ece-iv-control systems [10es43]-notes · pdf filecontrol systems 10es43 sjbit/ dept

Control Systems 10ES43

SJBIT/ Dept of ECE

CONTROL SYSTEMS

(Common to EC/TC/EE/IT/BM/ML)

Sub Code: 10ES43 IA Marks : 25

Hrs/ Week: 04 Exam Hours : 03

Total Hrs: 52 Marks : 100

UNIT 1:

Modeling of Systems: Introduction to Control Systems, Types of Control Systems, Effect of

Feedback Systems, Differential equation of Physical Systems -Mechanical systems, Friction,

Translational systems (Mechanical accelerometer, systems excluded), Rotational systems, Gear

trains, Electrical systems, Analogous systems. 7 Hrs

UNIT 2:

Block diagrams and signal flow graphs: Transfer functions, Block diagram algebra, Signal

Flow graphs (State variable formulation excluded), 6 Hrs

UNIT 3:

Time Response of feedback control systems: Standard test signals, Unit step response of First

and second order systems, Time response specifications, Time response specifications of second

order systems, steady123– state errors and error constants. Introduction to PID Controllers

(excluding design) 7 Hrs

UNIT 4:

Stability analysis: Concepts of stability, Necessary conditions for Stability, Routh- stability

criterion, Relative stability analysis; more on the Routh stability criterion. 6 Hrs

UNIT 5:

Root–Locus Techniques: Introduction, The root locus concepts,Construction of root loci 6 Hrs

UNIT 6:

Frequency domain analysis: Correlation between time and frequency response, Bode plots,

Experimental determination of transfer functions, Assessment of relative stability using Bode

Plots. Introduction to lead, lag and lead-lag compensating networks (excluding design). 7 Hrs

UNIT 7:

Stability in the frequency domain: Introduction to Polar Plots, (Inverse Polar Plots excluded)

Mathematical preliminaries, Nyquist Stability criterion, Assessment of relative stability using

Nyquist criterion, (Systems with transportation lag excluded). 7 Hrs

UNIT 8:

Introduction to State variable analysis: Concepts of state, state variable and state models for

electrical systems, Solution of state equations. 6 Hrs


SJBIT/ Dept of ECE

TEXT BOOK :

1. J. Nagarath and M.Gopal, ―Control Systems Engineering‖, New Age

International (P) Limited, Publishers, Fourth edition – 2005

REFERENCE BOOKS:

1. “Modern Control Engineering “, K. Ogata, Pearson Education Asia/

PHI, 4th Edition, 2002.

2. “Automatic Control Systems”, Benjamin C. Kuo, John Wiley India

Pvt. Ltd., 8th Edition, 2008.

3. “Feedback and Control System”, Joseph J Distefano III et al.,

Schaum‘s Outlines, TMH, 2nd Edition 2007.


SJBIT/ Dept of ECE

INDEX SHEET

SL.NO TOPIC PAGE

NO.

I UNIT 1:Modeling of Systems 1-22

1.1 Introduction to Control Systems, Types of Control Systems

1.2 Effect of Feedback Systems

1.3 Differential equation of Physical Systems -Mechanical systems, Friction

1.4 Translational systems (Mechanical accelerometer, systems excluded)

1.5 Rotational systems , Gear trains

1.6 Electrical systems, Analogous systems

II UNIT–2 : Block diagrams and signal flow graphs 23-42

2.1 Transfer functions

2.2 Block diagram algebra

2.3 Signal Flow graphs (State variable formulation excluded)

III UNIT – 3 :Time Response of feedback control systems

3.1 Standard test signals 43-84

3.2 Unit step response of first order systems

3.3 Unit step response of second order systems

3.4 Time response specifications

3.5 Time response specifications of second order systems

3.6 steady23 – state errors and error constants)

3.7 Introduction to PID Controllers (excluding design)

IV UNIT – 4 : Stability analysis 85-110

4.1 Concepts of stability, Necessary conditions for Stability

4.2 Routh- stability criterion

4.3 Relative stability analysis; More on the Routh stability criterion.

V UNIT – 5 : Root–Locus Techniques 111-133

5.1 Introduction, The root locus concepts

5.2 Construction of root loci (problems)

VI UNIT – 6 : Frequency domain analysis 134-169

6.1 Correlation between time and frequency response

6.2 Bode plots

6.3 Experimental determination of transfer functions

6.4 Assessment of relative stability using Bode Plots

6.5 Introduction to lead, lag and lead-lag compensating networks (excluding

design)

VII UNIT – 7 : Stability in the frequency domain 170-182

7.1 Introduction to Polar Plots,(Inverse Polar Plots excluded)

7.2 Mathematical preliminaries


SJBIT/ Dept of ECE

7.3 Nyquist Stability criterion

7.4 Assessment of relative stability using Nyquist criterion, (Systems with

transportation lag excluded)

VIII UNIT – 8 : Introduction to State variable analysis 183-201

8.1 Concepts of state, state variable

8.2 Concepts of state models for electrical systems

8.3 Solution of state equations


SJBIT/ Dept of ECE Page 1

UNIT-1

A control system is an arrangement of physical components connected or related in such a

manner as to command, direct, or regulate itself or another system, or is that means by which any

quantity of interest in a system is maintained or altered in accordance with a desired manner.

Any control system consists of three essential components namely input, system and out

put. The input is the stimulus or excitation applied to a system from an external energy source.

A system is the arrangement of physical components and output is the actual response obtained

from the system. The control system may be one of the following type.

1) man made

2) natural and / or biological and

3) hybrid consisting of man made and natural or biological.

Examples:

1) An electric switch is man made control system, controlling flow of electricity.

input : flipping the switch on/off

system : electric switch

output : flow or no flow of current

2) Pointing a finger at an object is a biological control system.

input : direction of the object with respect to some direction

system : consists of eyes, arm, hand, finger and brain of a man

output : actual pointed direction with respect to same direction

3) Man driving an automobile is a hybrid system.

input : direction or lane

system : drivers hand, eyes, brain and vehicle

output : heading of the automobile.

Classification of Control Systems

Control systems are classified into two general categories based upon the control action which is

responsible to activate the system to produce the output viz.

1) Open loop control system in which the control action is independent of the out put.

2) Closed loop control system in which the control action is some how dependent upon the

output and are generally called as feedback control systems.

Open Loop System is a system in which control action is independent of output. To each

reference input there is a corresponding output which depends upon the system and its operating

conditions. The accuracy of the system depends on the calibration of the system. In the presence

of noise or disturbances open loop control will not perform satisfactorily.



Speed of the

Prime mover

Induced Voltage

Output

Inputs

Time Cleanliness of clothes

EXAMPLE - 1 Rotational Generator

The input to rotational generator is the speed of the prime mover ( e.g steam turbine) in r.p.m.

Assuming the generator is on no load the output may be induced voltage at the output terminals.

Rotational Generator

Fig 1-2 Rotational Generator

EXAMPLE – 2 Washing machine

Most ( but not all ) washing machines are operated in the following manner. After the clothes to

be washed have been put into the machine, the soap or detergent, bleach and water are entered in

proper amounts as specified by the manufacturer. The washing time is then set on a timer and the

washer is energized. When the cycle is completed, the machine shuts itself off. In this example

washing time forms input and cleanliness of the clothes is identified as output.

Washing Machine

Fig 1-3 Washing Machine

EXAMPLE – 3 WATER TANK LEVEL CONTROL

To understand the concept further it is useful to consider an example let it be desired to maintain

the actual water level 'c ' in the tank as close as possible to a desired level ' r '. The desired level

will be called the system input, and the actual level the controlled variable or system output.

Water flows from the tank via a valve Vo , and enters the tank from a supply via a control valve

Vc. The control valve is adjustable manually.

Fig 1-4 b) Open loop control

Fig –1.4 a) Water level control

Actuating signal

nput

output input

Controller

System

WATER

TANK

Desired Water

level r

Actual

Water level c

Valve VC

Valve VO

Water in

Water

out

C

Actuating signal

nput

output input

Controller

System



A closed loop control system is one in which the control action depends on the output. In

closed loop control system the actuating error signal, which is the difference between the input

signal and the feed back signal (out put signal or its function) is fed to the controller.

Fig –1.5: Closed loop control system

EXAMPLE – 1 – THERMAL SYSTEM

To illustrate the concept of closed loop control system, consider the thermal system shown in fig-

6 Here human being acts as a controller. He wants to maintain the temperature of the hot water at

a given value ro C. the thermometer installed in the hot water outlet measures the actual

temperature C0

C. This temperature is the output of the system. If the operator watches the

thermometer and finds that the temperature is higher than the desired value, then he reduce the

amount of steam supply in order to lower the temperature. It is quite possible that that if the

temperature becomes lower than the desired value it becomes necessary to increase the amount

of steam supply. This control action is based on closed loop operation which involves human

being, hand muscle, eyes, thermometer such a system may be called manual feed back system.

Fig 1-6 a) Manual feedback thermal system b) Block diagram

EXAMPLE –2 HOME HEATING SYSTEM

The thermostatic temperature control in hour homes and public buildings is a familiar example.

An electronic thermostat or temperature sensor is placed in a central location usually on inside

Control

elements

System /

Plant

Feed back elements

controller

Forward path

Reference

input

Actuating

/ error

signal

Error

detector

Feed back signal

Controlled

output

Desired hot

water. temp

ro c

Brain of

operator (r-c)

Muscles

and Valve

Actual

Water temp

Co

C

+ C

Thermometer

+

Human operator

Thermometer

Hot water

Drain

Cold water

Steam

Steam



wall about 5 feet from the floor. A person selects and adjusts the desired room temperature ( r )

say 250

C and adjusts the temperature setting on the thermostat. A bimetallic coil in the

thermostat is affected by the actual room temperature ( c ). If the room temperature is lower than

the desired temperature the coil strip alters the shape and causes a mercury switch to operate a

relay, which in turn activates the furnace fire when the temperature in the furnace air duct system

reaches reference level ' r ' a blower fan is activated by another relay to force the warm air

throughout the building. When the room temperature ' C ' reaches the desired temperature ' r '

the shape of the coil strip in the thermostat alters so that Mercury switch opens. This deactivates

the relay and in turn turns off furnace fire, which in turn the blower.

Fig 1-7 Block diagram of Home Heating system.

A change in out door temperature is a disturbance to the home heating system. If the out side

temperature falls, the room temperature will likewise tend to decrease.

CLOSED- LOOP VERSUS OPEN LOOP CONTROL SYSTEMS

An advantage of the closed loop control system is the fact that the use of feedback makes the

system response relatively insensitive to external disturbances and internal variations in systems

parameters. It is thus possible to use relatively inaccurate and inexpensive components to obtain

the accurate control of the given plant, whereas doing so is impossible in the open-loop case.

From the point of view of stability, the open loop control system is easier to build

because system stability is not a major problem. On the other hand, stability is a major problem

in the closed loop control system, which may tend to overcorrect errors that can cause

oscillations of constant or changing amplitude.

It should be emphasized that for systems in which the inputs are known ahead of time and in

which there are no disturbances it is advisable to use open-loop control. closed loop control

systems have advantages only when unpredictable disturbances it is advisable to use open-loop

control. Closed loop control systems have advantages only when unpredictable disturbances and

/ or unpredictable variations in system components used in a closed –loop control system is more

than that for a corresponding open – loop control system. Thus the closed loop control system is

generally higher in cost.

Desired temp. ro c

Relay

switch

Actual

Temp.

Co C

Furnace Blower House +

Outdoor temp change

(disturbance)



Definitions:

Systems: A system is a combination of components that act together and perform a certain

objective. The system may be physical, biological, economical, etc.

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

to command, direct or regulate itself or another system.

Open loop: An open loop system control system is one in which the control action is

independent of the output.

Closed loop: A closed loop control system is one in which the control action is somehow

dependent on the output.

Plants: A plant is equipment the purpose of which is to perform a particular operation. Any

physical object to be controlled is called a plant.

Processes: Processes is a natural or artificial or voluntary operation that consists of a series of

controlled actions, directed towards a result.

Input: The input is the excitation applied to a control system from an external energy source.

The inputs are also known as actuating signals.

Output: The output is the response obtained from a control system or known as controlled

variable.

Block diagram: A block diagram is a short hand, pictorial representation of cause and effect

relationship between the input and the output of a physical system. It characterizes the functional

relationship amongst the components of a control system.

Control elements: These are also called controller which are the components required to

generate the appropriate control signal applied to the plant.

Plant: Plant is the control system body process or machine of which a particular quantity or

condition is to be controlled.

Feedback control: feedback control is an operation in which the difference between the output

of the system and the reference input by comparing these using the difference as a means of

control.

Feedback elements: These are the components required to establish the functional relationship

between primary feedback signal and the controlled output.

Actuating signal: also called the error or control action. It is the algebraic sum consisting of

reference input and primary feedback.

Manipulated variable: it that quantity or condition which the control elements apply to the

controlled system.

Feedback signal: it is a signal which is function of controlled output

Disturbance: It is an undesired input signal which affects the output.

Forward path: It is a transmission path from the actuating signal to controlled output

Feedback path: The feed back path is the transmission path from the controlled output to the

primary feedback signal.

Servomechanism: Servomechanism is a feedback control system in which output is some

mechanical position, velocity or acceleration.

Regulator: Regulator is a feedback system in which the input is constant for long time.

Transducer: Transducer is a device which converts one energy form into other

Tachometer: Tachometer is a device whose output is directly proportional to time rate of change

of input.

Synchros: Synchros is an AC machine used for transmission of angular position synchro motor-

receiver, synchro generator- transmitter.



Block diagram: A block diagram is a short hand, pictorial representation of cause and effect

relationship between the input and the output of a physical system. It characterizes the functional

relationship amongst the components of a control system.

Summing point: It represents an operation of addition and / or subtraction.

Negative feedback: Summing point is a subtractor.

Positive feedback: Summing point is an adder.

Stimulus: It is an externally introduced input signal affecting the controlled output.

Take off point: In order to employ the same signal or variable as an input to more than block or

summing point, take off point is used. This permits the signal to proceed unaltered along several

different paths to several destinations.

Time response: It is the output of a system as a function of time following the application of a

prescribed input under specified operating conditions.



DIFFERENTIAL EQUATIONS OF PHYSICAL SYSTEMS

The term mechanical translation is used to describe motion with a single degree of freedom or

motion in a straight line. The basis for all translational motion analysis is Newton‘s second law

of motion which states that the Net force F acting on a body is related to its mass M and

acceleration ‗a‘ by the equation F = Ma

‗Ma‘ is called reactive force and it acts in a direction opposite to that of acceleration. The

summation of the forces must of course be algebraic and thus considerable care must be taken in

writing the equation so that proper signs prefix the forces.

The three basic elements used in linear mechanical translational systems are ( i ) Masses (ii)

springs iii) dashpot or viscous friction units. The graphical and symbolic notations for all three

are shown in fig 1-8

M

Fig 1-8 a) Mass Fig 1-8 b) Spring Fig 1-8 c) Dashpot

The spring provides a restoring a force when a force F is applied to deform a coiled spring a

reaction force is produced, which to bring it back to its freelength. As long as deformation is

small, the spring behaves as a linear element. The reaction force is equal to the product of the

stiffness k and the amount of deformation.

Whenever there is motion or tendency of motion between two elements, frictional forces exist.

The frictional forces encountered in physical systems are usually of nonlinear nature. The

characteristics of the frictional forces between two contacting surfaces often depend on the

composition of the surfaces. The pressure between surfaces, their relative velocity and others.

The friction encountered in physical systems may be of many types

( coulomb friction, static friction, viscous friction ) but in control problems viscous friction,

predominates. Viscous friction represents a retarding force i.e. it acts in a direction opposite to

the velocity and it is linear relationship between applied force and velocity. The mathematical

expression of viscous friction F=BV where B is viscous frictional co-efficient. It should be

realized that friction is not always undesirable in physical systems. Sometimes it may be

necessary to introduce friction intentionally to improve dynamic response of the system. Friction

may be introduced intentionally in a system by use of dashpot as shown in fig 1-9. In

automobiles shock absorber is nothing but dashpot.



The basic operation of a dashpot, in which the housing is filled with oil. If a force f is applied to

the shaft, the piston presses against oil increasing the pressure on side ‗b‘ and decreasing

pressure side ‗a‘ As a result the oil flows from side ‗b‘ to side ‗a‘ through the wall clearance. The

friction coefficient B depends on the dimensions and the type of oil used.

Outline of the procedure

For writing differential equations

1. Assume that the system originally is in equilibrium in this way the often-troublesome

effect of gravity is eliminated.

2. Assume then that the system is given some arbitrary displacement if no distributing force

is present.

3. Draw a freebody diagram of the forces exerted on each mass in the system. There should

be a separate diagram for each mass.

4. Apply Newton‘s law of motion to each diagram using the convention that any force

acting in the direction of the assumed displacement is positive is positive.

5. Rearrange the equation in suitable form to solve by any convenient mathematical means.

Lever

Lever is a device which consists of rigid bar which tends to rotate about a fixed point

called ‗fulcrum‘ the two arms are called ―effort arm‖ and ―Load arm‖ respectively. The lever

bears analogy with transformer

a b

Applied force F

Piston

L1 L2 F2

F1 effort

Load

Fulcrum



It is also called ‗mechanical transformer‘

Equating the moments of the force

F1 L1 = F2 L 2

F 2 = F1 L1

L2

Rotational mechanical system

The rotational motion of a body may be defined as motion about a fixed axis. The variables

generally used to describe the motion of rotation are torque, angular displacement , angular

velocity ( ) and angular acceleration( )

The three basic rotational mechanical components are 1) Moment of inertia J

2 ) Torsional spring 3) Viscous friction.

Moment of inertia J is considered as an indication of the property of an element, which stores the

kinetic energy of rotational motion. The moment of inertia of a given element depends on

geometric composition about the axis of rotation and its density. When a body is rotating a

reactive torque is produced which is equal to the product of its moment

of inertia (J) and angular acceleration and is given by T= J = J d2

d t2

A well known example of a torsional spring is a shaft which gets twisted when a torque is

applied to it. Ts = K , is angle of twist and K is torsional stiffness.

There is viscous friction whenever a body rotates in viscous contact with another body. This

torque acts in opposite direction so that angular velocity is given by

T = f = f d2

Where = relative angular velocity between two bodies.

d t2

f = co efficient of viscous friction.

Newton‘s II law of motion states

T = J d2

.

d t2

Gear wheel

In almost every control system which involves rotational motion gears are necessary. It is often

necessary to match the motor to the load it is driving. A motor which usually runs at high speed

and low torque output may be required to drive a load at low speed and high torque.



Analogous Systems

Consider the mechanical system shown in fig A and the electrical system shown in fig B

The differential equation for mechanical system is

M + + B + K X = f (t) ---------- 1

The differential equation for electrical system is

L + + R + = e ---------- 2

Comparing equations (1) and (2) we see that for the two systems the differential equations are of

identical form such systems are called ― analogous systems and the terms which occupy the

corresponding positions in differential equations are analogous quantities‖

The analogy is here is called force voltage analogy

Table for conversion for force voltage analogy

Mechanical System Electrical System

Force (torque) Voltage

Mass (Moment of inertia) Inductance

Viscous friction coefficient Resistance

Spring constant Capacitance

d2x

dt2

dx

dt

N1

N2

Driving wheel

Driven wheel

d2q

dt2

d2q

dt2

q

c



Displacement Charge

Velocity Current.

Force – Current Analogy

Another useful analogy between electrical systems and mechanical systems is based on force –

current analogy. Consider electrical and mechanical systems shown in fig.

For mechanical system the differential equation is given by

M + + B + K X = f (t) ---------- 1

For electrical system

C

+ + + = I ( t )

Comparing equations (1) and (2) we find that the two systems are analogous systems. The

analogy here is called force – current analogy. The analogous quantities are listed.

Table of conversion for force – current analogy

Mechanical System Electrical System

Force( torque) Current

Mass( Moment of inertia) Capacitance

Viscous friction coefficient Conductance

Spring constant Inductance

Displacement Flux

( angular)

Velocity (angular) Voltage

Illustration 1:For a two DOF spring mass damper system obtain the mathematical model where

F is the input x1 and x2 are responses.

d2x

dt2

dx

dt

d2x

dt2

1

R

d

dt2

L



Figure 1.10 (a)

Figure 1.10 (b)

From NSL F= ma

For mass m1

m1x1 = F - b1 (x1-x2) - k1 (x1-x2) --- (a)

For mass m2

m2 x2 = b1 (x2-x1) + k1 (x2-x1) - b2 x2 - k2 x2 --- (b)

m2

m1

F

k2 x2 b2 x2

k1 x2 k1 x1 b1 x1 b1 x2 x2

k1 x2 k1 x1 b1 x1 b1 x2

.

. . .

m2

m1

F

k2 x2 b2 x2

k1 (x1-x2)

.

. b1 (x1-x2) .

x1

.

m2

m1

k2 b2 (Damper)

x2 (Response)

x1 (Response)

k1

F

b1

.. . .

.. . . .

Draw the free body diagram for mass

m1 and m2 separately as shown in figure

1.10 (b)

Apply NSL for both the masses

separately and get equations as given in

(a) and (b)



Illustration 2: For the system shown in figure 2.16 (a) obtain the mathematical model if x1 and

x2 are initial displacements.

Let an initial displacement x1 be given to mass m1 and x2 to mass m2.

Figure 1.11 (a)

K1

K2

K3

m2

m1

X1

X2



Figure 2.16 (b)

Based on Newton‘s second law of motion: F = ma

For mass m1

m1x1 = - K1x1 + K2 (x2-x1)

m1x1 + K1x1 – K2 x2 + K2x1 = 0

m1x1 + x1 (K1 + K2) = K2x2 ----- (1)

For mass m2

m2x2 = - K3x2 – K2 (x2 – x1)

m2x2 + K3 x2 + K2 x2 – K2 x1

m2 x2 + x2 (K2 + K3) = K2x1 ----- (2)

Mathematical models are:

K1 X1

K2 X2

K2 X2

K2 X1

K2 X1

K3 X2 K3 X2

K1 X1

X1 X1

m1 m1

m2

m2 X2

X2

K2 (X2 – X1)

K2 (X2 – X1)

..

..

..

..

..

..



m1x1 + x1 (K1 + K2) = K2x2 ----- (1)

m2x2 + x2 (K2 + K3) = K2x1 ----- (2)

1.Write the differential equation relating to motion X of the mass M to the force input u(t)

X

(output)

U(t)

(input)

2. Write the force equation for the mechanical system shown in figure

X (output)

X1

F(t)

(input)

3. Write the differential equations for the mechanical system shown in figure.

4. Write the modeling equations for the mechanical systems shown in figure.

M

K1 K2

M

K B2

B1

f12 K1

M1

f1 f(t)

M2

X2

f2

X1

..

..



5. For the systems shown in figure write the differential equations and obtain the transfer

functions indicated.

6. Write the differential equation describing the system. Assume the bar through which

force is applied is not flexible, has no mass or moment of inertia, and all

displacements are small.

7. Write the equations of motion in terms of given mechanical quantities.

M

Xi

Xo

M

B

force f(t)

X K

Xi Xo Xi Xo

Yk

C

K F

b

M

B

X K

f(t)

a

b a

X2

K1 Force f



8. Write the force equations for the mechanical systems shown in figure.

T(t)

9. Write the force equation for the mechanical system shown in figure.

10. Write the force equation for the mechanical system shown in figure.

11. Torque T(t) is applied to a small cylinder with moment of inertia J1 which rotates with in a

larger cylinder with moment of inertia J2. The two cylinders are coupled by viscous friction B1.

M1

B1

B1

J1

K J1 J2

1 2

T(t)

1 2

J1 J2 J3

K1 K2 K3

Torque T

B1 B2 B3

3



The outer cylinder has viscous friction B2 between it and the reference frame and is restrained by

a torsion spring k. write the describing differential equations.

12. The polarized relay shown exerts a force f(t) = Ki. i(t) upon the pivoted bar. Assume the relay

coil has constant inductance L. The left end of the pivot bar is connected to the reference frame

through a viscous damper B1 to retard rapid motion of the bar. Assume the bar has negligible

mass and moment of inertia and also that all displacements are small. Write the describing

differential equations. Note that the relay coil is not free to move.

13. Figure shows a control scheme for controlling the azimuth angle of an armature controlled

dc. Motion with dc generator used as an amplifier. Determine transfer function

L (s)

K

B2 J1

Torque T1, 1 B1

J2



. The parameters of the plant are given below.

u (s)

Motor torque constant = KT in N.M /amp

Motor back emf constant = KB in V/ rad / Sec

Generator gain constant = KG in v/ amp

Motor to load gear ratio = N2

N 1

Resistance of the circuit = R in ohms.

Inductance of the circuit = L in Henry

Moment of inertia of motor = J

Viscous friction coefficient = B

Field resistance = Rf

Field inductance = Lf

14. The schematic diagram of a dc motor control system is shown in figure where Ks is error

detector gain in volt/rad, k is the amplifier gain, Kb back emf constant, Kt is torque

constant, n is the gear train ratio = 2 = Tm Bm = motion friction constant

1 T2



Jm = motor inertia, KL = Torsional spring constant JL = load inertia.

15. Obtain a transfer function C(s) /R(s) for the positional servomechanism shown in figure.

Assume that the input to the system is the reference shaft position (R) and the system output is

the output shaft position ( C ). Assume the following constants.

Gain of the potentiometer (error detector ) K1 in V/rad

Amplifier gain ‗ Kp ‘ in V / V

Motor torque constant ‗ KT ‘ in V/ rad

Gear ratio N1 N2

Moment of inertia of load ‗J‘

Viscous friction coefficient ‗f‘



16. Find the transfer function E0 (s) / I(s)

C1

I E0

C2 R Output

input



Recommended Questions :

1. Name three applications of control systems.

2. Name three reasons for using feedback control systems and at least one reason for not

using them.

3. Give three examples of open- loop systems.

4. Functionally, how do closed – loop systems differ from open loop systems.

5. State one condition under which the error signal of a feedback control system would not

be the difference between the input and output.

6. Name two advantages of having a computer in the loop.

7. Name the three major design criteria for control systems.

8. Name the two parts of a system‘s response.

9. Physically, what happens to a system that is unstable?

10. Instability is attributable to what part of the total response.

11. What mathematical model permits easy interconnection of physical systems?

12. To what classification of systems can the transfer function be best applied?

13. What transformation turns the solution of differential equations into algebraic

manipulations ?

14. Define the transfer function.

15. What assumption is made concerning initial conditions when dealing with transfer

functions?

16. What do we call the mechanical equations written in order to evaluate the transfer

function ?

17. Why do transfer functions for mechanical networks look identical to transfer functions

for electrical networks?

18. What function do gears and levers perform.

19. What are the component parts of the mechanical constants of a motor‘s transfer function?



UNIT-2

Block Diagram:

A control system may consist of a number of components. In order to show the functions

performed by each component in control engineering, we commonly use a diagram called the

―Block Diagram‖.

A block diagram of a system is a pictorial representation of the function performed by

each component and of the flow of signals. Such a diagram depicts the inter-relationships which

exists between the various components. A block diagram has the advantage of indicating more

realistically the signal flows of the actual system.

In a block diagram all system variables are linked to each other through functional

blocks. The ―Functional Block‖ or simply ―Block‖ is a symbol for the mathematical operation on

the input signal to the block which produces the output. The transfer functions of the components

are usually entered in the corresponding blocks, which are connected by arrows to indicate the

direction of flow of signals. Note that signal can pass only in the direction of arrows. Thus a

block diagram of a control system explicitly shows a unilateral property.

Fig 2.1 shows an element of the block diagram. The arrow head pointing towards the block

indicates the input and the arrow head away from the block represents the output. Such arrows

are entered as signals.

X(s)

Y(s)

Fig 2.1

The advantages of the block diagram representation of a system lie in the fact that it is

easy to form the over all block diagram for the entire system by merely connecting the blocks of

the components according to the signal flow and thus it is possible to evaluate the contribution of

each component to the overall performance of the system. A block diagram contains information

concerning dynamic behavior but does not contain any information concerning the physical

construction of the system. Thus many dissimilar and unrelated system can be represented by the

same block diagram.

G(s

)



It should be noted that in a block diagram the main source of energy is not explicitly

shown and also that a block diagram of a given system is not unique. A number of a different

block diagram may be drawn for a system depending upon the view point of analysis.

Error detector : The error detector produces a signal which is the difference between the

reference input and the feed back signal of the control system. Choice of the error detector is

quite important and must be carefully decided. This is because any imperfections in the error

detector will affect the performance of the entire system. The block diagram representation of

the error detector is shown in fig2.2

R(s) C(s)

C(s)

Fig2.2

Note that a circle with a cross is the symbol which indicates a summing operation. The plus or

minus sign at each arrow head indicates whether the signal is to be added or subtracted. Note

that the quantities to be added or subtracted should have the same dimensions and the same units.

Block diagram of a closed loop system .

Fig2.3 shows an example of a block diagram of a closed system

Summing point

Branch point

R(s) C(s)

Fig. 2.3

The output C(s) is fed back to the summing point, where it is compared with reference input

R(s). The closed loop nature is indicated in fig1.3. Any linear system may be represented by a

block diagram consisting of blocks, summing points and branch points. A branch is the point

from which the output signal from a block diagram goes concurrently to other blocks or

summing points.

G(s)

+

-

+ -



When the output is fed back to the summing point for comparison with the input, it is

necessary to convert the form of output signal to that of he input signal. This conversion is

followed by the feed back element whose transfer function is H(s) as shown in fig 1.4. Another

important role of the feed back element is to modify the output before it is compared with the

input.

B(s)

R(s) C(s) C(s)

B(s)

Fig 2.4

The ratio of the feed back signal B(s) to the actuating error signal E(s) is called the open

loop transfer function.

open loop transfer function = B(s)/E(s) = G(s)H(s)

The ratio of the output C(s) to the actuating error signal E(s) is called the feed forward

transfer function .

Feed forward transfer function = C(s)/E(s) = G(s)

If the feed back transfer function is unity, then the open loop and feed forward transfer

function are the same. For the system shown in Fig1.4, the output C(s) and input R(s) are related

as follows.

C(s) = G(s) E(s)

E(s) = R(s) - B(s)

= R(s) - H(s)C(s) but B(s) = H(s)C(s)

Eliminating E(s) from these equations

C(s) = G(s)[R(s) - H(s)C(s)]

C(s) + G(s)[H(s)C(s)] = G(s)R(s)

C(s)[1 + G(s)H(s)] = G(s)R(s)

G(s

)

H(s

)

+ -



C(s) G(s)

=

R(s) 1 + G(s)H(s)

C(s)/R(s) is called the closed loop transfer function.

The output of the closed loop system clearly depends on both the closed loop transfer

function and the nature of the input. If the feed back signal is positive, then

C(s) G(s)

=

R(s) 1 - G(s)H(s)

Closed loop system subjected to a disturbance

Fig2.5 shows a closed loop system subjected to a disturbance. When two inputs are present in

a linear system, each input can be treated independently of the other and the outputs

corresponding to each input alone can be added to give the complete output. The way in

which each input is introduced into the system is shown at the summing point by either a plus

or minus sign.

Disturbance

N(s)

R(s)

C(s)

Fig2.5

Fig2.5 closed loop system subjected to a disturbance.

Consider the system shown in fig 2.5. We assume that the system is at rest initially with

zero error. Calculate the response CN(s) to the disturbance only. Response is

CN(s) G2(s)

=

R(s) 1 + G1(s)G2(s)H(s)

On the other hand, in considering the response to the reference input R(s), we may

assume that the disturbance is zero. Then the response CR(s) to the reference input R(s)is

G1(s) G2(s)

H(s

)

+ -

+ +



CR(s) G1(s)G2(s)

=

R(s) 1 + G1(s)G2(s)H(s).

The response C(s) due to the simultaneous application of the reference input R(s) and the

disturbance N(s) is given by

C(s) = CR(s) + CN(s)

G2(s)

C(s) = [G1(s)R(s) + N(s)]

1 + G1(s)G2(s)H(s)

Procedure for drawing block diagram :

To draw the block diagram for a system, first write the equation which describes the dynamic

behaviour of each components. Take the laplace transform of these equations, assuming zero

initial conditions and represent each laplace transformed equation individually in the form of

block. Finally assemble the elements into a complete block diagram.

As an example consider the Rc circuit shown in fig2.6 (a). The equations for the circuit

shown are

R

C

eoei i

Fig. 2.6a

ei = iR + 1/c∫ idt -----------(1)

And

eo = 1/c∫ idt ---------(2)

Equation (1) becomes

ei = iR + eo

ei - eo

= i --------------(3)

Laplace transforms of equations (2) & (3) are

R



Eo(s) = 1/CsI(s) -----------(4)

Ei(s) - Eo(s)

= I(s) -------- (5)

R

Equation (5) represents a summing operation and the corresponding diagram is shown in fig1.6

(b). Equation (4) represents the block as shown in fig2.6(c). Assembling these two elements, the

overall block diagram for the system shown in fig2.6(d) is obtained.

I(s) Eo(S)

Ei(s) + I(s)

_ Fig2.6(c)

Eo(s)

Eo(s) + I(s) Eo(s)

Fig2.6(b) _

Fig2.6(d)

SIGNAL FLOW GRAPHS

An alternate to block diagram is the signal flow graph due to S. J. Mason. A signal flow graph is

a diagram that represents a set of simultaneous linear algebraic equations. Each signal flow graph

consists of a network in which nodes are connected by directed branches. Each node represents a

system variable, and each branch acts as a signal multiplier. The signal flows in the direction

indicated by the arrow.

Definitions:

Node: A node is a point representing a variable or signal.

Branch: A branch is a directed line segment joining two nodes.

Transmittance: It is the gain between two nodes.

Input node: A node that has only outgoing branche(s). It is also, called as source and

corresponds to independent variable.

Output node: A node that has only incoming branches. This is also called as sink and

corresponds to dependent variable.

1/R 1/C

S

1/R 1/C

S



Mixed node: A node that has incoming and out going branches.

Path: A path is a traversal of connected branches in the direction of branch arrow.

Loop: A loop is a closed path.

Self loop: It is a feedback loop consisting of single branch.

Loop gain: The loop gain is the product of branch transmittances of the loop.

Nontouching loops: Loops that do not posses a common node.

Forward path: A path from source to sink without traversing an node more than once.

Feedback path: A path which originates and terminates at the same node.

Forward path gain: Product of branch transmittances of a forward path.

Properties of Signal Flow Graphs:

1) Signal flow applies only to linear systems.

2) The equations based on which a signal flow graph is drawn must be algebraic equations

in the form of effects as a function of causes.

Nodes are used to represent variables. Normally the nodes are arranged left to right,

following a succession of causes and effects through the system.

3) Signals travel along the branches only in the direction described by the arrows of the

branches.

4) The branch directing from node Xk to Xj represents dependence of the variable Xj on Xk

but not the reverse.

5) The signal traveling along the branch Xk and Xj is multiplied by branch gain akj and

signal akjXk is delivered at node Xj.

Guidelines to Construct the Signal Flow Graphs:

The signal flow graph of a system is constructed from its describing equations, or by direct

reference to block diagram of the system. Each variable of the block diagram becomes a node

and each block becomes a branch. The general procedure is

1) Arrange the input to output nodes from left to right.

2) Connect the nodes by appropriate branches.

3) If the desired output node has outgoing branches, add a dummy node and a unity gain

branch.

4) Rearrange the nodes and/or loops in the graph to achieve pictorial clarity.

Signal Flow Graph Algebra

Addtion rule



The value of the variable designated by a node is equal to the sum of all signals entering the

node.

Transmission rule

The value of the variable designated by a node is transmitted on every branch leaving the node.

Multiplication rule

A cascaded connection of n-1 branches with transmission functions can be replaced by a single

branch with new transmission function equal to the product of the old ones.

Masons Gain Formula

The relationship between an input variable and an output variable of a signal flow graph is given

by the net gain between input and output nodes and is known as overall gain of the system.

Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs.

Gain P is given by

kkkPP

1

Where, Pk is gain of kth forward path,

∆ is determinant of graph

∆=1-(sum of all individual loop gains)+(sum of gain products of all possible combinations of

two nontouching loops – sum of gain products of all possible combination of three

nontouching loops) + ∙∙∙

∆k is cofactor of kth

forward path determinant of graph with loops touching kth forward path. It is

obtained from ∆ by removing the loops touching the path Pk.

Example1

Draw the signal flow graph of the block diagram shown in Fig.2.7

Figure 2.7 Multiple loop system

Choose the nodes to represent the variables say X1 .. X6 as shown in the block diagram..

Connect the nodes with appropriate gain along the branch. The signal flow graph is shown in

Fig. 2.7

G2 G1 G3

H2

H1

−

−

R X1 X2 X3 X4 X5 X6 C



Figure 1.8 Signal flow graph of the system shown in Fig. 2.7

Example 2.9

Draw the signal flow graph of the block diagram shown in Fig.2.9.

Figure 2.9 Block diagram feedback system

The nodal variables are X1, X2, X3.

The signal flow graph is shown in Fig. 2.10.

R X1 X2 X3

X4 X5 X6

C G1

H1

-H2

G2 G3

-1

1 1 1 1

G1

G2

G3

G4

R

−

C

−

X1 X2

X3



Figure 2.10 Signal flow graph of example 2

Example 3

Draw the signal flow graph of the system of equations.

3332321313

223232221212

113132121111

XaXaXaX

ubXaXaXaX

ubXaXaXaX

The variables are X1, X2, X3, u1 and u2 choose five nodes representing the variables.

Connect the various nodes choosing appropriate branch gain in accordance with the equations.

The signal flow graph is shown in Fig. 2.11.


R

G4

G2

-G3

G1

C 1 1

X1

X2 X3

u1 b1

b2

a21

a12

a33

a31

X1

X2 a11

a13

a32

X3

u2

a23

a22



Example 4

LRC net work is shown in Fig. 2.12. Draw its signal flow graph.

Figure 2.12 LRC network

The governing differential equations are

3

2

11

tidt

deC

teeRidt

diL

or

teidtC

Ridt

diL

c

c

Taking Laplace transform of Eqn.1 and Eqn.2 and dividing Eqn.2 by L and Eqn.3 by C

51

0

411

0

sIC

essE

sEL

sEL

SIL

RissI

cc

c

Eqn.4 and Eqn.5 are used to draw the signal flow graph shown in Fig.7.

Figure 2.12 Signal flow graph of LRC system

Cs

1L

RsL

1

LRsL

1-

L R s

1

s

1

i(0+)

Ec(s)

I(s)

E(s)

ec(0+)

R L

C i(t)

−

ec(t)

−

e(t)



SIGNAL FLOW GRAPHS

The relationship between an input variable and an output variable of a signal flow graph is given

by the net gain between input and output nodes and is known as overall gain of the system.

Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs.

Masons Gain Formula

Gain P is given by

kkkPP

1

Where, Pk is gain of kth forward path,

∆ is determinant of graph




∆k is cofactor of kth

forward path determinant of graph with loops touching kth forward path. It is

obtained from ∆ by removing the loops touching the path Pk.

Example 1

Obtain the transfer function of C/R of the system whose signal flow graph is shown in Fig.2.13


There are two forward paths:

Gain of path 1 : P1=G1

Gain of path 2 : P2=G2

There are four loops with loop gains:

L1=-G1G3, L2=G1G4, L3= -G2G3, L4= G2G4

R

G4

G2

-G3

G1

C 1 1



There are no non-touching loops.

∆ = 1+G1G3-G1G4+G2G3-G2G4

Forward paths 1 and 2 touch all the loops. Therefore, ∆1= 1, ∆2= 1

The transfer function T = 42324131

212211

1 GGGGGGGG

GGPP

sR

sC

Example 2

Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in

Fig.2.14.


There is one forward path, whose gain is: P1=G1G2G3

There are three loops with loop gains:

L1=-G1G2H1, L2=G2G3H2, L3= -G1G2G3

There are no non-touching loops.

∆ = 1-G1G2H1+G2G3H2+G1G2G3

Forward path 1 touches all the loops. Therefore, ∆1= 1.


32111

1 GGGHGGHGG

GGGP

sR

sC

Example 3

Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in

Fig.2.15.

R(s) C(s) 1 1 1 G1 G2 G3

H1

-1

-H2




There are three forward paths.

The gain of the forward path are: P1=G1G2G3G4G5

P2=G1G6G4G5

P3= G1G2G7


L1=-G4H1, L2=-G2G7H2, L3= -G6G4G5H2 , L4=-G2G3G4G5H2

There is one combination of Loops L1 and L2 which are nontouching with loop gain product

L1L2=G2G7H2G4H1

∆ = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1

Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.

Forward path 3 is not in touch with loop1. Hence, ∆3= 1+G4H1.

The transfer function T =

2174225432254627214

14721654154321332211

1

1

HHGGGHGGGGHGGGHGGHG

HGGGGGGGGGGGGGPPP

sR

sC

Example 4

G1 C(s) R(s)

G7 G6

-H1

G2 G3 G4 G5

-H2

X1 X2 X3 X4 X5

1



Find the gains 1

3

2

5

1

6 ,,X

X

X

X

X

X for the signal flow graph shown in Fig.2.16.

Figure 2.16 Signal flow graph of MIMO system

Case 1: 1

6

X

X

There are two forward paths.

The gain of the forward path are: P1=acdef

P2=abef


L1=-cg, L2=-eh, L3= -cdei, L4=-bei


L1L2=cgeh

∆ = 1+cg+eh+cdei+bei+cgeh

Forward path 1 and 2 touch all the four loops. Therefore ∆1= 1, ∆2= 1.

The transfer function T = cgehbeicdeiehcg

abefcdefPP

X

X

1

2211

1

6

Case 2: 2

5

X

X

The modified signal flow graph for case 2 is shown in Fig.2.17.

b

a d c f e

-h

-g

-i

X1 X6 X5

X4 X3 X2



Figure 2.17 Signal flow graph of example 4 case 2

The transfer function can directly manipulated from case 1 as branches a and f are removed

which do not form the loops. Hence,

The transfer function T=cgehbeicdeiehcg

becdePP

X

X

1

2211

2

5

Case 3: 1

3

X

X

The signal flow graph is redrawn to obtain the clarity of the functional relation as shown in

Fig.2.18.

Figure 2.18 Signal flow graph of example 4 case 3

There are two forward paths.

The gain of the forward path are: P1=abcd

P2=ac

b

1 d c 1 e

-h

-g

-i

X2 X5 X5

X4 X3 X2

a e b f

-h

-g

-i

X1 X5

X4 X3

X2

c

d

X3 1



There are five loops with loop gains:

L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg


L1L2=ehcg

∆ = 1+eh+cg+bei+efd+befg+ehcg

Forward path 1 touches all the five loops. Therefore ∆1= 1.

Forward path 2 does not touch loop L1. Hence, ∆2= 1+ eh

The transfer function T = ehcgbefgefdbeicgeh

ehacabefPP

X

X

1

12211

1

3

Example 5

For the system represented by the following equations find the transfer function X(s)/U(s) using

signal flow graph technique.

uXaX

uXXaX

uXX

1122

22111

31

Taking Laplace transform with zero initial conditions

sUsXassX

sUsXsXassX

sUsXsX

1122

22111

31

Rearrange the above equation

sUs

sXs

asX

sUs

sXs

sXs

asX

sUsXsX

11

22

221

11

31

1

The signal flow graph is shown in Fig.2.19.



Figure 2.19 Signal flow grapgh of example 5

There are three forward paths.

The gain of the forward path are: P1= 3

P2= 1/ s2

P3= 2/ s

There are two loops with loop gains:

2

22

11

s

aL

s

aL

L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg

There are no combination two Loops which are nontouching.

2

211s

a

s

a

Forward path 1 does not touch loops L1 and L2. Therefore

2

211 1

s

a

s

a

Forward path 2 path 3 touch the two loops. Hence, ∆2= 1, ∆2= 1.


2

1221

2

3332211

1

3

asas

sasasPPP

X

X

U

X1

X2

3

s

a1

s

1

s

β2

s

a2

s

β1

1

X X



Recommended Questions:

1. Define block diagram & depict the block diagram of closed loop system.

2. Write the procedure to draw the block diagram.

3. Define signal flow graph and its parameters

4. Explain briefly Mason‘s Gain formula

5. Draw the signal flow graph of the block diagram shown in Fig below.

6. Draw the signal flow graph of the block diagram shown in Fig below

7. For the LRC net work is shown in Fig Draw its signal flow graph.

G2 G1 G3

H2

H1

−

−

R X1 X2 X3 X4 X5 X6 C

G1

G2

G3

G4

R

−

C

−

X1 X2

X3



Figur

8. Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in

Fig.

Q.9 For the system represented by the following equations find the transfer function X(s)/U(s)

using signal flow graph technique.

R L

C i(t)

−

ec(t)

−

e(t)

G1 C(s) R(s)

G7 G6

-H1

G2 G3 G4 G5

-H2

X

1

X

2

X3 X4 X5

1

uXaX

uXXaX

uXX

1122

22111

31



UNIT- 3

Time response analysis of control systems:

Introduction:

Time is used as an independent variable in most of the control systems. It is important to

analyse the response given by the system for the applied excitation, which is function of time.

Analysis of response means to see the variation of out put with respect to time. The output

behavior with respect to time should be within these specified limits to have satisfactory

performance of the systems. The stability analysis lies in the time response analysis that is when

the system is stable out put is finite

The system stability, system accuracy and complete evaluation is based on the time

response analysis on corresponding results.

DEFINITION AND CLASSIFICATION OF TIME RESPONSE

Time Response:

The response given by the system which is function of the time, to the applied excitation is

called time response of a control system.

Practically, output of the system takes some finite time to reach to its final value.

This time varies from system to system and is dependent on different factors.

The factors like friction mass or inertia of moving elements some nonlinearities present etc.

Example: Measuring instruments like Voltmeter, Ammeter.

Classification:

The time response of a control system is divided into two parts.

1 Transient response ct(t)

2 Steady state response css(t)

. . . c(t)=ct(t) +cSS(t)

Where c(t)= Time Response

Total Response=Zero State Response +Zero Input Response

Transient Response:

It is defined as the part of the response that goes to zero as time becomes very large. i,e,

Lim ct(t)=0

t

A system in which the transient response do not decay as time progresses is an Unstable

system.



C(t)

Ct(t) Css(t) Step ess

= study state

error

O Time

Transient time Study state

Time

2. Steady State Response:

It is defined the part of the response which remains after complete transient response

vanishes from the system output.

. i,e, Lim ct(t)=css(t)

t

The time domain analysis essentially involves the evaluation of the transient and

Steady state response of the control system.

Standard Test Input Signals

For the analysis point of view, the signals, which are most commonly used as reference

inputs, are defined as standard test inputs.

The performance of a system can be evaluated with respect to these test signals.

Based on the information obtained the design of control system is carried out.

The commonly used test signals are

1. Step Input signals.

2. Ramp Input Signals.

3. Parabolic Input Signals.

4. Impulse input signal.

Details of standard test signals

1. Step input signal (position function)

It is the sudden application of the input at a specified time as usual in the figure or

instant any us change in the reference input

Example :-

a. If the input is an angular position of a mechanical shaft a step input represent

the sudden rotation of a shaft.

b. Switching on a constant voltage in an electrical circuit.

The transient response may be experimental

or oscillatory in nature.



c. Sudden opening or closing a valve.

r(t)

A

O t

When, A = 1, r(t) = u(t) = 1

The step is a signal who‘s value changes from 1 value (usually 0) to another level A in

Zero time.

In the Laplace Transform form R(s) = A / S

Mathematically r(t) = u(t)

= 1 for t > 0

= 0 for t < 0

2. Ramp Input Signal (Velocity Functions):

It is constant rate of change in input that is gradual application of input as shown

in fig (2 b). r(t)

Ex:- Altitude Control

of a Missile

Slope = A

t

O

The ramp is a signal, which starts at a value of zero and increases linearly with

time.

Mathematically r (t) = At for t ≥ 0

= 0 for t≤ 0.

In LT form R(S) = A

S2

If A=1, it is called Unit Ramp Input

Mathematically

r(t) = t u(t)

{

In LT form R(S) = A = 1

S2

S2

t for t ≥ 0

0 for t ≤ 0 =



∆ t 0

At2

for t ≥ 0

= 2

0 for t ≤ 0

ie., t 0 (zero) applied

momentarily

3. Parabolic Input Signal (Acceleration function):

The input which is one degree faster than a ramp type of input as shown in fig (2 c) or

it is an integral of a ramp .

Mathematically a parabolic signal of magnitude

A is given by r(t) = A t2

u(t)

2

r(t)

t

In LT form R(S) = A

S3

If A = 1, a unit parabolic function is defined as r(t) = t2 u(t)

2

ie., r(t)

{

In LT for R(S) = 1

S3

4. Impulse Input Signal :

It is the input applied instantaneously (for short duration of time ) of very high amplitude

as shown in fig 2(d)

Eg: Sudden shocks i e, HV due lightening or short circuit.

It is the pulse whose magnitude is infinite while its width tends to zero.

r(t)

O t

Area of impulse = Its magnitude

If area is unity, it is called Unit Impulse Input denoted as (t)

Mathematically it can be expressed as

r(t) = A for t = 0

Slope = At

= t2

for t ≥ 0

2

0 for t ≤ 0

A



= 0 for t ≠ 0

In LT form R(S) = 1 if A = 1

Standard test Input Signals and its Laplace Transforms.

r(t) R(S)

Unit Step 1/S

Unit ramp 1/S2

Unit Parabolic 1/S3

Unit Impulse 1

First order system:-

The 1st order system is represent by the differential Eq:- a1dc(t )+aoc (t) = bor(t)------ (1)

dt

Where, e (t) = out put , r(t) = input, a0, a1 & b0 are constants.

Dividing Eq:- (1) by a0, then a1. d c(t ) + c(t) = bo.r (t)

a0 dt ao

T . d c(t ) + c(t) = Kr (t) ---------------------- (2)

dt

Where, T=time const, has the dimensions of time = a1 & K= static sensitivity = b0

a0 a0

Taking for L.T. for the above Eq:- [ TS+1] C(S) = K.R(S)

T.F. of a 1st order system is ; G(S) = C(S ) = K .

R(S) 1+TS

If K=1, Then G(S) = [ It‘s a dimensionless T.F.]

……I

This system represent RC ckt. A simplified bloc diagram is as shown.;

R(S)+ 1 C(S)

TS

-

1 .

1+TS



t

0.632

Unit step response of 1st order system:-

Let a unit step i\p u(t) be applied to a 1st order system,

Then, r (t)=u (t) & R(S) = 1 . ---------------(1)

S W.K.T. C(S)

= G(S). R(S)

C(S) = 1 . 1 . = 1 . T . ----------------- (2)

1+TS S S TS+1

Taking inverse L.T. for the above Eq:-

then, C(t)=u (t) – e –t/T

; t.>0.------------- (3) slope = 1 .

T

At t=T, then the value of c(t)= 1- e –1

= 0.632. c (t)

The smaller the time const. T. the

faster the system response. 1 – e –t/T

The slope of the tangent line at at t= 0 is 1/T.

From Eq:- (4) , We see that the slope of the response curve c(t) decreases monotonically from 1

. at t=0 to zero. At t=

T

Second order system:-

The 2nd

order system is defined as,

a2 d2

c(t) + a1 dc(t) + a0 c(t) = b0 .r(t)-----------------(1)

dt2 dt

Where c(t) = o/p & r(t) = I/p

-- ing (1) by a0,

a2 d2 c(t) + a1 . dc (t) + c(t) = b0 . r(t).

a0 dt2 a0 dt a0

a2 d2 c(t) + 2a1 . a2 . dc (t) + c(t) = b0 . r(t).

a0 dt2 2 a0 a0 . a2 dt a0

3) The open loop T.F. of a unity feed back system is given by G(S) = K . where,

S(1+ST)

Since dc = 1 .e -t/T

= 1 . at t .=0. ------------- (4)

dt T T

T



T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so

that (a) The peak overshoot of unity step response of the system is reduced from 75% to 25%

(b) The damping ratio increases from 0.1 to 0.6.

Solution: G(S) = K .

S(1+ST)

Let the value of damping ratio is, when peak overshoot is 75% & when peak

overshoot is 25%

Mp = .

e 1-

2

ln 0. 75 = . 0.0916 = .

1- 2

1- 2

1 = 0.091 (0.0084) (1-2) =

2

2 = 0.4037 (1.0084

2) = 0.0084

= 0.091

k .

S + S2T .

w.k.t. T.F. = G(S) =

1 + K . = K .

1+ G(S) . H(S) S + S2T S + S

2T+K

T.F. = K / T .

S2 + S + K .

T T

Comparing with std Eq :-

Wn = K . , 2 Wn = 1 .

T T

Let the value of K = K1 When = 1 & K = K2 When = 2.

Since 2 Wn = 1 . , = 1 . = 1 .

T 2TWn

2 KT

1 .

1 . = 2 K1T = K2 .

2 1 K1

2 K2T

0.091 = K2 . K2 . = 0.0508

0.4037 K1 K1



K2 = 0.0508 K1

a) The amplitude K has to be reduced by a factor = 1 . = 20

0.0508

b) Let = 0.1 Where gain is K1 and

= 0.6 Where gain is K2

0.1 = K2 . K2 . = 0.027 K2 = 0.027 K1

0.6 K1 K1

The amplitude gain should be reduced by 1 . = 36

0.027

4) Find all the time domain specification for a unity feed back control system whose open loop

T.F. is given by

G(S) = 25 .

S(S+6)

Solution:

25 .

G(S) = 25 . G(S) . = S(S+6) .

S(S+6) 1 + G(S) .H(S) 1 + 25 .

S(S+6)

= 25 .

S2 + ( 6S+25 )

W2n = 25 , Wn = 5, 2 Wn = 6 = 6 . = 0.6

2 x 5

Wd = Wn 1- 2

= 5 1- (0.6)2 = 4

tr = - , = tan-1

Wd = Wn = 0.6 x 5 = 3

Wd

= tan-1

( 4/3 ) = 0.927 rad.

tp = . = 3.14 = 0.785 sec.

Wd 4

MP = . = 0.6 . x3.4 = 9.5%

e

1- 2

e

1- 0.62

ts = 4 . for 2% = 4 . = 1.3 ………3sec.

Wn 0.6 x 5



5) The closed loop T.F. of a unity feed back control system is given by

C(S) = 5 .

R(S) S2 + 4S +5

Solution:

C(S) = 5 . , Wn2 = 5 Wn = 5 = 2.236

R(S) S2 + 4S +5

2 Wn = 4 = 4 . = 0.894. Wd = 1.0018

2 x 2.236

MP = . = 0.894 .

X 3.14 = 0.19%

e

1- 2

e

1-(0.894)2

W. K.T. C(t) = e- Wnt

Cos Wdtr + . sin wdtr

1-2

= e-0.894x2.236t

Cos 1.0018t + 0.894 . sin 1.0018t

1-(0.894)2

6) A servo mechanism is represent by the Eq:-

d2

+ 10 d = 150E , E = R- is the actuating signal calculate the

dt2 dt value of damping ratio, undamped and damped

frequency of ascillation.

Soutions:- d2

+ 10 d = 15 ( r - ) , = 150r – 150 .

dt2 dt

Taking L.T., [S2 + 10S + 150] (S) = 150 R (S).

(S) = 150 .

R(S) S2 + 10S + 15O

Wn2 = 150 Wn = 12.25. ………………………….rad sec .1

2 Wn = 10 = 10 . = 0.408.

2 x 12.25

Determine (1) Damping ratio (2) Natural

undamped response frequency Wn. (3) Percent

peak over shoot Mp (4) Expression for error

resoponse.



Wd = Wn 1 - 2 = 12.25 1- (0.408)2 = 11.18. rad 1sec.

7) Fig shows a mechanical system and the response when 10N of force is applied to the system.

Determine the values of M, F, K,.

K x(t)inmt

f(t) 0.00193 The T.F. of the mechanical system is ,

0.02 X(S) = 1 .

F(S) MS2 + FS = K

f(t) = Md2X + F dX + KX

F x dt2

dt

F(S) = (MS2 + FS + K) x (S)

1 2 3 4 5

Given :- F(S) = 10

S.

X(S) = 10 .

S(MS2 + FS + K)

SX (S) = 10 .

MS2 + FS + K

The steady state value of X is By applying final value theorem,

lt. SX(S) = 10 . = 10 = 0.02 ( Given from Fig.)

S O M(0) + F (0) + K K. ( K = 500.)

MP = 0.00193 = 0.0965 = 9.62%

0.02

MP = e . ln 0.0965 = .

1 - 2

1 - 2

= . 0.5539 = 2 . 0.744

1 - 2

1 - 2

0.5539 – 0.5539 2

= 2

= 0.597 = 0.6

tp = = .

Wn 1 – 2 Wd

3 = . Wn = 1.31…… rad / Sec.

M



Wn 1 – (0.6)2

Sx(S) = 10/ M .

(S2 + F S + K )

M M

Comparing with the std. 2nd

order Eq :-, then,

Wn2 = K Wn = K (1.31)

2 = 500 . M = 291.36 kg.

M M M

F = 2 Wn F = 2 x 0.6 x 291 x 1.31

M F = 458.7 N/M/ Sec.

8) Measurements conducted on sever me mechanism show the system response to be c(t) =

1+0.2e-60t

– 1.2e-10t

, When subjected to a unit step i/p. Obtain the expression for closed

loop T.F the damping ratio and undamped natural frequency of oscillation .

Solution:

C(t) = 1+0.2e-60t

–1.2e-10t

Taking L.T., C(S) = 1 . + 0.2 . – 1.2 .

S S+60 S+10

C(S) . = 600 / S .

S2 + 70S + + 600

Given that :- Unit step i/p r(t) = 1 R(S) = 1 .

C(S) . = 600 / S .

R(S) S2 + 70S + + 600

Comparing, Wn2 = 600, 24.4 …..rad / Sec

2 Wn =

70, = 70 . = 1.428

2 x 24.4

10) A feed back system employing o/p damping is as shown in fig.

1) Find the value of K1 & K2 so that closed loop system resembles a 2nd

order system with

= 0.5 & frequency of damped oscillation 9.5 rad / Sec.



2) With the above value of K1 & K2 find the % overshoot when i/p is step i/p

3) What is the % overshoot when i/p is step i/p, the settling time for 2% tolerance?

R + C

__

C . = K1 .

R S2 + ( 1 + K2 ) S + K1

Wn2 = K1 Wn = K1

2 Wn = 1 + K2 = 1 + K2

2 K1

Wn 1 - 2 Wn = 9.5 . 10.96 rad/Sec Wd =

1 – 0.52

K1 = (10.96)2 = 120.34

2 Wn = 1 + K2 , K2 = 9.97

MP = . = 16.3%

e 1 - 2

ts = 4 . = 4 . = 0.729 sec

Wn 0.5 x 10.97

Steady state Error :-

Steady state errors constitute an extremely important aspect of system

performance. The state error is a measure of system accuracy. These errors arise from the nature

of i/p‘s type of system and from non-linearties of the system components. The steady state

performance of a stable control system is generally judged by its steady state error to step, ramp

and parabolic i/p.

K1 1 .

S(1+S)

K2S



Consider the system shown in the fig.

R(S) E(S) C(S)

C(S) = G(S) . …………………………(1)

R(S) 1+G(S) . H(S)

The closed loop T.F is given by (1). The T.F. b/w the actuating error signal e(t) and the

i/p signal r(t) is,

E(S) = R(S) – C(S) H(S) = 1 – C(S) . H(S)

R(S) R(S) R(S)

= 1 – G(S) . H(S) . = 1 + G(S) . H(S) – G(S)H(S)

1 + G(S) . H(S) 1+G(S) . H(S)

= 1 .

1 + G(S) . H(S)

Where e(t) = Difference b/w the i/p signal and the feed back signal

E(S) = 1 . .R(S) ……………………….(1)

1 + G(S) . H(S)

The steady state error ess may be found by the use of final value theorem

and is as follows;

ess = lt e(t) = lt SE(S)

t S O

Substituting (1), ess = lt S.R(S) . ……………….(2)

S O 1+G(S) . H(S)

Eq :- (2) Shows that the steady state error depends upon the i/p R(S) and the forward T.F.

G(S) and loop T.F G(S) . H(S).

G(S)

H(S)



The expression for steady state errors for various types of standard test signals are

derived below;

1) Steady state error due to step i/p or position error constant (Kp):-

The steady state error for the step i/p is

I/P r(t) = u(t). Taking L.T., R(S) = 1/S.

From Eq:- (2), ess = lt S. R(s) . = 1 .

S O 1 +G(S). H.S 1 + lt G(S). H(S)

S O

lt G(S) . H(S) = Kp

(S O )

Where Kp = proportional error constant or position error const.

ess = 1 .

1 + Kp

(1 + Kp) ess = 1 Kp = 1 - ess

ess

Note :- ess = R . for non-unit step i/p

1 + Kp

2) Steady state error due to ramp i/p or static velocity error co-efficient (Kv) :-

The ess of the system with a unit ramp i/p or unit velocity i/p is given by,

r ( t) = t. u(t) , Taking L -T, R(S) = 1/S2

Substituting this to ess Eq:-

ess = lt S . . 1 . = lt 1 .

S O 1 + G(S) . H(s) S2 S O S +S G(S) H(s)S

lt = SG(S) . H(S) = Kv = velocity co-efficient then

S O

ess = lt 1 . ess = 1 .

S O S + Kv Kv

Velocity error is not an error in velocity , but it is an error in position error due to a ramp

i/p

3) Steady state error due to parabolic i/p or static acceleration co-efficient (Ka) :-

The steady state actuating error of the system with a unit parabolic i/p (acceleration i/p)

which is defined by r(t) + 1 . t2 Taking L.T. R(S)= 1 .

2 S3



ess = lt S . 1 . lt 1 .

S O 1 + G(S) . H(S) S3

S O S2

+ S2 G(S) . H(S)

lt S2 G(S) . H(S) = Ka.

S O

ess = lt 1 . = 1 .

S O S2 + Ka Ka

Note :- ess = R . for non unit parabolic.

Ka

Types of feed back control system :-

The open loop T.F. of a unity feed back system can be written in two std, forms;

1) Time constant form and 2) Pole Zero form,

G(S) = K(TaS +1) (TbS +1)…………………..

Sn(T1 S+1) (T2S + 1)……………….

Where K = open loop gain.

Above Eq:- involves the term Sn in denominator which corresponds to no, of

integrations in the system. A system is called Type O, Type1, Type2,……….. if n = 0, 1,

2, ………….. Respectively. The Type no., determines the value of error co-efficients. As

the type no., is increased, accuracy is improved; however increasing the type no.,

aggregates the stability error. A term in the denominator represents the poles at the origin

in complex S plane. Hence Index n denotes the multiplicity of the poles at the origin.

The steady state errors co-efficient for a given type have definite values. This is

illustration as follows.

1) Type – O system :- If, n = 0, the system is called type – 0, system. The

steady state error are as follows;

Let, G(S) = K . [ .. . H(s) = 1]

S + 1

ess (Position) = 1 . = 1 . = 1 .

1 + G(O) . H(O) 1 + K 1 + Kp

.. . Kp = lt G(S) . H(S) = lt K . = K

S O S O S + 1

ess (Velocity) = 1 . = 1 . =

Kv O



Kv = lt G(S) . H(S) = lt S K . = O.

S O S O S + 1

ess (acceleration) = 1 . = 1 . =

Ka O

Ka = lt S2 G(S) . H(S) = lt S

2 K . = O.

S O S O S + 1

2) Type 1 –System :- If, n = 1, the ess to various std, i/p, G(S) = K .

S (S + 1)

ess (Position) = 1 . = O

1 +

Kp = lt G(S) . H(S) = lt K . =

S O S O S( S + 1)

Kv = lt S K . = K

S O S(S+1)

ess (Velocity) = 1 .

K


O O

Ka = lt S2

K . = O.

S O S (S + 1)

3) Type 2 –System :- If, n = 2, the ess to various std, i/p, are , G(S) = K .

S2 (S + 1)

Kp = lt K . =

S O S2 (S + 1)

. .

. ess (Position) = 1 . = O

Kv = lt S K . =

S O S2 (S + 1)

. .

. ess (Velocity) = 1 . = O



Ka = lt S2 K . = K.

S O S2 (S + 1)

. .

. ess (acceleration) = 1 .

K

3) Type 3 –System :- Gives Kp = Kv = Ka = & ess = O.

(Onwards)

The error co-efficient Kp, Kv, & Ka describes the ability of the system to eliminate the steady

state error therefore they are indicative of steady state performance. It is generally described to

increase the error co-efficient while maintaining the transient response within an acceptable

limit.

PROBLEMS;

1. The unit step response of a system is given by

C (t) = 5/2 +5t – 5/2 e-2t

. Find the T. F of the system.

T/P = r(t) = U (t). Taking L.T, R(s) = 1/S.

Response C(t) = 5/2+5t-5/2 e-2t

Taking

L.T, C(s)

= 5 1 + 5 1 5 1 = 5

2 S S2 2 (S+2) 2

C(s) = 5 S(S+2)+2(S+2)-

S2 = 5

2 S2(

S+2)

2

= 10 (S+1)

S2 (S+2)

T.F = C (S) = 10 (S+1) S = 10 (S+1)

R (S) S2(S+2) S(S+2)

2. The open loop T F of a unity food back system is G(s) = 100

S (S+10)

1 + 2 - 1

S S2 S+2

S2+2S+2S+4—

s2 S2 (S+2)



Find the static error constant and the steady state error of the system when subjected to an i/p

given by the polynomial

R(t) = Po + p1t + P2 t2

2

G(s) = 100 position error co-efficient

KP = lt G(s) = lt 100 =

Similarly KV = lt SG(s) = lt 100 x s = 100 = 10

Ka = lt S2 G(S)

Given :- r(t) = Po+P1t +P2 t2

2

Therefore steady state error ess

ess

Ess = 0+0.1 P1 + =

3. Determine the error co-efficeint and static error for G(s) = 1

And H(s) = (S+2) S(S+1) (S+10)

S (S+10)

S 0 S 0 S (S+10)

S 0 S 0 S (S+10)

10

lt 100 x s2 = 0

S 0 S 0 S (S+10)

R1 R2 R3

+ +

1+Kp Kv Ka

R1 R2

R3+

+

1+ 10 0

P0 P1

P2

+ +

1+ 10 0

=



The error constants for a non unity feed back system is as follows

Ka = 0

Static Error:-

Steady state error for unit step i/p = 0

Unit ramp i/p

Unit parabolic i/p = 1/0 =

4. A feed back C.S is described as G(S) =

H(S)=1/s.For unit step i/p,cal steady state error

constant and errors.

(S+2)

G(S).H(S) =

S(S+1) (S+10)

Kp = lt G(S) H(S) =

lt

S 0 S

0

(0+2)

=

0(0+1) (0+10)

Kv = lt G(S) H(S) =

lt

S 0 S

0

(0+2)

= 1/5 = 0.2

0(0+1) (0+10)

1 1

= = 5

Kv 0.2

50

S (S+2) (S+5)

Kp = lt G(S) H(S) =

lt

S 0 S

0

50

=

S2 (S+2) (S+5)

Kv = lt G(S) H(S) =

lt

S 0 S

0

50 x S

=

S2 (S+2) (S+5)

Ka = lt G(S) H(S) = lt

S 0 S

0

S2 x 50 50

= = 5

S2 (S+2) (S+5) 10



The steady state error

= 0/50 = 0

5. A certain feed back C.S is described by following C.S G(S)

=

Determine steady state error co-efficient and also determine the value of K to limit the steady to

10 units due to i/p r(t) = 1 + 10 + t 20/2 t2.

Steady state error:-

Error due to unit step i/p

Error due o r(t) ramp i/p

Ess = lt S. 1/S

S 0 1+50

S2(S+2)(S+5)

Lt S2 (S+2) (S+5)

S 0 S2 (S+2) (S+5) + 50

K

H(S) = 1

S2 (S+20) (S+30)

Kp = lt G(S) H(S) =

lt

S 0 S

0

50

=

S2 (S+20) (S+30)

Kv = lt S K

S 0 S2 (S+20) (S+30) =

Ka = lt S2 K

S 0 S2 (S+20) (S+30)

K

600

1 1

+ = 0

1+Kp 1+

10 10

+ = 0

Kv



Error due to para i/p,

,

r (t) = 0+0 12000 = 10 = K = 1200

K

First order system:-

The 1st order system is represent by the differential Eq:- a1dc(t )+aoc (t) = bor(t)------ (1)

dt

Where, e (t) = out put , r(t) = input, a0, a1 & b0 are constants.

Dividing Eq:- (1) by a0, then a1. d c(t ) + c(t) = bo.r (t)

a0 dt ao

T . d c(t ) + c(t) = Kr (t) ---------------------- (2)

dt

Where, T=time const, has the dimensions of time = a1 & K= static sensitivity = b0

a0 a0

Taking for L.T. for the above Eq:- [ TS+1] C(S) = K.R(S)

T.F. of a 1st order system is ; G(S) = C(S ) = K .

R(S) 1+TS

If K=1, Then G(S) = [ It‘s a dimensionless T.F.]

……I

This system represent RC ckt. A simplified bloc diagram is as shown.;

R(S)+ 1 C(S)

TS

-

Unit step response of 1st order system:-

Let a unit step i\p u(t) be applied to a 1st order system,

Then, r (t)=u (t) & R(S) = 1 . ---------------(1)

S W.K.T. C(S)

= G(S). R(S)

20 40

= =

Ka 2Ka

20 x 600 12000

=

K K

1 .

1+TS



t

0.632

C(S) = 1 . 1 . = 1 . T . ----------------- (2)

1+TS S S TS+1

Taking inverse L.T. for the above Eq:-

then, C(t)=u (t) – e –t/T

; t.>0.------------- (3) slope = 1 .

T

At t=T, then the value of c(t)= 1- e –1

= 0.632. c (t)

The smaller the time const. T. the

faster the system response. 1 – e –t/T

The slope of the tangent line at at t= 0 is 1/T.

From Eq:- (4) , We see that the slope of the response curve c(t) decreases monotonically from 1

. at t=0 to zero. At t=

T

Second order system:-

The 2nd

order system is defined as,

a2 d2

c(t) + a1 dc(t) + a0 c(t) = b0 .r(t)-----------------(1)

dt2 dt

Where c(t) = o/p & r(t) = I/p

-- ing (1) by a0,

a2 d2 c(t) + a1 . dc (t) + c(t) = b0 . r(t).

a0 dt2 a0 dt a0

a2 d2 c(t) + 2a1 . a2 . dc (t) + c(t) = b0 . r(t).

a0 dt2 2 a0 a0 . a2 dt a0

Step response of 2nd

order system:

The T.F. = C(s) = Wn2 Based on value

R(s) 32+2 WnS+ Wn

2

The system may be,

Since dc = 1 .e -t/T

= 1 . at t .=0. ------------- (4)

dt T T

T



2) Under damped system (0< <1)

3) Critically damped system ( =1)

4) Over damped system ( >1)

1) Under damped system :- (0< <1)

In this case C(s) can be written as

R(s)

C(s) = Wn2

R(s) (S+ wn + jwd ) (S+ wn - jwd )

Where wd = wn 1- 2 The Freq. wd is called damped

natural frequency

For a unit step i/p :- [ R(t)= 1 R(S) = 1/S]

C(S) = Wn2 . X R (S) = Wn

2 . = 1 .

(S2+2 Wn S+ Wn

2) (S

2+2 WnS+ Wn

2) S

C(S) = 1 . S+2 Wn .

S S2+2 WnS+ Wn

2

= 1 . S+ Wn . __ Wn . --------------- (5)

S (S+ Wn)2

+ Wd2 (S+ Wn

2 + Wd

2)

C(S) = 1 . S+ Wn . __ . Wd .

S (S+ Wn) 2

+ Wd2 1-

2 (S+ Wn)

2 + Wd

2

Taking ILT, C(t) = 1-e- Wnt

COS Wdt + . Sin Wdt ------------ (6)

1- 2

The error signal for this system is the difference b/w the I/p & o/p.

e(t) = r(t) c(t) .

= 1 c(t)

= e- Wnt

COS Wdt + . Sin Wdt --------------------- (7)



1- 2 t > o.

At t = , error exists b/w the i/p & o/p.

If the damping ratio = O, the response becomes undamped & oscillations continues

indefinitely.

The response C(t) for the zero damping case is ,

c(t) =1-1(COS wnt) =1- COS wnt ; t > O --------------------- (8)

From Eq:- (8) , we see that the Wn represents the undamped natural frequency of the system. If

the linear system has any amount of damping the undamped natural frequency cannot be

observed experimentally. The frequency, which may be observed, is the damped natural

frequency.

Wd =wn 1 2

This frequency is always lower than the undamped natural frequency. An

increase in would reduce the damped natural frequency Wd . If is increased beyond unity,

the response over damped & will not oscillate.

Critically damped case:- ( =1).

If the two poles of C(S) are nearly equal, the system may be approximated by a

R(S)

Critically damped one.

For a step I/p R(S) = 1/S

C(S) = Wn2 . 1 .

S2+2 Wn S+ Wn

2 S

= 1 . 1 . Wn .

S (S + Wn) ( S+ Wn)2

= 1 . Wn2 .

S ( S + Wn )2S

Taking I.L.T.,

C(t) = 1 – e -Wnt

(1+wnt)

Over damped system :- ( > 1)

If this case, the two poles of C(S) are negative, real and unequal.

R(S)

For a unit step I/p R(S) = 1/S , then,

C(S) = Wn2

.



(S+ Wn + Wn 2 - 1 ) ( S+ Wn - Wn

2 – 1)

Taking ILT, C(t) = 1+ 1 . e ( + 2 – 1) Wn t.

S 2

– 1 ( + 2

– 1)

1 . e ( + 2 – 1) Wn t.

S 2 – 1 ( +

2 – 1)

C(t) = 1+ Wn . e-S

1t

. - e

-S2

t . ; t > O

S 2

– 1 S1 S2

Where S1 = ( + 2 – 1) Wn

S2 = ( - 2 – 1) Wn

Time response (Transient ) Specification (Time domain) Performance :-

The performance characteristics of a controlled system are specified in terms of the

transient response to a unit step i/p since it is easy to generate & is sufficiently drastic.

MP

The transient response of a practical C.S often exhibits damped oscillations before

reaching steady state. In specifying the transient response characteristic of a C.S to unit step

i/p, it is common to specify the following terms.

1) Delay time (td)



2) Rise time (tr)

Response curve

3) Peak time (tp)

4) Max over shoot (Mp)

5) Settling time (ts)

1) Delay time :- (td)

It is the time required for the response to reach 50% of its final value for the 1st

time.

2) Rise time :- (tr)

It is the time required for the response to rise from 10% and 90% or 0% to

100% of its final value. For under damped system, second order system the 0 to 100% rise

time is commonly used. For over damped system, the 10 to 90% rise time is commonly used.

3) Peak time :- (tp)

It is the time required for the response to reach the 1st of peak of the overshoot.

4) Maximum over shoot :- (MP)

It is the maximum peak value of the response curve measured from unity. The amount

of max over shoot directly indicates the relative stability of the system.

5) Settling time :- (ts)



It is the time required for the response curve to reach & stay with in a range about the

final value of size specified by absolute percentage of the final value (usually 5% to 2%).

The settling time is related to the largest time const., of C.S.

Transient response specifications of second order system :-

W. K.T. for the second order system,

T.F. = C(S) = Wn2 . ------------------------------(1)

R(S) S2+2 WnS+ Wn

2

Assuming the system is to be underdamped ( < 1)

Rise time tr

W. K.T. C(tr) = 1- e- Wnt


1-2

Let C(tr) = 1, i.e., substituting tr for t in the above Eq:

Then, C(tr) = 1 = 1- e- Wntr

Cos wdtr + . sin wdtr

1-2

Cos wdtr + . sin wdtr = tan wdtr = - 1-2 = wd .

1-2

jW

Thus, the rise time tr is , jWd

tr = 1 . tan-1

- w d = - secs Wn 1-2

Wn

Wd wd

When must be in radians. -

Wn

Peak time :- (tp)

Peak time can be obtained by differentiating C(t) W.r.t. t and equating that

derivative to zero.

dc = O = Sin Wdtp Wn . e- Wntp

dt t = t p 1-2

Since the peak time corresponds to the 1st peak over shoot.

Wdtp = = tp = .

Wd

The peak time tp corresponds to one half cycle of the frequency of damped oscillation.

S- Plane



Maximum overshoot :- (MP)

The max over shoot occurs at the peak time.

i.e. At t = tp = .

Wd

Mp = e –( / Wd)

or e –( / 1- 2)

Settling time :- (ts)

An approximate value of ts can be obtained for the system O < <1 by using the

envelope of the damped sinusoidal waveform.

Time constant of a system = T = 1 .

Wn

Setting time ts = 4x Time constant.

= 4x 1 . for a tolerance band of +/- 2% steady state.

Wn

Delay time :- (td)

The easier way to find the delay time is to plot Wn td VS . Then approximate

the curve for the range O< < 1 , then the Eq. becomes,

Wn td = 1+0.7

td = 1+0.7

Wn

PROBLEMS:

(1) Consider the 2nd

order control system, where = 0.6 & Wn = 5 rad / sec, obtain the

rise time tr, peak time tp, max overshoot Mp and settling time ts When the system is subject

to a unit step i/p.,

Given :- = 0.6, Wn = 5rad /sec, tr = ?, tp = ?, Mp = ?, ts = ?

Wd = Wn 1- 2 = 5 1-(0.6)

2 = 4

= Wn = 0.6 x 5 = 3.

tr = - , = tan-1

Wd = tan-1

4 = 0.927 rad

Wd 3

tr = 3.14 – 0.927 = 0.55sec.

4



tp = = 3.14 = 0.785 sec.

Wd 4

. = e / Wd

MP = e 1- 2

MP = e (3/4) x 3.14

= 0.094 x 100 = 9.4%

ts :- For the 2% criteria.,

ts = 4 . = 4 . = 1.33 sec.

Wn 0.6x5

For the 5% criteria.,

ts = 3 = 3 = 1 sec

3 EXERCISE:

(2) A unity feed back system has on open loop T.F. G(S) = K .

S ( S+10)

Determine the value of K so that the system has a damping factors of 0.5 For this value of K

determine settling time, peak over shoot & time for peak over shoot for unit step i/p

LCS

The error co-efficient Kp, Kv, & Ka describes the ability of the system to eliminate the steady

state error therefore they are indicative of steady state performance. It is generally described to

increase the error co-efficient while maintaining the transient response within an acceptable

limit.

PROBLEMS;

2. The unit step response of a system is given by

C (t) = 5/2 +5t – 5/2 e-2t

. Find the T. F of the system.

T/P = r(t) = U (t). Taking L.T, R(s) = 1/S.

Response C(t) = 5/2+5t-5/2 e-2t

Taking 1 + 2 - 1

S S2 S+2



L.T, C(s) = 5 1 + 5 1 - 5 1 = 5

2 S S2 2 (S+2) 2

C(s) = 5 S(S+2)+2(S+2)-

S2 = 5

3 S2(

S+2)

2

= 10 (S+1)

S2 (S+2)

T.F = C (S) = 10 (S+1) S = 10 (S+1)

R (S) S2(S+2) S(S+2)

2. The open loop T F of a unity food back system is G(s) = 100

Find the static error constant and the steady state error of the system when subjected to an i/p

given by the polynomial

R(t) = Po + p1t + P2 t2

2

G(s) = 100 position error co-efficient

KP = lt G(s) = lt 100 =

Similarly KV = lt SG(s) = lt 100 x s = 100 = 10

Ka = lt S2 G(S)

S (S+10)

S2+2S+2S+4—

s2 S2 (S+2)

S (S+10)

S 0 S 0 S (S+10)

S 0 S 0 S (S+10)

10

lt 100 x s2 = 0

S 0 S 0 S (S+10)



Given :- r(t) = Po+P1t +P2 t2

2

Therefore steady state error ess

ess

Ess = 0+0.1 P1 + =

3. Determine the error co-efficeint and static error for G(s) = 1

And H(s) = (S+2) S(S+1) (S+10)

The error constants for a non unity feed back system is as follows

R1 R2 R3

+ +

1+Kp Kv Ka

R1 R2 R3

+ +

1+ 10 0

P0 P1 P2

+ +

1+ 10 0

=

(S+2)

G(S).H(S) =

S(S+1) (S+10)

Kp = lt G(S) H(S) = lt

S 0 S

0

(0+2)

=

0(0+1) (0+10)

Kv = lt G(S) H(S) =

lt

S 0 S

0

(0+2)

= 1/5 = 0.2

0(0+1) (0+10)



Ka = 0

Static Error:-

Steady state error for unit step i/p = 0

Unit ramp i/p

Unit parabolic i/p = 1/0 =

4. A feed back C.S is described as G(S) =

H(S)=1/s.For unit step i/p,cal steady state error

constant and errors.

The steady state error

= 0/50 = 0

1 1

= = 5

Kv 0.2

50

S (S+2) (S+5)

Kp = lt G(S) H(S) =

lt

S 0 S

0

50

=

S2 (S+2) (S+5)

Kv = lt G(S) H(S) =

lt

S 0 S

0

50 x S

=

S2 (S+2) (S+5)

Ka = lt G(S) H(S) = lt

S 0 S

0

S2 x 50 50

= = 5

S2 (S+2) (S+5) 10

Ess = lt S. 1/S

S 0 1+50

S2(S+2)(S+5)

Lt S2 (S+2) (S+5)

S 0 S2 (S+2) (S+5) + 50

K

H(S) = 1

S2 (S+20) (S+30)



5. A certain feed back C.S is described by following C.S G(S) =

Determine steady state error co-efficient and also determine the value of K to limit the steady to

10 units due to i/p r(t) = 1 + 10 + t 20/2 t2.

Steady state error:-

Error due to unit step i/p

Error due o r(t) ramp i/p

Error due to para i/p,

,

r (t) = 0+0 12000 = 10 = K = 1200

K

3) The open loop T.F. of a unity feed back system is given by G(S) = K . where,

S(1+ST)

Kp = lt G(S) H(S) = lt

S 0 S

0

50

=

S2 (S+20) (S+30)

Kv = lt S K

S 0 S2 (S+20) (S+30) =

Ka = lt S2 K

S 0 S2 (S+20) (S+30)

K

600

1 1

+ = 0

1+Kp 1+

10 10

+ = 0

Kv

20 40

= =

Ka 2Ka

20 x 600 12000

=

K K



T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so

that (a) The peak overshoot of unity step response of the system is reduced from 75% to 25%

(b) The damping ratio increases from 0.1 to 0.6.

Solution: G(S) = K .

S(1+ST)

Let the value of damping ratio is, when peak overshoot is 75% & when peak

overshoot is 25%

Mp = .

e 1-

2

ln 0. 75 = . 0.0916 = .

1- 2

1- 2

1 = 0.091 (0.0084) (1-2) =

2

2 = 0.4037 (1.0084

2) = 0.0084

= 0.091

k .

S + S2T .

w.k.t. T.F. = G(S) =

1 + K . = K .

1+ G(S) . H(S) S + S2T S + S

2T+K

T.F. = K / T .

S2 + S + K .

T T

Comparing with std Eq :-

Wn = K . , 2 Wn = 1 .

T T

Let the value of K = K1 When = 1 & K = K2 When = 2.

Since 2 Wn = 1 . , = 1 . = 1 .

T 2TWn

2 KT

1 .

1 . = 2 K1T = K2 .

2 1 K1

2 K2T

0.091 = K2 . K2 . = 0.0508



0.4037 K1 K1

K2 = 0.0508 K1

a) The amplitude K has to be reduced by a factor = 1 . = 20

0.0508

b) Let = 0.1 Where gain is K1 and

= 0.6 Where gain is K2

0.1 = K2 . K2 . = 0.027 K2 = 0.027 K1

0.6 K1 K1

The amplitude gain should be reduced by 1 . = 36

0.027

4) Find all the time domain specification for a unity feed back control system whose open loop

T.F. is given by

G(S) = 25 .

S(S+6)

Solution:

25 .

G(S) = 25 . G(S) . = S(S+6) .

S(S+6) 1 + G(S) .H(S) 1 + 25 .

S(S+6)

= 25 .

S2 + ( 6S+25 )

W2n = 25 , Wn = 5, 2 Wn = 6 = 6 . = 0.6

2 x 5

Wd = Wn 1- 2

= 5 1- (0.6)2 = 4

tr = - , = tan-1

Wd = Wn = 0.6 x 5 = 3

Wd

= tan-1

( 4/3 ) = 0.927 rad.

tp = . = 3.14 = 0.785 sec.

Wd 4

MP = . = 0.6 . x3.4 = 9.5%

e

1- 2

e

1- 0.62

ts = 4 . for 2% = 4 . = 1.3 ………3sec.

Wn 0.6 x 5



5) The closed loop T.F. of a unity feed back control system is given by

C(S) = 5 .

R(S) S2 + 4S +5

Solution:

C(S) = 5 . , Wn2 = 5 Wn = 5 = 2.236

R(S) S2 + 4S +5

2 Wn = 4 = 4 . = 0.894. Wd = 1.0018

2 x 2.236

MP = . = 0.894 .

X 3.14 = 0.19%

e

1- 2

e

1-(0.894)2

W. K.T. C(t) = e- Wnt


1-2

= e-0.894x2.236t

Cos 1.0018t + 0.894 . sin 1.0018t

1-(0.894)2

6) A servo mechanism is represent by the Eq:-

d2




Soutions:- d2

+ 10 d = 15 ( r - ) , = 150r – 150 .

dt2 dt

Taking L.T., [S2 + 10S + 150] (S) = 150 R (S).

(S) = 150 .

R(S) S2 + 10S + 15O

Wn2 = 150 Wn = 12.25. ………………………….rad sec .1

2 Wn = 10 = 10 . = 0.408.




resoponse.



2 x 12.25

Wd = Wn 1 - 2 = 12.25 1- (0.408)2 = 11.18. rad 1sec.

7) Fig shows a mechanical system and the response when 10N of force is applied to the system.

Determine the values of M, F, K,.

K x(t)inmt

f(t) 0.00193 The T.F. of the mechanical system is ,

0.02 X(S) = 1 .

F(S) MS2 + FS = K

f(t) = Md2X + F dX + KX

F x dt2

dt

F(S) = (MS2 + FS + K) x (S)

1 2 3 4 5

Given :- F(S) = 10

S.

X(S) = 10 .

S(MS2 + FS + K)

SX (S) = 10 .

MS2 + FS + K

The steady state value of X is By applying final value theorem,

lt. SX(S) = 10 . = 10 = 0.02 ( Given from Fig.)

S O M(0) + F (0) + K K. ( K = 500.)

MP = 0.00193 = 0.0965 = 9.62%

0.02

MP = e . ln 0.0965 = .

1 - 2

1 - 2

= . 0.5539 = 2 . 0.744

1 - 2

1 - 2

0.5539 – 0.5539 2

= 2

= 0.597 = 0.6

tp = = .

Wn 1 – 2

Wd

M



3 = . Wn = 1.31…… rad / Sec.

Wn 1 – (0.6)2

Sx(S) = 10/ M .

(S2 + F S + K )

M M

Comparing with the std. 2nd

order Eq :-, then,

Wn2 = K Wn = K (1.31)

2 = 500 . M = 291.36 kg.

M M M

F = 2 Wn F = 2 x 0.6 x 291 x 1.31

M F = 458.7 N/M/ Sec.

9) Measurements conducted on sever me mechanism show the system response to be c(t) =

1+0.2e-60t

– 1.2e-10t

, When subjected to a unit step i/p. Obtain the expression for closed

loop T.F the damping ratio and undamped natural frequency of oscillation .

Solution:

C(t) = 1+0.2e-60t

–1.2e-10t

Taking L.T., C(S) = 1 . + 0.2 . – 1.2 .

S S+60 S+10

C(S) . = 600 / S .

S2 + 70S + + 600

Given that :- Unit step i/p r(t) = 1 R(S) = 1 .

S

C(S) . = 600 / S .

R(S) S2 + 70S + + 600

Comparing, Wn2 = 600, 24.4 …..rad / Sec

2 Wn = 70, = 70 . = 1.428

2 x 24.4

9) The C.S. shown in the fig employs proportional plus error rate control. Determine the value

of error rate const. Ke, so the damping ratio is 0.6 . Determine the value of settling time, max



overshoot and steady state error, if the i/p is unit ramp, what will be the value of steady state

error without error rate control.

R +

(S)

__

10) A feed back system employing o/p damping is as shown in fig.


order system with

= 0.5 & frequency of damped oscillation 9.5 rad / Sec.



R + C

__

C . = K1 .

R S2 + ( 1 + K2 ) S + K1

Wn2 = K1 Wn = K1

2 Wn = 1 + K2 = 1 + K2

2 K1

Wn 1 - 2 Wn = 9.5 . 10.96 rad/Sec Wd =

1 – 0.52

K1 = (10.96)2 = 120.34

2 Wn = 1 + K2 , K2 = 9.97

MP = . = 16.3%

e 1 - 2

1+SKe

S+2

10 .

S

K1 1 .

S(1+S)

K2S



ts = 4 . = 4 . = 0.729 sec

Wn 0.5 x 10.97

Types of feed back control system :-

The open loop T.F. of a unity feed back system can be written in two std, forms;

1) Time constant form and 2) Pole Zero form,

G(S) = K(TaS +1) (TbS +1)…………………..

Sn(T1 S+1) (T2S + 1)……………….

Where K = open loop gain.

Above Eq:- involves the term Sn in denominator which corresponds to no, of

integrations in the system. A system is called Type O, Type1, Type2,……….. if n = 0, 1,

2, ………….. Respectively. The Type no., determines the value of error co-efficients. As

the type no., is increased, accuracy is improved; however increasing the type no.,

aggregates the stability error. A term in the denominator represents the poles at the origin

in complex S plane. Hence Index n denotes the multiplicity of the poles at the origin.

The steady state errors co-efficient for a given type have definite values. This is

illustration as follows.

2) Type – O system :- If, n = 0, the system is called type – 0, system. The

steady state error are as follows;

Let, G(S) = K . [ .. . H(s) = 1]

S + 1

ess (Position) = 1 . = 1 . = 1 .

1 + G(O) . H(O) 1 + K 1 + Kp

.. . Kp = lt G(S) . H(S) = lt K . = K

S O S O S + 1

ess (Velocity) = 1 . = 1 . =

Kv O

Kv = lt G(S) . H(S) = lt S K . = O.

S O S O S + 1


Ka O

Ka = lt S2 G(S) . H(S) = lt S

2 K . = O.

S O S O S + 1



2) Type 1 –System :- If, n = 1, the ess to various std, i/p, G(S) = K .

S (S + 1)

ess (Position) = 1 . = O

1 +

Kp = lt G(S) . H(S) = lt K . =

S O S O S( S + 1)

Kv = lt S K . = K

S O S(S+1)

ess (Velocity) = 1 .

K


O O

Ka = lt S2

K . = O.

S O S (S + 1)

3) Type 2 –System :- If, n = 2, the ess to various std, i/p, are , G(S) = K .

S2 (S + 1)

Kp = lt K . =

S O S2 (S + 1)

. .

. ess (Position) = 1 . = O

Kv = lt S K . =

S O S2 (S + 1)

. .

. ess (Velocity) = 1 . = O

Ka = lt S2 K . = K.

S O S2 (S + 1)

. .

. ess (acceleration) = 1 .

K

3) Type 3 –System :- Gives Kp = Kv = Ka = & ess = O.

(Onwards)



Recommended Questions

1. Define and classify time response of a system.

2. Mention the Standard Test Input Signals and its Laplace transform

3. The open loop T.F. of a unity feed back system is given by G(S) = K . Where,

S(1+ST)

T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so that

(a) The peak overshoot of unity step response of the system is reduced from 75% to 25% (b)

The damping ratio increases from 0.1 to 0.6.

4. Find all the time domain specification for a unity feed back control system whose open loop

T.F. is given by

G(S) = 25 .

S(S+6)

5. The closed loop T.F. of a unity feed back control system is given by

C(S) = 5 .

R(S) S2 + 4S +5

6. A servo mechanism is represent by the Eq:-

d2




7. Measurements conducted on sever me mechanism show the system response to be c (t) =

1+0.2e-60t

– 1.2e-10t

, When subjected to a unit step i/p. Obtain the expression for closed loop T.F

the damping ratio and undamped natural frequency of oscillation .

8. A feed back system employing o/p damping is as shown in fig.


order system with =

0.5 & frequency of damped oscillation 9.5 rad / Sec.



R

+ C

__




resoponse.

K1 1 .

S(1+S)

K2S



UNIT-4

Stability Analysis

Every System, for small amount of time has to pass through a transient period. Whether system

will reach its steady state after passing through transients or not. The answer to this question is

whether the system is stable or unstable. This is stability analysis.

For example, we want to go from one station to other. The station we want to reach is

our final steady state. The traveling period is the transient period. Now any thing

may happen during the traveling period due to bad weather, road accident etc, there is a chance

that we may not reach the next station in time. The analysis of wheather the given system can

reach steady state after passing through the transients successfully is called the stability analysis

of the system.

In this chapter, we will steady

1. The stability & the factor on which system stability depends.

2. Stability analysis & location of closed loop poles.

3. Stability analysis using Hurwitz method.

4. Stability analysis using Routh-Hurwitz method.

5. Special cases of Routh‘s array.

6. Applications of Routh-Hurwitz method.

Concept of stability:

Consider a system i.e a deep container with an object placed inside it as shown in fig(1)

force „F‟

(a) fig(1) (b)

Now, if we apply a force to take out the object, as the depth of container is more, it will oscillate

& settle down again at original position.

Assume that force required to take out the object tends to infinity i.e always object will

oscillate when force is applied & will settle down but will not come out such a system is called

absolutely stable system. No change in parameters, disturbances, changes the output.



Now consider a container which is pointed one, on which we try to keep a circular object. In this

object will fall down without any external application of force. Such system is called Unstable

system.

(a) fig(2) (b)

While in certain cases the container is shallow then there exsists a critical value of

force for which the object will come out of the container.

F

F F

Fig(3) F<Fcritical F>Fcritical

As long as F<Fcritical object regains its original position but if F>Fcritical object will come out.

Stability depends on certain conditions of the system, hence system is called conditionally stable

system.

Pendulum where system keeps on oscillating when certain force is applied. Such

systems are neither stable nor unstable & hence called critically stable or marginally stable

systems

Stability of control systems:

The stability of a linear closed loop system can be determined from the locations of closed loop

poles in the S-plane.

If the system has closed loop T.F.

C(s) 10

R(s) (S+2) (S+4) 1

Output response for unit step input R(s) =

S

C(s) 10 A B C

S(S+2) (S+4) S S+2 S+4

Find out partial fractions



1 1 1

C(s) = 8 4 8 10

S S+2 S+4

= 1.25 2.5 1.25

S S+2 S+4

C(s) = 1.25 – 2.5e-2t

+1.25e-4t

=Css + Ct(t)

If the closed loop poles are located in left half of s-plane, Output response contains exponential

terms with negative indices will approach zero & output will be the steady state output.

i.e.

Ct (t) = 0

t

Transient output = 0

Such system are called absolutely stable systems.

Now let us have a system with one closed loop pole located in right half of s- plane

C(s) 10

R(s) S(S-2)(s+4)

A + B + C 10

= S S-2 S + 4

C(t) = - 1.25 + 0.833e 2t

+ 0.416e – 4 t

Here there is one exponential term with positive in transient output

Therefore Css = - 1.25

t C(t)

0 0

1 + 4.91

2 + 44.23

4 +2481.88

From the above table, it is clear that output response instead of approaching to steady state value

as t due to exponential term with positive index, transients go on increasing in amplitude.

So such system is said to be unstable. In such system output is uncontrollable & unbounded one.

Output response of such system is as shown in fig(4).



C(t) C(t)

OR

Steady state

output

t

(a) fig(4) t

(b)

For such unstable systems, if input is removed output may not return to zero. And if the input

power is turned on, output tends to . If no saturation takes place in system & no mechanical

stop is provided then system may get damaged.

If all the closed loop poles or roots of the characteristic equation lies in left of s-plane, then in the

output response contains steady state terms & transient terms. Such transient terms approach to

zero as time advances eventually output reaches to equilibrium & attains steady state value.

Transient terms in such system may give oscillation but the amplitude of such oscillation will

be decreasing with time & finally will vanish. So output response of such system is shown in

fig5 (a) & (b).

C(t) C(t) Damped oscillations

Steady state

----------------------- ---------------------------

OR

Steady state output

t t

(a) fig 5 (b)

BIBO Stability : This is bounded input bounded output stability.

Definition of stable system:

A linear time invariant system is said to be stable if following conditions are satisfied.

1. When system is excited by a bounded input, output is also bounded & controllable.

2. In the absence of input, output must tend to zero irrespective of the initial conditions.

Unstable system:

A linear time invariant system is said to be unstable if,

1. for a bounded input it produces unbounded output.

2. In the absence of input, output may not be returning to zero. It shows certain output

without input.

Besides these two cases, if one or more pairs simple non repeated roots are located on the

imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output



response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such

systems are said to be critically or marginally stable systems.

Critically or Marginally stable systems:

A linear time invariant system is said to be critically or marginally stable if for a bounded

input its output oscillates with constant frequency & Amplitude. Such oscillation of output are

called Undamped or Sustained oscillations.

For such system one or more pairs of non repeated roots are located on the imaginary axis as

shown in fig6(a). Output response of such systems is as shown in fig6(b).

C(t)

Constant Amplitude & frequency oscillations

X J 2 ------------------------------------

----------------------------- steady state output

X J 1 ------------------------------------

X - J 1

X - J 2

Fig 6(a)

non repeated poles on J axis. t



If there are repeated poles located purely on imaginary axis system is said to be unstable.

C(t)

J 1 x x

-----------------------------------------

steady state output

s-plane

J 1 x x

t

Conditionally Stable:

A linear time invariant system is said to be conditionally stable, if for

a certain condition if a particular parameter of the system, its output is bounded one. Otherwise if

that condition is violated output becomes unbounded system becomes unstable. i.e. Stability of

the system depends the on condition of the parameter of the system. Such system is called

conditionally stable system.

S-plane can be divided into three zones from stability point of view.

J axis

Left half of s-plane Right half of s-plane

Real

Stable Unstable

S-Plane

(repeated)

unstable (non repeated roots)

Marginally stable



3. Complex

conjugate with

negative real

part

J

x J 1

-a1

x

-J 2

C(t)

t

Damped oscillation

Absolutely stable.

4. Complex

conjugate with

positive real

part

J 1 x

-J 1 -x

Ct

t

oscillations with

increasing amplitude

Unstable.

Sl.

No

Nature of closed

loop poles.

Location of closed loop

poles in s-plane

Step response Stability condition

1. Real negative i.e

in LHS of splane

J

x x

-02 -01

C(t)

---------------------

t

Absolutely stable

2.

Real positive in

RHS of s-plane J

x

a1

C(t)

------------------------

t

increasing towards

Unstable



5. Non repeated

pair on

imaginary axis

J

x J 1

x -J 2

OR

J

X J 2

x J 1

x –J 1

x – J 2

Two non repeated pairs on

imaginary axis.

C(t)

t

C(t)

t

Sustained oscillations

with two frequencies 1 &

2

Marginally or

critically stable

Marginally or

critically stable

6. Repeated pair

on imaginary

axis

J

x x J 1

x x -J 1

C(t)

------------------------

t

oscillation of increasing

amplitude

Unstable



Relative Stability:

The system is said to be relatively more stable or unstable on the basis of

settling time. System is said to be more stable if settling time for that system is less than that of

other system.

The settling time of the root or pair of complex conjugate roots is inversely proportional

to the real part of the roots.

Sofar the roots located near the J axis, settling time will be large. As the roots move

away from J axis i.e towards left half of the s-plane settling time becomes lesser or smaller &

system becomes more & more stable. So the relative stability improves.

J C(t)

Stable for P1

x x ----------------------------------

P2 P1

Relatively more stable for P2

t

J

C(t) stable for 1

x x

2 1 --------------------------------------------------------------

x x

more stable for 2

J axis

t

Relative stability s-plane

improves

Routh – Hurwitz Criterion :



This represents a method of determining the location of poles of a characteristics equation

with the respect to the left half & right half of the s-plane without actually solving the equation.

The T.F.of any linear closed loop system can be represented as,

C(s) b0 sm

+ b1 sm-1

+….+ bm

=

R(s) a0 sn + a1 s

n-1 + …. + an

Where ‗a‘ & ‗b‘ are constants.

To find the closed loop poles we equate F(s) =0. This equation is called as Characteristic

Equation of the system.

F(s) = a0 sn + a1 s

n-1 + a2 s

n-2 + ….. + an = 0.

Thus the roots of the characteristic equation are the closed loop poles of the system which decide

the stability of the system.

Necessary Condition to have all closed loop poles in L.H.S. of s-plane.

In order that the above characteristic equation has no root in right of s-plane, it is necessary

but not sufficient that,

1. All the coefficients off the polynomial have the same sign.

2. Non of the coefficient vanishes i.e. all powers of ‗s‘ must be present in descending order

from ‗n‘ to zero.

These conditions are not sufficient.

Hurwitz‘s Criterion :

The sufficient condition for having all roots of characteristics equation in left half of s-plane

is given by Hurwitz. It is referred as Hurwitz criterion. It states that:

The necessary & sufficient condition to have all roots of characteristic equation in left half of

s-plane is that the sub-determinants DK, K = 1, 2,………n obtained from Hurwitz determinant

‗H‘ must all be positive.



Method of forming Hurwitz determinant:

a1 a3 a5 …….. a2n-1

a0 a2 a4 ..…… a2n-2

0 a1 a3 .……. a2n-3

0 a0 a2 …….. a2n-4

H =

0 0 a1 ……... a2n-5

- - ……... -

- - - ……... -

0 - - ……. an

The order is n*n where n = order of characteristic equation. In Hurwitz determinant all

coefficients with suffices greater than ‗n‘ or negative suffices must all be replaced by zeros.

From Hurwitz determinant subdeterminants, DK, K= 1, 2, ….n must be formed as follows:

a1 a3 a5

D1 = a1 D2 = a1 a3 D3 = a 0 a2 a4 DK = H

a0 a2 0 a1 a3

For the system to be stable, all above determinants must be positive.

Determine the stability of the given characteristics equation by Hurwitz,s method.

Ex 1: F(s)= s3 + s

2 + s

1 + 4 = 0 is characteristic equation.

a0 = 1, a1 = 1, a2 = 1, a3 = 4, n = 3

a1 a3 a5 1 4 0

H = a 0 a2 a4 = 1 1 0

0 a1 a3 0 1 4

D1 = 1 = 1

1 4

D2 = 1 1 = -3



1 4 0

D3 = 1 1 0 = 4 –16 = -12.

0 1 4

As D2 & D3 are negative, given system is unstable.

Disadvantages of Hurwitz‘s method :

1. For higher order system, to solve the determinants of higher order is very complicated &

time consuming.

2. Number of roots located in right half of s-plane for unstable system cannot be judged by

this method.

3. Difficult to predict marginal stability of the system.

Due to these limitations, a new method is suggested by the scientist Routh called Routh‘s

method. It is also called Routh-Hurwitz method.

Routh‘s Stability Criterion:

It is also called Routh‘s array method or Routh-Hurwitz‘s method

Routh suggested a method of tabulating the coefficients of characteristic equation

in a particular way. Tabulation of coefficients gives an array called Routh‘s array.

Consider the general characteristic equation as,

F(s) = a0 sn + a1 s

n-1 + a2 s

n-2 + ….. + an = 0.

Method of forming an array :

Sn

a0 a2 a4 a6 ……….

Sn-1

a1

a3 a5 a7

Sn-2

b1 b2 b3

Sn-3

c1 c2 c3

- - - -

- - - -

S0

an

Coefficients of first two rows are written directly from characteristics equation.

From these two rows next rows can be obtained as follows.



a1 a2 – a0 a3 a1 a4 – a0 a5 a1 a6 – a0 a7

b1 = , b2 = , b3 =

a1 a1 a1

From 2nd

& 3rd

row , 4th row can be obtained as

b1 a3 – a1 b2 b1 a5 – a1 b3

C1 = , C2 =

b1 b1

This process is to be continued till the coefficient for s0 is obtained which will be an. From this

array stability of system can be predicted.

Routh‘s criterion :

The necessary & sufficient condition for system to be stable is ― All the terms in the first

column of Routh‘s array must have same sign. There should not be any sign change in first

column of Routh‘s array‖.

If there are sign changes existing then,

1. System is unstable.

2. The number of sign changes equals the number of roots lying in the right half of the

s-plane.

Examine the stability of given equation using Routh‘s method :

Ex.2: s3+6s

2 + 11s + 6 =0

Sol: a0 = 1, a1 = 6, a2 =11, a3 = 6, n = 3

S3

1 11

S2

6 6

S1

11 * 6 – 6 =10 0

6

S0

6

As there is no sign change in the first column, system is stable.



Ex. 3 s3 + 4s

2 + s + 16 = 0

Sol: a0 =1, a1 = 4, a2 = 1, a3 = 16

S3

1

1

S2

+4

16

S1

4 - 16

= -3

0

4

S0

+16

As there are two sign changes, system is unstable.

Number of roots located in the right half of s-plane = number of sign changes = 2.

Special Cases of Routh‟s criterion :

Special case 1 :

First element of any of the rows of Routh‘s array is zero & same remaining rows contains at

least one non-zero element.

Effect : The terms in the new row become infinite & Routh‘s test fails.

e.g. : s5 + 2s

4 + 3s

3 + 6s

2 + 2s + 1 = 0

S5

1

3

2

S4

2 6 1 Special case 1 Routh‟s array failed

S3

0 1.5 0

S2

…. …

Following two methods are used to remove above said difficulty.

First method : Substitute a small positive number ‗ ‘ in place of a zero occurred as a first

element in the row. Complete the array with this number ‗ ‘. Then

examine

lim

Sign change by taking . Consider above Example.

0



S5

1

3

2

S4

2 6 1

S3

1.5 0

S2

6 - 3 1 0

S1 1.5(6 - 3) 0

-

(6 - 3)

S0

1

To examine sign change, Lim

Lim = 6 - 3 = 6 - 3

0 0

= 6 -

= - sign is negative.

Lim 1.5(6 – 3) - 2

= Lim 9 - 4.5 - 2

0 6 -3 0 6 - 3

= 0 – 4.5 – 0

0 -3

= + 1.5 sign is positive



Routh‘s array is,

S5

1

3

2

S4

2 6 1

S3

+

1.5 0

S2 -

1 0

S1 +1.5 0 0

S0

1

0 0


Second method : To solve the above difficulty one more method can be used. In this, replace

‗s‘ by ‗1/Z‘ in original equation. Taking L.C.M. rearrange characteristic equation in descending

powers of ‗Z‘. Then complete the Routh‘s array with this new equation in ‗Z‘ & examine the

stability with this array.

Consider F(s) = s5 + 2s

4 + 3s

3 + 6s

2 + 2s + 1 = 0

Put s = 1 / Z

1 + 2 + 3 + 6 + 2 + 1 = 0

Z5 Z

4 Z

3 Z

2 Z

Z5 + 2Z

4+ 6Z

3+3Z

2+2Z+ 1 = 0

Z5 1 6 2

Z4 2 3 1

Z3 4.5 1.5 0

Z2 2.33 1 0

Z1 - 0.429 0

Z0 1


Special case 2 :



All the elements of a row in a Routh‘s array are zero.

Effect : The terms of the next row can not be determined & Routh‘s test fails.

S5

a b c

S

4 d e f

S3 0 0 0 Row of zeros, special case 2

This indicates no availability of coefficient in that row.

Procedure to eliminate this difficulty :

1. Form an equation by using the coefficients of row which is just above the row of zeros.

Such an equation is called an Auxillary equation denoted as A(s). For above case such

an equation is,

A(s) = ds4 + es

2 + f

Note that the coefficients of any row are corresponding to alternate powers of ‗s‘ starting

from the power indicated against it.

So ‗d‘ is coefficient corresponding to s4

so first term is ds4 of A(s).

Next coefficient ‗e‘ is corresponding to alternate power of ‗s‘ from 4 i.e. s2 Hence the term es

2 & so on.

2. Take the derivative of an auxillary equation with respect to ‗s‘.

i.e. dA(s)

= 4d s3 + 2e s

ds

3. Replace row of zeros by the coefficients of dA(s)

ds

S5

a b c

S

4 d e f

S3 4d 2e 0

4. Complete the array of zeros by the coefficients.

Importance of auxillary equation :



Auxillary equation is always the part of original characteristic equation. This means the roots of

the auxillary equation are some of the roots of original characteristics equation. Not only this but

roots of auxillary equation are the most dominant roots of the original characteristic equation,

from the stability point of view. The stability can be predicted from the roots of A(s)=0 rather

than the roots of characteristic equation as the roots of A(s) = 0 are the most dominant from the

stability point of view. The remaining roots of the characteristic equation are always in the left

half & they do not play any significant role in the stability analysis.

e.g. Let F(s) = 0 is the original characteristic equation of say order n = 5.

Let A(s) = 0 be the auxillary equation for the system due to occurrence of special case 2

of the order m = 2.

Then out of 5 roots of F(s) = 0, the 2 roots which are most dominant (dominant means very

close to imaginary axis or on the imaginary axis or in the right half of s-plane) from the stability

point of view are the 2roots of A(s) = 0. The remaining 5 –2 = 3 roots are not significant from

stability point of view as they will be far away from the imaginary axis in the left half of s-plane.

The roots of auxillary equation may be,

1. A pair of real roots of opposite sign i.e.as shown in the fig. 8.10 (a).

j j

x

x x

x Fig. 8. 10 (b)

Fig 8. 10(a)

2. A pair of roots located on the imaginary axis as shown in the fig. 8.10(b).

3. The non-repeated pairs of roots located on the imaginary axis as shown in the

fig.8.10 (c).

j j

x xx

x

x

x xx

Fig. 8.10(c) Fig. 8.10(d).

4. The repeated pairs of roots located on the imaginary axis as shown in the Fig.8.10 (d).



Hence total stability can be determined from the roots of A(s) = 0, which can be out of

four types shown above.

Change in criterion of stability in special case 2 :

After replacing a row of zeros by the coefficients of dA(s) , complete the Routh‘s array.

ds

But now, the criterion that, no sign in 1st column of array for stability, no longer remains

sufficient but becomes a necessary. This is because though A(s) is a part of original

characteristic equation, dA(s) is not, which is in fact used to complete the array.

ds

So if sign change occurs in first column, system is unstable with number of sign changes

equal to number of roots of characteristics equation located in right half of

s-plane.

But there is no sign changes, system cannot be predicted as stable . And in such case

stability is to be determined by actually solving A(s) = 0 for its roots. And from the location

of roots of A(s) = 0 in the s-plane the system stability must be determined. Because roots

A(s) = 0 are always dominant roots of characteristic equation.

Application of Routh‘s of criterion :

Relative stability analysis :

If it is required to find relative stability of system about a line s = - . i.e. how many roots

are located in right half of this line s = - , the Routh‘s method can be used effectively.

To determine this from Routh‘s array, shift the axis of s – plane & then apply Routh‘s

array i.e. substitute s = s 1 - , ( = constant) in characteristic equation. Write polynomial in

terms of s1. Complete array from this new equation. The number of sign changes in first

column is equal to number of roots those are located to right of the vertical line

s = - .

Imaginary j

- 0

Determining range of values of K :

In practical system, an amplifier of variable gain K is introduced .

The closed loop transfer function is

C(s) KG(s)



=

R(s) 1+ KG(s) H(s)

Hence the characteristic equation is

F(s) = 1+ KG(s) H(s) = 0

So the roots of above equation are dependent on the proper selection of value of ‗K‘.

So unknown ‗K‘ appears in the characteristic equation. In such case Routh‘s array is to be

constructed in terms of K & then the range of values of K can be obtained in such away that it

will not produce any sign change in first column of the Routh‘s array. Hence it is possible to

obtain the range of values of K for absolute stability of the system using Routh‘s criterion. Such

a system where stability depends on the condition of parameter K, is called conditionally stable

system.

Advantages of Routh‘s criterion :

Advantages of routh‘s array method are :

1. Stability of the system can be judged without actually solving the characteristic equation.

2. No evaluation of determinants, which saves calculation time.

3. For unstable system it gives number of roots of characteristic equation having

positive real part.

4. Relative stability of the system can be easily judged.

5. By using the criterion, critical value of system gain can be determined hence

frequency of sustained oscillations can be determined.

6. It helps in finding out range of values of K for system stability.

7. It helps in finding out intersection points of roots locus with imaginary axis.

Limitation of Routh‘s criterion :

1. It is valid only for real coefficients of the characteristic equation.

2. It does not provide exact locations of the closed loop poles in left or right half of s-plane.

3. It does not suggest methods of stabilizing an unstable system.

4. Applicable only to linear system.

Ex.1. s6 + 4s

5 +3s

4 – 16s

2- 64s – 48 = 0 Find the number of roots of this equation with positive

real part, zero real part & negative real part

Sol: S6 1 3 -16 -48

S5 4 0 -64 0

S4 3 0 -48 0

S3 0 0 0



dA

A(s) = 3S4 – 48 = 0 = 12s

3

ds

S6 1 3 -16 -48

S5 4 0 -64 0

S4 3 0 -48 0

S3 12 0 0 0

S2 ( )0 -48 0 0

S1 576 0 0 0

S

0 -48

Lim 576

0 +

Therefore One sign change & system is unstable. Thus there is one root in R.H.S of the s – plane

i.e. with positive real part. Now

solve A(s) = 0 for the dominant roots

A(s) = 3s4 – 48 =0

Put S2 = Y

3Y2 = 48 Y

2 =16, Y = 16 = 4

S2 = + 4 S

2 = -4

S = 2 S = 2j

So S = 2j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S. indicated

by a sign change is S = 2 as obtained by solving A(s) = 0. Total there are 6 roots as n = 6.

Roots with Positive real part = 1

Roots with zero real part = 2

Roots with negative real part = 6 –2 – 1 = 3



Ex.2 : For unity feed back system,

k

G(s) = , Find range of values of K, marginal value of K

S(1 + 0.4s) ( 1 + 0.25 s) & frequency of sustained oscillations.

Sol : Characteristic equation, 1 + G (s) H (s) = 0 & H(s) = 1

K

1 + = 0

s(1 + 0.4s) ( 1 + 0.25s)

s [ 1 + 0.65s + 0.1s2} + K = 0

0.1s3 + 0.65s

2 +s + K = 0

S3

0.1 1 From s0, K > 0

S2

0.65 K from s1,

S1 0.65 – 0.1K 0 0.65 – 0.1K > 0

0.65 0.65 > 0.1 K

S0

K 6.5 > K

Range of values of K, 0 < K < 6.5

Now marginal value of ‗K‘ is that value of ‗K‘ for which system becomes marginally stable. For

a marginal stable system there must be row of zeros occurring in Routh‘s array. So value of ‗K‘

which makes any row of Routh array as row of zeros is called marginal value of ‗K‘. Now K = 0

makes row of s0 as row of zeros but K = 0 can not be marginal value because for K = 0, constant

term in characteristic equation becomes zeros ie one coefficient for s0 vanishes which makes

system unstable instead of marginally stable.

Hence marginal value of ‗K‘ is a value which makes any row other than s0 as row of zeros.

0.65 – 0.1 K mar = 0

K mar = 6.5

To find frequency, find out roots of auxiliary equation at marginal value of ‗K‘

A(s) = 0.65 s2 + K = 0 ;

0.65 s2 + 6.5 = 0 Because K = 6.5

s2 = -10

s = j 3.162



comparing with s = j

= frequency of oscillations = 3.162 rad/ sec.

Ex : 3 For a system with characteristic equation

F(s) = s5 + s4 + 2s

3 + 2s

2 + 3s +15 = , examine the stability

Solution :

S5 1 2 3

S4 1 2 15

S3 0 -12 0

S2

S1

S0

S5 1 2 3

S4 1 2 15

S3 -12 0

S2 (2 + 12) 15 0

S1

(2 + 12)( -12 ) – 15 0

2 + 12

S

0 15

Lim 2 + 12 12

0 = 2 + = 2 + = +

Lim (2 + 12)( -12 ) – 15 Lim -24 - 144 – 15 2

0 = 0 2 + 12

2 + 12

0 – 144 - 0

= = - 12

0 + 12



S5 1 2 3

S4 1 2 15

S3 -12 0

There are two sign changes, so

system is unstable.

S2 + 15 0

S1

- 12 0

S0

15

Ex : 4 Using Routh Criterion, investigate the stability of a unity feedback system

whose open loop transfer function is

e -sT

G(s) =

s ( s + 1 )

Sol : The characteristic equation is

1 + G(s) H(s) = 0

e -sT

1 + = 0

s ( s + 1 )

s2 + s + e

–sT = 0

Now e – sT

can be Expressed in the series form as

s2 T

2

e –sT

= 1 – sT + + ……

2!

Trancating the series & considering only first two terms we get

esT

= 1 – sT

s2 + s + 1 – sT = 0

s2 + s ( 1- T ) + 1 = 0



So routh‘s array is

S2 1 1

S 1-T 0

S0 1

1 – T > 0 for stability

T < 1

This is the required condition for stability of the system.

Ex : 5 Determine the location of roots with respect to s = -2 given that

F(s) = s4 + 10 s

3 + 36s

2 + 70s + 75

Sol : shift the origin with respect to s = -2

s = s1 – 2

(s – 2 ) 4 + 10 (s – 2)

3 + 36(s – 2 )

2 + 70 ( s –2) + 75 = 0

s4 + 2s

3 + 0s

2 + 14s + 15 = 0

S4 1 0 15

S3 2 14 0

S2 -7 15 0

S1

18.28 0 0

S 0

15

Two sign change, there are two roots to the right of s = -2 & remaining ‗2‘ are to the left of

the line s = -2. Hence the system is unstable.




1. Explain briefly how system depends on poles and zeros.

2. Mention the necessary condition to have all closed loop poles in LHS of S-Plane

3. Explain briefly the Hurwitz‘s Criterion.

4. Explain briefly the Routh‘s Stability Criterion.

5..Examine the stability of given equation using Routh‘s method

s3+6s

2 + 11s + 6 =0

6. Examine the stability of given equation using Routh‘s method

s5 + 2s

4 + 3s

3 + 6s

2 + 2s + 1 = 0

7. Using Routh Criterion, investigate the stability of a unity feedback system whose open loop

transfer function is

e -sT

G(s) =

s ( s + 1 )



UNIT-5

Root–Locus Techniques

The characteristics of the transient response of a closed loop control system are related to

location of the closed loop poles. If the system has a variable loop gain, then the location of the

closed loop poles depends on the value of the loop gain chosen. It is important, that the designer

knows how the closed loop poles move in the s-plane as the loop gain is varied. W. R. Evans

introduced a graphical method for finding the roots of the characteristic equation known as root

locus method. The root locus is used to study the location of the poles of the closed loop transfer

function of a given linear system as a function of its parameters, usually a loop gain, given its

open loop transfer function. The roots corresponding to a particular value of the system

parameter can then be located on the locus or the value of the parameter for a desired root

location can be determined from the locus. It is a powerful technique, as an approximate root

locus sketch can be made quickly and the designer can visualize the effects of varying system

parameters on root locations or vice versa. It is applicable for single loop as well as multiple loop

system.

ROOT LOCUS CONCEPT

To understand the concepts underlying the root locus technique, consider the second

order system shown in Fig. 1.

Fig. 3 Second order control system

The open loop transfer function of this system is

)1(a)s(s

KG(s)

Where, K and a are constants. The open loop transfer function has two poles one at origin s = 0

and the other at s = -a. The closed loop transfer function of the system shown in Fig.1 is

(2)Kass

K

G(s)H(s)1

G(s)

R(s)

C(s)2

The characteristic equation for the closed loop system is obtained by setting the

denominator of the right hand side of Eqn.(2) equal to zero. That is,

(3)0 KassG(s)H(s)1 2

R(s) E(s)

−

a)s(s

K

C(s)



The second order system under consideration is always stable for positive values of a and

K but its dynamic behavior is controlled by the roots of Eqn.(3) and hence, in turn by the

magnitudes of a and K, since the roots are given by

(4)K2

a

2

a

2a

4K)(a

2

as,s

22

21

From Eqn.(4), it is seen that as the system parameters a or K varies, the roots change.

Consider a to be constant and gain K to be variable. As K is varied from zero to infinity, the two

roots s1 and s2 describe loci in the s-plane. Root locations for various ranges of K are:

1) K= 0, the two roots are real and coincide with open loop poles of the system s1 =

0, s2 = -a.

2) 0 K < a2/4, the roots are real and distinct.

3) K= a2/4, roots are real and equal.

4) a2/4 < K < , the rots are complex conjugates.

The root locus plot is shown in Fig.2

Fig. 4 Root loci of s

2+as+K as a function of K

Figure 2 has been drawn by the direct solution of the characteristic equation. This

procedure becomes tedious. Evans graphical procedure helps in sketching the root locus quickly.

The characteristic equation of any system is given by

(5)0Δ(s)

Where, (s) is the determinant of the signal flow graph of the system given by Eqn.(5).




Or



(6)PPP1Δ(s)m

m3m

m2m

m1

Where, Pmr is gain product of mth

possible combination of r nontouching loops of the graph.

The characteristic equation can be written in the form

(7)0B(s)

KA(s)1

0P(s)1

For single loop system shown in Fig.3

(8)G(s)H(s)P(s)

Where, G(s)H(s) is open loop transfer function in block diagram terminology or transmittance in

signal flow graph terminology.

Fig. 5 Single loop feedback system

From Eqn.(7) it can be seen that the roots of the characteristic equation (closed loop

poles)occur only for those values of s where

(9)1P(s)

Since, s is a complex variable, Eqn.(9) can be converted into the two Evans conditions

given below.

)10(1)(sP

)11(2,1,0);12(180)( qqsP

Roots of 1+P(s) = 0 are those values of s at which the magnitude and angle condition

given by Eqn.(10) and Eqn.(11). A plot of points in the complex plane satisfying the angle

criterion is the root locus. The value of gain corresponding to a root can be determined from the

magnitude criterion.

To make the root locus sketching certain rules have been developed which helps in

visualizing the effects of variation of system gain K ( K > 0 corresponds to the negative feed

−

R(s) E(s)

G(s) C(s)

H(s)



back and K < 0 corresponds to positive feedback control system) and the effects of shifting

pole-zero locations and adding in anew set of poles and zeros.

GENERAL RULES FOR CONSTRUCTING ROOT LOCUS

1) The root locus is symmetrical about real axis. The roots of the characteristic equation are

either real or complex conjugate or combination of both. Therefore their locus must be

symmetrical about the real axis.

2) As K increases from zero to infinity, each branch of the root locus originates from an

open loop pole (n nos.) with K= 0 and terminates either on an open loop zero (m nos.)

with K = along the asymptotes or on infinity (zero at ). The number of branches

terminating on infinity is equal to (n – m).

3) Determine the root locus on the real axis. Root loci on the real axis are determined by

open loop poles and zeros lying on it. In constructing the root loci on the real axis choose

a test point on it. If the total number of real poles and real zeros to the right of this point

is odd, then the point lies on root locus. The complex conjugate poles and zeros of the

open loop transfer function have no effect on the location of the root loci on the real axis.

4) Determine the asymptotes of root loci. The root loci for very large values of s must be

asymptotic to straight lines whose angles are given by

)12(1-mn0,1,2,;mn

1)(2q180asymptotesofAngle

A

q



5) All the asymptotes intersect on the real axis. It is denoted by a , given by

)13(mn

)zz(z)pp(p

mn

zerosofsumpolesofsumσ

m21n21

a

6) Find breakaway and breakin points. The breakaway and breakin points either lie on the

real axis or occur in complex conjugate pairs. On real axis, breakaway points exist

between two adjacent poles and breakin in points exist between two adjacent zeros. To

calculate these polynomial 0ds

dK must be solved. The resulting roots are the breakaway

/ breakin points. The characteristic equation given by Eqn.(7), can be rearranged as

)z(s)z)(szK(sA(s)

and )p(s)p)(sp(s B(s) where,

(14)0KA(s)B(s)

m21

n21

The breakaway and breakin points are given by

)15(0Bds

dABA

ds

d

ds

dK

Note that the breakaway points and breakin points must be the roots of Eqn.(15), but

not all roots of Eqn.(15) are breakaway or breakin points. If the root is not on the root

locus portion of the real axis, then this root neither corresponds to breakaway or breakin

point. If the roots of Eqn.(15) are complex conjugate pair, to ascertain that they lie on

root loci, check the corresponding K value. If K is positive, then root is a breakaway or

breakin point.

7) Determine the angle of departure of the root locus from a complex pole

)16()zerosotherfromquestioninpolecomplexatovectorsofanglesof(sum

poles)otherfromquestioninpolecomplexatovectorsofanglesof(sum

180pcomplexafromdepartureofAngle

8) Determine the angle of arrival of the root locus at a complex zero

(17)poles)otherfromquestioninzerocomplexatovectorsofanglesof(sum

zeros)otherfromquestioninzerocomplexatovectorsofanglesof(sum

180zerocomplexatarrivalofAngle



9) Find the points where the root loci may cross the imaginary axis. The points where the

root loci intersect the j axis can be found by

a) use of Routh‘s stability criterion or

b) letting s = j in the characteristic equation , equating both the real part and

imaginary part to zero, and solving for and K. The values of thus found give

the frequencies at which root loci cross the imaginary axis. The corresponding K

value is the gain at each crossing frequency.

10) The value of K corresponding to any point s on a root locus can be obtained using the

magnitude condition, or

)18(zerostopointsbetweenlengthofproduct

polestopointsbetweenlengthsofproductK

PHASE MARGIN AND GAIN MARGIN OF ROOT LOCUS

Gain Margin

It is a factor by which the design value of the gain can be multiplied before the closed

loop system becomes unstable.

(19)KofvalueDesign

overcrossimaginaryatKofValueMarginGain

The Phase Margin

Find the point j 1 on the imaginary axis for which 1jHjG for the design value

of K i.e. design

Kj/AjB .

The phase margin is

(20))H(jωjωargG180φ11



Problem No 1

Sketch the root locus of a unity negative feedback system whose forward path transfer function

is s

KG(s) .

Solution:

1) Root locus is symmetrical about real axis.

2) There are no open loop zeros(m = 0). Open loop pole is at s = 0 (n = 1). One branch of

root locus starts from the open loop pole when K = 0 and goes to asymptotically when

K .

3) Root locus lies on the entire negative real axis as there is one pole towards right of any

point on the negative real axis.

4) The asymptote angle is A = .01,)12(180

mnqmn

q

Angle of asymptote is A = 180 .

5) Centroid of the asymptote is mn

zeros)of(sumpoles)of(sumσA

0.01

0

6) The root locus does not branch. Hence, there is no need to calculate the break points.

7) The root locus departs at an angle of -180 from the open loop pole at s = 0.

8) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross

over.




Figure 6 Root locus plot of K/s

Comments on stability:

The system is stable for all the values of K > 0. Th system is over damped.

Problem No 2

The open loop transfer function is 21)(s

2)K(sG(s) . Sketch the root locus plot

Solution:


2) There is one open loop zero at s=-2.0(m=1). There are two open loop poles at

s=-1, -1(n=2). Two branches of root loci start from the open loop pole when

K= 0. One branch goes to open loop zero at s =-2.0 when K and other goes to

(open loop zero ) asymptotically when K .

3) Root locus lies on negative real axis for s ≤ -2.0 as the number of open loop poles plus

number of open loop zeros to the right of s=-0.2 are odd in number.

4) The asymptote angle is A = .01,)12(180

mnqmn

q

Angle of asymptote is A = 180 .



0.01

)2()11(



6) The root locus has break points.

The root loci brakesout at the open loop poles at s=-1, when K =0 and breaks in onto the

real axis at s=-3, when K=4. One branch goes to open loop zero at s=-2 and other goes to

along the asymptotically.

7) The branches of the root locus at s=-1, -1 break at K=0 and are tangential to a line s=-

1+j0 hence depart at 90 .

8) The locus arrives at open loop zero at 180 .

9) The root locus does not cross the imaginary axis, hence there is no need to find the

imaginary axis cross over.

The root locus plot is shown in Fig.2.

Figure 7 Root locus plot of K(s+2)/(s+1)

2


System is stable for all values of K > 0. The system is over damped for K > 4. It is critically

damped at K = 0, 4.

4K3,s0;K1,s

02)(s

1)(s2)1)(s2(s

0ds

dKbygivenispointBreak

2)(s

1)(sK

21

2

2

2



Problem No 3

The open loop transfer function is 2)s(s

4)K(sG(s) . Sketch the root locus.

Solution:


2) There are is one open loop zero at s=-4(m=1). There are two open loop poles at s=0, -

2(n=2). Two branches of root loci start from the open loop poles when K= 0. One branch

goes to open loop zero when K and other goes to infinity asymptotically when K

.

3) Entire negative real axis except the segment between s=-4 to s=-2 lies on the root locus.

4) The asymptote angle is A = .01,1,0,)12(180

mnqmn

q

Angle of asymptote are A = 180 .



0.21

)4()2(

6) The brake points are given by dK/ds =0.

7) Angle of departure from open loop pole at s =0 is 180 . Angle of departure from pole at

s=-2.0 is 0 .

8) The angle of arrival at open loop zero at s=-4 is 180

9) The root locus does not cross the imaginary axis. Hence there is no imaginary cross over.

The root locus plot is shown in fig.3.

11.7K6.828,s

0.343;K1.172,s

04)(s

2s)(s4)2)(s(2s

ds

dK

4)(s

2)s(sK

2

1

2

2



Figure 3 Root locus plot of K(s+4)/s(s+2)


System is stable for all values of K.

0 > K > 0.343 : > 1 over damped

K = 0.343 : = 1 critically damped

0.343 > K > 11.7 : < 1 under damped

K = 11.7 : = 1 critically damped

K > 11.7 : >1 over damped.

Problem No 4

The open loop transfer function is 3.6)(ss

0.2)K(sG(s)

2. Sketch the root locus.

Solution:


2) There is one open loop zero at s = -0.2(m=1). There are three open loop poles at

s = 0, 0, -3.6(n=3). Three branches of root loci start from the three open loop poles when

K= 0 and one branch goes to open loop zero at s = -0.2 when K and other two go to

asymptotically when K .

3) Root locus lies on negative real axis between -3.6 to -0.2 as the number of open loop

poles plus open zeros to the right of any point on the real axis in this range is odd.

4) The asymptote angle is A = 1,01,)12(180

mnqmn

q



Angle of asymptote are A = 90 , 270 .



7.12

)2.0()6.3(

6) The root locus does branch out, which are given by dK/ds =0.

The root loci brakeout at the open loop poles at s = 0, when K =0 and breakin onto the

real axis at s=-0.432, when K=2.55 One branch goes to open loop zero at s=-0.2 and

other goes breaksout with the another locus starting from open loop ploe at s= -3.6. The

break point is at s=-1.67 with K=3.66. The loci go to infinity in the complex plane with

constant real part s= -1.67.

7) The branches of the root locus at s=0,0 break at K=0 and are tangential to imaginary axis

or depart at 90 . The locus departs from open loop pole at s=-3.6 at 0 .

8) The locus arrives at open loop zero at s=-0.2 at 180 .

9) The root locus does not cross the imaginary axis, hence there is no imaginary axis cross

over.


ly.respective 3.662.55,0,Kand1.670.432,0,s

01.44s4.8s2s

0.2)(s

)3.6s(s0.2)7.2s)(s(3s

ds

dK

0.2s

)3.6s(s-K

23

2

232

23



Figure 4 Root locus plot of K(s+0.2)/s

2(s+3.6)


System is stable for all values of K. System is critically damped at K= 2.55, 3.66. It is under

damped for 2.55 > K > 0 and K >3.66. It is over damped for 3.66 > K >2.55.

Problem No 5

The open loop transfer function is 25)6ss(s

KG(s) . Sketch the root locus.

Solution:


2) There are no open loop zeros (m=0). There are three open loop poles at s=-0,

-3 j4(n=3). Three branches of root loci start from the open loop poles when K= 0 and all

the three branches go asymptotically when K .

3) Entire negative real axis lies on the root locus as there is a single pole at s=0 on the real

axis.

4) The asymptote angle is A = .2,1,01,1,0,)12(180

mnqmn

q

Angle of asymptote are A = 60 , 180 , 300 .



0.23

)33(



6) The brake points are given by dK/ds =0.

j18.0434K

j2.0817and2s

02512s3sds

dK

25s)6s(s25)6ss(sK

1,2

1,2

2

232

For a point to be break point, the corresponding value of K is a real number greater than

or equal to zero. Hence, S1,2 are not break points.

7) Angle of departure from the open loop pole at s=0 is 180 . Angle of departure from

complex pole s= -3+j4 is

zeros) from inquestion polecomplex a to vectorsof angles theof (sum

poles)other fromquestion in polecomplex a to vectorsof angles theof (sum

180p

87.36)903

4tan180(180 1

p

Similarly, Angle of departure from complex pole s= -3-j4 is 36.87or323.13)270(233.13180φp

8) The root locus does cross the imaginary axis. The cross over point and the gain at the

cross over can be obtained by

Rouths criterion

The characteristic equation is 0K25s6ss 23 . The Routh‘s array is

For the system to be stable K < 150. At K=150 the auxillary equation is 6s2+150=0.

s = ±j5.

or

substitute s= j in the characteristic equation. Equate real and imaginary parts to zero. Solve

for and K.

The plot of root locus is shown in Fig.5.

6s6

K150s

K6s

251s

0

1

2

3

1500,Kj50,ω

025ωjωK)6ω(

0Kjω25jω6jω

0K25s6ss

22

23

23



Figure 5 Root locus plot of K/s(s

2+6s+25)


System is stable for all values of 150 > K > 0. At K=150, it has sustained oscillation of 5rad/sec.

The system is unstable for K >150.

Problem No 1

Sketch the root locus of a unity negative feedback system whose forward path transfer function

is j)3j)(s31)(s(s

2)K(sG(s)H(s) . Comment on the stability of the system.

Solution:


10) There is one open loop zero at s = -2 (m = 1). There are three open loop poles at

s = -1, -3 ± j (n=3). All the three branches of root locus start from the open loop

poles when K = 0. One locus starting from s = -1 goes to zero at s = -2 when

K , and other two branches go to asymptotically (zeros at ) when K .

11) Root locus lies on the negative real axis in the range s=-1 to s= -2 as there is one pole to

the right of any point s on the real axis in this range.

12) The asymptote angle is A = .0,11mnq,mn

1)(2q180

Angle of asymptote is A = 90 , 270 .





2.51

2)(3)31(

14) The root locus does not branch. Hence, there is no need to calculate break points.

15) The angle of departure at real pole at s=-1 is 180 . The angle of departure at the complex

pole at s=-3+j is 71.57 .



180p

The angle of departure at the complex pole at s=-3-j is -71.57 .

16) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross

over.


57.71135)90(153.43180

900

2tanθ,135or -45

1-

1tan

153.43)atan2(-2,1θ

153.43or 57.262-

1tanθ

p

1

3

1

1

1

1

57.71522)270(206.57180p



Figure 1 Root locus plot of K(s+2)/(s+1)(s+3+j)(s+3-j)


The system is stable for all the values of K > 0.

Problem No 2

The open loop transfer function is10)0.6s0.5)(ss(s

KG(s)H(s)

2 Sketch the root locus

plot. Comment on the stability of the system. .

Solution:


11) There are no open loop zeros (m=0). There are four open loop poles (n=4) at s=0,

-0.5, -0.3 ± j3.1480. Four branches of root loci start from the four open loop poles when

K= 0 and go to (open loop zero at infinity) asymptotically when K .

12) Root locus lies on negative real axis between s = 0 to s = -0.5 as there is one pole to the

right of any point s on the real axis in this range.

13) The asymptote angle is A = .3,2,1,01,)12(180

mnqmn

q

Angle of asymptote is A = 45 , 135 , 225 , ±315 .





275.04

)3.03.05.0(

The value of K at s=-0.275 is 0.6137.

15) The root locus has break points.

K = -s(s+0.5)(s2+0.6s+10) = -(s

4+1.1s

3+10.3s

2+5s)

Break points are given by dK/ds = 0

0520.6s3.3s4sds

dK 23

s= -0.2497, -0.2877 j 2.2189

There is only one break point at -0.2497. Value of K at s = -0.2497 is 0.6195.

16) The angle of departure at real pole at s=0 is 180 and at s=-0.5 is 0 . The angle of

departure at the complex pole at s = -0.3 + j3.148 is -91.8



180p

The angle of departure at the complex pole at s = -0.3 - j3.148 is 91.8

17) The root locus does cross the imaginary axis, The cross over frequency and gain is

obtained from Routh‘s criterion.

The characteristic equation is

s(s+0.5)(s2+0.6s+10)+K =0 or s4+1.1s3+10.3s2+5s+K=0

8.91)9086.4(95.4180

900

6.296tanθ, 4.68

0.2

3.148tan

4.95or6.840.3-

3.148tanθ

p

1

3

1

2

1

1

91.8 )270273.6(264.6180p



The Routh‘s array is

The system is stable if 0 < K < 26.13

The auxiliary equation at K 26.13 is 5.75s2+26.13 = 0 which gives s = ± j2.13 at

imaginary axis crossover.


Figure 8 Root locus plot of K/s(s+0.5)(s

2+0.6s+10)


System is stable for all values of 26.13 >K > 0. The system has sustained oscillat ion at =

2.13 rad/sec at K=26.13. The system is unstable for K > 26.13.

Problem No 3

The open loop transfer function is 20)4s)(s4s(s

KG(s)

2. Sketch the root locus.

Ks5.75

1.1K-28.75s

K5.75s

51.1s

K10.31s

0

1

2

3

4



Solution:


11) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = -0, -4, -

2 j4. Three branches of root loci start from the three open loop poles when K= 0 and to

infinity asymptotically when K .

12) Root locus lies on negative real axis between s = 0 to s = -4.0 as there is one pole to the

right of any point s on the real axis in this range.

13) The asymptote angle is A = 3,2,1,01,)12(180

mnqmn

q

Angle of asymptote are A = 45 , 135 , 225 , 315 .



0.24

)0.40.20.2(

15) The root locus does branch out, which are given by dK/ds =0.

The root loci brakeout at the open loop poles at s = -2.0, when K = 64 and breakin and

breakout at s=-2+j2.45, when K=100

16) The angle of departure at real pole at s=0 is 180 and at s=-4 is 0 . The angle of

departure at the complex pole at s = -2 + j4 is -90 .

100Kj2.45,2.0s

64;K2.0,s

40)16s2)(4s(s

08040s32s16s8s4s

08072s24s4s

0ds

dKbygivenispointBreak

80s)36s8s(s

20)4s4)(ss(sK

2

1

2

223

23

234

2





180p

The angle of departure at the complex pole at s = -2 – j4 is 90

17) The root locus does cross the imaginary axis, The cross over point and gain at cross over

is obtained by either Routh‘s array or substitute s= j in the characteristic equation and

solve for and gain K by equating the real and imaginary parts to zero.

Routh‟s array

The characteristic equation is 0K80s36s8ss 234

For the system to be stable K > 0 and 2080-8K > 0. The imaginary crossover is given by

2080-8K=0 or K = 260.

At K = 260, the auxiliary equation is 26s2+260 = 0. The imaginary cross over occurs at s=

j 10.

or

Ks26

8K2080s

K26s

808s

K361s

isarrayRouthsThe

0

1

2

3

4

260K0K36ωω

10js;10j0,ω080ω8ω

zerotopartsimaginaryandrealEquate

080ω8ωjK36ωω

0Kjω80jω36jω8jω

jωsput

0K80s36s8ss

24

3

324

234

234

90)90.463(116.6-180

900

8tanθ,4.36

2

4tanθ

116.6)atan2(4,-2θ

116.6or63.42-

4tanθ

p

1

3

1

2

1

1

1

90270

)270296.6(243.4-180p




Figure 9 Root locus plot of K/s(s+4)(s

2+4s+20)


For 260 > K > 0 system is stable

K = 260 system has stained oscillations of 10 rad/sec.

K > 260 system is unstable.




1. Give the general rules for constructing root locus.

2. Define Phase margin and Gain margin of root locus.

3. Sketch the root locus of a unity negative feedback system whose forward path transfer

function is s

KG(s) .

4. The open loop transfer function is 21)(s

2)K(sG(s) . Sketch the root locus plot.

5. The open loop transfer function is 2)s(s

4)K(sG(s) . Sketch the root locus.

6. The open loop transfer function is 25)6ss(s

KG(s) . Sketch the root locus.

7. The open loop transfer function is 20)4s)(s4s(s

KG(s)

2. Sketch the root



UNIT- 6

Stability in the frequency domain

Introduction

Frequency response of a control system refers to the steady state response of a system subject to

sinusoidal input of fixed (constant) amplitude but frequency varying over a specific range,

usually from 0 to ∞. For linear systems the frequency of input and output signal remains the

same, while the ratio of magnitude of output signal to the input signal and phase between two

signals may change. Frequency response analysis is a complimentary method to time domain

analysis (step and ramp input analysis). It deals with only steady state and measurements are

taken when transients have disappeared. Hence frequency response tests are not generally carried

out for systems with large time constants.

The frequency response information can be obtained either by analytical methods or by

experimental methods, if the system exits. The concept and procedure is illustrated in Figure 6.1

(a) in which a linear system is subjected to a sinusoidal input. I(t) = a Sin t and the

corresponding output is O(t) = b Sin ( t + ) as shown in Figure 6.1 (b).

Figure 6.1 (a) Figure 6.1 (b)

The following quantities are very important in frequency response analysis.

M ( ) = b/a = ratio of amplitudes = Magnitude ratio or Magnification factor or gain.

( ) = = phase shift or phase angle



These factors when plotted in polar co-ordinates give polar plot, or when plotted in rectangular

co-ordinates give rectangular plot which depict the frequency response characteristics of a

system over entire frequency range in a single plot.

Frequency Response Data

The following procedure can be adopted in obtaining data analytically for frequency response

analysis.

1. Obtain the transfer function of the system

)(

)()(

SI

SOSF , Where F (S) is transfer function, O(S) and I(S) are the Laplace transforms

of the output and input respectively.

2. Replace S by (j ) (As S is a complex number)

)(

)()(

jI

jOjF

)()()(

)(jBA

jI

jO (another complex number)

3. For various values of , ranging from 0 to ∞ determine M ( ) and .

jBAjBAjI

jOM )()(

)(

)()(

22 BAM

jBAjI

jO

)(

)(

A

B1tan



4. Plot the results from step 3 in polar co-ordinates or rectangular co-ordinates. These plots are

not only convenient means for presenting frequency response data but are also serve as a

basis for analytical and design methods.

Comparison between Time Domain and Frequency Domain Analysis

An interesting and revealing comparison of frequency and time domain approaches is based on

the relative stability studies of feedback systems. The Routh‘s criterion is a time domain

approach which establishes with relative ease the stability of a system, but its adoption to

determine the relative stability is involved and requires repeated application of the criterion. The

Root Locus method is a very powerful time domain approach as it reveals not only stability but

also the actual time response of the system. On the other hand, the Nyquist criterion (discussed

later in this Chapter) is a powerful frequency domain method of extracting the information

regarding stability as well as relative stability of a system without the need to evaluate roots of

the characteristic equation.

Graphical Methods to Represent Frequency Response Data

Two graphical techniques are used to represent the frequency response data. They are: 1) Polar

plots 2) Rectangular plots.

Polar Plot

The frequency response data namely magnitude ratio M( ) and phase angle ( ) when

represented in polar co-ordinates polar plots are obtained. The plot is plotted in complex plane

shown in Figure 6.2. It is also called Nyquist plot. As is varied the magnitude and phase angle

change and if the magnitude ratio M is plotted for varying phase angles, the locus obtained gives



the polar plot. It is easier to construct a polar plot and ready information of magnitude ratio and

phase angle can be obtained.

Figure 6.2: Complex Plane Representation

A typical polar plot is shown in Figure 6.3 in which the magnitude ratio M and phase angle at a

given value of can be readily obtained.

Figure 6.3: A Typical Polar Plot

M

Real

Img

0, + 360, -360

+90, -270

+180, -180

+270, -90

Positive angles

Negative angles



Rectangular Plot

The frequency response data namely magnitude ratio M( ) and phase angle ( ) can also be

presented in rectangular co-ordinates and then the plots are referred as Bode plots which will be

discussed in Chapter 7.

Illustrations on Polar Plots: Following examples illustrate the procedure followed in obtaining

the polar plots.

Illustration 1: A first order mechanical system is subjected to a input x(t). Obtain the polar plot,

if the time constant of the system is 0.1 sec.

x (t) (input)

y (t) (output)

C

·

K

Governing Differential Equation:

KxKydt

dyC. ÷ by K

xydt

dy

K

C.

xydt

dy. Take Laplace transform

τ SY(S) + Y(S) = X(S)

(τS+1) Y(S) = X(S)



Transfer function F(S) = 1

1

)(

)(

SSX

SY

Given τ = 0.1 sec

SSSX

SY

1.01

1

11.0

1

)(

)(

To obtain the polar plot (i.e., frequency response data) replace S by j .

)1.0(1

1

11.0

1

)(

)(

jjjX

jY

Magnification Factor M = )1.0(1

)0(1

)1.0(1

1

)(

)(

j

j

jjX

jY

2)1.0(1

1M

Phase angle = )1.01()0(1)()(

)(jXjjY

jX

jY

)1.0(tan1

0tan 11

)1.0(tan 1

Now obtain the values of M and for different values of ranging from 0 to ∞ as given in Table

6.1.

Table 6.1 Frequency Response Data

2)1.0(1

1M

)1.0(tan 1

0 1.00 0

2 0.98 -11.13

4 0.928 -21.8

5 0.89 -26.6

6 0.86 -30.9

10 0.707 -45

20 0.45 -63.4



40 0.24 -76

50 0.196 -78.69

100 0.099 -84.29

∞ 0 -90

The data from Table 6.1 when plotted on the complex plane with as a parameter polar plot is

obtained as given below.

Illustration 2: A second order system has a natural frequency of 10 rad/sec and a damping ratio

of 0.5. Sketch the polar plot for the system.

The transfer function of the system is given by

· · · ·

· · = 6

= 50

= 20

m

x (Input)

y (Response)

C

K



22

2

2)(

)(

nn

n

SSSX

SY Given n = 10 rad/sec and = 0.5

10010

100

)(

)(2 SSSX

SY

Replace S by j

10010

)0(100

10010)(

100

)(

)(22 j

j

jjjX

jY as 1j

222

2

2)10()100(

100

)10()100(

)0(100

)(

)(

j

j

jX

jYM

Magnification Factor = 222 )10()100(

100M

Phase angle = )10()100()0(100)10()100(

)0(100 2

2jj

j

j

2

11

100

10tan

100

0tan

Now obtain the values of M and for various value of ranging from 0 to ∞ as given in the

following Table 6.2.

Table 6.2 Frequency Response Data: Illustration 2

M( )

0 1.00 0.0

2 1.02 -11.8

5 1.11 -33.7

8 1.14 -65.8

10 1.00 -90.0

12 0.78 -110.1

15 0.51 -129.8

20 0.28 -146.3

40 0.06 -165.1

70 0.02 -171.7

∞ 0.00 -180.0



The data from Table 6.2 when plotted on the complex plane with as a parameter polar plot is

obtained as given below.

Note: The polar plot intersects the imaginary axis at a frequency equal to the natural frequency of

the system = n = 10 rad/sec.

Illustration 3: Obtain the polar plot for the transfer function

)1(

10)(

SSF Replace S by j

1

10)(

jjF

1

10)()(

2jFM

( ) = F (j ) = 10 - (j +1)

1tan 1

Table 6.3 Frequency Response Data: Illustration 3



M( )

0 1.00 0

0.2 9.8 -11

0.4 9.3 -21

0.6 8.6 -31

0.8 7.8 -39

2.0 4.5 -63

3.0 3.2 -72

4.0 2.5 -76

5.0 1.9 -79

10 0.99 -84

∞ 0.00 -90

Polar plot for Illustration 3

Guidelines to Sketch Polar Plots

Polar plots for some typical transfer function can be sketched on the following guidelines.

I )()(

))((SFjBA

jfe

jdcjba --- (Transfer function)

Magnitude Ratio = 22

2222 *

fe

dcba

jfe

jdcjbaM

Phase angle:

= ∞

M( ) = 0



)()()( jfejdcjbajfe

jdcjba

e

f

c

d

a

b 111 tantantan

II Values of tan functions

IQ: tan-1

(b/a) – Positive:

IIQ: tan-1

(b/-a) – Negative: 180 -

IIIQ: tan-1

(-b/-a) – Positive: 180 +

IVQ: tan-1

(-b/a) – Negative: 360 -

tan-1

(0) = 00 tan

-1 (-0) =180

0

tan-1

(1) = 450 tan

-1 (-1) = 135

0

tan-1

(∞) = 900 tan

-1 (-∞) = 270

0

III Let K = Constant

= K + j(0)

Therefore KKK 22 0 Applicable for both K>0 and K<0

Real

Img IQ

IIQ

IIIQ IVQ

(a+jb)

(-a-jb) (a-jb)

(-a+jb)



K = (K+ j0)

= tan-1

(0/K)

= tan-1

(0) = 0, if K>0

= tan-1

(-0) = 1800, if K<0

IV: Sn = (j )

n = (0+j )

n

= (0+j ) (0+j ) . . . . . . . . . . . . . n times

a. Magnitude

)0()0()( jjjS nn . . . . . . . . . . . . . n times

22 0()0( . . . . . . . . . . . . . n times

Therefore Sn =

n

b. Angle

)0()0()( jjjS nn. . . . . . . . . . . . n times

= tan-1

( /0) + tan-1

( /0)+ . . . . . . . . . . . . n times

= 900 + 90

0 + . . . . . . . . . . . . n times

Sn = n * 90

0

Real

Img

· · K+j0 -K+j0



Illustration 4: Sketch the polar plot for the system represented by the following open loop

transfer function.

)2)(10(

10)()(

SSSSHSG , obtain M and for different values of

i) As 0,0 S jS

0

5.05.0

2*10*

10)()( 0

SSSHSG S

S5.0

0900

090

ii) As S,

010

))((

10)()(

3SSSSSHSG S

310 S

027090*30

M( ) ( )



Illustration 5: Sketch the polar plots for the system represented by the following open loop

transfer function.

)5()()(

2 SS

KSHSG

i) As 0,0 S jS

, S is far lesser than 5 and can be neglected

22

5/

5 S

K

S

K

n

K

S

KM

2

5/)(

2

5/)(

S

K

2)5/( SK = 0 – 2*90 = 0 – 180 = - 1800

ii) As S,

0 ∞ -900

0 -2700

)5()()(

200

SS

KSHSG S

Real

Img

0

-90

-180

-270

= ∞, M = 0

= 0, M = ∞



32 )5()()(

S

K

SS

KSHSG S

0)(33

K

S

KM

3)( SK

= 0 - 3*90 = - 2700


transfer function.

)8)(5(

10)()(

2 SSSSHSG

i) As 0,0 S jS

2200

4/1

8.5.

10)()(

SSSHSG S

M( ) ( )

0 ∞ -1800

0 -2700

Real

Img

0

-90

-180

-270

= ∞, M = 0 = 0, M = ∞



0

4/14/14/1)(

22SM

24/1)( S

= 0 - 2*90

= - 2*900 = 180

0

ii) As S,

42

10

..

10)()(

SSSSSHSG S

44

1010)(

SM

as , M ( ) = 0

410)( S

= 0 – 4*90 = - 3600




transfer function.

)5)(4(

)2(10)()(

3 SSS

SSHSG

i) As 0,0 S jS

)5)(4(

)2(10)()(

300

SSS

SSHSG S

33

1

5.4.

2.10

SS

33

11)(

SM

31)( S = 0 – 3*90 = - 2700

ii) As S,

43

10

)..

.10)()(

SSSS

SSHSG S

01010

)(44S

M

M( ) ( )

0 ∞ -1800

0 -3600 Real

Img

0, -360

-90

-180

-270

= ∞, M = 0 = 0, M = ∞



410)( S = 0 – 4*90 = -3600


transfer function.

)4)(2(

10)()(

SSSSHSG

i) As 0,0 S jS

SSSHSG S

8/10

4).2(

10)()(

00

8/108/10)(

SM

S8/10)( = 1800 - 90

0 = 90

0

ii) As S,

3

10)()(

SSHSG S

01010

)(33S

M

M( ) ( )

0 ∞ -2700

0 -3600

Real

Img

0, -360

-90

-180

-270

= ∞, M = 0

= 0, M = ∞



310)( S = 0 – 3*900 = - 270

0

Illustration 9: Sketch the polar plot for the system represented by the following transfer function.

10010

100

)(

)(2 SSSX

SY

i) As 0,0 S jS

11)(M , for both K positive and negative

01801)(

ii) As S,

2

100

S 0

100)(

2 SS

M , 2

100)(

S = 0 – 2* = - 90

0*2 = -180

0

M( ) ( )

0 ∞ 900

0 -2700

M( ) ( )

0 1 1800

0 -1800

Real

Img

0, -360

+270

-90

+180

-180

-270

+90

= ∞, M = 0

= 0, M = ∞

Img

+270

+180

-180

-270

+90



Experimental determination of Frequency Response

Many a times the transfer function of a physical system may not be available in such

circumstances it is necessary to obtain frequency response information experimentally. Such data

may then be used to establish the transfer function. This method requires the actual system.

SYSTEM ANALYSIS USING POLAR PLOTS: NYQUIST CRITERION

System Analysis using Polar Plots: Nyquist Criterion

Polar plots can be used to predict feed back control system stability by the application of Nyquist

Criterion, and therefore are also referred as Nyquist Plots. It is a labor saving technique in the

analysis of dynamic behaviour of control systems in which the need for finding roots of

characteristic equation of the system is eliminated.

Consider a typical closed loop control system which may be represented by the simplified block

diagram as shown in Figure 6.4

Figure 6.4 Simplified System Block Diagram

Real

0, -360 = 0, M = 1

= ∞, M = 0

H(S)

G(S) C(S)

R(S) + -



The closed-loop transfer function or the relationship between the output and input of the system

is given by

)()(1

)(

)(

)(

SHSG

SG

SR

SC

The open-loop transfer function is G(S) H(S) (the transfer function with the feedback loop

broken at the summing point).

1+ G(S) H(S) is called Characteristic Function which when equated to zero gives the

Characteristic Equation of the system.

1 + G(S) H(S) = 0 Characteristic Equation

The characteristic function F(S) = 1 + G(S) H(S) can be expressed as the ratio of two factored

polynomials.

Let ).().........)()((

)....().........)(()()(1)(

321

21

n

k

n

ZSPSPSPSS

ZSZSZSKSHSGSF

The Characteristic equation in general can be represented as

F(S) = K (S+Z1) (S+Z2) (S+Z3) ………. (S+Zn) = 0

Then:

–Z1, -Z2, -Z3 …. –Zn are the roots of the characteristic equation

at S= -Z1, S= -Z2, S= -Z3, 1+ G(S) H(S) becomes zero.

These values of S are termed as Zeros of F(S)

Similarly:

at S= -P1, S= -P2, S= -P3 ……. Etc. 1+ G (S) H (S) becomes infinity.

These values are called Poles of F (S).



Condition for Stability

For stable operation of control system all the roots of characteristic equation must be

negative real numbers or complex numbers with negative real parts. Therefore, for a system

to be stable all the ―Zeros‖ of characteristic equation (function) should be either negative real

numbers or complex numbers with negative real parts. These roots can be plotted on a complex-

plane or S-plane in which the imaginary axis divides the complex plane in to two parts: right half

plane and left half plane. Negative real numbers or complex numbers with negative real parts lie

on the left of S-plane as shown Figure 6.5.

Figure 6.5 Two halves of Complex Plane

Therefore the roots which are positive real numbers or complex numbers with positive real parts

lie on the right-half of S-plane.

In view of this, the condition for stability can be stated as ―For a system to be stable all the

zeros of characteristic equation should lie on the left half of S-plane‖.

Therefore, the procedure for investigating system stability is to search for ‗Zeros‘ on the right

half of S-plane, which would lead the system to instability, if present. However, it is

impracticable to investigate every point on S-plane as to which half of S-plane it belongs to and

·

·

·

·

2+j1

2-j1

-3-j2

-3+j2

Real

Img Right half

of S Plane

Left half

of S Plane



so it is necessary to have a short-cut method. Such a procedure for searching the right half of S-

plane for the presence of Zeros and interpretation of this procedure on the Polar plot is given by

the Nyquist Criterion.

Nyquist Criterion: Cauchy‘s Principle of Argument:

In order to investigate stability on the Polar plot, it is first necessary to correlate the region of

instability on the S-plane with identification of instability on the polar plot, or 1+GH plane. The

1+GH plane is frequently the name given to the plane where 1+G(S) H(S) is plotted in complex

coordinates with S replaced by j . Likewise, the plot of G(S) H(S) with S replaced by j is often

termed as GH plane. This terminology is adopted in the remainder of this discussion.

The Nyquist Criterion is based on the Cauchy‘s principle of argument of complex variable

theory. Consider [F(S) = 1+G(S) H(S)] be a single valued rational function which is analytic

everywhere in a specified region except at a finite number of points in S-plane. (A function F(S)

is said to be analytic if the function and all its derivatives exist). The points where the function

and its derivatives does not exist are called singular points. The poles of a point are singular

points.

Let CS be a closed path chosen in S-plane as shown Figure 6.6 (a) such that the function F(S) is

analytic at all points on it. For each point on CS represented on S-plane there is a corresponding

mapping point in F(S) plane. Thus when mapping is made on F(S) plane, the curve CG mapped

by the function F(S) plane is also a closed path as shown in Figure 6.6 (b). The direction of

traverse of CG in F(S) plane may be clockwise or counter clockwise, depending upon the

particular function F(S).



Then the Cauchy principle of argument states that: The mapping made on F(S) plane will

encircle its origin as many number of times as the difference between the number of Zeros

and Poles of F(S) enclosed by the S-plane locus CS in the S-plane.


Figure 6.6 Mapping on S-plane and F(S) plane

Thus N = Z – P

N0+j0 = Z – P

Where N0+j0: Number of encirclements made by F(S) plane plot (CG) about its origin.

Z and P: Number of Zeros and Poles of F(S) respectively enclosed by the locus CS in the S-

plane.

Illustration: Consider a function F(S)

)25)(25)(5)(3(

)22)(22)(1()(

jSjSSSS

jSjSSKSF

Zeros: -1, (-2-j2), (-2+j2) indicated by O (dots) in the S-plane

Poles: 0, -3, -5, (-5 –j2), (-5 +j2) indicated by X (Cross) in S-plane: As shown in Figure 6.6 (c)

S-plane +j

-j

σ

S5

S1

S2

S3 S4

CS

-σ

+j

-j

σ -σ

GS1

GS2

GS3

GS4

GS5

(0+j )

CG

F(S) = 1+G(S) H(S) plane



Figure 6.6 (c) Figure 6.6 (d)

Now consider path CS1 (CCW) on S-plane for which:

Z: 2, P =1

Consider another path CS2 (CCW) in the same S plane for which:

Z = 1, P = 4

CG1 and CG2 are the corresponding paths on F(S) plane [Figure 6.6 (d)].

Considering CG1 [plot corresponding to CS1 on F (S) plane]

N0+j0 = Z – P = 2-1 = +1

CG1 will encircle the origin once in the same direction of CS1 (CCW)

Similarly for the path CG2

N0+j0 = Z – P = 1 – 4 = - 3

CG2 will encircle the origin 3 times in the opposite direction of CS2 (CCW)

Note: The mapping on F(S) plane will encircle its origin as many number of times as the

difference between the number of Zeros and Poles of F(S) enclosed by the S-plane locus.

From the above it can be observed that

In the expression

S-plane +j

-j

σ

CS2

-σ

CS1

O: ZEROS

X: POLES

+j

-j

σ

CG2

-σ

CG1

CG2

CG2

(0+j0)

F(S) = 1+G(S) H(S) plane

CS1

CS2



N= Z - P,

N can be positive when: Z>P

N = 0 when: Z = P

N can be negative when: Z<P

When N is positive the map CG encircles the origin N times in the same direction as that of

CS

When N = 0, No encirclements

N is negative the map CG encircles the origin N times in the opposite direction as that of CS

Nyquist Path and Nyquist Plot

The above Cauchy‘s principle of argument can be used to investigate the stability of control

systems. We have seen that if the Zeros of characteristic function lie on the right half of S-plane

it will lead to system instability. Now, to encircle the entire right half of S-plane, select a closed

path as shown in Figure 6.6 (e) such that all the Zeros lying on the right-half of S-plane will lie

inside this path. This path in S-plane is known as Nyquist path. Nyquist path is generally taken

in CCW direction. This path consists of the imaginary axis of the S-plane (S = 0+j , - < < )

and a closing semicircle of infinite radius. If the system being tested has poles of F(S) on the

imaginary axis, it is customary to modify the contour as shown Figure 6.6 (f) excluding these

poles from the path.



Figure 6.6 (e): Nyquist Path Figure 6.6 (f)

Corresponding to the Nyquist path a plot can be mapped on F(S) = 1+G(S) H(S) plane as shown

in Figure 6.6 (g) and the number of encirclements made by this F(S) plot about its origin can be

counted.

Figure 6.6 (g)

Now from the principle of argument

N0+j0 = Z-P

N0+j0 = number of encirclements made by F(S) plane plot

Z, P: Zeros and Poles lying on right half of S-plane

For the system to be stable: Z = 0

S-plane

+j

-j

σ -σ

-j∞

+j∞

0+j0

0-j0

r = ∞

+j

-j

σ -σ

S=+j0

S= -j0

r = ∞

S= -j∞

r ≈ 0

Real

Img

Real

Img

0+j0

1+ G(S) H(S) plane



N0+j0 = - P Condition for Stability

Apart from this, the Nyquist path can also be mapped on G(S) H(S) plane (Open-loop transfer

function plane) as shown in Figure 6.6 (h).

Now consider

F(S) = 1+ G(S) H(S) for which the origin is (0+j0) as shown in Figure 6.6 (g).

Therefore G(S) H(S) = F(S) – 1

= (0+j0) – 1 = (-1+j0) Coordinates for origin on G(S) H(S) plane as shown in

Figure 6.6(h)

Figure 6.6 (h)

Thus a path on 1+ G(S) H(S) plane can be easily converted to a path on G(S) H(S) plane or open

loop transfer function plane. This path will be identical to that of 1+G(S) H(S) path except that

the origin is now shifted to the left by one as shown in Figure 6.6 (h).

This concept can be made use of by making the plot in G(S) H(S) plane instead of 1+ G(S) H(S)

plane. The plot made on G(S) H(S) plane is termed as the Niquist Plot and its net encirclements

about (-1+j0) (known as critical point) will be the same as the number of net encirclements made

by F(S) plot in the F(S) = 1+G(S) H(S) plane about the origin.

Now, the principle of argument now can be re-written as

N-1+j0 = Z-P

Img

0+j0

(-1+j0)

Origin of the plot

for G(S) H(S)

G(S) H(S) plane



Where N-1+j0 = Number of net encirclements made by the G(S) H(S) plot (Nyquist Plot) in the

G(S) H(S) plane about -1+j0

For a system to be stable Z = 0

N-1+j0 = -P

Thus the Nyquist Criterion for a stable system can be stated as The number of net

encirclements made by the Nyquist plot in the G(S) H(S) plane about the critical point

(-1+j0) is equal to the number of poles of F(S) lying in right half of S-plane. [Encirclements

if any will be in the opposite direction. Poles of F(S) are the same as the poles of G(S) H(S)].

Thus the stability of closed-loop control system is determined from its open-loop transfer

function.

System Analysis using Nyquist Criterion: Illustrations

Illustration 1: Sketch the Nyquist plot for the system represented by the open loop transfer

function and comment on its stability.

0,0)(

)()( aKaSS

KSHSG Poles: S = 0 (on imaginary axis) and S = -a

Step 1: Define Nyquist path. Let the Nyquist path be defined as given below.



Section I: S = +j to S = +j0; Section II: S = +j0 to S = -j0

Section III: S = -j0 to S = -j ; Section IV: S = -j to S = +j

2. Corresponding to different sections namely I, II, III, and IV Obtain polar plots on G(S) H(S)

plane, which are nothing but Nyquist Plots.

Nyquist Plot for Section I: In S-plane section I runs from S= + j to S = +j0

To obtain polar plot in G(S) H(S) plane:

0,0)(

)()( aKaSS

KSHSG

(i) 2

)()(S

KSHSG jS

,

0)()(2S

KSHSG

2

2)()( SK

S

KSHSG

= 0 – 2*900 = - 180

0

+j

-j

S=+j0

S= -j0

r = ∞

S= -j∞

r ≈ 0

σ -σ

S= -j∞

Nyquist Path

S= +j∞



(ii) S

K

S

aK

aS

KSHSG jS

1

0.

)()( ,

G(S) H(S) = K/a – S = 0 - 900 = - 90

0

Nyquist Plot for Section II: In S-plane section II runs from S= + j0 to S = -j0

In this region 0S

In S-plane section II is a semicircle from S = + j0 to S = -j0 of radius r ≈ 00, covering an angle of

1800 in clockwise direction

S

K

aS

KSHSG

1

)()()( where K

1 = K/a

But S = re+j

equation of a circle in exponential form

wherere

KSHSG j

j,Re)()(

1

r

KR

1

G(S) H(S) = R.e-j

r

KR

1

(as r is very small)

S M( ) ( )

S 0 -1800

0S ∞ -900

S=j0

Img

Real

-900

S=j∞

-1800

Section I

G(S) H(S) Plane



This shows that G(S) H(S) plot of section II of Nyquist path is a circle of radius = starting

from S = +j0 and ending at a point S = -j0 covering an angle of 1800 in opposite direction of

section II of Nyquist path (CCW direction i.e., negative sign)

In general if G (S) H (S) = nS

K 1

jn

jnneR

er

KSHSG .)()(

1

Where R = nr

K 1

The G (S) H (S) plot will be a portion circle (part) of radius R , starting at a point S = +jθ

and ending at a point S = -jθ covering an angle of (n*1800) in the opposite direction (CCW)

(since sign is negative)

Nyquist Plot for Section III: In S-plane section III runs from S= -j0 to S = j∞

)()()( 0

aSS

KSHSG jS

SKaS

K/

.

1 where Kl = (K/a)

SKSHSG /)()( 1

G(S) H(S) = Kl – S, K

l is negative

= -Kl – S

G(S) H(S) = 1800 – 90

0 = 90

0

jSjSaSS

KSHSG

)()()( = K/S

2

S=j0

S= -j0

Real

Img

R=∞

G(S) H(S) Plane

Section II



0)()(2S

KSHSG

G(S) H(S) = K - S2 = 0 – 2 (-90

0) = 180

0 ; (‗S‘ is negative)

This Section III is the mirror image of section of the Section I.

Nyquist Plot for Section IV: In S-plane section IV runs from S= -j∞ to S = j∞

In this region S

In the S-plane it is a semicircle of radius R from S = - j ∞ to S = + j ∞ covering an angle of

1800 in the counter clockwise direction.

In the G(S) H(S) plane

2)()(

S

KSHSG S

But S = R e+j

Equation in circle in exponential form

2

22.

.)()( j

jer

eR

KSHSG where 0

2R

Kr

Thus G(S) H(S) plot for section IV is also a circle of radius 0r starting at S = - j ∞ and

ending at S = + j ∞ covering an angle of 2* 1800 (2 ) in the opposite direction (CW).

S= -j∞

S= -j0 +900

+1800

Section III

G(S) H(S) Plane

Real

Img



Now assemble the Nyquist plots of all the sections as given below to get the overall Nyquist

plot

From Nyquist Criterion:

No. of encirclements made by Nyquist plot about (– 1 +j0) = N-1+j0 = Z – P

P = No. of poles lying in the right half of S plane

S= +j∞

S= -j∞

Real

Img

Section IV

G(S) H(S) Plane

r ≈ 0

Real

Img G(S) H(S) Plane

(-1+j0)

j0

j∞

-j0

-j∞



For the function)(

)()(aSS

KSHSG , the Poles are: S = 0, S = - a = 0, which lie on the left half

of S-plane

Therefore P = 0: Number of poles on the right half of S-plane

N-1+j=0 = 0 as counted from the Nyquist plot

N-1+j0 = Z – P

0 = Z – 0

Therefore Z = 0

Number of zeros lying on the right half of S-plane is 0 and hence the system is stable

Illustration 2: Obtain the Nyquist diagram for the system represented by the block diagram given

below and comment on its stability

)1()(

2 SS

KSG

SSH )(

,)1(

*)1(

)()(2 SS

KS

SS

KSHSG Poles are S = 0, -1/ P = 0


R (S) + C (S)

)1(2 SS

K

S

-



1. Explain briefly the different Graphical Methods to Represent Frequency Response Data .

2. A second order system has a natural frequency of 10 rad/sec and a damping ratio of 0.5.

Sketch the polar plot for the system.

3. Obtain the polar plot for the transfer function

4. Sketch the polar plots for the system represented by the following open loop transfer function.

)5()()(

2 SS

KSHSG


)8)(5(

10)()(

2 SSSSHSG


)4)(2(

10)()(

SSSSHSG

7. Draw Nyquist path for the function F(s)

)25)(25)(5)(3(

)22)(22)(1()(

jSjSSSS

jSjSSKSF



UNIT-7

Frequency domain analysis

Frequency response of a control system refers to the steady state response of a system subject to

sinusoidal input of fixed (constant) amplitude but frequency varying over a specific range,

usually from 0 to ∞. For linear systems the frequency of input and output signal remains the

same, while the ratio of magnitude of output signal to the input signal and phase between two

signals may change. Frequency response analysis is a complimentary method to time domain

analysis (step and ramp input analysis). It deals with only steady state and measurements are

taken when transients have disappeared. Hence frequency response tests are not generally carried

out for systems with large time constants.

The frequency response information can be obtained either by analytical methods or by

experimental methods, if the system exits. The concept and procedure is illustrated in Figure 7.1

(a) in which a linear system is subjected to a sinusoidal input. I(t) = a Sin t and the

corresponding output is O(t) = b Sin ( t + ) as shown in Figure 6.1 (b).


The following quantities are very important in frequency response analysis.

M ( ) = b/a = ratio of amplitudes = Magnitude ratio or Magnification factor or gain.

( ) = = phase shift or phase angle



These factors when plotted in polar co-ordinates give polar plot, or when plotted in rectangular

co-ordinates give rectangular plot which depict the frequency response characteristics of a

system over entire frequency range in a single plot.

Frequency Response Data

The following procedure can be adopted in obtaining data analytically for frequency response

analysis.

5. Obtain the transfer function of the system

)(

)()(

SI

SOSF , Where F (S) is transfer function, O(S) and I(S) are the Laplace transforms

of the output and input respectively.

6. Replace S by (j ) (As S is a complex number)

)(

)()(

jI

jOjF

)()()(

)(jBA

jI

jO (another complex number)

7. For various values of , ranging from 0 to ∞ determine M ( ) and .

jBAjBAjI

jOM )()(

)(

)()(

22 BAM

jBAjI

jO

)(

)(

A

B1tan



8. Plot the results from step 3 in polar co-ordinates or rectangular co-ordinates. These plots are

not only convenient means for presenting frequency response data but are also serve as a

basis for analytical and design methods.

Comparison between Time Domain and Frequency Domain Analysis

An interesting and revealing comparison of frequency and time domain approaches is based on

the relative stability studies of feedback systems. The Routh‘s criterion is a time domain

approach which establishes with relative ease the stability of a system, but its adoption to

determine the relative stability is involved and requires repeated application of the criterion. The

Root Locus method is a very powerful time domain approach as it reveals not only stability but

also the actual time response of the system.

It was pointed out earlier that the performance of a feedback control system is more

preferably measured by its time domain response characteristics. This is in contrast to the

analysis & design of systems in the communication field, where the frequency response is of

more importance, since in this case most of the signals to be processed are either sinusoidal or

periodic in nature. However analytically, the time response of a control system is usually

difficult to determine, especially in the case of high order systems. In the design aspects, there

are no unified ways of arriving at a designed system given the time-domain specifications, such

as peak overshoot, rise time , delay time & setting time. On the other hand, there is a wealth of

graphical methods available in the frequency-domain analysis, all suitable for the analysis &

design of linear feedback control systems once the analysis & design are carried out in the

frequency domain, time domain behavior of the system can be interpreted based on the

relationships that exist between the time-domain & the frequency-domain properties. Therefore,

we may consider that the main purpose of conducting control systems analysis & design in

frequency domain is merely to use the techniques as a convenient vehicle toward the same

objectives as with time-domain methods.

The starting point in frequency-domain analysis is the transfer function.

For a single loop feed back system, the closed loop transfer function is written

C(s) G(s)



M(s) = = (1)

R(s) 1+ G(s) H(s)

Under the sinusoidal steady-state, we get s = j ; then equation(1.0) becomes,

C(j ) G(j )

M(j ) = = (1.1)

R(j ) 1+ G(j ) H(j )

The sinusoidal steady-state transfer relation M(j ), which is a complex function of , may be

expressed in terms of a real & an imaginary part; that is,

M((j ) = Re [ M (j )] + j Im [ M(j )] (1.2)

Or , M(j ) can be expressed in terms of its magnitude & phase as

M(j ) = M ( ) m( ) (1.3)

Where

G(j )

M(( ) = (1.4)

1 + G(j ) H(j )

And

G(j )

m( ) =

1 + G(j ) H(j )

(1.5)

= G(j ) - 1 + G(j ) H(j )



since the analysis is now in the frequency domain, some of the terminology used in

communication system may be applied to the present control system characterization. For

instance, M( ) of Eq. (1.4) may be regarded as the magnification of the feed back control system

is similar to the gain or amplification of an electronic amplifier. In an audio amplifier, for

instance, an ideal design criterion is that the amplifier must have a flat gain for all frequencies.

Of course, realistically, the design criterion becomes that of having a flat gain in the audio

frequency range. In control system the ideal design criterion is similar. If it is desirable to keep

the output C(j ) identical to the input R(j ) at all frequencies, M(j ) must be unity for all

frequencies. However, from Eq. (1.1) it is apparent that M(j ) can be unity only when G(j ) is

infinite, while H(j ) is finite & nonzero. An infinite magnitude for g(j ) is, of course,

impossible to achieve in practice, nor would it be desirable, since most control system become

unstable when its loop gain becomes very high. Further more, all control system are subject

noise. Thus in addition to responding to the input signal, the system should be able to reject &

suppress noise & unwanted signals. This means that the frequency response of a control system

should have a cutoff characteristic in general, & sometimes even a band-pass characteristic.

The phase characteristics of the frequency response are also of importance. The ideal situation

is that the phase must be a linear function of frequency within the frequency range of interest.

Figure 1.1 shows the gain & phase characteristics of an ideal low-pass filter, which is impossible

to realize physically. Typical gain & phase characteristics of a feedback control system are

shown in Fig. 1.2. The fact is that the great majority of control systems have the characteristics

of a low-pass filter, so the gain decreases as the frequency increases.

M( ) 0

1

m( )

0 c Deg

Fig. 1.1. Gain-phase characteristics of an ideal low-pass filter.



0

M( ) Mp

m( )

1 Deg

0

Fig.1.2. Typical gain & phase characteristics of a feedback control system.

Frequency-Domain characteristics:

If a control system is to be designed or analyzed using frequency-domain techniques, we need

a set of specification to describe the system performance. The following frequency-domain

specifications are often used in practice.

Peak response Mp :

The peak response Mp is defined as the maximum value of M( ) that is given in

Eq.(1.4). In general, the magnitude of Mp gives an indication of the relative stability of a feed

back control system. Normally, a large Mp corresponds to a large peak overshoot in the step

response. For most design problems it is generally accepted that an optimum value Mp of should

be somewhere between 1.1 & 1.5.

Resonant frequency p :

The resonant frequency p is defined as the frequency at which the peak resonance Mp occurs.

Bandwidth :

The bandwidth , BW, is defined as the frequency at which the magnitude of M(j ), M( ),

drops at 70.7 percent of its zero-frequency level, or 3 dB down from the zero-frequency gain. In

general, the bandwidth of a control system indicates the noise-filtering characteristics of the

system. Also, bandwidth gives a measure of the transient response properties, in that a large

bandwidth corresponds to a faster rise time, since higher-frequency signals are passed on to the

outputs. Conversely, if the bandwidth is small, only signals of relatively low frequencies are

passed, & the time response will generally be slow & sluggish.



Cutoff rate :

Often, bandwidth alone is inadequate in the indication of the characteristics of the system in

distinguishing signals from noise. Sometimes it may be necessary to specify the cutoff rate of the

frequency response at the higher frequencies. However, in general, a steep cutoff characteristics

may be accompanied by a large Mp, which corresponds to a system with a low stability margin.

The performance criteria defined above for the frequency-domain analysis are illustrated on

the closed-loop frequency response, as shown in Fig. 1.3.

There are other criteria defined that may be used to specify the relative stability &

performance of a feedback control system. These are defined in the ensuring sections of this

chapter.

M( )

Mp

1 Bandwidth

0

p BW

Fig.1.3. Typical magnification curve of a feedback control system.

Mp, , p & the bandwidth of a second-order system:

For a second-order feedback control system, the peak resonance Mp, the resonant frequency p,

& the bandwidth are all uniquely related to the damping ratio & the natural undamped

frequency n of the system. Consider the second-order sinusoidal steady-state transfer function

of a closed-loop system,

C(j ) 2

n

M(j ) = = (1.6)

R(j ) (j )2 + 2 n (j ) +

2n

1

=



1 + j 2 ( / n) - ( / n)2

We may simplify the last expression by letter u = / n. The Eq. (1.6) becomes

1

M(ju) = ( 1.7)

1 + j2 u - u2

The magnitude & phase of M (j ) are

1

M(ju) = M(u) = (1.8)

[( 1 – u2 )

2 + ( 2 u)

2]

½

And

2 u

M(ju) = m(u) = - tan -1

(1.9)

1 – u2

The resonant frequency is determined first by taking the derivative of M(u) with respect to u &

setting it equal to zero. Thus

dM(u) 1

= - [( 1 – u2 )

2 + ( 2 u)

2]

–3/2 ( 4 u

3 – 4u + 8u 2) = 0 ( 1.10)

du 2

from which

4u3 – 4u + 8u

2 = 0 ( 1.11)

The root of Eq. (1.11) are

up = 0 (1.12)

and

up = 1 - 22 ( 1.13)



The solution Eq. (1.12) merely indicates that the slope of the M( ) versus curve is zero at

= 0; it is not true maximum. The solution of eq. (1.13) gives the resonant frequency,

p = n 1 - 22 (1.14)

Since frequency is a real quantity, Eq. (1.14) is valid only for 1 22

or 0.707. This means

simply that for all values of greater than 0.707, the solution of p = 0 becomes the valid one, &

Mp = 1.

Substituting Eq. (1.13) into Eq. (1.8) & simplifying, we get

1

Mp = (1.15)

2 1 - 2

It is important to note that Mp is a function of only, whereas p is a function of & n

5

4

3

Mp

2

1

0

0.5 0.707 1.0 1.5 2.0

Damping ratio

1

Fig.1.4 Mp versus-damping ratio for a second – order system, Mp =

2 1 - 2



1.0

0.8

up = p / n

0.6

0.4

0.2

0

0.5 0.707 1.0

Damping ratio

Fig 1.5. Normalized resonant frequency- versus-damping ratio for a second order system,

Up = 1 - 2 .

Fig.1.4 & 1.5 illustrate the relationship between Mp & , & u = p / n & ,

respectively.

Bandwidth :

Bandwidth BW of a system is a frequency at which M( ) drops to 70.7% of its zero

frequency level or 3 dB down from the zero frequency gain. Equating the Eq.

1

M(u) = = 0.707

[( 1 –u2)

2 + ( 2 u)

2]

½

1

= [( 1 –u2)

2 + ( 2 u)

2]

½ =

0.707

= 2.



Squaring both sides

( 1 –u2)2 + 4

2u

2 = 2

1 + u4 – 2u

2 + 4

2u

2 =

2

Let u

2 = x

1 + x2 – 2x + 4

2 x =

2

x2 – 2x + 4

2 x =

1

x2 – x ( 2 - 4

2 ) =

1

x

2 – x ( 2 - 4

2 ) – 1 = 0

a = 1, b = - ( 2 - 4

2 ) , c = -1

-b b2 – 4ac

x =

2a

(2 – 42 ) (2 – 4

2 )

2 + 4

=

2

(2 – 42 ) (4 + 16

4 – 16

2 + 4

=

2

(2 – 42 ) 16

4 – 16

2 + 8

=

2

2 (1 – 22 ) 4 + 16

4 – 16

2 + 4

=

2

2 (1 – 22 ) 2 2 + 4

4 – 4

2

=

2



2 (1 – 22 ) 2 2 + 4

4 – 4

2

=

2

2 [(1 – 22 ) 2 + 4

4 – 4

2]

=

2

u2

= x = (1 – 22 ) 2 + 4

4 – 4

2

/ n = u = [(1 – 22 ) (2 + 4

4 – 4

2 ]

1/2

BW = n [ ( 1 – 22 ) + 4

4 - 4

2 +2]

1/2

For the second order system under consideration, we easily establish some simple relationship

between the time – domain response & the frequency-domain response of the system.

1. The maximum over shoot of the unit step response in the time domain depends upon

only.

2. The response peak of the closed - loop frequency response Mp depends upon only.

3. The rise time increases with , & the bandwidth decreases with the increase of

, for a

fixed n, therefore, bandwidth & rise time are inversely proportional to each other.

4. Bandwidth is directly proportional to n.

5. Higher bandwidth corresponds to larger Mp.




1. Give the Comparison between Time Domain and Frequency Domain Analysis

2. For the second order system under consideration, establish some simple relationship between

the time – domain response & the frequency-domain response of the system.

3. Define Peak response, resonant frequency, cutoff rate and bandwidth.



UNIT - 8

Introduction to State variable analysis:




1.Define state equation & write the mathematical form of it.

2. Define output variables & give the matrix form of it for n variables of the system.

3. Explain briefly the state equation based on modeling procedure.

4. Draw a block diagram for the general second-order, single-input single-output system:

5. Find the transfer function and a single first-order differential equation relating the output y(t)

to the input u(t) for a system described by the first-order linear state and output equations:

Q.6 Use the Laplace transform method to derive a single differential equation for the capacitor

voltage vC in the series R-L-C electric circuit shown in Fig below.

ece-iv-control systems [10es43]-notes · pdf filecontrol systems 10es43 sjbit/ dept

Documents