Control Systems. CONTROL SYSTEMS (Common to EC/TC/EE/IT/BM/ML) Sub Code: 10ES43 IA Marks : 25 Hrs/ Week: 04 Exam Hours : 03 Total Hrs: 52 Marks : 100

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1 CONTROL SYSTEMS (Common to EC/TC/EE/IT/BM/ML) Sub Code: IA Marks : 5 Hrs/ Week: 04 Exam Hours : 03 Total Hrs: 5 Marks : 00 UNIT : Modeling of Systems: Introduction to Control Systems, Types of Control Systems, Effect of Feedback Systems, Differential equation of Physical Systems -Mechanical systems, Friction, Translational systems (Mechanical accelerometer, systems excluded), Rotational systems, Gear trains, Electrical systems, Analogous systems. 7 Hrs UNIT : Block diagrams and signal flow graphs: Transfer functions, Block diagram algebra, Signal Flow graphs (State variable formulation excluded), 6 Hrs UNIT 3: Time Response of feedback control systems: Standard test signals, Unit step response of First and second order systems, Time response specifications, Time response specifications of second order systems, steady3 state errors and error constants. Introduction to PID Controllers (excluding design) 7 Hrs UNIT 4: Stability analysis: Concepts of stability, Necessary conditions for Stability, Routh- stability criterion, Relative stability analysis; more on the Routh stability criterion. 6 Hrs UNIT 5: Root Locus Techniques: Introduction, The root locus concepts,construction of root loci 6 Hrs UNIT 6: Frequency domain analysis: Correlation between time and frequency response, Bode plots, Experimental determination of transfer functions, Assessment of relative stability using Bode Plots. Introduction to lead, lag and lead-lag compensating networks (excluding design). 7 Hrs UNIT 7: Stability in the frequency domain: Introduction to Polar Plots, (Inverse Polar Plots excluded) Mathematical preliminaries, Nyquist Stability criterion, Assessment of relative stability using Nyquist criterion, (Systems with transportation lag excluded). 7 Hrs UNIT 8: Introduction to State variable analysis: Concepts of state, state variable and state models for electrical systems, Solution of state equations. 6 Hrs SJBIT/ Dept of ECE

2 TEXT BOOK :. J. Nagarath and M.Gopal, Control Systems Engineering, New Age International (P) Limited, Publishers, Fourth edition 005 REFERENCE BOOKS:. Modern Control Engineering, K. Ogata, Pearson Education Asia/ PHI, 4th Edition, 00.. Automatic Control Systems, Benjamin C. Kuo, John Wiley India Pvt. Ltd., 8th Edition, Feedback and Control System, Joseph J Distefano III et al., Schaum s Outlines, TMH, nd Edition 007. SJBIT/ Dept of ECE

3 INDEX SHEET PAGE SL.NO TOPIC NO. I UNIT :Modeling of Systems -. Introduction to Control Systems, Types of Control Systems. Effect of Feedback Systems.3 Differential equation of Physical Systems -Mechanical systems, Friction.4 Translational systems (Mechanical accelerometer, systems excluded).5 Rotational systems, Gear trains.6 Electrical systems, Analogous systems II UNIT : Block diagrams and signal flow graphs 3-4. Transfer functions. Block diagram algebra.3 Signal Flow graphs (State variable formulation excluded) III UNIT 3 :Time Response of feedback control systems 3. Standard test signals Unit step response of first order systems 3.3 Unit step response of second order systems 3.4 Time response specifications 3.5 Time response specifications of second order systems 3.6 steady3 state errors and error constants) 3.7 Introduction to PID Controllers (excluding design) IV UNIT 4 : Stability analysis Concepts of stability, Necessary conditions for Stability 4. Routh- stability criterion 4.3 Relative stability analysis; More on the Routh stability criterion. V UNIT 5 : Root Locus Techniques Introduction, The root locus concepts 5. Construction of root loci (problems) VI UNIT 6 : Frequency domain analysis Correlation between time and frequency response 6. Bode plots 6.3 Experimental determination of transfer functions 6.4 Assessment of relative stability using Bode Plots 6.5 Introduction to lead, lag and lead-lag compensating networks (excluding design) VII UNIT 7 : Stability in the frequency domain Introduction to Polar Plots,(Inverse Polar Plots excluded) 7. Mathematical preliminaries SJBIT/ Dept of ECE

4 7.3 Nyquist Stability criterion 7.4 Assessment of relative stability using Nyquist criterion, (Systems with transportation lag excluded) VIII UNIT 8 : Introduction to State variable analysis Concepts of state, state variable 8. Concepts of state models for electrical systems 8.3 Solution of state equations SJBIT/ Dept of ECE

5 UNIT- A control system is an arrangement of physical components connected or related in such a manner as to command, direct, or regulate itself or another system, or is that means by which any quantity of interest in a system is maintained or altered in accordance with a desired manner. Any control system consists of three essential components namely input, system and out put. The input is the stimulus or excitation applied to a system from an external energy source. A system is the arrangement of physical components and output is the actual response obtained from the system. The control system may be one of the following type. ) man made ) natural and / or biological and 3) hybrid consisting of man made and natural or biological. Examples: ) 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 ) 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. ) Open loop control system in which the control action is independent of the out put. ) 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. SJBIT/ Dept of ECE Page

6 input Controller Actuating signal System output Water in EXAMPLE - 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. Speed of the Induced Voltage Prime mover Inputs Rotational Generator Output EXAMPLE Washing machine Fig - Rotational Generator 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 -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 V o, and enters the tank from a supply via a control valve V c. The control valve is adjustable manually. Valve V C Valve V O Desired Water level r WATER TANK Actual Water level c C Water out Fig.4 a) Water level control Fig -4 b) Open loop control SJBIT/ Dept of ECE Page

7 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 input Error detector Actuating / error signal Control elements Forward path System / Plant Controlled output controller Feed back signal Feed back elements Fig.5: Closed loop control system EXAMPLE THERMAL SYSTEM To illustrate the concept of closed loop control system, consider the thermal system shown in fig- 6 Here human being acts as a controller. He wants to maintain the temperature of the hot water at a given value r o C. the thermometer installed in the hot water outlet measures the actual temperature C 0 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 Steam Steam Thermometer Hot water Desired hot water. temp r o c + Brain of operator (r-c) Muscles C + and Valve Actual Water temp C o C old water Drain Thermometer Fig -6 a) Manual feedback thermal system EXAMPLE HOME HEATING SYSTEM b) Block diagram 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 SJBIT/ Dept of ECE Page 3

8 wall about 5 feet from the floor. A person selects and adjusts the desired room temperature ( r ) say 5 0 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. Desired temp. r o c + Relay switch Outdoor temp change (disturbance) Furnace Blower House Actual Temp. C o C Fig -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. SJBIT/ Dept of ECE Page 4

9 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. SJBIT/ Dept of ECE Page 5

10 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. SJBIT/ Dept of ECE Page 6

11 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 -8 M Fig -8 a) Mass Fig -8 b) Spring Fig -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 -9. In automobiles shock absorber is nothing but dashpot. SJBIT/ Dept of ECE Page 7

12 Applied force F a 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 Lever. Assume that the system originally is in equilibrium in this way the often-troublesome effect of gravity is eliminated.. 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 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 L L F Load effort F Fulcrum SJBIT/ Dept of ECE Page 8

13 It is also called mechanical transformer Equating the moments of the force F L = F L F = F L L Rotational mechanical system The rotational motion of a body may be defined as motion about a fixed axis. The variables generally used to describe the motion of rotation are torque, angular displacement, angular velocity ( ) and angular acceleration( ) The three basic rotational mechanical components are ) Moment of inertia J ) 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 d A well known example of a torsional spring is a shaft which gets twisted when a torque is applied to it. T s = K, is angle of twist and K is torsional stiffness. d t 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 d Where = relative angular velocity between two bodies. d t f = co efficient of viscous friction. Newton s II law of motion states T = J d. d t Gear wheel In almost every control system which involves rotational motion gears are necessary. It is often necessary to match the motor to the load it is driving. A motor which usually runs at high speed and low torque output may be required to drive a load at low speed and high torque. SJBIT/ Dept of ECE Page 9

14 Driving wheel N N 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 d x dx M + + B + K X = f (t) dt dt The differential equation for electrical system is d q d q q L + dt + R + dt c = e Comparing equations () and () 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 Force (torque) Mass (Moment of inertia) Viscous friction coefficient Spring constant Electrical System Voltage Inductance Resistance Capacitance SJBIT/ Dept of ECE Page 0

15 Displacement Velocity Charge 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 d x M + + B + K X = f (t) dt dx dt For electrical system C d x dt = I ( t ) R d dt L Comparing equations () and () 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 Current Capacitance Conductance Inductance Displacement Flux ( angular) Velocity (angular) Voltage Illustration :For a two DOF spring mass damper system obtain the mathematical model where F is the input x and x are responses. SJBIT/ Dept of ECE Page

16 k k m b (Damper) b x (Response) Draw the free body diagram for mass m and m separately as shown in figure.0 (b) Apply NSL for both the masses separately and get equations as given in (a) and (b) m x (Response) F Figure.0 (a). k x b x. k x b x m.... k x k x b x b x k x k x b x b x x k (x -x ) m. b (x. -x ) m F x m Figure.0 (b) F From NSL F= ma For mass m.... m x = F - b (x -x ) - k (x -x ) --- (a) For mass m..... m x = b (x -x ) + k (x -x ) - b x - k x --- (b) SJBIT/ Dept of ECE Page

17 Illustration : For the system shown in figure.6 (a) obtain the mathematical model if x and x are initial displacements. Let an initial displacement x be given to mass m and x to mass m. K m K X m K 3 X Figure. (a) SJBIT/ Dept of ECE Page 3

18 K X K X m m K X K X X X K (X X ) K X K X K (X X ) m m K 3 X X K 3 X X Figure.6 (b) Based on Newton s second law of motion: For mass m.. m x = - K x + K (x -x ).. m x + K x K x + K x = 0.. m x + x (K + K ) = K x () F = ma For mass m.. m x = - K 3 x K (x x ).. m x + K 3 x + K x K x.. m x + x (K + K 3 ) = K x () Mathematical models are: SJBIT/ Dept of ECE Page 4

19 .. m x + x (K + K ) = K x ().. m x + x (K + K 3 ) = K x ().Write the differential equation relating to motion X of the mass M to the force input u(t) K K M X (output) U(t) (input). Write the force equation for the mechanical system shown in figure X K B M X F(t) (input) (output) 3. Write the differential equations for the mechanical system shown in figure. B X X K f f(t) M f M f 4. Write the modeling equations for the mechanical systems shown in figure. SJBIT/ Dept of ECE Page 5

20 X i K X M X o M force f(t) B 5. For the systems shown in figure write the differential equations and obtain the transfer functions indicated. K F Y k X i X o X i X o C 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 K X a M B Force f 7. Write the equations of motion in terms of given mechanical quantities. K SJBIT/ Dept of ECE Page 6 a b X

21 M B 8. Write the force equations for the mechanical systems shown in figure. B J T(t) 9. Write the force equation for the mechanical system shown in figure. T(t) K J J 0. Write the force equation for the mechanical system shown in figure. K K 3 K 3 J J J 3. Torque T(t) is applied to a small cylinder with moment of inertia J which rotates with in a larger cylinder with moment of inertia J. The two cylinders are coupled by viscous friction B. Torque T B B B 3 SJBIT/ Dept of ECE Page 7

22 The outer cylinder has viscous friction B between it and the reference frame and is restrained by a torsion spring k. write the describing differential equations. J K J B Torque T, B. The polarized relay shown exerts a force f(t) = K i. 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 B 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. 3. 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) SJBIT/ Dept of ECE Page 8

23 u. The parameters of the plant are given below. (s) Motor torque constant = KT in N.M /amp Motor back emf constant = K B in V/ rad / Sec Generator gain constant = K G in v/ amp Motor to load gear ratio = N N 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 = R f Field inductance = L f 4. The schematic diagram of a dc motor control system is shown in figure where K s is error detector gain in volt/rad, k is the amplifier gain, K b back emf constant, K t is torque constant, n is the gear train ratio = = T m B m = motion friction constant T SJBIT/ Dept of ECE Page 9

24 J m = motor inertia, K L = Torsional spring constant J L = load inertia. 5. 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 ) K in V/rad Amplifier gain K p in V / V Motor torque constant K T in V/ rad Gear ratio N N Moment of inertia of load J Viscous friction coefficient f SJBIT/ Dept of ECE Page 0

25 6. Find the transfer function E 0 (s) / I(s) I E 0 C R Output input C SJBIT/ Dept of ECE Page

26 Recommended Questions :. Name three applications of control systems.. 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? 0. Instability is attributable to what part of the total response.. What mathematical model permits easy interconnection of physical systems?. To what classification of systems can the transfer function be best applied? 3. What transformation turns the solution of differential equations into algebraic manipulations? 4. Define the transfer function. 5. What assumption is made concerning initial conditions when dealing with transfer functions? 6. What do we call the mechanical equations written in order to evaluate the transfer function? 7. Why do transfer functions for mechanical networks look identical to transfer functions for electrical networks? 8. What function do gears and levers perform. 9. What are the component parts of the mechanical constants of a motor s transfer function? SJBIT/ Dept of ECE Page

27 UNIT- 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. 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. 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. SJBIT/ Dept of ECE Page 3

28 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 fig. R(s) + - C(s) Fig. C(s) 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. Fig.3 shows an example of a block diagram of a closed system Summing point R(s) + - G(s) Branch point C(s) Fig..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 fig.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. SJBIT/ Dept of ECE Page 4

29 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.4. Another important role of the feed back element is to modify the output before it is compared with the input. B(s) R(s) + C(s) C(s) G(s - B(s) H(s Fig.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 Fig.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)[ + G(s)H(s)] = G(s)R(s) SJBIT/ Dept of ECE Page 5

30 C(s) G(s) = R(s) + 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) - G(s)H(s) Closed loop system subjected to a disturbance Fig.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) + - G(s) + + G (s) C(s) H(s Fig.5 Fig.5 closed loop system subjected to a disturbance. Consider the system shown in fig.5. We assume that the system is at rest initially with zero error. Calculate the response C N (s) to the disturbance only. Response is C N (s) G (s) = R(s) + G (s)g (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 SJBIT/ Dept of ECE Page 6

31 CR(s) R(s) = G (s)g (s) + G (s)g (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) = C R (s) + C N (s) G (s) C(s) = [G (s)r(s) + N(s)] + G (s)g (s)h(s) Procedure for drawing block diagram : To draw the block diagram for a system, first write the equation which describes the dynamic behaviour of each components. Take the laplace transform of these equations, assuming zero initial conditions and represent each laplace transformed equation individually in the form of block. Finally assemble the elements into a complete block diagram. As an example consider the Rc circuit shown in fig.6 (a). The equations for the circuit shown are R ei i C eo Fig..6a e i = ir + /c idt () And e o = /c idt () Equation () becomes e i = ir + e o e i - e o = i (3) R Laplace transforms of equations () & (3) are SJBIT/ Dept of ECE Page 7

32 Eo(s) = /CsI(s) (4) Ei(s) - Eo(s) = I(s) (5) R Equation (5) represents a summing operation and the corresponding diagram is shown in fig.6 (b). Equation (4) represents the block as shown in fig.6(c). Assembling these two elements, the overall block diagram for the system shown in fig.6(d) is obtained. Ei(s) + I(s) Eo(S) I(s) /C /R _ Fig.6(c) Eo(s) Eo(s) + I(s) Eo(s) Fig.6(b) _ /R /C Fig.6(d) SIGNAL FLOW GRAPHS An alternate to block diagram is the signal flow graph due to S. J. Mason. A signal flow graph is a diagram that represents a set of simultaneous linear algebraic equations. Each signal flow graph consists of a network in which nodes are connected by directed branches. Each node represents a system variable, and each branch acts as a signal multiplier. The signal flows in the direction indicated by the arrow. Definitions: Node: A node is a point representing a variable or signal. Branch: A branch is a directed line segment joining two nodes. Transmittance: It is the gain between two nodes. Input node: A node that has only outgoing branche(s). It is also, called as source and corresponds to independent variable. Output node: A node that has only incoming branches. This is also called as sink and corresponds to dependent variable. SJBIT/ Dept of ECE Page 8

33 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: ) Signal flow applies only to linear systems. ) 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 X k to X j represents dependence of the variable Xj on X k but not the reverse. 5) The signal traveling along the branch X k and X j is multiplied by branch gain a kj and signal a kj X k is delivered at node X j. 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 ) Arrange the input to output nodes from left to right. ) Connect the nodes by appropriate branches. 3) If the desired output node has outgoing branches, add a dummy node and a unity gain branch. 4) Rearrange the nodes and/or loops in the graph to achieve pictorial clarity. Signal Flow Graph Algebra Addtion rule SJBIT/ Dept of ECE Page 9

34 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- branches with transmission functions can be replaced by a single branch with new transmission function equal to the product of the old ones. Masons Gain Formula The relationship between an input variable and an output variable of a signal flow graph is given by the net gain between input and output nodes and is known as overall gain of the system. Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs. Gain P is given by P k P k k Where, P k is gain of k th forward path, is determinant of graph =-(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 k th forward path determinant of graph with loops touching k th forward path. It is obtained from by removing the loops touching the path P k. Example Draw the signal flow graph of the block diagram shown in Fig..7 H R X X X 3 X 4 X 5 X 6 C G G G 3 H Figure.7 Multiple loop system Choose the nodes to represent the variables say X.. X 6 as shown in the block diagram.. Connect the nodes with appropriate gain along the branch. The signal flow graph is shown in Fig..7 SJBIT/ Dept of ECE Page 30

35 -H R X X X 3 G G G 3 C X 4 X 5 X 6 H Figure.8 Signal flow graph of the system shown in Fig..7 - Example.9 Draw the signal flow graph of the block diagram shown in Fig..9. R G X X C G G 3 X 3 G 4 Figure.9 Block diagram feedback system The nodal variables are X, X, X 3. The signal flow graph is shown in Fig..0. SJBIT/ Dept of ECE Page 3

36 G R G X X 3 C X -G 3 G 4 Figure.0 Signal flow graph of example Example 3 Draw the signal flow graph of the system of equations. X X X 3 a a a 3 X X X a a a 3 X X X a a 3 a 3 33 X X 3 X 3 3 b u b u The variables are X, X, X 3, u and u 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... a 3 u a u b a a b X a 3 a 33 X a X 3 a 3 a 3 Figure. Signal flow graph of example SJBIT/ Dept of ECE Page 3

37 Example 4 LRC net work is shown in Fig... Draw its signal flow graph. R L e(t) i(t) C e c (t) Figure. LRC network The governing differential equations are di L Ri dt C idt e t or di L dt dec C dt Ri e c i t 3 e t Taking Laplace transform of Eqn. and Eqn. and dividing Eqn. by L and Eqn.3 by C R si s i 0 I S Ec s E s 4 L L L sec s ec 0 I s 5 C Eqn.4 and Eqn.5 are used to draw the signal flow graph shown in Fig.7. i(0 + ) e c (0 + ) L s E(s) R L s R L I(s) - L s R L Cs s E c (s) Figure. Signal flow graph of LRC system SJBIT/ Dept of ECE Page 33

38 SIGNAL FLOW GRAPHS The relationship between an input variable and an output variable of a signal flow graph is given by the net gain between input and output nodes and is known as overall gain of the system. Masons gain formula is used to obtain the over all gain (transfer function) of signal flow graphs. Masons Gain Formula Gain P is given by P k P k k Where, P k is gain of k th forward path, is determinant of graph =-(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 k th forward path determinant of graph with loops touching k th forward path. It is obtained from by removing the loops touching the path P k. Example Obtain the transfer function of C/R of the system whose signal flow graph is shown in Fig..3 G R G C -G 3 G 4 Figure.3 Signal flow graph of example There are two forward paths: Gain of path : P =G Gain of path : P =G There are four loops with loop gains: L =-G G 3, L =G G 4, L 3 = -G G 3, L 4 = G G 4 SJBIT/ Dept of ECE Page 34

39 There are no non-touching loops. = +G G 3 -G G 4 +G G 3 -G G 4 Forward paths and touch all the loops. Therefore, =, = The transfer function T = C R s s P P G G G G G G G G G G Example Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig..4. R(s) -H G G G 3 C(s) H - Figure.4 Signal flow graph of example There is one forward path, whose gain is: P =G G G 3 There are three loops with loop gains: L =-G G H, L =G G 3 H, L 3 = -G G G 3 There are no non-touching loops. = -G G H +G G 3 H +G G G 3 Forward path touches all the loops. Therefore, =. C s P GGG3 The transfer function T = R s G G H G G H G G G Example 3 Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig SJBIT/ Dept of ECE Page 35

40 G 6 G 7 R(s) G G G 3 G 4 G 5 C(s) X X X 3 X 4 X5 -H -H Figure.5 Signal flow graph of example 3 There are three forward paths. The gain of the forward path are: P =G G G 3 G 4 G 5 P =G G 6 G 4 G 5 P 3 = G G G 7 There are four loops with loop gains: L =-G 4 H, L =-G G 7 H, L 3 = -G 6 G 4 G 5 H, L 4 =-G G 3 G 4 G 5 H There is one combination of Loops L and L which are nontouching with loop gain product L L =G G 7 H G 4 H = +G 4 H +G G 7 H +G 6 G 4 G 5 H +G G 3 G 4 G 5 H + G G 7 H G 4 H Forward path and touch all the four loops. Therefore =, =. Forward path 3 is not in touch with loop. Hence, 3 = +G 4 H. The transfer function T = C s P P P3 3 GGG3G 4G5 GG4G5G6 GGG7 G4H R s G H G G H G G G H G G G G H G G G H H Example 4 SJBIT/ Dept of ECE Page 36

41 Find the gains X X 6 X, X 5 X, X 3 for the signal flow graph shown in Fig..6. b -h X a c d X 5 f X 6 e X X 3 X 4 -g -i Figure.6 Signal flow graph of MIMO system Case : X X 6 There are two forward paths. The gain of the forward path are: P =acdef P =abef There are four loops with loop gains: L =-cg, L =-eh, L 3 = -cdei, L 4 =-bei There is one combination of Loops L and L which are nontouching with loop gain product L L =cgeh = +cg+eh+cdei+bei+cgeh Forward path and touch all the four loops. Therefore =, =. X 6 P P cdef abef The transfer function T = X cg eh cdei bei cgeh X 5 Case : X The modified signal flow graph for case is shown in Fig..7. SJBIT/ Dept of ECE Page 37

42 b -h X X 5 X d e 5 c X X 3 X 4 -g -i Figure.7 Signal flow graph of example 4 case The transfer function can directly manipulated from case as branches a and f are removed which do not form the loops. Hence, The transfer function T= X X 5 P P cg cde be eh cdei bei cgeh X 3 Case 3: X The signal flow graph is redrawn to obtain the clarity of the functional relation as shown in Fig..8. -h c X a X b e X 5 f X 3 X 4 X 3 -i d Figure.8 Signal flow graph of example 4 case 3 -g There are two forward paths. The gain of the forward path are: P =abcd P =ac SJBIT/ Dept of ECE Page 38

43 There are five loops with loop gains: L =-eh, L =-cg, L 3 = -bei, L 4 =edf, L 5 =-befg There is one combination of Loops L and L which are nontouching with loop gain product L L =ehcg = +eh+cg+bei+efd+befg+ehcg Forward path touches all the five loops. Therefore =. Forward path does not touch loop L. Hence, = + eh The transfer function T = X X 3 P P eh abef ac cg bei efd eh 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 u 3 X a X X u X a X u Taking Laplace transform with zero initial conditions X s X s U s sx sx s s a a X X s 3 s X U s s U s Rearrange the above equation X s X s U s X X s s a X s a X s 3 s X s s s U s s s U s The signal flow graph is shown in Fig..9. SJBIT/ Dept of ECE Page 39

44 s β s a a X s s X X X U β s 3 Figure.9 Signal flow grapgh of example 5 There are three forward paths. The gain of the forward path are: P = 3 P = / s P 3 = / s There are two loops with loop gains: a L s a L s L =-eh, L =-cg, L 3 = -bei, L 4 =edf, L 5 =-befg There are no combination two Loops which are nontouching. a a s s Forward path does not touch loops L and L. Therefore a a s s Forward path path 3 touch the two loops. Hence, =, =. The transfer function T = X X 3 P P P s a s s a a s a s SJBIT/ Dept of ECE Page 40

45 Recommended Questions:. Define block diagram & depict the block diagram of closed loop system.. 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. H R X X X 3 X 4 X 5 X 6 C G G G 3 H 6. Draw the signal flow graph of the block diagram shown in Fig below R G X X C G G 3 X 3 G 4 7. For the LRC net work is shown in Fig Draw its signal flow graph. SJBIT/ Dept of ECE Page 4

46 R L e(t) i(t) C e c (t) Figur 8. Obtain the transfer function of C(s)/R(s) of the system whose signal flow graph is shown in Fig. G 6 G 7 R(s) G G G 3 G 4 G 5 C(s) X X X 3 X 4 X5 -H -H Q.9 For the system represented by the following equations find the transfer function X(s)/U(s) using signal flow graph technique. X X u 3 X a X X u X a X u SJBIT/ Dept of ECE Page 4

47 Time response analysis of control systems: Introduction: UNIT- 3 Time is used as an independent variable in most of the control systems. It is important to analyse the response given by the system for the applied excitation, which is function of time. Analysis of response means to see the variation of out put with respect to time. The output behavior with respect to time should be within these specified limits to have satisfactory performance of the systems. The stability analysis lies in the time response analysis that is when the system is stable out put is finite The system stability, system accuracy and complete evaluation is based on the time response analysis on corresponding results. DEFINITION AND CLASSIFICATION OF TIME RESPONSE Time Response: The response given by the system which is function of the time, to the applied excitation is called time response of a control system. Practically, output of the system takes some finite time to reach to its final value. This time varies from system to system and is dependent on different factors. The factors like friction mass or inertia of moving elements some nonlinearities present etc. Example: Measuring instruments like Voltmeter, Ammeter. Classification: The time response of a control system is divided into two parts. Transient response c t (t) Steady state response c ss (t)... c(t)=c t (t) +c SS (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 c t (t)=0 t A system in which the transient response do not decay as time progresses is an Unstable system. SJBIT/ Dept of ECE Page 43

48 C(t) Step C t (t) C ss (t) e ss = study state error O Transient time Time Study state Time The transient response may be experimental or oscillatory in nature.. 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 c t (t)=c ss (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. Step Input signals.. Ramp Input Signals. 3. Parabolic Input Signals. 4. Impulse input signal. Details of standard test signals. 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. SJBIT/ Dept of ECE Page 44

49 c. Sudden opening or closing a valve. r(t) A O t When, A =, r(t) = u(t) = The step is a signal who s value changes from value (usually 0) to another level A in Zero time. In the Laplace Transform form R(s) = A / S Mathematically r(t) = u(t) = for t > 0 = 0 for t < 0. Ramp Input Signal (Velocity Functions): It is constant rate of change in input that is gradual application of input as shown in fig ( b). r(t) Ex:- Altitude Control of a Missile Slope = A O t time. The ramp is a signal, which starts at a value of zero and increases linearly with Mathematically r (t) = At for t 0 = 0 for t 0. In LT form R(S) = A S If A=, it is called Unit Ramp Input Mathematically r(t) = t u(t) { In LT form R(S) t for = t A 0 = = 0 for t S 0 S SJBIT/ Dept of ECE Page 45

50 3. Parabolic Input Signal (Acceleration function): The input which is one degree faster than a ramp type of input as shown in fig ( c) or it is an integral of a ramp. Mathematically a parabolic signal of magnitude A is given by r(t) = A t u(t) r(t) At for t 0 = 0 for t 0 t Slope = At In LT form R(S) = A S 3 If A =, a unit parabolic function is defined as r(t) = t u(t) ie., r(t) { In LT for R(S) = S 3 4. Impulse Input Signal : = t for t 0 0 for t 0 It is the input applied instantaneously (for short duration of time ) of very high amplitude as shown in fig (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 t Area of impulse = Its magnitude If area is unity, it is called Unit Impulse Input denoted as (t) Mathematically it can be expressed as r(t) = A for t = 0 SJBIT/ Dept of ECE Page 46

51 = 0 for t 0 In LT form R(S) = if A = Standard test Input Signals and its Laplace Transforms. r(t) R(S) Unit Step /S Unit ramp /S Unit Parabolic /S 3 Unit Impulse First order system:- The st order system is represent by the differential Eq:- a dc(t )+aoc (t) = bor(t) () dt Where, e (t) = out put, r(t) = input, a0, a & b0 are constants. Dividing Eq:- () by a0, then a. d c(t ) + c(t) = bo.r (t) a0 dt ao T. d c(t ) + c(t) = Kr (t) () dt Where, T=time const, has the dimensions of time = a & K= static sensitivity = b0 a0 a0 Taking for L.T. for the above Eq:- [ TS+] C(S) = K.R(S) T.F. of a st order system is ; G(S) = C(S ) = K. R(S) +TS If K=, Then G(S) =. +TS [ It s a dimensionless T.F.] I This system represent RC ckt. A simplified bloc diagram is as shown.; R(S)+ C(S) TS - SJBIT/ Dept of ECE Page 47

52 Unit step response of st order system:- Let a unit step i\p u(t) be applied to a st order system, Then, r (t)=u (t) & R(S) = () S W.K.T. C(S) = G(S). R(S) C(S) =.. =. T () +TS S S TS+ Taking inverse L.T. for the above Eq:- then, C(t)=u (t) e t/t ; t.> (3) slope =. T At t=t, then the value of c(t)= - e = c (t) The smaller the time const. T. the faster the system response. The slope of the tangent line at at t= 0 is /T. Since dc =.e - t/t =. at t.= (4) dt T T 0.63 e t/t From Eq:- (4), We see that the slope of the response curve c(t) decreases monotonically from. at t=0 to zero. At t= T Second order system:- The nd order system is defined as, a d c(t) + a dc(t) + a 0 c(t) = b 0.r(t) () dt dt Where c(t) = o/p & r(t) = I/p -- ing () by a 0, a d c(t) + a. dc (t) + c(t) = b 0. r(t). a 0 dt a 0 dt a 0 a d c(t) + a. a. dc (t) + c(t) = b 0. r(t). a 0 dt a 0 a 0. a dt a 0 3) The open loop T.F. of a unity feed back system is given by G(S) = K. where, S(+ST) T t SJBIT/ Dept of ECE Page 48

53 T&K are constants having + Ve values.by what factor () the amplitude gain be reduced so that (a) The peak overshoot of unity step response of the system is reduced from 75% to 5% (b) The damping ratio increases from 0. to 0.6. Solution: G(S) = K. S(+ST) Let the value of damping ratio is, when peak overshoot is 75% & when peak overshoot is 5% Mp =. e - ln = =. - - = 0.09 (0.0084) (- ) = = (.0084 ) = = 0.09 k. S + S T. w.k.t. T.F. = G(S) = + K. = K. + G(S). H(S) S + S T S + S T+K T.F. = K / T. S + S + K. T T Comparing with std Eq :- Wn = K., Wn =. T T Let the value of K = K When = & K = K When =. Since Wn =., =. =. T TWn KT.. = K T = K. K K T 0.09 = K. K. = K K SJBIT/ Dept of ECE Page 49

54 K = K a) The amplitude K has to be reduced by a factor =. = b) Let = 0. Where gain is K and = 0.6 Where gain is K 0. = K. K. = 0.07 K = 0.07 K 0.6 K K The amplitude gain should be reduced by. = ) Find all the time domain specification for a unity feed back control system whose open loop T.F. is given by G(S) = 5. S(S+6) Solution: 5. G(S) = 5. G(S). = S(S+6). S(S+6) + G(S).H(S) + 5. S(S+6) = 5. S + ( 6S+5 ) W n = 5, Wn = 5, Wn = 6 = 6. = 0.6 x 5 Wd = Wn - = 5 - (0.6) = 4 tr = -, = tan - Wd = Wn = 0.6 x 5 = 3 Wd = tan - ( 4/3 ) = 0.97 rad. tp =. = 3.4 = sec. Wd 4 MP =. = 0.6. x3.4 = 9.5% e - e ts = 4. for % = 4. =.3 3sec. Wn 0.6 x 5 SJBIT/ Dept of ECE Page 50

55 5) The closed loop T.F. of a unity feed back control system is given by C(S) = 5. R(S) S + 4S +5 Determine () Damping ratio () Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse. Solution: C(S) = 5., Wn = 5 Wn = 5 =.36 R(S) S + 4S +5 Wn = 4 = 4. = Wd =.008 x.36 MP =. = X 3.4 = 0.9% e - e -(0.894) W. K.T. C(t) = e - Wnt Cos Wdt r +. sin wdt r - = e x.36t Cos.008t sin.008t -(0.894) 6) A servo mechanism is represent by the Eq:- + 0 d = 50E, E = R- is the actuating signal calculate the dt dt value of damping ratio, undamped and damped frequency of ascillation. d Soutions:- d + 0 d = 5 ( r - ), = 50r 50. dt dt Taking L.T., [S + 0S + 50] (S) = 50 R (S). (S) = 50. R(S) S + 0S + 5O Wn = 50 Wn =.5..rad sec. Wn = 0 = 0. = x.5 SJBIT/ Dept of ECE Page 5

56 Wd = Wn - =.5 - (0.408) =.8. rad sec. 7) Fig shows a mechanical system and the response when 0N of force is applied to the system. Determine the values of M, F, K,. M x(t)inmt K f(t) The T.F. of the mechanical system is, 0.0 X(S) =. F(S) MS + FS = K f(t) = Md X + F dx + KX F x dt dt F(S) = (MS + FS + K) x (S) Given :- F(S) = 0 S. X(S) = 0. S(MS + FS + K) SX (S) = 0. MS + FS + K The steady state value of X is By applying final value theorem, lt. SX(S) = 0. = 0 = 0.0 ( Given from Fig.) S O M(0) + F (0) + K K. ( K = 500.) MP = = = 9.6% 0.0 MP = e. ln = = = = = = 0.6 tp = =. Wd Wn 3 =. Wn =.3 rad / Sec. SJBIT/ Dept of ECE Page 5

57 Wn (0.6) Sx(S) = 0/ M. (S + F S + K ) M M Comparing with the std. nd order Eq :-, then, Wn = K Wn = K (.3) = 500. M = 9.36 kg. M M M M F = Wn F = x 0.6 x 9 x.3 F = N/M/ Sec. 8) Measurements conducted on sever me mechanism show the system response to be c(t) = +0.e -60t.e -0t, 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) = +0.e -60t.e -0t Taking L.T., C(S) = S S+60 S+0 C(S). = 600 / S. S + 70S Given that :- Unit step i/p r(t) = R(S) =. C(S). = 600 / S. R(S) S + 70S Comparing, Wn = 600, 4.4..rad / Sec 70, = 70. =.48 x 4.4 Wn = 0) A feed back system employing o/p damping is as shown in fig. ) Find the value of K & K so that closed loop system resembles a nd order system with = 0.5 & frequency of damped oscillation 9.5 rad / Sec. SJBIT/ Dept of ECE Page 53

58 ) With the above value of K & K 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 % tolerance? K. S(+S) R + C K S C. = K. R S + ( + K ) S + K Wn = K Wn = K Wn = + K = + K K Wd = Wn - Wn = rad/sec 0.5 K = (0.96) = 0.34 Wn = + K, K = 9.97 MP =. = 6.3% e - ts = 4. = 4. = 0.79 sec Wn 0.5 x 0.97 Steady state Error :- Steady state errors constitute an extremely important aspect of system performance. The state error is a measure of system accuracy. These errors arise from the nature of i/p s type of system and from non-linearties of the system components. The steady state performance of a stable control system is generally judged by its steady state error to step, ramp and parabolic i/p. SJBIT/ Dept of ECE Page 54

59 Consider the system shown in the fig. G(S) R(S) E(S) C(S) H(S) C(S) = G(S). () R(S) +G(S). H(S) The closed loop T.F is given by (). 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) = C(S). H(S) R(S) R(S) R(S) = G(S). H(S). = + G(S). H(S) G(S)H(S) + G(S). H(S) +G(S). H(S) =. + G(S). H(S) Where e(t) = Difference b/w the i/p signal and the feed back signal E(S) =..R(S).() + G(S). H(S) The steady state error e ss may be found by the use of final value theorem and is as follows; e ss = lt e(t) = lt SE(S) t S O Substituting (), e ss = lt S.R(S)..() S O +G(S). H(S) Eq :- () 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). T.F. SJBIT/ Dept of ECE Page 55

60 The expression for steady state errors for various types of standard test signals are derived below; ) Steady state error due to step i/p or position error constant (K p ):- The steady state error for the step i/p is I/P r(t) = u(t). Taking L.T., R(S) = /S. From Eq:- (), e ss = lt S. R(s). =. S O +G(S). H.S + lt G(S). H(S) S O lt G(S). H(S) = Kp (S O ) Where Kp = proportional error constant or position error const. e ss =. + Kp ( + Kp) e ss = Kp = - e ss e ss Note :- e ss = R. for non-unit step i/p + Kp ) Steady state error due to ramp i/p or static velocity error co-efficient (Kv) :- The e ss 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) = /S Substituting this to e ss Eq:- e ss = lt S... = lt. S O + G(S). H(s) S S O S +S G(S) H(s)S lt S O = SG(S). H(S) = Kv = velocity co-efficient then e ss = lt. e ss =. S O S + Kv Kv Velocity error is not an error in velocity, but it is an error in position error due to a ramp i/p 3) Steady state error due to parabolic i/p or static acceleration co-efficient (K a ) :- The steady state actuating error of the system with a unit parabolic i/p (acceleration i/p) which is defined by r(t) +. t Taking L.T. R(S)=. S 3 SJBIT/ Dept of ECE Page 56

61 e ss = lt S.. lt. S O + G(S). H(S) S 3 S O S + S G(S). H(S) lt S S G(S). H(S) = Ka. O e ss = lt. =. S O S + Ka Ka Note :- e ss = R. for non unit parabolic. Ka Types of feed back control system :- The open loop T.F. of a unity feed back system can be written in two std, forms; ) Time constant form and ) Pole Zero form, G(S) = K(TaS +) (TbS +).. S n (T S+) (T S + ). Where K = open loop gain. Above Eq:- involves the term S n in denominator which corresponds to no, of integrations in the system. A system is called Type O, Type, Type,.. if n = 0,,,.. 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. ) Type O system :- If, n = 0, the system is called type 0, system. The steady state error are as follows; Let, G(S) = K. [... H(s) = ] S + e ss (Position) =. =. =. + G(O). H(O) + K + K p... K p = lt G(S). H(S) = lt K. = K S O S O S + e ss (Velocity) =. =. = Kv O SJBIT/ Dept of ECE Page 57

62 Kv = lt G(S). H(S) = lt S K. = O. S O S O S + e ss (acceleration) =. =. = Ka O Ka = lt S G(S). H(S) = lt S K. = O. S O S O S + ) Type System :- If, n =, the e ss to various std, i/p, G(S) = K. S (S + ) e ss (Position) =. = O + Kp = lt G(S). H(S) = lt K. = S O S O S( S + ) Kv = lt S K. = K S O S(S+) e ss (Velocity) =. K e ss (acceleration) =. =. = O O Ka = lt S K. = O. S O S (S + ) 3) Type System :- If, n =, the e ss to various std, i/p, are, G(S) = K. S (S + ) Kp = lt K. = S O S (S + )... e ss (Position) =. = O Kv = lt S K. = S O S (S + )... e ss (Velocity) =. = O SJBIT/ Dept of ECE Page 58

63 Ka = lt S K. = K. S O S (S + )... e ss (acceleration) =. K 3) Type 3 System :- Gives Kp = Kv = Ka = & e ss = O. (Onwards) The error co-efficient Kp, Kv, & Ka describes the ability of the system to eliminate the steady state error therefore they are indicative of steady state performance. It is generally described to increase the error co-efficient while maintaining the transient response within an acceptable limit. PROBLEMS;. The unit step response of a system is given by C (t) = 5/ +5t 5/ e -t. Find the T. F of the system. T/P = r(t) = U (t). Taking L.T, R(s) = /S. Response C(t) = 5/+5t-5/ e -t = = 5 S S (S+) + - S S S+ Taking L.T, C(s) S = 5 S +S+S+4 s S (S+) C(s) = 5 S(S+)+(S+)- S ( S+) = 0 (S+) S (S+) R (S) S (S+) S(S+) T.F = C (S) = 0 (S+) S = 0 (S+). The open loop T F of a unity food back system is G(s) = 00 S (S+0) SJBIT/ Dept of ECE Page 59

64 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 + pt + P t G(s) = 00 position error co-efficient S (S+0) S 0 S 0 S (S+0) KP = lt G(s) = lt 00 = Similarly KV = lt SG(s) = lt 00 x s = 00 = 0 S 0 S 0 S (S+0) 0 lt 00 x s = 0 S 0 S 0 S (S+0) Ka = lt S G(S) Given :- r(t) = Po+Pt +P t Therefore steady state error ess R R R Kp Kv Ka ess R R R = P0 P P Ess = 0+0. P + = 3. Determine the error co-efficeint and static error for G(s) = And H(s) = (S+) S(S+) (S+0) SJBIT/ Dept of ECE Page 60

65 The error constants for a non unity feed back system is as follows Kp = lt G(S) H(S) = (0+) lt (S+) G(S).H(S) = 0(0+) (0+0) S(S+) (S+0) = Kv = lt G(S) H(S) = lt (0+) 0(0+) (0+0) = /5 = 0. Ka = 0 Static Error:- Steady state error for unit step i/p = 0 Unit ramp i/p = = 5 Kv 0. Unit parabolic i/p = /0 = 4. A feed back C.S is described as G(S) = H(S)=/s.For unit step i/p,cal steady state constant and errors. 50 S (S+) (S+5) error Kp = lt G(S) H(S) = lt S 50 (S+) (S+5) = Kv = lt G(S) H(S) = lt S 50 x S (S+) (S+5) = Ka = lt G(S) H(S) = lt S 0 S 0 S x = = 5 S (S+) (S+5) 0 SJBIT/ Dept of ECE Page 6

66 The steady state error = 0/50 = 0 Ess = lt S. /S S S (S+)(S+5) 5. A certain feed back C.S is described by following C.S G(S) = Lt S (S+) (S+5) S 0 S (S+) (S+5) + 50 S K H(S) = (S+0) (S+30) Determine steady state error co-efficient and also determine the value of K to limit the steady to 0 units due to i/p r(t) = t 0/ t. Kp = lt G(S) H(S) = lt S 0 S S 50 = (S+0) (S+30) Kv = lt S K S 0 S (S+0) (S+30) = Ka = lt S K S 0 S (S+0) (S+30) K 600 Steady state error:- Error due to unit step i/p + = 0 +Kp + Error due o r(t) ramp i/p = 0 Kv SJBIT/ Dept of ECE Page 6

67 Error due to para i/p,, 0 40 = = Ka Ka 0 x = K K r (t) = = 0 = K = 00 K First order system:- The st order system is represent by the differential Eq:- a dc(t )+aoc (t) = bor(t) () dt Where, e (t) = out put, r(t) = input, a0, a & b0 are constants. Dividing Eq:- () by a0, then a. d c(t ) + c(t) = bo.r (t) a0 dt ao T. d c(t ) + c(t) = Kr (t) () dt Where, T=time const, has the dimensions of time = a & K= static sensitivity = b0 a0 a0 Taking for L.T. for the above Eq:- [ TS+] C(S) = K.R(S) T.F. of a st order system is ; G(S) = C(S ) = K. R(S) +TS If K=, Then G(S) =. +TS [ It s a dimensionless T.F.] I This system represent RC ckt. A simplified bloc diagram is as shown.; R(S)+ C(S) TS - Unit step response of st order system:- Let a unit step i\p u(t) be applied to a st order system, Then, r (t)=u (t) & R(S) = () S W.K.T. C(S) = G(S). R(S) SJBIT/ Dept of ECE Page 63

68 C(S) =.. =. T () +TS S S TS+ Taking inverse L.T. for the above Eq:- then, C(t)=u (t) e t/t ; t.> (3) slope =. T At t=t, then the value of c(t)= - e = c (t) The smaller the time const. T. the faster the system response. The slope of the tangent line at at t= 0 is /T. Since dc =.e - t/t =. at t.= (4) dt T T 0.63 e t/t From Eq:- (4), We see that the slope of the response curve c(t) decreases monotonically from. at t=0 to zero. At t= T Second order system:- The nd order system is defined as, a d c(t) + a dc(t) + a 0 c(t) = b 0.r(t) () dt dt T t Where c(t) = o/p & r(t) = I/p -- ing () by a 0, a d c(t) + a. dc (t) + c(t) = b 0. r(t). a 0 dt a 0 dt a 0 a d c(t) + a. a. dc (t) + c(t) = b 0. r(t). a 0 dt a 0 a 0. a dt a 0 Step response of nd order system: The T.F. = C(s) = W n R(s) 3 + W n S+ W n Based on value The system may be, SJBIT/ Dept of ECE Page 64

69 ) Under damped system (0< <) 3) Critically damped system ( =) 4) Over damped system ( >) ) Under damped system :- (0< <) In this case C(s) can be written as R(s) C(s) = W n R(s) (S+ w n + jw d ) (S+ w n - jw d ) Where w d = w n - The Freq. w d is called damped natural frequency For a unit step i/p :- [ R(t)= R(S) = /S] C(S) = W n. X R (S) = W n. =. (S + W n S+ W n ) (S + W n S+ W n ) S C(S) =. S+ W n. S S + W n S+ W n =. S+ W n. W n (5) S (S+ W n ) + W d (S+ W n + W d ) C(S) =. S+ W n.. W d. S (S+ W n ) + W d - (S+ W n ) + W d Taking ILT, C(t) = -e - Wnt C OS W dt +. Sin W dt (6) - The error signal for this system is the difference b/w the I/p & o/p. e(t) = r(t) c(t). = c(t) = e - Wnt C OS W dt +. Sin W dt (7) SJBIT/ Dept of ECE Page 65

70 - t > o. At t =, error exists b/w the i/p & o/p. If the damping ratio = O, the response becomes undamped & oscillations continues indefinitely. The response C(t) for the zero damping case is, c(t) =-(C OS w nt ) =- C OS w nt ; t > O (8) From Eq:- (8), we see that the W n represents the undamped natural frequency of the system. If the linear system has any amount of damping the undamped natural frequency cannot be observed experimentally. The frequency, which may be observed, is the damped natural frequency. Wd =w n This frequency is always lower than the undamped natural frequency. An increase in would reduce the damped natural frequency W d. If is increased beyond unity, the response over damped & will not oscillate. Critically damped case:- ( =). If the two poles of C(S) R(S) Critically damped one. are nearly equal, the system may be approximated by a For a step I/p R(S) = /S C(S) = W n.. S + W n S+ W n S =.. W n. S (S + W n ) ( S+ W n ) Taking I.L.T., =. W n. S ( S + W n ) S C(t) = e -Wnt (+w n t) Over damped system :- ( > ) If this case, the two poles of C(S) are negative, real and unequal. R(S) For a unit step I/p R(S) = /S, then, C(S) = W n. SJBIT/ Dept of ECE Page 66

71 (S+ W n + W n - ) ( S+ W n - W n ) Taking ILT, C(t) = +. e ( + S ( + ) ) W n t. S ( + ). e ( + ) W n t. C(t) = + W n. e -S t. - e -S t. ; t > O S S S Where S = ( + S = ( - ) W n ) W n Time response (Transient ) Specification (Time domain) Performance :- The performance characteristics of a controlled system are specified in terms of the transient response to a unit step i/p since it is easy to generate & is sufficiently drastic. MP The transient response of a practical C.S often exhibits damped oscillations before reaching steady state. In specifying the transient response characteristic of a C.S to unit step i/p, it is common to specify the following terms. ) Delay time (td) SJBIT/ Dept of ECE Page 67

72 ) Rise time (tr) Response curve 3) Peak time (tp) 4) Max over shoot (Mp) 5) Settling time (ts) ) Delay time :- (t d ) It is the time required for the response to reach 50% of its final value time. for the st ) Rise time :- (t r ) It is the time required for the response to rise from 0% and 90% or 0% to 00% of its final value. For under damped system, second order system the 0 to 00% rise time is commonly used. For over damped system, the 0 to 90% rise time is commonly used. 3) Peak time :- (t p ) It is the time required for the response to reach the st of peak of the overshoot. 4) Maximum over shoot :- (MP) It is the maximum peak value of the response curve measured from unity. The amount of max over shoot directly indicates the relative stability of the system. 5) Settling time :- (t s ) SJBIT/ Dept of ECE Page 68

73 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 %). 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) = W n () R(S) S + W n S+ W n Assuming the system is to be underdamped ( < ) Rise time tr W. K.T. C(t r ) = - e - Wnt Cos Wdt r +. sin wdt r - Let C(t r ) =, i.e., substituting t r for t in the above Eq: Then, C(t r ) = = - e - Wntr Cos wdt r +. sin wdt r - Cos wdt r +. sin wdt r = tan wdt r = - - = wd. - jw Thus, the rise time t r is, jwd t r =. tan - - w d = - secs Wn - Wn Wd wd When must be in radians. - Wn Peak time :- (tp) S- Plane 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 - Since the peak time corresponds to the st peak over shoot. Wdtp = = tp =. Wd The peak time tp corresponds to one half cycle of the frequency of damped oscillation. SJBIT/ Dept of ECE Page 69

74 Maximum overshoot :- (MP) The max over shoot occurs at the peak time. i.e. At t = tp =. Wd Mp = e ( / Wd) ( / - ) or e Settling time :- (ts) An approximate value of ts can be obtained for the system O < < by using the envelope of the damped sinusoidal waveform. Time constant of a system = T =. Wn Setting time ts = 4x Time constant. = 4x. for a tolerance band of +/- % steady state. Wn Delay time :- (td) The easier way to find the delay time is to plot Wn td VS the curve for the range O< <, then the Eq. becomes, Wn td = +0.7 td = +0.7 Wn. Then approximate PROBLEMS: () Consider the nd order control system, where = 0.6 & Wn = 5 rad / sec, obtain the rise time tr, peak time tp, max overshoot Mp and settling time ts When the system is subject to a unit step i/p., Given :- = 0.6, Wn = 5rad /sec, tr =?, tp =?, Mp =?, ts =? Wd = Wn - = 5 -(0.6) = 4 = Wn = 0.6 x 5 = 3. tr = -, = tan - Wd = tan - 4 = 0.97 rad Wd 3 tr = = 0.55sec. 4 SJBIT/ Dept of ECE Page 70

75 tp = = 3.4 = sec. Wd 4 MP = e -. = e / Wd EXERCISE: MP = e (3/4) x 3.4 = x 00 = 9.4% ts :- For the % criteria., ts = 4. = 4. =.33 sec. Wn 0.6x5 For the 5% criteria., ts = 3 = 3 = sec 3 () A unity feed back system has on open loop T.F. G(S) = K. S ( S+0) Determine the value of K so that the system has a damping factors of 0.5 For this value of K determine settling time, peak over shoot & time for peak over shoot for unit step i/p LCS The error co-efficient Kp, Kv, & Ka describes the ability of the system to eliminate the steady state error therefore they are indicative of steady state performance. It is generally described to increase the error co-efficient while maintaining the transient response within an acceptable limit. PROBLEMS;. The unit step response of a system is given by C (t) = 5/ +5t 5/ e -t. Find the T. F of the system. T/P = r(t) = U (t). Taking L.T, R(s) = /S. Response C(t) = 5/+5t-5/ e -t + - S S S+ Taking SJBIT/ Dept of ECE Page 7

76 L.T, C(s) = = 5 S S (S+) S = 5 S +S+S+4 s S (S+) C(s) = 5 S(S+)+(S+)- 3 S ( S+) = 0 (S+) S (S+) R (S) S (S+) S(S+) T.F = C (S) = 0 (S+) S = 0 (S+). The open loop T F of a unity food back system is G(s) = 00 S (S+0) 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 + pt + P t G(s) = 00 position error co-efficient S (S+0) S 0 S 0 S (S+0) KP = lt G(s) = lt 00 = Similarly KV = lt SG(s) = lt 00 x s = 00 = 0 S 0 S 0 S (S+0) 0 lt 00 x s = 0 S 0 S 0 S (S+0) Ka = lt S G(S) SJBIT/ Dept of ECE Page 7

77 Given :- r(t) = Po+Pt +P t Therefore steady state error ess R R R Kp Kv Ka R R R ess P0 P P + = Ess = 0+0. P + = 3. Determine the error co-efficeint and static error for G(s) = And H(s) = (S+) S(S+) (S+0) The error constants for a non unity feed back system is as follows G(S).H(S) = (S+) S(S+) (S+0) Kp = lt G(S) H(S) = lt S 0 S 0 (0+) 0(0+) (0+0) = Kv = lt G(S) H(S) = lt (0+) 0(0+) (0+0) = /5 = 0. SJBIT/ Dept of ECE Page 73

78 Ka = 0 Static Error:- Steady state error for unit step i/p = 0 Unit ramp i/p = = 5 Kv 0. Unit parabolic i/p = /0 = 4. A feed back C.S is described as G(S) = H(S)=/s.For unit step i/p,cal steady state constant and errors. 50 S (S+) (S+5) error Kp = lt G(S) H(S) = lt S 0 S S 50 (S+) (S+5) = Kv = lt G(S) H(S) = lt S 0 S S 50 x S (S+) (S+5) = Ka = lt G(S) H(S) = lt S 0 S 0 S x = = 5 S (S+) (S+5) 0 The steady state error = 0/50 = 0 Ess = lt S. /S S S (S+)(S+5) Lt S (S+) (S+5) S 0 S (S+) (S+5) + 50 S K H(S) = (S+0) (S+30) SJBIT/ Dept of ECE Page 74

79 5. A certain feed back C.S is described by following C.S G(S) = Determine steady state error co-efficient and also determine the value of K to limit the steady to 0 units due to i/p r(t) = t 0/ t. Kp = lt G(S) H(S) = lt S 0 S 0 S 50 = (S+0) (S+30) Kv = lt S K S 0 S (S+0) (S+30) = Ka = lt S K S 0 S (S+0) (S+30) K 600 Steady state error:- Error due to unit step i/p + = 0 +Kp + Error due o r(t) ramp i/p = 0 Kv Error due to para i/p,, 0 40 = = Ka Ka 0 x = K K r (t) = = 0 = K = 00 K 3) The open loop T.F. of a unity feed back system is given by G(S) = K. where, S(+ST) SJBIT/ Dept of ECE Page 75

80 T&K are constants having + Ve values.by what factor () the amplitude gain be reduced so that (a) The peak overshoot of unity step response of the system is reduced from 75% to 5% (b) The damping ratio increases from 0. to 0.6. Solution: G(S) = K. S(+ST) Let the value of damping ratio is, when peak overshoot is 75% & when peak overshoot is 5% Mp =. e - ln = =. - - = 0.09 (0.0084) (- ) = = (.0084 ) = = 0.09 k. S + S T. w.k.t. T.F. = G(S) = + K. = K. + G(S). H(S) S + S T S + S T+K T.F. = K / T. S + S + K. T T Comparing with std Eq :- Wn = K., Wn =. T T Let the value of K = K When = & K = K When =. Since Wn =., =. =. T TWn KT.. = K T = K. K K T 0.09 = K. K. = SJBIT/ Dept of ECE Page 76

81 K K K = K a) The amplitude K has to be reduced by a factor =. = b) Let = 0. Where gain is K and = 0.6 Where gain is K 0. = K. K. = 0.07 K = 0.07 K 0.6 K K The amplitude gain should be reduced by. = ) Find all the time domain specification for a unity feed back control system whose open loop T.F. is given by G(S) = 5. S(S+6) Solution: 5. G(S) = 5. G(S). = S(S+6). S(S+6) + G(S).H(S) + 5. S(S+6) = 5. S + ( 6S+5 ) W n = 5, Wn = 5, Wn = 6 = 6. = 0.6 x 5 Wd = Wn - = 5 - (0.6) = 4 tr = -, = tan - Wd = Wn = 0.6 x 5 = 3 Wd = tan - ( 4/3 ) = 0.97 rad. tp =. = 3.4 = sec. Wd 4 MP =. = 0.6. x3.4 = 9.5% e - e ts = 4. for % = 4. =.3 3sec. Wn 0.6 x 5 SJBIT/ Dept of ECE Page 77

82 5) The closed loop T.F. of a unity feed back control system is given by C(S) = 5. R(S) S + 4S +5 Determine () Damping ratio () Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse. Solution: C(S) = 5., Wn = 5 Wn = 5 =.36 R(S) S + 4S +5 Wn = 4 = 4. = Wd =.008 x.36 MP =. = X 3.4 = 0.9% e - e -(0.894) W. K.T. C(t) = e - Wnt Cos Wdt r +. sin wdt r - = e x.36t Cos.008t sin.008t -(0.894) 6) A servo mechanism is represent by the Eq:- + 0 d = 50E, E = R- is the actuating signal calculate the dt dt value of damping ratio, undamped and damped frequency of ascillation. d Soutions:- d + 0 d = 5 ( r - ), = 50r 50. dt dt Taking L.T., [S + 0S + 50] (S) = 50 R (S). (S) = 50. R(S) S + 0S + 5O Wn = 50 Wn =.5..rad sec. Wn = 0 = 0. = SJBIT/ Dept of ECE Page 78

83 x.5 Wd = Wn - =.5 - (0.408) =.8. rad sec. 7) Fig shows a mechanical system and the response when 0N of force is applied to the system. Determine the values of M, F, K,. M x(t)inmt K f(t) The T.F. of the mechanical system is, 0.0 X(S) =. F(S) MS + FS = K f(t) = Md X + F dx + KX F x dt dt F(S) = (MS + FS + K) x (S) Given :- F(S) = 0 S. X(S) = 0. S(MS + FS + K) SX (S) = 0. MS + FS + K The steady state value of X is By applying final value theorem, lt. SX(S) = 0. = 0 = 0.0 ( Given from Fig.) S O M(0) + F (0) + K K. ( K = 500.) MP = = = 9.6% 0.0 MP = e. ln = = = = = = 0.6 tp = =. Wd Wn SJBIT/ Dept of ECE Page 79

84 3 =. Wn =.3 rad / Sec. Wn (0.6) Sx(S) = 0/ M. (S + F S + K ) M M Comparing with the std. nd order Eq :-, then, Wn = K Wn = K (.3) = 500. M = 9.36 kg. M M M F = Wn F = x 0.6 x 9 x.3 M F = N/M/ Sec. 9) Measurements conducted on sever me mechanism show the system response to be c(t) = +0.e -60t.e -0t, 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) = +0.e -60t.e -0t Taking L.T., C(S) = S S+60 S+0 C(S). = 600 / S. S + 70S Given that :- Unit step i/p r(t) = R(S) =. S C(S). = 600 / S. R(S) S + 70S Comparing, Wn = 600, 4.4..rad / Sec Wn = 70, = 70. =.48 x 4.4 9) The C.S. shown in the fig employs proportional plus error rate control. Determine the value of error rate const. Ke, so the damping ratio is 0.6. Determine the value of settling time, max SJBIT/ Dept of ECE Page 80

85 overshoot and steady state error, if the i/p is unit ramp, what will be the value of steady state error without error rate control. R + (S) +SKe 0. S 0) A feed back system employing o/p damping is as shown in fig. 4) Find the value of K & K so that closed loop system resembles a nd order system with = 0.5 & frequency of damped oscillation 9.5 rad / Sec. 5) With the above value of K & K find the % overshoot when i/p is step i/p 6) What is the % overshoot when i/p is step i/p, the settling time for % tolerance? K. S(+S) R + C K S C. = K. R S + ( + K ) S + K Wn = K Wn = K Wn = + K = + K K Wd = Wn - Wn = rad/sec 0.5 K = (0.96) = 0.34 Wn = + K, K = 9.97 MP =. = 6.3% e - SJBIT/ Dept of ECE Page 8

86 ts = 4. = 4. = 0.79 sec Wn 0.5 x 0.97 Types of feed back control system :- The open loop T.F. of a unity feed back system can be written in two std, forms; ) Time constant form and ) Pole Zero form, G(S) = K(TaS +) (TbS +).. S n (T S+) (T S + ). Where K = open loop gain. Above Eq:- involves the term S n in denominator which corresponds to no, of integrations in the system. A system is called Type O, Type, Type,.. if n = 0,,,.. 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. ) Type O system :- If, n = 0, the system is called type 0, system. The steady state error are as follows; Let, G(S) = K. [... H(s) = ] S + e ss (Position) =. =. =. + G(O). H(O) + K + K p... K p = lt G(S). H(S) = lt K. = K S O S O S + e ss (Velocity) =. =. = Kv O Kv = lt G(S). H(S) = lt S K. = O. S O S O S + e ss (acceleration) =. =. = Ka O Ka = lt S G(S). H(S) = lt S K. = O. S O S O S + SJBIT/ Dept of ECE Page 8

87 ) Type System :- If, n =, the e ss to various std, i/p, G(S) = K. S (S + ) e ss (Position) =. = O + Kp = lt G(S). H(S) = lt K. = S O S O S( S + ) Kv = lt S K. = K S O S(S+) e ss (Velocity) =. K e ss (acceleration) =. =. = O O Ka = lt S K. = O. S O S (S + ) 3) Type System :- If, n =, the e ss to various std, i/p, are, G(S) = K. S (S + ) Kp = lt K. = S O S (S + )... e ss (Position) =. = O Kv = lt S K. = S O S (S + )... e ss (Velocity) =. = O Ka = lt S K. = K. S O S (S + )... e ss (acceleration) =. K 3) Type 3 System :- Gives Kp = Kv = Ka = & e ss = O. (Onwards) SJBIT/ Dept of ECE Page 83

88 Recommended Questions. Define and classify time response of a system.. Mention the Standard Test Input Signals and its Laplace transform 3. The open loop T.F. of a unity feed back system is given by G(S) = K. Where, S(+ST) T&K are constants having + Ve values.by what factor () the amplitude gain be reduced so that (a) The peak overshoot of unity step response of the system is reduced from 75% to 5% (b) The damping ratio increases from 0. to Find all the time domain specification for a unity feed back control system whose open loop T.F. is given by G(S) = 5. S(S+6) 5. The closed loop T.F. of a unity feed back control system is given by C(S) = 5. R(S) S + 4S +5 Determine () Damping ratio () Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse. 6. A servo mechanism is represent by the Eq:- + 0 d = 50E, E = R- is the actuating signal calculate the dt dt value of damping ratio, undamped and damped frequency of ascillation. d 7. Measurements conducted on sever me mechanism show the system response to be c (t) = +0.e -60t.e -0t, When subjected to a unit step i/p. Obtain the expression for closed loop T.F the damping ratio and undamped natural frequency of oscillation. 8. A feed back system employing o/p damping is as shown in fig. ) Find the value of K & K so that closed loop system resembles a nd order system with = 0.5 & frequency of damped oscillation 9.5 rad / Sec. ) With the above value of K & K 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 % tolerance? R K. S(+S) K S + C SJBIT/ Dept of ECE Page 84

89 UNIT-4 Stability Analysis Every System, for small amount of time has to pass through a transient period. Whether system will reach its steady state after passing through transients or not. The answer to this question is whether the system is stable or unstable. This is stability analysis. For example, we want to go from one station to other. The station we want to reach is our final steady state. The traveling period is the transient period. Now any thing may happen during the traveling period due to bad weather, road accident etc, there is a chance that we may not reach the next station in time. The analysis of wheather the given system can reach steady state after passing through the transients successfully is called the stability analysis of the system. In this chapter, we will steady. The stability & the factor on which system stability depends.. Stability analysis & location of closed loop poles. 3. Stability analysis using Hurwitz method. 4. Stability analysis using Routh-Hurwitz method. 5. Special cases of Routh s array. 6. Applications of Routh-Hurwitz method. Concept of stability: Consider a system i.e a deep container with an object placed inside it as shown in fig() force F (a) fig() (b) Now, if we apply a force to take out the object, as the depth of container is more, it will oscillate & settle down again at original position. Assume that force required to take out the object tends to infinity i.e always object will oscillate when force is applied & will settle down but will not come out such a system is called absolutely stable system. No change in parameters, disturbances, changes the output. SJBIT/ Dept of ECE Page 85

90 Now consider a container which is pointed one, on which we try to keep a circular object. In this object will fall down without any external application of force. Such system is called Unstable system. (a) fig() (b) While in certain cases the container is shallow then there exsists a critical value of force for which the object will come out of the container. F F F Fig(3) F<F critical F>F critical As long as F<F critical object regains its original position but if F>F critical object will come out. Stability depends on certain conditions of the system, hence system is called conditionally stable system. Pendulum where system keeps on oscillating when certain force is applied. Such systems are neither stable nor unstable & hence called critically stable or marginally stable systems Stability of control systems: The stability of a linear closed loop system can be determined from the locations of closed loop poles in the S-plane. If the system has closed loop T.F. C(s) 0 R(s) (S+) (S+4) Output response for unit step input R(s) = S C(s) 0 A B C S(S+) (S+4) S S+ S+4 Find out partial fractions SJBIT/ Dept of ECE Page 86

91 C(s) = S S+ S+4 = S S+ S+4 C(s) =.5.5e -t +.5e -4t =Css + C t (t) If the closed loop poles are located in left half of s-plane, Output response contains exponential terms with negative indices will approach zero & output will be the steady state output. i.e. C t (t) = 0 t Transient output = 0 Such system are called absolutely stable systems. Now let us have a system with one closed loop pole located in right half of s- plane C(s) 0 R(s) S(S-)(s+4) A + B + C 0 = S S- S + 4 C(t) = e t e 4 t Here there is one exponential term with positive in transient output Therefore Css = -.5 t C(t) From the above table, it is clear that output response instead of approaching to steady state value as t due to exponential term with positive index, transients go on increasing in amplitude. So such system is said to be unstable. In such system output is uncontrollable & unbounded one. Output response of such system is as shown in fig(4). SJBIT/ Dept of ECE Page 87

92 C(t) C(t) Steady state output OR t (a) fig(4) t (b) For such unstable systems, if input is removed output may not return to zero. And if the input power is turned on, output tends to. If no saturation takes place in system & no mechanical stop is provided then system may get damaged. If all the closed loop poles or roots of the characteristic equation lies in left of s-plane, then in the output response contains steady state terms & transient terms. Such transient terms approach to zero as time advances eventually output reaches to equilibrium & attains steady state value. Transient terms in such system may give oscillation but the amplitude of such oscillation will be decreasing with time & finally will vanish. So output response of such system is shown in fig5 (a) & (b). C(t) C(t) Damped oscillations Steady state OR Steady state output t (a) fig 5 (b) t BIBO Stability : This is bounded input bounded output stability. Definition of stable system: A linear time invariant system is said to be stable if following conditions are satisfied.. When system is excited by a bounded input, output is also bounded & controllable.. In the absence of input, output must tend to zero irrespective of the initial conditions. Unstable system: A linear time invariant system is said to be unstable if,. for a bounded input it produces unbounded output.. In the absence of input, output may not be returning to zero. It shows certain output without input. Besides these two cases, if one or more pairs simple non repeated roots are located on the imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output SJBIT/ Dept of ECE Page 88

93 response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such systems are said to be critically or marginally stable systems. Critically or Marginally stable systems: A linear time invariant system is said to be critically or marginally stable if for a bounded input its output oscillates with constant frequency & Amplitude. Such oscillation of output are called Undamped or Sustained oscillations. For such system one or more pairs of non repeated roots are located on the imaginary axis as shown in fig6(a). Output response of such systems is as shown in fig6(b). C(t) Constant Amplitude & frequency oscillations X J steady state output X J X - J X - J Fig 6(a) non repeated poles on J axis. t SJBIT/ Dept of ECE Page 89

94 If there are repeated poles located purely on imaginary axis system is said to be unstable. C(t) J x x s-plane J x x steady state output Conditionally Stable: A linear time invariant system is said to be conditionally stable, if for a certain condition if a particular parameter of the system, its output is bounded one. Otherwise if that condition is violated output becomes unbounded system becomes unstable. i.e. Stability of the system depends the on condition of the parameter of the system. Such system is called conditionally stable system. S-plane can be divided into three zones from stability point of view. J axis t Left half of s-plane Right half of s-plane Stable Unstable Real S-Plane (repeated) unstable Marginally stable (non repeated roots) SJBIT/ Dept of ECE Page 90

95 Sl. Nature of closed No loop poles.. Real negative i.e in LHS of splane Location of closed loop poles in s-plane J Step response C(t) Stability condition x x Absolutely stable t. Real positive in RHS of s-plane J x a C(t) Unstable increasing towards t 3. Complex conjugate with negative real part J x J -a x -J C(t) Damped oscillation t Absolutely stable. 4. Complex conjugate with positive real part J x Ct -J -x Unstable. oscillations increasing amplitude t with SJBIT/ Dept of ECE Page 9

96 5. Non repeated pair on imaginary axis J x J C(t) Marginally critically stable or x -J t OR C(t) J X J t x J x J Sustained oscillations with two frequencies & Marginally critically stable or x J 6. Repeated pair on imaginary axis Two non repeated pairs on imaginary axis. J C(t) x x J Unstable t x x -J oscillation of increasing amplitude SJBIT/ Dept of ECE Page 9

97 Relative Stability: The system is said to be relatively more stable or unstable on the basis of settling time. System is said to be more stable if settling time for that system is less than that of other system. The settling time of the root or pair of complex conjugate roots is inversely proportional to the real part of the roots. Sofar the roots located near the J axis, settling time will be large. As the roots move away from J axis i.e towards left half of the s-plane settling time becomes lesser or smaller & system becomes more & more stable. So the relative stability improves. J C(t) Stable for P x x P P Relatively more stable for P t J x x C(t) stable for x x J axis more stable for t Relative stability improves s-plane Routh Hurwitz Criterion : SJBIT/ Dept of ECE Page 93

98 This represents a method of determining the location of poles of a characteristics equation with the respect to the left half & right half of the s-plane without actually solving the equation. The T.F.of any linear closed loop system can be represented as, C(s) b 0 s m + b s m- +.+ b m = R(s) a 0 s n + a s n a n Where a & b are constants. To find the closed loop poles we equate F(s) =0. This equation is called as Characteristic Equation of the system. F(s) = a 0 s n + a s n- + a s n a n = 0. Thus the roots of the characteristic equation are the closed loop poles of the system which decide the stability of the system. Necessary Condition to have all closed loop poles in L.H.S. of s-plane. In order that the above characteristic equation has no root in right of s-plane, it is necessary but not sufficient that,. All the coefficients off the polynomial have the same sign.. Non of the coefficient vanishes i.e. all powers of s must be present in descending order from n to zero. These conditions are not sufficient. Hurwitz s Criterion : The sufficient condition for having all roots of characteristics equation in left half of s-plane is given by Hurwitz. It is referred as Hurwitz criterion. It states that: The necessary & sufficient condition to have all roots of characteristic equation in left half of s-plane is that the sub-determinants D K, K =,, n obtained from Hurwitz determinant H must all be positive. SJBIT/ Dept of ECE Page 94

99 Method of forming Hurwitz determinant: a a 3 a 5.. a n- a 0 a a 4.. a n- 0 a a 3.. a n-3 H = 0 a 0 a.. a n a... a n a n The order is n*n where n = order of characteristic equation. In Hurwitz determinant all coefficients with suffices greater than n or negative suffices must all be replaced by zeros. From Hurwitz determinant subdeterminants, D K, K=,,.n must be formed as follows: a a 3 a 5 D = a D = a a 3 D 3 = a 0 a a 4 D K = H a 0 a 0 a a 3 For the system to be stable, all above determinants must be positive. Determine the stability of the given characteristics equation by Hurwitz,s method. Ex : F(s)= s 3 + s + s + 4 = 0 is characteristic equation. a 0 =, a =, a =, a 3 = 4, n = 3 a a 3 a H = a 0 a a 4 = 0 0 a a D = = 4 D = = -3 SJBIT/ Dept of ECE Page 95

100 4 0 D 3 = 0 = 4 6 = As D & D 3 are negative, given system is unstable. Disadvantages of Hurwitz s method :. For higher order system, to solve the determinants of higher order is very complicated & time consuming.. Number of roots located in right half of s-plane for unstable system cannot be judged by this method. 3. Difficult to predict marginal stability of the system. Due to these limitations, a new method is suggested by the scientist Routh called Routh s method. It is also called Routh-Hurwitz method. Routh s Stability Criterion: It is also called Routh s array method or Routh-Hurwitz s method Routh suggested a method of tabulating the coefficients of characteristic equation in a particular way. Tabulation of coefficients gives an array called Routh s array. Consider the general characteristic equation as, F(s) = a 0 s n + a s n- + a s n a n = 0. Method of forming an array : S n a 0 a a 4 a 6. S n- a a 3 a 5 a 7 S n- b b b 3 S n-3 c c c S 0 a n Coefficients of first two rows are written directly from characteristics equation. From these two rows next rows can be obtained as follows. SJBIT/ Dept of ECE Page 96

101 a a a 0 a 3 a a 4 a 0 a 5 a a 6 a 0 a 7 b =, b =, b 3 = a a a From nd & 3 rd row, 4 th row can be obtained as b a 3 a b b a 5 a b 3 C =, C = b b This process is to be continued till the coefficient for s 0 is obtained which will be a n. From this array stability of system can be predicted. Routh s criterion : The necessary & sufficient condition for system to be stable is All the terms in the first column of Routh s array must have same sign. There should not be any sign change in first column of Routh s array. If there are sign changes existing then,. System is unstable.. The number of sign changes equals the number of roots lying in the right half of the s-plane. Examine the stability of given equation using Routh s method : Ex.: s 3 +6s + s + 6 =0 Sol: a 0 =, a = 6, a =, a 3 = 6, n = 3 S 3 S 6 6 S * 6 6 =0 0 6 S 0 6 As there is no sign change in the first column, system is stable. SJBIT/ Dept of ECE Page 97

102 Ex. 3 s 3 + 4s + s + 6 = 0 Sol: a 0 =, a = 4, a =, a 3 = 6 S 3 S +4 6 S 4-6 = S 0 +6 As there are two sign changes, system is unstable. Number of roots located in the right half of s-plane = number of sign changes =. Special Cases of Routh s criterion : Special case : First element of any of the rows of Routh s array is zero & same remaining rows contains at least one non-zero element. Effect : The terms in the new row become infinite & Routh s test fails. e.g. : s 5 + s 4 + 3s 3 + 6s + s + = 0 S 5 3 S 4 6 Special case Routh s array failed S S. Following two methods are used to remove above said difficulty. First method : Substitute a small positive number in place of a zero occurred as a first element in the row. Complete the array with this number. Then examine lim Sign change by taking. Consider above Example. 0 SJBIT/ Dept of ECE Page 98

103 S 5 3 S 4 6 S S S.5(6-3) 0 - (6-3) S 0 To examine sign change, Lim Lim = 6-3 = = 6 - = - sign is negative. Lim.5(6 3) - = Lim = = +.5 sign is positive SJBIT/ Dept of ECE Page 99

104 Routh s array is, S 5 3 S 4 6 S S - 0 S S As there are two sign changes, system is unstable. Second method : To solve the above difficulty one more method can be used. In this, replace s by /Z in original equation. Taking L.C.M. rearrange characteristic equation in descending powers of Z. Then complete the Routh s array with this new equation in Z & examine the stability with this array. Consider F(s) = s 5 + s 4 + 3s 3 + 6s + s + = 0 Put s = / Z = 0 Z 5 Z 4 Z 3 Z Z Z 5 + Z 4 + 6Z 3 +3Z +Z+ = 0 Z 5 6 Z 4 3 Z Z.33 0 Z Z 0 As there are two sign changes, system is unstable. Special case : SJBIT/ Dept of ECE Page 00

105 All the elements of a row in a Routh s array are zero. Effect : The terms of the next row can not be determined & Routh s test fails. S 5 a b c S 4 d e f S Row of zeros, special case This indicates no availability of coefficient in that row. Procedure to eliminate this difficulty :. Form an equation by using the coefficients of row which is just above the row of zeros. Such an equation is called an Auxillary equation denoted as A(s). For above case such an equation is, A(s) = ds 4 + es + f Note that the coefficients of any row are corresponding to alternate powers of s starting from the power indicated against it. So d is coefficient corresponding to s 4 so first term is ds 4 of A(s). Next coefficient e is corresponding to alternate power of s from 4 i.e. s Hence the term es & so on.. Take the derivative of an auxillary equation with respect to s. i.e. da(s) = 4d s 3 + e s ds 3. Replace row of zeros by the coefficients of da(s) ds S 5 a b c S 4 d e f S 3 4d e 0 4. Complete the array of zeros by the coefficients. Importance of auxillary equation : SJBIT/ Dept of ECE Page 0

106 Auxillary equation is always the part of original characteristic equation. This means the roots of the auxillary equation are some of the roots of original characteristics equation. Not only this but roots of auxillary equation are the most dominant roots of the original characteristic equation, from the stability point of view. The stability can be predicted from the roots of A(s)=0 rather than the roots of characteristic equation as the roots of A(s) = 0 are the most dominant from the stability point of view. The remaining roots of the characteristic equation are always in the left half & they do not play any significant role in the stability analysis. e.g. Let F(s) = 0 is the original characteristic equation of say order n = 5. Let A(s) = 0 be the auxillary equation for the system due to occurrence of special case of the order m =. Then out of 5 roots of F(s) = 0, the roots which are most dominant (dominant means very close to imaginary axis or on the imaginary axis or in the right half of s-plane) from the stability point of view are the roots of A(s) = 0. The remaining 5 = 3 roots are not significant from stability point of view as they will be far away from the imaginary axis in the left half of s-plane. The roots of auxillary equation may be,. A pair of real roots of opposite sign i.e.as shown in the fig. 8.0 (a). j j x x x Fig 8. 0(a) x Fig (b). A pair of roots located on the imaginary axis as shown in the fig. 8.0(b). 3. The non-repeated pairs of roots located on the imaginary axis as shown in the fig.8.0 (c). j j x x xx x x xx Fig. 8.0(c) Fig. 8.0(d). 4. The repeated pairs of roots located on the imaginary axis as shown in the Fig.8.0 (d). SJBIT/ Dept of ECE Page 0

107 Hence total stability can be determined from the roots of A(s) = 0, which can be out of four types shown above. Change in criterion of stability in special case : After replacing a row of zeros by the coefficients of da(s), complete the Routh s array. ds But now, the criterion that, no sign in st column of array for stability, no longer remains sufficient but becomes a necessary. This is because though A(s) is a part of original characteristic equation, da(s) is not, which is in fact used to complete the array. ds So if sign change occurs in first column, system is unstable with number of sign changes equal to number of roots of characteristics equation located in right half of s-plane. But there is no sign changes, system cannot be predicted as stable. And in such case stability is to be determined by actually solving A(s) = 0 for its roots. And from the location of roots of A(s) = 0 in the s-plane the system stability must be determined. Because roots A(s) = 0 are always dominant roots of characteristic equation. Application of Routh s of criterion : Relative stability analysis : If it is required to find relative stability of system about a line s = -. i.e. how many roots are located in right half of this line s = -, the Routh s method can be used effectively. To determine this from Routh s array, shift the axis of s plane & then apply Routh s array i.e. substitute s = s -, ( = constant) in characteristic equation. Write polynomial in terms of s. Complete array from this new equation. The number of sign changes in first column is equal to number of roots those are located to right of the vertical line s = -. Imaginary j - 0 Determining range of values of K : In practical system, an amplifier of variable gain K is introduced. The closed loop transfer function is C(s) KG(s) SJBIT/ Dept of ECE Page 03

108 R(s) = + KG(s) H(s) Hence the characteristic equation is F(s) = + KG(s) H(s) = 0 So the roots of above equation are dependent on the proper selection of value of K. So unknown K appears in the characteristic equation. In such case Routh s array is to be constructed in terms of K & then the range of values of K can be obtained in such away that it will not produce any sign change in first column of the Routh s array. Hence it is possible to obtain the range of values of K for absolute stability of the system using Routh s criterion. Such a system where stability depends on the condition of parameter K, is called conditionally stable system. Advantages of Routh s criterion : Advantages of routh s array method are :. Stability of the system can be judged without actually solving the characteristic equation.. No evaluation of determinants, which saves calculation time. 3. For unstable system it gives number of roots of characteristic equation having positive real part. 4. Relative stability of the system can be easily judged. 5. By using the criterion, critical value of system gain can be determined hence frequency of sustained oscillations can be determined. 6. It helps in finding out range of values of K for system stability. 7. It helps in finding out intersection points of roots locus with imaginary axis. Limitation of Routh s criterion :. It is valid only for real coefficients of the characteristic equation.. It does not provide exact locations of the closed loop poles in left or right half of s-plane. 3. It does not suggest methods of stabilizing an unstable system. 4. Applicable only to linear system. Ex.. s 6 + 4s 5 +3s 4 6s - 64s 48 = 0 Find the number of roots of this equation with positive real part, zero real part & negative real part Sol: S S S S SJBIT/ Dept of ECE Page 04

109 da A(s) = 3S 4 48 = 0 = s 3 ds S S S S S ( ) S S 0-48 Lim Therefore One sign change & system is unstable. Thus there is one root in R.H.S of the s plane i.e. with positive real part. Now solve A(s) = 0 for the dominant roots A(s) = 3s 4 48 =0 Put S = Y 3Y = 48 Y =6, Y = 6 = 4 S = + 4 S = -4 S = S = j So S = j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S. indicated by a sign change is S = as obtained by solving A(s) = 0. Total there are 6 roots as n = 6. Roots with Positive real part = Roots with zero real part = Roots with negative real part = 6 = 3 SJBIT/ Dept of ECE Page 05

110 Ex. : For unity feed back system, G(s) = k, Find range of values of K, marginal value of K S( + 0.4s) ( s) & frequency of sustained oscillations. Sol : Characteristic equation, + G (s) H (s) = 0 & H(s) = K + = 0 s( + 0.4s) ( + 0.5s) s [ s + 0.s } + K = 0 0.s s +s + K = 0 S 3 0. From s 0, K > 0 S 0.65 K from s, S K K > > 0. K S 0 K 6.5 > K Range of values of K, 0 < K < 6.5 Now marginal value of K is that value of K for which system becomes marginally stable. For a marginal stable system there must be row of zeros occurring in Routh s array. So value of K which makes any row of Routh array as row of zeros is called marginal value of K. Now K = 0 makes row of s 0 as row of zeros but K = 0 can not be marginal value because for K = 0, constant term in characteristic equation becomes zeros ie one coefficient for s 0 vanishes which makes system unstable instead of marginally stable. Hence marginal value of K is a value which makes any row other than s 0 as row of zeros 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 s + K = 0 ; 0.65 s = 0 Because K = 6.5 s = -0 s = j 3.6 SJBIT/ Dept of ECE Page 06

111 comparing with s = j = frequency of oscillations = 3.6 rad/ sec. Ex : 3 For a system with characteristic equation F(s) = s 5 + s4 + s 3 + s + 3s +5 =, examine the stability Solution : S 5 3 S 4 5 S S S S 0 S 5 3 S 4 5 S 3-0 S ( + ) 5 0 S ( + )( - ) S 0 5 Lim + 0 = + = + = + Lim ( + )( - ) 5 Lim = = = SJBIT/ Dept of ECE Page 07

112 S 5 3 S 4 5 S 3-0 S There are two sign changes, so system is unstable. S - 0 S 0 5 Ex : 4 Using Routh Criterion, investigate the stability of a unity feedback system whose open loop transfer function is e -st G(s) = s ( s + ) Sol : The characteristic equation is + G(s) H(s) = 0 e -st + = 0 s ( s + ) s + s + e st = 0 Now e st can be Expressed in the series form as s T e st = st + +! Trancating the series & considering only first two terms we get e st = st s + s + st = 0 s + s ( - T ) + = 0 SJBIT/ Dept of ECE Page 08

113 So routh s array is S S -T 0 S 0 T > 0 for stability T < This is the required condition for stability of the system. Ex : 5 Determine the location of roots with respect to s = - given that F(s) = s s s + 70s + 75 Sol : shift the origin with respect to s = - s = s (s ) (s ) (s ) + 70 ( s ) + 75 = 0 s 4 + s 3 + 0s + 4s + 5 = 0 S S S S S 0 5 Two sign change, there are two roots to the right of s = - & remaining are to the left of the line s = -. Hence the system is unstable. SJBIT/ Dept of ECE Page 09

114 Recommended Questions:. Explain briefly how system depends on poles and zeros.. Mention the necessary condition to have all closed loop poles in LHS of S-Plane 3. Explain briefly the Hurwitz s Criterion. 4. Explain briefly the Routh s Stability Criterion. 5..Examine the stability of given equation using Routh s method s 3 +6s + s + 6 =0 6. Examine the stability of given equation using Routh s method s 5 + s 4 + 3s 3 + 6s + s + = 0 7. Using Routh Criterion, investigate the stability of a unity feedback system whose open loop transfer function is e -st G(s) = s ( s + ) SJBIT/ Dept of ECE Page 0

115 Root Locus Techniques UNIT-5 The characteristics of the transient response of a closed loop control system are related to location of the closed loop poles. If the system has a variable loop gain, then the location of the closed loop poles depends on the value of the loop gain chosen. It is important, that the designer knows how the closed loop poles move in the s-plane as the loop gain is varied. W. R. Evans introduced a graphical method for finding the roots of the characteristic equation known as root locus method. The root locus is used to study the location of the poles of the closed loop transfer function of a given linear system as a function of its parameters, usually a loop gain, given its open loop transfer function. The roots corresponding to a particular value of the system parameter can then be located on the locus or the value of the parameter for a desired root location can be determined from the locus. It is a powerful technique, as an approximate root locus sketch can be made quickly and the designer can visualize the effects of varying system parameters on root locations or vice versa. It is applicable for single loop as well as multiple loop system. ROOT LOCUS CONCEPT To understand the concepts underlying the root locus technique, consider the second order system shown in Fig.. R(s) E(s) K s(s a) C(s) Fig. 3 Second order control system The open loop transfer function of this system is G(s) K s(s a) () Where, K and a are constants. The open loop transfer function has two poles one at origin s = 0 and the other at s = -a. The closed loop transfer function of the system shown in Fig. is C(s) R(s) G(s) G(s)H(s) s K as K () The characteristic equation for the closed loop system is obtained by setting the denominator of the right hand side of Eqn.() equal to zero. That is, G(s)H(s) s as K 0 (3) SJBIT/ Dept of ECE Page

116 The second order system under consideration is always stable for positive values of a and K but its dynamic behavior is controlled by the roots of Eqn.(3) and hence, in turn by the magnitudes of a and K, since the roots are given by s,s a (a 4K) a a a K (4) From Eqn.(4), it is seen that as the system parameters a or K varies, the roots change. Consider a to be constant and gain K to be variable. As K is varied from zero to infinity, the two roots s and s describe loci in the s-plane. Root locations for various ranges of K are: ) K= 0, the two roots are real and coincide with open loop poles of the system s = 0, s = -a. ) 0 K < a /4, the roots are real and distinct. 3) K= a /4, roots are real and equal. 4) a /4 < K <, the rots are complex conjugates. The root locus plot is shown in Fig. Fig. 4 Root loci of s +as+k as a function of K Figure has been drawn by the direct solution of the characteristic equation. This procedure becomes tedious. Evans graphical procedure helps in sketching the root locus quickly. The characteristic equation of any system is given by Δ(s) 0 (5) Where, (s) is the determinant of the signal flow graph of the system given by Eqn.(5). =-(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) + Or SJBIT/ Dept of ECE Page

117 Δ(s) P P P (6) m m m m m m3 Where, P mr is gain product of m th possible combination of r nontouching loops of the graph. The characteristic equation can be written in the form P(s) KA(s) B(s) 0 0 (7) P(s) For single loop system shown in Fig.3 G(s)H(s) (8) Where, G(s)H(s) is open loop transfer function in block diagram terminology or transmittance in signal flow graph terminology. R(s) E(s) G(s) C(s) H(s) Fig. 5 Single loop feedback system From Eqn.(7) it can be seen that the roots of the characteristic equation (closed loop poles)occur only for those values of s where P(s) (9) Since, s is a complex variable, Eqn.(9) can be converted into the two Evans conditions given below. P (s) P ( s) 80 (q ); q 0,, (0) () Roots of +P(s) = 0 are those values of s at which the magnitude and angle condition given by Eqn.(0) and Eqn.(). A plot of points in the complex plane satisfying the angle criterion is the root locus. The value of gain corresponding to a root can be determined from the magnitude criterion. To make the root locus sketching certain rules have been developed which helps in visualizing the effects of variation of system gain K ( K > 0 corresponds to the negative feed SJBIT/ Dept of ECE Page 3

118 back and K < 0 corresponds to positive feedback control system) and the effects of shifting pole-zero locations and adding in anew set of poles and zeros. GENERAL RULES FOR CONSTRUCTING ROOT LOCUS ) The root locus is symmetrical about real axis. The roots of the characteristic equation are either real or complex conjugate or combination of both. Therefore their locus must be symmetrical about the real axis. ) As K increases from zero to infinity, each branch of the root locus originates from an open loop pole (n nos.) with K= 0 and terminates either on an open loop zero (m nos.) with K = along the asymptotes or on infinity (zero at ). The number of branches terminating on infinity is equal to (n m). 3) Determine the root locus on the real axis. Root loci on the real axis are determined by open loop poles and zeros lying on it. In constructing the root loci on the real axis choose a test point on it. If the total number of real poles and real zeros to the right of this point is odd, then the point lies on root locus. The complex conjugate poles and zeros of the open loop transfer function have no effect on the location of the root loci on the real axis. 4) Determine the asymptotes of root loci. The root loci for very large values of s must be asymptotic to straight lines whose angles are given by Angle of asymptotes A 80 (q n m ) ; q 0,,, n m - () SJBIT/ Dept of ECE Page 4

119 5) All the asymptotes intersect on the real axis. It is denoted by a, given by σ a sum of poles sum of zeros n m (p p p ) (z z n n m z m ) (3) 6) Find breakaway and breakin points. The breakaway and breakin points either lie on the real axis or occur in complex conjugate pairs. On real axis, breakaway points exist between two adjacent poles and breakin in points exist between two adjacent zeros. To dk calculate these polynomial 0 must be solved. The resulting roots are the breakaway ds / breakin points. The characteristic equation given by Eqn.(7), can be rearranged as where, B(s) A(s) (s K(s p )(s z )(s B(s) p ) (s KA(s) z ) (s p n z 0 ) and The breakaway and breakin points are given by m ) (4) dk ds d ds A B A d ds B 0 (5) Note that the breakaway points and breakin points must be the roots of Eqn.(5), but not all roots of Eqn.(5) are breakaway or breakin points. If the root is not on the root locus portion of the real axis, then this root neither corresponds to breakaway or breakin point. If the roots of Eqn.(5) are complex conjugate pair, to ascertain that they lie on root loci, check the corresponding K value. If K is positive, then root is a breakaway or breakin point. 7) Determine the angle of departure of the root locus from a complex pole Angle of departure from a complex p 80 (sum of angles of vectors to a complex pole in question from other poles) (sum of angles of vectors to a complex pole in question from other zeros) (6) 8) Determine the angle of arrival of the root locus at a complex zero Angle of arrival at complex zero 80 (sum of angles of vectors to a complex zero in question from other zeros) (sum of angles of vectors to a complex zero in question from other poles) (7) SJBIT/ Dept of ECE Page 5

120 9) Find the points where the root loci may cross the imaginary axis. The points where the root loci intersect the j axis can be found by a) use of Routh s stability criterion or b) letting s = j in the characteristic equation, equating both the real part and imaginary part to zero, and solving for and K. The values of thus found give the frequencies at which root loci cross the imaginary axis. The corresponding K value is the gain at each crossing frequency. 0) The value of K corresponding to any point s on a root locus can be obtained using the magnitude condition, or K product of lengths between points to poles product of length between points to zeros (8) PHASE MARGIN AND GAIN MARGIN OF ROOT LOCUS Gain Margin It is a factor by which the design value of the gain can be multiplied before the closed loop system becomes unstable. Value of K at imaginary cross over Gain Margin Design value of K The Phase Margin (9) of K i.e. φ 80 Find the point j on the imaginary axis for which G j H j for the design value B j /A j K. design The phase margin is argg jω H(jω ) (0) SJBIT/ Dept of ECE Page 6

121 Problem No Sketch the root locus of a unity negative feedback system whose forward path transfer function K is G(s). s Solution: ) Root locus is symmetrical about real axis. ) There are no open loop zeros(m = 0). Open loop pole is at s = 0 (n = ). One branch of root locus starts from the open loop pole when K = 0 and goes to asymptotically when K. 3) Root locus lies on the entire negative real axis as there is one pole towards right of any point on the negative real axis. 4) The asymptote angle is A = 80 (q ), n m q n m 0. Angle of asymptote is A = 80. 5) Centroid of the asymptote is σ A (sum of poles) n (sum of m zeros) 6) The root locus does not branch. Hence, there is no need to calculate the break points. 7) The root locus departs at an angle of -80 from the open loop pole at s = 0. 8) The root locus does not cross the imaginary axis. Hence there is no imaginary axis cross over. The root locus plot is shown in Fig. SJBIT/ Dept of ECE Page 7

122 Figure 6 Root locus plot of K/s Comments on stability: The system is stable for all the values of K > 0. Th system is over damped. Problem No K(s ) The open loop transfer function is G(s). Sketch the root locus plot (s ) Solution: ) Root locus is symmetrical about real axis. ) There is one open loop zero at s=-.0(m=). There are two open loop poles at s=-, -(n=). Two branches of root loci start from the open loop pole when K= 0. One branch goes to open loop zero at s =-.0 when K and other goes to (open loop zero ) asymptotically when K. 3) Root locus lies on negative real axis for s -.0 as the number of open loop poles plus number of open loop zeros to the right of s=-0. are odd in number. 4) The asymptote angle is A = 80 (q ), n m q n m 0. Angle of asymptote is A = 80. 5) Centroid of the asymptote is σ A (sum of ( poles) n ) ( (sum of m ) 0.0 zeros) SJBIT/ Dept of ECE Page 8

123 6) The root locus has break points. K dk Break point is given by ds (s )(s ) (s ) 0 (s ) s (s (s, K ) ) 0; s 3, K 4 0 The root loci brakesout at the open loop poles at s=-, 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=- and other goes to along the asymptotically. 7) The branches of the root locus at s=-, - break at K=0 and are tangential to a line s=- +j0 hence depart at 90. 8) The locus arrives at open loop zero at 80. 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.. Figure 7 Root locus plot of K(s+)/(s+) 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. SJBIT/ Dept of ECE Page 9

124 Problem No 3 The open loop transfer function is Solution: G(s) K(s s(s 4). Sketch the root locus. ) ) Root locus is symmetrical about real axis. ) There are is one open loop zero at s=-4(m=). There are two open loop poles at s=0, - (n=). 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=- lies on the root locus. 80 (q ) 4) The asymptote angle is A =, q 0,, n m 0. n m Angle of asymptote are A = 80. 5) Centroid of the asymptote is σ A (sum of poles) n ( ) ( 4) (sum of m.0 zeros) 6) The brake points are given by dk/ds =0. K dk ds s s s(s (s (s ) 4).7, K 6.88, K )(s 4) (s (s 4) 0.343;.7 s) 0 7) Angle of departure from open loop pole at s =0 is 80. Angle of departure from pole at s=-.0 is 0. 8) The angle of arrival at open loop zero at s=-4 is 80 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. SJBIT/ Dept of ECE Page 0

125 Figure 3 Root locus plot of K(s+4)/s(s+) Comments on stability: System is stable for all values of K. 0 > K > : > over damped K = : = critically damped > K >.7 : < under damped K =.7 : = critically damped K >.7 : > over damped. Problem No 4 The open loop transfer function is Solution: K(s 0.) G(s). Sketch the root locus. s (s 3.6) ) Root locus is symmetrical about real axis. ) There is one open loop zero at s = -0.(m=). 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. when K and other two go to asymptotically when K. 3) Root locus lies on negative real axis between -3.6 to -0. 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. 80 (q ) 4) The asymptote angle is A =, q n m 0, n m SJBIT/ Dept of ECE Page

126 Angle of asymptote are A = 90, 70. 5) Centroid of the asymptote is σ A (sum of poles) n ( 3.6) ( 0.) (sum of m.7 zeros) 6) The root locus does branch out, which are given by dk/ds =0. K dk ds 3 (s 3.6s ) - s 0. (3s 7.s)(s (s 0.) 0.) (s 3 3.6s ) s s 3 0, 4.8s 0.43,.44s 0.67 and K 0,.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=-0.43, when K=.55 One branch goes to open loop zero at s=-0. and other goes breaksout with the another locus starting from open loop ploe at s= The break point is at s=-.67 with K=3.66. The loci go to infinity in the complex plane with constant real part s= ) 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. at 80. 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. SJBIT/ Dept of ECE Page

127 Figure 4 Root locus plot of K(s+0.)/s (s+3.6) Comments on stability: System is stable for all values of K. System is critically damped at K=.55, It is under damped for.55 > K > 0 and K >3.66. It is over damped for 3.66 > K >.55. Problem No 5 The open loop transfer function is Solution: G(s) s(s K 6s. Sketch the root locus. 5) ) Root locus is symmetrical about real axis. ) 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. 80 (q ) 4) The asymptote angle is A =, q 0,, n m 0,,. n m Angle of asymptote are A = 60, 80, ) Centroid of the asymptote is σ A (sum of poles) n ( 3 3) 3 (sum of m.0 zeros) SJBIT/ Dept of ECE Page 3

128 6) The brake points are given by dk/ds =0. 3 K s(s 6s 5) (s 6s 5s) dk ds s K,, 3s 34 s j.087and j For a point to be break point, the corresponding value of K is a real number greater than or equal to zero. Hence, S, are not break points. 7) Angle of departure from the open loop pole at s=0 is 80. Angle of departure from complex pole s= -3+j4 is 80 p (sum of the angles of (sum of the angles of vectors to a complex pole in question from other poles) vectors to a complex pole inquestion from zeros) p φ p 4 80 (80 tan 90 ) Similarly, Angle of departure from complex pole s= -3-j4 is 80 ( ) 33.3 or ) 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 3 The characteristic equation is s 6s 5s K 0. The Routh s array is 3 s 5 s s s K 6 6 For the system to be stable K < 50. At K=50 the auxillary equation is 6s +50=0. s = ±j5. or substitute s= j in the characteristic equation. Equate real and imaginary parts to zero. Solve for and K. s ( 3 ω jω 6ω 6s The plot of root locus is shown in Fig , 6 K) j5 5s jω K jω ω K K 0 5 jω 5 0,50 K 0 0 SJBIT/ Dept of ECE Page 4

129 Figure 5 Root locus plot of K/s(s +6s+5) Comments on stability: System is stable for all values of 50 > K > 0. At K=50, it has sustained oscillation of 5rad/sec. The system is unstable for K >50. Problem No Sketch the root locus of a unity negative feedback system whose forward path transfer function K(s ) is G(s)H(s). Comment on the stability of the system. (s )(s 3 j)(s 3 j) Solution: 9) Root locus is symmetrical about real axis. 0) There is one open loop zero at s = - (m = ). There are three open loop poles at s = -, -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 = - goes to zero at s = - when K, and other two branches go to asymptotically (zeros at ) when K. ) Root locus lies on the negative real axis in the range s=- to s= - as there is one pole to the right of any point s on the real axis in this range. ) The asymptote angle is A = 80 (q ), n m q n m 0,. Angle of asymptote is A = 90, 70. SJBIT/ Dept of ECE Page 5

130 3) Centroid of the asymptote is σ A (sum of ( 3 poles) n 3) ( (sum of zeros) m ).5 4) The root locus does not branch. Hence, there is no need to calculate break points. 5) The angle of departure at real pole at s=- is 80. The angle of departure at the complex pole at s=-3+j is p 80 (sum of the angles of (sum of the angles of vectors to a complex pole in question from other poles) vectors to a complex pole inquestion from zeros) θ θ p tan - atan(-,) tan or 35, ( or ) θ tan The angle of departure at the complex pole at s=-3-j is ) 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. p 80 ( ) SJBIT/ Dept of ECE Page 6

131 Figure Root locus plot of K(s+)/(s+)(s+3+j)(s+3-j) Comments on stability: The system is stable for all the values of K > 0. Problem No The open loop transfer function is G(s)H(s) K s(s 0.5)(s 0.6s 0) Sketch the root locus plot. Comment on the stability of the system.. Solution: 0) Root locus is symmetrical about real axis. ) There are no open loop zeros (m=0). There are four open loop poles (n=4) at s=0, -0.5, -0.3 ± j 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. ) 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. 80 (q ) 3) The asymptote angle is A =, q n m 0,,,3. n m Angle of asymptote is A = 45, 35, 5, ±35. 4) Centroid of the asymptote is σ A (sum of poles) n (sum of m zeros) SJBIT/ Dept of ECE Page 7

132 The value of K at s=-0.75 is ( ) ) The root locus has break points. K = -s(s+0.5)(s +0.6s+0) = -(s 4 +.s s +5s) Break points are given by dk/ds = 0 dk 3 4s 3.3s ds 0.6s 5 0 s= , j.89 There is only one break point at Value of K at s = is ) The angle of departure at real pole at s=0 is 80 and at s=-0.5 is 0. The angle of departure at the complex pole at s = j3.48 is p (sum of the angles of vectors to a complex pole in question from other poles) (sum of the angles of vectors to a complex pole inquestion from zeros) 3.48 θ tan 84.6 or tan 86.4, θ tan ( ) 9.8 p The angle of departure at the complex pole at s = j3.48 is 9.8 p 80 ( ) 9.8 7) 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)(s+0.6s+0)+k =0 or s4+.s3+0.3s+5s+k=0 SJBIT/ Dept of ECE Page 8

133 The Routh s array is s s s s s K 5.75 K K K The system is stable if 0 < K < 6.3 The auxiliary equation at K 6.3 is 5.75s +6.3 = 0 which gives s = ± j.3 at imaginary axis crossover. The root locus plot is shown in Fig.. Figure 8 Root locus plot of K/s(s+0.5)(s +0.6s+0) Comments on stability: System is stable for all values of 6.3 >K > 0. The system has sustained oscillation at =.3 rad/sec at K=6.3. The system is unstable for K > 6.3. Problem No 3 The open loop transfer function is K G(s). Sketch the root locus. s(s 4)(s 4s 0) SJBIT/ Dept of ECE Page 9

134 Solution: 0) Root locus is symmetrical about real axis. ) There are no open loop zeros (m=0). There are three open loop poles (n=3) at s = -0, -4, - j4. Three branches of root loci start from the three open loop poles when K= 0 and to infinity asymptotically when K. ) 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. 80 (q ) 3) The asymptote angle is A =, q n m 0,,, 3 n m Angle of asymptote are A = 45, 35, 5, 35. 4) Centroid of the asymptote is σ A (sum of poles) (sum of zeros) n m ( ) 4.0 5) The root locus does branch out, which are given by dk/ds =0. K dk Break point is given by ds 3 4s 4s 7s s (s s s 3 s(s (s 8s 4 )(4s.0, K.0 4)(s 8s 3 6s 6s 64; 4s 36s 3s j.45, K 40) 0) 80s) 40s The root loci brakeout at the open loop poles at s = -.0, when K = 64 and breakin and breakout at s=-+j.45, when K=00 6) The angle of departure at real pole at s=0 is 80 and at s=-4 is 0. The angle of departure at the complex pole at s = - + j4 is -90. SJBIT/ Dept of ECE Page 30

135 p 80 (sum of the angles of (sum of the angles of vectors to a complex pole in question from other poles) vectors to a complex pole inquestion from zeros) θ θ θ p 4 tan 63.4 or6.6 - atan(4,-) tan 63.4, θ ( tan 0 90 ) The angle of departure at the complex pole at s = - j4 is 90 7) 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 4 3 The characteristic equation is s 8s 36s 80s K 0 For the system to be stable K > 0 and 080-8K > 0. The imaginary crossover is given by 080-8K=0 or K = 60. At K = 60, the auxiliary equation is 6s +60 = 0. The imaginary cross over occurs at s= j 0. or The Rouths array is s s s s s s K 6 K 4 put p K K jω 8 jω 36 jω 80 jω K 0 SJBIT/ Dept of ECE 4 3 Page 3 ω 36ω K j 8ω 80ω 0 Equate real and 8ω 8s 4 3 s 80 - ( s jω 80ω s imaginary parts to zero ω 96.6 K 0, 0 j 70 ) 0;s j 0

136 The root locus plot is shown in Fig.3. Figure 9 Root locus plot of K/s(s+4)(s +4s+0) Comments on stability: For 60 > K > 0 system is stable K = 60 system has stained oscillations of 0 rad/sec. K > 60 system is unstable. SJBIT/ Dept of ECE Page 3

137 Recommended Questions:. Give the general rules for constructing root locus.. 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 ). Sketch the root locus plot. (s ) 5. The open loop transfer function is G(s) K(s 4). Sketch the root locus. s(s ) 6. The open loop transfer function is G(s) K. Sketch the root locus. s(s 6s 5) 7. The open loop transfer function is G(s) K. Sketch the root s(s 4)(s 4s 0) SJBIT/ Dept of ECE Page 33

138 Stability in the frequency domain UNIT- 6 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. (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. (b). Figure 6. (a) Figure 6. (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 SJBIT/ Dept of ECE Page 34

139 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.. Obtain the transfer function of the system O( S) F ( S), Where F (S) is transfer function, O(S) and I(S) are the Laplace transforms I( S) of the output and input respectively.. Replace S by (j ) (As S is a complex number) F ( j ) O( j I( j ) ) O( j I( j ) ) A( ) B( j ) (another complex number) 3. For various values of, ranging from 0 to determine M ( ) and. O( j ) M ( ) A( ) B( j ) I( j ) A jb M A B O( j I( j ) ) A jb tan B A SJBIT/ Dept of ECE Page 35

140 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: ) Polar plots ) 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.. 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 SJBIT/ Dept of ECE Page 36

141 the polar plot. It is easier to construct a polar plot and ready information of magnitude ratio and phase angle can be obtained. +90, -70 Img Positive angles M +80, -80 0, + 360, -360 Real Negative angles Figure 6.: Complex Plane Representation +70, -90 A typical polar plot is shown in Figure 6.3 in which the magnitude ratio M and phase angle at a given value of can be readily obtained. Figure 6.3: A Typical Polar Plot SJBIT/ Dept of ECE Page 37

142 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 : 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. sec. x (t) (input) K y (t) (output) C Governing Differential Equation: dy C. Ky Kx by K dt C dy. y x K dt dy. y x Take Laplace transform dt τ SY(S) + Y(S) = X(S) (τs+) Y(S) = X(S) SJBIT/ Dept of ECE Page 38

143 Transfer function F(S) = Y( S) X ( S) S Given τ = 0. sec Y( S) X ( S) 0.S 0.S To obtain the polar plot (i.e., frequency response data) replace S by j. Y( j X ( j ) ) 0. j j(0. ) Magnification Factor M = Y( j X ( j ) ) j(0. ) j(0) j(0. ) M (0. ) Y( j ) Phase angle = Y( j ) Xj j(0) ( 0. ) X ( j ) tan 0 tan (0. ) tan (0. ) Now obtain the values of M and for different values of ranging from 0 to as given in Table 6.. Table 6. Frequency Response Data M (0. ) tan (0. ) SJBIT/ Dept of ECE Page 39

144 The data from Table 6. when plotted on the complex plane with as a parameter polar plot is obtained as given below. = 50 = 6 = 0 Illustration : A second order system has a natural frequency of 0 rad/sec and a damping ratio of 0.5. Sketch the polar plot for the system. x (Input) K m y (Response) C The transfer function of the system is given by SJBIT/ Dept of ECE Page 40

145 Y( S) X ( S) S n n S n Given n = 0 rad/sec and = 0.5 Y( S) X ( S) S 00 0S 00 Replace S by j Y( j X ( j ) ) ( j ) 00 0 j j(0) 0 j 00 as j M Y( j X ( j ) ) 00 (00 j(0) ) j(0 ) (00 00 ) (0 ) Magnification Factor = M (00 00 ) (0 ) 00 j(0) Phase angle = 00 j(0) (00 ) j(0 ) (00 ) j(0 ) tan 0 00 tan 0 00 Now obtain the values of M and for various value of ranging from 0 to as given in the following Table 6.. Table 6. Frequency Response Data: Illustration M( ) SJBIT/ Dept of ECE Page 4

146 The data from Table 6. 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 = 0 rad/sec. Illustration 3: Obtain the polar plot for the transfer function 0 F ( S) Replace S by j ( S ) F ( j ) 0 j M ( ) F( j ) 0 ( ) = F (j ) = 0 - (j +) tan Table 6.3 Frequency Response Data: Illustration 3 SJBIT/ Dept of ECE Page 4

147 M( ) = 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. ( a jb)( c jd) I A jb F( S) ( e jf ) --- (Transfer function) Magnitude Ratio = M a jb c e jf jd a b e * f c d Phase angle: SJBIT/ Dept of ECE Page 43

148 a jb e c jf jd ( a jb) ( c jd) ( e jf ) tan b a tan d c tan f e II Values of tan functions IQ: tan - (b/a) Positive: IIQ: tan - (b/-a) Negative: 80 - IIQ (-a+jb) Img (a+jb) IQ IIIQ: tan - (-b/-a) Positive: 80 + IVQ: tan - (-b/a) Negative: (-a-jb) (a-jb) Real IIIQ IVQ tan - (0) = 0 0 tan - (-0) =80 0 tan - () = 45 0 tan - (-) = 35 0 tan - ( ) = 90 0 tan - (- ) = 70 0 III Let K = Constant = K + j(0) Therefore K K 0 K Applicable for both K>0 and K<0 SJBIT/ Dept of ECE Page 44

149 K = (K+ j0) Img = tan - (0/K) = tan - (0) = 0, if K>0 -K+j0 K+j0 Real = tan - (-0) = 80 0, if K<0 IV: S n = (j ) n = (0+j ) n = (0+j ) (0+j ) n times a. Magnitude S n n ( j ) (0 j ) (0 j ) n times ( 0 ) ( n times Therefore S n = n b. Angle S n n ( j ) (0 j ) (0 j ) n times = tan - ( /0) + tan - ( /0) n times = n times S n = n * 90 0 SJBIT/ Dept of ECE Page 45

150 Illustration 4: Sketch the polar plot for the system represented by the following open loop transfer function. 0 G ( S) H( S), obtain M and for different values of S( S 0)( S ) i) As 0, S 0 S j G S ( S) H( S) 0 S 0 *0* 0.5 S S ii) As, S 0 0 ( S) H( S) 3 S( S)( S) S G S S 0 3* M( ) ( ) SJBIT/ Dept of ECE Page 46

151 Img =, M = 0 0 Real = 0, M = -90 Illustration 5: Sketch the polar plots for the system represented by the following open loop transfer function. K G ( S) H ( S) S ( S 5) i) As 0, S 0 S j K G( S) H( S) S 0 0 S ( S 5), S is far lesser than 5 and can be neglected K K / 5 5S S M ( ) K / 5 S K n ( ) K / 5 S ( K / 5) S = 0 *90 = 0 80 = ii) As, S SJBIT/ Dept of ECE Page 47

152 G( S) H( S) S S K ( S 5) K 3 S K K M ( ) 3 3 S 0 ( ) K S 3 = 0-3*90 = Img M( ) ( ) = 0, M = -80 =, M = 0 0 Real -90 Illustration 6: Sketch the polar plots for the system represented by the following open loop transfer function. G ( S) H( S) S ( S 0 5)( S 8) i) As 0, S 0 S j G( S) H( S) S S.5.8 / 4 S SJBIT/ Dept of ECE Page 48

153 / 4 / 4 M ( ) S / 4 0 ( ) / 4 S = 0 - *90 = - *90 0 = 80 0 ii) As, S G( S) H( S) S 0 S. S. S 0 4 S M ( ) 0 4 S 0 4 as, M ( ) = 0 ( ) 0 S 4 = 0 4*90 = SJBIT/ Dept of ECE Page 49

154 -70 Img M( ) ( ) = 0, M = =, M = 0 0, Real -90 Illustration 7: Sketch the polar plots for the system represented by the following open loop transfer function. 0( S ) G ( S) H( S) 3 S ( S 4)( S 5) i) As 0, S 0 S j 0( S ) ( S) H( S) S ( S 4)( S G S 5) 0. 3 S.4.5 S 3 M ( ) S ( ) S = 0 3*90 = ii) As, S G( S) H( S) S 0. S 3 S. S. S) 0 4 S 0 0 M ( ) 4 4 S 0 SJBIT/ Dept of ECE Page 50

155 4 ( ) 0 S = 0 4*90 = M( ) ( ) -70 Img = 0, M = =, M = 0 0, -360 Real -90 Illustration 8: Sketch the polar plots for the system represented by the following open loop transfer function. G ( S) H( S) S( S 0 )( S 4) i) As 0, S 0 S j G( S) H( S) S S( ).4 0/8 S M ( ) 0/8 S 0/8 ( ) 0/8 S = = 90 0 ii) As, S G( S) H( S) S 0 3 S 0 0 M ( ) 3 3 S 0 SJBIT/ Dept of ECE Page 5

156 +70 Control Systems 3 ( ) 0 S = 0 3*90 0 = Img = 0, M = M( ) ( ) =, M = 0 0, -360 Real Illustration 9: Sketch the polar plot for the system represented by the following transfer function. Y( S) X ( S) S 00 0S 00 i) As 0, S 0 S j M ( ), for both K positive and negative ( ) 0 80 ii) As, S 00 S M ( ) 0 S, ( ) S S = 0 * = * = M( ) ( ) Img SJBIT/ Dept of ECE Page 5

157 = 0, M = =, M = 0 0, -360 Real Experimental determination of Frequency Response Many a times the transfer function of a physical system may not be available in such circumstances it is necessary to obtain frequency response information experimentally. Such data may then be used to establish the transfer function. This method requires the actual system. SYSTEM ANALYSIS USING POLAR PLOTS: NYQUIST CRITERION System Analysis using Polar Plots: Nyquist Criterion Polar plots can be used to predict feed back control system stability by the application of Nyquist Criterion, and therefore are also referred as Nyquist Plots. It is a labor saving technique in the analysis of dynamic behaviour of control systems in which the need for finding roots of characteristic equation of the system is eliminated. Consider a typical closed loop control system which may be represented by the simplified block diagram as shown in Figure 6.4 R(S) + - G(S) C(S) H(S) Figure 6.4 Simplified System Block Diagram SJBIT/ Dept of ECE Page 53

158 The closed-loop transfer function or the relationship between the output and input of the system is given by C( S) R( S) G( S) G( S) H( S) The open-loop transfer function is G(S) H(S) (the transfer function with the feedback loop broken at the summing point). + G(S) H(S) is called Characteristic Function which when equated to zero gives the Characteristic Equation of the system. + G(S) H(S) = 0 Characteristic Equation The characteristic function F(S) = + G(S) H(S) can be expressed as the ratio of two factored polynomials. Let F ( S) G( S) H ( S) S k K( S ( S P )( S Z )( S P )( S Z )...( S P )...( S 3 Z n ) Z n ) The Characteristic equation in general can be represented as F(S) = K (S+Z ) (S+Z ) (S+Z 3 ). (S+Z n ) = 0 Then: Z, -Z, -Z 3. Z n are the roots of the characteristic equation at S= -Z, S= -Z, S= -Z 3, + G(S) H(S) becomes zero. These values of S are termed as Zeros of F(S) Similarly: at S= -P, S= -P, S= -P 3. Etc. + G (S) H (S) becomes infinity. These values are called Poles of F (S). SJBIT/ Dept of ECE Page 54

159 Condition for Stability For stable operation of control system all the roots of characteristic equation must be negative real numbers or complex numbers with negative real parts. Therefore, for a system to be stable all the Zeros of characteristic equation (function) should be either negative real numbers or complex numbers with negative real parts. These roots can be plotted on a complexplane or S-plane in which the imaginary axis divides the complex plane in to two parts: right half plane and left half plane. Negative real numbers or complex numbers with negative real parts lie on the left of S-plane as shown Figure 6.5. Left half of S Plane -3+j Img Right half of S Plane +j Real -j -3-j Figure 6.5 Two halves of Complex Plane Therefore the roots which are positive real numbers or complex numbers with positive real parts lie on the right-half of S-plane. In view of this, the condition for stability can be stated as For a system to be stable all the zeros of characteristic equation should lie on the left half of S-plane. Therefore, the procedure for investigating system stability is to search for Zeros on the right half of S-plane, which would lead the system to instability, if present. However, it is impracticable to investigate every point on S-plane as to which half of S-plane it belongs to and SJBIT/ Dept of ECE Page 55

160 so it is necessary to have a short-cut method. Such a procedure for searching the right half of S- plane for the presence of Zeros and interpretation of this procedure on the Polar plot is given by the Nyquist Criterion. Nyquist Criterion: Cauchy s Principle of Argument: In order to investigate stability on the Polar plot, it is first necessary to correlate the region of instability on the S-plane with identification of instability on the polar plot, or +GH plane. The +GH plane is frequently the name given to the plane where +G(S) H(S) is plotted in complex coordinates with S replaced by j. Likewise, the plot of G(S) H(S) with S replaced by j is often termed as GH plane. This terminology is adopted in the remainder of this discussion. The Nyquist Criterion is based on the Cauchy s principle of argument of complex variable theory. Consider [F(S) = +G(S) H(S)] be a single valued rational function which is analytic everywhere in a specified region except at a finite number of points in S-plane. (A function F(S) is said to be analytic if the function and all its derivatives exist). The points where the function and its derivatives does not exist are called singular points. The poles of a point are singular points. Let C S be a closed path chosen in S-plane as shown Figure 6.6 (a) such that the function F(S) is analytic at all points on it. For each point on C S represented on S-plane there is a corresponding mapping point in F(S) plane. Thus when mapping is made on F(S) plane, the curve C G mapped by the function F(S) plane is also a closed path as shown in Figure 6.6 (b). The direction of traverse of C G in F(S) plane may be clockwise or counter clockwise, depending upon the particular function F(S). SJBIT/ Dept of ECE Page 56

161 Then the Cauchy principle of argument states that: The mapping made on F(S) plane will encircle its origin as many number of times as the difference between the number of Zeros and Poles of F(S) enclosed by the S-plane locus C S in the S-plane. +j S-plane G S5 F(S) = +G(S) H(S) plane +j S G S4 G S S 5 S -σ C S σ -σ G S3 (0+j ) σ S 4 S 3 C G G S -j -j Figure 6.6 (a) Figure 6.6 (b) Figure 6.6 Mapping on S-plane and F(S) plane Thus N = Z P N 0+j0 = Z P Where N 0+j0 : Number of encirclements made by F(S) plane plot (C G ) about its origin. Z and P: Number of Zeros and Poles of F(S) respectively enclosed by the locus C S in the S- plane. Illustration: Consider a function F(S) F ( S) K( S )( S S( S 3)( S 5)( S j)( S 5 j)( S j) 5 j) Zeros: -, (--j), (-+j) indicated by O (dots) in the S-plane Poles: 0, -3, -5, (-5 j), (-5 +j) indicated by X (Cross) in S-plane: As shown in Figure 6.6 (c) SJBIT/ Dept of ECE Page 57

162 +j S-plane +j F(S) = +G(S) H(S) plane C S C S C G C G -σ C S σ -σ (0+j0) C G σ O: ZEROS X: POLES C G C S -j Figure 6.6 (c) -j Figure 6.6 (d) Now consider path C S (CCW) on S-plane for which: Z:, P = Consider another path C S (CCW) in the same S plane for which: Z =, P = 4 C G and C G are the corresponding paths on F(S) plane [Figure 6.6 (d)]. Considering C G [plot corresponding to C S on F (S) plane] N 0+j0 = Z P = - = + C G will encircle the origin once in the same direction of C S (CCW) Similarly for the path C G N 0+j0 = Z P = 4 = - 3 C G will encircle the origin 3 times in the opposite direction of C S (CCW) Note: The mapping on F(S) plane will encircle its origin as many number of times as the difference between the number of Zeros and Poles of F(S) enclosed by the S-plane locus. From the above it can be observed that In the expression SJBIT/ Dept of ECE Page 58

163 N= Z - P, N can be positive when: Z>P N = 0 when: Z = P N can be negative when: Z<P When N is positive the map C G encircles the origin N times in the same direction as that of C S When N = 0, No encirclements N is negative the map C G encircles the origin N times in the opposite direction as that of C S Nyquist Path and Nyquist Plot The above Cauchy s principle of argument can be used to investigate the stability of control systems. We have seen that if the Zeros of characteristic function lie on the right half of S-plane it will lead to system instability. Now, to encircle the entire right half of S-plane, select a closed path as shown in Figure 6.6 (e) such that all the Zeros lying on the right-half of S-plane will lie inside this path. This path in S-plane is known as Nyquist path. Nyquist path is generally taken in CCW direction. This path consists of the imaginary axis of the S-plane (S = 0+j, - < < ) and a closing semicircle of infinite radius. If the system being tested has poles of F(S) on the imaginary axis, it is customary to modify the contour as shown Figure 6.6 (f) excluding these poles from the path. SJBIT/ Dept of ECE Page 59

164 +j +j S-plane +j r = -σ 0+j0 0-j0 σ -σ S=+j0 S= -j0 r 0 r = σ -j -j S= -j -j Figure 6.6 (e): Nyquist Path Figure 6.6 (f) Corresponding to the Nyquist path a plot can be mapped on F(S) = +G(S) H(S) plane as shown in Figure 6.6 (g) and the number of encirclements made by this F(S) plot about its origin can be counted. Img + G(S) H(S) plane Real 0+j0 Real Img Figure 6.6 (g) Now from the principle of argument N 0+j0 = Z-P N 0+j0 = number of encirclements made by F(S) plane plot Z, P: Zeros and Poles lying on right half of S-plane For the system to be stable: Z = 0 SJBIT/ Dept of ECE Page 60

165 N 0+j0 = - P Condition for Stability Apart from this, the Nyquist path can also be mapped on G(S) H(S) plane (Open-loop transfer function plane) as shown in Figure 6.6 (h). Now consider F(S) = + G(S) H(S) for which the origin is (0+j0) as shown in Figure 6.6 (g). Therefore G(S) H(S) = F(S) = (0+j0) = (-+j0) Coordinates for origin on G(S) H(S) plane as shown in Figure 6.6(h) G(S) H(S) plane Img Origin of the plot for G(S) H(S) (-+j0) 0+j0 Figure 6.6 (h) Thus a path on + G(S) H(S) plane can be easily converted to a path on G(S) H(S) plane or open loop transfer function plane. This path will be identical to that of +G(S) H(S) path except that the origin is now shifted to the left by one as shown in Figure 6.6 (h). This concept can be made use of by making the plot in G(S) H(S) plane instead of + G(S) H(S) plane. The plot made on G(S) H(S) plane is termed as the Niquist Plot and its net encirclements about (-+j0) (known as critical point) will be the same as the number of net encirclements made by F(S) plot in the F(S) = +G(S) H(S) plane about the origin. Now, the principle of argument now can be re-written as N -+j0 = Z-P SJBIT/ Dept of ECE Page 6

166 Where N -+j0 = Number of net encirclements made by the G(S) H(S) plot (Nyquist Plot) in the G(S) H(S) plane about -+j0 For a system to be stable Z = 0 N -+j0 = -P Thus the Nyquist Criterion for a stable system can be stated as The number of net encirclements made by the Nyquist plot in the G(S) H(S) plane about the critical point (-+j0) is equal to the number of poles of F(S) lying in right half of S-plane. [Encirclements if any will be in the opposite direction. Poles of F(S) are the same as the poles of G(S) H(S)]. Thus the stability of closed-loop control system is determined from its open-loop transfer function. System Analysis using Nyquist Criterion: Illustrations Illustration : Sketch the Nyquist plot for the system represented by the open loop transfer function and comment on its stability. K G ( S) H( S) K 0, a 0 Poles: S = 0 (on imaginary axis) and S = -a S( S a) Step : Define Nyquist path. Let the Nyquist path be defined as given below. SJBIT/ Dept of ECE Page 6

167 +j S= +j -σ S=+j0 S= -j0 r = r 0 σ S= -j Nyquist Path -j Section I: S = +j to S = +j0; Section II: S = +j0 to S = -j0 Section III: S = -j0 to S = -j ; Section IV: S = -j to S = +j. Corresponding to different sections namely I, II, III, and IV Obtain polar plots on G(S) H(S) plane, which are nothing but Nyquist Plots. Nyquist Plot for Section I: In S-plane section I runs from S= + j to S = +j0 To obtain polar plot in G(S) H(S) plane: K G ( S) H( S) K 0, a S( S a) 0 K (i) G( S) H( S) j S S, K G ( S) H( S) S 0 K G ( S) H( S) K S S = 0 *90 0 = - 80 S= 0 -j SJBIT/ Dept of ECE Page 63