Ece IV Control Systems [10es43] Notes

7/13/2019 Ece IV Control Systems [10es43] Notes

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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 secondorder systems, steady123 state errors and error constants. Introduction to PID Controllers(excludingdesign) 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:

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

UNIT 8:

Introduction to State variable analysis: Concepts of state, state variable and state models forelectrical systems, Solution of state equations. 6 Hrs


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SJBIT/ Dept of ECE

TEXT BOOK :

1. J. Nagarath and M.Gopal, Control Systems Engineering, New AgeInternational (P) Limited, Publishers, Fourth edition2005

REFERENCE BOOKS:1. Modern Control Engineering , K. Ogata, Pearson Education Asia/PHI, 4th Edition, 2002.

2. Automatic Control Systems, Benjamin C. Kuo, John Wiley IndiaPvt. Ltd., 8th Edition, 2008.

3. Feedback and Control System, Joseph J Distefano III et al.,Schaums Outlines, TMH, 2nd Edition 2007.


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SJBIT/ Dept of ECE

INDEX SHEET

SL.NO TOPICPAGE

NO.

I UNIT 1:Modeling of Systems 1-22

1.1 Introduction to Control Systems, Types of ControlSystems

1.2 Effect of Feedback Systems

1.3 Differential equation of PhysicalSystems -Mechanical systems, Friction

1.4 Translational systems (Mechanicalaccelerometer, systems excluded)

1.5 Rotational systems , Gear trains

1.6 Electricalsystems, Analogous systems

II UNIT2 : 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 UNIT3 :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 steady23state errors and error constants)3.7 Introduction to PID Controllers (excluding design)

IV UNIT4 : 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 UNIT5 : RootLocus Techniques 111-133

5.1 Introduction, The root locus concepts

5.2 Construction of root loci (problems)

VI UNIT6 : 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

6.5Introduction to lead, lag and lead-lag compensating networks (excludingdesign)

VII UNIT7 : Stability in the frequency domain170-182

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

7.2 Mathematical preliminaries


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SJBIT/ Dept of ECE

7.3 Nyquist Stability criterion

7.4Assessment of relative stability using Nyquist criterion, (Systems withtransportation lag excluded)

VIII UNIT8 : Introduction to State variable analysis 183-201

8.1 Concepts of state, state variable

8.2 Concepts of state models for electrical systems8.3 Solution of state equations


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SJBIT/ Dept of ECE Page 1

UNIT-1

A control system is an arrangement of physical components connected or related in such amanner 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 output. 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 obtainedfrom the system. The control system may be one of the following type.

1) man made2) natural and / or biological and3) 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/offsystem : electric switch

output : flow or no flow of current2) Pointing a finger at an object is a biological control system.

input : direction of the object with respect to some directionsystem : consists of eyes, arm, hand, finger and brain of a man

output : actual pointed direction with respect to same direction3) Man driving an automobile is a hybrid system.

input : direction or lanesystem : 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 isresponsible 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 operatingconditions. 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.


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Speed of thePrime mover

Induced Volta e

OutputInputs

TimeCleanliness 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

EXAMPLE2 Washing machine

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

EXAMPLE3 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 levelwill 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 valveVc. The control valve is adjustable manually.

Fig 1-4 b) Open loop control

Fig1.4 a) Water level control

Actuatin si nal out utin ut

Controller System

WATER

TANK

Desired Waterlevel r

Actual

Water level c

Valve VC

Valve VO

Waterin

Waterout

C


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

Fig1.5: Closed loop control system

EXAMPLE1THERMAL 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 theamount 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 amountof 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

EXAMPLE2 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

Referencein ut

Actuating/ error

Errordetector

Feed back si nal

Controloutput

Desired hotwater. temp

roc

Brain ofoperator (r-c)

Musclesand Valve

ActualWater temp

Co

CC

Thermometer

+

Human operator

Thermometer

Hot water

Drain

d water

Steam

Steam


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wall about 5 feet from the floor. A person selects and adjusts the desired room temperature ( r )say 25

0C 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 thanthe 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 airthroughout 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 sidetemperature 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 systemsparameters. 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 problemin 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 controlsystems 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 closedloop control system is morethan that for a corresponding open loop control system. Thus the closed loop control system isgenerally higher in cost.

esired temp. roc

Relay

switch

ActualTemp.

CoC

Furnace Blower House+

Outdoor temp change

(disturbance)


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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 isindependent 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 functionalrelationship amongst the components of a controlsystem.

Control elements: These are also called controller which are the components required togenerate the appropriate control signal applied to the plant.

Plant: Plant is the control system body process or machine of which a particular quantity orcondition is to be controlled.

Feedback control:feedback control is an operation in which the difference between the outputof 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 theprimary feedback signal.

Servomechanism: Servomechanism is a feedback control system in which output is somemechanical position, velocity or acceleration.

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

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

Tachometer: Tachometeris a device whose output is directly proportional to time rate of changeof input.

Synchros: Synchrosis an AC machine used for transmission of angular position synchro motor-receiver, synchro generator- transmitter.


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Block diagram:A block diagram is a short hand, pictorial representation of cause and effectrelationship 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 severaldifferent paths to several destinations.

Time response:It is the output of a system as a function of time following the application of aprescribed input under specified operating conditions.


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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 Newtons second lawof 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. Thesummation of the forces must of course be algebraic and thus considerable care must be taken inwriting 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 threeare 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 issmall, 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 thecomposition 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 tothe 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 berealized 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. Frictionmay be introduced intentionally in a system by use of dashpot as shown in fig 1-9. In

automobiles shock absorber is nothing but dashpot.


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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 decreasingpressure side a As a result the oil flows from side b toside a through the wall clearance. Thefriction 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-troublesomeeffect of gravity is eliminated.

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

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

4. Apply Newtons law of motion to each diagram using the convention that any forceacting 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 pointcalled fulcrum the two arms are called effort arm and Load arm respectively. The leverbears analogy with transformer

a b

Applied forceF Piston

L1 L2

F2

F1effort

Load

Fulcrum


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It is also called mechanical transformer

Equating the moments of the force

F1 L1 = F2 L2

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 thekinetic 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 areactive 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.

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


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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 ofidentical form such systems are called analogous systems and the terms which occupy thecorresponding positions in differential equations are analogous quantities

The analogy is here is called force voltage analogyTable 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

d x

dt2

dx

dt

N1

N2

Driving wheel

Driven wheel

d q

dt2

d q

dt2

q

c


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Displacement Charge

Velocity Current.

ForceCurrent 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 forcecurrent analogy. The analogous quantities are listed.

Table of conversion for forcecurrent analogyMechanical 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 x1and x2are responses.

d x

dt2

dx

dt

d x

dt2

1

R

d

dt2

L


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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

m2x2= b1(x2-x1) + k1(x2-x1) - b2x2- k2x2 --- (b)

m2

m1

F

k2x2 b2x2

k1x2 k1x1 b1x1 b1x2x2

k1x2 k1x1 b1x1 b1x2

.

m2

m1

F

k2x2 b2x2

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

m1and m2separately as shown in figure1.10 (b)

Apply NSL for both the masses

separately and get equations as given in

(a) and (b)


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Illustration 2: For the system shown in figure 2.16 (a) obtain the mathematical model if x 1and

x2are initial displacements.

Let an initial displacement x1be given to mass m1and x2to mass m2.

Figure 1.11 (a)

K1

K2

K3

m2

m1

X1

X2


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Figure 2.16 (b)

Based on Newtons second law of motion: F = maFor mass m1

m1x1= - K1x1+ K2(x2-x1)

m1x1+ K1x1K2x2+ K2x1= 0

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

For mass m2

m2x2= - K3x2K2(x2x1)

m2x2+ K3x2+ K2x2K2x1

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

Mathematical models are:

K1X1

K2X2

K2X2

K2X1

K2X1

K3X2K3X2

K1X1

X1 X1

m1m1

m2

m2X2

X2

K2(X2X1)

K2(X2X1)

..

..

..

..

..

..


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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

K K

M

K B2

B1

f12K1M1

f1f(t)

M2

X2

f2

X1

..

..


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

6. Write the differential equation describing the system. Assume the bar through whichforce 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)

XK

Xi Xo Xi Xo

Yk

C

K F

b

M

B

XK

f t

a

ba

X2

K1Force f


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

M1

B1

B1

J1

KJ J2

1 2T(t)

1 2J1 J2 J3

K1 K2 K3

Torque T

B1 B2 B3

3


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The outer cylinder has viscous friction B2between it and the reference frame and is restrained bya 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 relaycoil 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 negligiblemass 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

B2J1Torque T1, 1 B1

J2


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. 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/ ampMotor to load gear ratio = N2

N1

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 errordetector 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


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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


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

C1

I E0

C2 R Output

input


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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 closedloop 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 systems 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 transferfunction ?

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

18.What function do gears and levers perform.19.What are the component parts of the mechanical constants of a motors transfer function?


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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 onthe 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 informationconcerning 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


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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 thereference 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. Notethat 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 systemSumming point

Branch pointR(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)

+

-

+-


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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

+-


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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 transferfunction 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 ina 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)

Fig2.5Fig2.5 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 G s

H s

+-

+

+


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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 zeroinitial 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


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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 isa 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 asystem 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


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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 Xkto Xjrepresents 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 akjXkis 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 nodeand 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


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

Transmission ruleThe 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 singlebranch 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 givenby 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

k

kkPP 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 oftwo nontouching loops sum of gain products of all possible combination of threenontouching loops) +

kis cofactor of kthforward path determinant of graph with loops touching k

thforward 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 .. X6as shown in the block diagram..Connect the nodes with appropriate gain along the branch. The signal flow graph is shown in

Fig. 2.7

G2G1 G3

H2

H1

RX1 X2 X3 X4 X5 X6 C


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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

X4X5 X6

CG

1

H1

-H2

G2 G

3

-1

11 1 1

G1

G2

G3

G4

R

C

X1 X2

X3


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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, u1and u2choose 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

C1 1

X1

X2 X3

u b1

b2

a21

a12

a33

a31

X1

X2a11

a13

a32

X3

u2

a23

a22


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Example 4

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

Figure 2.12 LRC networkThe governing differential equations are

3

2

11

tidt

deC

teeRidt

diL

or

teidt

C

Ri

dt

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

1

L

RsL

1

LRsL

1-

LRs

1

s

1

i(0+)

Ec(s)

I s

E s

ec(0+)

R L

Ci(t)

ec(t)

e(t)


38/205



SIGNAL FLOW GRAPHS

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

=1-(sum of all individual loop gains)+(sum of gain products of all possible combinations oftwo nontouching loops sum of gain products of all possible combination of three

nontouching loops) +

kis cofactor of kthforward path determinant of graph with loops touching kthforward path. It is

obtained from by removing the loops touching the path Pk.

Example 1Obtain 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=G1Gain 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

C1 1


39/205



There are no non-touching loops. = 1+G1G3-G1G4+G2G3-G2G4Forwardpaths 1 and 2 touch all the loops. Therefore, 1= 1, 2= 1

The transfer function T =42324131

212211

1 GGGGGGGG

GGPP

sR

sC

Example 2Obtain 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=G1G2G3There are three loops with loop gains:L1=-G1G2H1, L2=G2G3H2, L3= -G1G2G3There are no non-touching loops. = 1-G1G2H1+G2G3H2+G1G2G3Forward path 1 touches all the loops. Therefore, 1= 1.


32111

1 GGGHGGHGG

GGGP

sR

sC

Example 3Obtain 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 1G1 G2 G3

H1

-1

-H2


40/205




There are three forward paths.

The gain of the forward path are: P1=G1G2G3G4G5P2=G1G6G4G5P3=G1G2G7

There are four loops with loop gains:

L1=-G4H1, L2=-G2G7H2, L3= -G6G4G5H2 , L4=-G2G3G4G5H2There is one combination of Loops L1 and L2 which are nontouching with loop gain product

L1L2=G2G7H2G4H1 = 1+G4H1+G2G7H2+G6G4G5H2+G2G3G4G5H2+ G2G7H2G4H1Forward 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

G1C(s)R(s)

G7G6

-H1

G2 G3 G4 G5

-H2

X1 X2 X3 X4 X5

1


41/205



Find the gains1

3

2

5

1

6 ,,X

X

X

X

X

Xfor 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=abefThere are four loops with loop gains:L1=-cg, L2=-eh, L3= -cdei, L4=-beiThere is one combination of Loops L1 and L2 which are nontouching with loop gain productL1L2=cgeh

= 1+cg+eh+cdei+bei+cgehForward 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 dc fe

-h

-g

-i

X1 X6X5

X4X3X2


42/205



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 removedwhich 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 inFig.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

c1

e

-h

-g

-i

X2 X5X5

X4X3X2

a eb f

-h

-g

-i

X1 X5

X4 X3

X2

c

d

X31


43/205



There are five loops with loop gains:L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befgThere is one combination of Loops L1 and L2 which are nontouching with loop gain productL1L2=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) usingsignal 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.


44/205



Figure 2.19 Signal flow grapgh of example 5

There are three forward paths.

The gain of the forward path are: P1= 3P2= 1/ s

2

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 L1and L2. Therefore

2

21

1 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

s2

s

a2

s

11

X X


45/205



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 Masons 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.

G2G1 G3

H2

H1

RX1 X2 X3 X4 X5 X6 C

G1

G2

G3

G4

R

C

X1 X2

X3


46/205



Figur

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

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

Ci(t)

ec(t)

e(t)

G1C sR(s)

G7G6

-H1

G2 G3 G4 G5

-H2

X

X

X3 X4 X5

1

uXaX

uXXaX

uXX

1122

22111

31


47/205



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 toanalyse 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 satisfactoryperformance 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 timeresponse 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 ResponseTotal 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)=0t

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

system.


48/205



C(t)Ct(t) Css(t)

Stepess

= study state

error

O Time

Transienttime Study stateTime

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 referenceinputs, 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 are1. 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 orinstant 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 responsemay be experimentalor oscillatory in nature.


49/205



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 whos value changes from 1 value (usually 0) to another level A inZero time.

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

t

O

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

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 = 1S

2S

2t for t 00 for t 0

=


50/205



t 0

At 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)

2r(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) = 1S

3

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

=t for t 020 for t 0

A


51/205



= 0 for t 0In LT form R(S) = 1 if A = 1

Standard test I nput Signals and its Laplace Transforms.

r(t) R(S)

Unit Step 1/S

Unit ramp 1/S

Unit Parabolic 1/S

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 1storder system is ; G(S) = C(S ) = K .R(S) 1+TS

If K=1, Then G(S) = [ Its 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


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t

0.632

Unit step response of 1storder system:-

Let a unit step i\p u(t) be applied to a 1storder 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- e1 = 0.632. c (t)

The smaller the time const. T. thefaster the system response. 1e 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,a2d

2c(t) + a1dc(t) + a0c(t) = b0 .r(t)-----------------(1)

dt2 dt

Where c(t) = o/p & r(t) = I/p-- ing (1) by a0,

a2 d2c(t) + a1 . dc (t) + c(t) = b0. r(t).

a0 dt2 a0 dt a0

a2 d2c(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


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T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced sothat (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 (1.00842) = 0.0084

= 0.091

k .S + S2T .

w.k.t. T.F. = G(S) = 1 + K . = K .1+ G(S) . H(S) S + S

2T 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 = K1When = 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 . = 0.05080.4037 K1 K1


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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 K10.6 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 . = 0.62 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


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

C(S) = 5 .R(S) S

2+ 4S +5

Solution:

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

R(S) S2+ 4S +5

2 Wn = 4 = 4 . = 0.894. Wd = 1.00182 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 thedt

2 dt value of damping ratio, undamped and damped

frequency of ascillation.

Soutions:- d2

+ 10 d = 15 ( r - ) , = 150r150 .

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) Percentpeak over shoot Mp (4) Expression for error

resoponse.


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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,.

Kx(t)inmt

f(t) 0.00193The 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

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

1 2 3 4 5

Given :- F(S) = 10

S.

X(S) = 10 .S(MS

2+ FS + K)

SX (S) = 10 .MS

2+ 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

0.55390.5539 2 = 2= 0.597 = 0.6

tp = = .

Wn 1 2Wd

3 = . Wn = 1.31 rad / Sec.

M


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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

-60t1.2e-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-60t1.2e-10tTaking 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) S

2+ 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& K2so that closed loop system resembles a 2 ndorder system with= 0.5 & frequency of damped oscillation 9.5 rad / Sec.


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2) With the above value of K1& K2 find the % overshoot when i/p is step i/p3) 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/SecWd =

10.52

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/ps type of system and from non-linearties of the system components. The steady stateperformance of a stable control system is generally judged by its steady state error to step, rampand parabolic i/p.

K1 1 .

S(1+S)

K2S


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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) = 1C(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 essmay be found by the use of final value theoremand 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)


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The expression for steady state errors for various types of standard test signals arederived 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/p1 + 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 . t2Taking L.T. R(S)= 1 .

2 S3


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ess = lt S . 1 . lt 1 .

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

S O S2

+ S2G(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(T1S+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. Asthe 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 isillustration as follows.

1) TypeO system :- If, n = 0, the system is called type0, system. Thesteady 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


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

S O S O S + 1

ess (acceleration) = 1 . = 1 . =

Ka O

Ka = lt S2G(S) . H(S) = lt S

2 K . = O.

S O S O S + 1

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

ess (Position) = 1 . = O1 +

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

ess (acceleration) = 1 . = 1 . =O O

Ka = lt S2

K . = O.

S O S (S + 1)

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

2(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


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Ka = lt S2 K . = K.

S O S2(S + 1)

..

. ess (acceleration) = 1 .K

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

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

PROBLEMS;

1.

The unit step response of a system is given by

C (t) = 5/2 +5t5/2 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 = 52 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


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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

2G(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) = 1And 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)

lt 100 x s2 = 0

S 0 S 0 S (S+10)

R1 R2 R3+ +

1+Kp Kv Ka

R1 R2

R3++

P0 P1

P2+ +

=


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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 errorconstant and errors.

(S+2)

G(S).H(S) =

S(S+1) (S+10)

Kp = lt G(S) H(S) = (0+2)

=

Kv = lt G(S) H(S) = (0+2)

= 1/5 = 0.2

1 1

= = 5

Kv 0.2

50

Kp = lt G(S) H(S) = 50

=2

Kv = lt G(S) H(S) = 50 x S

=2

Ka = lt G(S) H(S) = ltS 0 S

Sx 50 50

= = 5

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


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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 to10 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

Lt S (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

50

=

S2 (S+20) (S+30)

Kv = lt S K

S 0 S2 S+20 S+30 =

Ka = lt S K

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

K

600

1 1+ = 0

1+Kp 1+

10 10+ = 0

Kv


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Error due to para i/p,,

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

First order system:-

The 1storder 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 = b0a0 a0

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

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

R(S) 1+TS

If K=1, Then G(S) = [ Its 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 1storder system:-

Let a unit step i\p u(t) be applied to a 1storder 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=

1 .

1+TS


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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- e1

= 0.632. c (t)

The smaller the time const. T. thefaster the system response. 1e 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,a2d

2c(t) + a1dc(t) + a0c(t) = b0 .r(t)-----------------(1)

dt2 dt

Where c(t) = o/p & r(t) = I/p-- ing (1) by a0,

a2 d2c(t) + a1 . dc (t) + c(t) = b0. r(t).

a0 dt2 a0 dt a0

a2 d2c(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


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


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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(COSwnt) =1- COSwnt; t > O --------------------- (8)

From Eq:- (8) , we see that the Wnrepresents the undamped natural frequency of the system. Ifthe linear system has any amount of damping the undamped natural frequency cannot be

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

Wd =wn 12

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+ Wn2 S

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

2

= 1 . Wn2 .

S ( S + Wn )2S

Taking I.L.T.,

C(t) = 1e -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

.


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(S+ Wn + Wn2 - 1 ) ( S+ Wn - Wn

21)

Taking ILT, C(t) = 1+ 1 . e ( +21) Wn t.

S21 ( + 21)

1 . e ( +2

1) Wn t.S 21 ( + 21)

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

1t

.- e

-S2

t. ; t > O

S 21 S1 S2

Where S1 = ( +21) Wn

S2 = ( -21) Wn

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

The performance characteristics of a controlled system are specified in terms of thetransient response to a unit step

Documents

Ece IV Control Systems [10es43] Notes