Kenyatta University: April 10, 2019

Kenyatta University
BSc Mechanical Engineering
EMM 416: CONTROL SYSTEMS I
April 10, 2019

EMM 416: CONTROL SYSTEMS I
Prerequisites
EMM 311 Mechanics of machines II

EMM313 Measurement and Instrumentation
Objective
To equip the student with the knowledge on planning and control of production for maximum
productivity and profitability.
Expected Outcomes
At the end of this course, the student should be able to;
1. Distinguish between open loop and closed loop control systems
2. Determine the transfer function for a given system.
3. Describe various control elements
4. Assess the stability of a system
5. Describe the various control actions in control systems
Course Outline
Introduction to Automatic Control Systems: Dynamic models and dynamic responses: models of
dynamic systems in differential equation form. Linearization, amplitude and time scaling; Transfer
function representation of models; Time-domain effects such as rise time, overshoot, settling time;
Feedback control system concepts and stability: essential principles of feedback, direct block di-
agram modeling of feedback systems, effect of feedback on parameter sensitivity and disturbance
response, steady-state errors in feedback systems, transient response versus steady-state errors,
stability, Routh-Hurwitz stability criterion, relative stability of feedback, Root Locus method: root
loci, plotting of root loci, system design using root loci, phase lead and lag compensation using root
loci, system design using root loci, phase lead and lag compensation using root loci, computer-aided
plotting of root loci; Frequency-response methods: Bode plots, M and N-circles, lead-lag compen-
sation, frequency response performance specifications; Nyquist stability criterion, Nyquist diagram
and stability, gain and phase margins, closed-loop frequency response, stability of control systems
with time delays; Examples of frequency response design and analysis using a computer-aided con-
trol engineering tool such as MATLABâS Control System Toolbox.
i
Teaching Methodology
2 hour lectures and 1 hour tutorial per week, and at least five 3-hour laboratory sessions per
semester organized on a rotational basis.
Instructional Materials/Equipment
1. Control Engineering laboratories;
2. Computer laboratory;
3. Overhead projector;
Prescribed Text Books
1. Distefano J. J., Stubberud A. R. Williams I. J. (1994) Feedback and Control Systems: Theory
and Problems (Schaum’s Outline Series), McGraw-Hill, 2 Ed.
2. Ogata K. (1996) Modern Control Engineering, Prentice Hall, 3rd Ed.
References
1. Kuo B. C. Farid G. (2002) Automatic Control Systems, Wiley, 8th Ed.
2. Gene F. (2005) Feedback Control of Dynamic Systems, Prentice Hall, 5th Ed.
3. Journal of Dynamic Systems, Measurement, and Control
ii
Contents
1 Introduction to Control Systems 1
1.1 Definitions: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Requirements of a good control system . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.3 Classification of Control systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Transfer Function and Mathematical Modelling 7

2.1 Transfer Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Modeling of Electrical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Modelling of Mechanical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.4 Analogous Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3 Block Diagram Representation 30

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.2 Canonical form of Feedback Control System . . . . . . . . . . . . . . . . . . . . . . . 31
3.3 Rules for Block Diagram Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4 PID Controllers 43
4.1 Proportional (P) Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
4.2 Integral (I) Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
4.3 Derivative (D) Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
4.4 Proportional plus Integral (PI) Control . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4.5 Proportional plus Derivative (PD) Control . . . . . . . . . . . . . . . . . . . . . . . . 49
4.6 Proportional plus Integral plus Derivative (PID) Control . . . . . . . . . . . . . . . . 50
5 Time Domain Analysis 53

5.1 Transient Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.2 Steady State Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.3 Analysis of first order System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
5.4 Analysis of Second order system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
5.5 Time domain specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
6 Stability Analysis 77
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
6.2 Hurwitz Stability Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
6.3 Routh’s Stability Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
iii
7 Root Locus 85
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
7.2 Angle and Magnitude Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
7.3 Procedure for Drawing Root Locus . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
8 Frequency Domain Analysis 100
9 Bode Plots 108

9.1 Bode plots of standard factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
9.1.1 Bode gain K or Constant factor . . . . . . . . . . . . . . . . . . . . . . . . . 109
1
9.1.2 Poles at the origin of order k or integral factor . . . . . . . . . . . . . . 110
(jω)k
9.1.3 Zeros at the origin(jω)g of order g or derivative factor . . . . . . . . . . . . . 111
1
9.1.4 Finite poles at or first order integral factor . . . . . . . . . . . . 112
(1 + jω/p1 )
9.1.5 Finite zeros at (1 + jω/z1 ) or first order derivative factor . . . . . . . . . . . 113
9.1.6 Second order poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
9.1.7 Second order zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
9.2 Frequency Domain Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
9.2.1 Gain Margin (GM) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
9.2.2 Phase Margin (φP M ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
9.2.3 Relative Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
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1 Introduction to Control Systems
1.1 Definitions:
System: An arrangement of physical components connected or related in such a manner as to

form and/or act as an entire unit that attains a certain objective
Control: to mean regulate, direct, or command a system so that the desired objective is achieved
Control system: an arrangement of physical components connected or related in such a manner

as to command, direct, or regulate itself or another system.
Input: the stimulus, excitation or command applied to a control system, typically from an exter-
nal energy source, usually in order to produce a specified response from the control system.
Output: the actual response obtained from a control system. It may or may not be equal to the
specified response implied by the input.
Thus a control system must have
• An input or inputs
• An output or outputs
• An arrangement to achieve this input-output combination
1.2 Requirements of a good control system
Accuracy: accuracy should be very high as any error arising should be detected. Can be improved
using feedback element.
Sensitivity: any control system should be very sensitive to the input signals and insensitive to
any other disturbances Noise: A good system should be able to reduce the effects of noise or
undesired inputs
Stability: stability means bounded inputs and bounded outputs. In the absence of the input,
the output should tend to zero as time increases.
Bandwidth: Operating frequency range decides the bandwidth of any system. For frequency
response of good control system, bandwidth should be large. The required output means maximum
possible output without overshoots and it should be stable for the required input frequency range.
1
Speed: a good control system should have high speed i.e. the output of the system should be as
fast as possible.
Oscillation: for a good control system, oscillations of the output should be constant or sustained
oscillations.
1.3 Classification of Control systems
Control action is that quantity responsible for actuating the system to produce the output. De-
pending on whether such a control action is dependent on the status of the output, control systems
are classified as
1. Open loop systems
2. Closed looped systems
Open loop systems
This is a control system in which the output has no effect on the control action i.e. the control
action is totally independent of the output of the system. The output is neither measured nor fed
back for comparison with the input. To each reference input, there corresponds a fixed operating
condition and as a result the accuracy of the system depends on the calibration. Open loop control
can be used, in practice, if the relationship between the input and output is known and if there are
neither internal nor external disturbances.
A general open loop control system is shown in Figure 1 below.
Figure 1: Open Loop Control System
The reference input r(t) is applied to the controller which generates the actuating signal u(t) to
give the controlled output c(t).
Examples of open loop control systems
1. Automatic washing machine
2. Automatic hand drier
3. Traffic control by means of signal operated on time basis.
4. Bread toaster.
2
Advantages of open loop systems
1. They are simple in construction and design
2. They are easy to maintain
3. Convenient to use when the output is difficult to measure
4. Not much problems of stability
Disadvantages of open loop systems
1. Inaccurate and unreliable
2. Inaccurate results obtained with parameter variations, internal disturbances
3. Recalibration of the controller from time to time is necessary to maintain quality and accuracy
Closed loop control system
This is a control system in which the control action is somehow dependent on the output of the
system. There is a comparison of the state of the output and the reference state. This property is
known as feedback and is the main difference between open loop and closed loop systems.
Feedback is the property of the system which permits the output to be compared with the reference
input so that appropriate control action is formed
The generalized closed loop control system is shown in Figure 2 below
Figure 2: Closed Loop Control System
r (t) - reference input

u (t) - actuating signal
e (t) - error signal
c (t) - controlled output
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b (t) - feedback signal
Part of the output is fed back to the input. The comparison between the reference input r (t) and
the feedback b (t) gives the error e (t) .
When feedback signal is positive, the system is called positive feedback system and e (t) = r (t) +
b (t).
When the feedback signal is negative, the system is called negative feedback system and e (t) =
r (t) + b (t).
The error is applied to the controller which gives the actuating signal u (t). The action of the
controller will be to drive the controlled output in such a manner so that the error is reduced to
zero i.e. the feedback signal is equal to the input signal.
Examples of closed looped control systems include
1. Automatic electric iron
2. DC motor speed controlled by tacho feedback
3. Sun-seeker solar systems
4. Automatic water level controller.
Advantages of closed loop systems
1. Accuracy is very high since any error arising is corrected.
2. Reduced effects of non-linearities
3. It senses changes in input due to parameter changes, internal disturbances etc and corrects
the same
4. High bandwidth
5. Facilitates automation
Disadvantages of closed loop systems
1. Complicated in design and maintenance costlier
2. System may become unstable
4
Comparison of open loop and closed loop systems
Table 1: Open loop and closed loop systems

Sr. No. Open Loop System Closed Loop System
1. No feedback. Hence feedback Feedback exists. Hence feedback
elements absent elements exists
2. No error detector Error detector present
3. It is inaccurate It is accurate
4. Highly sensitive to parameter Less sensitive to parameter
changes changes
5. Small bandwidth Large bandwidth
6. Stable May become unstable
7. Economical Costly
8. Example: automatic toaster, Examples: Temperature control
hand drier oven, guided missile
Servomechanisms
Out of a number of closed loop systems in the industry, a particular group of applications are
called servomechanisms. Here the controlled output is position, speed, velocity, acceleration etc.
i.e. position or time derivative of position.
A servomechanism is a power amplifying feedback control system in which the controlled variable
is mechanical position or its time derivative, such as velocity, acceleration.
Example: an automobile steering mechanism where the angular position of wheels on the road is
controlled as shown in Figure 3 below
Figure 3: Servomechanism of automobile steering system
Linear and Non-linear Control Systems
Linear systems are systems which obey the principle of superposition and proportionality.
If an input x1 produces output y1 by proportionality theorem an input αx1 produces an output
αy1 . By superposition theorem, an input αx1 + βx2 will produce an output αy1 + βy2 .
Linear systems are systems where mathematical tools like Laplace, Fourier etc. can be used. These
systems can be analysed mathematically or graphically.
5
• Practical systems are not fully linear but exhibit certain non-linearities. The common exam-
ples of non-linearity are saturation, friction, deadzone, backlash etc.
• Non-linear systems exhibit peculiar behaviour
• They do not obey the law of superposition.
• Standard test signals approach cannot be used here
• The stability of linear systems depends on the root location and is independent of the initial
state. In non-linear systems, stability depends on root location as well as the initial condition
and type of input.
• Non-linear systems exhibit self sustained oscillations of fixed frequency and amplitude called
Limit Cycles. Linear systems do not have this feature
• Non linear systems may exhibit hysteresis while linear systems do not show this behaviour
Table 2: Linear and Non Linear Systens

SR. NO Linear Systems Non Linear Systems
1 Obey superposition theorem Do not obey superposition theorem
2 Can be analysed by standard Cannot be analysed by standard
test signals test signals
3 Stability depends only on the Stability depends on the root location,
root location initial condition and type of input
4 Do not exhibit Limit Cycles Exhibit Limit Cycles
5 Do not exhibit Hysteresis / Exhibit Hysteresis /Jump Resonance
Jump Resonance
6 Can be analysed by Laplace, Cannot be analysed by these methods
Fourier and Z-transforms
6
2 Transfer Function and Mathematical Modelling
2.1 Transfer Function
The relationship between the input and output of a system is given by the transfer function.
Definitions: The ratio of the Laplace transform of the output (or response) to the Laplace
transform of the input (or excitation) under the assumption of zero initial conditions is defined as
the transfer function of the given system.
For the system shown in Figure below
where
c (t) - output
r (t) - input
g (t)- system
Taking the Laplace transforms
Lc (t) = C (s)
Lr (t) = R (s)
Lg (t) = G (s)
Therefore the transfer function G (s)for Figure above is
Laplace of output C (s)

G (s) = = (1)
Laplace of input R (s)
Example 2.1
Consider a differential equation which gives the relationship between the input x(t) and the output
y(t)
d2 y (t) dy (t) dx (t)
+3 + 2y (t) = 3 − x (t)
dt 2 dt dt
Obtain the transfer function
Solution
Taking Laplace transform and assuming zero initial conditions
s2 Y (s) + 3sY + 2Y (s) = 3sX (s) − X (s)
7
And the transfer function is obtained as
Y (s) 3s − 1
=
X (s) 2
s + 3s + 2
Hence the transfer function is defined as the ratio of the Laplace transform of the output to the
Laplace transform of the input under the assumption that the initial conditions are zero
Laplace transform of three basic inputs
The three basic inputs considered in control systems are:
1. Unit step input:
Figure 4: Unit step input
the unit step input is mathematically defined as


1 t≥0
r (t) =
0 t<0
The Laplace transform of the unit step signal is given by
1
R (s) =
s
2. Ramp input:
Figure 5: Ramp input
8
Ramp signal is mathematically defined as

kt t>0
r (t) =
0 t<0
where k is a constant
The Laplace transform of the ramp signal is given by
k
R (s) =
s2
3. Impulse input:
Figure 6: Unit impulse input
Impulse signal is mathematically defined as


1 t=0
r (t) =
0 t 6= 0
The Laplace transform of the impulse signal is given by
R (s) = 1
Impulse Response and Transfer Function
Let us assume that the Laplace of an impulse input δ (t) is unity

C(s)
From equation (1) G (s) = then
R(s)
C (s) = G (s) R (s) (2)
Equation (2) gives the relation for finding the response C (s)
Consider an impulse input applied to Fig 6
r (t) = δ (t) = impulse input
9
Therefore
R (s) = Lr (t) = Lδ (t) = 1
Substituting for R (s) in (2) gives
C (s) = G (s) or c (t) = L−1 C (s) = L−1 G (s) = g (t) (3)
From (3), the impulse response C (s) equals the transfer function G (s)
In other words, if the transfer function is not given but the impulse response is known, then the
transfer function is as good as known.
Definition: The transfer function of a Linear Time Invariant System is defined as the Laplace
transform of the impulse response, with all initial conditions set to zero.
Example 2.2
The impulse response of a system is e−2t . Determine the transfer function of the system.
Solution
The Laplace transform of e−2t is given by
1
L e−2t =
n o
s+2
From the above theory then the transfer function of the system is
1
G (s) =
s+2
Example 2.3
Given the input to the system of example 2.2 as e−3t determine the output of the system.
Solution
1
R (s) = L e−3t =
n o
s+3
1
G (s) =
s+2
1 1
C (s) = G (s) × R (s) = ×
s+2 s+3
Using partial fractions
10
1 1
C (s) = −
s+2 s+3
Taking inverse Laplace transform gives
c (t) = e−2t − e−3t
Properties of Transfer Function
1. The transfer function of a system is the Laplace transform of its impulse response for zero
initial conditions
2. Conversely the transfer function can be determined from the system input-output pair taking
the ratio of the Laplace of the output to the Laplace of the input
3. The system differential equation can be obtained from the transfer function by replacing
s-variable with the Linear Differential Operator D defined by D = d
dt
4. The transfer function is independent of the inputs to the system
5. The system poles/zeros can be found from the transfer function (discussed in the next section)
6. Stability can be determined from the characteristic equation (discussed later)
7. The transfer function is defined only for linear time invariant systems. It is not defined for
non-linear systems
Proper and improper transfer functions
A transfer function is said to be proper if the order of the denominator is greater than the order of
the numerator i.e. if the system has more poles than zeros
Example
1 1 s+1
, ,
s (s + 2) (s + 3) s2 + 6s + 4
A transfer function is said to be improper if the order of the numerator is greater than the order
of the denominator i.e. if the system has more zeros than poles
Advantages of Transfer function
1. It is a mathematical model that gives the gain of the given block/system
2. Integral differential equations are converted to simple algebraic equations
3. Once transfer function is known, any output for any given input, can be known
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4. From the knowledge of the transfer function the poles and zeros of a system is found out.
Both poles and poles of a system play a very important role in response of the system.
5. From the characteristic equation, system stability can be inferred.
6. System differential equation can be obtained by replacement of variable 0 s0 by d

dt
7. Transfer function is the characteristic and property of the system itself
8. The value of the transfer function is dependent on the parameters of the system and inde-
pendent of the input applied
Disadvantages of Transfer function
1. Transfer function is valid only for liner time invariant systems
2. It does not take into account the initial conditions. Initial conditions loses its significance
3. It does not give any idea about how the present output is progressing. No idea about the
physical structure of the system is immediately known
Poles and Zeros of a transfer function
The transfer function is given by
C (s)
G (s) =
R (s)
Both C (s) and R (s) are polynomials in s, thus
bm sm + bm−1 sm−1 + · · · + b0
G (s) =
sn + an−1 sn−1 + · · · + an
k (s − b1 ) (s − b2 ) (s − b3 ) . . . (s − bm )
= (4)
(s − a1 ) (s − a2 ) (s − a3 ) . . . (s − an )
k is the system gain factor
Poles
The values of s for which the transfer function magnitude |G (s)| becomes infinite after substitution
in the denominator of the transfer function are called the poles of the transfer function.
For equation (4),s = a1 , a2 , a3 etc. are the poles of G (s). In order to get these poles, equate the
denominator of the transfer function to zero.
When the poles are not repeated, such poles are called simple poles. If repeated, such poles are
called multiple poles of the order equal to the number of times they are repeated.
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Zeros
The values of s for which the transfer function magnitude |G (s)| becomes zero after substitution
in the numerator of the transfer function are called the poles of the transfer function.
For equation (4), the values of zeros are s = b1 , b2 , b3 etc. In order to get these poles, equate the
numerator of the transfer function to zero.
When the zeros are not repeated, such zeros are called simple zeros. If repeated, such zeros are
called multiple zeros of the order equal to the number of times they are repeated.
Pole-zero plot
This is the diagram obtained by locating all poles and zeros of the transfer function in the s-plane.
The s-plane has two axis, real and imaginary axis. Simple poles are shown by X and simple zeros
by a circle O.
Repeated poles are shown by repeated X (equal to the order of multiplicity) while repeated zeros
are shown by overlapping circles or circles side by side (equal to the order of multiplicity).
Example 2.4
Given the transfer function
3 (s + 3) (s + 1.5)3
G (s) =
(s + 5) (s + 7)2
Determine and plot the poles and zeros of the transfer function
Solution
Poles are at s = −5 (simple pole) and s = −7 (repeated pole of 2nd order multiplicity)
Zeros are at s = −3 (simple zero) and s = −1.5 (repeated zero of 3rd order multiplicity)
The pole-zero plot is as shown in Figure 7
Figure 7: Pole Zero plot
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2.2 Modeling of Electrical Systems
The basic passive elements of electrical systems are resistance, inductance and capacitance as shown
in Figure 8
Figure 8: Basic electrical elements
For a resistive element, Ohm’s law can be written as
(v1 (t) − v2 (t)) = Ri (t)
(V1 (s) − V2 (s)) = RI (s) (1)
For an inductive element, the relationship between voltage and current is
di
(v1 (t) − v2 (t)) = L
dt
(V1 (s) − V2 (s)) = LsI (s) (2)
For a capacitive element, the electrostatic equation is
Q (t) = C (v1 (t) − v2 (t))
Differentiating both sides yields
dq d
= i (t) = C (v1 (t) − v2 (t))
dt dt
sQ (s) = I (s) = Cs (V1 (s) − V2 (s)) (3)
14
Laplace transform of Electrical Networks
Element Time-domain expression s - domain expression

R v (t) = i (t) R V (s) = I (s) R
L v (t) = L di V (s) = LsI (s)
dt
C v (t) = C1 R i (t) dt 1 I (s)
V (s) = SC
Example 2.5
Obtain the transfer function for the circuit of Figure 9
Figure 9: Example 2.5
Solution
Applying KVL to this circuit
1
Z
vi (t) = Ri (t) + i (t) dt..........(i)
C
and
1
Z
v0 (t) = i (t) dt.................(ii)
C
Taking Laplace transform of both equations yields
1
Vi (s) = RI (s) + I (s) .......(iii)
sC
1
V0 (s) = I (s) ..............(iv)
sC
From (iv)
I (s) = sCV0 (s) .............(v)
Substituting (v) into (iii)
1

Vi (s) = R+ sCV0 (s) .............(vi)
sC
15
The transfer function is obtained from (vi) as
V (s) 1
G (s) = 0 =
Vi (s) 1

sC R + sC
Example 2.6
Determine the transfer function of the network of Figure 10. Hence determine the voltage U2 (t)
when the inputs are
1. U1 (t) = 5
2. U1 (t) = 5t
3. U1 (t) = 5t2
Solution
The Laplace of the given circuit is as shown in Figure 11
Figure 11: Laplace form of the circuit
Applying KVL to the circuit of Figure 11
1

U1 (s) = I (s) + 2 ..................(i)
s
U2 (s) = 2I (s) .................(ii)
From (ii)
16
U2 (s)
I (s) = ...............(iii)
2
Substituting (iii) into (i)
U2 (s) 1

U1 (s) = +2
2 s
The transfer function is obtained as
U2 (s) s
= ..............(iv)
U1 (s) s + 0.5
1. U1 (t) = 5
∴U1 (s) = 5s
Substituting this value into (iv)
U2 (s) = 5
(s+0.5)
Taking Inverse Laplace Transform
U2 (t) = 5e−0.5t
2. U1 (t) = 5t
∴U1 (s) = 52
s
U2 (s) = 5
s(s+0.5)
By partial fractions
U2 (s) = 10 10
s − (s+0.5)
U2 (t) = 10 − 10e−0.5t
17
3. U1 (t) = 5t2
∴U1 (s) = 10
s3
U2 (s) = 2 10
s (s+0.5)
By partial fractions
U2 (s) = 20 40 40
s − s + (s+0.5)
U2 (t) = 20t − 40 + 40e−0.5t
2.3 Modelling of Mechanical Systems
Mechanical systems are of two types:
1. Translational
The motion of the body is along a straight line or curved path
2. Rotational
The motion of the body is about its own axis
Mechanics of translational motion
There are three elements here
1. Mass
Figure 12: Mass
18
The physical model of a mass element assumes that mass is concentrated at the center of the
body. When a block of mass M is subjected to a force F, it undergoes a displacement. The
relationship between the force and mass is
F ∝ acceleration
F = M a where a is the acceleration
2
∴F = M dv = M d 2x
dt dt
where v - velocity and x - displacement
2. Spring
Figure 13: Spring
Here the body subjected to force undergoes elastic deformation. The relationship is
F ∝displacement
F = Kx where K is the spring constant
3. Damper
Figure 14: Damper
Viscous friction: it is between moving surfaces separated by a viscous fluid or between a solid
body and a fluid medium. It is proportional to velocity. While friction opposes motion, it is
not always undesirable. Viscous is sometimes introduced intentionally also and is shown by
a dashpot
Here F ∝ velocity
19
F = B dx = B d (x1 − x2 )
dt dt
x1 and x2 being the displacement of the two ends of a dashpot e.g. piston
Translational elements
Sr.No. Element Relation Remarks
2
1 F = M d 2x Mass
dt
2 F = B dx = B d (x1 − x2 ) Dumper
dt dt
3 F = K (x1 − x2 ) Spring
Mechanics of Rotational motion
There is a lot of similarity between translational motion and rotational motion as far as modeling
is concerned. Table below is a list of counter parts of these systems
Sr. No. Translational Motion Rotational Motion

1 Mass(M ) Inertia(J)
2 Damper (B) Damper (B)
3 Spring (K) Spring (K)
4 Force (F ) Torque (T )
5 Displacement(x) Angular displacement (θ)
6 Velocity v = dx Angular velocity ω = dθ
dt dt
There are three elements here
1. Inertial J
Figure 15: Inertia
20
The relation is
2
T = J d 2θ = J dω
dt dt
Inertia has only one displacement
2. Damper B
Figure 16: Damper
The relation is
T = B dθ = B d (θ2 − θ1 )
dt dt
= B (ω2 − ω1 )
3. Spring K
Figure 17: Spring
Here the relation is
T = K (θ1 − θ2 ) = Kθ
Rotational elements
Sr. No. Element Relation Remarks
2
1 T = J d 2θ Inertia
dt
2 T = B d (θ2 − θ1 ) Damper
dt
3 T = K (θ1 − θ2 ) Spring
21
2.4 Analogous Systems
There are two main analogous systems
1. Force-voltage analogy
2. Force-current analogy
Force-Voltage analogy
The force is analogous to voltage. Consider the circuit of Figure 18 below
Figure 18: Series RLC Circuit
Applying KVL
1
Z
di
V (t) = Ri + L + idt
dt C
but
dq
i=
dt
therefore
dq d2 i 1
V (t) = R +L + q
dt dt2 C
Laplace transform gives
1
V (s) = sRQ (s) + s2 LQ (s) + Q (s)
C
1
= s2 LQ (s) + sRQ (s) + Q (s) (1)
C
Comparing this with the Translational system
d2 x dx
F =M +B + Kx
dt2 dt
Laplace gives
F (s) = s2 M X (s) + sBX (s) + KX (s) (2)
Comparing (1) and (2) gives the following analogies
22
F → V, M → L, B → R, K → C1 ,X → Q
Comparing with the Rotational System
d2 θ dθ
T =J +B + Kθ
dt2 dt
Laplace transform gives
T (s) = s2 Jθ (s) + sBθ (s) + Kθ (s) (3)
Comparing (1) and (3) gives the following analogies

T → V, J → L, B → R, K → C 1 ,θ → Q
Table 3: Force-voltage analogy

Translational Electrical Rotational
Force F Voltage V Torque T
Mass M Inductance L Inertia J
Damper B Resistance R Damper B
Spring K Elasticity D = C1 Spring K
Displacement x Charge q Displacement θ
dq
Velocity v Current i = Velocity ω
dt
Force-Current Analogy
Here force is analogous to current. Consider the circuit of Figure 19 below
Figure 19: Parallel RLC Circuit
Applying KCL
1
Z
V dv
i= + V dt + C
R L dt
dϕ
But V =
dt
therefore
1 dq 1 d2 ϕ
i= + ϕ+C
R dt L dt2
23
Laplace gives
1 1
I (s) = s2 Cϕ (s) + sϕ (s) + ϕ (s) (4)
R L
Comparing equations (2) and (4)
F → I, M → C, B → R 1 ,K → 1,X → ϕ
L
Comparing equations (3) and (4)
T → I, J → C, B → R 1 ,K → 1,θ → ϕ
L
Table 4: Force-Current Analogy

Translational Electrical Rotational
Force F Currenti Torque T
MassM Capacitance C Inertia J
Spring K 1
Reciprocal of Inductance L Spring K
Damper B Conductance R 1 Damper B
Displacement x Flux linkage ϕ Displacement θ
dx dϕ
Velocity Voltage V = Velocity dθ
dt dt dt
Example 2.7
For the system shown in Fig below , write the system equations
Solution
Step 1:
There is one mass, hence number of displacement = 1
Take one reference node in addition. Let the displacement be x at M
Step 2:
Mass M has a displacement x. connect between node x and reference Spring constant is connected
to displacement x on one end and reference. Damper B is similarly connected between elements
M and reference Force F is applied to M i.e. displacement x
Step 3:
the node diagram is as shown in Fig below
24
Step 4:
At x incoming force = F
Outgoing force
d2 x dx
F =M +B + Kx
dt2 dt
F (s) = s2 M X (s) + sBX (s) + KX (s)
= s2 M + sB + K X (s)

Example 2.8
Draw the Force-current analogy for the Fig in example 2.7 above and draw the direct analogous
circuit
Solution
In force-current analogy
1 ,K → 1,X → ϕ
F → I, M → C, B → R L
Therefore
F (s) = s2 M X (s) + sBX (s) + KX (s)
Its analogy is
1 1
I (s) = s2 Cϕ (s) + sϕ (s) + ϕ (s)
R L
Using
V (s) = sϕ (s)
1 1
I (s) = sCV (s) + V (s) + V (s)
R sL
This is the force current analogous equation. To draw the circuit, replace every element in previous
Figure with its analog value to obtain figure below
Example 2.9
For the mechanical system of Figure 21 below
25
1. Draw the mechanical network
2. Write the differential equations of performance
3. Draw the force voltage and force current analogous networks
Solution
Step 1:
number of displacement = number of masses = number of nodes = 2
Take a reference node in addition. Assume displacements x1 and x2
Step 2:
For Mass
M1 is connected to x1 and M2 to x2
For spring
K1 is connected between x1 and reference
K2 is connected betweenx1 and x2
For damper
B1 is connected between x1 and reference
Step 3: Draw the nodes x1 and x2 and the reference and connect each element to obtain Figure 22
below
Figure 22: Mechanical diagram
26
Step 4: Writing equations at nodes x1 and x2
At node x1 :
d2 x1 dx
F = M1 + B1 1 + K1 x1 + K2 (x1 + x2 )
dt2 dt
F (s) = s2 M1 X1 + sB1 X1 (s) + K1 X1 (s) + K2 (X1 (s) − X2 (s)) .........(i)
At node x2 :
d2 x1
K2 (x1 + x2 ) = M2
dt2
K2 (X1 (s) − X2 (s)) = s2 M2 X2 (s) .............(ii)
(ii) Force-Current analogy:

F → I, M → C, B → R 1 ,K → 1,X → ϕ
L
Replace equation (i) and (ii) term by term
1 1 1
I (s) = s2 C1 ϕ1 (s) + s ϕ (s) + ϕ (s) + (ϕ (s) − ϕ2 (s))
R1 1 L1 1 L2 1
But V (s) = sϕ (s)
1 1 1
I (s) = s2 C1 V1 (s) + V (s) + V (s) + (V (s) − V2 (s)) ........(iii)
R1 1 sL1 1 sL2 1
Equation (ii) becomes
1
(ϕ (s) − ϕ2 (s)) = s2 C2 ϕ2 (s)
L2 1
1
(V (s) − V2 (s)) = s2 C2 V2 (s) ...............(iv)
sL2 1
Equations (iii) and (iv) are F − I analogy equations

Draw the circuit, by replacing every element in previous Figure with its analog value to obtain
Figure 23 below
(iii) Force-Voltage analogy:

F → V, M → L, B → R, K → C 1 ,X → Q
Putting I (s) = sQ (s) into (i) and (ii) gives
1 1
V (s) = sL1 I1 (s) + I (s) + R1 I1 (s) + (I (s) − I2 (s)) ..........(v)
sC1 1 sC2 1
27
Figure 23: Force-Current analogy
1
(I (s) − I2 (s)) = sL2 I2 (s) ..............(vi)
sC2 1
The F − V circuit is as shown in Figure 24 below
Figure 24: Force-Voltage analogy
Exercise
1. Obtain the differential equations describing the mechanical system shown in Figure 25 and
draw the electric network using force-voltage analogy.
Figure 25: Exercise 1
2. Determine the transfer function for the electrical circuit shown in Figure 26 below
3. Write differential equations for the mechanical system shown in Figure 27 below
4. For the circuit shown in Figure 28 below C = 1µF . What values of R1 and R2 will give
28
T = 0.6sec and a = 0.1 given the expression for the transfer function as
V0 (s) a (1 + sT )
=
Vi (s) 1 + aST
29
3 Block Diagram Representation
3.1 Introduction
A block diagram is a pictorial representation of the system representing the relationship between
input and output of the system.
Each element of a practical system is represented as a block. The transfer function of that element is
inserted inside the block. The different blocks are interconnected to each other as per the sequence
of operation of the system.
Lines drawn between the blocks indicate the connection of the blocks. An arrow indicates the
direction of the flow of signals from one block to another.
Some important definitions
Block diagram: a pictorial representation of the cause and effect relationship between input and
output of the system
Figure 29: Block diagram
Output: The value of the input is multiplied with the value of block gain to get the output
Output = Gain × Input
R (s)input C (s) Output
Summing point: more than one signal can be added or subtracted at the summing point
Take off point: The point at which the output is taken for feedback purpose.
30
Forward path: The direction of flow of signal is from input to output
Feedback path: The direction of flow of signal is from output to input
Advantages of Block Diagram
1. The functional operation of the system can be observed from the block diagram
2. The diagram gives the information about performance of the system
3. Block diagram is used for analysis and design of control systems
4. It is very simple to construct block diagram for big and complicated systems
Disadvantages of Block Diagrams
1. Block diagram for given system are non unique
2. Source of energy in the system is not shown in the diagram
3. In the procedure of reduction of block diagram algebra some important functions may be
omitted or hidden. There is no check for it.
4. The block diagram does not give any information about the physical construction of the
system.
3.2 Canonical form of Feedback Control System
The canonical form consists of
1. One forward path
31
2. One feedback path
3. One summing point
4. One take-off point
Figure 30: Canonical form of block diagram
The following terms are used with reference to Figure 30 above
1. R (s) = Laplace transform of input, r (t)
2. C (s) = Laplace transform of controlled output, c (t)
3. E (s) = Laplace transform of error signal, e (t)
4. B (s) =Laplace transform of feedback signal, b (t)
5. G (s) = Forward transfer function
6. H (s) = Feedback transfer function

C(s)
7. = Control ratio = Closed loop transfer function
R(s)
E(s)
8. =Error ratio = Actuating signal ratio
R(s)
B(s)
9. = Primary feedback ratio
R(s)
Derivation of Transfer Function of Canonical Closed Loop System
Referring to Figure 31 below
E (s) = R (s) ± B (s) (1)
32
Figure 31: Closed loop system
B (s) = C (s) H (s) (2)
C (s) = E (s) H (s) (3)
Substituting (2) into (1)
E (s) = R (s) ± C (s) H (s)
C (s)
E (s) = = R (s) ± C (s) H (s)
G (s)
C (s) = R (s) G (s) ± C (s) G (s) H (s)
C (s) G (s)
=
R (s) 1 ± G (s) H (s)
Figure 32: Closed loop system transfer function
Positive sign means negative feedback

Negative sign means positive feedback
This can be used as a standard result to eliminate such simple loop in a complicated system
reduction procedure.
33
3.3 Rules for Block Diagram Reduction
Any complicated system can be converted into canonical form by using block diagram reduction
rules.
The following rules are widely used in block diagram reduction:
Rule 1: Blocks in Series
Any finite number of blocks in series may be algebraically combined by multiplication.
Figure 33: Blocks in Series
The blocks of Figure 33 (a) can be reduced by an equivalent single block as shown in Figure 33(b).
If there is a summing point or take-off point between the blocks, the blocks cannot be said to be
in series.
Rule 2: Blocks in Parallel
The blocks which are connected in parallel get added algebraically (considering the sign of the
signal).
Figure 34: Blocks in Parallel
Considering the system of Figure 34 above
C (s) = R (s) G1 + R (s) G2 − R (s) G3
= R (s) [G1 + G2 − G3 ]
All the three blocks can be replaced by an equivalent single block as shown in the figure below
34
This rule cannot be applied directly when a take-off point occurs as shown in the figure below
For a parallel combination, the direction of flow of signals through the blocks must be the same.
This rule cannot be applied in the system of figure below
Rule 3: Eliminate Feedback Loop
Eliminate feedback loop as follows
Figure 35: Feedback Loop
For negative feedback
C (s) G (s)
=
R (s) 1 + G (s) H (s)
For positive feedback
C (s) G (s)
=
R (s) 1 − G (s) H (s)
Now consider a system shown in Figure 36 below.
Here G1 andG2 are in series. First the series combination is obtained. Then the feedback loop is
eliminated.
35
Figure 36: Feedback loop
Rule 4: Associative Law for Summing Point
The order of summing point can be changed if two or more summing points are in series (output
remains the same).
Figure 37: Summing point
For Figure 37(a) above,
y = R (s) − B1
C (s) = y − B2 = R (s) − B1 − B2
For Figure 37(b) above,
y = R (s) − B2
C (s) = y − B1 = R (s) − B2 − B1
Which means output is same. Associative law for summing point is applicable only to summing
points which are directly connected to each other.
Rule 5: Shifting a summing point before a block
In Figure 38(a) below a summing point needs to be shifted before G as shown by the dotted line.
Now C (s) = X + GR (s)
36
Figure 38: Summing point before a block
To get the same output, the transformation is shown in Figure 38(b)

X
C (s) = G R + = X + GR (s)
G
which is the same output

NB: To shift a summing point before a block in the system, another block having a block function
1/G is added before the summing point as shown in Figure 38(b)
Rule 6: Shifting a summing point after a block
Here our aim is to shift the summing point after G as shown by the dotted line.
Figure 39: Summing point after a block
Here the output of the system is C (s) = (R + X) G

Shifting the summing point without making any change as shown in Figure 39 (b) above , the
output becomes C (s) = RG + X
This output is not the same as the original output. To obtain the same output X is multiplied
by G. This means to shift a summing point after a block, add a block having the same transfer
function at the summing point.
37
Rule 7: Shifting a take-off point before a block
To shift a take-off point before a block, add a block having the transfer function G in series with
the signal taking-off from the take-off point.
Figure 40: Take-off point before a block
From Figure 40(a) above , C = X = RG

Shifting the take-off point without making any change as shown in Figure 40(b), C = RG andX = R
To satisfy the condition of Figure 40(a), a block having the transfer function G is added in series
with X as shown in Figure 40(c).
Rule 8: Shifting a take-off point after a block
1 in series with
To shift a take-off point after a block, add a block having the transfer function G
the signal taking-off from the take-off point.
Figure 41: Take-off point after a block
From Figure 41(a), C = RG and X = R

Shifting the take-off point after the block without doing any modifications as shown in Figure 41(b)
yields C = RG and X = RG
1 is added in series with X as
To satisfy the original system, a block with the transfer function G
shown in Figure 41(c).
38
Rule 9: Shifting a take-off point after a summing point
To shift a take-off point after a summing point, one more summing point is added in series with
take-off point.
Suppose we have to shift a take-off point as shown with the dotted line in Figure 42(a)
Figure 42: Take-off point after a summing point
Here Z = R and C (s) = R ± Y

Shifting the take-off point after the summing point directly as shown in Figure 42(b) yields C (s) =
R ± Y andZ = R ± Y
We achieve the original value of Z by the operation
Z =R±Y ±Y
This operation is shown in Figure 42(c) below
Rule 10: Shifting a take-off point before a summing point
To shift a take-off point before a summing point, one more summing point is added in series with
take-off point.
Suppose we have to shift a take-off point as shown with the dotted line in Figure 43(a)
Here Z = R ± Y and C(s) = R ± Y
Shifting the take-off point after the summing point directly as shown in Figure 43(b) yields C(s) =
R ± Y and Z = R
We achieve the original value of Z by adding a summing point in series with Z as shown in Figure
43(c)
39
Figure 43: Take-off point before a summing point
Procedure to solve block diagram reduction problem
Step 1: Reduce the blocks in series

Step 2: Reduce the blocks in parallel
Step 3: Eliminate feedback path
Step 4: Always try to shift take-off points towards right and summing points towards left.
Step 5: Repeat above procedure till getting simplest form
Step 6: Find out transfer function for the whole system. The transfer function of the overall system
is given by
C (s)
TF =
R (s)
Example 3.1
Find the single block equivalent by block diagram reduction for the given system
Solution
40
G4 , G3 and G5 are in parallel. Apply rule 2
Blocks G2 and G3 + G4 + G5 are in cascade. Apply rule 1
Rule 3 can be applied to feedback loop G1 and H1
Now rule 1 can be applied to cascade blocks in the forward path

G1
1+G1 H1 and G2 (G3 + G4 + G5 )
Rule 3 is applied to the resulting feedback loop
41
Rule 1 is applied to the resulting block and G6
42
4 PID Controllers
An automatic controller determines the value of controlled variable, compare the actual value to
the desired value, determines the deviation and produces a control action signal that will reduce
the deviation to zero or to a smallest possible value.
The method by which the automatic controller produces the control signal is called mode of control
or control action. A Controller consists of an error detector and amplifier.
4.1 Proportional (P) Control
The control action or signal u (t) is proportional to the error signal e (t) i.e
u (t) = kp e (t) (1)
where kp is the proportional gain constant
Figure 44: General Proportional Controller
The controller is an amplifier with a gain kp (the gain may be adjustable)
Electronic Proportional controller can be realized using inverting Op-amp circuit which is an am-
plifier with adjustable gain as shown in Figure 45 below
Figure 45: Electronic Proportional Controller
43
Z i = Ri Z o = Rf
Vo (s) Zo Rf
== =−
Vi (s) Zi Ri
Therefore
Rf
Vo (s) = − V (s) = −Kp Vi (s)
Ri i
Where
Rf
Kp =
Ri
Advantages of proportional control action
1. This type of controller increases the forward path gain which in turn reduces the damping
ratio ξ. So the response becomes faster and therefore reduces the rise time. It means that an
immediate proportional output is produced as soon as an error signal exists at the controller.
2. The proportional controller is considered a fast-acting device. This immediate output change
enables the proportional controller to reposition the final control element within a relatively
short period of time in response to the error.
3. Also as the proportional constant increases Kp , the steady state error ess is reduced. This is
a stable control as long as is not too high.
Limitations of proportional control action
1. The main disadvantage of the proportional control mode is that a residual offset error exists
between the measured variable and the set point for all set of system conditions. It will
not return the measured variable to set point, but to a value that is within a defined span
(Proportional Band) around the set point.
2. Proportional action responds only to a change in the magnitude of the error and also increases
the overshoot and the settling time.
4.2 Integral (I) Control
The Integral Control action is shown in Figure 46 below

For integral control, the control signal u (t) is given by
Z t
u (t) = KI e (t) dt (1)
0
where KI is the integral gain constant
converting equation 1 to frequency domain
44
Figure 46: Integral Controller
KI
U (s) = E (s)
s
The inverse of KI is called the integral time and is defined as the time of change of output caused
by a unit change of actuating error signal. The integral control action is also known as reset control.
The integral controller eliminates the steady state error ess , reduces the rise time but increases the
overshoot and the settling time.
Electronic Integral controller can be realized using integrator circuit of Figure 47 below.
Figure 47: Electronic Integral Controller
1
Z i = Ri Z o =
sCf
Vo (s) Zo 1
=− =−
Vi (s) Zi sCf Ri
Therefore
1 K
Vo (s) = − V (s) = − I Vi (s)
sCf Ri i s
1
KI =
Cf Ri
45
4.3 Derivative (D) Control
The Derivative Control action is as shown in Figure 48 below
Figure 48: Derivative Controller
The control signal u (t)is given by
d
u (t) = KD e (t)
dt
where KD is the derivative gain
Derivative action will minimize the deviation from the set point when there is a change in the
process condition. It is interesting to note that derivative action will only apply itself when there
is a change in error signal. If the error is steady, whatever the offset, then derivative action does
not occur.
The differentiator acts to transform a changing signal to a constant magnitude signal. As long as
the input rate of change is constant, the magnitude of the output is constant. A new input rate of
change would give a new output magnitude.
Rate action is not usually employed with fast responding processes such as flow control or noisy
processes because derivative action responds to any rate of change in the error signal, including the
noise.
Electronic implementation of Derivative Control action is as shown in Figure 49
sCi Rd + 1
Zi = Z0 = Rf
sCi
sCi Rf
Vo (s) = − V (s) = −sKKD Vi (s)
sCi Rd + 1 i
Where
KD = Ci Rf
46
Figure 49: Electronic Derivative Controller
4.4 Proportional plus Integral (PI) Control
The PI control action is as shown in Figure 50 below
Figure 50: Proportional plus Integral Controller
The control signal u (t) is given by
Z t
u (t) = Kp e (t) + KI e (t) dt
0
Kp t
Z
= Kp e (t) + e (t) dt
TI 0
where
Kp
= KI
TI
" Z t #
1
⇒ u (t) = Kp e (t) + e (t) dt (1)
TI 0
Kp - proportional gain
KI - Integral gain
TI - Integral time
Taking Laplace transform of equation (1)
47
1

U (s) = Kp 1 + E (s)
sTI
Electronic Implementation of the PI controller is as shown in Figure 51
Figure 51: Electronic Proportional plus Integral Controller
Ri 1
Zi = Zo =
sCi Ri + 1 sCf
" #
Vo (s)

Ci Ci Kp KI
=− + = − Kp +
Vi (s) Cf sCf Ci Ri s

K
V0 (s) = −Kp 1 + I Vi (s)
s
Where
Ci 1
Kp = and KI =
Cf Ci Ri
Alternatively
Figure 52: Electronic Proportional plus Integral Controller
48
Where
Rf 1
Kp = and KI =
Ri Cf Rf
The following are advantages of PI controller
1. ess is reduced which means accuracy is improved
2. The order and type of the system is increased by 1. That improves the damping and reduces
overshoot
3. Noise is filtered
4. Offset is eliminated
4.5 Proportional plus Derivative (PD) Control
The Proportional plus Derivative Control action is as shown in Figure 53
Figure 53: Proportional plus Derivative Controller
The control signal u (t)is given by
d
u (t) = Kp e (t) + KD e (t) (1)
dt
Taking Laplace transform of (1)
U (s) = Kp E (s) + sKD E (s)
= Kp + sKD E (s)

= Kp 1 + sTD E (s)

where
Kp TD = kD
49
Figure 54: Electronic Proportional plus Derivative Controller
Electronic implementation of PD controller is as shown in Figure 54

PD control does not improve the steady state performance of a system.
In general PD control increases the damping of the system and this reduces the overshoots. The
D component of the controller also makes the system faster. Reduce settling time and improve the
transient response
PD control adds a zero to the system pulls root-locus to the left - improves the stability margin of
the system
4.6 Proportional plus Integral plus Derivative (PID) Control
The Proportional plus Integral plus Derivative Control action is as shown in Figure 55 The control
Figure 55: Proportional plus Integral plus Derivative Controller
signal u (t) is given by
Z t
d
u (t) = Kp e (t) + KI e (t) dt + KD e (t) (1)
0 dt
Taking Laplace transform of (1)
50
KI
U (s) = Kp E (s) + E (s) + sKD E (s)
s

K
= E (s) Kp + I + sKD
s
1

= Kp E (s) 1 + + sTD
sTI
s2 TI TD + sTI + 1
" #
= Kp E (s)
sTI
Electronic implementation of PID control is as shown in Figure 56
Figure 56: Electronic Proportional plus Integral plus Derivative Controller
PID adds two zeros and a pole to the system. It increases the system type by 1
PID controller is used so as to utilize the best properties of PID controller.
However, PID controller requires more components to implement and therefore is the most expensive
Example 2.1
For the control system of Figure 57, determine
1. The damping factor and natural undamped frequency with KD = 0
2. The derivative constant KD for a damping factor of 0.5
Solution
51
1. With KD = 0
C(s) 50
=
R(s) s2 + 3s + 50
Comparing with the equation of a general second order system
C(s) 2
ωn
=
R(s) 2
s2 + 2ξωn s + ωn
√
ωn = 50 = 7.071 rad/s
3
ξ= = 0.2121
2ωn
2. With KD 6= 0
C(s) 50
=
R(s) s2 + s(3 + kD ) + 50
2ξωn = 3 + kD
kD = 2 × 0.5 × 7.071 − 3 = 4.071
52
5 Time Domain Analysis
In time domain analysis , time is the independent variable. When a system is given an excitation
(input) , there is a response (output). This response varies with time and is called the time response
Generally the response of any system has two parts
1. Transient response
2. Steady state response
5.1 Transient Response
This is the variation of output response during the time it takes to reach its final value is called
transient period ts
After completion of transient response, the output of a system becomes stable i.e transient response
becomes zero
5.2 Steady State Response
This is the final value achieved by the system output. It starts when the transient response com-
pletely dies out denoted by css (t)
Total response: addition of transient response and steady state response
c (t) = ct (t) + css (t)
Steady state error: The difference between the desired output and actual output of the system.
Illustration
Figure 58: Transient and steady state response
53
Steady State error
This is the difference between desired output and actual output denoted by ess and its Laplace
transform as E (s)
Consider the closed loop system below
E (s) = R (s) − B (s) (1)
= R (s) − C (s) H (s)
but
C (s) = E (s) G (s)
⇒ E (s) = R (s) − E (s) G (s) H (s)
E (s) [1 + G (s) H (s)] = R (s)
R (s)
E (s) = (2)
1 + G (s) H (s)
In time domain , error signal is denoted by e (t) and is obtained as
e (t) = L−1 E (s) (3)
Steady state error ess is given by

ess = lim e (t) (4)
t→∞
Applying final value theorem in Laplace transform
lim f (t) = lim sF (s)

t→∞ s→0
⇒ ess = lim e (t)

t→∞
54
= lim sE (s)
s→0
R (s)
= lim s (5)
s→0 1 + G (s) H (s)
This shows that the steady state error depends on
1. Types of input - R (s)
2. Types of system - G (s) H (s)
State error Coefficients
Control system has steady state errors for changes in position, velocity, and acceleration
kp - position error constant
kv - velocity error constant
ka - acceleration error constant
Then error constants are called static error constants
Step Input
Consider a step of magnitude A applied as a reference input to the system


A t>0
r (t) =
0 t<0
A
R (s) =
s
From (5)
A 1
ess = lim s
s→0 s 1 + G (s) H (s)
A
= lim
s→0 1 + G (s) H (s)
A
(6)
1 + lims→0 G (s) H (s)
Defining position error constant kp as
kp = lim G (s) H (s) (7)

s→0
55
A
ess = (8)
1 + kp
This indicates that the steady state error is controlled by position error coefficient kp and magnitude
of input A
Ramp Input
Consider a ramp input of slope A
r (t) = At
A
R (s) = (9)
s2
From (5)
A
s2
ess = lim s
s→0 1 + G (s) H (s)
A
= lim
s→0 s + sG (s) H (s)
A
=
lims→0 s + lims→0 sG (s) H (s)
A
(10)
lims→0 sG (s) H (s)
Introducing the term velocity error coefficient kv defined by
kv = lim sG (s) H (s)

s→0
A
⇒ ess = (11)
kv
Thus for ramp input , the velocity error coefficient kv will control the steady state error
Parabolic Input
Consider a parabolic reference input of slope A

2
A 2
r (t) = t
2
56
A
R (s) =
s3
A
s3
ess = lim s
s→0 1 + G (s) H (s)
A
= lim
s→0 s2 + s2 G (s) H (s)
A
=
lims→0 s2 G (s) H (s)
Defining acceleration error coefficient ka defined by
ka = lim s2 G (s) H (s)

s→0
A
⇒ ess =
ka
5.3 Analysis of first order System
The order of a system is the highest power of s in the denominator of the closed loop transfer
function
Consider a first order system shown below
Figure 59: First order system
1
C (s) Ts
=
R (s) 1 + T1
s
1
=
1 + Ts
Unit Step response of first order system
For unit step input to the first order system
57
1
R (s) =
s
C (s) 1
=
R (s) 1 + Ts
1
1 1 T
C (s) = =
s 1 + Ts s s + T1

In partial fraction form

A B
C (s) = +
s s + T1
1 1
= −
s s+ 1
T
Taking inverse Laplace transform
1 1
c (t) = L−1 −
s s+ 1
T
−1 t
= 1 − exp T
which can be plotted as
The poles of the closed loop transfer function are given by
1 + Ts = 0
1
s=−
T
58
5.4 Analysis of Second order system
Definitions
Damping is the tendency of the system to oppose an oscillatory behavior

Damping is measured by a factor called damping ratio denoted by ξ
When ξ is maximum, it provides maximum opposition to oscillatory behavior of the system
Natural Frequency of Oscillation
When ξ is zero , the system oscillates naturally with maximum frequency called natural frequency
of oscillation ωn
Consider the general second order system below
C (s) 2
ωn
=
R (s) 2
s2 + 2ξωn s + ωn
where
ωn - undamped natural frequency
ξ - damping factor
this is the standard form of the closed loop transfer function of a second order system
The characteristic equation of the second order system is given by
s2 + 2ξωn s + ωn
2 =0
The poles of the transfer function are given by
59
q
±2ξωn ± 2 − 4ω 2
4ξ 2 ωn n
s=
2
q
= ξωn ± ωn ξ2 − 1
1. When ξ = 0
s = ±jωn ⇒ the roots are purely imaginary, Undamped system
2. When ξ = 1
s = −ξωn = −ωn roots are real and equal, Critically damped system
3. When ξ > 1
q
s = −ξωn ± ωn ξ 2 − 1 roots are real and unequal , Overdamped system
4. When 0 < ξ < 1

q
s = −ξωn ± jωn ξ 2 − 1 roots are complex conjugates , Under damped system
Undamped 2nd order system for unit step input ξ = 0
C (s) 2
ωn
=
R (s) 2
s2 + ωn
For unit step input
1
R (s) =
s
1 ωn 2
C (s) =
2
s s2 + ωn
A Bs
= +
s s2 + ωn 2
2

ωn
A= =1
s2 + ωn2
s=0
2

ωn
B= = −1
s2 + ωn2
s=0
1 s
C (s) = −
2
s s2 + ωn
" #
−1 1 s
c (t) = L −
2
s s2 + ωn
60
= 1 − cos ωn t
The response of undamped second order system for unit step input is completely oscillatory
Figure 60: Undamped second order system response
Under-damped 2nd order system with unit step input 0 < ξ < 1
C (s) 2
ωn
=
R (s) 2
s2 + 2ξωn s + ωn
with R (s) = 1s
1 2
ωn
C (s) =
2
s s2 + 2ξωn s + ωn
A Bs + C
+
s 2 2
s + 2ξωn s + ωn
(A + B) s2 + (2Aξωn + C) s + Aωn
2 = ω2
n
A=1 B = −1 C = −2ξωn
1 s + 2ξωn
C (s) = −
2
s s2 + 2ξωn + ωn
" #
−1 1 s + 2ξωn
c (t) = L −
2
s s2 + 2ξωn + ωn
s + 2ξωn s + 2ξωn
=
2 2
(s + ξωn )2 + ωn2 − ξ2 ω2

s + 2ξωn + ωn n
s + 2ξωn
=
(s + ξωn )2 + ωn
2 1 − ξ2

61
s + 2ξωn
= q 2
(s + 2ξωn )2 + ωn 1 − ξ 2
q
let ωd = ωn 1 − ξ2
s + 2ξωn
⇒
(s + ξωn )2 + ω 2
d
s + ξωn ξωn
= +
2
(s + ξωn ) + ω 2 (s + ξωn )2 + ω 2
d d
1 s + ξωn ξωn
C (s) = − −
2
s (s + ξωn ) + ω 2 (s + ξωn )2 + ω 2
d d
from Laplace transform tables
ω
⇔ exp−at sin ωt
2
(s + a) + ω 2
s+a
⇔ exp−at cos ωt
(s + a)2 + ω 2

−ξω n t ξωn −ξω n t
c (t) = 1 − exp cos ωd t − exp sin ωd t
ωd

−ξω n t ξωn
c (t) = 1 − exp cos ωd t − sin ωd t
ωd
exp−ξωn t
q
=1− q 2
1 − ξ cos ωd t + ξ sin ωd t
1 − ξ2
q
q 1−ξ 2
sin θ = 1 − ξ 2 tan θ = cos θ = ξ
ξ
exp−ξωn t
c (t) = 1 − q sin θ cos ωd t + cos θ sin ωd t

1 − ξ2
62
sin (x + y) = sin x cos y + cos x sin y
exp−ξωn t
c (t) = 1 − q sin θ + ωd t

1 − ξ2
Figure 61: Under-damped second order system response
The response of under damped 2nd order system with unit step input oscillates before settling to
a final value. The oscillations depend on the value of the damping ratio.
Critically damped 2nd order system with unit step input ξ = 1
C (s) 2
ωn
=
R (s) 2
s2 + 2ωn s + ωn
1
R (s) =
s
1 2
ωn
C (s) =
2
s s2 + 2ξωn s + ωn
2
ωn
s (s + ωn )2
A B C
= + +
s (s + ωn ) (s + ωn )2
A=1 B = −1 C = −ωn
1 1 ωn
C (s) = − −
s s + ωn (s + ωn )2
63
c (t) = 1 − exp−ωn t −ωn t exp−ωn t
= 1 − exp−ωn t [1 + ωn t]
Figure 62: Critically damped second order system response
The response of critically damped 2nd order system has no oscillations
Over damped 2nd order system with unit step input ξ > 1
C (s) 2
ωn
=
R (s) 2
s2 + 2ξωn s + ωn
1
R (s) =
s
the roots of s2 + 2ξωn s + ωn
2 = 0 are
q
−ξωn ± ωn ξ 2 − 1
q
sa = −ξωn + ωn ξ 2 − 1
q
sb = −ξωn − ωn ξ 2 − 1
let s1 = −sa and s2 = −sb
C (s) ωn 2
=
R (s) (s + s1 ) (s + s2 )
A B C
= + +
s s + s1 s + s2
2 2

ωn ωn
A= =

(s + s1 ) (s + s2 ) s1 s2

s=0
ω2 ω2
⇒A= n = n =1
2
q q
−ξωn + ωn ξ 2 − 1 −ξωn − ωn ξ 2 − 1 ωn
64
2

ωn
B=

s (s + s2 )

s=−s1
2
ωn 2
−ωn
= q q q = q
ξωn − ωn 2 2 2
ξ − 1 ξωn − ωn ξ − 1 + −ξωn − ωn ξ − 1 2
2 ξ − 1 s1
ωn
C= q
2 ξ 2 − 1 s2
1 ω2 1 ωn 1
C (s) = − q n + q
+ +

s s s1 s s2
2 ξ 2 − 1 s1 2 ξ 2 − 1 s2
exp−s1 t exp−s2 t
" #
ωn
c (t) = 1 − q −
s1 s2
2 ξ2 − 1
q q
where s1 = ξωn − ωn ξ 2 − 1 and s2 = ξωn + ωn ξ 2 − 1
Figure 63: Over damped second order system response
The response of over damped 2nd order system has no oscillations but it takes longer time for the
response to reach the final steady state value
5.5 Time domain specifications
The transient response of a practical control system often exhibits damped oscillations before
reaching steady state.
A typical damped oscillating response of a system is as shown in Figure 64
Definitions
1. Decay time td - time taken for the response to reach 50% of the final value for the first time
2. Rise time tr - time taken for the response to raise from 0 to 100% for the very first time
65
Figure 64: Time domain specifications
3. Peak time tp - time taken for the response to reach the peak value for the very first time
or time taken for the response to reach the peak overshoot mp
4. Peak overshoot mp - ratio of the maximum peak value measured from the final value to
the maximum value
final value = c (∞)
max value = c tp

c tp −c(∞)
mp =
c(∞)
5. Settling time - time taken by the response to reach and stay within a specified error.
It is usually expressed as a % of final value
66
Usual tolerance error is 2% or 5% of the final value
Expression for time domain specifications
The response of 2nd order system under damped case is given by
exp−ξωn t
c (t) = 1 − q sin ωd t + θ

1 − ξ2
at t = tr , c (t) = c (tr ) = 1
exp−ξωn tr
⇒1− q sin ωd tr − θ = 1

1 − ξ2
exp−ξωn tr
sin ωd tr − θ = 0

⇒ q
1 − ξ2
Since exp−ξωn tr 6= 0 , then
sin ωd tr + θ = 0

⇒ ωd tr + θ = π
π−θ
tr =
ωd
q
1 − ξ2
θ = tan−1
ξ
q
ωd = ωn 1 − ξ2
q
1−ξ 2
π − tan−1
ξ
tr = q
ωn 1 − ξ 2
Peak time
Differentiating c (t)w.r.t. t and equating to 0
exp−ξωn t
c (t) = 1 − q sin ωd t + θ

1 − ξ2
67
q
ωd = ωn 1 − ξ2
q
d ξωn ω n 1 − ξ2
c (t) = q exp−ξωn t sin ωd t + θ − q exp−ξωn t cos ωd t + θ

dt
1 − ξ2 1 − ξ2
ξωn
exp−ξωn t sin ωd t + θ − ωn exp−ξωn t cos ωd t + θ = 0

⇒q
1 − ξ2

ωn
q
exp−ξωn t ξ sin ωd t + θ − 1 − ξ 2 cos ωd t + θ = 0

⇒q
1 − ξ2
q
but cos θ = ξ and sin θ = 1 − ξ 2
ωn
exp−ξωn t cos θ sin ωd t + θ − sin θ cos ωd t + θ = 0

⇒q
1 − ξ2
sin (x − y) = sin x cos y − cos x sin y
ωn
exp−ξωn t sin ωd t + θ − θ

⇒q
1 − ξ2
ωn
=q exp−ξωn t sin ωd t
1 − ξ2
d
c (t) = 0

dt tp
ωn exp−ξωn t
⇒ q sin ωd tp = 0
1 − ξ2
sin ωd tp = 0
⇒ ωd t p = π
π
tp =
ωd
π
= q
ωn 1 − ξ 2
68
Peak overshoot
% overshoot = %mp
c tp − c (∞)

%mp = ∗ 100
c (∞)
where
c tp = peak response at t = tp

c (∞) = final steady state value

at t = ∞
c (∞) = 1
exp−ξωn t
1− q sin ωd t + θ = 1

1 − ξ2
at t = tp with tp = ωπ
d
exp−ξωn tp
c tp = 1 − q sin ωd tp + θ

1 − ξ2
−ξωn ωπ
exp d

π
c tp = 1 − q sin ωd +θ

ωd
1 − ξ2
sin (π + θ) = − sin θ
q
ωd = ωn 1 − ξ2
q
−ξω n π/ω n 1−ξ 2
exp
c tp = 1 − (− sin θ)

q
1 − ξ2
q
−ξπ/ 1−ξ 2
exp
c tp = 1 + sin θ

q
1 − ξ2
but q
sin θ = 1 − ξ2
69
q
−ξπ/ 1−ξ 2
c tp = 1 + exp

c (∞) − c tp

% overshoot = ∗ 100
c (∞)
q
−ξπ/ 1−ξ 2
= exp ∗100
Settling time t (s)

The response of 2nd order system has 2 components
exp−ξωn t
1. decaying eonential component q
1−ξ 2
2. sinusoidal component sin ωd t + θ

The decaying component term reduces oscillations produced by the sinusoidal component. Hence
the settling time is decided by the eonential component.
The settling time can be found by equating the percentage tolerance error to the eonential term
For 2% tolerance
exp−ξωn ts
q = 0.02
1−ξ 2
For least value of ξ
exp−ξωn ts = 0.02
−ξωn ts = ln (0.02)
ln(0.02)
ts =
−ξωn
= 4
ξωn
= 4T for 2% error
For 5% error
exp−ξωn ts = 0.05
−ξωn ts = ln (0.05)
ts = 3
ξωn
70
= 3T
In general for a specified % error, ts is given by
ln (% error)
ts =
ξωn
Example 5.1
A control system having unity feedback has an open loop transfer function
20
G (s) =
s (1 + 4s) (1 + s)
Determine
1. Different static error coefficient

2
2. Steady state error if input = r (t) = 2 + 4t + t2
Solution
(1)
kp = lim G (s) H (s)
s→0
20
= lim =∞
s→0 s (1 + 4s) (1 + s)
kr = lim s G (s) H (s)

s→0
20
= lim s = 20
s→0 s (1 + 4s) (1 + s)
ka = lim s2 G (s) H (s)

s→0
20
= lim s2 =0
s→0 s (1 + 4s) (1 + s)
(2)
t2
r (t) = 2 + 4t +
2
2 4 1
R (s) = L {r (t)} = + +
s s2 s3
R (s)
ess = lim s
s→0 1 + G (s) H (s)
71
 
2 4 1  1

= lim s + +
s s2 s3 20

s→0 1+
s(1+4s)(1+s)
2 + 4 + 1 (1 + 4s) (1 + s) s2
h i
s s2 s3
= lim s
s→0 s (1 + 4s) (1 + s) + 20
ess = ∞
Example 5.2
The open loop transfer function of a unity feedback control system is given by
25
G (s) =
s (s + 5)
Determine
1. maximum overshoot
2. peak time
3. rise time
4. settling time
Solution
25
G (s) = H (s) = 1
s (s + 5)
closed loop transfer function
C (s) G (s)
=
R (s) 1 + G (s) H (s)
25
=
s2 + 5s + 25
general second order equation
2
ωn
H (s) =
2
s2 + 2ξωn s + ωn
2 = 25
ωn ωn = 5rad/s
5
2ξωn = 5 ξ= = 0.5
2∗5
72
q
ωd = ωn 1 − ξ2
√
= 5 1 − 0.25 = 4.33
peak time q
1−ξ 2
π − tan−1
ξ
tr = q
ωn 1 − ξ 2
π − 1.04
= = 0.485sec
4.33
settling time
for 2% error
4 4
ts = = = 1.6sec
ξωn 0.5 ∗ 5
overshoot q
−ξπ/ 1−ξ 2
%mp = exp ∗100
1−0.52 ∗100 = 16.3%

p
= exp−0.5π/
Example 5.3
A unity feedback system has a forward path transfer function
s+2
G (s) =
s (s + 1)
Determine
1. Rise time
2. Peak time
3. Peak overshoot
4. Settling time (2% tolerance)
5. delay time
6. output response to a unit step input
Solution
s+2
G (s) = H (s) = 1
s (s + 1)
73
G (s)
C.L.T.F ==
1 + G (s) H (s)
s+2
=
2
s + 2s + 2
compare with general second order system
s2 + 2ξωn s + ωn
2 =0
s2 + 2s + 2 = 0
2 =2
⇒ ωn ωn = 1.414rad/s
1
2ξωn = 2 ξ= = 0.707
1.414
q
ωd = ωn 1 − ξ 2 = 0.999rad/s
q
1 − ξ2
θ = tan−1 = 0.785
ξ
π−θ
tr = = 2.36sec
ωd
π
tp = = 3.14sec
ωd
q
−ξπ/ 1−ξ 2
%mp = exp ∗100 = 4.32%
4
ts = = 4sec
ξωn
exp−ξωn t
c (t) = 1 − q sin ωd t + θ

1 − ξ2
exp−t
=1− sin (0.999t + 0.785)
0.707
Example 5.4
74
For the control system below, find the values of k1 and k2 so that mp = 25% and Tp = 4sec .
Assume step input
Solution
k
G (s) = 1
s2
H (s) = 1 + k2 s
C (s) G (s)
=
R (s) 1 + G (s) H (s)
k1
=
s2 + k1 k2 s + k1
comparing the denominator with
s2 + 2ξωn s + ωn
2
2 =k
ωn 1
k1 k2 = 2ξωn
2ξωn 2ξ
k2 = =
k1 ωn
mp = 25%
q
−ξπ/ 1−ξ 2
exp ∗100 = 25
q
−ξπ/ 1−ξ 2
exp = 0.25
75
ξπ
q = 1.386
1 − ξ2
ξ2 π2
= 1.92
1 − ξ2
ξ 2 = 0.1628
ξ = 0.4
Tp = 4sec
π π
= = q =4
ωd
ωn 1 − ξ 2
π
=4
ωn ∗ 0.9165
ωn = 0.857
2 = 0.73
k1 = ωn
2ξ
k2 = = 0.93
ωn
76
6 Stability Analysis
6.1 Introduction
A stable system is one that will have a bounded response for all possible bounded inputs A linear
system is stable if and only if all the poles of its transfer function are located on the left of the jω
axis
It is not necessary to determine the actual location of the poles of the transfer function for investi-
gating the stability of a linear system. We only need to find out if the number of poles on the right
of the s-plane is zero or not
Stability Criterion
Simple criterion are used to judge the location the poles of a characteristics equation with respect
to the left or right of the s-plane without actually solving the equation They are;
1. Hurwitz Stability Criterion
2. Routh’s Stability Criterion
6.2 Hurwitz Stability Criterion

The transfer function
C (s) b sm + b1 sm−1 + · · · + bm N (s)

= 0 =
R (s) n
a0 s + a1 sn−1 + · · · + bn D (s)
The necessary and sufficient condition for the polynomial D (s) = a0 sn + a1 sn−1 + · · · + bn = 0
to have all roots on the left half of the s-plane is that the n sub-determinants of the Hurwitz
determinant H should be positive where

a
1 a3 a5 − − −
a0 a2 a4 − − −

0

a1 a3 − − −
H=0 a0 a2 − − −

− − − − − −

− − − − − −

0 0 0

− − an
D1 = |a1 |

a a3
D2 = 1

a0 a2
77

a a3 a5
1
D3 = a0 a2 a4

0

a1 a3

Dn = H

For a stable system, all the sub-determinants should be positive

Example 6.1
Check for stability of the system whose characteristic equation is
s3 + 8s2 + 14s + 24 = 0
Solution

a a3 a5
1
H = a0 a2 a4

0

a1 a3
8 24 0

D3 = 1 14 0 = 2112

0 8 24

D1 = 8 = 8

8 24
D2 = = 88

1 14
D1 , D2 and D3 are positive, hence the system is stable
6.3 Routh’s Stability Criterion
For the transfer function
C (s) b sm + b1 sm−1 + · · · + bm N (s)

= 0 =
R (s) n
a0 s + a1 sn−1 + · · · + bn D (s)
The coefficients of D (s) are arranged in an array called Routh’s array as shown below
78
sn a0 a2 a4 a6
sn−1 a1 a3 a5 a7
sn−2 b1 b2 b3
sn−3 c1 c2 c3
| |
| |
| |
| |
s0 an
The first 2 rows are obtained from the coefficients of D (s).
Elements of the following rows are obtained as below
a a − a0 a3
b1 = 1 2
a1
a 4 − a0 a5
b2 = 1
a1
a a − a0 a7
b3 = 1 6
a1
b a − a1 b2
c1 = 1 3
b1
b a − a1 b3
c2 = 1 5
b1
b a
c3 = 1 7
b1
and so on
Each column will reduce by one as we move down the array. This process is obtained till s0 row is
obtained. The coefficient will be an for this row
Routh’s Criterion
The necessary and sufficient condition for a system to be stable is that all the terms in the first
column of the Routh’s array should have the same sign.
There should be no any sign change in the first column. When there are sign changes in the first
column, then the system is unstable. There are routes on the right hand of the s-plane.
The number of sign changes equals to the number of roots on the right hand side of the s-plane.
79
Example 6.2
D (s) = s4 + 5s3 + 20s2 + 40s + 50 = 0
Solution
s4 1 20 50
s3 5 40 0
s2 12 50 0
s1 230 0 0
12
s0 50 0 0
No sign changes in the first column hence no roots on the right half of the s-plane and therefore
the system is stable.
Example 6.3
D (s) = s3 + s2 + 2s + 24 = 0
Solution
s3 1 2
s2 1 24
s1 −22
s0 24
There are two sign changes in the first column hence two roots on the right half of the s-plane and
therefore the system is unstable.
Special cases
1. If the first element of any row is zero and the rest are non zeros
The next row cannot be formed as division by zero will take place. There are two methods
of this problem
(a) Method 1
It is replaced by a small positive number . The sign of the element of the first column
is then examined as
Example 6.4
80
D (s) = s5 + s4 + 2s3 + 2s2 + 3s + 5 = 0
Solution
s5 1 2 3
s4 1 2 5
s3 0 () −2 0
s2 2+2 5

s1 −4−4−52
2+2
s0 5
Let → 0 from the right side
2 + 2 2
lim = 2 + lim = +∞
→0 →0
No sign changes
−4 − 4 − 52
lim = −2
→0 2 + 2
Sign changes
There are two sign in the first column hence two roots on the right half of the s-plane
and therefore the system is unstable.
(b) Method 2
1 and complete the Routh’s test for z.
Replace s with Z
Example 6.5
D (s) = s5 + s4 + 2s3 + 2s2 + 3s + 5 = 0
Solution
Replacing s with z1
5 4 3 2
1 1 1 1 1

+ + +2 +3 +5=0
Z Z Z Z Z
5Z 5 + 3Z 4 + 2Z 3 + 2Z 2 + Z + 1 = 0
81
Z5 5 2 1
Z4 3 2 1
Z3 − 43 − 23
Z2 1 1
2
Z1 2
Z0 1
There are two sign in the first column hence two roots on the right half of the s-plane
and therefore the system is unstable.
2. When all elements in any one row are zeros
(a) Method 1
Form an auxiliary equation with the help of the coefficients of the row just above the
row of zeros
Take the derivatives of this equation and replace its coefficients in the present row of
zeros. Proceed with the Routh test
Example 6.6
D (s) = s6 + 2s5 + 8s4 + 12s + 20s2 + 16s + 16 = 0
Solution
s6 1 8 20 16
s5 2 12 16 0
s4 2 12 16
s3 0 0 0
s3 row is zero
The auxiliary equation for s4 row is
A (s) = 2s4 + 12s2 + 16
dA
= 8s3 + 24s
dS
Completed array
82
s6 1 8 20 16
s5 2 12 16 0
s4 2 12 16
s3 8 24 0
s2 6 16
s1 2.67
s0 16
No sign change on the first column hence the system may be stable
Solving for auxiliary equation
A (s) = 2s4 + 12s2 + 16 = 0
Let s2 = y
2y 2 + 12y + 16 = 0
y = −4 and y = −2
√
s = ±j2 and s = ±j 2
Non repeated roots on the jω axis give marginally stable system.
(b) Method 2
Form an auxiliary polynomial from the preceding row. The auxiliary polynomial is a
factor of the characteristic polynomial.
Obtain the number of right half plane roots in the auxiliary polynomial
Divide the characteristic polynomial by the auxiliary polynomial and obtain the number
of right half plane roots in the polynomial obtained.
The number of roots of the characteristic polynomial in the right half of the s-plane is
the sum of the two obtained above
Example 6.7
D (s) = s6 + 2s5 + 8s4 + 12s3 + 20s2 + 16s + 16 = 0
Solution
s3 row is zero
The auxiliary equation for s4 row is
83
A (s) = 2s4 + 12s2 + 16 = 0s
√
s = ±j2 and s = ±j 2
Two pairs of roots on the imaginary axis

With D (s) divided by A (s)
Q (s) = s2 + 2s + 2 = 0
Routh array for Q (s)
s2 1 2
s1 2 0
s0 2
No sign change in the first column

All roots of D (s) are on the left half of the s-plane with 2 pairs on the imaginary axis.
The system is therefore marginally stable
84
7 Root Locus
7.1 Introduction
This is a graphical procedure for sketching the movement in the s-plane of the poles of the transfer
function of the closed loop system by varying the loop gain.
Consider the canonical closed loop system
C (s) G (s)
=
R (s) 1 + G (s) H (s)
The closed loop transfer function poles are given by the characteristic equation
1 + G (s) H (s) = 0
The system gain is varied and therefore the roots of the characteristic equation depends on the
system gain k. k is varied from 0 to ∞
If cell location are joined , the resulting diagram or locus is called the Root locus
Example 7.1
For a unity feedback system with G (s) = ks . Obtain the root locus
Solution
k
G (s) = H (s) = 1
s
closed loop poles are given by
1 + G (s) H (s) = 0
k
1+ =0
s
s+k =0
85
s = −k
for
k=0 s=0
k=∞ s = −∞
as k is varied from 0 to ∞ , the root locus varies from 0 to ∞
The root locus starts from the poles of the open loop transfer function and ends on the zeros of the
open loop transfer function
Example 7.2
Given that
ks
G (s) =
s+3
determine when the root locus begins and ends
Solution
O.L.T.F pole
s+3=0
s = −3
O.L.T.F zero
ks = 0
s=0
Hence the root locus begins at s = −3 for k = 0 and ends at s = 0 for k = ∞
86
7.2 Angle and Magnitude Criterion
The closed loop transfer function are given by
1 + G (s) H (s) = 0
G (s) H (s) = −1
= 1∠180° (1)
Since s is a complex number , the above equation has both angle and magnitude and both must
be satisfied as for equation (1)
i.e
|G (s) H (s)| = 1 (2)
angle G (s) H (s) = ±180 (3)
Any point s that lie on the root locus must satisfy equations (2) and (3) which is the magnitude
and angle criterion respectively for k > 0
Example 7.3
Given that
k
G (s) H (s) =
s (s + 2)
check whether points s = −1 + j and s = −2 + j lie on the root locus and determine the value of k
Solution
(1)
s = −1 + j

k
G (s) H (s) =

s=−1+j (−1 + j) (−1 + j + 2)
k
=
(−1 + j) (1 + j)
87
k k
= =
2
j −1 −2
k∠0
=
2∠180
k
= ∠ − 180
2
angle is −180 , point lies on the root locus
also from magnitude criterion
k
=1
2
⇒k=2
no other gain is possible

(2)

k
G (s) H (s) =

s=−2+j (−2 + j) (−2 + j + 2)
k k
= =
j (−2 + j) −1 − j2
k∠0
=√
5∠63.43
k
√ ∠ − 63.43
5
angle criterion not satisfied hence the point does not lie on the root locus
7.3 Procedure for Drawing Root Locus
1. The root locus is always symmetrical about the real axis
2. Number of loci
let n = number of O.L.T.F poles
m = number of O.L.T.F zeros
(a) if m < n
number of loci = number of poles = n
each loci will start from an O.LT.F pole and end at O.L.T.F zero
(n − m) loci will end at infinity
88
(b) if m > n
number of loci = number of zeros = m
each loci will start from an O.LT.F pole and end at O.L.T.F zero
(m − n)loci will end at these zeros from ∞
Example 7.4
for
k (s + 1)
G (s) =
s (s + 2) (s + 3)
n=3 m=1
n−m=2
loci ends at infinity

one loci ends at s = −1 (zero at −1)
3. Real axis Loci

Determined by open loop transfer function poles and zeros. Any point on the real axis is a
part of the loci iff the number of poles and zeros to its right is odd.
k (s + 1)
G (s) H (s) =
s (s + 2) (s + 3)
the poles and zeros are plotted and the loci analyzed as follows
(a) All points between s = 0 and s = −1 not (inclusive) have one pole at s = 0 (odd). Hence
loci is present in the region
89
(b) between s = −1 and s = −2 (not inclusive) , there are two (one pole at s = 0 , zero at
s = −1) to the right .Hence loci absent
(c) loci present between s = −2 and s = −3 (not inclusive), 3 elements (2 poles s = 0, −2
and one zero s = −1) to the right
(d) no loci beyond s = −3. four elements (3 poles, one zero to the right)
4. Angle of asymptotes
A line which the locus touches at infinity. The loci of these branches that **** to infinity
move along the asymptotes
Number of asymptotes = n − m
angle of asymptotes is given by
(2k + 1)
β= 180
n−m
for k = 0, 1, 2, . . . . . . n − m − 1
from
k (s + 1)
G (s) H (s) =
s (s + 2) (s + 3)
number of asymptotes = n − m = 3 − 1 = 2
(2k + 1)
β= 180 k = 0, 1
2
β = 90° , 270°
3 loci started at s = 0, −2, −3

one loci end at s = −1
The loci starting at s = −2 and s = −3 do not have any zero at real negative axis and
therefore they must break the real axis and move out
The point they break away is called break away point, they will be guided by asymptotes
which they should touch at k = ∞
5. center of asymptotes (centroid)

The asymptotes touch the real axis at a point called centroid σc where
P P
real parts of poles of OLT F − real parts of zeros of OLT F
σc =
n−m
The centroid σc and angle of asymptotes , fixes the location of asymptotes
From previous example
poles at s = −2, −3 and 0
90
zero at s = −1
− (0 + 2 + 3) −
P P
−1
σc =
3−1
−5 + 1
= = −2
2
6. Breakaway point
The point at which the root locus comes out of the real axis.
Obtained from
dk
=0
ds
7. Angle of arrival/departure
The root locus arrives at a complex zero at k = ∞. Similarly, the root locus departs from a
complex pole at k = 0 . These angles are known as angle of arrival and departure respectively
angle of departure from poles φD is given by
φD = 180° + arg 6 H
where arg 6 H → angle of G (s) H (s) excluding the pole when angle is calculated
angle of arrival φA is given by
φA = 180° − arg 6 H
8. jω crossover or intersection of imaginary axis

we make use of Routh criterion
91
9. Sketch the root loci
10. Comment on stability of the system
Example 7.5
Draw the root locus of the system
k
G (s) H (s) =
s (s + 3) (s + 6)
obtain the value of k when ξ = 0.6 from the root locus
Determine the value of k for marginal stability and critical damping
Solution
1. number of loci
n=3
m=0
number of loci is 3
loci ending at infinity is 3
There are three loci all proceeding to end at infinity
2.
3. Real axis loci
(a) present between −3 < σ < 0

(b) absent between −6 < σ < 3
92
(c) present between −∞ < σ < −6
three loci starting at s = 0 , s = −3 , s = −6 all proceed to infinity along 3 asymptotes
between s = 0 and s = −3 , two loci start towards each other. They must break away
between 0 and −3
4. number of asymptotes and their angles
n−m=3
(2k + 1)
β= 180 k = 0, 1, 2
n−m
(2k + 1)
180
3
= 60° , 180° , 300°
5. centroid
P P
σc =
n−m
(0 + −3 + −6) − 0
=
3−0
= −3
6. Breakaway point
k = − s3 + 9s2 + 18s

dk
=0
ds
0 = − s3 + 9s2 + 18s

s2 + 6s + 6 = 0
s = −4.73 s = −1.27
93
taking s = −1.27 (real axis loci present below)
s=0 and s = −3
therefore
k = − (−1.27)3 + 9 (−1.27)2 + 18 (1.27)
h i
= 10.39
critical damping k = 10.39
7. angle of arrival
Not needed : no complex poles at zeros are present
8. Intersection of imaginary axis

Characteristic equation
s (s + 3) (s + 6) + k = 0
s3 + 9s2 + 18s + k = 0
Routh array
s3 1 18
s2 9 k
s1 162−k
9
s0 k
for stability column 1 should be position
162 − k
>0
9
k < 162 k>0
⇒ 0 < k < 162
94
jω axis cross over s = ±j4.24
Example 7.6
The open loop transfer function of a system is
k
G (s) H (s) =
s (s + 2 + j2) (s + 2 − j2)
determine the complete root locus and comment on the stability of the closed loop system
Solution
1. number of loci
n=3
95
m=0
number of loci is 3
2. Real axis loci

present between 0 < σ < −∞
3. number of asymptotes and their angles
n−m=3
(2k + 1)
β= 180 k = 0, 1, 2
n−m
(2k + 1)
180
3
= 60° , 180° , 300°
4. centroid
P P
σc =
n−m
(−2 − 2) − 0
= = −1.33
3
5. Breakaway point
1 + G (s) H (s) = 0
s3 + 4s2 + 8s + k = 0
96
k = −s3 − 4s2 − 8s
dk
= −3s2 − 8s − 8 = 0
ds
3s2 + 8s + 8 = 0
s = −1.33 ± j0.94
6. Angle of departure
k
σH =
s (s + 2 + j2) (s + 2 − j2)

k
σH 0 =

s (s + 2 + j2) s=−2+j2
k
=
(−2 + j2) (−2 + j2 + 2 + j2)
k
=
(−2 + j2) j4
k∠0
√
8∠135 4∠90
arg σH 0 = −225°
φD = 180° + arg σH 0 = −45°
φD = −45 at s = 2 + j2 and
φD = 45° at s = −2 − j2
7. jω crossover
1 + G (s) H (s) = 0
s3 + 4s2 + 8s + k = 0
97
s3 1 8
s2 4 k
s1 32−k
4
s0 k
32 − k = 0 kmax = 32 k=0
for stability 0 < k < 32
4s2 + k = 0
4s2 + 32 = 0
s = ±j2.82
system unstable for k > 32

system marginally stable for k = 32
system stable at 0 < k < 32
98
99
8 Frequency Domain Analysis
Frequency response is the steady state response of a system to a sinusoidal input.
Sinusoidal Response of a Linear System
Consider a system as shown in Figure below where

r(t)- input function(sinusoidal)
y (t)- output function
g (t)- Transfer function
c (t) = g (t) r (t)
Taking Laplace transform
C (s) = G (s) R (s)
and
C (s) N (s) N (s)

G (s) = = =
R (s) D (s) (s + a) (s + b) (s + c) + · · ·
N (s)
C (s) = R (s) (1)
(s + a) (s + b) (s + c) + · · ·
Consider r (t) = A sin ωt

Thus
Aω
R (s) =
s + ω2
2
N (s) Aω
C (s) = (2)
(s + a) (s + b) (s + c) + · · · s + ω 2
2
eanding equation (2) in partial fraction form
A1 A2 A3 B1 B2
C (s) = + + + ··· + +
s+a s+b s+c s − jω s + jω
The inverse Laplace becomes
100
c(t) = A1 exp−at +A2 exp−bt +A3 exp−ct + · · · + B1 expjωt +B2 expjωt (3)
The terms with coefficients A1 to An contribute to decaying eonents and therefore the transient
response. The terms with coefficients B1 and B2 contribute the steady state response and thus
css (t) = B1 expjωt +B2 expjωt (4)
Thus
Aω
C (s) = G (s)
s + ω2
2
The coefficients B1 and B2 are calculated as

Aω
B1 = G (s) (s − jω) |s=jω
s2 + ω 2
A
= G (jω)
2j

Aω
B2 = G (s) (s + jω) |s=jω
s2 + ω 2
A
= G (−jω)
2j
G (jω) is a complex quantity written as
G (jω) = |G (jω)| expjφ
Also
G (−jω) = |G (−jω)| exp−jφ
Putting the values ofB1 and B2 into equation (4)
A A
css (t) = |G (jω)| expjφ expjωt + |G (−jω)| exp−jφ expjωt
2j 2j
expj(ωt+φ) − exp−j(ωt+φ)
" #
= A |G (jω)|
2j
= A |G (jω)| sin (ωt + φ) (5)
101
therefore
css (jω) = |C (jω)|
where
|C (jω)| = A |G (jω)|
= |G (jω)| |R (jω)|
Conclusions
1. The sinusoidal output has same frequency as input
2. The output is sinusoidal and has a phase difference of φ and a different magnitude
3. The system merely changes the amplitude and phase of the output with respect to the input
Methods used in Frequency Response

Frequency response characteristics have two plots
1. Magnitude function
2. Phase function
For both plots, angular frequency ω is the variable
Performance Specifications
Consider a closed loop system of Figure below
The transfer function for the system is given by
C (s) G (s)
M (s) = =
R (s) 1 + G (s) H (s)
C (jω) G (jω)
M (jω) = =
R (jω) 1 + G (jω) H (jω)
1. Response Peak Mr
This is the maximum value of M (jω) as ω is varied It indicates relative stability of the closed
loop system. A larger value of Mr corresponds to large maximum overshoot in a step response
102
2. Resonance frequency ωr
This is the frequency at which peak response Mr occurs
3. Phase Margin φP M
It is a measure of relative stability. Let φ1 be the phase angle of a system at unity gain. The
phase margin is given by
φpM = 180 + φ1
The frequency at which the gain is unity, ω1 is called crossover frequency.
4. Gain Margin
The frequency ω2 where the phase angle of transfer function is −180° is called phase crossover
frequency.
The gain at this frequency is given by|G (jω2 )| |H (jω2 )|
The gain margin is given by
1
Gain M argin =
|G (jω2 )| |H (jω2 )|
The phase margin indicates how much the system angle can be increased to cause the system
to become unstable from a stable condition
The gain margin indicates how much gain can be increased to cause system instability
Correlation between Time and Frequency Response Specifications
For a canonical second order system
2
ωn
G (s) =
s2 + 2ξωn s + ωn
With usual notation

2
ωn
G (jω) =
(jω)2 + 2ξωn (jω) + ωn
2
Dividing the numerator and denominator by ωn
1
G (jω) = 2 (6)
2ξω

1 − ωωn + j ωn
a) Resonant peak
Let u = ωωn
103
1
G (jω) = h
1 − u2 + j2ξu
i
therefore
1
|G (jω)| = r 2 =M (7)
1−u2 + (2ξu) 2
2ξu
∠G (jω) = tan−1 (8)
1 − u2
For resonant frequency ωr
d
|G (jω)| = 0
du
1 2 1 − u2 (−2u) + 8ξ 2 u

d − 2
|G (jω)| = 3/2 =0
du 2
1−u 2 + (2ξu) 2
−4u + 4u3 + 8ξ 2 u = 0
u2 + 2ξ 2 − 1 = 0
q
u= 1 − 2ξ 2
therefore
ω
q
= 1 − 2ξ 2
ωn
At ω = ωr
q
ωr = ωn 1 − 2ξ 2
1
Mr = r 2
1 − u2r + (2ξur )2
1
= q (9)
2ξ 1 − ξ 2
Mr is a function of the damping factor ξonly
104
Similarly
2ξu
φr = tan−1
1 − u2
√
1 − 2ξ
= tan−1
ξ
Conclusion
1. As ξapproaches zero, Mr approaches infinity and ωr approaches ωn
2. For 0 < ξ < 0.707, Mr is greater than unity and ωr is less than ωn
3. For ξ > 0.707, there is no resonant peak and the maximum value Mr of is unity and ωr = 0
b) Cut off frequency ωc and Bandwidth (BW)

Cut off frequency ωc is the point where M has a value of 0.707.
Usually control systems are low pass filters, hence M = 1 for zero frequency, bandwidth becomes
equal to ωc
Let uc = ωωc
n
Substituting into equation (7)
1
0.707 = r 2
1 − u2c + (2ξuc )2
2
1 − u2c + (2ξuc )2 = 2

2 2
q
uc = 1 − 2ξ ± 4ξ 4 − 4ξ 2 + 2

1
2

1 − 2ξ 2 ± 4ξ 4 − 4ξ 2 + 2
q
BW = ωc = ωn (10)
Conclusions
From (11) BW is a function of ωn and ξ
1. BW is directly proportional to ωn
2. BW and Mr are directly proportional for 0 < ξ < 0.707
3. As time increases, BW decreases
105
Example
For a second order system with unity feedback
200
G (s) =
s (s + 8)
Determine the frequency domain specifications

Solution
200
M (s) =
2
s + 8s + 200
2
ωn
s2 + 2ξωn s + ωn
ωn = 14.14 and ξ = 0.283
1
Mr = q = 1.842
2ξ 1 − ξ 2
q
ωr = ωn 1 − 2ξ 2 = 12.96rad/s
1/2
1 − 2ξ 2 ± 4ξ 4 − 4ξ 2 + 2
q
BW = ωc = ωn = 30.25rad/s
Example
Consider a closed loop unity feedback system with
k
G (s) =
s (s + a)
Determine the values of k and a that will satisfy the conditions Mr = 1.04 and ωr = 11.55rad/s.
Determine also the settling time and bandwidth.
Solution
M (s) = 2 k
s +as+k
√ a
ωn = k and ξ = √
2 k
1 2k
Mr = = p
a 4k − a2
q
2ξ 1 − ξ2
106
Substituting forMr = 1.04
1.04a 4k − a2 = 2k
p
4k 2 − 4.32a2 k + 1.08a4 = 0
k = 0.686a2 or k = 0.393a2
From ωr = 11.55rad/s condition

q
ωr = ωn 1 − 2ξ 2
1p
= 2k − a2
2
2k − a2 = 266.7
For k = 0.686a2
1.372a2 − a2 = 266.7
a = 26.77 and k = 491.61
For k = 0.393a2
0.786a2 − a2 = 266.7
This does not give real value of a
a = 26.77 and k = 491.61
√
ωn = k = 22.17
and
a
ξ = √ = 0.604
2 k
4
ts = = 0.298
ξωn
107
1/2
1 − 2ξ 2 ± 4ξ 4 − 4ξ 2 + 2
q
BW = ωc = ωn = 28.15rad/s
9 Bode Plots
The frequency response of any linear system is indicated by plotting two curves or graphs
• Graph 1- The magnitude (M) of the transfer function

The magnitude of the open loop transfer function is plotted in decibels on Y-axis. On the
X-axis frequency is plotted on a logarithmic scale.
The logarithm of the magnitude of the transfer function G(jω) is expressed in decibels as
M = |G(jω)| = 20log|G(jω)dB
• Graph 2- The phase angle (φ) as function of frequency (ω)

The phase angle of the open loop transfer function is plotted on the Y-axis. On the X-axis,
same as in graph 1, frequency is plotted on a logarithmic scale
These characteristics are called Bode plots. Both the graphs are plotted in one semilog paper with
same X-axis scale.
Two units used to express frequency ratios are the octave and decade.
f2
An octave is a frequency band from f1 to f2 where = 2. The number of octaves in the frequency
f1
range from f1 to f2 is
log(f2 /f1 ) f
= 3.32log 2
log2 f1
There is an increase of 1 decade from f1 to f2 when f 2/f 1 = 10. The number of decades in the
frequency range from f1 to f2 is
f
log 2 decades
f1
As a number doubles, the decibel value increases by 6 dB. As a number increases by a factor of 10
the decibel value increases by 20 dB.
Advantages of Bode Plot
1. Both low and high frequency characteristics of a transfer function can be shown in one diagram
2. Gain margin and phase margin can be obtained with minimum computational efforts from
the plots
3. The can be plotted with ease using asymptotic approximations
108
4. It indicates clearly relative stability of the system
5. Data for constructing polar plots and Nyquist plots of complex Transfer function can be
obtained from bode plots
6. Frequency domain specifications can be obtained easily from bode plots
7. Stability of open loop transfer function can be obtained using bode plots
8. Transfer function of the system can be obtained from bode plots
The frequency transfer function can be written in a generalized form as
N (s) b sm + b1 sm−1 + ... + bm

G(jω) = = 0
D(s) a0 sn + a1 sn−1 + ... + bn
This must be factorized to general type
K1 sg (s + z1 )(s + z2 )...
G(jω) =
sk (s + p1 )(s + p2 )...
NB: either sg or sk is present, not both
Ksg (1 + s/z1 )(1 + s/z2 )...

=
sk (1 + s/p1 )(1 + s/p2 )...
Note:
K1 z1 z2 ...
K=
p1 p2 ...
Putting s = jω
K(jω)g (1 + jω/z1 )(1 + jω/z2 )...
G(jω) =
(jω)k (1 + jω/p1 )(1 + jω/p2 )...
where K is the gain constant or Bode gain.
In general a transfer function can be considered to be a combination of the following terms
• Constant term
• Poles and Zeros at the origin
• Poles and Zeros not at the origin
• Complex Poles and Zeros
9.1 Bode plots of standard factors

9.1.1 Bode gain K or Constant factor
For the magnitude in dB

20logG(jω) = 20logK = K2 dB
109
is plotted on the Y-axis
For the phase, K is either positive or negative. Thus for positive value of K, the bode plot is a
straight horizontal line at 0o for all ω and the phase angle is −180o for all ω for negative value of
K
Both the plots are as shown in Figure 65
Figure 65: Magnitude and Phase Plots for Constant factor
1
9.1.2 Poles at the origin of order k or integral factor
(jω)k
Consider
1
G(jω) = = (jω)−k
(jω)k
The magnitude in dB is obtained as
20logG(jω) = 20log(jω)−k
= −20klog(jω) dB
The quantity plotted on the X-axis is log(jω) and on Y-axis in this case is
Y = −20klog(jω) = −kX
This corresponds to a graph of Y = M X where the slope is −20k and is a straight line passing
through the origin given by (X = 0, Y = 0). Since X = log(jω) = 0 at ω = 1
Note:
• ω = 0 is not defined as log(jω) = −∞
• If k = 1, the straight line passes through (0 dB, ω = 1) at slope = −20 db/dec
• If k = 2, the straight line passes through (0 dB, ω = 1) at slope = −40 db/dec and so on
For the phase
• For k = 1 argG(jω) = −90o
• For k = 2 argG(jω) = −180o
• For k = 3 argG(jω) = −270o
• For k = k argG(jω) = −90k o
Both the magnitude and phase plots are as shown in Figure 66
110
Figure 66: Magnitude and Phase Plots for Integral factor
9.1.3 Zeros at the origin(jω)g of order g or derivative factor
Consider
G(jω) = (jω)g
The Magnitude in dB is obtained as
20logG(jω) = 20log(jω)g
= 20glog(jω)
This is a plot at a slope of 20g dB/dec

If g = 1, the straight line passes through (0 dB, ω = 1) at slope= 20db/dec
If g = 2, the straight line passes through (0 dB, ω = 1) at slope= 40db/dec and so on
The Phase is obtained as

For g = 1 argG(jω) = 90o
For g = 4 argG(jω) = 9g0o
Both the magnitude and phase plots are as shown in Figure 67
111
Figure 67: Magnitude and Phase Plots for Derivative factor
1
9.1.4 Finite poles at or first order integral factor
(1 + jω/p1 )
Consider
1
G(jω) = ω
(1 + j )
p1
The quantity plotted on the Y-axis is
1 ω −1
Y = 20logG(jω) = 20log ω = 20log(1 + j p )
(1 + j ) 1
p1
ω
= −20log(1 + j )
p1
For ω p1 , the equation becomes
Y = −20log(1) = 0 dB
For ω p1 then equation becomes
ω
Y = −20log( ) dB
p1
ω ω ω
If we change X-axis to log( ) instead of logω then the graph of −20log( ) versus log( ) is a
p1 p1 p1
straight line of slope −20 dB/dec. Then this case can be approximated into two parts.
ω ω
For 1,the slope= 0 dB and for 1, the slope= −20 dB. Also the change in slope of
p1 p1
0 dB to −20 dB is assumed to occur at ω = p1 . This can be summarized as follows
ω
• The graph has a slope of 0 dB for 1
p1
112
ω
• The graph has a slope of −20 dB for 1
p1
ω
• The change of slope occurs at = 1 i.e. ω = p1
p1
Error in approximation
The actual magnitude curve is
1
|G(jω)| = v
u 2
t1 + ω
u
p21
The actual magnitude is given by

1
|G(jω)| = √
1+1
Therefore 20log|G(jω)| = −3 dB
The phase is obtained as

ω
argG(jω) = −tan−1
p1
As ω becomes larger and larger the phase angle approaches −90o from 0o
Example
Plot the Magnitude and Phase plots for
1
G(jω) = ω
1+
50
Solution
Magnitude plot
|G(jω)| = 0 dB f or ω < 50
= −20 dB/dec f or ω > 50
Phase plot
ω
50
for ω = 0 argG(jω) = 0o
as ω → ∞ argG(jω) → −90o
Figure 68: Magnitude and Phase Plots for First order integral factor
9.1.5 Finite zeros at (1 + jω/z1 ) or first order derivative factor
For the Magnitude plot
113
The analysis has the same treatment as that of first order poles
ω
• The graph has a slope of 0 for 1
z1
ω
• The graph has a slope of 20 dB for 1
z1
ω
• The change of slope occurs at = 1 i.e. ω = z1
z1
For Phase plot

ω
argG(jω) = tan−1
z1
for ω = 0 argG(jω) = 0o
as ω → ∞ argG(jω) → 90o
Error in approximation
At ω = z1
ω2
s
|G(jω)| = 1+
z1
In dB this gives 3 dB at ω = z1
ω
The magnitude and phase plots for (1 + j ) is as shown in Figure 69
z1
Figure 69: Magnitude and Phase Plots for First order derivative factor
9.1.6 Second order poles
Consider
ωn
G(s) =
2 2
s + 2ζωn s + ωn
Then
2
ωn 1
G(jω) = =
(jω)2 + j2ζωn ω + ω 2 (1 − (
ω 2
) + j2ζ
ω
ωn ωn
For the magnitude plot
1
|G(jω)| = v"
2 #2
ω 2
u
t 1− ω
u
+ 2ζ
ωn ωn
As G(s) is a second order factor approximate, Bode magnitude plot is drawn as

0 dB line for ω ωn
−40 dB/dec line for ω ωn
Also do necessary error correction at ω = ωn depending on the value of ζ
114
The phase angle is given by
ω
2ζ
ωn
ω 2

1−
ωn
Error in Approximation
1
At ω = ωn the value of |G(jω)| =
2ζ
1
∴ |G(jω| in dB = 20log
2ζ
This error depends on ζ
9.1.7 Second order zeros
ω 2

ω
G(jω) = 1 − + j2z
ωn ωn
The magnitude is approximated as

0 dB line for ω ωn
40 dB/dec line for ω ωn
The nature of the error has just reversed as compared to earlier case.
The phase is approximated as

ω
2ζ
ω
argG(jω) = tan−1 n
ω 2

1−
ωn
The phase plots are functions of both ω and ζ.
9.2 Frequency Domain Specifications
9.2.1 Gain Margin (GM)
It is a measure of relative stability and is the reciprocal of the open-loop transfer function, evaluated
at the phase crossover frequency ωpc , the frequency at which the phase is −180o .
The gain margin in dB is the distance on the Bode magnitude plot from the amplitude at the phase
crossover frequency up to the 0 dB point.
115
9.2.2 Phase Margin (φP M )
As a measure of relative stability, it is defined as 180o plus the phase angle φ1 of the open loop
transfer function at unity gain. That is
φP M = 180 + arg|G(jωgc )|degrees
Where G(jωgc ) = 1 and ωgc is the gain crossover frequency which is the point on the magnitude
plot where the magnitude is unity.
9.2.3 Relative Stability
The relative stability of a system is indicated by the gain and phase margin. The parameters are
easily determined from the Bode plots.
In most cases, positive gain and phase margins will ensure stability of the system.
For stable systems ωgc < ωpc i.e. GM and P M are positive
For unstable systems ωgc > ωpc i.e. GM and P M are negative
When GM is +∞, then the system is absolute stable.
When both GM and P M are zero, then the system is marginally stable
Figure 70: Gain Margin and Phase Margin
Example
For the system with open loop transfer function
10
G(s) =
s(s + 1)(s + 5)
Determine the stability of the system by plotting the Bode plot of the system
Solution
Writing G(s) in standard form
2
G(s) = s
s(1 + s)(1 + )
5
Substituting for s = jω
2
G(jω) =
jω
jω(1 + jω)(1 + )
5
In the given transfer function, the following four factors are present
1. Constant term k = 2
The magnitude is obtained as
116
20 log 2 = 6.02 dB
The Phase is 0o
1
2. Pole at the origin
jω
The magnitude plot is a straight line of slope −20 db/dec crossing the 0 dB line at ω = 1
The phase is −90o
1
3. First order pole The magnitude plot consists of
1 + jω
a line of slope 0 dB/dec for ω < 1 and a line of slope −20 dB/dec for ω > 1
The magnitude plot terminates at ω = 5
The phase is given by −tan−1 ω
1
4. First order pole ω The magnitude plot consists of
1+j
5
a line of slope 0 dB/dec for ω < 5 and a line of slope −20 dB/dec for ω > 5
The magnitude plot terminates at ω = ∞
ω
The phase is given by −tan−1
5
Preparing a table for the magnitude plot

Factor Resultant slope Start point (ω) End point (ω)
k=2 straight line at 6.02 dB 0.1 ∞
1
−20 dB/dec 0.1 1
jω
1
−40 db/dec 1 5
1 + jω
1
ω −60 dB/dec 5 ∞
1+j
5
The resultant phase is obtained as
ω
φ(ω) = −90 − tan−1 (ω) − tan−1 ( )
5
Preparing the phase angle table
117
ω
ω −900 −tan−1 ω −tan−1 φ(ω)
5
0.01 −900 −0.57 −0.11 −90.68
0.05 −900 −2.86 −0.57 −93.43
0.1 −900 −5.71 −1.15 −96.86
0.5 −900 −26.56 −5.71 −122.27
1 −900 −45.00 −11.31 −146.31
10 −900 −84.29 −63.43 −237.72
100 −900 −89.42 −87.14 −266.56
500 −900 −89.89 −89.42 −269.31
1000 −900 −89.94 −89.71 −269.65
The bode plot is as shown in Figure 71
Figure 71: Bode Plot
From the Bode plot Gain margin = 12 dB
Phase margin = 15o
118
Gain crossover frequency ωgc = 1.4 rad sec
Phase crossover frequency ωpc = 2.7 rad sec
ωgc < ωpc and therefore the system is stable
119

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Kenyatta University

BSc Mechanical Engineering

EMM 416: CONTROL SYSTEMS I

April 10, 2019

EMM 311 Mechanics of machines II

At the end of this course, the student should be able to;

1. Distinguish between open loop and closed loop control systems

2. Determine the transfer function for a given system.

3. Describe various control elements

4. Assess the stability of a system

5. Describe the various control actions in control systems

1. Control Engineering laboratories;

Prescribed Text Books

2. Ogata K. (1996) Modern Control Engineering, Prentice Hall, 3rd Ed.

1. Kuo B. C. Farid G. (2002) Automatic Control Systems, Wiley, 8th Ed.

3. Journal of Dynamic Systems, Measurement, and Control

2 Transfer Function and Mathematical Modelling 7

3 Block Diagram Representation 30

5 Time Domain Analysis 53

8 Frequency Domain Analysis 100

9 Bode Plots 108

System: An arrangement of physical components connected or related in such a manner as to

Control system: an arrangement of physical components connected or related in such a manner

• An arrangement to achieve this input-output combination

1.2 Requirements of a good control system

1.3 Classification of Control systems

1. Open loop systems

2. Closed looped systems

Open loop systems

Figure 1: Open Loop Control System

1. Automatic washing machine

2. Automatic hand drier

3. Traffic control by means of signal operated on time basis.

1. They are simple in construction and design

2. They are easy to maintain

3. Convenient to use when the output is difficult to measure

4. Not much problems of stability

Disadvantages of open loop systems

1. Inaccurate and unreliable

2. Inaccurate results obtained with parameter variations, internal disturbances

Closed loop control system

Figure 2: Closed Loop Control System

r (t) - reference input

1. Automatic electric iron

2. DC motor speed controlled by tacho feedback

3. Sun-seeker solar systems

4. Automatic water level controller.

Advantages of closed loop systems

1. Accuracy is very high since any error arising is corrected.

2. Reduced effects of non-linearities

Disadvantages of closed loop systems

1. Complicated in design and maintenance costlier

2. System may become unstable

Table 1: Open loop and closed loop systems

Figure 3: Servomechanism of automobile steering system

Linear and Non-linear Control Systems

• Non-linear systems exhibit peculiar behaviour

• They do not obey the law of superposition.

• Standard test signals approach cannot be used here