FEE 411: CONTROL SYSTEMS A
Contents
1 Introduction to Control Systems
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Basic Concepts and Terminologies . . . . . . . . . . . . . . . . . .
1.2.1 Control Systems Models / Representation . . . . . . . . .
1.3 Concept of Feedback . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Open Loop System . . . . . . . . . . . . . . . . . . . . . .
1.3.2 Closed Loop (Feedback) System . . . . . . . . . . . . . . .
1.3.3 Comparison of Open Loop and Closed Loop Systems . . .
1.3.4 Effects of Feedback . . . . . . . . . . . . . . . . . . . . . .
1.3.5 System Classification . . . . . . . . . . . . . . . . . . . . .
1.3.6 Time varying and time invariant . . . . . . . . . . . . . . .
1.4 The Concept of Transfer Function . . . . . . . . . . . . . . . . . .
1.4.1 Poles (x) and Zeros (o) of the Transfer Function . . . . . .
1.4.2 Complex s-Plane and the Locations of Poles and Zeros . .
1.4.3 Laplace Transform . . . . . . . . . . . . . . . . . . . . . .
1.4.4 System Representation in Block Diagram . . . . . . . . . .
1.4.5 Transfer Function Model of Electrical Network . . . . . . .
1.4.6 Transfer Function Model of Mechanical Systems . . . . . .
1.4.7 Transfer Function Model for Mechanical Rotational System
1.4.8 Transfer Function Model of Electro-Mechanical Systems .
1.4.9 Transfer Function Model of field controlled D.C. Motor . .
1.4.10 Transfer Function of Armature Controlled D.C. Motor . .
1.4.11 State Variable Models . . . . . . . . . . . . . . . . . . . .
1.4.12 Block Diagram . . . . . . . . . . . . . . . . . . . . . . . .
1.4.13 Block Diagram Reduction . . . . . . . . . . . . . . . . . .
2 Time Response Analysis
2.1 Test Input Signals . . . . . . . . . . . . . . . .
2.2 First-Order System . . . . . . . . . . . . . . .
2.2.1 First-Order System Modelling . . . . .
2.2.2 Step Response of a First Order System
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Contents
2.3
Second Order System Modelling . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Time Response of a Second Order System . . . . . . . . . . . . .
2.3.2 Time Domain Specifications of a Second Order System . . . . . .
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3 Error Analysis
3.1 Steady state error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.1.1 Static Error Coefficient Constants . . . . . . . . . . . . . . . . .
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4 Concept of Stability & Routh-Hurwitz Criterion
4.1 The Concept of Stability . . . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Bounded Input Bounded Output (BIBO) (External) Stability
4.1.2 BIBO Stability via Impulse Response Function . . . . . . . . .
4.1.3 BIBO Stability and the Characteristic Roots . . . . . . . . . .
4.1.4 Effects Of Location Of Poles On Stability . . . . . . . . . . .
4.1.5 Routh–Hurwitz Criterion . . . . . . . . . . . . . . . . . . . . .
4.1.6 Routh–Hurwitz Criterion to Determine Absolute Stability . .
4.1.7 Formation of Routh’s Array . . . . . . . . . . . . . . . . . . .
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5 Root Locus
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5.0.1 Advantages of Root Locus Method . . . . . . . . . . . . . . . . . 101
5.1 The Concept of Root Locus . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.1.1 Properties of the Root Loci (Rules of the Root Loci) . . . . . . . 105
6 Frequency Response
6.1 Time Domain Analysis Vs Frequency Domain Analysis . . . . . . . .
6.1.1 Relationship between s and jω . . . . . . . . . . . . . . . . .
6.1.2 Frequency-Domain Specifications . . . . . . . . . . . . . . . .
6.1.3 Correlation Between Time Response and Frequency Response
6.2 Plotting Frequency Response . . . . . . . . . . . . . . . . . . . . . . .
6.2.1 Bode Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2.2 Bode Asymptotic Magnitude Plot . . . . . . . . . . . . . . . .
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Chapter 1
Introduction to Control Systems
Learning Outcomes
After completing this section, student should be able to:
• Explain and represent the system in open and closed-loop forms.
• Define various control system terminologies..
• Develop dynamic equations of mechanical, electrical and electromechanical systems
and derive their transfer functions.
• Derive the transfer function of commonly used control system components.
• Represent the system in block diagram and perform block diagram reduction.
1.1 Introduction
Control system is an interdisciplinary field. Control engineering is not limited to any
particular engineering but it is applicable to various fields of engineering such as chemical,
aeronautical, mechanical, electrical, civil and environmental engineering. It finds wide
applications in space / military such as; space vehicle systems, missile guidance systems,
robotic systems, anti-aircraft gun control and radar systems. It also finds applications
in manufacturing and industrial process control, autopilot systems, aerospace industries,
design of cars, trucks and war weapons, and in industrial operations such as controlling
pressure, temperature, humidity, viscosity, flow etc.
Why Control System
We build control systems for four primary reasons:
• Power amplification
1
1.2. BASIC CONCEPTS AND TERMINOLOGIES
Figure 1.1: (a) A plant to be controlled and (b) A plant with a controller
• Remote Control
• Convenience of input form
• Compensation for disturbances
1.2 Basic Concepts and Terminologies
The portion of a system that we want to control is called a plant or process. The input
signal which is applied to the system interacts with the plant and produces the output.
The objective of control engineer is to get desired output by providing appropriate input
for the plant under consideration, this is depicted in Fig. 1.1(a).
The task of the control system engineer is to ensure that the plant operates as intended.
When there is a disturbance in the system, a controller is placed between the plant /
process and the input. The controller output signals cause the plant / process to produce
the desired output. This is represented in Fig. 1.1(b).
Fig. 1.1 shows block diagram of interconnected control system components used in analysis
and design. The Arrows indicate the direction of signal flow. Complex systems are easier
to understand using block diagrams.
In discussing control systems, numerous terms are encountered, some of which are widely
employed. The definitions of these terminologies are provided below.
Definition 1. A system is defined as a set of interconnected objects with a definite
relationship between the objects and attributes.
Definition 2. Control systems are the systems that implement certain objectives.
The interconnection of connection of components forming a system configuration that
provides the desired output is also called control system.
2
1.2. BASIC CONCEPTS AND TERMINOLOGIES
Definition 3. Variables are the attributes that are present at different parts of the
system.
Definition 4. Input of a system is an attribute of a system which is varied directly
and independently by the operator.
Definition 5. Output of a system is an attribute that varies as a result of variation
in the input. It cannot be independently varied by the operator.
Definition 6. Output of a system is an attribute that varies as a result of variation in
the input. It cannot be independently varied by the operator.
Definition 7. System Parameters
Parameters are the properties of the given system which is inherently present in the
system.
Definition 8. Plant or Process
The device which is under control is called plant or process.
Definition 9. Automation
The control of a process by automatic means is called automation.
Definition 10. Servomechanism
The control system which follows the frequently changing reference signal is called servomechanism.
It is usually referred to system whose output is mechanical position or velocity.
Definition 11. Regulator System
If the reference or set point signal is kept at a steady value for a long period, and the
output remains at the prescribed set level rejecting disturbance signals, it is called a
regulator system.
Example 1.1. Define the following terms as used in control systems.
• System Ans: A system is defined as the interconnections of objects which have a
definite relationship between objects and attributes (variables)
• Control system Ans: A system, which is formed by interconnection of components
when it provides a desired response is called a control system
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1.3. CONCEPT OF FEEDBACK
1.2.1 Control Systems Models / Representation
Three basic models (representations) of components and systems are used in study of
control systems.
• Mathematical Models
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–
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Linear Differential Equations
Difference Equations
Laplace Transform
State-Space
Z- Transform
• Block diagram
• Signal Flow Graphs
1.3 Concept of Feedback
1.3.1 Open Loop System
Consider a DC motor speed control system schematic and its block diagram representation
in Fig. 1.2. The motor speed (ω) depends on the armature voltage (Va ), which is
adjusted via potentiometer. When motor load (L) increases, the speed (ω) drops, hence
to maintain the desired shaft speed, the armature voltage (Va ) needs to be raised. When
load (L) decreases, the speed (ω) increases, therefore armature voltage Va , is required
to be decreased so as to maintain the desired speed. Thus, adjusting the potentiometer
setting according to the load variations requires manual operation which may require
constant vigil on the part of the human operator which may sometime be not accurate.
This type of the system is called open loop system. In the above system Va is the
input and it causes control action, and ω, the speed is the output. In this system
when there is a change in the output, it does not have any influence on the control action.
We can define the open loop system as,
Definition 12. An open loop system is defined as the system in which the control action
(actuating signal) is independent of the output or the desired result.
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1.3. CONCEPT OF FEEDBACK
Figure 1.2: (a) D.C. motor speed control system and (b) Block diagram representation
of speed control system
1.3.2 Closed Loop (Feedback) System
Consider DC motor speed control with feedback mechanism employed by tachogenerator
(T) in Fig. 1.3. Tachogenerator is connected to the motor and load shaft, and generates a
voltage proportional to motor speed (ω). The output of T is compared with the reference
voltage Va , the difference of Va and VT (Va − VT ) is shown in Fig. 1.3(a). For this system
we can say that the output is feedback and compared with the reference input. This
kind of system is termed as feedback or closed loop system. For connection shown in the
figure, the system has a negative feedback.
The working principle of the feedback
Under no load condition and with feedback loop not connected (open), the refference
input (Va ) is adjusted using potentionmeter in order to set the desired motor speed
(ω).
When the load is connected,
• ω decreases
• VT decreases, since tachonegerator volatge is proportional to motor speed
• (Va − VT ) which is applied to the armature increases, this leads to increase in shaft
speed and is brought to the desired speed.
Suppose now, a portion of the load is removed,
• ω will increase
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1.3. CONCEPT OF FEEDBACK
Figure 1.3: (a) Closed loop D.C. motor speed control system and (b) Block diagram of
closed loop D.C. motor speed control system
• This increase will also cause an increase in VT
• Control action (Va − VT ) decreases which in turn decreases the shaft speed ω and
brought back to the desired speed.
From above setup and discussion we note that, disturbance in the load side, is rejected
automatically by the feedback connection and the desired output is restored may be with
marginal acceptable deviation. The control action (Va − VT ) is also called by the name
Actuating Signal or Error Signal.
Definition 13. A closed loop system is defined as a system in which the measured
output is compared with the reference input. Here the control action depends upon
the changes in the output.
1.3.3 Comparison of Open Loop and Closed Loop Systems
Open Loop Systems
Advantages
• Simple easy to build.
6
1.3. CONCEPT OF FEEDBACK
• Cheaper as less number component.
• Usually stable.
• Cost of maintenance is low.
Disadvantages
• Less accurate.
• If external disturbance are present, output differs significantly from the desired
value.
• If there are variations in the parameters of the system the output changes.
Close Loop Systems
Advantages
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•
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•
More accurate.
Effect of External disturbances signal can be very small.
The variations in parameters of the system do not affect the output of the system.
Speed of the response can be greatly increased.
Disadvantages
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•
•
•
Complex and Expensive.
Requires high forward path gains.
The systems are prone to instability. Oscillations in the output may occur.
Cost of maintenance is high.
Open-Loop systems Examples
• A traffic Light system: the signals change according to present time and are not
affected by the density of traffic on any road.
• A washing machine: quality of wash is not measured, every cycle like wash, rinse
and dry cycle goes according to preset timing.
• Bread toaster, humidity check, field control dc motor
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1.3. CONCEPT OF FEEDBACK
1.3.4 Effects of Feedback
We can summarize the effects of feedback as below,
• Negative feedback reduces the overall gain of the system while positive feedback
increases it. Further, in one frequency range, the overall gain is increased while in
another frequency range, the overall gain is reduced for the given feedback.
• Properly designed feedback system will improve the stability. Improper design
will make the system unstable.
• The sensitivity function can be made small by increasing the feedback gain. This
will remove the harmful effect of parameter variation.
• Feedback reduces the effect of noise and disturbances.
• Feedback increases the bandwidth of the system. This increases the fastness of
time response of the system in the time domain and increases the frequency range
of operation in the frequency domain
Exercise 1.2. The student - teacher learning process is inherently a feedback process
intended to reduce the system error to a minimum. With the aid of block diagram,
construct a feedback model of the learning process and identify each block of the system
Exercise 1.3. Draw the block diagram of a closed loop control system and indicate the
following on it: Plant, command input, controlled output, actuating signal and feedback
& control element.
Exercise 1.4. Given a sports team with the coaches, players etc., from a control point
of view discuss the different constituents of this system and their roles.
Exercise 1.5. Given an organization with its president, employees, etc., from a control
point of view discuss the different constituents of this system and their roles.
Exercise 1.6. Briefly explain the following
1. Open loop system
2. Closed loop system
3. Control action
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1.3. CONCEPT OF FEEDBACK
1.3.5 System Classification
For convenience of description and mathematical analysis, systems are classified according
to the nature of inputs, number of inputs/outputs, inherent characteristics, design solution
components used etc. Systems are broadly classified as given below:
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Linear and non-linear.
Time varying and time invariant.
Continuous time system and discrete time system.
Single input–single output (SISO) and multi-input–multi-output (MIMO).
Deterministic and stochastic system.
Lumped parameter and distributed parameter system.
Optimal/adaptive/fuzzy type/robust control systems.
Continuous time system and discrete time system
If the signals in all parts of a control system are continuous functions of time, the system
is classified as continuous time feedback control system. 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.
Linear and non-linear
If a system obeys superposition principle, the system is said to be a linear system. Let
x1 (t) and x2 (t) be two inputs to a system and y1 (t) and y2 (t) be the corresponding
outputs. For arbitrary real constants k1 and k2 , and for input k1 x1 (t) + k2 x2 (t), if the
output of the system is given by k1 y1 (t) + k2 y2 (t), then the system is said to be a linear
system. There are several simple techniques available for the analysis and design of
linear control systems. The principal of superposition and homogeneity is shown in
fig. 1.4. Any system which does not obey superposition principle is said to be a nonlinear system. Physical systems are in general non-linear and analysis of such systems is
very complicated.
x1 (t)
y1 (t) x2 (t)
G1
y2 (t)
G1
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1.3. CONCEPT OF FEEDBACK
k1 x1 (t) + k2 x2 (t)
k1 y1 (t) + k2 y2 (t)
G1
Figure 1.4: Linear System
1.3.6 Time varying and time invariant
Systems can be further classified depending on whether the parameters of the system are
constants, or varying with respect to time. When the input to a system is delayed by T
time, if the output is also delayed by the same time T , the system is said to be a time
invariant system.
y(t) x(t − T )
x(t)
G1
y(t − T )
G1
Figure 1.5: Time - Invariant System
Short Answer Type Questions
Define a system A system is defined as the interconnections of objects which have a
definite relationship between objects and attributes (variables).
What is a control system? A system, which is formed by interconnection of components
when it provides a desired response is called a control system.
What is an open loop system? An open loop system is one in which the control action
is independent of the output.
What is a closed loop system? A closed loop system is one in which the control action
depends upon the output.
What is control action? Actuating signal also otherwise called error signal is called
control action which actuates the power device directly or sometimes indirectly.
The actuating signal is the difference of the reference input and part of the output
or full output.
What are the advantages of open loop system? Open loop systems are simple to design.
Component requirements are less and therefore less costly. Almost all physically
realizable open loop systems are stable.
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1.4. THE CONCEPT OF TRANSFER FUNCTION
What are the drawbacks of open loop systems? Open loop systems are manually controlled
and hence accuracy is less or it depends upon the ingenuity of the human operator.
The time response of the system is poor (sluggish operation). Its operation requires
close manual inspection during any internal and external disturbances.
What are the advantages of closed loop systems? Closed loop systems are very fast
acting and more accurate. It rejects internal and external disturbances automatically
without human involvement. Since the bandwidth is high, the range of frequency
over which the system responds is high. If the open loop system is unstable, it is
possible to make the closed loop system stable by proper feedback.
What are the drawbacks of closed loop systems? Closed loop systems tempt to oscillate.
System complexity is increased by addition of components. This increases the cost
and reduces the stability limit. The over all gain of the system is reduced.
What is an input? Input is an attribute of the system which can be varied independently.
What is an output? Output is an attribute which can be varied by varying the input.
Why positive feedback is not used in closed loop system? For positive feedback systems,
the characteristic equation is 1 − GH(s) = 0 which will yield poles in the right half
s-plane. This makes the closed loop system unstable. That is why positive feedback
is not employed in closed loop system. However, positive feedback may be employed
in the design of oscillators which has sustained oscillations.
1.4 The Concept of Transfer Function
The dynamic behavior of a simple system such as RC circuit, may be represented as
differential equation relating input and output. However, this representation is not
suitable for system analysis when subsystems are connected in cascade. To convert the
time domain model to the frequency domain model of a linear time-invariant continuous
time system, the Laplace transform1 is used. This transform converts differential
equations into algebraic expressions, making them easier to manipulate. The Laplace
transform converts functions with a real dependent variable such as time into functions
with a complex dependent variable such as frequency which is often represented by s
which is a complex variable and expressed as s = α + jω. The transfer function is defined
as,
1
https://en.wikipedia.org/wiki/Laplace_transform
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1.4. THE CONCEPT OF TRANSFER FUNCTION
Definition 14. The transfer function of a linear time invariant 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.
Consider electrical network in fig. 1.6, by KVL we have,
Z
1
vi (t) = (R1 + R2 )i(t) +
i(t) dt
C
Z
1
vo (t) = R1 i(t) +
i(t) dt
C
(1.1)
(1.2)
Taking Laplace transform on both side of eqs. (1.1) and (1.2), we get
Vi (s) = (R1 + R2 )I(s) +
Vo (s) = R1 I(s) +
I(s)
sC
(1.3)
I(s)
sC
(1.4)
Simplifying above,
Vo (s)
= G(s)
Vi (s)
Vo (s)
sR2 C + 1
=
Vi (s)
s(R1 + R2 )C + 1
(1.5)
(1.6)
The transfer function G(s) of system described by eq. (1.5) is given in eq. (1.6) and is
represented in the block diagram as shown in fig. 1.7.
R1
+
R2
vi (t)
+
vo (t)
C
−
−
Figure 1.6: Electrical Network
Vi (s)
Vo (s)
G(s)
Figure 1.7: Block diagram representation of transfer function
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1.4. THE CONCEPT OF TRANSFER FUNCTION
Note:
From fig. 1.7, it is noted that;
• the transfer function is independent of the input.
• the transfer function it is not a function of the real variable, time or any other
independent variable.
• transfer function of a continuous data system is expressed as the function of the
complex variable s together with the system parameter,
• the transfer function of a discrete time system (described by difference equation) is
the function of z.
Consider transfer function eq. (1.7), where numerator and denominator are in the form
of polynomials in s,
G(s) =
am sm + am−1 sm−1 + am−2 sm−2 + · · · + a0
bn sn + bn−1 sn−1 + bn−2 sn−2 + · · · + b0
(1.7)
Note:
• If the order of the denominator polynomial is greater than that of the numerator
polynomial (n > m) the transfer function is said to be strictly proper.
• If the order of numerator polynomials is the same as that of the denominator
polynomial (n = m), is called proper transfer function.
• If the order of the denominator polynomial is less than that of the numerator
(n < m), such a transfer function is called improper.
• when the denominator polynomial is set to zero, the equation so obtained is called
characteristic equation.
bn sn + bn−1 sn−1 + bn−2 sn−2 + · · · + b0 = 0
(1.8)
Equation (1.8) is known as characteristic equation since it determines the stability of
the system, the transient and steady state responses, also depends upon the characteristic
equation.
1.4.1 Poles (x) and Zeros (o) of the Transfer Function
The transfer function in eq. (1.7), it also be written in factor form as,
G(s) =
(s + z1 )(s + z2 ) . . . (s + zm )
(s + p1 )(s + p2 ) . . . (s + pn )
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(1.9)
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.8: s-plane
Note: Zeros
In eq. (1.9),
• the factors in the numerator are called zeros of the transfer function and located
in the complex s-plane at s = −z1 , s = −z2 , . . . .
• the zero locations are marked with a small circle ◦.
• at s = −z1 , s = −z2 etc, the transfer function becomes zero.
• Hence, the factors of the numerator of the transfer function which are called zeros
are defined as “the points in the complex s-plane at which the transfer
function becomes zero”
Note: Poles
In eq. (1.9),
• the factors in the denominator are called poles of the transfer function and located
in the complex s-plane at s = −p1 , s = −p2 , . . . .
• the poles locations are marked with a small cross ×.
• at s = −p1 , s = −p2 etc, the transfer function becomes infinitive.
• Hence, the factors of the denominator of the transfer function which are called poles
are defined as “the points in the complex s-plane at which the transfer
function becomes infinitive”.
1.4.2 Complex s-Plane and the Locations of Poles and Zeros
Figure 1.8 shows complex variable s = α + jω and the s-plane.
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1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.9: s-plane
•
•
•
•
The horizontal axis represents the positive and negative real parts
The vertical axis represents positive and negative imaginary parts.
It s-plane contains four quadrants as shown in fig. 1.8.
The infinite space complex s-plane is divided by the vertical imaginary as Left Half
Plane (LHP) and Right Half Plane (RHP).
• This division plays significant role in the system stability and the transient response
studies which.
Consider transfer function, eq. (1.10),
G(s) =
10(s − 2 + j3)(s − 2 − j3)(s + 1)
s(s + 2)2 (s − 3 + j)(s − 3 − j)
(1.10)
From eq. (1.10) we note,
• Three zeros and five poles.
• The order of the numerator polynomial is three and that of the denominator is five.
Hence, system eq. (1.10) is strictly proper.
• In all practically realizable system the order of the denominator polynomial is higher
than that of the numerator polynomial.
• The characteristic equation is s(s + 2)2 (s − 3 + j)(s − 3 − j) = 0.
• System eq. (1.10) is called fifth order system.
The order of the system is defined as the highest order of the denominator polynomial
or the numerator polynomial whichever is greater. Figure 1.9 show location of
poles and zeros in s-plane for eq. (1.10). The pole s = 0, which is called a pure integrator
is located at the origin in the s-plane. By the number of poles located at the origin, the
systems are designated by its Type. Since there is one pole located at the origin of the
s-plane, system eq. (1.10) is called Type one system. If no pole is located at the origin
the system is designated as Type zero and so on.
15
1.4. THE CONCEPT OF TRANSFER FUNCTION
Definition 15. The Type of the system is defined by the number which represents
the presence of pure integrators in the transfer function, alternatively, defined
as the number of poles of the transfer function located at the origin of the
s-plane.
Table 1.1: Common Laplace transform pairs
f (t)
F (s)
1
Unit impulse δ(t)
2
Unit step u(t) = 1
3
Unit ramp r(t) = t
5
dn f (t)
Derivative
Z dtn
Integral f (t)
6
tn
7
e−at
8
1 − e−at
9
sin ωt
10
cos ωt
11
e−at sin ωt
12
e−at (cos ωt −
4
1
1
s
1
s2
sn F (s)
F (s)
s
n!
sn+1
1
s+a
a
s(s + a)
ω
(s2 + ω 2 )
s
(s2 + ω 2 )
ω
(s + a)2 + ω 2
s
(s + a)2 + ω 2
a
sin ωt)
ω
1.4.3 Laplace Transform
Laplace transform is a mathematical tool which transform a function from time domain
to frequency domain and vice versa. Laplace transform of a function f (t) is defined
as,
Z
∞
f (t)e−st dt
L[f (t)] = F (s) =
0
where s = α + jω is a complex variable.
Laplace transform simplifies the solution of a constant coefficient differential equations,
since it reduces it’s solution to that of solving linear algebraic equations. Transient
and steady states components of the solution can be determined from the transformed
equation. Common Laplace Transform Pairs2 are given in table 1.1. Among the properties
of Laplace transform that we will apply in coming chapters are Initial Value Theorem
(IVT) and Final Value Theorem given (FVT) as,
2
https://en.wikipedia.org/wiki/List_of_Laplace_transforms
16
1.4. THE CONCEPT OF TRANSFER FUNCTION
Initial Value Theorem (IVT)
f (0) = lim f (t) = lim sF (s)
t→0
(1.11)
s→∞
Final Value Theorem (FVT)
f (∞) = lim f (t) = lim sF (s)
t→∞
(1.12)
s→0
1.4.4 System Representation in Block Diagram
Block diagram provides a pictorial representation of a system. Transfer function of any
system can be represented as a block diagram, these operational blocks are unidirectional.
Systems are often represented in a simple block diagram as shown in fig. 1.10 where the
symbols are defined as,
R(s)
+
−
E(s)
C(s)
G(s)
GH(s)
H(s)
Figure 1.10: System representation in simple block diagram
R(s)
G(s)
1+G(s)H(s)
C(s)
Figure 1.11: Simplified block diagram
•
•
•
•
•
•
R(s) = Laplace transform of the input variable r(t).
C(s) = Laplace transform of the output variable c(t).
E(s) = Laplace transform of the error signal e(t).
G(s) = Forward path or open loop transfer function.
H(s) = Feedback path transfer function.
GH(s) = Loop transfer function.
17
1.4. THE CONCEPT OF TRANSFER FUNCTION
The transfer function in fig. 1.11 can obtained from fig. 1.10 as follows,
E(s) = R(s) − C(s)H(s)
E(s)G(s) = C(s)
{R(s) − C(s)H(s)} G(s) = C(s)
Simplifying,
G(s)
C(s)
=
R(s)
1 + G(s)H(s)
(1.13)
The transfer function relating R(s) and C(s) in eq. (1.13) is termed the Closed-loop
transfer function.
1.4.5 Transfer Function Model of Electrical Network
The transfer function of electric circuits is determined using differential equations, which
connect input and output using Kirchhoff’s laws. These equations are written for the sum
of voltages around loops or currents at nodes, and equated to zero. The transfer function
can be obtained using Laplace transforms. Table 1.2 provides components, voltage and
current relationships.
Table 1.2: Voltage-current relationship
Component
Capacitor, C
Inductor, L
Resistor, R
v−i
Z
1
v(t) =
i(t) dt
C
Ldi(t)
v(t) =
dt
v(t) = Ri(t)
i−v
i(t) =
i(t) =
1
L
Cdv(t)
Z dt
i(t) =
v(t) dt
v(t)
R
Example 1.7. Find the T.F. of the network shown in fig. 1.12.
Solution example 1.7
Using v − i relationship from table 1.2 and applying KVL to mesh 1 and mesh 2, we get
mesh equations as,
18
1.4. THE CONCEPT OF TRANSFER FUNCTION
R1
R2
+
+
vi (t) i1 (t)
i2 (t)
C
L
−
vo (t)
−
Figure 1.12: Circuit for example 1.7
Mesh 1
1
vi (t) = R1 i1 (t) +
C
Z
(i1 (t) − i2 (t)) dt
(1.14)
Mesh 2
1
Ldi2 (t)
+
0 = R2 i2 (t) +
dt
C
Z
(i2 (t) − i1 (t)) dt
(1.15)
Taking Laplace transform of eqs. (1.14) and (1.15) we get,
1
I2 (s)
V1 (s) = R1 +
I1 (s) −
sC
sC
(1.16)
1
I1 (s)
I2 (s) R2 + sL +
=
sC
sC
(1.17)
and
respectively. Rearranging eq. (1.17) as I2 (s) sR2 C + s2 LC + 1 = I1 (s) and substituting
I1 (s) into eq. (1.16), we get
I2 (s)
1
V1 (s) = R1 +
I2 (s) sR2 C + s2 LC + 1 −
sC
sC
2
{(sR1 C + 1)(s LC + sR2 C + 1) − 1} I2 (s)
=
sC
(1.18)
(1.19)
Output Voltage is given as,
di2 (t)
dt
(1.20)
Vo (s) = sLI2 (s)
(1.21)
vo (t) = L
The Laplace transform
19
1.4. THE CONCEPT OF TRANSFER FUNCTION
Substituting I2 (s) from eq. (1.21) into eq. (1.18), we obtain the transfer function as,
Vo (s)
sL
= 2
Vi (s)
s R1 LC + s(R1 R2 C + L) + (R1 + R2 )
Exercise 1.8. Figure 1.13 shows a band-pass filter, obtain transfer function
(1.22)
Vo
(s).
Vi
R
+
vi (t)
+
L
C
−
vo (t)
−
Figure 1.13: Circuit for exercise 1.8
Answer to exercise 1.8
Vo
(s) =
Vi
sL
R
sL
s2 LC +
+1
R
1.4.6 Transfer Function Model of Mechanical Systems
Mechanical systems, like electrical networks, consist of three passive linear components.
Mechanical Systems are classified into mechanical translational and mechanical rotational
systems. In the translational system, linear displacement, velocity, and acceleration are
the variables of motion, with force as the input. The translational system uses mass,
spring, and a viscous damper to store energy, while the rotational system uses angular
displacement, velocity, and acceleration as variables. Gear arrangements are used as
matching devices to alter torque, angular speed, and displacement, allowing energy to
be transmitted between parts for maximum power transfer. Both systems utilize passive
elements like levers and matching devices to achieve optimal performance.
Transfer Function Model for Mechanical Translational System
The basic elements of the translational systems are mass, spring, dash-pot and lever.
Newton law of motion is applied according to which the algebraic sum of the forces
acting at a point is zero.
20
1.4. THE CONCEPT OF TRANSFER FUNCTION
•
•
•
•
•
•
•
Mass, M (kg).
Viscous friction coefficient, B, N/m/sec.
Spring constant, K, N m−1 .
Force, f (t), (N).
Displacement, x(t), (m).
Velocity, v(t), m/sec.
Acceleration, a(t), m/sec2 .
Table 1.3: mass, spring, dash-pot relation
Component
Relation
M d2 x(t)
= f (t)
dt2
Bdx(t)
= f (t)
dt
Mass (M )
Dash - pot (B)
Spring (K)
Kx(t) = f (t)
Example 1.9. Consider the mechanical system shown in fig. 1.14. The applied force
f (t) is the input and the displacement x(t) of mass M is the output variable. Obtain the
T.F.
Solution to example 1.9
f (t) =
M d2 x(t) Bdx(t)
+
+ Kx(t)
dt2
dt
Taking Laplace transform
F (s) = s2 M X(s) + sBX(s) + KX(s)
X(s)
1
= 2
F (s)
s M + sB + K
1.4.7 Transfer Function Model for Mechanical Rotational System
The behaviour of mechanical rotational system resembles with that of translational
system. The basic difference between the two is that rotational system has angular
displacement while translational system has linear displacement. Further, instead of
21
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.14: (a) Translational mechanical system and (b) Free-body diagram
force, torque is applied in the rotational system. Mass is replaced by inertia. The basic
elements of the rotational system consists of inertia, rotational dash-pot and spring. The
gear arrangements are used as matching device in rotational system which is used to alter
torque, angular speed, angular displacement and transfer maximum power.
The units of the above elements and the associated variables are given below:
•
•
•
•
•
•
•
Inertia, J = Kg − m2 .
Viscous friction coefficient, B = N − m/rad/s.
Torsional spring, K = N − m/rad.
Torque, T (t) = N − m.
Angular displacement, θ(t) = radians(rad).
Angular velocity, ω(t) = rad/s.
Angular acceleration, a(t) = rad./s2 .
The dynamic behaviour of these elements due to the torque applied are explained below
in fig. 1.15.
Example 1.10. Consider the mechanical rotational translational system shown in Fig.1.35a.
Find the T.F. relating the angular displacement of inertia J and the applied torque T (t)
and angular displacement of θ1 and T (t).
Solution for example 1.10
T1 (t) = J
dθ(t)
d2 θ(t)
+
B
+ Kθ(t)
dt2
dt
Laplace transform
T1 (s) = s2 Jθ(s) + sBθ(s) + Kθ(s)
22
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.15: Torque acting on mechanical rotation system components and their freebody diagrams
Figure 1.16: (a) Mechanical system with gear arrangement for example 1.10. (b) Freebody diagrams of (a)
23
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.17: (a) Solenoid coil for example 1.11, (b) Equivalent electrical circuit and
mechanical freebody diagram of (a)
also
T1 N1 = T N2 → T1 =
N2
T
N1
N2
T = s2 Jθ(s) + sBθ(s) + Kθ(s)
N1
N1
T =
θ(s) s2 J + sB + K
N
2
N2
1
θ(s)
=
2
T (s)
N1 (s J + sB + K)
also
N1
N2
N1 θ1 = N2 θ → θ(s) =
θ1 (s)
2
N2
1
θ1 (s)
=
2
T (s)
N1
(s J + sB + K)
1.4.8 Transfer Function Model of Electro-Mechanical Systems
Example 1.11. Consider the solenoid shown in fig. 1.17. The force of attraction generated
is Kf i(t) N. The back emf constant of the solenoid coil is Kb V/m/sec. Derive the T.F.
relating the displacement x(t) of the M of the solenoid core and the applied voltage e(t).
24
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.18: (a) Solenoid coil for exercise 1.12
Solution for example 1.11
di(t)
+ Ri(t) + eb (t)
dt
di(t)
dx(t)
=L
+ Ri(t) + Kb
dt
dt
E(s) = sLI(s) + RI(s) + sKb X(s) = (sL + R)I(s) + sKb X(s)
e(t) = L
also
M d2 x(t) Bdx(t)
+
+ Kx(t) = Kf i(t)
dt2
dt
s2 M X(s) + sBX(s) + KX(s) = Kf I(s)
(s2 M + sB + K)
+ sKb X(s)
E(s) = (sL + R)
Kf
X(s)
Kf
=
2
E(s)
(sL + R)(s M + sB + K) + sKf Kb
Exercise 1.12. Consider the solenoid coil shown in fig. 1.18. The coil is connected to the
voltage source via an R-C filter as shown. Derive the T.F. relating X(s)/E(s). Solution
Kf
X(s)
= 2
2
E(s)
(s M + sB + K) [LCR1 s + s(R1 RC + L) + (R1 + R) + sKf Kb (1 + sR1 C)]
1.4.9 Transfer Function Model of field controlled D.C. Motor
The field controlled D.C. motor is shown in fig. 1.13. The symbols and notations have
the following meanings:
• J = Moment of inertia of the motor in kg m−2 .
• f = Friction coefficients N m rad−1 s−1 .
25
1.4. THE CONCEPT OF TRANSFER FUNCTION
•
•
•
•
•
•
Rf , Lf = Field coil resistance and inductance respectively.
Ra , La = Armature coil resistance and inductance respectively.
ia (t) = Armature current
vf (t) = Voltage applied across the field coil to produce flux.
= Motor shaft speed in rad s−1 .
ω(t) = dθ(t)
dt
θ(t) = Motor shaft position in rad.
Figure 1.19: Field controlled D.C. motor
Air-gap flux of the motor is proportional to the field current if (t) and is given by,
ϕ = Kf if (t)
Torque developed by the motor which is proportional to the field flux and the armature
current is given by,
Td = Ka ϕia (t) = Ka Kf if (t)ia (t)
For field control DC motor, armature current is constant, therefore,
Td = Ka Kf ia if (t) = Kt if (t)
Td = Kt if (t)
This developed torque is used to drive the system having the total inertia J and to
overcome the friction f . This torque is expressed as,
TL (t) = J
dθ(t)
d2 θ(t)
+f
2
dt
dt
Assuming no loss, then Td (t) = TL (t).
d2 θ(t)
dθ(t)
Kt if (t) = J
+f
2
dt
dt
Taking Laplace transform,
Kt If (s) = s2 Jθ(s) + sf θ(s) = s (sJ + f ) θ(s)
26
1.4. THE CONCEPT OF TRANSFER FUNCTION
Kt If (s) = s (sJ + f ) θ(s)
Writing KVL equation of the electrical side of DC field-controlled motor and apply
Laplace transform, we get,
Vf (s)
If (s) =
Rf + sLf
Therefore,
Kt
Vf (s)
= s (sJ + f ) θ(s)
Rf + sLf
Rearranging above,
θ(s)
Kt
=
Vf (s)
s(Rf + sLf )(sJ + f )
Abvoe can further simplified by factoring Rf and f resulting,
K
θ(s)
=
Vf (s)
s(1 + sTe )(1 + sTm )
Where
K=
Lf
J
Kt
, Te =
and Tm =
f Rf
Rf
f
Te and Tm are known as Electrical and Mechanical time constants respectively. The block
diagram representation of field controlled D.C. motor is shown in fig. 1.20.
Vf
1
Rf + sLf
If
Kf
Td (s)
1
f + sJ
ω(s)
1
s
θ(s)
Figure 1.20: Block diagram of field controlled D.C. motor
Exercise 1.13. A simple diagram of a DC motor controlled by its stator (i.e., by its
field) is given in 1.19 . Obtain (a) differential equations, (b) transfer function and (c)
block diagram governing dynamics of the system.
1.4.10 Transfer Function of Armature Controlled D.C. Motor
For the armature controlled D.C. motor, the output variable is angular displacement θ(t)
and the input variable is the applied voltage va (t). The ratio of the Laplace transform
of these variables which is defined as the T.F is obtained as follows, In an armaturecontrolled DC motor, the input is the armature voltage, va (t),and the output is motor
speed, ω(t), or the shaft angular position, θ(t).
In order to develop a model of the DC motor, let ia (t) denote the armature current, and La
and Ra denote the electrical side inductance and the coil resistance, respectively.
27
1.4. THE CONCEPT OF TRANSFER FUNCTION
The mechanical side inertia and friction are denoted as J and f , respectively. Let kt
denote the torque constant and kb the motor constant; then, the dynamic equations of
the DC motor for the electrical and mechanical sides are given as,
La
dia (t)
+ Ra ia (t) + kb ω(t) = va (t)
dt
J
dω(t)
+ f ω(t) − kt ia (t) = 0
dt
By applying the Laplace transform, these equations are transformed into algebraic equations
as
(sLa + Ra )Ia (s) + kb ω(s) = Va (s)
(sJ + f )ω(s) − kt Ia (s) = 0
In order to obtain a single input–output relation for the DC motor, we may solve the
first equation for Ia (s) and substitute in the second equation.
Alternatively, we multiply the first equation by kt , the second equation by (Ls + R), and
add them together to obtain,
(sLa + Ra )(sJ + f )ω(s) + kt kb ω(s) = kt Va (s)
From the above equation, the transfer function of the DC motor with voltage input and
angular velocity output is derived as,
kt
ω(s)
=
Va (s)
(Ls + R)(Js + b) + kt kb
Figure 1.21: Armature controlled D.C. motor
28
1.4. THE CONCEPT OF TRANSFER FUNCTION
The angular position θ(s) is obtained by integrating the angular velocity ω(s);
hence, the transfer function from Va (s) to the angular displacement θ(s) is given as,
θ(s)
kt
=
Va (s)
s [(sLa + Ra )(sJ + f ) + kt kb ]
The denominator polynomial in the motor transfer function typically has real and distinct
roots, which are reciprocal of the time constants contributed by the electrical and the
mechanical sides of the motor (Te , Tm ).
In terms of the time constants, the DC motor model is alternatively described as,
kt
ω(s)
JLa
=
Va (s)
(s + T1e )(s +
1
)
Tm
The block diagram of armature controlled D.C. motor is shown in fig. 1.22.
Vf
+
−
Kt
Ra + sLa
Td (s)
1
sJ + f
ω(s)
1
s
θ(s)
Kb
Figure 1.22: Block diagram of armature controlled D.C. motor
1.4.11 State Variable Models
State variable models are time-domain models that express system behavior as time
derivatives of a set of state variables.
State-space equations, or simply state equations may be applied to a very wide category of
systems, such as linear and nonlinear systems, time-invariant and time-varying systems,
systems with non-zero initial conditions, and others. The term state of a system refers to
the past, present, and future of the system. From the mathematical point of view, the
state of a system is expressed by its state variables. Usually, a system is described by a
finite number of state variables, which are designated by x1 (t) ,x2 (t), ... , xn (t) and are
defined as follows:
29
1.4. THE CONCEPT OF TRANSFER FUNCTION
Definition 16 (state variables). The state variables x1 (t) ,x2 (t), ... , xn (t); of a system
are defined as a (minimum) number of variables such that if we know (a) their values at
a certain moment t0 , (b) the input function applied to the system for t ≥ 0, and (c) the
mathematical model which relates the input, the state variables, and the system itself,
then the determination of the system’s states for t > t0 is guaranteed.
The state variables are often the natural variables associated with the energy storage
elements appearing in the system. The system order equals the number of such elements
in the system. In the case of electrical circuits, capacitor voltage and inductor currents
serve as natural state variables. In the case of mechanical systems modeled with inertial
elements, position and velocity of the inertial mass serve as natural state variables.
The State Differential Equation
The state of a system is described by the set of first-order differential equation written
as,
ẋ = Ax + Bu
(1.23)
The differential equation 1.23 is commonly known as State Equation. In general the
outputs of a linear system can be related to the state variables and the input signals by
the Output Equation,
y = Cx + Du
(1.24)
Example 1.14. The governing equation of a series RLC circuit in Fig.1.23 driven by a
constant voltage source, Vs, with mesh current used as the circuit variable is given as,
1
di(t)
+
Ri(t) + L
dt
C
R
i(t)
Z
i(t)dt = vs
L
+
vs (t) −
C
Figure 1.23: RLC Series Circuit
30
vc (t)
1.4. THE CONCEPT OF TRANSFER FUNCTION
The circuit contains two energy storage elements: an inductor and a capacitor. Accordingly,
let the inductor current, i(t), and the capacitor voltage, vc (t), serve as state variables for
the circuit. The state equations represent time derivatives of the state variables, expressed
as,
dvc
= i(t)
C
dt
and
di(t)
L
= vs − vc − Ri(t)
dt
We may note that the right-hand sides expressions in both equations contain state and
input variables. In vector-matrix form, these equations are given as,
  
   
1
v
d vc   0
C   c
0
 =
   +   vs
dt
1
i
− L1 − R
i
L
L
Let vc denote the RLC circuit output; then, an output equation is formed as,
 
vc 
vc = [1 0]  
i
Exercise 1.15. A bandpass RLC network is shown in Figure 1.13. Obtain
• differential equations,
• state equations,
• transfer function governing dynamics of the system.
1.4.12 Block Diagram
In the block diagram representation while the blocks represent the transfer function,
the direction of the signal flow is represented by arrow. The signal can only pass in
the direction of the arrows. The dimensions of the output signal from the block are
the dimensions of the input signals multiplied by the dimension of the transfer function
G(s).
The technique to reduce the interconnected subsystem to a single block diagram is known
as block diagram reduction technique.
31
1.4. THE CONCEPT OF TRANSFER FUNCTION
Advantages of Block Diagram Representation
• It is easy to form the overall block diagram for the entire system by merely connecting
the blocks of the components according to the signal flow.
• It is easy to evaluate the contribution of each component to the overall performance
of the system.
• In block diagram, the functional operation of the system can be visualized more
readily by examining the individual blocks rather than examining the physical
system itself.
Disadvantage of Block Diagram Representation
• The main source of energy flow in the system is not shown in block diagram.
• For a particular system, different block diagram is drawn. It is not unique.
• It is difficult to identify the nature of physical elements connected in the system.
The second order electrical or mechanical or hydraulic systems are all represented
by the same block diagram.
• Important functions are concealed within the walls of block diagram.
• Block diagram representation does not take into account the effect of interaction
between blocks.
• Passive symmetrical π networks are reciprocal since the input and output are interchangeable. However, in an amplifier the input and output are not inter-changeable.
This property is not considered in block diagram representation.
Block diagram consists of the following basic elements:
•
•
•
•
Blocks (rectangle) - operator.
Line (arrow) - representing unidirectional signal flow.
Circle - summation points.
Takeoff point.
Figure 1.24 illustrates implementation of the above elements.
Figure 1.24: Block diagram elements
32
1.4. THE CONCEPT OF TRANSFER FUNCTION
• The flow of system variable from one block to another is represented by the arrow.
• Sum or difference of the signal is represented by a summing point (junction) as in
fig. 1.25a.
• Application of one input to two or more block is represented by a takeoff point, as
shown in fig. 1.25b.
Tak e- of f poi nt
(a) Summing junction.
(b) Take-off point.
Figure 1.25: Take-off point.
33
1.4. THE CONCEPT OF TRANSFER FUNCTION
1.4.13 Block Diagram Reduction
A block diagram with several summers and pick off points can be reduced to a single
block, by using block diagram algebra, consisting of the rules;
Transformation 1 Combining blocks in cascade.
Figure 1.26: Combining blocks in cascade
Transformation 2 Moving summing point behind the block.
Figure 1.27: Moving summing point behind the block
Transformation 3 Moving a takeoff point ahead of a block.
Figure 1.28: Moving a takeoff point ahead of a block.
Transformation 4 Moving a takeoff point behind a block.
Figure 1.29: Moving a takeoff point behind a block.
34
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.30: Moving summing point ahead of a block.
Transformation 5 Moving summing point ahead of a block.
Transformation 6 Eliminating feedback loop.
Figure 1.31: Eliminating feedback loop.
Example 1.16. Reduce the block diagram shown in fig. 1.32 and obtain its closed loop
transfer function CR (s)
Figure 1.32: example 1.16
Solution example 1.16
G2
1 + G2 H2
(Transformation 6). The summers S1 and S2 can be interchanged, so that the blocks G1
G1
and H1 come under close-loop, they are transformed as
(Transformation 6) as
1 + G1 H1
in fig. 1.34.
In fig. 1.33 G2 and H2 blocks are connected in close-loop and their equivalence is
35
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.33: Example 1.16
Figure 1.34: Example 1.16
36
1.4. THE CONCEPT OF TRANSFER FUNCTION
The two blocks in the forward path (fig. 1.34) are in cascade and they are transformed
G1 G2
as,
and is shown in fig. 1.35. Figure 1.35 is in feedback mode and
(1 + G1 H1 )(1 + G2 H2 )
Figure 1.35: Example 1.16
is reduced using (Transformation 6) with the following result,
C
G1 G2
(s) =
R
(1 + G1 H1 )(1 + G2 H2 ) + G1 G2
Figure 1.36: Required solution (Example 1.16)
Example 1.17. Using block diagram reduction technique find the closed loop T.F. of
the system whose block diagram is given in fig. 1.37.
Figure 1.37: (Example 1.17)
37
1.4. THE CONCEPT OF TRANSFER FUNCTION
Solution example 1.17
Steps in solving example 1.17 is shown in fig. 1.38
Figure 1.38: Steps in solving (Example 1.17)
The reduced block diagram in as shown in fig. 1.39
Figure 1.39: Required solution (Example 1.17)
38
1.4. THE CONCEPT OF TRANSFER FUNCTION
Exercise 1.18. The block diagram of a closed loop system is shown in Figure 1.40. Using
block diagram reduction technique determine the closed loop T.F.
Answer
C
G2 (1 + G1 )
(s) =
R
(1 + G2 H2 + G1 G2 H1 )
Figure 1.40: (Exercise 1.18)
Exercise 1.19. The block diagram of a closed loop system is shown in Figure 1.41. Using
block diagram reduction technique determine the closed loop T.F.
Answer
G2 (G1 + G3 )
C
(s) =
R
(1 + G1 G2 H1 )
Figure 1.41: (Exercise 1.19)
Exercise 1.20. For the block diagram shown in Figure 1.41. Using block diagram
reduction technique determine the closed loop T.F.
Answer
C
G2 G1 G3 G4
(s) =
R
[1 + G2 G3 H2 + G3 G4 H1 + +G1 G2 G3 G4 H3 ]
39
1.4. THE CONCEPT OF TRANSFER FUNCTION
Figure 1.42: (Exercise 1.20)
Exercise 1.21. A servo mechanism is represented by the equation,
d2 y
dy
+ 4.8 = 144e
2
dt
dt
where e = c−0.5y is the actuating signal. Draw the block diagram of the system described
by the equation.
Exercise 1.22. Obtain the input-output transfer function of the following system in
fig. 1.43
Figure 1.43: (Exercise 1.22)
Answer
C
(G1 + G2 )G3
(s) =
R
(G1 + G2 )G3 G4 + 1 + G1 G3
Short Answer Questions
Why positive feedback is not used in closed loop system? For positive feedback systems,
the characteristic equation is 1 − GH(s) = 0 which will yield poles in the right half
s-plane. This makes the closed loop system unstable. That is why positive feedback
is not employed in closed loop system. However, positive feedback may be employed
in the design of oscillators which has sustained oscillations.
40
1.4. THE CONCEPT OF TRANSFER FUNCTION
What is characteristic equation and why is it so called? The denominator of a transfer
function when equated to zero is called the characteristic equation. Since the roots
of the characteristic equation determine the character of the transient response and
the stability character it is called characteristic equation.
Define transfer function of a system. Does it exit for non-linear systems? Transfer function
is defined as the ratio of the Laplace transform of output variable to the Laplace
transform of input variable with all initial conditions being zero. Laplace transform
does not exist for non-linear system.
What do you understand by the T.F being strictly proper? A transfer function is said
to be strictly proper if the order of the denominator polynomial is greater than the
order of the numerator polynomial.
What is a proper T.F? A T.F where numerator and denominator polynomials have the
same order is called a proper T.F.
What is an improper T.F? A transfer function whose order of the numerator polynomial
is greater than the order of the denominator polynomial is called improper T.F.
Define zero of a T.F.? The points at which the T.F. becomes zero in the s-plane are
called zeros of the T.F. They are also the roots of the numerator polynomial of the
T.F.
Define pole of a T.F.? The points at which the T.F. becomes infinity in the s-plane are
called poles of the T.F. They are also the roots of the denominator polynomial of
the T.F.
What are the advantages of armature controlled D.C motor? The electrical time constant
of the armature controlled D.C motor is small and hence its time response is very
fast. The back emf developed in the armature circuits provides additional damping
which eliminates unwanted oscillations during transients.
What are the advantages of field controlled D.C motor? In field controlled D.C motor,
the power requirement is less and there is no need for any power amplifier and it
requires voltage amplifier only which is not difficult to design.
What are the disadvantages of armature controlled D.C motor? Armature controlled
D.C motor requires power amplifier which is somewhat difficult to design and costly
too.
What are the disadvantages of field controlled D.C motor? The electrical time constant
of field circuit of the D.C motor is high and hence the time response is very sluggish.
41
1.4. THE CONCEPT OF TRANSFER FUNCTION
Further it requires a constant current source in the armature circuit which is some
what difficult to get.
What is a block diagram? The pictorial representation of a system is called a block
diagram. System inputs and outputs are represented by drawing a single line
towards the block and away from the block while the block represents the functional
relationship between input and output.
What are the basic connections of block diagram? Series (cascade), parallel and feedback
connections of block diagram are the basic connections.
What are the basic operations performed in block diagram reduction? The basic operations
performed in block diagram reduction include shifting the blocks before and after
the summer, moving a take off point before and after a block and before and after
summers, interchanging of summers and reducing blocks in parallel, cascade and
feedback.
State the block diagram simplification rule for removing the feedback. If G(s) is the
T.F of the forward path and H(s) is that of the feedback path, than the overall
G(s)
= 1+GH(s)
The sign of GH(s) will be
T.F is simplified for negative feedback as C(s)
R(s)
negative for positive feedback.
What are the advantages of block diagram representation? By connecting the blocks
of individual components, the block diagram for the overall system can be easily
formed. Further, it is easy to visualize the actual system by examining the block
diagram.
What are the disadvantages of block diagram representation? The disadvantages of
block diagram are that it is not unique for a given system. The main source of energy
flow in the system is not shown in the block diagram. Some important functions
are often concealed inside the four walls of the blocks. From the block diagram it
is difficult to understand what type of system the block diagram represents since
no information about the construction is known. Further, block diagram is drawn
under the assumption that what is inside the block is not affected by what is outside
of the block which is not true.
42
Chapter 2
Time Response Analysis
Learning objectives
After completing this chapter, student should be able to:
• Model first and second order systems.
• Find the time response of first order systems for various test signal as inputs.
• Derive expressions for time domain specifications of first order system.
• Derive the transfer function of second order system in terms of damping ratio and natural
frequency of oscillation.
• Classify the second order systems as under damped, critically damped and over damped
systems.
• Find the step response of second order system for different damping ratios.
• Derive expressions for time domain specifications of second order system for step input.
• Find the steady state error using static and dynamic error constants for unity and nonunity feedback systems.
• Plot the dynamic response of the system using Scilab1
The manner in which a dynamic system responds to an input, expressed as a function of
time, is called the time response. The theoretical evaluation of this response is said to
be undertaken in the time domain, and is referred to as time domain analysis. The
problem of time-domain analysis may be briefly stated as follows: given the system
(i.e., given a specific description of the system) and its input, determine the timedomain behavior of the output of the system. In Most control-system problems, the
final evaluation of the performance of the system is based on the time response. Time
response of a control system is usually divided into two parts:
• the transient and
• the steady state response.
1
https://www.scilab.org/
43
2.1. TEST INPUT SIGNALS
To analyse and design a control system, we we must define and measure its performance.
Since control systems are dynamic, their performance is usually specified in terms of both
the transient response and steady state response. Their definitions are given as,
Definition 17. The transient response ct (t) of a system is that particular part of the
response of the system which tends to zero as time increases. This means the ct (t) has
the property,
lim ct (t) = 0
t→∞
Definition 18. The steady-state response css (t) of a system is that particular part of
the response of the system which remains after the transient part has reached zero.
From the above two definitions it is clear that the total solution c(t) is the sum of the
transient response and the steady-state response, i.e.,
c(t) = ct (t) + css (t)
2.1 Test Input Signals
The time-domain performance specifications are important indices because control systems
are inherently time-domain system. Response to a specific input signal provides several
measures of the system performance. However, because the actual input signal of the
system is usually unknown, a standard test input signal is normally chosen. The
standard test signal commonly used are;
•
•
•
•
impulse
step input,
the ramp input, and
the parabolic input.
These inputs are shown in figure 2.1.A unit impulse function is also useful for test signal
purposes. The equations representing these test signals and Laplace transforms are given
in figure table 2.1 The commonly used test input signals are step functions,ramp functions,
acceleration functions, impulse functions, and sinusoidal functions.
• Input signal provide measures of the performance.
• The actual input signal of the system is usually unknown, hence standard test input
signal are used.
44
2.1. TEST INPUT SIGNALS
Figure 2.1: Test input signals:(a) impulse (b) step, (c) ramp (d) parabolic.
• Mathematical and Experimental analysis of CS can carried out easily with test
input signals.
If input to a control system is,
• Gradually changing function of time, then (Ramp is a good test signal)
• Subjected to sudden disturbances, then (We Use Step Signal)
• Subjected to shock inputs, then we use (impulse)
Table 2.1: Test Signal Input Equations
Input
Test Signal
r(t)
R(s)
Impulse
δ(t)
1
A
s
A
s2
2A
t3
Step
A
Ramp
At
Parabolic
At2
Time response of a system can be found by,
• Apply input to transfer function
• Simplify
• Use Inverse Laplace Transform tables to convert from s-domain back to t- domain
45
2.2. FIRST-ORDER SYSTEM
2.2 First-Order System
2.2.1 First-Order System Modelling
Consider a First order system with unity feedback as shown in fig. 2.2. The closed loop
transfer,
R(s)
C(s)
1
sτ
+
−
Figure 2.2: System representation in simple block diagram
1
C(s)
G
sτ
=
=
R(s)
1 + GH
1 + sτ1 (1)
=
1
sτ
(2.1)
(2.2)
1 + sτ1
1
C(s)
=
R(s)
sτ + 1
(2.3)
Equation (2.3) gives the transfer function of a first order system. This is represented in
block diagram and is shown in fig. 2.3.
R(s)
C(s)
1
1+sτ
Figure 2.3: Block diagram representation of first order system
2.2.2 Step Response of a First Order System
For a unit step input R(s) = 1s . The output is given by
C(s) =
R(s)
1
=
(1 + sτ )
s(1 + sτ )
(2.4)
Using partial fraction method on 2.4 and Inverse Laplace transformation yields,
t
c(t) = 1 − e− τ
When t = τ (Time Constant), c(t) = 1−e−1 = 0.632, or 63.2% of the final value.
46
(2.5)
2.2. FIRST-ORDER SYSTEM
Definition 19 (Time constant τ ). The time required for signal to attain 63.2% of final
or steady state value.
Time constant indicates how fast the system reaches the final value. Smaller time
constant faster system response. A large time constant corresponds to a sluggish system
(slow moving). The plot of c(t) against t is shown in Figure 2.4. Another important
1
0.8
0.632
c(t)
0.6
0.4
0.2
τ
0
0
0.5
1
1.5
2
2.5
3
3.5
t
Figure 2.4: Unit step response of a first order system.
characteristic of the system is the error between the desired value and the actual value
under steady state conditions. This error is known as steady state error of the system
and is denoted as ess . We know that error function as,
E(s)
1
=
R(s)
1 + G(s)H(s)
For our Fisrt order system with unity feedback, G(s) =
input R(s) = 1s , Equation 2.6 becomes
E(s) =
(2.6)
1
,
sτ
H(s) = 1, and Unit step
1
s
1+
1
sτ
(2.7)
Steady state error ess can be found using Final Value Theorem (FVT),
ess = lim e(t) = lim sE(s)
t→∞
ess = lim s
s→0
s→0
1
s+
1
τ
=0
(2.8)
(2.9)
Thus, for first order system the output approaches the reference input which is the desired
output, without any error. We say first order system tracks the step input without any
47
2.2. FIRST-ORDER SYSTEM
steady state error.
Exercise 2.1. The open loop plant transfer function of a unity feedback Linear time
1
. Use Laplace transform to derive an
invariant control system is given as, Gp (s) = s+2
expression for the time response and sketch the response for the unit step input and unit
ramp input.
1
C(s)
=
, to impulse input and input
R(s)
sτ + 1
Exercise 2.2. Derive system response of
ramp (r(t) = t).
Time Domain Specifications
The following time domain specifications are defined for a first order system:
•
•
•
•
Time constant τ .
Rise time tr .
Time delay td .
Settling time ts .
Consider a first order system with transfer function
G(s) =
K
s+a
as shown in Figure 2.5 For unit step input (R(s) = 1s ), the output is
R(s)
K
s+a
C(s)
Figure 2.5: Generalized first order system
C(s) =
K
s(s + a)
(2.10)
Using partial fraction on eq. (2.10) we have,
A
B
+
s s+a
K 1
1
=
−
a s s+a
C(s) =
48
(2.11)
(2.12)
2.2. FIRST-ORDER SYSTEM
Taking inverse Laplace transform of eq. (2.12),
c(t) =
K
1 − e−at
a
(2.13)
Time constant τ
The time constant τ is defined as the time taken for the step response to reach 63.2%
of its final value for the first time. From eq. (2.13), 63.2% output is 0.632 × Ka and
t = τ.
K
K
0.632
=
1 − e−at
(2.14)
a
a
e−at = 0.368
(2.15)
Taking natural log (loge ) on both sides and setting t = τ of eq. (2.15)
τ=
1
a
Rise time tr
The rise time tr is defined as the time taken for the response to go from 10% to 90% of
its final value. From Equation (2.13), at t = τ1 , let the output be 10% and t = τ2 the
output be 90%. Therefore,
K
0.1
=
a
K
0.9
=
a
K
a
K
a
1 − e−aτ1
(2.16)
1 − e−aτ2
(2.17)
Dividing the eq. (2.17) by the eq. (2.16), we get
9=
1 − e−aτ2
1 − e−aτ1
Taking natural log (loge ) on both sides and setting tr = τ2 −τ1 , The rise time
tr = 2.2τ
49
(2.18)
2.2. FIRST-ORDER SYSTEM
Time delay td
Time delay td is defined as the time taken for the response c(t) to reach 50% of its final
value for the first time. From Equation (2.13), for 50% output, we may write the following
equation with t = td .
td
0.5 = 1 − e− τ
td
e− τ = 0.5
(2.19)
(2.20)
Solving for td
(2.21)
td = 0.693τ
Settling Time ts
Settling time ts is defined as the time taken for the response c(t) to reach and stay within
2% of its final value for the first time. From Equation (2.13) for 98% output we may
write the following equation with t = ts .
ts
0.98 = 1 − e− τ
td
e− τ = 0.02
(2.22)
(2.23)
Solving for ts
ts = 3.91τ
(2.24)
Exercise 2.3. Show that settling time (ts ) of first order system for error tolerance of
95% and 97%, are 3τ and 2.667τ respectively.
Example 2.4. Consider the following T.F. of a certain first order system:
G(s) =
10
(s + 10)
Derive an expression for the response of the system for r(t) = 5u(t). Find the time
constant, settling time, time delay and rise time.
50
2.3. SECOND ORDER SYSTEM MODELLING
Solution example 2.4
C(s)
10
= G(s) =
R(s)
(s + 10)
5
R(s) = L [5u(t)] =
s
10
50
C(s) = R(s) ×
=
(s + 10)
s(s + 10)
1
1
=5
−
s s + 10
(2.25)
(2.26)
(2.27)
(2.28)
Taking inverse Laplace transform of Equation (2.28),
c(t) = 5 1 − e−10t
(2.29)
Therefore from Equation (2.29), Time constant,
τ=
1
sec
10
Settling time,
ts = 3.91τ = 0.391sec
Rise time,
tr = 2.2τ = 0.22sec
Time delay,
td = 0.693τ = 0.0693sec
Exercise 2.5. A glass bulb thermometer reads 98.2% of its final value of temperature
1min after immersing it in hot water. Determine the time constant, rise time, time delay
and settling time for 5% error tolerance. Ans:τ = 15sec, tr = 33sec, td = 10.4sec and ts =
45sec (Hint:mercury thermometer is a first order system)
2.3 Second Order System Modelling
Consider a Type 1, second order system as shown in Figure 2.6.
51
2.3. SECOND ORDER SYSTEM MODELLING
R(s)
+
−
K
s(sτ +1)
C(s)
Figure 2.6: Second Order System
K
C(s)
G
s(sτ +1)
=
=
R(s)
1 + GH
1 + s(sτK+1) (1)
K
s(sτ + 1) + K
K
C(s)
= 2
R(s)
τs + s + K
K
C(s)
= 2 τs K
R(s)
s +τ+ τ
C(s)
ωn2
= 2
R(s)
s + 2δωn + ωn2
=
(2.30)
where,
r
• ωn =
K
Natural frequency of oscillation in rad/sec.
τ
1
• δ= √
Damping ratio and
2 Kτ
• δωn Damping factor
The denominator of eq. (2.30) is a second degree polynomial in s and therefore eq. (2.30)
describes the dynamics of a second order system. Equation (2.30) is also called prototype
second order system equation. Thus, natural frequency of oscillation ωn and damping
ratio δ are the two parameters of a generalized second order system. Equation (2.30) is
also known as standard equation for a second order system.
Definition 20. The natural frequency ωn is defined as the frequency of oscillation of a
second order system without damping.
2.3.1 Time Response of a Second Order System
Importance of Step Response
Step response of a system is important for the following reasons:
• It is easy to generate step signal and test the system in the laboratory.
52
2.3. SECOND ORDER SYSTEM MODELLING
Figure 2.7: (a) Pole locations and transient response of a second order system. (b) Effect
of roots on damping ratio
• The step signal is sufficiently drastic and if satisfactory step response is obtained,
then the system is likely to give satisfactory performance for other types of inputs.
• From step response impulse response can be obtained by differentiating it and useful
information may be derived. Similarly from step response, ramp response can be
obtained by integrating it.
• The application of step input is equivalent to the application of numerous sinusoidal
signals with a wide range of frequencies.
classification of Time Response for second Order control system
According to different values of δ, the time response of a second order control system can
be classified as follows.
• Underdamped: For 0 < δ < 1, the transient response is oscillatory in nature, but
decays exponentially to give a stable response.
• Critically damped: For δ = 1, the response just becomes non-oscillatory and
gives a stable response after transient disappears.
• Overdamped: For δ > 1, the response is non-oscillatory and gives a somewhat
delayed stable response after transient disappears.
• Undamped: For δ = 0, the transient does not disappear and the response gives a
sustained oscillation. The time response c(t) of a second-order control system for
different values of damping ratio δ are shown in Figure 2.7.
53
2.3. SECOND ORDER SYSTEM MODELLING
Step Response of a Second Order System
Consider a standard equation for a second order system Equation (2.30),
ωn2
C(s)
= 2
R(s)
s + 2δωn s + ωn2
1
For unit step input, R(s) = , then
s
C(s) =
ωn2
s (s2 + 2δωn s + ωn2 )
(2.31)
Step response of a second order system can be evaluated for three scenario, depending
on the value of damping ratio δ.
• Under damped case (δ < 1)
• Critically damped case (δ = 1)
• Over damped case (δ > 1)
Under damped response δ < 1
For δ < 1, Equation (2.31) is written as,
C(s) =
ωn2
s (s + δωn + jωd ) (s + δωn − jωd )
Using partial fractions,
C(s) =
A1
A2
A3
+
+
s
(s + δωn + jωd ) (s + δωn − jωd )
(2.32)
Determining the Residues A1 , A2 and A3
A1 =
ωn2 × s
s (s2 + 2δωn s + ωn2 )
ωn2 × (s + δωn + jωd )
A2 =
s (s + δωn + jωd ) (s + δωn − jωd )
√
where ϕ =
1−δ 2
δ
=
s=−δωn −jωd
=1
s=0
1
1
π
√
√
= √
∠−ϕ −
2
2
2
2
j2 1 − δ (δ + j 1 − δ )
2 1−δ
1
π
A3 = A∗2 = √
∠ϕ +
2
2 1 − δ2
54
2.3. SECOND ORDER SYSTEM MODELLING
Substituting the above residues in Equation (2.32) we get,
1
C(s) = 1 + √
2 1 − δ2
π
π
e−j(ϕ+ 2 )
ej(ϕ+ 2 )
+
s + δωn + jωd s + δωn − jωd
Taking inverse Laplace Transform,
π
π
e−δωn t
e−j( 2 +ϕ+ωd t) + ej( 2 +ϕ+ωd t)
c(t) = 1 + √
2 1 − δ2
π
e−δωn t
cos
+ ϕ + ωn t
=1+ √
2
1 − δ2
e−δωn t
c(t) = 1 − √
sin (ϕ + ωn t)
1 − δ2
(2.33)
Note:
•
•
•
•
Frequency of transient oscillation is the damped natural frequency ωd and
Thus varies with the damping ratio δ
The error signalfor this system is e(t)
= r(t) − c(t) and given as,
√
e√−δωn t
−1 1−δ 2
e(t) = 1−δ2 sin ωd t + tan
δ
– Error signal exhibits a damped sinusoidal oscillation.
– At steady state, or as t → ∞, no error exists between the input and output.
• This response is plotted in Figure 2.8.
• The response is oscillatory and as t → ∞, it approaches unity.
Figure 2.8: Step response of an under damped second order system
Critically damped response δ = 1
For δ = 1, Equation (2.31) is written as,
C(s) =
ωn2
s (s + ωn )2
55
2.3. SECOND ORDER SYSTEM MODELLING
Using partial fractions,
C(s) =
A3
A1
A2
+
2 +
s
(s + ωn )
(s + ωn )
(2.34)
Determining the Residues A1 , A2 and A3
A1 =
ωn2 × s
s(s + ωn )2
=1
s=0
× (s + ωn )2
= −ωn
s(s + ωn )2
s=−ωn
2
d ωn × (s + ωn )2
= −1
A3 =
ds
s(s + ωn )2
A2 =
ωn2
s=−ωn
Substituting the above residues in Equation (2.35) we get,
C(s) =
1
ωn
1
−
2 −
s (s + ωn )
(s + ωn )
(2.35)
Taking inverse Laplace Transform,
c(t) = 1 − tωn eωn t − e−ωn t
(2.36)
This is plotted in Figure 2.9. As the damping is increased from a value less than unity,
the oscillations decrease and when the damping factor equals unity the oscillations just
disappear, as depicted in Figure 2.9
Figure 2.9: Step response of critically damped second order system
If δ is increased beyond unity, the roots of the characteristic equation are real and negative
and hence, the response approaches unity in an exponential way, as shown in fig. 2.10
56
2.3. SECOND ORDER SYSTEM MODELLING
Figure 2.10: Step response of an over damped second order system
2.3.2 Time Domain Specifications of a Second Order System
The unit step response is easy to generate and mathematically the response to any input
can be derived if the response to a step input is known. Therefore, the performance
characteristics of a control system are described in terms of transient response to a unit
step input; with standard initial conditions of output and all time derivatives being zero
when the system is at rest. The time response of second and higher order control systems
to a unit step input is generally damped oscillatory in nature before reaching steady state.
The performance of a system is usually evaluated in terms of the following qualities.
• How fast it is able to respond to the input,
• How fast it is reaching the desired output,
• What is the error between the desired output and the actual output, once the
transients die down and steady state is achieved,
• Does it oscillate around the desired value,
These four questions can be answered in terms of time domain specifications of the system
based on its response to a unit step input.The step response of a typical under damped
(δ<1 ) second order system is plotted in Figure 2.11.
Figure 2.11: Transient time response specifications to unit step input
57
2.3. SECOND ORDER SYSTEM MODELLING
The performance of a second order system is measured by the following specifications:
•
•
•
•
•
•
Peak overshoot Mp and %peak overshoot %Mp .
Time at which the peak overshoot occurs peak time tp .
Time constant τ .
Rise time tr .
Settling time ts .
Time delay td .
Expressions for the above specifications are derived in terms of the second order system
parameters δ and ωn .
Rise time tr
It is the time required for the response to reach 100% of the steady state value for under
damped systems. However, for overdamped systems, it is taken as the time required for
the response to rise from 10% to 90% of the steady state value. Consider time response
for under damped 2nd system,
e−δωn t
c(t) = 1 − √
sin (ϕ + ωd t)
1 − δ2
(2.37)
From definition of rise time tr , c(t) = 1 therefore above eq. (2.37) is reduced to,
e−δωn tr
sin (ϕ + ωd tr )
1=1− √
1 − δ2
e−δωn tr
0= √
sin (ϕ + ωd tr )
1 − δ2
0 = sin (ϕ + ωd tr )
sin π = sin (ϕ + ωd tr )
tr =
where ϕ = tan−1
√
1−δ 2
δ
π−ϕ
ωd
√
and ωn = ωn 1 − δ 2
58
(2.38)
2.3. SECOND ORDER SYSTEM MODELLING
Peak time tp
Peak time is obtained by differentiating c(t) (eq. (2.37)) with respect to t and equating
dc(t)
to zero. At maxima the slope is zero. Therefore peak time is obtained from
.
dt t=tp
dc(t)
δωn e−δωn tp
e−δωn tp
=0+ √
sin(ωd tp + ϕ) − √
ωd cos(ωd t + ϕ) = 0
dt
1 − δ2
1 − δ2
δωn sin(ωd tp + ϕ) = ωd cos(ωd t + ϕ)
√
δ sin(ωd tp + ϕ) = 1 − δ 2 cos(ωd t + ϕ)
√
It is noted that tan ϕ =
1 − δ2
sin ϕ
=
δ
cos ϕ
cos ϕ sin(ωd tp + ϕ) = sin ϕ cos(ωd t + ϕ)
cos ϕ sin(ωd tp + ϕ) − sin ϕ cos(ωd t + ϕ) = 0
sin(ωd tp + ϕ − ϕ) = 0
sin(ωd tp ) = 0
sin(ωd tp ) = sin nπ
Therefore,
tp =
tp =
nπ
ωd
nπ
√
ωn 1 − δ 2
Peak time (tp ),
tp =
ωn
√
π
1 − δ2
Also, for first undershoot,
tp =
2π
√
ωn 1 − δ 2
tp =
3π
√
ωn 1 − δ 2
Second overshoot,
59
2.3. SECOND ORDER SYSTEM MODELLING
Maximum overshoot (Mp )
Maximum overshoot Mp is found by substituting the value of tp in the expression for c(t)
and subtracting the steady state response value from it.
c(tp ) − c(∞)
∞
Mp =
Peak overshoot is obtained as,
Mp = c(tp ) − 1
e−δωn tp
Mp = 1 − √
sin (ϕ + ωd tp ) − 1
1 − δ2
setting tp =
π
ωd
→ tp ωd = π
π
−δω
e n ωd
Mp = − √
sin (ϕ + π)
1 − δ2
−δωn ωπ
e
Mp = √
But sin ϕ =
d
1 − δ2
sin (ϕ)
√
√
1 − δ 2 and ωd = ωn 1 − δ 2
− √ δπ
Mp = e
1−δ 2
− √ δπ
Mp = e
1−δ 2
(2.39)
Percent Overshoot,
Mp = 100 × e
− √ δπ
1−δ 2
We notice that Mp is independent of ωn and is dependent on δ. As the value of damping
ratio increases the peak overshoot decreases.
Settling time (ts )
− √δωn t
The transient response curve c(t) always lies in between its two envelop curves 1±e 1−δ2
whose time constant is T = δω1n . So, the settling time (ts ) for the tolerance band of x
per cent may be obtained from the given equation as follows. For response to get into
60
2.3. SECOND ORDER SYSTEM MODELLING
tolerance band of say 2%, the expression for settilng time is obtained as,
√ n ts
− −δω
2
1+e
= 1.02
1−δ
√ n ts
− −δω
1−δ 2
= 0.02
e
for small values of δ
e−δωn ts = 0.02
3.91
4
ts =
≈
δωn
δωn
ts ≈
4
δωn
ts ≈
3
δωn
Similarly for tolerance of 4%
Exercise 2.6. Derive the time response of a second order system subjected to impulse
input function.
Time delay td
Time delay is defined as the time taken for the transient response to reach 50% of the
final value for the first time. The expression for the time delay is given by the following
empirical formula:
1 + 0.7δ
td =
ωn
Time constant τ
Time constant is defined as the time taken for the transient response to reach 63.2% of
its final value for the first time.
1
td =
δωn
Example 2.7. Obtain the impulse and step responses of the following unity feedback
control system with open loop transfer function
G(s) =
6
s(s + 5)
61
2.3. SECOND ORDER SYSTEM MODELLING
solution
C
G
=
R
1+G
=
1
6
s(s+5)
6
+ s(s+5)
=
6
s(s + 5) + 6
C
6
= 2
R
s + 5s + 6
C(s) = R(s) ×
For unit step input R(s) =
s2
6
+ 5s + 6
1
s
C(s) =
6
s(s + 3)(s + 2)
C(s) =
A2
A3
A1
+
+
s
s+3 s+2
C(s) =
1
2
3
+
−
s s+3 s+2
By partial fractions
Residues are found to be
Taking inverse Laplace
c(t) = 1 + 2e−3t − 3e−2t
Impulse response is found by integrating step response
h(t) =
d
c(t) = −6e−3t + 6e−2t
dt
It is to be noted that the poles of the closed loop transfer function lie on the -ve real axis
of the s-plane and hence the system is over-damped.
Example 2.8. Reduce the block diagram in fig. 2.12 to derive the transfer function for
the system. Find the value of K that will result in a critically damped case.
62
2.3. SECOND ORDER SYSTEM MODELLING
Solution:
G
Y (s)
=
X(s)
1+G
=
1
K
s(s+20)
K
+ s(s+20)
Y (s)
K
= 2
X(s)
s + 20s + K
Comparing above with standard 2nd order system we get,
K = ωn2 and 2δωn = 20
For critically damped case δ = 1
2δωn = 20
2ωn = 20
ωn = 10
Therefore
K = ωn2 = 102
K = 100
Figure 2.12: Block diagram for example 2.8
Exercise 2.9. Considers the circuit shown in Figure 2.13.
(s)
• Calculate the transfer function VVoi (s)
.
• Let L = 0.01H. Choose the remaining parameters of the circuit such that ωn =
105 rad/sec and δ = 0.05. Plot the poles of the system, indicating the pole positions
in s-plane.
63
2.3. SECOND ORDER SYSTEM MODELLING
• For these parameters, make a carefully-dimensioned sketch of the response to a unit
step in input voltage. Specifically, show the overshoot value, the time to the peak,
and the time-scale for each of the next 3 peaks in the response. What is the time
required to settle to 5%?
R
L
+
vi (t) −
C
vo (t)
Figure 2.13: RLC Network.
Exercise 2.10. Consider the system in fig. 2.14.
• Suppose that C(s) = 1. Find the response of the system for a step input.
• With this controller find its rise time, peak time, maximum percent over-shoot,
settling time.
Figure 2.14: Block diagram elements
Exercise 2.11. A certain unity negative feedback control system has an open loop
transfer function.
10
G(s) =
s(s + 2)
Find the rise time (tr ), percentage overshoot (Mp ), peak time (tp ), delay time td , and
settling time (ts ) for a step input of 12 units.
Answers Mp = 4.2units; %Mp = 35%; tp = 1.05s; tr = 0.63s; τ = 1s; ts = 3.91s; td =
0.3862s
64
2.3. SECOND ORDER SYSTEM MODELLING
Example 2.12. Figure 2.15 shows a unity feedback system. Calculate δ and ωn when
K = 0. Also calculate K when δ = 0.6.
Solution
Applying block diagram reduction technique, we get overal transfer function as,
64
C(s)
= 2
R(s)
s + (K + 4) + 64
For K = 0
C(s)
64
= 2
R(s)
s + 4 + 64
Comparing above with the following standard second order equation,
ωn2 = 64 → ωn = 8rad/sec
2δωn = 4 → δ =
For δ = 0.6
4
= 0.25
2ωn
64
C(s)
= 2
R(s)
s + (K + 4) + 64
Comparing above with the following standard second order equation,
ωn2 = 64 → ωn = 8rad/sec
2δωn = K + 4 → K = 2δωn − 4 = 2 × 0.6 × 8 − 4 = 5.6
K = 5.6
Figure 2.15: Block diagram of Example 2.12
Exercise 2.13. Figure 1 shows armature controlled DC motor schematic, where the
following parameter values are assumed:R = 1Ω, L = 10mH, J = 0.01kg m2 , b =
0.1N sec/rad, Kt = Kb = 0.02. Obtain,
• the differential equations for the DC motor.
• the motor transfer functions with angular velocity and displacement as outputs.
• a state variable model of the motor with {ia , ω, θ} as state variables.
65
2.3. SECOND ORDER SYSTEM MODELLING
• the transfer function for the DC motor from armature voltage to motor speed ( Vω(s)
)
a (s)
by ignoring armature inductance.
• Solve and sketch the step response from the transfer function obtained above.
Exercise 2.14. A certain second order unity feedback system is described by the following
forward path T.F.
25
G(s) =
s(s + 6)
Derive an expression for the output response of the system for unit step input. Also find
tr , tp , τ, Mp and ts .
(Ans: c(t) = 1−1.25e−3t sin(4t+0.928rad)tr = 0.5535sec; tp = 0.785sec; τ = 0.333sec; ts =
1.3sec; Mp = 9.5%.)
66
2.3. SECOND ORDER SYSTEM MODELLING
Short Answer Questions
• What do you understand by time response of a system?
The output response, starting at some initial condition expressed as a function of
time when the input is applied, is called time response of the system. It is divided
into two parts namely transient response and steady state response.
• What is transient response?
Transient response is the time response which goes from initial state to final state
as time becomes large. The variation is expressed as a function of independent
variable time t.
• What is steady state response?
The manner in which the output behaves as time t approaches infinity is called
steady state response.
• How the impulse, step and ramp signals are related? The relationship
between impulse, step and ramp signals are as indicated below,
– Impulse → Integrate Step → Integrate Ramp
– Ramp → Differentiate Step → Differentiate Impulse
• How impulse response of a system is important?
– If the area under the impulse response curve is finite, the system is stable.
– By integrating the impulse response, step response can be obtained.
– If the impulse response function is known, then by taking the Laplace transform, the T.F. of the system can be obtained.
• Why systems are tested with step input?
It is easy to generate a step function. If the step response is satisfactory, the given
system is likely to give satisfactory performance to other commonly used signals. If
the step response is known, impulse response is obtained by differentiating it and
velocity response is obtained by integrating it.
• What are the performance specifications of a first order system?
The performance specifications of first order system are
– Time constant T ,
– Rise time tr ,
– Setting time ts , and
– Time delay td
• Define time constant of the first order system. How is it related to the
location of the pole?
67
2.3. SECOND ORDER SYSTEM MODELLING
Time constant τ of a first order system is defined as the time taken for the step
response to reach 63.2% of its final value for the first time. It is expressed as τ = a1
where a is the location of the pole on the negative real axis in the s-plane.
• Define rise time,time delay, and settline time of a first order system.
– Rise time tr is defined as the time taken for the step response to go from 10%
to 90% of its final value. It is expressed as tr = 2.2τ .
– Time delay td of a first order system is defined as the time taken for the step
response to reach 50% of its final value for the first time. It is expressed as
td = 0.693τ .
– The settling time ts of a first order system is defined as the time taken for the
step response to reach and stay within 2% of its final value for the first time.
It is expressed as ts = 3.91τ . For 5% error tolerance ts = 3τ .
• What information we get from performance specifications of first order
system?
The time constant of a first order system is inversely proportional to the pole located
on the negative real axis of the s-plane. If the pole is moved away from the origin,
the time constant is less. The rise time, settling time and the time delay are all
directly proportional to the time constant. Hence, the time response is faster. The
time response will be slower if the pole is moved towards the origin.
• What do you understand by prototype second order system?
For a unity feedback system, the forward path T.F. and the closed loop T.F are
respectively writing as
ωn2
s(s + 2δωn )
C(s)
T (s) =
R(s)
ωn2
= 2
s + 2δωn s + ωn
G(s) =
where δ and ωn are system parameters. The system with the above equations is
called prototype second order system.
• What is characteristic equation? Why is it called so?
The dominator of the closed loop T.F of a system when equated to zero is called
the characteristic equation.The corresponding polynomial is called characteristic
polynomial. These polynomials contain the closed loop poles whose locations in
the s-plane decide the stability and the time response of the system. Since it
characterizes the system response, it is called characteristic equation.
68
2.3. SECOND ORDER SYSTEM MODELLING
• List the time domain specifications of a prototype second order system?
Peak overshoot, The time at which the peak overshoot occurs, Rise time, Settling
time, Time delay, and Time constant.
• How maximum or peak overshoot is defined? What is the percentage maximum
overshoot and how it is related to the damping ratio?
Maximum overshoot Mp is found by substituting the value of tp in the expression
for c(t) and subtracting the steady state response value from it.
Mp =
c(tp ) − c(∞)
∞
Peak overshoot is obtained as,
− √ δπ
Mp = 100 × e
1−δ 2
• What pole location characterize: undamped, the over damped, the under
damped and critically damped and negatively damped?
–
–
–
–
–
Undamped, δ = 0; s1 , s2 = ±jω
Over damped, δ > 1; s1 , s2 = −δω ± ωn
Under damped, δ < 1; s1 , s2 = −δωn ± jωn
Critically damped, δ = 1; s1 , s2 = −ωn
Negatively damped, δ < 0; s1 , s2 = δωn ± jωn
69
Chapter 3
Error Analysis
Control systems are crucial for accurately controlling physical variables, ensuring they
closely match the desired value. A standard measure of performance is steady state
error, which is the difference between the input and output, and can be a function of
time, known as transient error. This allows for accurate measurement and prediction of
system performance.
3.1 Steady state error
The expression in Equation (2.39) which is the peak overshoot is the transient error. This
error when t tends to ∞ is called the steady state error and is denoted by ess . The
system under test is simulated with standard inputs such as step, ramp and acceleration.
The steady state error of certain control system for these signals is shown in Figure 3.1.
The steady state error in control systems mainly come from non-linearities of some of the
Figure 3.1: (a) Steady state error for a Step input; (b) Velocity input; and (c)
Acceleration input
70
3.1. STEADY STATE ERROR
components connected in the systems. The non-linearities include the backlash in gear
or a motor that will not move unless the input voltage exceeds a threshold. However,
here we will cover the steady state errors which are due to system configuration and the
type of input that is applied.
Steady state error is given mathematically as,
ess = lim e(t) = lim sE(s)
s→∞
s→0
(3.1)
Consider closed loop control system depicted in Figure 3.2 E(s) is the error which is the
Figure 3.2
difference between the input and the output and can be written as
1
E(s)
=
R(s)
1 + G(s)
R(s)
E(s) =
1 + G(s)
ess = lim e(t) = lim sE(s)
s→∞
s→0
sR(s)
s→0 1 + G(s)
ess = lim
(3.2)
Using Equation (3.2), the steady state errors are calculated for the step, ramp and
parabolic inputs. It is to be noted that final value theorem as applied to Equation (3.2)
is valid iff the closed loop system is stable and the transfer function G(s) is proper.
71
3.1. STEADY STATE ERROR
3.1.1 Static Error Coefficient Constants
Steady State Error Due to Step Input - Position Error Constant Kp
sR(s)
s→0 1 + G(s)
ess = lim
For unit step
R(s) =
1
s
s. 1s
s→0 1 + G(s)
1
ess = lim
s→0 1 + G(s)
1
ess =
1 + lims→0 G(s)
ess = lim
we define,
Kp = lim G(s)
s→0
ess =
1
1 + Kp
Kp is called the position error constant. Consider the system with the following T.F.,
G(s) =
K(1 + sz1 )(1 + sz2 ) . . .
sn (1 + sp1 )(1 + sp2 ) . . .
For n = 0, the system is Type 0 system. Kp for Type 0 system is calculated as,
Kp = lim G(s)
s→0
K(1 + sz1 )(1 + sz2 ) . . .
=K
s→0 (1 + sp1 )(1 + sp2 ) . . .
= lim
The steady state error is,
ess =
1
1+K
For Type 1 system,
Kp = lim G(s)
s→0
K(1 + sz1 )(1 + sz2 ) . . .
=∞
s→0 s(1 + sp1 )(1 + sp2 ) . . .
= lim
The steady state error is,
ess =
1
=0
1+∞
72
3.1. STEADY STATE ERROR
For Type 2 and Type 3 systems the steady state error will similarly be zero. Thus, when
unit step input is applied, for Type 0 system the steady state error is finite and is given
1
. For systems of Type 1 and above the steady state error is zero.
by
1+K
Example 3.1. Consider the closed loop system with the following forward path T.F.
G(s) =
200(s + 2)(s + 5)
(s + 4)(2s + 6)
A step input of height 12 size is applied. Find the steady state error.
Solution
12
s
200(s + 2)(s + 5)
200(2)(5)
1000
Kp = lim G(s) = lim
=
=
s→0
s→0 (s + 4)(2s + 6)
(4)(6)
12
12
12
ess =
=
1 + Kp
1 + 1000
12
144
ess =
1012
r(t) = 12u(t) → R(s) =
Steady State Error Due to Velocity (Ramp) Input - Velocity Error Constant Kv
sR(s)
s→0 1 + G(s)
ess = lim
For unit ramp
R(s) =
1
s2
s2 . 1s
s→0 1 + G(s)
1
ess = lim
s→0 s(1 + G(s))
1
ess =
lims→0 (s + sG(s))
ess = lim
we define,
Kv = lim sG(s)
s→0
ess =
1
Kv
Kv is called the velocity error constant and has the unit sec. Consider the system with
73
3.1. STEADY STATE ERROR
the following T.F.,
G(s) =
K(1 + sz1 )(1 + sz2 ) . . .
sn (1 + sp1 )(1 + sp2 ) . . .
For Type 0 system,
Kv = lim sG(s)
s→0
sK(1 + sz1 )(1 + sz2 ) . . .
=0
s→0 (1 + sp1 )(1 + sp2 ) . . .
1
ess = = ∞
0
= lim
For Type 1 system,
Kv = lim sG(s)
s→0
= lim
s→0
ess =
sK(1 + sz1 )(1 + sz2 ) . . .
=K
s(1 + sp1 )(1 + sp2 ) . . .
1
K
For Type 2 system,
Kv = lim sG(s)
s→0
sK(1 + sz1 )(1 + sz2 ) . . .
=∞
s→0 s2 (1 + sp1 )(1 + sp2 ) . . .
1
ess =
=0
∞
= lim
Therefore,
• For Type 0 system ess = ∞
• For Type 1 system ess = K
• For Type 2 system ess = 0
For Type 2 and higher Type system ess = 0.
Example 3.2. A negative feedback control system has the following loop T.F.
G(s) =
10(2s + 3)(s − 4)
s(3s + 2)(4s + 5)
The system is subjected to ramp input of slope 2rad/sec. Find the steady state error.
Solution
74
3.1. STEADY STATE ERROR
r(t) = 2t → R(s) =
2
s2
Kv = lim sG(s) = lim s
s→0
s→0
10(3)(−4)
10(2s + 3)(s − 4)
=
= −12
s(3s + 2)(4s + 5)
(2)(5)
2
2
1
=− =−
Kv
12
6
1
ess = − rad
6
ess =
The negative sign indicates that the output is higher than the input at steady state.
Exercise 3.3. Consider a unity negative feedback system with the following open loop
T.F.
K
G(s) =
s(s + 1)(s + 4)
The steady state error ess ≤ 2rad for a velocity input of 2 rad/sec. Find Kv and hence
K. (Ans: K = 4)
Steady State Error Due to Parabolic Input - Acceleration Error Constant Ka
sR(s)
s→0 1 + G(s)
ess = lim
For parabolic r(t) =
t2 u(t)
2
R(s) =
1
s3
s3 . 1s
s→0 1 + G(s)
1
ess = lim 2
s→0 s (1 + G(s))
1
ess =
2
lims→0 (s + s2 G(s))
ess = lim
we define,
Ka = lim s2 G(s)
s→0
ess =
1
Ka
Ka is called acceleration or parabolic error constant. Consider the system with the
following T.F.,
K(1 + sz1 )(1 + sz2 ) . . .
G(s) = n
s (1 + sp1 )(1 + sp2 ) . . .
75
3.1. STEADY STATE ERROR
For n = 0, the system is Type 0 system. Ka for Type 0 system is calculated as,
Ka = lim s2 G(s) =
s→0
ess =
s2 K(1 + sz1 )(1 + sz2 ) . . .
=0
(1 + sp1 )(1 + sp2 ) . . .
1
1
=
Ka
0
ess = ∞
For n = 1, the system is Type 1 system. Ka for Type 1 system is calculated as,
Ka = lim s2 G(s) =
s→0
ess =
s2 K(1 + sz1 )(1 + sz2 ) . . .
=0
s(1 + sp1 )(1 + sp2 ) . . .
1
1
=
Ka
0
ess = ∞
For n = 2, the system is Type 2 system. Ka for Type 2 system is calculated as,
Ka = lim s2 G(s) =
s→0
s2 K(1 + sz1 )(1 + sz2 ) . . .
=K
s2 (1 + sp1 )(1 + sp2 ) . . .
1
1
=
Ka
K
1
ess =
K
ess =
Therefore,
• For Type 0 system ess = ∞
• For Type 1 system ess = ∞
• For Type 2 system ess = K
For Type 3 and higher Type system ess = 0.
Example 3.4. The Open -loop transfer function of unity feedback system is given by,
G(s) =
50
(1 + 0.1s)(s + 10)
Determine the static error coefficients Kp , Kv , and Ka
Solution
76
3.1. STEADY STATE ERROR
Unity Feedback H(s) = 1.
GH(s) = G(s)
G(s) =
50
(1 + 0.1s)(s + 10)
50
50
=
=5
s→0
s→0 (1 + 0.1s)(s + 10)
10
s50
0 × 50
Kv = lim sG(s) = lim
=
=0
s→0
s→0 (1 + 0.1s)(s + 10)
10
s2 50
0 × 50
Ka = lim s2 G(s) = lim
=
=0
s→0
s→0 (1 + 0.1s)(s + 10)
10
Kp = lim G(s) = lim
Example 3.5. A certain control system has the following T.F.
G(s) =
(s + 2)
s2 (2s + 3)(s + 5)
3t2
The input applied to the system is r(t) =
u(t). Find the steady state error.
2
Solution
r(t) =
3t2
3
→ R(s) = 3
2
s
s2 (s + 2)
2
=
2
s→0 s (2s + 3)(s + 5)
3×5
Ka = lim s2 G(s) = lim
s→0
2
15
3
3
= 2
ess =
Ka
15
45
ess =
2
Ka =
By using the static error constants Kp , Kv and Ka , the steady state error can be obtained
as described. When these test signals are simultaneously applied to the system, the steady
state error can be derived for the individual signals and using superposition theorem, the
total steady state error can be determined.
Example 3.6. The open loop transfer function of a unity feedback system is given by
G(s) =
s2 (s
108
+ 4)(s2 + 3s + 12)
Find the static coefficients and steady state error of the system when subjected to an
77
3.1. STEADY STATE ERROR
input given by
r(t) = 2 + 5t + 2t2
Solution
108
=∞
s→0
+ 4)(s2 + 3s + 12)
s108
=∞
Kv = lim sG(s) = lim 2
s→0
s→0 s (s + 4)(s2 + 3s + 12)
s2 108
108
108
Ka = lim s2 G(s) = lim 2
=
=
2
s→0
s→0 s (s + 4)(s + 3s + 12)
(4)(12)
48
2
5
2
r(t) = 2 + 5t + 2t2 → L [r(t)] = R(s) = + 2 + 2 3
s s
s
5
4
2
+
+
ess =
1 + Kp Kv Ka
2
5
4
ess =
+
+ 108
1+∞ ∞
48
4 × 48
= 1.77
ess =
108
Kp = lim G(s) = lim
s→0 s2 (s
Exercise 3.7. A certain unity feedback system has the following forward path T.F.
G(s) =
K(s + a)
s(s + b)
where K, a and b are constants and greater than zero. It is required that the steady state
error for unit ramp input is less than 0.1. To improve transient response, the closed loop
system poles are to be placed at s1,2 = −2 ± j4. Find the values of K, a and b. (Ans:
K = 2; a = 10; b = 2)
Exercise 3.8. A unity feedback control system has the following forward path T.F.
G(s) =
K
s(1 + sτ )
where K and τ are constants and greater than zero. Determine the factor by which K
should be multiplied to reduce the overshoot for step input from 85% to 35%.
(Ans: K2 = 0.02658K1 where K1 corresponds to 85% overshoot and K2 corresponds to
35% overshoot. The gain has to be reduced by a factor 0.02658.)
Exercise 3.9. Consider a unity feedback control system with the closed loop transfer
function
C(s)
Ks + b
= 2
R(s)
s + as + b
78
3.1. STEADY STATE ERROR
Determine the open loop transfer function. Show that the steady state error in the unit
ramp input response is given by
a−k
ess =
b
79
Chapter 4
Concept of Stability & Routh-Hurwitz Criterion
System stability is one of the most important performance specification of a control
system. A system is considered unstable if it does not return to its initial position
but continues to oscillate after it is subjected to any change in input or is subjected to
undesirable disturbance. For any time invariant control system to be stable the following
two conditions need to be satisfied. These are:
• The system will produce a bounded output for every bounded input;
• If there is no input, the output should tend to be zero, irrespective of any initial
conditions.
Stability of a system may be referred to as absolute stability or in terms of relative
stability. The term relative stability is used in relation to comparative analysis of stability
of systems and their operating conditions. Absolute stability refers to the condition
whether the given system is stable or not. If the system is stable, the closeness of stability
is measured by relative stability.
4.1 The Concept of Stability
Stable systems are defined as the systems which will return to a stable equilibrium position
after some small disturbance. On the other hand, systems which are given small initial
disturbance, move off position and do not return to stable equilibrium position are termed
as unstable systems.
Linear system stability is defined in terms of the system time response. The time response
consists of two parts. One is due to the initial condition and the other is the response due
to the input. The total response = zero state response + zero input response.
• Zero input response: The zero input response is due to the initial conditions
only. All the inputs are zero.
80
4.1. THE CONCEPT OF STABILITY
• Zero state response: The zero state response is the response of the system due
to the input only. All the initial conditions are zero.
• Zero input stability: It refers to the stability condition due to zero input response
which is due to the initial conditions only and all the inputs are zero. This stability
also depends on the roots of the characteristic equation.
A linear time invariant system is said to be zero input stable if the zero input response
of the system with finite initial conditions reaches zero as time t tends to infinity.
4.1.1 Bounded Input Bounded Output (BIBO) (External) Stability
The bounded input bounded output (BIBO) stability is defined with respect to the total
response which is the sum of the zero input and zero state responses.
A linear time invariant system is said to be bounded input bounded output (BIBO) stable
if every bounded input yields a bounded output. BIBO stability is also called as external
stability.
4.1.2 BIBO Stability via Impulse Response Function
Consider the system with T.F. G(s) represented as in Figure 4.1, where g(t) is the impulse
response function, r(t) is the input and c(t) is the external output. The external output
c(t) can obtained using the convolution integral,
c(t) = g(t) ∗ r(t)
Z ∞
g(τ )r(t − τ ) dτ
=
0
Z ∞
|c(t)| =
g(τ )r(t − τ ) dτ
(4.1)
0
eq. (4.1) can be written as,
Z
∞
Z
∞
g(τ )r(t − τ ) dτ ≤
|g(τ )| |r(t − τ )| dτ
0
0
Z
|c(t)| ≤
∞
|g(τ )| |r(t − τ )| dτ
(4.2)
0
For the input r(t) being bounded, → |r(t)| < M , therefore eq. (4.2) is written as
Z
|c(t)| ≤ M
∞
|g(τ )| dτ
0
81
(4.3)
4.1. THE CONCEPT OF STABILITY
For the output c(t) to be bounded, it can be expressed as
c(t) ≤ N < ∞
(4.4)
Substituting Equation (4.4) in Equation (4.3),
Z
∞
|g(τ )| dτ ≤
0
N
<∞
M
(4.5)
Equation (4.5) gives the area of the impulse response curve. For the system to be BIBO
stable, the area of the impulse response curve should be finite in the time interval
0 < t < ∞.
R(s)
C(s)
G(s)
Figure 4.1: Block diagram representation of TF G(s)
4.1.3 BIBO Stability and the Characteristic Roots
It is not always possible to test the system with impulse input and measure the output
with respect to time to asses the BIBO stability requirement. It can be easily determined
by knowing the characteristic roots of the system. Consider the system shown in Figure 4.1.
By Laplace transform definition, G(s) can be written as,
∞
Z
g(t)e−st dt
G(s) =
Z0 ∞
|G(s)| ≤
Z0 ∞
≤
Z0 ∞
≤
|g(t)||e−st | dt
|g(t)||e−σt ||e−jωt | dt
|g(t)||e−σt | dt
(4.6)
0
G(s) → ∞ at ′ s′ poles of G(s), therefore Equation (4.6) becomes,
Z
∞≤
∞
|g(t)||e−σt | dt
(4.7)
0
For one / more number of poles in Right Half Plane (RHP) of s-plane or in jω axis, σ > 0
and
|e−σt | ≤ A or e−σt = 1
(4.8)
82
4.1. THE CONCEPT OF STABILITY
Substituting Equation (4.8) into Equation (4.7)
Z
∞
|g(t)|A dt ≥ ∞
0
∞
Z
g(t) dt ≥ ∞
(4.9)
0
From Equation (4.9),
• The system is BIBO unstable when atleast one pole lies in RHP or on jω axis.
• The stability condition (Equation (4.9)) does not depend upon the input applied.
• A linear time invariant continuous time system is said to be BIBO stable if all the
roots are located in the left half s-plane.
• A system is unstable if it is not BIBO stable.
The nature of impulse response depends on the location of poles of the transfer function.
4.1.4 Effects Of Location Of Poles On Stability
Poles on Negative Real axis
s = −a
g(t) = L
−1
K
= Ke−at
s+a
As t → ∞, the response approaches zero and system is STABLE.
Figure 4.2
83
4.1. THE CONCEPT OF STABILITY
Poles on Positive Real axis
s=a
g(t) = L
−1
K
= Keat
s−a
The response increases exponentially with time, hence the system is UNSTABLE.
Figure 4.3
Poles at Origin
s=0
g(t) = L
−1
K
=K
s
Figure 4.4
If there are double poles at origin.
s=0
g(t) = L
−1
84
K
= Kt
s2
4.1. THE CONCEPT OF STABILITY
The Constant value, indicates the system is MARGINALLY STABLE.
Figure 4.5
Complex Pole in the Left Half of s-Plane
s = −α ± jω
K
K
−1
g(t) = L
+
s + σ + jω s + σ − jω
2A(s + σ)
−1
g(t) = L
= 2Ae−σt cos ωt
(s + σ)2 + ω 2
As t → ∞, g(t) approaches zero and system is STABLE.
Figure 4.6
Complex Pole in the Right Half of s-Plane
s = σ ± jω
K
K
−1
g(t) = L
+
s − σ + jω s − σ − jω
2A(s − σ)
−1
g(t) = L
= 2Aeσt cos ωt
(s − σ)2 + ω 2
85
(4.10)
(4.11)
(4.12)
4.1. THE CONCEPT OF STABILITY
As response increases exponentially sinusoid with time and therefore the system is UNSTABLE.
Figure 4.7
Poles on the jω
s = ±jω
K
K
g(t) = L
+
s + jω s − jω
2As
−1
= 2A cos ωt
g(t) = L
s2 + ω 2
−1
Response is MARGINALLY STABLE. Sustained oscillations of constant amplitude.
Figure 4.8
86
4.1. THE CONCEPT OF STABILITY
In Figure 4.9, the stable region, unstable region and the marginally stable region in the splane are sketched and shown. Characteristic roots in LHP indicate asymptotic stability,
the roots in RHP region indicates instability and the presence of single root in the jω
axis indicates marginal stability. Presence of repeated roots in the jω axis indicates
instability. The necessary and sufficient condition that a feedback system is STABLE is
Figure 4.9: Characteristic root location in the complex s-plane and system stability.
that all the roots of the characteristic equation ∆ = 1 + GH(s) have negative real part.
OR In terms of poles we can say that the necessary and sufficient condition that a feedback
system be STABLE is that all poles of overall transfer function have negative real part.
4.1.5 Routh–Hurwitz Criterion
Routh–Hurwitz criterion is a time domain approach to determine the number of roots
of the characteristic polynomial with constant real coefficients with respect to the LHP,
RHP and the jω axis of the s-plane. The method requires two steps. For the given
characteristic equation, using the coefficients, the Routh table is prepared. From the
Routh table, the information about how many roots of the characteristic polynomial are
in the LHP, RHP and the jω axis is interpreted. Even though this information can be
easily obtained using modern calculators and computers, and also determine the exact
location of the closed loop system poles, the power of this method lies in design rather
than analysis.
87
4.1. THE CONCEPT OF STABILITY
The Routh–Hurwitz criterion is applied to determine the following:
• Absolute stability of the system.
• Relative stability of the system in the time domain.
• The gain of the system can be increased before the closed loop system becomes
unstable.
• To factorize the given polynomial in some special cases and locate the roots in the
s-plane.
• To find the number of LHP, RHP and jω axis roots.
• Stability of the system is determined without solving the characteristic equation.
4.1.6 Routh–Hurwitz Criterion to Determine Absolute Stability
Consider the system shown in Figure 4.10, its closed loop T.F. is given by,
G(s)
C(s)
=
R(s)
1 + GH(s)
For a strictly proper T.F., the above equation in polynomial form is as given as,
G(s) =
am sm + am−1 sm−1 + am−2 sm−2 + · · · + a0
bn sn + bn−1 sn−1 + bn−2 sn−2 + · · · + b0
(4.13)
where n ≥ m and a′ s and b′ s are constants and real. The characteristic equation of
Equation (4.13) is,
bn sn + bn−1 sn−1 + bn−2 sn−2 + · · · + b0 = 0
(4.14)
The roots of the characteristic Equation (4.14) should all be in the LHP for the system
Figure 4.10
to be stable. For this, the following conditions are necessary before proceeding further to
investigate the absolute stability using Routh–Hurwitz stability criterion.
88
4.1. THE CONCEPT OF STABILITY
• The coefficient a0 , a1 , . . . , an−1 , an must be positive.
• All the coefficients in the characteristic equation must be present. If any one
coefficient is zero, there will be a root or roots in the imaginary axis or in the
RHP, which means the system will be unstable
.
If above two conditions are NOT satisfied then the system will be UNSTABLE. But if all
the coefficients have same SIGN and there is NO MISSING TERM, we have no guarantee
that the system will be STABLE. The above are ONLY NECESSARY CONDITION but
NOT SUFFICIENT. All stable system must satisfy these conditions but the system
satisfying all these conditions need not to be stable. For example consider characteristic
equation given by polynomial,
s3 + s2 + 3s + 24 = 0
√
√
(s − 1 + j2 2)(s − 1 − j2 2)(s + 3) = 0
Note that complex roots have real positive parts, even though all coefficients have same
sign and there is no missing term.
4.1.7 Formation of Routh’s Array
Routh (1874) developed a a necessary and sufficient condition for stability based on Routh
array, which states:
Routh’s criterion: A system is stable if and only if all the elements in the first column
of the Routh array are positive.
Routh array: The first two rows of the Routh array are composed of the even and
odd coefficients of the characteristic polynomial, respectively, while the remaining rows
89
4.1. THE CONCEPT OF STABILITY
are composed of elements derived from the first two rows:
a0
a2
a4
···
row n − 1 a1
a3
a5
···
row n − 2 b1
b2
b3
···
row n − 3 c1
c2
c3
···
row n
···
··· ··· ··· ···
row 2
∗
row 1
∗
···
row 0
∗
···
∗
···
The elements of the third row are computed as follows:


 a0 a2 
det 

a1 a3
b1 = −
a1

=
a1 a2 − a0 a3
a1
=
a1 a4 − a0 a5
a1
=
a1 a6 − a0 a7
a1

 a0 a4 
det 

a1 a5
b2 = −
a1


 a0 a6 
det 

a1 a7
b3 = −
a1
The elements of the forth row are computed as follows:


 a1 a3 
det 

b1 b2
c1 = −
=
b1
90
b 1 a3 − a1 b 2
b1
4.1. THE CONCEPT OF STABILITY


 a1 a5 
det 

b1 b3
c2 = −
b1

=
b 1 a5 − a1 b 3
b1
=
b 1 a7 − a1 b 4
b1

 a1 a7 
det 

b1 b4
c3 = −
b1
The elements from the third row on are computed based on the determinant of a 2 by
2 array composed of the two elements of the first column of the previous two rows and
the two elements of the subsequent columns. Any missing coefficient is represented by a
zero.
Example 4.1.
a0 s 3 + a1 s 2 + a2 s + a3 = 0
s3
a0 a2
s2
a1 a3
s1
(a2 a1 − a0 a3 )
a1
s0
a3
For this system to be stable, we must have
a2 a1 − a0 a3
> 0.
a1
Example 4.2. Check stability of the system whose characteristic equation is given by,
s4 + 2s3 + 6s2 + 4s + 1 = 0
s4
1
6 1
s3
2
4
s2
s1
s0
(2×6−1×4)
2
(4×4−2×1)
4
=4
(2×1−1×0)
2
=1
= 3.5
1
Since all coefficient in the first column of Routh array are of the same sign, the equations
has NO ROOTS with positive real parts, hence the system is STABLE.
91
4.1. THE CONCEPT OF STABILITY
Example 4.3. Determine stability of the system whose characteristic equation is given
by,
2s4 + 2s3 + s2 + 3s + 2 = 0
s4
2 1 2
s3
2 3
(2−6)
2
s2
= −2 2
s1
5
s0
2
There are two changes of SIGN in the first column (from +2 to -2, and -2 to +5), hence
there are two positive real part roots, therefore the system is UNSTABLE.
Example 4.4.
s6 + 4s5 + 3s4 + 2s3 + s2 + 4s + 4 = 0
with a0 = 1, a1 = 4, a2 = 3, a3 = 2, a4 = 1, a5 = 4, a6 = 4. First we find the Routh array:
s6
1
3
1 4
s5
4
2
4
s4
(4×3−1×2)
4
s3
(2.5×2−4×0)
2.5
=2
s2
(2×0−2.5×(−2.4))
2
=3
s1
s0
(3×(−2.4)−2×4)
3
=
( −76
×4−0×3)
15
−76
15
(4×1−4×1)
4
= 2.5
(2.5×4−4×4)
2.5
=0
(4×4−1×0)
4
=4
= −2.4
(2×4−2.5×0)
2
=4
−76
15
=4
The elements in the first column are: 1, 4, 2.5, 2, 3, -76/15, 4 with two sign changes (3
to -76/15 and -76/15 to 4), there are two poles on the RP and the system is not stable.
Solving the characteristic equation, we can get the five roots: −3.26, 0.68±j0.75, −0.60±
j0.99, −0.89.
Example 4.5. The transfer function of the feedforward pass of a feedback system is
G(s) = K
s+1
s(s − 1)(s + 6)
92
4.1. THE CONCEPT OF STABILITY
and the feedback gain is H(s) = 1 (negative feedback). The overall transfer function is
therefore:
s+1
K s(s+1)(s+2)
G(s)
K(s + 1)
T (s) =
=
=
s+1
1 + G(s)H(s)
s(s − 1)(s + 6) + K(s + 1)
1 + s(s+1)(s+2)
The characteristic equation is
s3 + 5s2 + (K − 6)s + K = 0
Routh’s criterion is used To find the range of the gain K for stability:
s3
1 K −6
s2
5
K
s1
(4K−30)
5
0
s0
K
0
For all elements of the first column to be positive, we have K > 0 and K > 7.5. Consider
the following three cases:
• When K = 9 > 7.5, the roots of the characteristic equation are p0 = −4.77,
p1,2 = −0.17 ± j1.37, representing respectively an exponentially decaying term and
a sinusoid also exponentially decaying.
• When K = 7.5, the characteristic equation is
√
√
D(s) = s3 + 5s2 + 1.5s + 7.5 = (s + 5)(s2 + 1.5) = (s + 5)(s + j 1.5)(s − j 1.5) = 0
with three poles p0 = −5 representing an exponential decay and a complex conjugate
√
pair p1,2 = ±j 1.5 representing a sinusoidal oscillation.
• When K = 5 < 7.5, the roots of the characteristic equation are p0 = −5.36,
p1,2 = 0.18 ± j0.95, representing respectively an exponentially decaying term and a
sinusoid which grows exponentially.
Exercise 4.6. The transfer function of the feedforward pass of a feedback system is
G(s) =
K
s(s + 1)(s + 2)
93
4.1. THE CONCEPT OF STABILITY
and the feedback gain is just H(s) = 1. The overall transfer function is therefore:
G(s)
K
=
1 + G(s)H(s)
s(s + 1)(s + 2) + K
T (s) =
and the characteristic equation is
D(s) = s3 + 3s2 + 2s + K = 0
Routh’s criterion is used To find the range of the gain K for stability:
s3
1
s2
3 K
s1 (6 − K)/3
s0
2
0
K
For all elements of the first column to be positive, we have K > 0 and K < 6, i.e.,
0 < K < 6. Consider the following five cases:
• When K = −1 < 0, we have
T (s) =
s3
−1
−1
=
2
+ 3s + 2s − 1
(s − 0.32)(s + 1.66 + j0.56)(s + 1.66 − j0.56)
with three poles p0 = 0.32 representing an exponentially growing term, and p1,2 =
1.66 ± j0.56 representing a decaying sinusoid. The system is unstable.
• When K = 0, we have
T (s) =
0
=0
s3 + 3s2 + 2s
• When 0 < K = 3 < 6, we have
T (s) =
s3
+
3s2
3
6
√
√
=
+ 2s + 3
(s + 3)(s + j 2)(s − j 2)
with three poles p0 = −2.67 representing an exponential decay and a complex
conjugate pair p1,2 = −0.16 ± j1.05 representing a decaying sinusoid. The system
is stable.
• When K = 6, we have
T (s) =
s3
+
3s2
6
6
√
√
=
+ 2s + 6
(s + 3)(s + j 2)(s − j 2)
94
4.1. THE CONCEPT OF STABILITY
with three poles p0 = −3 representing an exponential decay and a complex conjugate
√
pair p1,2 = ±j 2 representing a sinusoidal oscillation.
• When K = 8 > 6, we have
T (s) =
s3
+
3s2
8
8
=
+ 2s + 8
(s − 3.17)(s − 0.08 + j1.59)(s − 0.08 − j1.59)
with three poles p0 = 3.17 representing an exponentially growing term and a
√
complex conjugate pair p1,2 = −0.08 ± j 2 representing a decaying sinusoid. The
system is unstable.
Exercise 4.7. Find the range of parameters such that the system s3 + as2 + bs + c = 0
is stable
Exercise 4.8. Determine the number of unstable poles, if any, of the system with
characteristic equation s6 + 3s5 + 2s4 + 9s3 + 5s2 + 12s + 20 = 0
Special Cases
The derivation of the Routh’s tabular form fails in two special cases in which indefinite
results are obtained. These are as follows:
1. There is a zero coefficient in the first column (but the corresponding row has at
least one nonzero element.)
2. There is a complete zero row.
It must be mentioned that if (2) occurs, it is in an odd-numbered row, never in an evennumbered one. However (1) can happen in both even- and odd-numbered rows. In the
sequel we discuss possible ways of addressing the above problems.
There are three typical approaches to tackle the first problem, as follows.
• Substitute s with x1 . Clearly, this method does not work if the coefficients of the
original characteristic equation are symmetric.
• Multiply by (s + a), s > a, e.g., (s + 1) and carry out the original method.
• Substitute 0 with ϵ > 0. The number of sign changes in the first column is equal to
the number of RHP poles. If the elements above and below ϵ have the same sign,
there is a pair of imaginary roots.
95
4.1. THE CONCEPT OF STABILITY
As for the second special case, a recipe is as follows. Substitute the zero row with the
coefficients of the derivative of the auxiliary equation obtained from the previous row.
Then, resume the usual procedure. The following points are noteworthy for Auxiliary
equations:
• The auxiliary equation is always even.
• The roots of the auxiliary equation are the roots of the original characteristic
equation.
• These roots occur in pairs and are the negative of each other. That is, they have
these general forms: ±α, ±jβ, ±(α ± jβ).
The subsequent examples demonstrate the above cases and the proposed solutions.
Example 4.9. Determine the stability of the equation
s4 + 2s3 + 2s2 + 4s + 3 = 0
s4 1 2 3
s3 3 4
s2 0 3
In forming the Routh’s table we face a row with a zero in its first element but the row is
not identically zero. We try all three proposed methods to address it.
Method One
By the application of the first method we arrive at
3x4 + 4x3 + 2x2 + 2x + 1 = 0
whose Routh’s tabular procedure can be completed as follows:
x4
3 2 1
x3
4 2
x2
0.5 1
x1 −6
x0
1
As is seen there are two sign changes in the first column and thus there are two RHP
roots.
96
4.1. THE CONCEPT OF STABILITY
Method Two
Through the second method we obtain
s5 + 3s4 + 4s3 + 6s2 + 7s + 3 = 0
whose Routh’s tabular form is completed as follows:
s5
1 4 7
s4
3 6 3
s3
2 6
s2 −3 3
s1
8
s0
3
Likewise, here we conclude that there are two RHP poles.
Method Three
As for the third method, the table is completed as follows.
s5
1 2 3
s4
2 4
s3
0(ϵ) 3
s2
−3 3
s1 4 − 6/ϵ < 0
s0
3
It is observed that with small ϵ, the sign of 4 − 6ϵ is negative. There are two sign changes
in the first column and thus there are two RHP poles.
Example 4.10. Given
D(s) = s4 + 3s3 + 6s2 + 12s + 8 = 0.
97
4.1. THE CONCEPT OF STABILITY
Determine its stability.
s4
1
s3
3 12
s2
2
6 8
8
s1 0(4)
s0
8
In constructing the table we encounter a row which is identically zero. We thus replace
its zeros with the coefficients of the derivative of the auxiliary equation made from the
previous row.The auxiliary equation is A(s) = 2s2 + 8 = 0 and its derivative is 4s = 0. So
we replace 0 with 4, shown in parentheses. Then we continue forming the table as usual.
It is observed that there is no sign change in the first column and thus there is no root
in RHP. This also means that the roots of the auxiliary equation are on the j-axis. In
fact they are ±j2. Also note that roots of the auxiliary equation are roots of the original
equation.
Exercise 4.11. Determine stability of the system whose characteristic equation is given
by,
s6 + s5 + 6s4 + 5s3 + 10s2 + 5s + 5 = 0
Disadvantages of Routh’s Method
• If the system is stable, Routh’s criterion does not tell how stable it is.
• If the system is unstable, the method does not suggest how to improve the stability
of the system.
• In the preparation of Routh’s array, when the complete row contains zero coefficients,
and there is no sign change in the first column, the hasty conclusion that the system
is stable is erroneous. In such cases, careful evaluation of roots in the jω axis is very
much essential. Presence of simple roots in the jω axis indicates that the system
is neither stable nor unstable. It is called marginally stable. However, if repeated
roots are identified in the jω axis, the system is no more marginally stable but it
is unstable. Routh–Hurwitz criteria will not reveal this form of instability which is
a major disadvantage of this method.
• Routh–Hurwitz criterion is applicable for the characteristic polynomial with real
coefficients. For complex coefficient polynomial, it is not applicable. Further, if the
characteristic equation is not algebraic, such as containing exponential functions or
sinusoidal functions of s, the Routh–Hurwitz criterion cannot be applied.
98
4.1. THE CONCEPT OF STABILITY
• jω axis is considered as the stability boundary when the Routh–Hurwitz criterion
is applied. The method cannot be applied to any other stability boundaries in a
complex plane. Thus, for a discrete time system, where the unit circle is the stability
boundary, Routh–Hurwitz criterion cannot be applied. By linear transformation,
the unit circle has to be transformed to jω axis and then Routh criterion is applied.
99
Chapter 5
Root Locus
Differential equations can be used to model physical systems by considering the interconnection
of components and their behavior laws. However, this method is tedious and impractical
for practical systems. Three major tools for analyzing and designing control systems are
the root locus, Bode, and Nyquist plots. These tools reveal important characteristics of
the system and can be easily used for system analysis and design.
Both the transient and steady-state responses of the system are important in control
system analysis. The transient response is composed of exponential of the form e−pt ,
where t is the independent variable and p are the characteristic modes (the roots of the
characteristic equation of the closed-loop transfer function) of the system. The roots
are the poles of the system, which vary with the value of the gain K. Assume that
K is positive. The system is asymptotically stable only if its poles are located in the
left-half of the s-plane. That is, the real part of the root must be negative. The gain
can be increased until the roots are located in the right-half of the s-plane. As the roots
primarily determine the transient response, it helps to know the variations in the location
of the poles as the system gain is varied. The gain adjustment along with a compensator
is required to get a desirable response from a feedback control system.
The trajectories of the roots of the characteristic equation when investigated carefully as
the gain of the system is varied give all information about absolute and relative stabilities
like the other methods discussed in previous chapters. In addition, the method gives the
exact root locations of the closed loop system. The method is called the root locus
method, and the trajectories of the closed loop poles in the s-plane are called root loci.
The root locus is the path of the characteristic equation traced out in the
s-plane as the system parameter is varied.
A root locus plot is simply a plot of the s zeros and the s poles on a graph with real
and imaginary coordinates. It represents a curve of the location of the poles of a transfer
function as some parameter is varied. The locus of the roots of the characteristic equation
of the closed loop system as the gain varies from −∞ to ∞ gives the name of the method.
100
5.1. THE CONCEPT OF ROOT LOCUS
Such a plot shows clearly the contribution of each open loop pole or zero to the locations
of closed loop poles. General rules to construct the root locus exist and if the designer
follows them, sketching of the root loci becomes a simple matter.
5.0.1 Advantages of Root Locus Method
• Absolute and relative stability can be determined using root locus.
• For the given system, the gain of the system can be determined for the system to be
marginally stable. The corresponding frequency of oscillation can be determined.
• For a given value of the system gain K, the closed loop pole locations can be
determined. Similarly, for the required damping ratio of the dominant closed loop
system, the closed loop pole locations can be determined. From this, the time
response of the closed loop system for the given input can be obtained. Further, it
is possible to determine the closed loop frequency response also.
• For the given gain of the system, time domain specifications can be determined.
• From the root locus, the frequency domain specifications such as phase margin and
gain margin can be obtained, which are the measures of relative stability.
• The root locus shows clearly the contribution of each open loop pole or zero to
the locations of the closed loop poles which ultimately decide the absolute and
relative stability and also give information about the time and frequency domain
specifications.
• Like other methods, a very simple MATLAB, OCTAVE or scilab program is available
to draw root locus which makes the drawing of root loci simpler. That is why root
locus is extensively used in the design of compensators in the time domain.
5.1 The Concept of Root Locus
Consider the closed loop system shown in Figure 5.1. The closed loop T.F. is
T (s) =
KG(s)
1 + KG(s)H(s)
(5.1)
where K is a real constant and G(s) and H(s) are function of s with numerator and
denominator polynomials of any order, the characteristic equation of Equation (5.1) is
101
5.1. THE CONCEPT OF ROOT LOCUS
given as
L(s) = 1 + KG(s)H(s) = 0
(5.2)
KG(s)H(s) = −1
|KG(s)H(s)| = 1
(5.3)
∠KG(s)H(s) = ±180◦ (2n + 1)
(5.4)
n = 0, 1, 2, . . .
The root locus is the path of the roots of the characteristic equation (L(s))
that moves in the complex s-plane as the system parameter (K) varies in the
range −∞ ≤ K ≤ ∞. The root locus should always satisfy Equations (5.3) and (5.4),
R(s)
+
−
G(s)
K
C(s)
H(s)
Figure 5.1: System representation in simple block diagram
where Equation (5.3) is called magnitude criterion and Equation (5.4) is called angle
criterion. It is evident from Equations (5.3) and (5.4) that angle criterion is used to
trace the root locus while magnitude criterion is used to calibrate the root locus in terms
of K.
The angle criterion states that for any point s1 in the s-plane to be a point on the root
locus, the sum of the angles of the vectors drawn from all the zeros of G(s)H(s) minus
the sum of the angles of the vectors drawn from all the poles of G(s)H(s) to the point
s1 should be an odd multiple of 180◦ .
According to the magnitude condition, at any point s1 in the s-plane where root locus
exists, the value of K is calculated as,
K=
Product of the lengths of the vectors drawn from the poles of GH(s) to s1
Product of the lengths of the vectors drawn from the zeros of GH(s) to s1
Now consider the following open loop T.F
KGH(s) =
K
s(s + 2)
102
5.1. THE CONCEPT OF ROOT LOCUS
Characteristic equation is
K
=0
s(s + 2)
L(s) = 1 +
s2 + 2s + K = 0
(5.5)
The roots of Equation (5.5) are real for K ≤ 1 and are complex for K > 1, where the
complex roots are given as: has to roots given as,
s1 = −1 +
s2 = −1 −
√
1−K
√
1−K
The closed-loop roots for a range of K are tabulated in Table 5.1. The loci of these
Table 5.1: text
K
Roots of L(s)
0
0, −2
0.5
−0.29, −1.71
1.0
−1, −1
1.5
−1 ± j0.71
2.0
−1 ± j
2.5
−1 ± j1.22
3.0
−1 ± j1.41
roots, as K varies from 0 → ∞, comprise two branches that start at the open-loop (OL)
poles located at 0, −2, proceed inward along the real-axis, meet in the middle at σ = −1,
then split and extend along the σ = −1 line in both upward and downward directions to
s = −1 ± j∞. These directions are called the RL asymptotes. The RL plot is shown
in Figure 5.2.
103
5.1. THE CONCEPT OF ROOT LOCUS
Figure 5.2: The root locus plot for GH(s) =
104
K
s(s+2)
5.1. THE CONCEPT OF ROOT LOCUS
5.1.1 Properties of the Root Loci (Rules of the Root Loci)
The construction of root locus manually is made simpler using the properties of the root
loci. These properties are also called rules of root loci. Using these rules, certain key
points in the s-plane are identified. These points lie on the root loci. The connection
of these points gives approximate or asymptotic root loci. However, by smoothening the
asymptotic plots, root loci which are closer to the actual root loci are drawn. These rules
are discussed below.
Rule 1: The Starting Points of the Root Loci
Consider the characteristic equation,
1 + KGH(s) = 0
GH(s) =
−1
K
As the magnitude K approaches zero, |GH(s)| = ∞. This implies that s must approach
the poles of GH(s). Since
GH(s) =
(s + z1 )(s + z2 ) . . . (s + zm )
(s + p1 )(s + p2 ) . . . (s + pn )
the root loci start from the poles of GH(s) where the value of K = 0.
Rule 2: The Ending Points of Root Loci
Starting from the poles of GH(s), as the value of K increases, the root loci move in the
s-plane. For K = ∞, GH(s) becomes zero. This is possible if s approaches the zeros
of GH(s) or ∞ of the s-plane. The root loci end at the zeros of GH(s) or at ∞
where the value of K = ∞.
Rule 3: Number of Root Loci of 1 + GH(s)
The number of root loci of the characteristic equation 1 + GH(s) = 0 is equal to the
order of the polynomial of OL TF (GH(s)). Consider the following loop T.F.
GH(s) =
K(s + 1)(s + 3)
(s − 1)(s + 2)(s + 4)
105
5.1. THE CONCEPT OF ROOT LOCUS
The numerator polynomial is of order two while the denominator polynomial is of order
three, hence, three root loci as depicted in Figure 5.3.
Figure 5.3: Root locus showing the number of root loci, starting and ending parts and
the value of K
Rule 4: Symmetry of the Root Loci
The root loci are symmetrical with respect to the real axis of the s-plane. The rule is
derived based on the property that for real coefficients of characteristic polynomial, the
roots must be real or in complex conjugate form. Consider the following loop T.F
GH(s) =
K
s(s + 4)(s + 2)
The root locus is shown in Figure 5.4 Loci 1 and 2 are symmetrical with complex conjugate
root loci with respect to the real axis.
Figure 5.4: Root locus which illustrates complex symmetry
Rule 5: Angle of the Asymptote of the Root Loci
If there are Np number of poles and Nz number of zeros of GH(s), and if Np = Nz , all
the Np root loci originating from their respective poles will end at the zeros where the
106
5.1. THE CONCEPT OF ROOT LOCUS
value of K = ∞. If Np > Nz , then (Np − Nz ) loci will end at ∞ where K = ∞. The
root loci near ∞ in the s-plane are described by the asymptotes of the root loci. They
are giving guidance to the actual root locus. The angles of the asymptotes are given by
θ=
(2n + 1)π
Np − Nz
where Np and Nz = Finite number of poles and zeros of GH(s) respectively. For example
in Figure 5.4, the angles of the asymptotes are θ = 60◦ , 180◦ and − 60◦ .
Rule 6: Intersect of the Asymptote (Centroid)
The intersect of the asymptotes of the root loci lies on the real axis of the s-plane at
P
σc =
P
P− Z
Np − Nz
The value of σc is always a real number. For example in Figure 5.4, the centroid is
σc =
0 + (−2) + (−4) − 0
= −2
3−0
Rule 7: Root locus on the real axis
The root locus on the real axis always lies in a section of the real axis to the left of an
odd number of poles and zeros. Consider the following loop T.F.
GH(s) =
K(s + 1)(s + 3)
(s − 1)(s + 2)(s + 4)
The root loci on real-axis is illustrated in Figure 5.3.
Rule 8: Angle of Departure of the Root Locus
The angle of departure of root locus at a pole of GH(s) is nothing but the angle of the
tangent to the locus near the pole. The angle of departure is given by
θd = (2n + 1) × 180◦ − (θp − θz )
(5.6)
where θp and θz are the net angle contribution to the concerned pole by all other poles
and zeros of GH(s), respectively. This is illustrated in Figure 5.5 for the root locus drawn
107
5.1. THE CONCEPT OF ROOT LOCUS
for the loop T.F.
GH(s) =
K(s + 6)
K(s + 6)
=
(s + 3)(s2 + 2s + 2)
(s + 3)(s + 1 + j)(s + 1 − j)
The root locus plot is shown in Figure 5.5. The root loci from s = −1 + j and s = −1 − j
Figure 5.5: Illustration of angle of departure for GH(s) =
K(s+6)
(s+3)(s2+2s+2)
depart with a curved line. The angle of departure at s = −1 + j is calculated by drawing
vectors from poles and zeros of GH(s) as shown in Figure 5.5. Using Equation (5.6), we
get
θd = 180◦ + (90◦ + 26.6◦ − 11.3◦ ) = 285.3◦ = −74.7◦
Similarly, for the conjugate pole at s = −1 − j, we get θd = 74.7◦
Analytically the angle of departure θd can be evaluated as,
θd = arg GH(s) × (s − pi ) + 180◦
The departure angle of the root locus from a complex pole pi is given by the θd , as
above where arg GH(s) × (s − pi ) computed at the complex pole pi , but ignoring the
contribution of that particular pole.
θd at s = −1 + j
108
5.1. THE CONCEPT OF ROOT LOCUS
θd = arg GH(s) × (s − pi ) + 180◦
K(s + 6)(s + 1 − j)
(s + 3)(s + 1 + j)(s + 1 − j)
= 180◦ + arg
s=−1+j
K(−1 + j + 6)
(s
+1−j)
= 180 + arg
(−1 + j + 3)(−1 + j + 1 + j)
(s
+1−j)
◦
◦
s=−1+j
◦
= 180 − 105
= 74.7◦
θd at s = −1 − j
θd = arg GH(s) × (s − pi ) + 180◦
= 180◦ + arg
= 180◦ + arg
◦
= 180 + 105
K(s + 6)(s + 1 + j)
(s + 3)(s + 1 + j)(s + 1 + j)
s=−1−j
+ 6)
(s
+1+j)
K(−1 − j
(−1 − j + 3)(1 − j + 1 − j)
(s
+1+j)
s=−1−j
◦
= 285.3◦
= −74.7◦
Rule 9: Breakaway and Break-in Points
The point at which more than one root loci meet and then depart is called the breakaway point. The breakaway point is obtained by satisfying the condition
dK
=0
ds
(5.7)
in the characteristic equation 1+GH(s) and solving for s. Further, the point s so obtained
must also be a point on the root loci for some real value of K. Consider the loop T.F.
GH(s) =
K
s(s + 4)(s + 2)
1 + GH(s) = s3 + 6s2 + 8s + K = 0
109
5.1. THE CONCEPT OF ROOT LOCUS
Differentiating the above equation with respect to s and equating
dK
ds
= 0, we get
dK
=0
ds
3s2 + 12s + 8 = 0
3s2 + 12s + 8 +
s1 = −3.33 and s2 = −0.667. The root locus is shown in Figure 5.4. The breakaway
point s1 = −3.33 is not a point on the root locus, and hence it is not a breakaway
point. However, s2 = −0.667 is a point on the root locus. Hence, the breakaway point
is at s2 = −0.667.
Rule 10: Intersection of Root Loci with the Imaginary Axis
For some systems, as the parameter K is increased, the root loci cross the imaginary
axis. The value of K at this point c is calculated using the Routh–Hurwitz criterion.
The system is now marginally stable and has sustained oscillations, and the oscillation
frequency can also be determined by solving the auxiliary equation in Routh’s array.
Consider the following loop T.F.:
GH(s) =
K(s + 6)
(s + 3)(s2 + 2s + 2)
The root locus for the above T.F. is shown in Figure 5.5. The critical value or marginal
value of K when the root loci cross from LHP to RHP is calculated by forming Routh’s
array for the following characteristic equation:
1 + KGH(s) = 0
1+
K(s + 6)
=0
(s + 3)(s2 + 2s + 2)
(s + 3)(s2 + 2s + 2) + K(s + 6) = 0
s3 + 5s2 + (K + 8)s + 6(K + 1) = 0
s3
1
s2
5 6(K + 1)
s1 −K + 34
s0
K +1
110
K +8
5.1. THE CONCEPT OF ROOT LOCUS
For the system to be marginally stable, −K + 34 = 0 → K = 34. The auxiliary equation
from Routh’s array is written as
5s2 + 6(34 + 1) = 0
In the jω axis, s = jω and substituting this above equation we get 5ω 2 = 42 → ω =
6.48rad/sec This is represented in Figure 5.5.
Example 5.1. A certain negative feedback control system has the following loop T.F.
GH(s) =
K
s(s + 1)(s + 2)
Draw the root locus for 0 ≤ K ≤ ∞.
Solution
• For the OL TF GH(s) =
•
•
•
•
•
K
, there are 3 poles at s = 0, s = −1 and s =
s(s + 1)(s + 2)
−2 and no zeros.
The order of the numerator polynomial of GH(s) is zero and that of the denominator
polynomial is 3. Hence, there should be three separate root loci.[Rule 3]
The root loci on the real axis are identified between s = 0 and s = −1 and between
s = −2 and s = −∞.[Rule 7]
The root loci start from the poles of GH(s) where K = 0 and end at ∞ since there
are no zeros of GH(s). At s = ∞, the value of K = ∞.[Rule 1 and 2]
Number of asymptotes are Np − Nz = 3 − 0 = 3
The angle of the asymptote is given by
θn =
(2n + 1) × 180◦
(Np − Nz )
where Np = 3 and Nz = 0; n = 0, 1, · · · , |Np − Nz | − 1
θn = 60◦ , 180◦ and 300◦
• The centroid (intersection of the asymptote with the real axis) is calculated from
P
σc =
P
P− Z
Np − Nz
0 + (−1) + (−2) − 0
= −1
3−0
The two asymptotes with θ0 = 60◦ and θ2 = 300◦ or − 60◦ are drawn in dashed
lines. The third asymptote with θ1 = 180◦ coincides with the negative real axis.
σc =
111
5.1. THE CONCEPT OF ROOT LOCUS
• The root loci starting from s = 0 and s = −1 move on the negative real axis
between s = 0 and s = −1, meet each other and breakaway. The breakaway point
is calculated as,
1 + GH(s) = 0
K
=0
1+
s(s + 1)(s + 2)
s3 + 3s2 + 2s + K = 0
Differentiating the above equation with respect to s, we get
3s2 + 6s + 2 +
Setting
dK
ds
dK
=0
ds
= 0, we get
3s2 + 6s + 2 = 0
Solving for s, we get the breakaway points at s = −1.58 and s = −0.423 Since at
s = −1.58, there is no root locus, the point s = −0.423 alone is considered.
At the breakaway point, the gain is calculated as
K = 0.4226 × 1.5774 × 0.5774 = 0.385
• The critical value of K which is denoted as Kc when the root loci cross from LHP
to RHP is determined along with the frequency ω at that point as follows. The
characteristic equation of the system is
s3 + 3s2 + 2s + K = 0
The following Routh’s array is formed.
s3
1
s2
3 K
s1
6−K
3
s0
K
2
The system becomes oscillatory when K = 6. Hence, the critical value of K is
Kc = 6. The auxiliary equation is
3s2 + Kc = 0
112
5.1. THE CONCEPT OF ROOT LOCUS
Substituting s = jω and Kc = 6 in the above equation, we get
−3ω 2 + 6 = 0 → ω =
√
2
• The third root loci which starts from s = −2 moves towards the left on the −ve
real axis and goes to −∞, where K = ∞.
• The asymptotes, the breakaway point and the intersection of the root loci with the
imaginary axes give guidance for the movements of the root loci. They are joined
by a smooth curve and the complete root locus diagram is shown in Figure 5.6.
Figure 5.6: Root Locus of Example 5.1
Example 5.2. A unity feedback control system has an open loop transfer function.
G(s) =
s(s2
K
+ 4s + 13)
Draw the root locus as the value of K changes from 0 to ∞. Also find the value of K
and the frequency at which the root loci crosses the jω axis.
Solution
• OL TF
G(s) =
•
•
•
•
s(s2
K
+ 4s + 13)
Np = 3 and Nz = 0
Poles are at s = 0, s = −2 + j3, and s = −2 − j3
Order of 1 + GH(s) = 0 is three, therefore Root Locus has three branches.
Number of asymptotes is Np − Nz = 3
113
5.1. THE CONCEPT OF ROOT LOCUS
• Centroid of asymptotes
σc =
−4
0 + (−2 + j3) + (−2 − j3) − 0
=
= −1.333
3
3
• The angles the asymptotes make with the real axis
θn =
(2n + 1) × 180
where n = 0, 1, . . . |Np − Nz | − 1
|Np − Nz |
(2 × 0 + 1) × 180
= 60◦
3−0
(2 × 1 + 1) × 180
θ1 =
= 180◦
3−0
(2 × 2 + 1) × 180
θ2 =
= 300◦
3−0
θ0 =
• Angle of departure θd from complex pole p = −2 + j3 is found as,
θd = arg GH(s) × (s − p)|s=p + 180◦
= 180◦ + arg
= 180◦ + arg
◦
K(s + 2 − j3)
s(s + 2 + j3)(s + 2 − j3)
s=−2+j3
K(s + 2 − j3)
(−2 + j3)(−2 + j3 + 2 + j3)(s + 2 − j3)
s=−2+j3
◦
= 180 + 146.3
= −33.7◦
• Similarly angle of departure θd from complex pole p = −2 − j3 is
θd = −33.7◦
• Angle of departure from complex conjugate pole can also be found graphically, and
by using Equation (5.6). This is illustrated in example 5.2
θd = 180◦ − (θp − θz ) = 180◦ − (123.7 + 90 − 0) = −33.7◦
θd = −33.7◦
The critical value of K which is denoted as Kc when the root loci cross from LHP to
RHP is determined along with the frequency ω at that point as follows. The characteristic
equation of the system is
s3 + 4s2 + 13s + K = 0
114
5.1. THE CONCEPT OF ROOT LOCUS
The following Routh’s array is formed.
s3
1 13
s2
4
s1
52−K
4
s0
K
K
The system becomes oscillatory when K = 52. Hence, the critical value of K is Kc = 52.
The auxiliary equation is
4s2 + Kc = 0
Substituting s = jω and Kc = 52 in the above equation, we get
−4ω 2 + 52 = 0 → ω =
√
13 = 3.6rad/sec
The asymptotes and the intersection of the root loci with the imaginary axes give guidance
for the movements of the root loci. They are joined by a smooth curve and the complete
root locus diagram is shown in Figure 5.7.
Example 5.3. The transfer function of a unity feedback system is
G(s) =
K
s(s + 5)
Draw the root locus plot and determine the value of K for a damping ratio of 0.707.
Solution
115
5.1. THE CONCEPT OF ROOT LOCUS
Figure 5.7: Root Locus plot of GH(s) =
K
s(s2 +4s+13)
Example 5.2
• The two poles are at s = 0 and s = −5. There is no zero. Number of root locus
branches are 2. Since there is no open loop zero, both the root loci will terminate
at ∞.
• The angle of the asymptotes,
(2n + 1)180
2−0
(2 × 0 + 1)180
= 90◦
θ0 =
2−0
(2 × 1 + 1)180
= 270◦
θ1 =
2−0
θn =
• The asymptotes intersect at the negative real axis as the controid,σc
σc =
0 + (−5) − 0
= −2.5
2−0
• The breakaway point is calculated as,
1 + GH(s) = 0
K
1+
=0
s(s + 5)
s(s + 5) + K = 0
s2 + 5s + K = 0
dK
2s + 5 +
=0
ds
Setting
dK
ds
=0
→ s = −2.5
116
5.1. THE CONCEPT OF ROOT LOCUS
• The breakaway point and the point of intersection of asymptotes on real axis is the
same.
• From s-plane
ωn cos θ = δωn → θ = cos−1 (0.7071) = 45◦
We will draw the δ-line at 45◦ with the negative real axis as shown in Figure 5.8.
The δ-line intersects the root locus at R.
Figure 5.8: Root locus plot of GH(s) =
ratio of 0.707
K
s(s+5)
and determination of gain K for a damping
• At point R, the value of s = −2.5 + j2.5 Since R is on the root locus, the magnitude
criterion of |G(s)H(s)| = 1 will be satisfied. Therefore,
|G(s)H(s)| = 1
K
=1
s(s + 5)
K
=1
(−2.5 + j2.5)[(−2.5 + j2.5) + 5]
K
p
=1
[(6.25 + 6.25)(6.25 + 6.25)]
K
= 1 → K = 12.5
12.5
117
5.1. THE CONCEPT OF ROOT LOCUS
Short Answer Type Questions
• What is root locus? Root locus is the trajectory of the closed loop poles as the
gain of the system is varied in the range 0 ≤ K ≤ ∞.
• What is complementary root locus? The root locus drawn for the variation of
the system parameter K in the range −∞ ≤ K ≤ 0 is called complementary root
locus.
• What criteria are used to construct root locus? Magnitude and angle criteria
are used to construct root locus.
• What criterion decides the shape of the root locus trajectories? Angle
criterion decides the shape of the root locus trajectories.
• How magnitude criterion is used in the root locus plot? The root loci are
drawn for the variation of the gain parameter K. The values of K at different
points of the root loci are determined using magnitude criterion. At any point s1
in the s-plane, the K value is calculated from
K=
Product of directed distances drawn from all the poles to the point s1
Product of directed distances drawn from all the zeros to the point s1
• What is the nature of symmetry of root locus in the complex s-plane?
Root loci are symmetrical with respect to the real axis of the complex s-plane. They
have the complex conjugate property.
• What are the important properties of root loci? Some of the important
properties of root loci include the starting of the root loci, the ending of the root
loci, the number of root loci, the centroid, the angle of intersection of the asymptote,
breakaway and break-in points, the points at which root loci cross the jω axis, root
loci on the real axis, angle of departure and angle of arrival.
• Given the loop T.F. GH(s), how do you determine the total number of
root loci? For the given GH(s), the number of root loci is determined from the
number of poles or zeros of GH(s) whichever is greater.
• Given the loop T.F. GH(s), how do you determine the total number of
root loci? For the given GH(s), the number of root loci is determined from the
number of poles or zeros of GH(s) whichever is greater.
118
5.1. THE CONCEPT OF ROOT LOCUS
• What are the starting and ending points of the root loci? The root loci
start from poles of GH(s) where the value of K = 0 and end at zeros or ∞ where
the value of K = ∞ if Np , the number of poles of GH(s) is greater than or equal
to Nz the number of zeros of GH(s).
• There are three poles and five zeros. From where do the root loci start
and where do they end? There will be five root loci. Three of them will start
from the three poles and end at three zeros. The remaining two root loci start from
∞ and end at the remaining two zeros.
• Do the zeros of the system change their positions as a result of variation
of the gain K? If the system gain K is varied, the positions of the zeros of the
system in the s-plane are unaffected.
• What are asymptotes of root loci? The asymptotes of the root loci give
guidance to the actual root loci at s → ∞.
• How do you get the asymptotes for the given GH(s)? The asymptotes are
identified by their intersection and the angles with the real axis. The intersect is
given by
P
P
P− Z
σc =
Np − Nz
The angles of the asymptotes are given by
θ=
(2n + 1)π
Np − Nz
• What are the necessary conditions that the breakaway points must satisfy?
Will it be sufficient also? The necessary condition that the breakaway point
must satisfy is that
d[GH(s)]
dK
=0=
ds
ds
The above condition is not sufficient because the point obtained may not be a
point in the root locus for some real value of K. Hence, it is further necessary and
sufficient that the value of s so obtained is a point on the root locus.
• What are the methods available to determine the point of intersection
of the root loci with the jω-axis?
– By using the Routh–Hurwitz stability criterion, the critical value Kc can be
determined. Using this value in the auxiliary equation, the point of intersections
of the root loci with the jω axis can be determined.
119
5.1. THE CONCEPT OF ROOT LOCUS
– By trial and error, any point ω1 in the jω axis is chosen that satisfies the angle
criterion. By using the magnitude criterion, the critical value Kc is obtained.
• How is the absolute stability of the closed loop system determined using
root locus plot? For the given value of K, the roots of the characteristic equation
are found in their respective root locus. If all the roots so found are in LHP, the
closed loop system is absolutely stable.
• Write down the expression for the angle of departure and angle of arrival.
The angle of departure is given by
θd = 180◦ − (θp − θz )
θa = 180◦ − (θz − θp )
where θp is the sum of angles of vectors drawn from other poles to the pole concerned
and θz is the sum of angle of vectors drawn from other zeros to the pole concerned.
• For the given K, how do you find the closed loop poles and the closed
loop T.F.? If the value of K is given, in each root locus, using magnitude criterion,
deter- mine the roots by trial and error. These roots are nothing but the roots of
the characteristic equation or the poles of the closed loop system. The closed loop
T.F. is determined using the equation
G(s)
C(s)
=
R(s)
1 + G(s)H(s)
The numerator of [1 + GH(s)] is represented by the poles of the characteristic
equation.
• For a given damping ratio δ, how do you find the closed loop poles from
root locus? For the dominant complex conjugate poles, the damping factor is
defined as ϕ = cos−1 δ. Draw a radial line from the origin of the s-plane with an
angle (180◦ − ϕ) which cuts the dominant root loci. Using magnitude criterion, find
the value of K at this point. This point and the conjugate of this give the closed
loop roots. For the remaining root loci, the roots are found, by trial and error for
the value of K fixed on the dominant root loci. All these poles represent closed
loop poles.
• A certain closed loop unity feedback control system has the following open loop
T.F.:
K
G(s) =
s(s + 5)(s + 10)
120
5.1. THE CONCEPT OF ROOT LOCUS
Draw the root locus for 0 ≤ K ≤ ∞. For K = 70, find the closed loop roots. For
this value of K, determine the closed loop T.F.
Ans: Closed loop poles are at s1 = −11.05; s2,3 = −2 ± j1.6, RL is shown in
Figure 5.9
Figure 5.9: RL for GH(s) =
K
s(s+5)(s+10)
Example 5.4. A feedback control system has the following open loop transfer function,
G(s) =
K
s(s2 + 4s + 13)
Using asymptotes, sketch the root locus diagram for the closed loop system and find,
• The angle of departure from any complex open – loop poles.
• The frequency of transient oscillations at the onset of instability.
• The value of K to give dominant closed loop poles a damping ratio δ = 0.25
Solution
•
•
•
•
Open-loop poles: s = 0, s = −2 ± j3 → Np = 3
Open-loop zeroes: None → Nz = 0
Number of Root Loci branches: 3
Number of asymptotes Np − Nz = 3 − 0 = 3
121
5.1. THE CONCEPT OF ROOT LOCUS
• Asymptotes angles
(1 + 2n)180◦
Np − Nz
(1 + 0)180◦
180◦
θ0 =
=
= 60◦
3−0
3
3 × 180◦
(1 + 2)180◦
=
= 180◦
θ1 =
3−0
3
(1 + 4)180◦
5 × 180◦
θ2 =
=
= 300◦
3−0
3
θn =
• Asymptote Intersection
σc =
(0 + (−2 + j3) + (−2 − j3)) − (0)
= −1.333
3−0
• Breakaway points: None
• Imaginary axis crossover
Characteristic Equation
1 + GH(s) = 0
K
1+
=0
s(s2 + 4s + 13)
s(s2 + 4s + 13) + K = 0
s3 + 4s2 + 13s + K = 0
set s = jω
(jω)3 + 4(jω)2 + 13(jω) + K = 0
−jω 3 − 4ω 2 + j13ω + K = 0
(K − 4ω 2 ) + j(13ω − ω 3 ) = 0
K − 4ω 2 = 0 → K = 4ω 2 and 13ω − ω 3 = 0 → ω 2 = 13
Imaginary axis intercept
√
ω = ± 13 = ±3.61
K = 4ω 2 = 4 × 13 = 52
• Angle of departure
122
5.1. THE CONCEPT OF ROOT LOCUS
From
s = −2 + j3
K(s + 2 − j3)
+ 180◦
θd = arg
s(s + 2 − j3)(s + 2 + j3) s=−2+j3
K
θd = arg
+ 180◦
s(s + 2 + j3) s=−2+j3
K
θd = arg
+ 180◦
(−2 + j3)(−2 + j3 + 2 + j3)
K
+ 180◦
θd = arg
(−2 + j3)(j6)
θd = 146.3◦ + 180◦ = 326.3◦ = −33.7◦
• Angle of departure from s = −2 − j3 is 33.7◦
• Point that correspond to δ = 0.25
cos β = δ → β = cos−1 0.25 = 75.6◦
β = 75.6◦
Draw a damping ratio line as in Figure 5.10 and read the point of intersection with
complex conjugate RL.
Figure 5.10
• Applying angle criterion
K = 3.0 × 6.0 × 1.25 = 22.5
123
Chapter 6
Frequency Response
There are two methods of analysis of systems to determine certain properties, so that
design procedures can be developed, using these properties as performance measures. If
a time signal like step and ramp are used to excite the system and its time response
is obtained, we call it a time response analysis or time domain analysis. On the other
hand, if a sinusoidal signal of variable frequency is used to excite the system and the
magnitude and phase of the steady state output from the system is measured, we call it
frequency response analysis or frequency domain analysis. Both the methods have their
own advantages and disadvantages.
The frequency response is defined as the plot of magnitude and phase difference
between the input and output sinusoids over a range of frequencies.
6.1 Time Domain Analysis Vs Frequency Domain
Analysis
In the following a comparison of time domain and frequency domain analysis is given.
• Variable frequency, sinusoidal signal generators are readily available and precision
measuring instruments are available for measurement of magnitude and phase angle.
The time response for a step input is more difficult to measure with accuracy.
• It is easier to obtain the transfer function of a system by a simple frequency domain
test. Obtaining transfer function from the step response is more tedious.
• If the system has large time constants, it makes more time to reach steadystate
at each frequency of the sinusoidal input. Hence time domain method is preferred
over frequency domain method in such systems.
124
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
• In order to do a frequency response test on a system, the system has to be isolated
and the sinusoidal signal has to be applied to the system. This may not be possible
in systems which can not be interrupted. In such cases, a step signal or an impulse
signal may be given to the system to find its transfer function. Hence for systems
which cannot be interrupted, time domain method is more suitable.
• The design of a controller is easily done in the frequency domain method than in
time domain method. For a given set of performance measures in frequency domain.
the parameters of the open loop transfer function can be easily adjusted.
• The effect of noise signals can be assessed easily in frequency domain rather than
time domain.
• The most important advantage of frequency domain analysis is the ability to obtain
the relative stability of feedback control systems. The Routh Hurwitz criterion is
essentially a time domain method which determines the absolute stability of a
system. Nyquist criterion, which will be described in coming sections, will not only
give stability but also relative stability of the system without actually finding the
roots of the characteristic equation.
6.1.1 Relationship between s and jω
Consider sinusoidal input,
r(t) = A sin ωt In complex form,
r(t) = Aejωt
dr(t)
= jω Aejωt = jωr(t) Taking Laplace trasnform,
dt
sR(s) = jωR(s) or
s = jω
Hence, for sinusoidal input, the steady-state system response may be calculated by
substituting s = jω into the transfer function and using the laws of complex algebra
to calculate magnitude and phase angle.
Consider the linear time invariant system whose transfer function is G(s). The system is
excited with a sine wave
r(t) = R sin ωt
125
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
The system representation and input and output wave forms are shown in Figure 6.1
representing the steady state response of the system to a sinusoidal input which can be
directly obtained from G(s) by replacing s by jω.
C
= G(jω)
R(jω)
is called the sinusoidal transfer function, which is the ratio of C(jω) to R(jω). Thus,
by knowing the T.F. G(s), the magnitude characteristic |G(jω)| and phase characteristic
∠(jω) completely describe the steady state frequency response of the system. Given the
poles and zeros of G(s), the frequency response at any frequency can be evaluated.
Figure 6.1: (a) Linear system with sine wave input and sine wave output. (b) Inputoutput wave forms with their amplitude and phase
Example 6.1. A certain control system has the following transfer function
G(s) =
5(s + 2)
(s + 1)(s + 4)
Find the frequency response at ω = 5rad/sec.
Solution
For any frequency, the response can be obtained by setting s = jω
5(s + 2)
(s + 1)(s + 4)
5(jω + 2)
G(jω) =
(jω + 1)(jω + 4)
p
5 (ω 2 + 4)
p
|G(jω)| = p
at ω = 5 → |G(jω)| = 0.8247
(ω 2 + 1) (ω 2 + 16)
−1 ω
−1 ω
−1 ω
∠G(jω) = tan
− tan
− tan
at ω = 5 → ∠G(jω) = −61.83◦
2
1
4
G(s) =
126
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
6.1.2 Frequency-Domain Specifications
In the design of linear control systems using the frequency-domain methods, it is necessary
to define a set of specifications so that the performance of the system can be identified.
Specifications such as the maximum overshoot, damping ratio, and the like used in the
time domain can no longer be used directly in the frequency domain. The following
frequency domain specifications are often used in practice.
Definition 21 (Resonant Peak Mr ). The resonant peak Mr is the maximum value of
M (jω).
The magnitude of Mr gives indication on the relative stability of a stable closed-loop
system. A large Mr corresponds to a large maximum overshoot of the step response. For
most control systems, it is generally accepted in practice that the desirable value of Mr
should be between 1.1 and 1.5.
Definition 22 (Resonant Frequency ωr ). The resonant frequency ωr , is the frequency at
which the peak resonance Mr occurs.
Definition 23 (Bandwidth BW). The bandwidth BW is the frequency at which M (jω)
drops to 70.7% of, or 3dB down from, its zero-frequency value.
In general, the bandwidth of a control system gives indication on the transient response
properties in the time domain. A large bandwidth corresponds to a faster rise time,
since higher-frequency signals are more easily passed through the system. Conversely,
if the bandwidth is small, only signals of relatively low frequencies are passed, and the
time response will be slow and sluggish. Bandwidth also indicates the noise-filtering
characteristics and the robustness of the system. The robustness represents a measure
of the sensitivity of a system to parameter variations. A robust system is one that is
insensitive to parameter variations.
Resonant Peak and Resonant Frequency of second order system
Consider a second -order system, with transfer function given as,
M (s) =
ωn 2
C(s)
= 2
R(s)
s + 2δωn s + ωn 2
The steady state sinusoidal response is obtained by substituting s = jω in above equation,
127
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
M (jω) =
Setting u =
ω
,
ωn
C(jω)
ωn 2
=
R(jω)
(jω)2 + 2δωn (jω) + ωn 2
(6.1)
then the above equation simplifies to
M (ju) =
1
1 + j2uδ − u2
(6.2)
The magnitude and phase of M (ju) are given,
1
|M (ju)| =
1
(6.3)
[(1 − u)2 + (2δu)2 ] 2
and
∠M (ju) = − tan−1
2δu
1 − u2
(6.4)
Resonant Peak and Resonant Frequency
The resonant frequency is determined by setting the derivative of |M (ju)| with respect
to u to zero. Thus,
− 3
1
d|M (ju)|
= − (1 − u2 )2 + (2δu)2 2 (4u3 − 4u + 8uδ 2 ) = 0
du
2
(6.5)
4u3 − 4u + 8uδ 2 = 4u(u2 − 1 + 2δ 2 ) = 0
(6.6)
Simplifies to,
In normalized frequency, the roots of equation 7.13, ur = 0, and
√
1 − 2δ 2
(6.7)
√
ωr
= 1 − 2δ 2
ωn
(6.8)
ur =
Since frequency is a real quantity, then, 2δ 2 ≤ 1 or δ ≤ 0.707. Substituting Equation 6.7
into Equation 6.3 for u and simplifying, we arrive at,
Mr =
1
forδ ≤ 0.707
2δ 1 − δ 2
√
128
(6.9)
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
For the second-order system, Mr is a function of the damping ratio δ only, and ωr , is a
function of both δ and ωn .
Bandwidth - BW
In accordance with the definition of bandwidth, we set the value of |M (ju)| to
|M (ju)| =
Thus,
1
1
=√
2
[(1 − u2 ) + (2δu)2 ]
1
2
1 √
(1 − ub 2 ) + (2δub )2 2 = 2
√1
2
∼
= 0.707
(6.10)
(6.11)
Squaring on both sides of Equation 6.11 and simplifying, we have
ub 4 − 2ub 2 + 4δ 2 ub 2 = 2
(6.12)
ub 4 − 2(1 − 2δ 2 )ub 2 − 1 = 0
(6.13)
Solving for ub we arrive at,
s
p
4(1 − 2δ 2 )2 + 4
ub =
2
q
√
ub = (1 − 2δ 2 ) + 2 − 4δ 2 + 4δ 4
2(1 − 2δ 2 ) ±
(6.14)
(6.15)
Note that only positive sign is considered in Equation above, since ub must be positive
and real, ub - is the normalized bandwidth. From equation above we get bandwidth ωb
as,
q
√
(6.16)
ωb = ωn (1 − 2δ 2 ) + 2 − 4δ 2 + 4δ 4
6.1.3 Correlation Between Time Response and Frequency Response
From previous section derivations, we have established some simple relationships between
the time-domain response and the frequency-domain characteristics of the second-order
system. The time doman specifications are obtained by subjecting the second order
system to a unit step input. The important time domain specification are,
129
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
Peak Overshoot
For 0 ≤ δ ≤ 1
Mp = 100e
− √ πδ
1−δ 2
%
(6.17)
Damped frequency of oscillation
√
ωd = ωn 1 − δ 2
(6.18)
Settling Time
For tolerance of ±0.02%
ts ≃
4
.
δωn
(6.19)
In frequency domain, the second order system is subjected to a constant amplitude,
variable frequency. sinusoidal input and the magnitude and phase response are obtained.
The important frequency domain specifications are,
Resonant peak
For δ < 1
1
2δ 1 − δ 2
(6.20)
√
ωr
= 1 − 2δ 2
ωn
(6.21)
Mr =
√
Resonance frequency
Bandwidth
q
ωb = ωn
(1 − 2δ 2 ) +
130
√
2 − 4δ 2 + 4δ 4
(6.22)
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
Correlation between Mp and Mr
From Equation 6.17 and 6.20, we see that both Mp and Mr are dependent on the damping
factor δ only and hence they are both indicative of damping in the system. Given Mp , the
resonant peak Mr can be evaluated provided δ < 0.707.This condition is usually satisfied
by many practical control systems. Thus the resonant peak Mr and peak overshoot are
well correlated. Comparison of Equation 6.18 and 6.21 reveals the correlation between
the time domain specification of damped natural frequency and the frequency domain
specification of resonant frequency. For a given damping factor δ, the ratio ωωdr is fixed and
given the frequency domain specification, the corresponding time domain specification
and vice versa, can be easily obtained.
Settling Time and Bandwidth
The speed of response is indicated by settling time in time domain as given in Equation
6.19.The bandwidth, a frequency domain concept, given by Equation 6.22, is also indicative
of speed of response. Thus we can see that there is a perfect correlation between time
domain and frequency domain performance measures : given one, the other can be
obtained easily. Note that the correlation is valid only for δ < 0.707, which is usually
satisfied in many practical control systems.
Example 6.2. A certain unity feedback control system when subjected to sinusoidal
input has a resonant peak of 1.2 and the corresponding frequency is 8rad s−1 . Determine
the bandwidth of the closed loop system. Also determine the % overshoot, rise time and
settling time when subjected to step input.
Solution
1
= 1.2( given )
Mr = √
2δ 1 − δ 2
1
√
2δ 1 − δ 2
1
(Mr )2 = 2
4δ (1 − δ 2 )
1
1
δ 2 (1 − δ 2 ) =
=
2
4(Mr )
4 × 1.22
Mr =
δ 4 − δ 2 + 0.1736 = 0
δ 2 = 0.7764 → δ = 0.882
131
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
or
δ 2 = 0.2236 → δ = 0.473
Resonance occurs for δ < 0.707
∴ δ = 0.473
√
ωr = ωn 1 − 2δ 2
√
8 = ωn 1 − 2 × 0.4732
ωn = 10.76rad s−1
Bandwidth
q
√
ωb = ωn (1 − 2δ 2 ) + 2 − 4δ 2 + 4δ 4
q
√
ωb = 10.76 (1 − 2 × 0.4732 ) + 2 − 4 × 0.4732 + 4 × 0.4734
ωb = 14rad s−1
% overshoot
Mp = e
Mp = e
− √ δπ
1−δ 2
− √ 0.473π
1−0.4732
Mp = 18.52%
Time rise tr
π−ϕ
ωd
√
√
ωd = ωn 1 − δ 2 = 10.76 1 − 0.4732 = 9.481
tr =
ϕ = cos−1 δ = cos−1 0.473 = 61.77◦ or 1.078rad
π − 1.078
∴ tr =
= 0.2177s
9.481
132
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
Figure 6.2: Exercise 6.5
Settling time ts
tr =
3.91
δωn
tr =
3.91
= 0.767s
0.473 × 10.76
Exercise 6.3. The open loop transfer function of certain control system is given as
G(s) =
64
s(s + 8)
Determine the resonant peak, resonant frequency and the bandwidth of the closed loop
system.
(Answer: ωn = 8rad s−1 , δ = 0.5, Mr = 1.1547, ωr = 5.7rad s−1 , ωb = 10.2rad s−1 )
Exercise 6.4. A certain second order closed loop system when subjected to step input
exhibits a peak overshoot of 16.3% and the rise time is observed as 0.2s. Find the
values of resonant peak Mr , resonant frequency ωr and the bandwidth ωb . (Answer:
Mr = 1.155, ωr = 8.556rad s−1 , ωb = 14.82rad s−1 )
Exercise 6.5. For the system shown in Figure 6.2 find the value of K and b to satisfy
the following frequency domain specifications. Mr = 1.2; ωr = 14rad s−1 . For these
values of K and b, calculate the settling time and bandwidth.(Answer: K = 354.72, b =
17.817, ωb = 25.673rad s−1 )
Exercise 6.6. Construct the Bode plot for a unity feedback control system having
G(s) =
10
s(s + 1)(s + 10)
From the plot obtain the gain and phase margin and their corresponding frequencies.
Comment on the stability of the system.
Answer
133
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
Figure 6.3: Exercise 6.6
Exercise 6.7. Construct the Bode plot for a unity feedback control system having
G(s) =
80
s(s + 2)(s + 20)
From the plot obtain the gain and phase margin and their corresponding frequencies.
Comment on the stability of the system.
Answer
134
6.1. TIME DOMAIN ANALYSIS VS FREQUENCY DOMAIN ANALYSIS
Figure 6.4: Exercise 6.7
135
6.2. PLOTTING FREQUENCY RESPONSE
6.2 Plotting Frequency Response
The frequency response of a system can be plotted in several ways using the sinusoidal
transfer function G(jω). As we vary input frequency over a range, we get a vector of
frequency response data in the form of a complex vector. The complex vector can be
displayed in a variety of ways as given below.
• Polar (Nyquist) Plot: This is the plot of the imaginary versus the real part of
G(jω) drawn in a polar graph with ω as the varying parameter.
• Bode Plots: It has two separate plots one for the magnitude and another for phase
angle of G(jω). The magnitude is expressed in decibel (20 log 10|G(jω)|). By this
very large magnitude can be accommodated. The horizontal axes are logarithmic
by which we can accommodate larger frequency scale.
• Log Magnitude versus Phase Plot (Nichols Plot)
• This is the plot of the magnitude of G(jω) in dB versus the phase in degrees with
ω as the varying parameter.
6.2.1 Bode Plots
Bode plot contains two separate plots and they are called magnitude and phase plots.
In magnitude plot the gain |G(jω)| is expressed in decibel (dB) and displayed in the
vertical axis. 20 log10 |G(jω)| is called a decibel which is written in the abbreviated
form as dB. The horizontal axis represents frequency on a log scale. In the second plot
which is called the phase plot ∠G(jω) is plotted in the vertical axis while frequency
ω is represented in the horizontal axis which is graduated on a log scale. These plots
give useful information for designing a variety of control systems. Stability criteria can
be interpreted on Bode plots, and furthermore, there are numerous design techniques
available which are based on Bode plots.
Advantages of Bode Plots
• Bode plots are sketched by approximating the magnitude and phase with straight
lines which are easily sketched in Bode plot.
• Since the horizontal axis is on log scale, it takes care of logarithmic value of
frequency ω. Thus, a wide range of frequency study is possible with clarity at
all frequencies.
136
6.2. PLOTTING FREQUENCY RESPONSE
• The vertical axis represents the magnitude of G(jω) which is expressed in logarithmic
form. Very large amplitude of |G(jω)| can be represented in this way.
• The magnitude of G(jω) is expressed in dB. The multiplications and divisions in
|G(jω)| become addition and subtraction in Bode plot display which can be easily
manipulated. Similarly, phase angles of zeros and poles of G(jω) are added and
subtracted from each other algebraically.
• Frequency response specifications such as phase margin, gain margin, phase crossover
and gain crossover frequencies are more easily determined from the Bode plot than
the polar plot.
• In the design using Bode plots, the effects of adding controllers and their parameters
can be easily visualized than on the polar plot.
Disadvantages of Bode Plot
• If the Bode plots are manually drawn without the use of computers, the plots
become approximate. Unless otherwise, necessary corrections are made in appropriate
places in the Bode plots, which is very tedious to accomplish, the results obtained
deviate from the actual ones.
• Absolute and relative stability of the system are obtained from Bode plots provided
the system under study is of minimum phase. For non-minimum phase transfer
function, the Bode plot cannot be used to assess the absolute and relative stability.
6.2.2 Bode Asymptotic Magnitude Plot
Consider the following transfer function
G(s) =
K(1 + sτ1 )(1 + sτ2 ) . . . (1 + sτm )
sk (1 + sτa )(1 + sτb ) . . . (1 + sτn )
(6.23)
Equation (6.23) is said to be written in Time Constant Form. The sinusoidal T.F. of the
eq. (6.23) is written as
137
6.2. PLOTTING FREQUENCY RESPONSE
K(1 + sτ1 )(1 + sτ2 ) . . . (1 + sτm )
sk (1 + sτa )(1 + sτb ) . . . (1 + sτn )
K(1 + jωτ1 )(1 + jωτ2 ) . . . (1 + jωτm )
G(jω) =
(jω)k (1 + jωτa )(1 + jωτb ) . . . (1 + jωτn )
p
2)
K (1 + ω 2 τ12 )(1 + ω 2 τ22 ) . . . (1 + ω 2 τm
|G(jω)| = p
ω k (1 + ω 2 τa2 )(1 + ω 2 τb2 ) . . . (1 + ω 2 τn2 )
(6.24)
∠G(jω) = −k × 90◦ + (tan−1 ωτ1 + tan−1 ωτ2 + · · · + tan−1 ωτm )
(6.26)
G(s) =
− (tan−1 ωτa + tan−1 ωτb + · · · + tan−1 ωτn )
(6.25)
(6.27)
Bode magnitude plot is drawn for log10 |G(jω)| versus log10 ω and phase plot is drawn for
∠G(jω) versus log10 ω.
In Equation (6.24), G(jω) can contain just five simple factors
• G(jω) = K (a constant factor)
• G(jω) = K(jω)±p (poles or zeros at the origin of order p)
• G(jω) = (1 + jωτ )±q (poles or zeros at s = − τ1 of order q)
"
#±r
2
jωn
• G(jω) =
+ j2δωn ω + ωn2
(complex poles and zeros of order r)
ω
A constant factor G(jω) = K
G(s) = K
G(jω) = K
|G(jω)| = K
dB = 20 log K
The magnitude of G(jω) is independent of frequency and hence dB is constant at all
frequency. For K = 1 → dB = 20 log 1 = 0. For K > 1, dB is +ve and for K < 1 , dB
is -ve. The Bode magnitude plot is shown in Figure 6.5
138
6.2. PLOTTING FREQUENCY RESPONSE
Figure 6.5
Zero at the origin G(s) = Ks
G(s) = Ks
G(jω) = jKω
|G(jω)| = Kω
dB = 20 log Kω
1
, dB = 20 log 1 = 0
K
10
at ω =
, dB = 20 log 10 = 20
K
0.1
at ω =
, dB = 20 log 10−1 = −20
K
at ω =
This is a straight line. The Bode plot is shown in Figure 6.6(a). Figure 6.6(a) is drawn
using the following steps:
• Find ωl , the frequency at which the magnitude of G(jω) = 1. In the above case
|G(jω)| = Kω and hence ωl = K1 . At this frequency the dB is 20 log 1 = 0. This
corresponds to point A in Figure 6.6(a).
• Choose ω = 10ωl =
to point B.
10
.
K
At this point, the dB is 20 log 10K
= 20. This corresponds
K
• Join points A and B and extend it as much as you want. The straight line drawn
corresponds to Bode plot for G(jω) = jKω.
• The above straight line is called +20dB/decade line. This is the slope of the straight
line. The frequency ωl is calculated from Kωl = 1.
139
6.2. PLOTTING FREQUENCY RESPONSE
Figure 6.6
Now consider the following T.F.
G(s) = Ks2
|G(jω)| = Kω 2
Kωl2 = 1
1
ωl = √
K
At ω = 10ωl , dB = 20 log K
10
√
K
2
= 40dB
Thus, the straight line has a slope of +40dB/dec slope as shown in Figure 6.7 . In general
Figure 6.7
for G(s) = Ksp , the Bode plot is a straight line with a slope of p × 20dB/dec slope and
1
ωl = √
.
p
K
Pole at the origin G(s) =
K
s
Now consider the following T.F.
K
s
K
|G(jω)| =
ω
G(s) =
140
6.2. PLOTTING FREQUENCY RESPONSE
|G(jω)| = 1 at ω = K or ωl = K. At this frequency the dB = 20 log 1 = 0. And at
ω = 10ωl = 10K
K
K
= 20 log
= −20
dB = 20 log
ω
10K
Thus, the slope m here is −20dB/dec and the Bode plot is a straight line. In general,
√
for G(s) = sKp , ωl = p K and the slope is −p × 20dB/dec. These plots are shown in
Figure 6.8.
Figure 6.8
Zero on real axis G(s) = (1 + sτ )
G(s) = (1 + sτ )
G(jω) = (1 + jωτ )
p
|G(jω)| = (1 + ω 2 τ 2 )
(6.28)
Magnitude plot of Equation (6.28) can be approximated by considering two cases;
For ωτ ≪ 1
|G(jω)| = 1
dB = 20 log 1 = 0
For ωτ ≫ 1
|G(jω)| = ωτ
dB = 20 log Kω
The above plot is similar to Section 6.2.2 . It is a straight line with +20dB/dec slope
and ωl = τ1 . Thus, an approximate plot is drawn with 0dB up to ω = ωl = τ1 . Another
approximate plot is drawn with +20dB/dec slope for the frequency range ωl ≤ ω ≤
141
6.2. PLOTTING FREQUENCY RESPONSE
∞. These approximate plots are called Asymptotes. The points at which these two
asymptotes meet in the horizontal axis and the corresponding frequency is called Corner
frequency ωc . Thus, ωl is now replaced by the conventional symbol ωc when asymptotic
plots are drawn. The corner frequency of a Bode plot is therefore defined as the
frequency at which the two asymptotes (approximate plots) meet. The Bode
1
is the
plot for G(s) = (1 + sτ ) is shown in Figure 6.9. The Bode plot for G(s) = (1+sτ
)
Figure 6.9
mirror image of Figure 6.9 and is shown in fig. 6.10.
Figure 6.10
142
6.2. PLOTTING FREQUENCY RESPONSE
Complex quadratic factors
ωn2
G(s) = 2
s + 2δωn s + ωn2
ωn2
G(jω) =
−ω 2 + 2jδωn ω + ωn2
1
=
G(jω) = ω2
− ω2 + 2jδ ωωn + 1
1−
n
|G(jω)| = r
1
2
1−
ω2
2
ωn
ω2
ωn2
ω2
2
ωn
1
+ 2jδ ωωn
2
+ 4δ 2 ωω2
n
For ω ≪ ωn
|G(jω)| = 1
dB = 0
For ω ≫ ωn
|G(jω)| =
ω2
ωn2
dB = 20 log
= −40dB/dec at ωc = ωn
For the quadratic factor, the Bode asymptotic plot is drawn with −40dB/dec at straight
line with corner frequency ωc = ωn . This is depicted in Figure 6.11.
Note Phase
Figure 6.11
143
6.2. PLOTTING FREQUENCY RESPONSE
Table 6.1: Summary of Bode magnitude plot
Transfer function G(s)
K
Ks
Corner Freq. ωc
−
K
Slope dB/dec
0
−20
Transfer function G(s)
(1 + sτ )
(1 + sτ )2
1
τ
20
1
τ
40
Corner Freq. ωc
Slope dB/dec
K
s
1
K
K
s2
1
√
K
−40
20
40
1
(1 + sτ )
1
τ
−20
1
(1 + sτ )2
1
τ
−40
ωn2
(s2 + 2δωn s + ωn2 )
Ks2
√
K
ωn
−40
plots are normally drawn for the actual phase angle which is calculated at each ω. The
asymptotic phase plot is usually avoided.
Step by Step Procedure to Draw Bode Magnitude Plot
• The poles and zeros of the given T.F. is written in the time constant form as given
in Equation (6.23).
• Identify the low frequency term in the transfer function. This may be K, Ks, Ks2 , Ks ,
etc. The Bode plot starts at low frequency from any of these terms. If it contains
pure constant, the low frequency response starts with 20 log K. Otherwise find ωl
where Ks or Ks , etc. magnitude becomes 1 or its decibel is zero.
• Identify the corner frequencies of each pole and zero and write in the ascending
order. Locate these corner frequencies in the log ω scale in the horizontal axis.
Draw vertical dotted lines passing through these corner frequencies. Also locate ωl
in the log ω axis.
• Start the Bode plot for the low frequency part of the T.F. which will pass through
ωl had there been no other pole or zeros to interfere. (If there is no pure integrator
or differentiator, the low frequency response is due to K only and a horizontal line
is drawn).
• When a pole or a zero interferes with the low frequency magnitude response, stop
and calculate the dB at that point. If a zero interferes, the slope is changed by
+20dB/dec. If two zero interfere the slope is changed by +40dB/dec and so on.
On the other hand if a pole interferes, the slope of the straight line is changed by
-20dB/dec. When two poles interfere, the slope is changed by -40dB/dec and so
144
K
,
s2
6.2. PLOTTING FREQUENCY RESPONSE
on. Thus, with the new slope the magnitude plot is extended up to the next corner
frequency where poles or zeros interfere. At that point, the Bode plot is to be
stopped and the dB is calculated. Depending upon whether the zero interferes or
the pole interferes, the new slope of the magnitude plot is decided. The procedure
is repeated till the Bode plot crosses past all the corner frequencies and the dB
either goes to ∞, 0, finite or −∞ depending upon G(s).
Example 6.8. Consider the following T.F.
G(s) =
3000
s
(s + 2)(1 + 10
)(s + 50)
Sketch the Bode magnitude plot. What is the gain crossover frequency? Also determine
the position error constant Kp .
Solution
• Write T.F in Time Constant Form
3000
s
s
2 × + 1)(1 + 10
)( 50
+ 1) × 50
30
=
s
s
s
)(1 + 50
)
(1 + 2 )(1 + 10
G(s) =
( 2s
Write the factors of G(s) in ascending order of corner frequency as below;
G1 (s) = 30
1
(1 + 2s )
1
G3 (s) =
s
(1 + 10
)
1
G4 (s) =
s
(1 + 50
)
G2 (s) =
The range of corner frequencies and the corresponding slopes of the straight lines
are given in Table 6.2. The corner frequencies are ωc = 2, 10 and 50.
Table 6.2
Corner frequency range
ω≤2
2 ≤ ω ≤ 10
10 ≤ ω ≤ 50
ω ≥ 50
Slope dB/dec
0
−20
−40
−60
• The corner frequencies are identified as ωc = 2, 10 and 50. and marked in Figure 6.12.
145
6.2. PLOTTING FREQUENCY RESPONSE
• The low frequency factor of the T.F. is 30. Taking decibel for this we get
dB = 20 log 30 = 29.5
A horizontal line a-b starting from 29.5dB is drawn.
• At point b, a pole starts and changes the slope from 0 to −20dB/dec. Line bc is
drawn to represent this. The height h in the triangle bgc is obtained as
(higher frequency)
(lower frequency)
10
= 20 log
= 14
2
|h| = 20 log
The point c is located at (29.5 − 14) = 15.5dB.
Figure 6.12
• At point c, where the corner frequency is 10, a pole interferes and changes the slope
from -20dB/dec to -40dB/dec. Straight line c-d-e is drawn. The decibel at e is
calculated as in step above.
|h| = 40 log
50
= 28
10
The decibel at e equals 15.5 − 28 = −12.4.
• The frequency where the Bode magnitude plot crosses the 0dB line is
called gain crossover frequency. The gain crossover frequency is calculated as
40 log
ωgc
= 15.5 → ωgc = 24.4rad s−1
10
146
6.2. PLOTTING FREQUENCY RESPONSE
• At point e, another pole interferes and changes the slope from -40dB/dec to 60dB/dec. Any frequency greater than 50 is selected (say ω = 70). At this frequency
|h| = 60 log
70
= 8.8
50
The decibel at f equals −12.4 − 8.8 = −21.2. The straight line connecting e and f
is drawn and extended. This completes the Bode magnitude plot.
• The position error constant is calculated from the zero dB slope line.
20 log Kp = 29.5 → Kp = 30
Figure 6.12 is transferred to semi-log graph and is shown in fig. 6.13.
Figure 6.13
Exercise 6.9. Consider the following transfer function
s
15(1 + 10
)
G(s) =
s
s
s(1 + 4 )(1 + 20
)
Sketch the Bode magnitude plot. Find the velocity error constant. What is the gain
crossover frequency?
Answer Magnitude plot is shown in Figure 6.14
147
6.2. PLOTTING FREQUENCY RESPONSE
Figure 6.14
148
6.2. PLOTTING FREQUENCY RESPONSE
Exercise 6.10. Consider the unity feedback control system having
G(s) =
80
s(s + 2)(s + 20)
Find Phase margin ϕm , Gain margin Gm , phase cross over frequency ωpc and gain cross
over frequency ωgc using Bode diagram. Is the closed loop system stable?
Solution
ω
−90◦
− tan−1 (0.5ω)
− tan−1 (0.05ω)
Total
0.1
−90◦
−2.9◦
−0.3◦
−93◦
1
−90◦
−26.6◦
−2.9◦
−120◦
2
−90◦
−45◦
−5.7◦
−140◦
5
−90◦
−68.2◦
−14◦
−172◦
10
−90◦
−78.7◦
−26.6◦
−195◦
20
−90◦
−84.3◦
−45◦
−219◦
30
−90◦
−86.2◦
−56.3◦
−233◦
100
−90◦
−88.9◦
−78.7◦
−258◦
Gm = 20.8dB;ϕm = 47.4◦ ;ωpc = 6.32rad s−1 ;ωgc = 1.57rad s−1 .Both Gm and ϕm are
positive. ωpc > ωgc . The closed loop system is stable.
149
6.2. PLOTTING FREQUENCY RESPONSE
R(s)
+
−
2000(s+1)
s(s+10)(s+40)
C(s)
Figure 6.15: System representation in simple block diagram
150
6.2. PLOTTING FREQUENCY RESPONSE
Short Answer Questions
• Define frequency response. The frequency response is the steady state response
of the system to a sinusoidal input whose frequency is varied over a certain range.
• What are the two components of frequency response? The two components
of frequency response are the magnitude component and phase component.
• What is a sinusoidal transfer function? In the transfer function of a stable
system if s is replaced by jω, then G(s) becomes G(jω). G(jω) is called the
sinusoidal T.F.
• How magnitude and phase components of frequency response is obtained
for the given T.F.? In the T.F. G(s), if s is replaced by jω, then the sinusoidal
T.F. is obtained as G(jω) which is a complex quantity. —G(jω)— which is the
ratio of the output amplitude to input amplitude gives the magnitude component.
∠G(jω) gives the phase shift of the output from the input and this represents the
phase component of the frequency response.
• Mention any three major advantage of frequency response method. The
advantages of frequency response methods are as follows:
– It is easy to generate sine wave and collect data to draw frequency response
which can be used for design purpose.
– The frequency response tests involve measurement under steady state conditions
which are simpler to analyse.
– It can be applied to a certain class of non-linear system also.
• Mention any two disadvantages of frequency response method. The disadvantages
of frequency response methods are as follows:
– It is not often easy to correlate the results of the frequency response characteristics
of higher order system (order greater than two) to that of time domain characteristics.
– In frequency response method, the analysis is done at one frequency at a time
which requires very long test time.
• What are the methods available to represent the frequency response in
graph? The methods available to represent the frequency response in graph are as
follows:
– Linear frequency response plots vs. ω
– Linear frequency response plots vs. log ω
– Polar (Nyquist) plot
– Bode plot
– Log magnitude vs. phase (Nichols) plot.
151
6.2. PLOTTING FREQUENCY RESPONSE
• What is a polar (Nyquist) plot? Polar plot is the plot of the imaginary versus
the real part of the complex sinusoidal T.F. G(jω) drawn in a polar graph with ω
as the varying parameter.
• What are the advantages and disadvantages of polar plot representation
of frequency response? The advantage of polar plot representation of frequency
response is that both amplitude and phase angle are displayed on a single plot over
a wide frequency range. It has the property to show clearly, from the open loop
frequency response plot whether the closed loop system is stable or unstable. The
famous Nyquist stability criterion has been developed using polar plot only. The
only disadvantage of polar plot is that it does not clearly indicate the contributions
made by each pole and zero of the T.F.
• What are Bode plots or Bode diagrams? The log-magnitude and phase
frequency response curves plotted as a function of logarithmic frequency are called
Bode plots or Bode diagrams.
• Mention any three advantages of Bode plots. In Bode magnitude and phase
plots, the horizontal axis is on log scale. Since the horizontal axis represents
frequency, a wide range of frequency can be included with clarity at all the frequencies.
The vertical axis of the magnitude plot represents magnitude of G(jω) in dB. This
enables to provide large amplitude of G(jω) being represented. Multiplication and
division in the sinusoidal T.F. become addition and subtraction when magnitude is
expressed in dB. This provides clear information about the magnitude contribution
of each pole and zero at any frequency range.
• What are the disadvantages of Bode plots? Bode plots are sketched using
straight line approximation which introduces error. The estimation of relative
stability using phase margin and gain margin is applicable for minimum phase
T.F. and not for non-minimum phase T.F.
• What are asymptotes in Bode magnitude plot? Bode magnitude plot
is sketched by approximating the magnitude with straight line segments. These
approximate straight line segments are called the asymptotes.
• What is break frequency or corner frequency in Bode plot? The break
frequency or corner frequency which is denoted by ωc is the frequency at which the
low frequency asymptote and high frequency asymptote meet.
• Bode diagram is more suitable for what applications? Bode plot is an ideal
one for the determination of phase margin, gain margin, phase crossover and gain
crossover frequencies of minimum phase systems. It gives ωpc and ωgc very explicitly.
It is also widely used in the design of compensator where frequent adjustment of
poles and zeros of the compensators is required.
• Define bandwidth. The bandwidth BW is the frequency ωb at which |M (jω)|
drops to 70.7% of, or 3dB down from its zero frequency value.
152
6.2. PLOTTING FREQUENCY RESPONSE
• What are frequency domain specifications? The frequency domain specification
are as follows:
– Resonant peak,
– Resonant frequency,
– Bandwidth,
– Phase margin,
– Gain margin,
– Phase crossover frequency, and
– Gain crossover frequency.
• Define gain margin in Bode plot? In Bode plot, the gain margin is defined
as the amount of gain in dB that can be added to the loop before the closed loop
system becomes unstable.
• Define phase margin in Bode plot. In Bode plot, the phase margin is defined
as the amount of the phase angle that can be added to the loop before the closed
loop system becomes unstable.
• What is a polar (Nyquist) plot? Polar plot is the plot of the imaginary versus
the real part of the complex sinusoidal T.F. G(jω) drawn in a polar graph with ω
as the varying parameter.
• What are the advantages and disadvantages of polar plot representation
of frequency response? The advantage of polar plot representation of frequency
response is that both amplitude and phase angle are displayed on a single plot over
a wide frequency range. It has the property to show clearly, from the open loop
frequency response plot whether the closed loop system is stable or unstable. The
famous Nyquist stability criterion has been developed using polar plot only. The
only disadvantage of polar plot is that it does not clearly indicate the contributions
made by each pole and zero of the T.F.
153