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CONTROL SYSTEMS ENGINEERING (III B.Tech I SEM, JNTUH R13)

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Control Systems 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 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 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.

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Control Systems

input

Actuating signal nput

output

Controller

System

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. Induced Voltage Speed of the Rotational Generator Prime mover Inputs Output 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. Cleanliness of clothes Time 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 V

Actual Water level c

W at er in

level r Valve VO C

Water out

WATER TANK Fig 1-4 b) Open loop control

Fig -1.4 a) Water level control

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Control Systems 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.

Reference i n p u t

Forward path

Error detector Actuating / error



signal

Control elements

Controlled output

System / Plant

controller

Feed back elements Feed back signal 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 fig6 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. Human operator Thermometer Desired hot Steam Hot water

old water

ro c

CoCC

Muscles

water. temp +

Steam

Actual Water temp

Brain of operator (r-c)

+

and Valve Thermometer

Drain 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 ECE/ jntuworldupdates.org

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Control Systems wall ab0out 5 feet from the floor. A person selects and adjusts the desired room temperature ( r ) say 25 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. Outdoor temp change (disturbance) ro Desired temp.

c

+

Relay switch

Furnace

Actual Temp. Co C

House 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. ECE/ jntuworldupdates.org

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Control Systems 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 motorreceiver, synchro generator- transmitter. ECE/

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Control Systems 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.

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Control Systems 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.

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Control Systems

a Applied force F

b Piston

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

F2 Load L1

L2

Fulcrum effort F1 ECE/ Control Systems jntuworldupdates.org

It is also

12 called ‗mechanical transformer'

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Equating the moments of the force F1 L1 = F2 L 2 F2 =

Rotational mechanical system

F1 L1

L2 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.

dt2

f = co efficient of viscous friction.

Newton's II law of motion states  T = J d2. d t2 Gear wheel

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Control Systems 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.

Driving wheel N1

N2 Driven wheel

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 d2x dx dt2 + K X = f (t) ---------- 1 +B M+ dt The differential equation for electrical system is d2q d2q q +R L+ d t

dt+

2

2

c = 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

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Electrical System

Force (torque)

Voltage

Mass (Moment of inertia)

Inductance Specworld.in

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Control Systems

Viscous friction coefficient

Resistance

Spring constant

Capacitance

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

d2x M+

dt

2

dx +B

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

dt

For electrical system C

d2x dt

2

+

1

R

+

d

+





= I(t)

L

dt2 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) Mass( Moment of inertia) Viscous friction coefficient Spring constant Displacement ( angular)

Current Capacitance Conductance Inductance Flux Voltage

Velocity (angular)

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Control Systems Illustration 1:For a two DOF spring mass damper system obtain the mathematical model where F is the input x1 and x2 are responses.

b2 (Damper)

k2

Draw the free body diagram for mass m1 and m2 separately as shown in figure 1.10 (b)

x2 (Response)

m 2

Apply NSL for both the masses separately and get equations as given in (a) and (b)

b1

k1 x1 (Response)

m 1

F

.

Figure 1.10 (a) k x b2 x2 2 2

b2

m2

k1 x2

k

k1 x1

bx 1

.

x2

.

1

.

b1 x2

.

k2 x2

m2

..

k1 (x1-x2) b1 (x1-x2)

bx

k1 x1

.2x

1

x 2

m1

1

F

1

b1 x 2

x1

m1

F Figure 1.10 (b) From NSL jntuworldupdates.org

 F= ma

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For mass m1

..

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Control Systems m1x1 = F - b1 (x1-x2) - k1 (x1-x2)

--- (a)

For mass m2 m2..2 = b1 (x2 .1) + k1 (x2-x1) - b2 .2 - k2 x2 x x .x

--- (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.

K1 m 1

K2

m

K3

2

X 1

X 2

Figure 1.11 (a)

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m

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X1 K 1 X1

K 2 X2 K 2 X2

K1 X1 X1 K 2 X1

m1 K2 (X2 - X1)

K 2 X1

K2 (X2 - X1)

m2

m2 X2 K 3 X2

K 3 X2

X2

Figure 2.16 (b) Based on Newton's second law of motion: F = ma For mass m1 m1..1 = - K1x1 + K2 (x2-x1) x m1..1 + K1x1 - K2 x2 + K2x1 = 0 x m1..1 + x1 (K1 + K2) = K2x2 x

----- (1)

For mass m2 m2..2 = - K3x2 - K2 (x2 - x1) x m2..2 + K3 x2 + K2 x2 - K2 x1 x m2

..2 + x2 (K2 + K3) = K2x1 x

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----- (2)

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Control Systems Mathematical models are: m1.. + x1 (K1 + K2) = K2x2 x

----- (1)

m2..2 + x2 (K2 + K3) = K2x1 x

----- (2)

1

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

K2

K

U(t) (input)

M

1

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

K

X1

B

F(t) (input)

M

2

(output)

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

B1 X1 f12

K1

M2

M1 f(t)

X2

f1

f2 19

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Control Systems

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

K

X i

X

M

M

force f(t)

Xo

B

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

K

X

Xo

Xi

F

Yk

C

Xo

i

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.

f ( t )

b

a

K

X

M

B jntuworldupdates.org

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

F

X2

K

orce f

a

1

b

M1

M2

B1

K2

X1 8. Write the force equations for the mechanical systems shown in figure.

B1 J1

T(t) 9. Write the force equation for the mechanical system shown in figure.

T ( t )

J1

K

2

 1

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

J2

Torque S IT T ECE

JB Specworld.in

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1 K1 J1

2 K2 J2 B1

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3 B2

K3 J3 21 B3

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Control Systems 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. 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.

J2

J1 Torque T1,1

K

B2 B1

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.

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Control Systems 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) . The parameters of the plant are given below. u Motor torque constant Motor back emf constant Generator gain constant Motor to load gear ratio

(s) = = = =

KT in N.M /amp KB in V/ rad / Sec KG in v/ amp 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

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Control Systems 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

T2 1 JL = load inertia. Jm = motor inertia, KL = Torsional spring constant

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'

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Control Systems Viscous friction coefficient ‗f'

16. Find the transfer function E0 (s) / I(s) C1 I

E0 C2

R

Output

input

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Control Systems 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?

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Control Systems

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) G(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. ECE/

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Control Systems 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 R(s)

G(s)

+

-

Branch point 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. ECE/

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Control Systems 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)

G( C B(s)

(ss) ) H(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)

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Control Systems

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)

+ +

-

)

)

G1(s

G2(s

C(s)

H(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 ECE/

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Control Systems

CR(s) = R(s)

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

[G1(s)R(s) + N(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

ei

i

eo C

Fig. 2.6a

ei = iR + 1/c idt

-----------(1)

And ---------(2)

eo = 1/c idt Equation (1) becomes ei = iR + eo ei - eo R

ECE/

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=i

--------------(3)

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Control Systems Laplace transforms of equations (2) & (3) are 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.

Ei(s)

+

1/R _ Eo(s)

I(s)

I(s)

1/C

Eo(s) + 1/

Fig2.6(b)

_

R

Eo(S)

FiS .6(c) g2 I(s)

Eo(s)

1/C S

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. ECE/ jntuworldupdates.org

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Control Systems Output node: A node that has only incoming branches. This is also called as sink and corresponds to dependent variable. 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 X k 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.

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Control Systems 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. Mason‟s 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 P 1 Pkk k 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 X1 R



X2  

X3 G1

 

H2 X5

X4 G2

X6

C

G3



H1 Figure 2.7 Multiple loop system ECE/

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Control Systems 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 -H2 R

1

X1

X2

1

G1

X3

1

G2

G3

X4

1

X5

C

X6

H1

-1 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. G1 X1

X2

R

C



  

G2

G3 X3   G4

Figure 2.9 Block diagram feedback system ECE/

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Control Systems The nodal variables are X1, X2, X3. The signal flow graph is shown in Fig. 2.10. G1 G2

R

X2

1

X3

1

C

X1 -G3 G4

Figure 2.10 Signal flow graph of example 2 Example 3 Draw the signal flow graph of the system of equations. X1 a11X1 a12 X 2 a13X 3 b1u1 X 2 a21X1 a22 X 2 a23X 3 b2u2 X 3 a31X1 a32 X 2 a33X 3 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.

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Control Systems a a12

u

1

2

u

b

a11

b

2

1

1

X2

a33

a32

a21

X 1

X 3

a23

a31 Figure 2.11 Signal flow graph of example 2 Example 4 LRC net work is shown in Fig. 2.12. Draw its signal flow graph. R

L 

e(t) 

i(t)

C

ec(t)



Figure 2.12 LRC network The governing differential equations are L di Ri 1 idt et1 dt C or L di Ri ec et2 dt C dec it3 dt Taking Laplace transform of Eqn.1 and Eqn.2 and dividing Eqn.2 by L and Eqn.3 by C ECE/ jntuworldupdates.org

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Control Systems sIs i 0 R IS 1 Ecs 1 Es4





L

L

L

sEcs ec 0 1 Is5





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

i(0+)

ec(0+)

1 LsR



1 L



1 Cs

s

1 s Ec(s)

I E(s) L

L



s  R



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ntrol Systems 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. Mason‟s Gain Formula Gain P is given by P

1

 P 

k

k

k

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. Example 1 Obtain the transfer function of C/R of the system whose signal flow graph is shown in Fig.2.13

G1 R

G2

1

-G3

1

C

G4

Figure 2.13 Signal flow graph of example 1 There are two forward paths: Gain of path 1 : P1=G1 Gain of path 2 : P2=G2

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Control Systems There are four loops with loop gains: L1=-G1G3, L2=G1G4, L3= -G2G3, L4= G2G4 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 G1 G2

The transfer function T = Cs  P11 P22  Rs



1 G1G3 G1G4 G2G3 G2G4

Example 2 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig.2.14.

R ( s )

1

1

G1

G2

H1

-H2

G3

1

C(s)

-1

Figure 2.14 Signal flow graph of example 2 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. The transfer function T = Cs  P11 

jntuworldupdates.org

G1G2G3

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Rs



1 G1G2 H1 G1G3H 2 G1G2G3

Example 3 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig.2.15.

EControl Systems

G6

R(s)

G1

G3

G2 X1

G7

X2

G4 X3

G5 X4

1 C(s) X5

-H1

-H2

Figure 2.15 Signal flow graph of example 3 There are three forward paths. The gains of the forward path are: P1=G1G2G3G4G5 P2=G1G6G4G5 P3= G1G2G7 There are four loops with loop gains: 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= C(s) / R(s)

Cs

P11 P22 P33

R









s

jntuworldupdates.org

G1G2G3G4G5 G1G4G5G6 G1G2G71 G4H1 

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1

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G

4

H1 G2G7 H2 G6G4G5H2 G2G3G4G5 H2 G2G4G7 H1H 2

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example 4 Find the gains X 6 , X 5 , X 3 for the signal flow graph shown in Fig.2.16. X1 X 2 X1

b

X1

a

-h

c

d

e

X3

X2

X5 f

X6

X4

-g -i

Figure 2.16 Signal flow graph of MIMO system

Case 1: X 6 X1 There are two forward paths. The gain of the forward path are: P1=acdef P2=abef There are four loops with loop gains: L1=-cg, L2=-eh, L3= -cdei, L4=-bei There is one combination of Loops L1 and L2 which are nontouching with loop gain product 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 = X 6 P11

P22cdef abef

ECE/ jntuworldupdates.org

X1



1 cg eh cdei bei cgeh

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Control Systems

Case 2: X 5 X2 The modified signal flow graph for case 2 is shown in Fig.2.17. b X2

-h

c

1

d

X2

e

X3

X5

X5 1

X4

-g -i

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= X 5 P11 P22cde

be

X2



1 cg eh cdei bei cgeh

Case 3: X 3 X1 The signal flow graph is redrawn to obtain the clarity of the functional relation as shown in -h Fig.2.18. c a X2 b e f 1 X X3 5 X1 X4

-i

d

X3

-g Figure 2.18 Signal flow graph of example 4 case 3 ECE/ Control Systems

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here are two forward paths. The gain of the forward path are: P1=abcd P2=ac There are five loops with loop gains: L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg There is one combination of Loops L1 and L2 which are nontouching with loop gain product 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 = X 3 P11

P22abef ac1 eh

X1



1 eh cg bei efd befg ehcg

Example 5 For the system represented by the following equations find the transfer function X(s)/U(s) using signal flow graph technique. X X 1 3u 

X1a1 X1 X 22u 

X 2 a 2 X 1 1u Taking Laplace transform with zero initial conditions Xs X1s3Us sX1sa1 X1s X 2s2Us sX 2sa2 X1s1Us

Rearrange the above equation Xs X1s3Us X1s a1 X1s 1 X 2s  s 2 Us s

s

X 2s a2 X1s Us 1 s  s The signal flow graph is shown in Fig.2.19. Control Systems 1 U

s

s

1s 

jntuworldupdates.org

2 s

X 1

a1 s

 a 2

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1 X

X

X2

3

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: L1 a1 s L2 a 2 s2 L1=-eh, L2=-cg, L3= -bei, L4=edf, L5=-befg There are no combination two Loops which are nontouching.  1 a1 a22 ss Forward path 1 does not touch loops L1 and L2. Therefore 1 1 a1 a22 ss Forward path 2 path 3 touch the two loops. Hence, ∆2= 1, ∆2= 1.

 The transfer function T =



X 3 P11 P22 P333 s2 a1s a22s1 X1  s2 a1s a2

ECE/ Control Systems

45

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. jntuworldupdates.org

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X1 R

X2 G1

 



X3 



X4

H2

X5

X6

G2

C

G3

H1 6. Draw the signal flow graph of the block diagram shown in Fig below

G1

X2

X1 R 

C 





G2

G3 

X3

 G4

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Control Systems 7.

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

L  i(t)

e(t) 

C

ec(t)



Figure

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

G6

R(s)

G1

G7

G3

G2 X

X1

G4 X3

2

G5 X4

-H1

1

C(s)

X5

-H2

Q.9 For the system represented by the following equations find the transfer function X(s)/U(s) using signal flow graph technique. X X13u 

X1a1X1 X 22u 

X 2a2 X11u

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Control Systems 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 analyze 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.

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Control Systems

C(t) Ct(t)

Step

Css(t) ess = study state error

O

Transient time

Time Study state Time

The transient response may be experimental or oscillatory in nature.

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 signal. 2. Ramp Input Signals. 3. Parabolic Input Signal. 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. ECE/

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Control Systems 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) = A t for t  0 = 0 for t 0. In LT form R(S) = A2 S If A=1, it is called Unit Ramp Input

Mathematically r(t) = t u(t) {

t for t  0

In LT form R(S) = A2 = 1 2 S

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S

=

0 for t  0 50

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Control Systems 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 2 r At for t  0 ( t ) =

2

Slope = At

0 for t  0

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) = 13 = t2 for t  0 S 4. Impulse Input Signal :

2 0 for t  0

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) ie., t 0 (zero) applied momentarily A O ∆t

0

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 = 0 for t  0

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Control Systems In LT form R(S) = 1 if A = 1 Standard test Input Signals and its Laplace Transforms. R(S) r(t) 1/S Unit Step 1/S2 Unit ramp 1/S3 Unit Parabolic 1 Unit Impulse

First order system:The 1st order system is represent by the differential Eq:- a1dc(t )+ao c (t) = bo r(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 [ It's a dimensionless T.F.] 1. If K=1, Then G(S) = 1+TS I This system represent RC ckt. A simplified block diagram is as shown.; R(S)+

1 TS

C(S)

-

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Control Systems 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:slope = 1 . T

then, C(t)=u (t) - e -t/T ; t.>0.------------- (3) c (t)

At t=T, then the value of c(t)= 1- e -1 = 0.632.

0.632

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. Since dc = 1 .e -t/T = 1 . at t .=0. ------------- (4)

d t

T

T

t 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 d22c(t) + a1 dc(t) + a0 c(t) = b0 .r(t)-----------------(1) d dt t Where c(t) = o/p & r(t) = I/p -- ing (1) by a0, a2 d2 c(t) + a1 . a0 dt2 a0 a2 d2 c(t) + a0 dt2

dc (t) + c(t) = b0 . r(t). dt a0

2a1 . a2 . 2a0 a0 . a2

dc (t) + c(t) = b0 . r(t). dt a0

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

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K . where, S(1+ST) 53

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Control Systems 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 =  2

.  1-2



0.0916 =

(0.0084) (1-2) =2 (1.00842) = 0.0084  = 0.091

1 = 0.091 2 = 0.4037

w.k.t. T.F. =

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

T.F. =

.  1-

=

K/( S + S2T ) . 1+ K . .= 2 (S+S T)

K S + S T+K 2

K/T . S +S+K. TT 2

Comparing with std Eq :Wn =



K. T

, 2Wn = 1 . 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

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Control Systems

0.091 =

K2 . K2 . = 0.0508 K1 0.4037 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 K1 0.6 K 1 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: G(S) = =

25 . S(S+6)

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

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

25 . S2 + ( 6S+25 )

W2n = 25 , Wn = 5,

2Wn = 6 = 6 . = 0.6 2x5 Wd = Wn 1- = 5 1- (0.6) = 4 2

2

tr = - , = tan -1 Wd Wd

 =Wn = 0.6 x 5 = 3

  = tan-1 ( 4/3 ) = 0.927 rad. tp = MP =

ECE/

 . = 3.14 = 0.785 sec. Wd 4  e  1-2

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=

- 0.6 . x3.4 = 9.5% e1- 0.62 55

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Control Systems ts =

4. Wn

4 . = 1.3 3sec. 0.6 x 5

for 2% =

5) The closed loop T.F. of a unity feed back control system is given by C(S) = 2 5 . S + 4S +5 R ( S )

Determine (1) Damping ratio (2) Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse.

Solution: C(S) = 2 5 . , Wn 2 = 5 Wn =5 = 2.236 R S + 4S +5 ( S ) 2Wn = 4 = MP =

4 . = 0.894. Wd = 1.0018 2 x 2.236

= e  1-2

W. K.T. C(t) = e-Wnt

0.894 . e

X 3.14 = 0.19%

 1-(0.894)2

Cos Wdt r + . sin wdt r  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:-

Soutions:-

d2 + 10 d = 150E , E = R- is the actuating signal calculate the dt2 dt value of damping ratio, undamped and damped frequency of ascillation. = 150r - 150. d22 + 10 d = 15 ( r - ) , d t

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Taking L.T., [S2 + 10S + 150] (S) = 150 R (S). ECE/ Control Systems

56 .

 (S) = 2 150 R ( S )

S + 10S + 15O

Wn2 = 150 Wn = 12.25. sec .1 2Wn = 10 = 0.408.

.rad

10

.=

2x 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)inm

t f ( t )

0.0 2

0.0019 3

M

The T.F. of the mechanical system is , X(S) = F( S)

F

x dt

. M 21 + FS = S K f(t) = Md2X + F dX + KX dt2 F(S) = (MS2 + FS + K)

x (S) 12345 Given :- F(S) = 10 S.  X(S) = 10 . S(MS2 + FS +

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K) SX (S) =

10 . 2 MS + FS +

K The steady state value of X is By applying final value theorem, lt. SX(S) = Fig.)

10 S O = 500.)

. = 10 = 0.02 ( Given from M(0) + F (0) + K

K.

(K

MP = 0.00193 = 0.0965 = 9.62% 0.02 Mp =  2

e 1-

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Control Systems

0.744 = .  1 -2

 0.5539 =

2 .  1 -2

0.5539 - 0.55392 =2  = 0.597 = 0.6 t p = =  . Wd Wn 1 -2  . Wn = 1.31 rad / Sec. 3= Wn (1 - (0.6) 2

S x(S) =

2

)

10/ M

.

(S + F S + K ) M M Comparing with the std. 2nd order Eq :-, then, Wn2 = K M

 Wn =

F = 2Wn M

K M

(1.31)2 = 500 . M

M = 291.36 kg.

F = 2 x 0.6 x 291 x 1.31 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 C(S) . = ECE/ jntuworldupdates.org

 R(S) = 1 .

600 / S .R(S) 2 S + 70S + + 600 58 Specworld.in

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Control Systems

24.4 ..rad / Sec

Comparing, Wn2 = 600,

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? K1

1. S(1+S)

R

+C

__ K2S

K 1

C.= 2 R

.

S + ( 1 + K2 ) S + K1

Wn2 = K1 Wn = 2Wn = 1 + K2 = Wd

 K1 1 + K2 2 K1 Wn 1 -2  Wn =

=

9.5 . 10.96 rad/Sec  1 - 0.52

K1 = (10.96)2 = 120.34 2Wn = 1 + K2 , K2 = 9.97

Mp =

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 e 1-2

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Control Systems Mp Ts =

4.= Wn

= 16.3% 4. 0.5 x 10.97

= 0.729 sec

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. Consider the system shown in the fig.

G(S) R(S)

E(S)

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

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

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

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Control Systems 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;  E(S) =

ess = lt t

e(t) = lt SE(S) S O ess = lt S.R(S) . .(2) S O 1+G(S) . H(S)

Substituting (1),

Eq :- (2) Shows that the steady state error depends upon the i/p R(S) and the forward G(S) and loop T.F 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) .= S O 1 +G(S). H.S 1 + lt SO

T.F.

1 . G(S). H(S)

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O lt ECE/

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S . . 1 . = lt 1 . 2 1 + G(S) . H(s) S S O S +S G(S) H(s)S

= SG(S) . H(S) = Kv = velocity co-efficient then 61

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Control Systems SO ess = lt

1. (S + Kv)

SO

 ess

=

1.

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

S

= lt

.1.

lt S

S  

3

1 . 2 2 SO S + S G(S) . H(S)

O 1 + G ( S ) . H ( S ) 2

lt

SO  ess

= lt

Note :-

S

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

1.

SO

=

1. S + Ka 2

Ka

ess = R . for non unit parabolic. Ka

Types of feed back control system :jntuworldupdates.org

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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 +n1) (TbS +1).. S (T1 S+1) (T2S + 1). Where K = open loop gainn. Above Eq:- involves the term S 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.

ECE/

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Control Systems 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. S+1

ess (Position) = .. .

Kp =

lt

[

.. .

H(s) = 1]

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

S0

ess (Velocity) = 1 . = Kv

= 0.

ess (acceleration) = 1 . = Ka

1 .= 0

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

S2

If, n = 1, the ess to various std, i/p, G(S) =

ess (Position) = Kp =

K. S (S + 1)

1 .=O 1+ lt

S 0

lt

Kv =

K . =0 S+1

S 0

2) Type 1 -System :-

=K

1 .= 0

Kv = lt G(S) . H(S) = lt S K . S0 S 0 S + 1

S0

K. S+1

S0

1. 1 + Kp

S 0

ess (Velocity) = 1 .

G(S) . H(S) =

lt SO

SK. S(S+1)

=K

K . = S( S + 1)

K

ess (acceleration) = 1 . = 0 ECE/

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Control Systems

Ka = lt S0

S2

K . = 0. S (S + 1)

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

..

K. S (S + 1) 2

lt K .= 2 S 0 S (S + 1)

.

ess (Position) = 1 . = 0 

Kv = lt ..

.

S 2 K .= S (S + 1) ess (Velocity) = 1 . = 0  S0

Ka = lt ..

.

2

S 2 K . = K. S0 S (S + 1) ess (acceleration) = 1 . K

3) Type 3 -System :- Gives Kp = Kv = Ka = & ess = 0. (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 1+2-1 S S2

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Taking L.T, C(s) = 5

1 +5

15 S

2

S2

=

1 2 (S+2)

= 5 2

C(s) = 5

S2+2S+2S+4— s2 S2 (S+2)

5

2

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

S2(S+2)

10 (S+1)

S2 (S+2)

=

S2

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) 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

G(s) =

100 S (S+10)

2 position error co-efficient

KP = lt S

0

S

Similarly KV = lt S 10

0

SG(s) = lt 0

S

lt S

0

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S

100 x s 0

100 x s2 0

G(s) = lt

100

=

S (S+10)

S (S+10)

= 100

= 10

S (S+10)

=

0

Ka = lt S2 G(S)

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Control Systems Given

:-

r(t)

=

Po+P1t +P2 t2 2

R1

Therefore steady state error ess

R1 R3+

ess

1+Kp

R2 +

1+ 

10

R2 +

R3

Kv

+

Ka

P1

P0 P2

= 0



1+

+

10

+

0

Ess = 0+0.1 P1 + = 3. Determine the error co-efficeint and static error for G(s) And H(s) = (S+2)

= S(S+1) (S+10)

The error constants for a non unity feed back system is as follows Kp = lt G(S) H(S) = (0+2) lt (S+2) S 0 G(S).H(S) =S 0(0+1) (0+10) 0 S(S+1) (S+10)

Kv = lt lt S 0

G(S) H(S) =

1

=

(0+2) = 1/5 = 0.2

0

S

0(0+1) (0+10)

Ka = 0 Static Error:Steady state error for unit step i/p = 0 Unit ramp i/p

1

=

Kv

1

=5

0.2

Unit parabolic i/p = 1/0 = ECE/ jntuworldupdates.org

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Control Systems

50 4. A feed back C.S is described as G(S) = H(S)=1/s.

S (S+2) (S+5)

For unit step i/p,cal steady state error constant and errors.

Kp = lt

G(S) H(S) =

S

0

S

S2

0

50 (S+2) (S+5)

Kv = lt S 0

G(S) H(S) = 0

lt

=

50 x S

S

= S2 (S+2) (S+5)

Ka = lt S 0

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

S2 x 50 = S2 (S+2) (S+5)

Ess = lt S

The steady state error

0

Lt

S. 1/S 1+50

S

50 =5 10 S2 (S+2) (S+5)

0

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

S2(S+2)(S+5)

= 0/50 = 0

K H(S) = 1

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

S2 (S+20) (S+30)

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. Kp = lt S

G(S) H(S) = lt 0

S

50 0

= S (S+20) (S+30) 2

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Control Systems

Kv = lt S

Ka = lt S

S 0

K S2 (S+20) (S+30)

0

=

S22 K K S (S+20) (S+30) 600

Steady state error:-

1

Error due to unit step i/p

1+Kp +

Error due o r(t) ramp i/p

1 =0

1+ 

10

10 +

=0

Kv  20 Error due to para i/p, ,

=

Ka

40 2Ka

=

20 x 600 K

12000 = K

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

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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 [ It's a dimensionless T.F.] 1. If K=1, Then G(S) = 1+TS I This system represent RC ckt. A simplified bloc diagram is as shown.; R(S)+

1 TS

C(S)

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+TS

1. = S

1. S

Taking inverse L.T. for the above Et/T:- q t e h e n ,

T. TS+1

----------------- (2)

; t.>0.------------(3)

C ( t ) = u ( t ) jntuworldupdates.org

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slope = 1 . T At t=T, then the value of c(t)= 1- e -1 = 0.632. The smaller the time const. T. the faster the system response. The slope of the tangent line at at t= 0 is 1/T.

c (t) 1 - e -t/T

0.632

Since dc = 1 .e -t/T = 1 . at t .=0. ------------- (4)

d t

T

T

t 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= ECE/

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Control Systems T Second order system:The 2nd order system is defined as, a2 d22c(t) + a1 dc(t) + a0 c(t) = b0 .r(t)-----------------(1) dt d t

Where c(t) = o/p & r(t) = I/p -- ing (1) by a0, a2 d2 c(t) + a1 . a0 dt2 a0 a2 d2 c(t) + a0 dt2

dc (t) + c(t) = b0 . r(t). dt a0

2a1 . a2 . 2a0 a0 . a2

dc (t) + c(t) = b0 . r(t). dt a0

Step response of 2nd order system: The T.F. = C(s) = Wn2 2+2W S+ W R(s) 3 n n2

Based on value

The system may be, 2) Under damped system (0 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) . e ( +2 - 1) Wn 1 2 2 - 1 (+ t. 1)

T

1 .e ( + (2 - 1)) Wn t. S (2 - 1 )[(+ (2 - 1)] C(t) = 1+ Wn . S 2 - 1

e-S1t . - e-S2t . S1 S2

;t>O

Where S1 = ( +2 - 1) Wn 2 S2 = ( - - 1) Wn

Time response (Transient ) Specification (Time domain) Performance :ECE/ jntuworldupdates.org

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

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 time.

for the 1st

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. ECE/ jntuworldupdates.org

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Control Systems

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) S 2+2 W S+ W 2 n

n

Assuming the system is to be underdamped (< 1) Rise time tr W. K.T. C(tr) = 1- e-Wnt Cos Wdtr + . sin wdtr  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 =  1-2 

wd .  jW

Thus, the rise time tr is , tr =

1 . tan-1 - w d =- secs Wd  wd When must be in radians.

jWd Wn 1-2

 Wn

-  W n S- Plane

Peak time :- (tp)

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Control Systems 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-st2 Since the peak time corresponds to the 1 peak over shoot.  Wdtp = = tp = . Wd The peak time tp corresponds to one half cycle of the frequency of damped oscillation. Maximum overshoot :- (MP) The max over shoot occurs at the peak time. i.e. At t = tp = . Wd  M -(/ Wd) or e ( / 1-2) p = e

Settling time :- (ts) An approximate value of ts can be obtained for the system O 0

S2

0.65

K

from s1,

S1

0.65 - 0.1K 0.65 K

0

0.65 - 0.1K > 0  0.65 > 0.1 K  6.5 > K

S0

 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 ECE/

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Control Systems makes row of s0 as row of zeros but K = 0 can not be marginal value bec0ause for K = 0, constant term in characteristic equation becomes zeros ie one coefficient for s 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 ; s  s c

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 = frequency of oscillations = 3.162 rad/ sec. Ex : 3 For a system with characteristic equation F(s) = s5 + s4 + 2s3 + 2s2 + 3s +15 =, examine the stability Solution : S5 1 2 3 S4

1

2

15

S3

0

-12

0

S2 S1 S0

S5

1 1

2

15

S3



-12

0

S1

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3

S4

S2

ECE/

2

(2 + 12) 15  (2 + 12)( -12 ) - 15  2 + 12 

0

0

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Control Systems S0

15 2 + 12

Lim  0

12 =2 +

=2+ = +

 Lim 



(2 + 12)( -12 ) - 15  2 + 12 

0

Lim =

-24 - 144 - 152 0 2 + 12

0 - 144 - 0 =

= - 12 0 + 12

S5

1

S4

1

S3



2

3

2

15

-12

0 There are two sign changes, so system is unstable.

S2

+

15

S1

- 12

0

S0

0

15

Ex : 4

Using Routh Criterion, investigate the stability of a unity feedback system

whose open loop transfer function-isT s e G(s) = s(s+1) So l :

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 ECE/

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Control Systems

e

-sT

s2 T2 = 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 0. Th system is over damped. Problem No 2 The open loop transfer function is G(s) K(s 2) . Sketch the root locus plot (s1)2 Solution: 1) Root locus is symmetrical about real axis. 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 180 (2q1) , q n m1 0.  = nm Angle of asymptote isA = 180. 5) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm  (11 ) (2) 0.0 1

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6) The root locus has break points. 2

K (s 1) (s 2) Break point is given by dK 0 ds 2(s 1)(s 2) (s 1) 0 2 (s 2)2 s11, K 0; s23, K 4 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

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Control Systems Comments on stability: 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. Problem No 3 The open loop transfer function is G(s) K(s 4) . Sketch the root locus. s(s 2) Solution: 1) Root locus is symmetrical about real axis. 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 180 (2q1) , q 0,1,n m1 0.  = nm Angle of asymptote areA = 180.

5) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm  (2) (4) 2.0 1

6) The brake points are given by dK/ds =0. K s(s 2) (s 4) dK (2s 2)(s 4) (s2 2s) 0 ds (s 4)2 s11.172, K 0.343; s26.828, K 11.7 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 ECE/ jntuworldupdates.org

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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.

Figure 3 Root locus plot of K(s+4)/s(s+2) Comments on stability: 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 G(s) K(s 0.2) . Sketch the root locus. s2 (s 3.6) Solution: 1) Root locus is symmetrical about real axis. 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 .

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Control Systems 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 180 (2q1) , q n m1 0,1  = nm Angle of asymptote areA = 90, 270. 5) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm  (3.6) (0.2)1.7 2

6) The root locus does branch out, which are given by dK/ds =0. K - (s 3.6s ) 3

s 0.2

2

dK (3s2 7.2s)(s 0.2) (s3 3.6s2 ) ds (s 0.2)2 2s3 4.8s2 1.44s 0 s 0, 0.432,1.67 and K 0, 2.55,

3.66 respectively.

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. The root locus plot is shown in Fig.4.

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Figure 4 Root locus plot of K(s+0.2)/s2(s+3.6) Comments on stability: 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 G(s)

K s(s 6s 25)

. Sketch the root locus.

Solution: 1) Root locus is symmetrical about real axis. 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. 180 (2q1) , q 0,1,n m1 0, 1, 2. nm 4) The asymptote angle isA = Angle of asymptote areA = 60, 180, 300. 5) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm  (3 3)2.0 3

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6) The brake points are given by dK/ds =0. Ks(s2 6s 25) (s3 6s2 25s) dK 3s2 12s 25 0 ds s1,2 2 j2.0817and K1,2 34 j18.04 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 p 180  (sum of the angles of vectors to a complex polein question from other poles)  (sum of the angles of vectors to a complex poleinquestion from zeros)

p 180 (180 tan1 4 90 )36.87

3 Similarly, Angle of departure from complex pole s= -3-j4 is p 180 (233.13 270 )323.13 or 36.87

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 s3 6s2 25s K 0 . The Routh's array is s3 1 25 s2 6 K 150 K 1 s 6 s0 6 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. s3 6s2 25s K 0

j3 6j2 25j K 0 (62 K) j 2 25 0





 0, j5 K 0, 150

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Control Systems The plot of root locus is shown in Fig.5.

Figure 5 Root locus plot of K/s(s2+6s+25) Comments on stability: 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 K(s 2) is G(s)H(s) . Comment on the stability of the system. (s1)(s 3 j)(s 3 j) Solution: 9) Root locus is symmetrical about real axis. 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 180 (2q1) , q n m1 0,1.  = nm Angle of asymptote isA = 90, 270.

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13) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm  (1 3 3) (2)2.5 1

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.

p 180  (sum of the angles of vectors to a complex polein question from other poles)  (sum of the angles of vectors to a complex poleinquestion from zeros)

1 tan1 1 26.57 or 153.43 -2   atan2(-2,1) 153.43 1

 tan1 1 -45 or 135 ,   tan1 2 90 3 0 1   180 (153.43 90 ) 135 71.57 p









The angle of departure at the complex pole at s=-3-j is -71.57.

p 180 (206.57 270 ) 22571.57 16) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross over. The root locus plot is shown in Fig.1

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Control Systems

Figure 1 Root locus plot of K(s+2)/(s+1)(s+3+j)(s+3-j) Comments on stability: The system is stable for all the values of K > 0. Problem No 2 The open loop transfer function is G(s)H(s)

K

Sketch the root locus

s(s 0.5)(s  0.6s10)

plot. Comment on the stability of the system.

2

.

Solution: 10) Root locus is symmetrical about real axis. 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 180 (2q1) , q n m1 0,1,2,3.  = nm Angle of asymptote isA = 45, 135, 225, ±315. 133 ECE/ jntuworldupdates.org

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Control Systems 14) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm

The value of K at s=-0.275 is 0.6137.

 (0.5 0.3 0.3)0.275 4

15) The root locus has break points. K = -s(s+0.5)(s2+0.6s+10) = -(s4+1.1s3+10.3s2+5s) Break points are given by dK/ds = 0 dK 4s3 3.3s2 20.6s 5 0 ds

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 p 180  (sum of the angles of vectors to a complex polein question from other poles)  (sum of the angles of vectors to a complex poleinquestion from zeros) 1 tan1 3.14884.6 or 95.4 - 0.3

  tan1 3.148 86.4 ,   tan1 6.296 90 2 3 0 0 . 2 The angleofdepart (95.4th 86.4pe90polt91.= -0.3 - j3.148 is 91.8 180 ure at e com l x ) e a s 8 p









p 180 (264.6 273.6 270 ) 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

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134

The Routh's array is s 1 4

3

s 1.1 2 s 5.75 s1 s0 K

10.3 K 5 K

28.75 -1.1K 5.75

The system is stable if 0 < K < 26.13 The auxiliary equation at K 26.13 is 5.75s 2+26.13 = 0 which gives s = ± j2.13 at imaginary axis crossover. The root locus plot is shown in Fig.2.

Figure 8 Root locus plot of K/s(s+0.5)(s2+0.6s+10) Comments on stability: System is stable for all values of 26.13 >K > 0. The system has sustained oscillation 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 G(s)

ECE/ jntuworldupdates.org

K s(s 4)(s  4s 20) 2

. Sketch the root locus.

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Control Systems Solution: 10) Root locus is symmetrical about real axis. 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 180 (2q1) , q n m1 0, 1, 2, 3  = nm Angle of asymptote areA = 45, 135, 225, 315. 14) Centroid of the asymptote is A (sum of poles) (sum of zeros) nm  (2.0 2.0 4.0)2.0 4

15) The root locus does branch out, which are given by dK/ds =0. Ks(s 4)(s2 4s 20)  (s4 8s3 36s2 80s) Break point is given by dK 0 ds 4 s 24s 72s 80 0 3

2

4s  8s 16s2 32s 40s 80 0 3

2

(s 2)(4s216s 40) s12.0, K 64; s22.0 j2.45, K 100 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.

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Control Systems

  180 p

 (sum of the angles of vectors to a complex polein question from other poles)  (sum of the angles of vectors to a complex poleinquestion from zeros) 1 tan1 4 63.4 or 116.6 -2   atan2(4,-2) 116.6 1

  tan1 4 63.4,   tan1 8 90 2 3 2 0   180 - (116.6 63.4 90 )90 p









The angle of departure at the complex pole at s = -2 - j4 is 90

p 180 - (243.4 296.6 270 ) 270 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 s4 8s3 36s2 80s K 0 The Rouths array is s 4

1

36 K 80 K

s3 8 s2 26 s1 2080 8K 26 s0 K 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

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Control Systems s4 8s3 36s2 80s K 0 put s j

j4 8j3 36j2 80j K 0



4

 362 K j 83 80 0





  

Equate real and imaginary parts to zero  83 80 0  0, j 10;s j 10 4 362 K 0 K 260 The root locus plot is shown in Fig.3.

Figure 9 Root locus plot of K/s(s+4)(s2+4s+20) Comments on stability: For 260 > K > 0 system is stable K = 260 system has stained oscillations of10 rad/sec. K > 260 system is unstable.

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Control Systems

Recommended Questions: 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 G(s) K . s 4. The open loop transfer function is G(s) K(s 2) . Sketch the root locus plot. (s1)2 5. The open loop transfer function is G(s) K(s 4) . Sketch the root locus. s(s 2) K 6. The open loop transfer function is G(s) . Sketch the root locus. s(s 6s 25) 7. The open loop transfer function is G(s)

ECE/

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K s(s 4)(s2 4s 20)

. Sketch the root

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Control Systems UNIT: 06 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 ECE/

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Control Systems 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 F ( S ) O( S ) , Where F (S) is transfer function, O(S) and I(S) are the Laplace transforms I (S ) of the output and input respectively. 2. Replace S by (j) (As S is a complex number)  F( j) O( j) I ( j)

 O( j) A() B( j) (another complex number) I ( j) 3. For various values of, ranging from 0 to  determine M () and. O( j) M ()  A() B( j) A jB I ( j) M

A2 B 2

 O( j)A jB I ( j)  tan1 B

A

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Control Systems 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

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Control Systems the polar plot. It is easier to construct a polar plot and ready information of magnitude ratio and phase angle can be obtained. Img

+90, -270

Positive angles M 

+180, -180

0, + 360, -360 Real

Negative angles +270, -90 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.

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Control Systems Figure 6.3: A Typical Polar Plot 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)

K

•

y (t) (output)

C

Governing Differential Equation: C. dy Ky Kx dt

÷ by K

C . dy y x K dt



y

.

 

d y

x

  jntuworldupdates.org

T

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a k e 144

L a p l a c e t r a n s f o r m dt

 SY(S) + Y(S) = X(S) (S+1) Y(S) = X(S) ECE/

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Control Systems

Transfer function F(S) = Y (S) 1 X (S)S1 Given  = 0.1 sec 1   Y (S )  X (S) 0.1S1 1 0.1S

1

To obtain the polar plot (i.e., frequency response data) replace S by j. 1

Y

1



( j) X ( j) 0.1 j1 1 j(0.1)

Magnification Factor M =

M

Y ( j) 1  X  1 j(0) 1 j(0.1) 1 j(0.1) ( j)

1 1 (0.1)2

Phase angle = Y ( j)Y ( j)Xj1 j(0)(1 0.1) X ( j) 1

 tan1 0 tan (0.1)  1

1

 tan (0.1) 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  0 2 4 5 6 jntuworldupdates.org

M

1.00 0.98 0.928 0.89 0.86

1

1 1 (0.1)2

 tan (0.1) 0 -11.13 -21.8 -26.6 -30.9 Specworld.in

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Control Systems 10 20 40 50 100 

0.707 0.45 0.24 0.196 0.099 0

-45 -63.4 -76 -78.69 -84.29 -90

The data from Table 6.1 when plotted on the complex plane with as a parameter polar plot is obtained as given below.

• •

• • =6



=

5

0

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 = 20

•·

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. x (Input)

K y (Response)

m C

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Control Systems The transfer function of the system is given by n Givenn = 10 rad/sec and = 0.5 2 Y (S )  X (S) S 2 2 nSn2  Y (S) 2 100 X (S) S10S100 Replace S by j 100

Y



100 j(0)

as j1

( j) X ( j) ( j)10 j100 210 j100 2

100 j(0)

1002



 M Y ( j) X ( j) (100 ) j(10)

(100 )2 (10)2

2

Magnification Factor = M

2

100 (100 )  (10)2 2 2

100 j(0) Phase angle = (100 ) j(10) 2  tan1 0 tan1 10 2  

100 j(0)(100 ) j(10) 2



1 0 0    



100

Now obtain the values of M and for various value of ranging from 0 to  as given in the following Table 6.2. jntuworldupdates.org

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Table 6.2 Frequency Response Data: Illustration 2 M(  ) 0 1.00 2 1.02 5 1.11 8 1.14 10 1.00 12 0.78 15 0.51 20 0.28 40 0.06 70 0.02  0.00 ECE/

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 0.0 -11.8 -33.7 -65.8 -90.0 -110.1 -129.8 -146.3 -165.1 -171.7 -180.0

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Control Systems

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 10 F (S ) (S1)

Replace S by j

F( j) 10 j1 M () F( j) 10 21  () =F (j) =10 -(j+1)  tan1 1

Table 6.3 Frequency Response Data: Illustration 3

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Control Systems 

M()  1.00 9.8 9.3 8.6 7.8 4.5 3.2 2.5 1.9 0.99 0.00

0 0.2 0.4 0.6 0.8 2.0 3.0 4.0 5.0 10 

0 -11 -21 -31 -39 -63 -72 -76 -79 -84 -90

= M() = 0

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 (a jb)(c jd ) A jB F(S) (e jf ) Magnitude Ratio = M

--- (Transfer function)

a jb c jd e jf

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

a2 b2 * c2 d 2 e2 f

2

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Control Systems Phase angle:

a jb

c jd(a jb)(c jd )(e jf )  

e jf

 tan1 b tan1 d tan1 f a c

e

II Values of tan functions IQ:

tan-1 (b/a) - Positive:

IIQ:

tan (b/-a) - Negative: 180 -

(-a+jb)

IQ (a+jb)

-1

IIIQ: tan-1 (-b/-a) - Positive: 180 +

(-a-jb)

IVQ: tan-1 (-b/a) - Negative: 360 - IIIQ

tan-1 (0) = 00

tan-1 (-0) =1800

tan-1 (1) = 450

tan-1 (-1) = 1350

tan-1 () = 900

tan-1 (-) = 2700

III

Img

IIQ

Real (a-jb) IVQ

Let K = Constant = K + j(0)

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150

Therefore K K 2 02 K

Applicable for both K>0 and K0

-1

= tan-1 (-0) = 1800, IV:

• •

Real K+j0

if K