. Two-position or on-off control is relatively simple and inexpensive and, for this
reason, is very widely used in both industrial and domestic control systems.
Let the output signal from the controller be m(t) and the actuating error signal be e(t). In two position
control, the signal m(t) remains at either a maximum or minimum value, depending on whether the
actuating error signal is positive or negative, so that
𝑚 𝑡 = 𝑀1 𝑓𝑜𝑟 𝑒(𝑡) > 0
= 𝑀2 𝑓𝑜𝑟 𝑒(𝑡) < 0

+
- e(s)
y
ysp H(s)
Plant
Controller
G(s)

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UiTM Shah Alam
Where 𝑀1and 𝑀2, are constants. The minimum value 𝑀2, is usually either zero or −𝑀1. Two-position
controllers are generally electrical devices, and an electric, solenoid-operated valve is widely used in
such controller. Pneumatic proportional controller with very high gain act as two-position controller and
are sometimes called pneumatic two-position controller.
Figure 2.10 show the block diagrams for two-position controller. The range through which the actuating
error signal must move before the switching occurs is called the differential gap.
Figure 2.10: Two-position controller
2.4.2 Proportional controller
For a controller with proportional control action, the relationship between the output of the controller
m(t) and the actuating error signal e(t) is
𝑚 𝑡 = 𝐾𝑝𝑒(𝑡)
or, in Laplace Transform
𝑀(𝑠)
𝐸(𝑠)
= 𝐾𝑝
Where 𝐾𝑝, is termed the proportional sensitivity or the gain.

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Whatever the actual mechanism may be and whatever the form of the operating power, the
proportional controller is essentially an amplifier with and adjustable gain.
The proportional action has the following two properties:
1. Reduce rise time
2. Does not eliminate steady state error
Example 2.1:
Given a system consist of mass-spring and damper
a) The second order PDE is:
b) Taking the LT
c) The TF is therefore:
d) Let M=1kg, b=10N.s/m, k=20 N/m & F(s)=1, therefore X(s) / F(s):
e) From the Transfer Function, the DC gain is:
f) Corresponding to the steady state error of:
g) The settling time is:
b
M
x
F
k
Open Loop Response
Time (sec)
Displacement
(m)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0.05

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UiTM Shah Alam
P control (K) reduces the rise time, increases the overshoot and reduces the steady state error.
h) The closed-loop transfer function of the system with P controller is X(s)/F(s)=G/(1+G):
i) Let the P gain (K) equal 300
Rise time and ss error reduced, slightly reduced settling time but increased overshoot.
2.4.3 Integral controller
In a controller with integral control action, the value of the controller output m(t) is changed at a rate
proportional to, the actuating error signal e(t). That is
𝑑𝑚(𝑡)
𝑑𝑡
= 𝐾𝑖𝑒(𝑡)
Therefore; 𝑚 𝑡 = 𝐾𝑖 𝑒 𝑡 𝑑𝑡
𝑡
0
Where 𝐾𝑖 is an adjustable constant. The transfer function of the integral controller is
𝑀(𝑠)
𝐸(𝑠)
=
𝐾𝑖
𝑠
If the value of e(t) is doubled, then the value of m(t) varies twice as fast. For zero actuating error, the
value of m(t) remains stationary.
Closed Loop Step : K = 300
Time (sec)
Displacement
(m)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4

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UiTM Shah Alam
The integral controller has the following properties:
1. Proportional controllers often give a steady-state error. Integral controller arose from trying to
add a “reset” term to the control signal to eliminate steady state error. In other words, the
integral controller “resets” the bias error from the P controller.
2. Gives large gain at low frequencies resulting in “beating down” load disturbances.
3. May make the transient response worse.
4. Controller phase starts out at -90° and increases to 0° at the break frequency. This phase lag can
be compensated by derivative action.
The integral controller act as “automatic reset” as shown in figure 2.11
Figure 2.11: Automatic reset action
Almost always used in conjunction with P control.
Figure 2.12: PI control
The integral term may be expressed in (i) 𝑇𝑖 and (ii) 𝑘𝑖
The integral term 𝑇𝑖 is known as the integral time constant. 𝑇𝑖 = ∞ corresponds to pure (proportional)
gain.
The integral term 𝑘𝑖 is known as integral gain (e.g: in MATLAB)
The relationship between 𝑇𝑖 and 𝑘𝑖 is as follows:
𝑘𝑖
𝑠
=
𝐾
𝑇𝑖𝑠

+
-
ysp y
plant
K
load disturbance
1
sTi
u

e

+
-
ysp y
plant
K
load disturbance
1
sTi
u

e K

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UiTM Shah Alam
Example 2.2:
a) I control reduces the rise time, increases both settling time and overshoot, and eliminates the
steady-state error
b) The closed-loop transfer function of the system with a PI controller is: X(s)/F(s) =
______________ .
c) Let k = 30 and ki = 70. P gain (k) was reduced because the I controller also reduces the rise time
and increases the overshoot as does the P controller (double effect).
2.4.4 Derivative controller
Introducing a derivative controller will add damping and in doing so:
1. increases system stability (add phase lead)
2. reduces overshoot
3. generally improves transient response
A derivative controller may able to provide anticipative action but derivative action can make the system
become noisy.
Almost always used in conjunction with P control.
Figure 2.12: PD control
The integral term may be expressed in (i) 𝑇𝑑 and (ii) 𝑘𝑑
The integral term 𝑇𝑑 is known as the derivative time constant.
Closed Loop Step : K = 30, Ki = 70
Time (sec)
Displacement
(m)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4

+
-
ysp y
plant
c 
load disturbance
KTd s
sTd /N
1+

32
UiTM Shah Alam
The integral term 𝑘𝑑 is known as derivative gain (e.g: in MATLAB)
The relationship between 𝑇𝑑 and 𝑘𝑑 is as follows:
𝑘𝑑 𝑠 = 𝐾𝑇𝑑 𝑠
Example 2.3:
a) D control reduces both settling time and overshoot.
b) The closed-loop transfer function of the system with a PD controller is:
X(s)/F(s)=______________
c) Let k = 300 and kd = 10.
d) Reduced overshoot and settling time, small effect on rise time and ss error
Closed Loop Step : K = 300, Kd = 10
Time (sec)
Displacement
(m)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
1.4

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UiTM Shah Alam
2.4.5 PID controller
In some system the commonly implemented controller consist of the P, I and D control action. We call
this type of controller as PID controller.
Figure 2.13: PID control
The standard form of PID controller according to ISA (Instrument Society of America) is as follows:
𝐺𝑐 𝑠 = 𝐾(1 +
1
𝑠𝑇𝑖
+ 𝑇𝑑 𝑠)
Or 𝐺𝑐 𝑠 = 𝐾 +
𝑘𝑖
𝑠
+ 𝑘𝑑 𝑠
Closed Loop Step : K = 350, Ki = 300, Kd = 50
Time (sec)
Displacement
(m)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2
K
1/( )
Ts
i
ysp
T s
d
G s
( )
u


-
+ e y

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Example 2.4:
a) The closed-loop transfer function of the system with a PID controller is:
X(s)/F(s) = (kd s2
+ks+ki )/(s3
+ (10+kd)s2
+ (20+k)s + ki )
b) Let k = 350, ki = 300 and kd = 50.
c) No overshoot, fast rise and settling time and no steady-state error
2.4.6 PID tuning
Introducing the P, I and D controller has certainly proven to contribute some effect to our system’s
response. These effects are summarized as in table below.
CLOSED LOOP
RESPONSE
RISE TIME OVERSHOOT SETTLING
TIME
SS ERROR
K Decrease Increase Small change Decrease
𝑘𝑖 =
𝐾
𝑇𝑖
Decrease Increase Increase Eliminate
𝑘𝑑 = 𝐾𝑇𝑑 Small change Decrease Decrease Small change
When you are designing a PID controller for a given system, follow the steps shown below to obtain a
desired response.
1. Obtain an open-loop response and determine what needs to be improved
2. Add a proportional control to improve the rise time
3. Add a derivative control to improve the overshoot
4. Add an integral control to eliminate the steady-state error
5. Adjust each of K, Ki, and Kd until you obtain a desired overall response referring to the table
shown previously to find out which controller controls what characteristics.
Closed Loop Step : K = 350, Ki = 300, Kd = 50
Time (sec)
Displacement
(m)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.2
0.4
0.6
0.8
1
1.2

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6. It is not necessary to implement all three controllers (P, I & D) into a single system. For example,
if a PI controller gives a good enough response, then you don't need to add D control to the
system. Simple is better.
2.5 BLOCK DIAGRAM & REDUCTION METHODS
A block diagram is used to describe the composition and interconnection of a system, or it can be used
together with the transfer functions to describe the cause-and-effect relationships throughout the
system. For instance, Figure 2.14 (a) shows a dc motor wiring diagram, (b) sketch, and (c) shows the
block diagram with transfer function.
Figure 2.14: A dc motor: (a) wiring diagram (b) sketch
Figure 2.14 (c): A dc motor: Block diagram with transfer functions

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2.5.1 Block diagram reduction method
We shall now define the block diagram elements used frequently in linear control systems and the
related algebra. All component parts of a block diagram for linear time-invariant systems are shown in
Figure 2.15.
The characteristic of the summing junction as shown in Figure 2.15 (c) is that the output signal, )
(s
C , is
the algebraic sum of the input signals. The figure shows three inputs, but any number can be presented.
A pickoff point, as shown in Figure 2.15 (d), distributes the input signal, )
(s
R , undiminished, to several
output points.
Figure 2.15: Components of a block diagram for LTI systems
Figure 2.16 shows the block diagram of a linear feedback control system. The following terminology is
defined with reference to the diagram.
Figure 2.16: Feedback control system

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)
(
),
( t
r
s
R = reference input (command)
)
(
),
( t
c
s
C or )
(
),
( t
y
s
Y = output (controlled variable)
)
(
),
( t
b
s
B = feedback signal
)
(
),
( t
e
s
E = actuating signal = error signal
)
(s
H = feedback transfer function
)
(
)
( s
H
s
G = )
(s
L = loop transfer function
)
(s
G = forward-path transfer function
)
(
)
(
)
( s
R
s
C
s
M  or )
(
)
( s
R
s
Y = closed-loop transfer function or system transfer function.
)
(s
M can be expressed as a function of )
(s
G and )
(s
H . From Figure 2.16, we write
)
(
)
(
)
(
)
(
)
(
)
(
s
Y
s
H
s
B
s
E
s
G
s
Y


The actuating signal is written as
)
(
)
(
)
( s
B
s
R
s
E 

Thus,
)
(
)
(
1
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
s
H
s
G
s
G
s
R
s
Y
s
M
s
B
s
G
s
R
s
G
s
Y





The block diagram representation of a given system often can be reduced by block diagram reduction
techniques to a simplified block diagram with fewer blocks than the original diagram. Table below shows
some of the block diagram reduction techniques.
The block diagram reduction technique is based on the utilization of rule 6 in which eliminates feedback
loops. Therefore, the other transformations are used to transform the diagram to a form ready for
eliminating feedback loops.

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For parallel subsystems as shown below in (a), the reduction technique is shown in (b).

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Example 2.5: Block Diagram Reduction.
A block diagram of a multiple-loop feedback control system is shown in Figure 2-5. It is interesting to
note that the feedback signal H1(s)Y(s) is a positive feedback signal, and the loop G3(s)G4(s)H1(s) is called
a positive feedback loop. First, to eliminate the loop G3G4H1, we move H2 behind block G4 by using rule
4, and therefore obtain Figure 2-6 (a).
Multiple-loop feedback control system

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Eliminating the loop G3G4H1 by using rule 6, we obtain Figure 2-6 (b). Then, eliminating the inner loop
containing H2/G4, we obtain Figure 2-6 (c). Finally, by reducing the loop containing H3, we obtain the
closed-loop system transfer function as shown in Figure 2-6 (d).
Block diagram reduction of the system

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Example 2.6: Reduce the system shown to a single transfer function.
Block diagram for Example 2.6

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Steps in the block diagram reduction for Example 2.6
The block diagram representation of feedback control systems is a valuable and widely used approach.
The block diagram provides the analyst with a graphical representation of the interrelationships of
controlled and input variables. Furthermore, the designer can readily visualize the possibilities for
adding blocks to the existing system block diagram to alter and improve the system performance. The
transition from the block diagram method to a method utilizing a line path representation instead of a
block representation is readily accomplished and is presented in the following section.

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2.6 SIGNAL FLOW DIAGRAM & REDUCTION METHODS
Block diagrams are adequate for the representation of the interrelationships of controlled and input
variables. However, for a system with reasonably complex interrelationships, the block diagram
reduction technique is cumbersome and often quite difficult to complete. An alternative method for
determining the relationship between system variables has been developed by Mason and is based on a
representation of the linear system by line segments called Signal-Flow Graph (SFG).
The advantage of the SFG method is the availability of a flow graph gain formula, which provides the
relation between system variables without requiring any reduction procedure or manipulation of the
flow graph.
2.6.1 Basic elements of SFG
When constructing a SFG, junction points or nodes are used to represent variables. The nodes are
connected by line segments, called branches. A signal can transmit through a branch only in the
direction of the arrow.
For instance, consider that a linear system is represented by a simple algebraic equation
1
12
2 y
a
y 
where y1 is the input, y2 the output, and a12 the gain between two variables. The SFG is shown in Figure
2-9.
Figure 2.17: Signal-flow graph of 1
12
2 y
a
y 
Example 2.7: Consider the following set of algebraic equations:
4
45
2
25
5
4
44
3
34
2
24
4
4
43
2
23
3
3
32
1
12
2
y
a
y
a
y
y
a
y
a
y
a
y
y
a
y
a
y
y
a
y
a
y









Y1 Y2
a12

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The SFG for these equations is constructed, step by step, as shown:
Step-by-step construction of the SFG of Example 2.7

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2.6.2 Summary of the basic properties of SFG
The important properties of the SGF are summarized as follows:
1. SFG applies only to linear systems.
2. Nodes are used to represent variables. Normally, the nodes are arranged from left to right,
from input to output.
3. Signals travel along branches only in the direction described by the arrows of the branches.
2.6.3 Definitions of the SFG terms
Input Node
(source)
- An input node is a node that has only outgoing
branches.
Output Node
(sink)
- An output node is a node that has only incoming
branches. In general, we can make any non input node
an output node, simply by connecting a branch with
unity gain from the existing node to a new node with
the same name (Example: node y2 in Figure 2.18(b)). If
we attempt to convert y2 into input node, by using the
same unity gain branch (Figure 2.18 (c)), then y2 output
will differ from the original (y2 = y2 + a12y1 + a32y3).
Path - A path is any collection of a continuous succession of
branches traversed in the same direction.
Forward Path - A forward path is a path that starts at an input node
and ends at an output node, and along which no node is
traversed more than once.
Loop - A loop is a path that originates and terminates on the
same node and along which no other node is
encountered more than once. For example, there are
four loops in the SFG of Example 2.7. These are shown
in Figure 2.19
Path Gain - The product of the branch gains encountered in
traversing a path is called the path gain
Loop Gain - The loop gain is the path gain of a loop
Non-touching
Loops
- Two parts of a SFG are non-touching if they do not
share a common node. For example, the loop y2-y3-y2
and y4-y4 of the SFG in Figure (d) of Example 2.7 are
non-touching loops.

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Figure 2.18 (a & b): Modification of SFG so that y2 become output node
(c): Erroneous way to make node y2 an input node

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Figure 2.19: Four loops in the signal-flow graph of Example 2.7
2.6.4 SFG Algebra
Based on the properties of the SFG, we can outline the following manipulation rules and algebra of SFG.
1. The value of the variable represented by a node is equal to the sum of all the signals
entering the node. For the SFG of Figure 2.20 (a),
5
51
4
41
3
31
2
21
1 y
a
y
a
y
a
y
a
y 



2. The value of the variable represented by a node is transmitted through all branches
leaving the node. In Figure 2.20 (a), we have
1
18
8
1
7
1
7
1
16
6
y
a
y
y
a
y
y
a
y




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Figure 2.20
3. Parallel branches in the same direction connecting two nodes can be replaced by a
single branch with the gain equal to the sum of gains of the parallel branches. Example:
Figure 2.20 (b).
4. A series connection of unidirectional branches can be replaced by one branch with gain
equal to the product of branch gains.

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2.6.5 Gain formula for SFG (Mason’s Rule)
The overall gain between the input node yin and output node yout of a SFG with N forward paths and L
loops is given by
N
k
P
y
y
M k
k
k
in
out

,
2
,
1
, 





where
yin = input-node variable
yout = output-node variable
M = gain between yin and yout
N = total number of forward paths between yin and yout
Pk = kth
forward-path gain
 = 1 – (sum of all individual loop gains)
+ (sum of all gain products of two non-touching loops)
– (sum of all gain products of three non-touching loops) + …
k = , which is evaluated by eliminating all loops that touch kth
forward-path
Procedures to solve SFG by using Mason’s rule:
1. Identify the no. of forward paths and determine the forward-path gains.
2. Identify the no. of loops and determine the loop gains.
3. Identify the non-touching loops taken two at a time, three at a time and so on. Determine the
product of the non-touching loop gains.
4. Determine  and k.
5. Substitute all of the above information into the gain formula:
N
k
P
y
y
M k
k
k
in
out

,
2
,
1
, 






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Care must be taken when applying the gain formula to ensure that it is applied between an input node
and an output node.
Example 2.8: Consider the SFG of a closed loop control system as given in Figure below. By using the
gain formula, find the transfer function )
(
)
( s
R
s
Y .
SFG of a feedback control system
1. There is only one forward path between )
(s
R and )
(s
Y , and the forward-path gain is
P1 = )
(s
G .
2. There is only one loop; the loop gain is L1 = )
(
)
( s
H
s
G
 .
3. There are no non-touching loops.
4.  = 1 - L1 = )
(
)
(
1 s
H
s
G
 and 1 = 1.
5. Thus,
)
(
)
(
1
)
(
)
(
)
( 1
1
s
H
s
G
s
G
P
s
R
s
Y






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Example 2.9: For the system shown in Figure below, determine the gain between y1 and y5.
SFG for Example 2.9
1. There are three forward paths
Path 1: y1 – y2 – y3 – y4 – y5 P1 = a12 a23 a34 a45
Path 2: y1 – y2 – y4 – y5 P2 = a12 a24 a45
Path 3: y1 – y2 – y5 P3 = a12 a25
2. There are four loops
Loop 1: y2 – y3 – y2 L1 = a23 a32
Loop 2: y3 – y4 – y3 L2 = a34 a43
Loop 3: y2 – y4 – y3 – y2 L3 = a24 a43 a32
Loop 4: y4 – y4 L4 = a44
3. Non-touching loops: y2 – y3 – y2 and y4 – y4
Thus the product of the gains of the two non-touching loops:
L1L4 = a23 a32 a44
4.  = 1 – (L1 + L2 + L3 + L4) + L1L4
= 1 – (a23 a32 +a34 a43 + a24 a43 a32 + a44) + a23 a32 a44

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All the loops are in touch with forward path P1, thus 1 = 1.
Two loops (y3 – y4 – y3 and y4 – y4) are not touching with forward path P3.
Thus, 3 = 1 - a34a43 – a44.
5. Thus,
44
32
23
44
43
32
24
43
34
32
23
44
43
34
25
12
45
24
12
45
34
23
12
3
3
2
2
1
1
1
5
)
(
1
)
1
)(
(
)
(
)
(
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
P
P
P
y
y
M


















Example 2.10: Consider the SFG as shown in the figure. The following input-output relation is obtained
by use of the gain formula:









)
1
( 2
3
5
1
4
3
2
1
2
2
1
1
1
7 H
G
G
G
G
G
G
G
P
P
y
y
where
4
2
1
3
1
4
3
3
2
1
4
2
3
4
1
1
2
1
3
1
4
3
3
2
1
2
3
1
1
1
H
H
H
G
G
H
H
G
G
G
H
H
G
H
H
G
H
H
G
G
H
H
G
G
G
H
G
H
G











SFG for Example 2.10

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2.7 CONVERSION FROM BLOCK DIAGRAMS TO SFG
An equivalent SFG for a block diagram can be drawn by performing the following steps:
1. Identify the input/output signals, summing junctions & pickoff points → they are replaced with
nodes.
2. Interconnect the nodes & indicate the directions of signal flow by using arrows.
3. Identify the blocks - they are replaced with branches.
For each negative sum, a negative sign is included with the branch.
4. Add unity branches as needed for clarity or to make connections.
5. Simplify the SFG → eliminate redundant nodes/branches (only if the node is connected to branches
of a single flow in & a single flow out with unity gain).
6. Label the input/output signals and the branches accordingly.
Example 2.11: Convert the block diagram in the figure to a signal flow graph and determine the transfer
function using Mason’s gain formula.

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The equivalent SFG:
1. There are two forward paths; the forward-path gains are:
P1 = G1G2G3
P2 = G1G4
2. There are five individual loops; the loop gains are:
L1 = −G1G2H1
L2 = −G2G3H2
L3 = −G1G2G3
L4 = −G1G4
L5 = −G4H2
3. There are no non-touching loops.
4. ∆ = 1 – (L1 + L2 + L3 + L4 + L5)
= 1 + G1G2H1 + G2G3H2 + G1G2G3 + G1G4 + G4H2

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All the loops are in touch with forward path P2, thus 2 = 1
5. Thus,
2
4
4
1
3
2
1
2
3
2
1
2
1
4
1
3
2
1
2
2
1
1
1 H
G
G
G
G
G
G
H
G
G
H
G
G
G
G
G
G
G
P
P
R
Y












2.8 STATE SPACE EQUATIONS
State space approach is an alternative method for representing physical system. In order to use this
approach, we have to limit our approach to linear, time-invariant systems or system that can be
linearized by the methods we have covered previously.
In state space method, the models are constructed in the time domain. This means we can work directly
with the governing differential equations to model, analyze and design a wide range of system. In
contrast, classical control design practices looking at the frequency domain output to interpret system’s
physical dynamics. With the arrival of space exploration, requirements for control systems increased in
scope. Hence the use of classical control design seems inadequate.
Many systems do not have just a single input. Multiple-input, multiple-output systems can be compactly
represented in state space with a model similar in form and complexity to that used for single-input,
single-output systems. To address the multiple input and output system a convenient matrix based is
used in representing the state space. In addition, the state space approach is also attractive because of
the availability of numerous state-space software packages for the personal computer

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The table below outlines the advantages and disadvantages of state space models
ADVANTAGES DISADVANTAGES
Multiple input / output models are now
possible
Difficult to examine robustness (stability
margins)
Possible to minimize “error critera”
(optimal control)
More work than classical control for
“simple” problems
Possible to examine stability in more
depth “optimal” systems require “optimal” error
criteria
Ideally suited to computer-based design
and analysis
2.8.1 Definition of state space terms
State of a
system
- A set of quantities which completely determine the
evolution of the response of a system (in the absence of
external inputs)
State Variables - Set of variables that define the state. These variables
are not unique. For example x1, x2,….
State Vector - The (column) vector of the nth state variables:
𝑥 𝑡 = [𝑥1 𝑡 𝑥2 𝑡 … 𝑥𝑛 𝑡 ]
Note: system is of order n (i.e it is described by an nth
order D.E.
State Space - The n-dimensional space in which the components of
the state vector are the co-ordinate axes.
State trajectory - The path in state space produced by the state vector as
it changes with time.
Note: The selection of state variables is not unique. In the first instance, it is often reasonable to
choose something with “physical meaning”, often something associated with system “energy”

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2.8.2 State space model
We begin our state space equation with a state equation. A state equation consist of the state equation
and output equation as follows:
𝑥 = 𝐴𝑥 + 𝐵𝑢 State Equation
𝑦 = 𝐶𝑥 + 𝐷𝑢 Output Equation
Now A, B, C and D are all matrices involved in a state space equation
A = (n x n) state matrix that describes “internal (homogenous) motion
B = (n x r) input matrix that describes how r inputs affect n states
C = (m x n) output matrix that describes how n states contribute to m outputs
D = (m x r) direct transmission matrix that describes how r inputs are fed through to m outputs.
LECTURER’S NOTES:

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LECTURER’S NOTES:

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CHAPTER 3.0 SYSTEM PERFORMANCE ANALYSIS
The ability to adjust the transient and steady-state response of a control system is a beneficial outcome
of the design of feedback systems. Since time is used as an independent variable in most of control
systems, it is usually of interest to evaluate the state and output responses with respect to time, or
simply the time response.
In the analysis problem, we will use selected input signals to test the response of control systems. This
response will be characterized by a selected set of response measures. In this chapter, we will strive to
delineate a set of quantitative performance measures that adequately represent the performance of the
control systems.
3.1 Time Response and Test Signals
The time response of a control system is usually divided into two parts: the transient response and the
steady-state response. Let y(t) denote the time response of a continuous-data system; then, in general,
it can be written as
y(t) = yt(t) + yss(t) (3.1)
where yt(t) denotes the transient response and yss(t) denotes the steady-state response.
In control systems, the transient response is defined as the part of the time response that goes to zero
as time becomes very large. Thus yt(t) has the property
0
)
(
lim 


t
yt
t
(3.2)
The steady-state response is simply the part of the total response that remains after the transient has
died out. All real stable systems exhibit transient phenomena to some extent before the steady state is
reached.
In the design problem, specifications are usually given in terms of the transient and steady-state
performance, and controllers are designed so that the specifications are all met by the design system.

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Since it is difficult to design a control system that will perform satisfactorily for all possible forms of
input signals, it is necessary, for the purpose of analysis and design, to assume some basic types of test
signals properly for the prediction of the system's performance to other more complex inputs.
3.1.2 Step-Input Function
The step-function input represents an instantaneous change in the reference input. The mathematical
representation of a step function of magnitude A is






0
0
0
)
(
t
t
A
t
r
Mathematically, r(t) = Aus(t), where us(t) is the unit-step function. The step function is shown in Fig.
3.1(a).
3.1.3 Ramp-Input Function
The ramp function is a signal that changes constantly with time. Mathematically, a ramp function is
represented by
)
(
)
( t
Atu
t
r s

where A is a real constant. The ramp function is shown in Fig. 3.1(b).
3.1.4 Parabolic-Input Function
The parabolic function represents a signal that is one order faster than the ramp function.
Mathematically, it is represented by
)
(
2
)
(
2
t
u
At
t
r s

The parabolic function is shown in Fig. 3.1(c).

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Fig. 3.1 shows the three time-domain test signals.
Figure 3.1: Test input signals: (a) Step, (b) Ramp, (c) Parabolic.
3.2 First & Second Order System: Transient & Steady State Response
For linear control systems, the time response is characterized by using the unit-step input. The response
of the control system to the unit-step input is called the unit-step response. Fig. 3.2 illustrates a typical
unit-step response of a linear control system. With reference to the unit-step response, the following
performance criteria (parameters) are defined:
1. Maximum overshoot: Let ymax denotes the maximum value of y(t) and yss be the steady-state value of
y(t) and ymax  yss. The maximum overshoot of y(t) is defined as,
Maximum overshoot = ymax − yss
%
100
overshoot
maximum
overshoot
maximum
of
Percentage 

ss
y
(3.3)
2. Delay time: The delay time, td is defined as, the time required for the step response to reach 50% of
its final value.
3. Rise time: The rise time, tr is defined as, the time required for the step response to rise from 10 to 90
percent of its final value.
4. Settling time: The settling time, ts is defined as, the time required for the step response to reach and
stay within a specified percentage (5%) of its final value.

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Figure 3.2: Step response of a control system.
Analytically, these quantities are difficult to establish, except for simple systems that are lower than the
third order.
3.2.1 Transient Response of a Prototype of Second-Order Systems
Although it is true that second-order control systems are rare in practice, their analysis generally helps
to form a basis for the understanding of analysis and design of higher-order systems, especially the ones
that can be approximated by second-order systems.
Consider that a second-order control system with unity feedback is represented by the block diagram
shown in Fig. 3.3. The open-loop transfer function of the system is
 
n
n
s
s
s
G


2
)
(
2

 (3.5)

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where ζ and n are real constants. The closed-loop transfer function of the system is
2
2
2
2
)
(
)
(
n
n
n
s
s
s
R
s
Y





 (3.6)
The characteristic equation of the prototype of the second-order system is obtained by setting the
denominator of Eq. (3.6) to zero
0
2
)
( 2
2




 n
ns
s
s 
 (3.7)
The system is stable (Bounded output for bounded input) if the roots of the characteristic equation is
located on the left half of s-plane, and marginally stable (Oscillation for a bounded input) if the
characteristic equation has simple roots on the imaginary axis with all other roots on the left half of s-
plane. For an unstable (Unbounded output for any bounded input) system, the characteristic equation
has at least one root on the right half of the s-plane or it has a repeated j roots.
Figure 3.3: A prototype of a second-order control system.
For a unit-step input, R(s) = 1/s, the output response is given as
)
2
(
)
( 2
2
2
n
n
n
s
s
s
s
Y





 (3.8)
By taking inverse Laplace transform, we obtain the unit step response of the control system
  0
cos
1
sin
1
1
)
( 1
2
2





 

t
t
e
t
y n
t
n





(3.9)

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Fig. 3.4 shows the unit-step response of the second-order system for various values of . It may be
noted that the response becomes more oscillatory with larger overshoot as  decreases.
Figure 3.4: Unit-step response of a second-order system with various ζ values.
3.2.2 Damping Ratio and Damping Factor
The effects of the system parameters ζ and n on the step response y(t) can be studied by referring to
the roots of the characteristic equation in Eq. (4.7). The roots can be expressed as





j
j
s
s n
n






 1
, 2
2
1
(3.10)
where
 = ζn (3.11)
and

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2
1 

 
 n (3.12)
The physical significance of ζ and  is now investigated. As seen from Eq. (4.9), the factor  = n
appears as a constant multiplied by t in the exponential term of the response y(t). Therefore,  controls
the rate of rise or decay of the unit-step response y(t). In other words,  controls the "damping" of the
system and is called the damping factor.
The inverse of , 1/ is proportional to the time constant of the system. When  = 1, the oscillations
disappear and the system is said to be critically damped. Under this condition,  = n. Thus, we can
regard  as
damping
critical
the
at
factor
damping
factor
damping
actual
n





ratio,
Damping (3.13)
When  < 1, the system is under-damped and when  > 1, the system is over-damped.
3.2.3 Natural Undamped Frequency
The parameter n is defined as the natural undamped frequency. As seen from Eq. (3.10), when  = 0,
the roots of the characteristic equation are imaginary. Thus, the unit-step response of the system
becomes purely oscillatory with angular frequency of n. For 0 <  < 1, the imaginary parts of the roots
have the magnitude of the actual (damped) frequency of oscillation. Thus
2
1 

 
 n
Fig. 3.5 illustrates the relationships between the location of the roots of the characteristic equation and
, ζ, and n.

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Figure 3.5: The relationships between the location of the roots of the characteristic equation and , ζ,
and n.
The effect of the roots of the characteristic equation on the damping of the second-order system is
illustrated in Fig. 3.6.

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Figure 3.6: Step-response comparison for various locations of the roots of the characteristic equation in
the s-plane.

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3.2.4 Analytical Expression for Maximum Overshoot
By taking the derivative of Eq. (3.9) with respect to time t and setting the result to zero, we get
t
e
dt
t
dy
n
t
n n
.
1
.
sin
1
)
( 2
2



 


 
(3.14)
,...
2
,
1
,
0
1 2


 n
n
t
n 


From which we get
,...
2
,
1
,
0
1 2


 n
n
t
n 


(3.15)
For the unit-step responses shown in Fig. 3.4, the first overshoot is the maximum overshoot. This
corresponds to n = 1 in Eq. (4.15). Thus, the time at which the maximum overshoot occurs is
2
max
1 




n
t (3.16)
With reference to Fig. 3.4, the overshoots occur at odd values of n, that is, n =1, 3, 5, …, and
undershoots occur at even values of n.
The magnitude of the overshoots and undershoots can be determined by subistituting Eq. (3.14) into Eq.
(3.9). This results in y(t)max or min . Therefore
2
1
/
max 1
overshoot
maximum 
 



 e
y (3.17)
and the percentage of maximum overshoot is
2
-
1
/
-
100e
overshoot
maximum
of
percentage 

 (3.18)
The relationship between the percent maximum overshoot and the damping ratio, as given in Eq. (3.18),
is plotted in Fig. 3.7.

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Figure 3.7: The relationship between the percent maximum overshoot and the damping ratio.
3.2.5 Delay Time and Rise Time
It is more difficult to determine the exact analytical expressions of the delay time td, rise time tr, and
settling time ts. However, we can utilize the linear approximation
1.0
0
7
.
0
1



 


n
d
t (3.19)
The plot of ntr versus ζ is shown in Fig. 3.8. This relation can be approximated by a straight line over a
limited range of ζ.
1
0
16
.
2
60
.
0



 


n
r
t (3.20)

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Figure 3.8: Normalized rise time versus ζ for the prototype second-order system.
From this discussion, the following conclusions can be made:
1. tr and td are proportional to ζ and inversely proportional to n.
2. Increasing (decreasing) the natural undamped frequency n will reduce (increase) tr and td.
The settling time ts can be approximated as
n
s
t

3
 (3.21)
We can summarize the relationships between ts and the system parameters as follows:
1. For ζ < 0.69, the settling time is inversely proportional to ζ and n. A practical way of reducing
the settling time is to increase n while holding ζ constant.
2. For ζ > 0.69, the settling time is proportional to ζ and inversely proportional to n. Again, ts can
be reduced by increasing n.

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3.3 STABILITY & PERFORMANCE SPECIFICATIONS – ROUTH-HURWITZ STABILITY TEST
The discussions in the preceding sections lead to the conclusion that the stability of a linear time-
invariant system can be determined by checking on the location of the roots of the characteristic
equation. When the system parameters are all known, the roots of the characteristic equation can be
solved by means of a root-finding computer program.
For design purposes, there will be unknown or variable parameter embedded in the characteristic
equation, and it will be feasible to use the root-finding programs. The method outlined below is well
known for the determination of stability of a LTI system without involving root solving.
3.3.1 Routh-Hurwitz Criterion
The Routh-Hurwitz criterion represents a method of determining the location of zeros of a polynomial
with constant real coefficients with respect to the left and right half of the s-plane, without actually
solving for the zeros.
Consider that the characteristic equation of a linear time-invariant SISO system is of the form
0
)
( 0
1
1
1 




 
 a
s
a
s
a
s
a
s
F n
n
n
n  (3.25)
where all the coefficients are real. In order that Eq. (3.25) does not have roots in the right half of s-
plane, it is necessary and insufficient that the following conditions hold:
1. All the coefficients of the equation have the same sign
2. None of the coefficients vanishes
However, these conditions are not sufficient, for it is quite possible that an equation with all its
coefficients nonzero and of the same sign still will not have all the roots in the left half of the s-plane.
The first step in the Routh-Hurwitz criterion is to arrange the coefficients of the Eq. (3.25) as follows:

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

5
3
1
4
2
1






n
n
n
n
n
n
n
n
a
a
a
a
a
a
s
s
Further rows of the schedule are then completed as follows:
1
5
3
1
5
3
1
5
3
1
4
2
0
3
2
1















n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
h
c
c
c
b
b
b
a
a
a
a
a
a
s
s
s
s
s









where
3
1
3
1
1
1
5
1
4
1
3
3
1
2
1
1
1
1
1






















n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
b
b
a
a
b
c
a
a
a
a
a
b
a
a
a
a
a
b
and so on.
Once the Routh's tabulation has been completed, we investigate the signs of the coefficients in the first
column of the tabulation. The roots of the equation are all in the left half of the s-plane if all the
elements of the first column of the Routh's tabulation are of the same sign. The number of changes of
signs in the elements of the first column equal the number of roots with positive real parts or in the right-
half s-plane.

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Example 3.1: Consider the equation
    0
6
4
3
1
2 2
3







 s
s
s
s
s
s
This equation has one negative coefficient. Thus, we know without applying Routh's test that not all the
roots of the equation are in the left-half s-plane. In fact, from the factored form of the equation, we
know that there are two roots in the right-half s-plane, at s = 2 and s = 3. For the purpose of illustrating,
the Routh's tabulation is made as follows:
0
6
0
5
.
2
6
4
1
1
0
1
2
3
s
s
s
s

Since there are two sign changes in the first column of the tabulation, the equation has two roots
located in the right-half s-plane.
Example 3.2: Consider the equation
0
10
5
3
2 2
3
4




 s
s
s
s
Since this equation has no missing terms and the coefficients are all of the same sign, it satisfies the
necessary conditions for not having roots in the right half or on the imaginary axis of the s-plane.
However, since these conditions are necessary but not sufficient, we have to check the Routh's
tabulation.
0
0
10
0
0
43
.
6
0
10
7
0
5
1
10
3
2
0
1
2
3
4
s
s
s
s
s

Since there are two changes in the first column of the tabulation, the equation has two roots in the right
half of the s-plane.

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 Special Cases When Routh's Tabulation Terminates Prematurely
Depending on the coefficients of the equation, the following difficulties may occur that prevent the
Routh's tabulation from completing properly:
1. The first element in any one row of Routh's tabulation is zero, but the others are not.
2. The elements in one row of Routh's tabulation are all zero.
In the first case, we replace the zero element in the first column by an arbitrary small positive number ,
and then proceed with Routh's tabulation. This is illustrated by the following example:
Example 3.3: Consider the characteristic equation of a linear system:
0
3
2
2 2
3
4




 s
s
s
s
Since all the coefficients are nonzero and of the same sign, we need to apply the Routh-Hurwitz
criterion. Routh's tabulation is carried out as follows:
3
0
0
2
1
3
2
1
2
3
4
s
s
s
Since the first element of the s2
row is zero, the element in the s1
row would all be infinite. To overcome
this difficulty, we replace the zero in the s2
row by a small positive number  and then proceed with the
tabulation.
0
3
0
3
3
0
1
2
s
s
s




Since there are two sign changes in the first column of Routh's tabulation, the equation has two roots in
the right-half s-plane.

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In the second special case, when all the elements in one row of Routh's tabulation are zeros before the
tabulation is properly terminated, it indicates that one or more of the following conditions may exist.
1. The equation has at least one pair of real roots with equal magnitude but opposite signs.
2. The equation has one or more pairs of imaginary roots.
3. The equation has pairs of complex-conjugate roots forming symmetry about the origin of the s-
plane (e.g. s = -1  j1, s = 1  j1).
The situation with the entire row of zeros can be remedied by using the auxiliary equation A(s) = 0,
which is formed from the coefficients of the row just above the row of zeros in Routh's tabulation. The
roots of the auxiliary equation also satisfy the original equation. To continue with Routh's tabulation
when a row of zeros appears, we conduct the following steps:
1. Form the auxiliary equation A(s) = 0 by use of the coefficients from the row just preceding the
row of zeros.
2. Take the derivative of the auxiliary equation with respect to s; this gives dA(s)/ds = 0.
3. Replace the row of zeros with the coefficients of dA(s)/ds = 0.
4. Continue with Routh's tabulation in the usual manner.
Example 3.4: Consider the following characteristic equation of a linear control system:
0
4
7
8
8
4 2
3
4
5





 s
s
s
s
s
The Routh's tabulation is
0
0
4
4
0
6
6
4
8
4
7
8
1
1
2
3
4
5
s
s
s
s
s

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A(s) = 4s2
+ 4 = 0
The derivative of A(s) with respect to s is
dA(s)/ds = 8s = 0
From which the remaining portion of the Routh's tabulation is
4
0
8
0
1
s
s
Since there are no sign changes in the first column, the system is stable. Solving the auxiliary equation
A(s) = 0, we get the two roots at s = j and s = -j, which are also two of the roots of the characteristic
equation. Thus, the equation has two roots on the j-axis, and the system is marginally stable. These
imaginary roots caused the tabulation to have an entire row of zeros in the s1
row.
Example 3.5: Consider that a third-order control system has the characteristic equation
0
10
5
.
1
10
1204
3
.
3408 7
3
2
3





 k
s
s
s
Determine the crucial value of k for stability.
7
0
3
7
1
7
2
3
3
10
5
.
1
0
3408
10
1204
3408
10
5
.
1
10
5
.
1
3
.
3408
10
1204
1








s
k
s
k
s
s
For the system to be stable, all the coefficients in the first column must have the same sign. This lead to
the following conditions:
0
3408
10
36
.
410
10
5
.
1 7
7





k
Therefore, the condition of k for the system to be stable is
57
.
273
0 
 k

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If we let k = 273.57, the characteristic equation will have two roots on the j-axis. To find these roots,
we substitute k = 273.57 in the auxiliary equation, as follows:
0
10
1036
.
4
3
.
3408
)
( 9
2



 s
s
A
which has roots at s = j1097.27 and s = -j1097.27.
Thus if the system operate with k = 273.57, the system response will be an undamped sinusoid with a
frequency of 1097.27 rad/sec.
3.4 STEADY STATE RESPONSE – STEADY STATE ERROR
One of the objectives of most control systems is that the system output response follows a specific
reference signal accurately in the steady state. Steady-state error is the difference between the output
and the reference in the steady state. Steady-state errors in control systems are almost unavoidable and
generally derive from the imperfections, frictions, and the natural composition of the system. In the
design problem, one of the objectives is to keep the steady-state error below a certain tolerable value.
3.4.1 Definition of the Steady-State Error with respect to System Configuration
Let us refer to the closed-loop system shown in Fig. 3.11, where r(t) is the input, e(t) the actuating signal,
and y(t) is the output. The error of the system may be defined as:
Figure 3.11: Closed-Loop Control System.

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)
(
signal
reference
)
( t
y
t
e 
 (3.26)
where the reference signal is the signal that the output is to track. When the system has unity feedback
(i.e. H(s) = 1), the error is simply
)
(
)
(
)
( t
y
t
r
t
e 

The steady-state error is defined as
)
(
1
)
(
lim
)
(
lim
)
(
lim
0
s
0
s
G
s
sR
s
sE
t
e
e
s
t
ss








(3.27)
Clearly, ess depends on the characteristics of G(s). More specifically, ess depends on the number of poles
that G(s) has at s = 0. This number is known as the system type. Fig. 3.12 shows steady state errors for
different input functions.

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Figure 3.12: Steady-state errors (a) step input, (b) ramp input
Now let us investigate the effects of the types of inputs on the steady-state error.

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3.4.2 Steady-State Error of System with a Step-Input Function
When the input r(t) to a control system with unity-feedback is a step function with magnitude A, then
R(s) = A/s and the steady-state error is written from Eq. (4.27),
)
(
lim
1
)
(
1
lim
)
(
1
)
(
lim
0
0
0 s
G
A
s
G
A
s
G
s
sR
e
s
s
s
ss


 




 (3.28)
For convenience, we define
)
(
lim
0
s
G
k
s
p


as the step-error constant. Then Eq. (4.28) becomes
p
ss
k
A
e


1
(3.29)
We can summarize the steady-state error due to a step-function input as follows:
 Type 0 system:
p
ss
k
A
e


1
= constant
 Type 1 or higher system: ess = 0
3.4.3 Steady-State Error of System with a Ramp-Input Function
When the input to the unity-feedback control system is a ramp function with amplitude A,
)
(
)
( t
Atu
t
r s

where A is a real constant, the Laplace transform of r(t) is
2
)
(
s
A
s
R 
The steady-state error is written using Eq. (4.27) as follows:

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)
(
lim
)
(
lim
0
0 s
sG
A
s
sG
s
A
e
s
s
ss




 (3.30)
We define the ramp-error constant as
)
(
lim
0
s
sG
k
s
v


Then Eq. (3.30) becomes
v
ss
k
A
e  (3.31)
The following conclusions may be stated with regard to the steady-state error of a system with ramp
input:
 Type 0 system: ess = 
 Type 1 system: ess =A/kv = constant
3.4.4 Steady-State Error of System with a Parabolic Input
When the input is described by the standard parabolic form
)
(
2
)
(
2
t
u
At
t
r s

The Laplace transform of r(t) is
3
)
(
s
A
s
R 
The steady-state error of the system is
)
(
lim 2
0
s
G
s
A
e
s
ss

 (3.32)

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UiTM Shah Alam
Defining the parabolic-error constant as
)
(
lim 2
0
s
G
s
k
s
a

 (3.33)
the steady-state error becomes
a
ss
k
A
e  (3.34)
The following conclusions are made with regard to the steady-state error of a system with parabolic
input:
 Type 2 system: ess = A/ka = constant
Example 3.5: Find the steady state errors of the following system
1
)
5
.
0
)(
5
.
1
(
)
15
.
3
(
)
( H(s)
s
s
s
s
k
s
G 




It is clear that this system is a type 1 system. The steady-state errors are:
Step input Step-error constant, kp =  ess = A/1+kp = 0
Ramp input Ramp-error constant, kv = 4.2k ess= A/kv = A/(4.2k)
Parabolic input Parabolic-error constant, ka = 0 ess = A/ka = 
3.4.5 Steady-State Error for Non-unity Feedback System
For non-unity feedback control, we usually find the equivalent unity-feedback system, as shown in Fig.
3.13.

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Figure 3.13: Forming an equivalent unity feedback for nonunity feedback system.
We have to take into consideration, that the above steps require that input and output of the same
units. The following example summarizes the concepts of steady-state error, system type, and the
steady state errors.
Example 3.6: For the system shown in Fig. 4.14, find the system type and the steady state error for the
unit step function. Assume input and output units are the same.

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UiTM Shah Alam
Figure 3.14: Nonunity feedback control system for Example 3.6.
The first step in solving the problem is to convert the system of Fig. 3.14 into an equivalent unity
feedback system. Using the equivalent forward transfer function of Fig. 3.13(e) along with
)
10
(
100
)
(


s
s
s
G
and
5
1
)
(


s
s
H
we find
400
50
15
)
5
(
100
)
(
)
(
)
(
1
)
(
)
( 2
3








s
s
s
s
s
G
s
H
s
G
s
G
s
Ge
Thus, the system is type 0, and
4
5
400
5
100
)
(
lim
0







s
G
k e
s
p
The steady-state error is
4
1
1




p
ss
k
e

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3.5 FREQUENCY RESPONSE ANALYSIS
In practice, the performance of a control system is measured more realistically by its time-domain
characteristics. The reason is that the performance of most control systems is judged based on the time
response due to certain test signals.
In design problems, there are no unified methods of arriving at a designated system that meets time-
domain performance specifications. On the other hand, in frequency domain, a wealth of graphical and
other techniques are available that are useful for system analysis and design, irrespective of the order of
the system.
It is important to realize that there are correlating relations between the frequency- and time-domain
performances in linear system so that time-domain properties of the system can be predicted based on
the frequency–domain characteristics. With these in mind, we shall study the frequency response
analysis of control systems.
3.5.1 Frequency Response of a System
It is well known from linear system theory that, when the input to a linear time invariant system is
sinusoidal with amplitude R and frequency o, i.e.,
t
ω
R
r(t) o
sin

the steady-state output of the system, y(t), will be a sinusoid with the same frequency o, but possibly
with different amplitude and phase; i.e.
φ)
t
(ω
Y
y(t) o 
 sin
where Y is the amplitude of the output sine wave and  is the phase shift.
Let the transfer function of a SISO system be M(s); the output Y(s) and the input R(s) are related through
M(s)R(s)
Y(s) 

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For sinusoidal steady-state analysis, we replace s by j, and equation (6.3) becomes





 )
) 

 j
R
j
Μ
)
Y(j
By writing Y(j) and M(j) as (similar expression for )

j
R also):
)
(
)
(
)
( 

 j
Y
j
Y
j
Y 

)
(
)
(
)
( 

 j
M
j
M
j
M 

)
(
)
(
)
( 

 j
R
j
M
j
Y 
and the phase relation:
)
(
)
(
)
( 

 j
R
j
M
j
Y 




Thus, for the input and output signals described by equations (6.1) and (6.2),
R
j
M
Y o )
( 

)
( o
j
M 
 

Thus, by knowing the transfer function M(s), the frequency response of the system can be obtained.
The frequency response of the loop transfer function G(s)H(s) [G(s) if H(s) is unity] can be plotted in
several ways. The two commonly used representations are:
a. Bode diagram, or Logarithmic plot.
b. Polar plot, or Nyquist plot.

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3.5.2 Frequency Response – Bode Diagram
A Bode diagram consists of two graphs. One is a plot of the logarithm of the magnitude of a sinusoidal
transfer function; the other is a plot of the phase angle; both are plotted against the frequency on a
logarithmic scale.
The standard representation of the logarithmic magnitude of G(j) is 20 log |G(j)|, where the base of
the logarithm is 10. The unit used in this representation is the decibel (dB). The curves are drawn on a
semilog paper, using the log scale for frequency and linear scale for either magnitude (in dB) or phase
angle (degrees).
The main advantage of Bode diagrams is that the multiplication of magnitudes can be converted into
addition. Furthermore, a simple asymptotic method is available for sketching the approximate curve.
Should the exact curve be desired, corrections could be made easily to these basic asymptotic plots.
In Bode diagrams, the frequency ratios are expressed in terms of octaves or decades. An octave is a
frequency band from 1 to 21, where 1 is any frequency. A decade is a frequency band from 1 to
101, where 1 is any frequency.
Basic Factors of G(j)H(j):
The basic factors that very frequently occur in an arbitrary open-loop transfer function G(j)H(j) are:
a. Constant gain, K.
b. Zeros and poles at the origin,   n
j

 .
c. Simple zeros and poles,   1
1

 T
j .
d. Quadratic factors,    
 1
2
/
/
2
1


 n
n j
j 



 .
a. Real Constant: G(s)H(s) = K
G(jω)H(jω) = K
Magnitude: |G(jω)H(jω)| (dB) = 20 log10 |K| (dB).
Phase angle: G(jω)H(jω) = 0°.

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The Bode plot for any value of K is shown in Fig. 6.1.
Figure 6.1: Bode plot for gain K.
b. Poles and zeros at the origin: G(s)H(s) = s±n
For sn
:
G(jω)H(jω)=(jω)n
Magnitude: |G(jω)H(jω)| (dB) = 20n log10 |jω| (dB) = 20n log10 ω (dB) (6.11)
Phase angle: G(jω)H(jω) = 90n° (a constant).
For s-n
:
G(jω)H(jω)=(jω)-n
Magnitude: |G(jω)H(jω)| (dB) = −20n log10 |jω| (dB) = −20n log10 ω (dB) (6.12)
Phase angle: G(jω)H(jω) = −90n° (a constant).
The Bode magnitude plots are a straight line in semi log coordinate. The slope of the line is ±20n
dB/decade i.e. the magnitude change by ±20n dB for the frequency change of 10 times. The straight line
passes through 0 dB at  = 1. The phase angle () of ±j is constant and equal to ±900
.The Bode plots
are shown in Fig. 6.2.

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UiTM Shah Alam
Figure 6.2: Bode diagrams for (a)G(j) = 1/j (b)G(j) = j.
c. Simple zeros and poles: G(s)H(s) = (1+sT)±1
G(jω)H(jω)= (1+ jωT)±1
Magnitude:
|G(jω)H(jω)| (dB) = ±20 log10 |1 + jωT| (dB)
= ±20 log10 √*1 + ω2
T2
] (dB) (6.11)

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UiTM Shah Alam
To obtain asymptotic approximation we consider both very large and very small values of . For low
frequencies, such that T << 1, the log magnitude may be approximated by
dB
T 0
1
log
20
1
log
20 2
2




 
For high frequencies, such that T >> 1,
dB
T
T log
20
1
log
20 2
2

 



At =1/T, log magnitude = 0 dB while at =10/T, log magnitude = ±20 dB. Thus, the value of
T

log
20
 increases/decreases with 20 dB/decade. Hence, the magnitude plot can be approximated
by two straight-line asymptotes, one a straight line at 0 dB for the frequency range 0 <  < 1/T and the
other a straight line with slope ±20 dB/decade for the frequency range 1/T <  < . The frequency,
=1/T, at which the two asymptotes meet is called the corner frequency or break frequency.
Phase angle: G(jω)H(jω) = T

1
tan
 .
At corner frequency, G(jω)H(jω) = ±45. The phase plot can be approximated by a straight line passing
through 0 at one decade below corner frequency and ±90 at one decade above corner frequency. The
Bode plots are shown in Fig. 6.3 and Fig. 6.4.
An advantage of the Bode diagram is that for reciprocal factors, for example the factor 1/(1+jT), the
log-magnitude and phase angle curves need only be changed in sign, since
T
j
T
j






1
log
20
1
1
log
20 and phase angle of 1/(1+jT) = T

1
tan
 =  (phase angle of
(1+jT).

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UiTM Shah Alam
Figure 6.3: Bode plot for (1+jT).

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UiTM Shah Alam
Figure 6.4: Log-magnitude curve (with asymptotes) for 1/(1+jT).
d. Quadratic factors:    
 1
2
/
/
2
1
)
(
)
(



 n
n s
s
s
H
s
G 


   
 1
2
/
/
2
1
)
(
)
(



 n
n j
j
j
H
j
G 






Magnitude:
|G(jω)H(jω)| (dB) = ±20 log10 |1 + 2ζ(jω/ωn)+ (jω/ωn)2
| (dB)
= ±20 log10
2
2
2
2
2
1 

















n
n 




(dB) (6.12)

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UiTM Shah Alam
If ζ > 1, this quadratic factor can be expressed as a product of two first-order factors with real
zeros/poles. If 0 < ζ < 1, this quadratic factor is the product of two complex-conjugate zeros/poles. The
asymptotic frequency response curves can be obtained as follows.
For low frequencies such that /n << 1, the log magnitude becomes ±20 log 1 = 0 dB. The low
frequency asymptote is thus a horizontal line at 0 dB. For high frequencies such that /n >> 1, the log-
magnitude becomes
n
n 



log
40
log
20 2
2


 dB. The equation for the high frequency asymptote is a
straight line with a slope of ±40 dB/decade.
The frequency n is the corner frequency. The two asymptotes just derived are independent of the value
of ζ. Fig. 6.5 shows exact curves with the straight-line asymptotes and the exact phase angle curves.
Phase angle of the quadratic factor is:
























 
2
1
1
2
tan
n
n





 (6.13)

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Figure 6.5: Bode plot for Eqn. (6.12) and (6.13).

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Example 6.1: Sketch the Bode plot for the following function:
2
1
)
(


s
s
G
Solution:
2
1
)
(


s
s
G ,
















2
1
1
2
1
2
1
)
(



j
j
j
G
Magnitude:
2
1
20log
-
2
1
log
20
)
(
log
20
2
















j
G
2
tan
)
( 1 
 


 j
G
Figure 6.6: Bode plots for the system in Example 6.1.
Frequency (rad/sec)
Phase
(deg);
Magnitude
(dB)
-25
-20
-15
-10
-5
10
-1
10
0
10
1
-80
-60
-40
-20
0
To:
Y(1)

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General Procedure for Plotting the Bode Diagrams:
 Rewrite the sinusoidal transfer function as a product of the basic factors discussed above.
 Identify the corner frequencies associated with these basic factors.
 Draw the asymptotic log-magnitude curves with proper slopes between the corner frequencies
considering all the basic factors together. The exact curve, which lies very close to the
asymptotic curve, can be obtained by adding contributions from all the factors and proper
corrections.
 Phase-angle curve can be drawn by adding the phase-angle curves of individual factors.
3.5.3 Polar Plot (Nyquist Plot)
The polar plot of a sinusoidal transfer function G(j) is a plot of the magnitude of G(j) versus the phase
angle of G(j) on polar coordinates as  is varied from zero to infinity. Note that, in polar plots, a
positive (negative) phase angle is measured counterclockwise (clockwise) from the positive real axis. The
polar plot is very often called the Nyquist plot in control system engineering. An example of such a plot
is shown in Fig. 6.7. Each point on the polar plot of G(j) represents the terminal point of a vector at a
particular value of . In the polar plot, it is important to show the frequency graduation of the locus.

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UiTM Shah Alam
Example of a Polar Plot.
a. Poles and zeros at the origin: G(s)H(s) = s±n
  1
)
(
)
(

 

 j
j
H
j
G
The polar plot of G(j)H(j) = 1/j is the negative imaginary axis since
  0
90
1
1
)
( 










j
j
j
H
j
G (6.14)
The polar plot of G(j)H(j) = j is the positive imaginary axis.

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UiTM Shah Alam
b. Simple zeros and poles: G(s)H(s) = (1+sT)±1
  1
1
)
(
)
(


 T
j
j
H
j
G 


For the sinusoidal transfer function
  T
T
T
j
j
H
j
G 



 1
2
2
tan
1
1
1
1
)
( 





 (6.15)
the values of G(j)H(j) at  = 0 and at  = 1/T are, respectively,
  0
0
1
)
0
(
0 

j
H
j
G and 0
45
2
1
1
1















T
j
H
T
j
G
If  approaches infinity, the magnitude approaches 0 and the phase angle approaches –900
. The polar
plot of this transfer function is a semicircle as the frequency is varied from 0 to . It is shown in Fig. 6.8.
The center is located at 0.5 in the real axis and the radius is equal to 0.5. The lower semicircle
corresponds to 


0 , and the upper semicircle corresponds to 0



  .
Figure 6.8: (a) Polar plot of 1/(1+jT) ; (b)Same plot in X-Y plane.

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The polar plot of the transfer function 1+jT is simply the upper half of the straight line passing through
the point (1, 0) in the complex plane and parallel to the imaginary axis as shown in Fig. 6.7.
Figure 6.7: Polar plot of 1+jT.
c. Quadratic factors:    
 1
2
/
/
2
1
)
(
)
(



 n
n s
s
s
H
s
G 


   
 1
2
/
/
2
1
)
(
)
(



 n
n j
j
j
H
j
G 






The low and high frequency portions of the polar plot of the following transfer function
 
   2
/
/
2
1
1
)
(
n
n j
j
j
H
j
G










are given, respectively, by
  0
0
0
1
)
(
lim 





j
H
j
G and   0
180
0
)
(
lim 







j
H
j
G
Thus, the high frequency portion is tangent to the negative real axis. The polar plots are shown in Fig.
6.8.
*Phase angle of the quadratic factor is the same as Eqn. (6.13):

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























 
2
1
1
2
tan
n
n






Figure 6.8: Polar plots of
   2
/
/
2
1
1
n
n j
j 



 

for ζ > 0.
Next, consider the following transfer function:
     






















n
n
n
n
j
j
j
j
H
j
G











2
1
/
/
2
1
)
(
2
2
2
The low-frequency portion of the curve is:
  0
0
0
1
)
(
lim 





j
H
j
G
and the high-frequency portion of the curve is:

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  0
180
)
(
lim 






j
H
j
G
The general shape of the polar plot is shown in Fig. 6.9.
Figure 6.9: Polar plot of    2
/
/
2
1 n
n j
j 



 
 for ζ > 0.
Example 6.2: Draw polar plot of
2
1
)
(


s
s
G
Solution: First substitute s = j in G(s).
2
1
)
(




j
j
G
2
tan
)
(
4
1
)
( 1
2



 



 j
G
j
G

102
UiTM Shah Alam
Figure 6.10: Polar plot of G(s) in Example 6.2.
Example 6.3: Draw polar plot for the system with
)
2
)(
1
(
10
)
(



s
s
s
s
G
Solution:
)
2
)(
1
(
10
)
(







j
j
j
j
G
)
2
(
)
1
(
10
)
(
2
2







j
G


 1
1
tan
2
tan
90
)
( 





 
j
G

103
UiTM Shah Alam
The plot is shown in Fig. 6.11.
Figure 6.11: Polar plot for the system in Example 6.3.
NYQUIST STABILITY TEST – THE CAUCHY CRITERION
The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex
plane, and mapping it through a complex function G(s), the number of times, N, that the plot of G(s)
encircles the origin is equal to the number of zeros, Z, of G(s) enclosed by the frequency contour minus
the number of poles, P, of G(s) enclosed by the frequency contour.
N = Z – P
Encirclements of the origin are counted as positive if they are in the same direction as the original closed
contour or negative if they are in the opposite direction.
When studying feedback control, we are not as interested in G(s)H(s) as in the closed-loop transfer
function G(s)H(s)/[1+G(s)H(s)]
If 1+G(s)H(s) encircles the origin, then G(s)H(s) will enclose the point -1. Since we are interested in the
closed-loop stability, we want to know if there are any closed-loop poles (zeros of 1+G(s)H(s)) in the
right-half plane.

104
UiTM Shah Alam
The Nyquist Stability Criterion usually written as Z = P + N, where
> Z is the number of right hand plane poles for the closed loop system (or zeros of
1+G(s)H(s))
> P is the number of open-loop poles (in the RH side of the s-plane) of G(s)H(s) (or poles of
1+G(s)H(s)), and
> N is the number of clockwise encirclements of (-1,0)
“A feedback control system is stable if and only if the number of counter-clockwise encirclements of the
critical point (-1,0) by the GH polar plot is equal to the number of poles of GH with positive real parts.”
(Nyquist Stability Criterion Definition)
Example:
• Consider the unity feedback applied to the following system
G(s)=K/[s(s+3)(s+5)]
• The loop transfer function is
G(j)H(j)=K/[s(s+3)(s+5)]|K=1,s= j
• The number of open-loop poles in the RH side of the s-plane, P = __

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