Control System

CONTROL SYSTEMS Introduction to Control Systems In this chapter we attempt to familiarize the reader with the following subjects: 1. What a control system is 2. Why control systems are important 3. What the basic components of a control system are 4. Why feedback is incorporated into most control systems 5. Types of control systems Let's begin with a simple question. When did you last use the word " Control" ? Perhaps one may have to think for a while. But the paradox is that one invariably uses this word almost in every walk of life but fails to take notice. The following are a few common phrases we come across all the time. •

He has no control over his expenditure.

•

I could not control my tears.

•

The law and order situation in the city is out of control.

•

pest control in orchards.

•

The doctor suggested strict diet control.

There are many new products and services being introduced every day that depend on control systems yet they are not identified as control systems!. The user of the system does not focus on the control system but on the results. With regard to the first two items above we cite the example of the human being as perhaps the most sophisticated and the most complex control system in existence. An average human being is capable of performing a wide range of tasks, including decision making. Some of these tasks, such as picking up objects, or walking from one point to another, are normally carried out in a routine fashion. Under certain conditions some of the tasks are to be performed on the best possible way. For instance an athlete running a 100 yard dash has the objective of running that distance in the shortest possible time. A marathon runner on the other hand, not only must run the distance as quickly as possible, but in doing so, he or she must control the consumption of energy, so that the best result can be achieved. Therefore, we can state that in general that in life there are numerous objectives that need to be accomplished and the means of achieving the objectives usually involve the need for control systems. In recent years control system have assumed an increasingly important role in the development and advancement of modern civilization and technology. Particularly every aspect of our day to day activities is affected by some type of control system. For example in the domestic domain, automatic controls in heating and air-conditioning systems regulate the temperature and humidity of homes and buildings for comfortable living. To achieve maximum efficiency in energy consumption many modern heating and air conditioning systems in large office and factory buildings are computer controlled. The principles of control system can be illustrated in many fields.

•

In a simple transistor amplifier a low level signal applied to the base will control a relatively large level signal on the collector.

•

By turning a key the driver of an automobile can start a large H.P engine.

•

A person can lower the temperature in the room simple by turning a knob on the air conditioner.

•

The driver of several tonne automobile can control as motion by the simple use of steering wheel, accelerator, and brake pedal.

Control systems are found in abundance in all sectors of industry such as quality control of manufactured products, automatic assembly line, machine tool control, space technology and weapon systems, computer control, transportation systems, computer control, transportation systems, robotics and many others. Definition of control system A control system can be defined as an interconnection of several components all working together to perform a certain function. In most cases this function is the control of physical variable, such as temperature voltage, frequency, flowrate, current, position, hp speed, illumination, altitude etc., These are called controlled variables. Regardless of what type of control system we have, the basic ingredients of the system can be described by 1. Objectives of the control 2. Control system components 3. Results Objectives Inputs U

CONTROL SYSTEM (a) CONTROL SYSTEM (b)

Results Outputs C

Fig 1.1 Basic components of control systems In block diagram form, the basic relation between these three basic ingredients is illustrated in fig 1-1 (a) In more scientific terms, these three basic ingredients can be identified with inputs, system components, and outputs, respectively as shown in fig 1-1(b) In general, the objective of the control system is to control the outputs c in some prescribed manner by the inputs U through the elements of the control system. The inputs of the system are also called actuating signals, and outputs are known as controlled variables.

OPEN LOOP CONTROL SYSTEMS (NON FEED BACK SYSTEMS) Those systems in which the output has no effect on the control action are called open loop control systems. In other words, in an open - loop control systems the output is neither measured nor feedback for comparison with the input. Thus to each reference input there corresponds a fixed operating condition; as a result, the accuracy of the system depends on calibration. Open loop control can be used, in practice, only if the relationship between input and output is known, and if there are neither internal nor external disturbances. Note that any control system that operates on a time basis is open loop. We shall go through examples and try to identify the inputs (objectives) and outputs (effects). EXAMPLE - 1 Rotational Generator The input to rotational generator is the speed of the prime mover ( e.g steam turbine) in r.p.m. Assuming the generator is on no load the output may be induced voltage at the output terminals. Speed of the Induced Voltage Prime mover Rotational Generator Output Inputs Fig 1-2 Rotational Generator EXAMPLE – 2 washing machine Most ( but not all ) washing machines are operated in the following manner. After the clothes to be washed have been put into the machine, the soap or detergent, bleach and water are entered in proper amounts as specified by the manufacturer. The washing time is then set on a timer and the washer is energized. When the cycle is completed, the machine shuts itself off. In this example washing time forms input and cleanliness of the clothes is identified as output. Cleanliness of clothes Time Washing Machine Fig 1-3 Washing Machine EXAMPLE – 3 WATER TANK LEVEL CONTROL To understand the concept further it is useful to consider an example let it be desired to maintain the actual water level 'c ' in the tank as close as possible to a desired level ' r '. The desired level will be called the system input, and the actual level the controlled variable or system output. Water flows from the tank via a valve Vo , and enters the tank from a supply via a control valve Vc. The control valve is adjustable manually. Desired Water level r

Valve VC

WATER TANK

Water in Fig 1-4 b) Open loop control Valve VO C

Water out

Fig –1.4 a) Water level control In this form of control, the valves are adjusted to make output c equal to input r but not readjusted continually to keep the two equal. For this system, this form of control will normally not yield high performance. A difference between input and output, a system error e= r-c would be expected to develop, due to two major effects. 1. Disturbance acting on the system 2. Parameter variations of the system These are prime motivations for the use of feed back control. For example, Pressure variations upstream of Vc and downstream of VO can be important disturbances affecting inflow and output flow and hence level. A sudden change or gradual change of flow resistance of the valves due to foreign matter or valve deposits represents a system parameter variation. Open loop control systems are control systems in which the output has no effect upon the control action. The accuracy of the system depends on the calibration. Open loop control systems must be carefully calibrated and must maintain that calibration in order to be useful. In the presence of disturbances an open loop control system will not perform the desired task. As a last example consider a sprinkle used to water the lawn. The system is adjusted to water a given area by opening the water valve and observing the resulting pattern. When the pattern is considered satisfactory, the system is calibrated and no further valve adjustment is necessary. The pattern will be maintained reasonably well if there is no change in water pressure when a tap is opened inside the house, reducing pressure, the pattern changes; i.e the open loop control steady state condition. CLOSED LOOP CONTROL SYSTEMS (FEEDBACK CONTROL SYSTEMS) Referring back to water tank level example of open- loop control system, the system is going possess error when actual water level (c ) In the tank differs from desired level ( r ) To improve performance, the operator could continuously readjust the valves based on system error e=r-c what is missing in open loop control system for more accurate is a link or feed back from the output to the input of the system A feedback control system in effect automates this action, as follows: The output c is measured continuously and fed back to be compared with the input r . The error e = r-c is used to adjust the control valve by means of an actuator ( no shown in fig ) the feed back loop causes the system to take corrective action if output c ( actual level ) deviates from input 'r ' ( desired level ) whatever the reason.

Closed loop or feed back control operates according to a very simple principle. 1. Measure the variable to be controlled. 2.

Compare this measured valve with the desired value and determine the difference.

3. Use this difference to adjust the controlled variable so as to reduce the difference. (error)

-

Electronic thermostat

+

+ Desired temp. ro c -

Controlled output C Forward path element

Controller

C

Feed back path element Fig. 1-5 General block diagram of feedback system EXAMPLE – 1 – THERMAL SYSTEM To illustrate the concept of closed loop control system, consider the thermal system shown in fig-6 Here human being acts as a controller. He wants to maintain the temperature of the hot water at a given value ro C. the thermometer installed in the hot water outlet measures the actual temperature C0 C. This temperature is the output of the system. If the operator watches the thermometer and finds that the temperature is higher than the desired value, then he reduce the amount of steam supply in order to lower the temperature. It is quite possible that that if the temperature becomes lower than the desired value it becomes necessary to increase the amount of steam supply. This control action is based on closed loop operation which involves human being, hand muscle, eyes, thermometer such a system may be called manual feed back system. Human operator Thermometer Steam Steam

Hot water

Desired hot water. temp + ro c

Brain of operator (r-c)

+

Muscles and Valve Thermometer

Cold water Drain Fig 1-6 a) Manual feedback thermal system

b) Block diagram

Actual Water temp Co C C

EXAMPLE –2 HOME HEATING SYSTEM The thermostatic temperature control in hour homes and public buildings is a familiar example. An electronic thermostat or temperature sensor is placed in a central location usually on inside wall about 5 feet from the floor. A person selects and adjusts the desired room temperature ( r ) say 250 C and adjusts the temperature setting on the thermostat. A bimetallic coil in the thermostat is affected by the actual room temperature ( c ). If the room temperature is lower than the desired temperature the coil strip alters the shape and causes a mercury switch to operate a relay, which in turn activates the furnace fire when the temperature in the furnace air duct system reaches reference level ' r ' a blower fan is activated by another relay to force the warm air throughout the building. When the room temperature ' C ' reaches the desired temperature ' r ' the shape of the coil strip in the thermostat alters so that Mercury switch opens. This deactivates the relay and in turn turns off furnace fire, which in turn the blower. Electronic thermostat

o

Desired temp. r c

+

Relay switch

Outdoor temp change (disturbance) Furnace

Blower

House

Actual Temp. Co C

Fig 1-7 Block diagram of Home Heating system. A change in out door temperature is a disturbance to the home heating system. If the out side temperature falls, the room temperature will likewise tend to decrease. CLOSED- LOOP VERSUS OPEN LOOP CONTROL SYSTEMS An advantage of the closed loop control system is the fact that the use of feedback makes the system response relatively insensitive to external disturbances and internal variations in systems parameters. It is thus possible to use relatively inaccurate and inexpensive components to obtain the accurate control of the given plant, whereas doing so is impossible in the open-loop case. From the point of view of stability, the open loop control system is easier to build because system stability is not a major problem. On the other hand, stability is a major problem in the closed loop control system, which may tend to overcorrect errors that can cause oscillations of constant or changing amplitude. It should be emphasized that for systems in which the inputs are known ahead of time and in which there are no disturbances it is advisable to use open-loop control. closed loop control systems have advantages only when unpredictable disturbances it is advisable to use open-loop control. Closed loop control systems have advantages only when

unpredictable disturbances and / or unpredictable variations in system components used in a closed –loop control system is more than that for a corresponding open – loop control system. Thus the closed loop control system is generally higher in cost.

Session 4 -28.03.2005 REQUIREMENTS FOR THE CONTROL SYSTEM 4

Input command

Steady State error

Transient response

Floor

0

Time Fig. 1-8 Elevator input and output

Speed of response, accuracy and stability are the requirements demanded of every control system. We shall understand the significance of the above taking the example of an elevator. As noted earlier, a control system provides an output or response for a given input or stimulus. The input represents a desired response; the output is the actual response. Take the case of elevator. For example when the fourth- floor button of an elevator is pushed on the ground floor, the elevator rises to the fourth- floor with a speed and floor leveling accuracy designed for passenger comfort. Fig 1 below shows input and output for the elevator system. the push of the input and output for the elevator system. The push of the input and output for the elevator system. The push of the fourth floor button forms the input and is represented by a step command. Note that in the interest of the passenger comfort, we would not want the elector to mimic the suddenness of the input. The input represents what we would like the output to be after the elevator has stopped; the elevator itself follows the displacement described by the curve marked elevator response. Two factors make the output different from the input. First compare the instantaneous change of the input against the gradual change of the output in fig. 1 – physical entities ( position or velocity ) cannot change their states instantaneously. Thus, the elevator undergoes a gradual change as it rises from ground floor to the fourth floor. We call this part of the response 'transient response'. Transient response is important. In the case of an elevator, a slow transient response makes passengers impatient, where as an excessively or design components are adjusted to yield a desired transient response. After the transient response elevator approaches its steady state response, which is its approximation to the commanded or desired response. The accuracy of the elevator's leveling with the floor is a second factor that could make the output different from the input. An elevator must be level enough with the fourth floor for the passenger to exit.

MODELING IN FREQUENCY DOMAIN The two important topics in the study of control systems are 1. Control system analysis 2. Control system design By control system analysis we mean the investigation under specified conditions of the performance of the system. By control system design we mean to find out one which accomplishes given task. If the performance is unsatisfactory it can be improved with the help of design. Whether it is control A control system is a physical system as it is a collection of physical objects connected through to serve an objective. The system can be electrical, mechanical or electromechanical. Examples of physical system can be cited from laboratory, industrial plant- an electronic amplifier composed of many components, the governing mechanism of a steam turbine or communication satellite orbiting the earth are all examples physical systems. No physical system can be represented in its full physical intricacies and therefore idealizing assumption are always made for the purpose of analysis and synthesis of systems. An idealized physical system is called physical model. A physical system can be modeled in a number of ways depending up on specific problem to be dealt with and desired accuracy. For example an electronic amplifier may be modelled as an interconnection of linear lumped elements or some of these may be pictured as nonlinear elements in case the stress is on the study of distortion. Once a physical model of a physical system is obtained, the next step is to obtain a mathematical model which is the mathematical representation of the physical model through the use of appropriate physical laws ( Ohm’s law, kirchoff’s law, Newton’s Law, Hooke’s Law etc). Depending upon the choice of variables and the coordinate system, a given physical model may lead to different mathematical models. An electrical network, for example, may be modelled as a set of nodal equations using kirchoff’s current law or a set of mesh equations using using kirchoff’s voltage law. A control system may be modelled as a scalar differential equation.The particular mathematical model which gives a greater insight into the dynamic behaviour of physical system is selected. When the mathematical model of a physical system is solved for given input, the result represents the dynamic response of the system. Linear Systems

A system is called linear if the principle of superposition applies the principle of superposition states that the response ( output) produced by the simultaneous application of two different inputs is the sum of two individual responses ( outputs). Hence for the $ linear system the response to several inputs can be calculated by treating one input at a time and adding the results Linear time – invariant system and linear time- varying systems A differential equation is linear if the co-efficient are constants or functions only of independent variable. If the coefficients of the describing differential equations are constants, the model is linear time- invariant. 2

6

Example:

dx 2

dt

+ 3

dx dt

+ X =F

On the other hand if the coefficients of the coefficients of the describing differential equations are functions of time ‘t’ ( the independent variable ) then the mathematical model is linear time – variant. An example is a missile. The mass of a missile changes due to fuel consumption. 2 dx 2 dx t + X =F t + Transfer function2 dt dt The differential equation describing a linear time invariant system can be reshaped into different forms for the convenience of analysis. For single- input- single output linear system, the transfer function representation forms useful. On the other hand, when a system has multiple inputs and outputs, the vector- matrix notation may be more convenient.

The transfer function of a linear time- invariant system is defined as the ratio of the laplace transform of the output (response) to the laplace transform of the input (driving function) under the assumption that all initial conditions are zero. Consider the linear time invariant system defined by the following differential equation. a0

dnc dtn

+

a1

dn-1c dtn-1

+ ………+

an-1

dc + an C = bo dt

dm r + b1 dtm + bm-1

dr dt

dm-1 r dtm-1 + bm r

for n > m Taking Laplace transform on both sides and assuming zero initial conditions,

C(s) R(s)

=

bosm + b1sm-1 + …………+ bm aosn + a1sn-1 + …………+ bn

Comments on transfer function 1. The transfer function is an expression relating the output and input of a linear time invariant system in terms of the system parameters and is a property of the system itself independent of the input. 2. It does not provide any information concerning the physical structure of the system ( the transfer functions of many different physical systems can be identical). 3. The highest power of

in the denominator of the transfer function is equal to the

the order of the system. 4. The transfer function between an input and output of a system is defined as the laplace transform of impulse.

Session 5 – 30.03.2005 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 Netforce F acting on a body is related to its mass M and acceleration ‘a’ by the equation Σ F = Ma ‘Ma’ is called reactive force and it acts in a direction opposite to that of acceleration. The summation of the forces must of course be algebraic and thus considerable care must be taken in writing the equation so that proper signs prefix the forces. The three basic elements used in linear mechanical translational systems are ( i ) Masses (ii) springs iii) dashpot or viscous friction units. The graphical and symbolic notations for all three are shown in fig 1-9

M Fig 1-9 a) Mass

Fig 1-9 b) Spring

Fig 1-9 c) Dashpot

The spring provides a restoring a force when a force F is applied to deform a coiled spring a reaction force is produced, which to bring it back to its freelength. As long as deformation is small, the spring behaves as a linear element. The reaction force is equal to the product of the stiffness k and the amount of deformation. Whenever there is motion or tendency of motion between two elements, frictional forces exist. The frictional forces encountered in physical systems are usually of nonlinear nature. The characteristics of the frictional forces between two contacting surfaces often depend on the composition of the surfaces. The pressure between surfaces, their relative velocity and others. The friction encountered in physical systems may be of many types ( coulomb friction, static friction, viscous friction ) but in control problems viscous friction, predominates. Viscous friction represents a retarding force i.e. it acts in a direction opposite to the velocity and it is linear relationship between applied force and velocity. The mathematical expression of viscous friction F=BV where B is viscous frictional co-efficient. It should be realized that friction is not always undesirable in physical systems. Sometimes it may be necessary to introduce friction intentionally to improve dynamic response of the system. Friction may be introduced intentionally in a system by use of dashpot as shown in fig 1-10. In automobiles shock absorber is nothing but dashpot.

a Applied force F

b Piston

The basic operation of a dashpot, in which the housing is filled with oil. If a force f is applied to the shaft, the piston presses against oil increasing the pressure on side ‘b’ and decreasing pressure side ‘a’ As a result the oil flows from side ‘b’ to side ‘a’ through the wall clearance. The friction coefficient B depends on the dimensions and the type of oil used. Outline of the procedure For writing differential equations

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.

•

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

•

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.

•

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 F2 Load L2

L1

Fulcrum effort F 1 It is also called ‘mechanical transformer’ Equating the moments of the force F1 L1 = F2 L 2 F2 =

F1 L1 L2

Rotational mechanical system The rotational motion of a body may be defined as motion about a fixed axis. The variables generally used to describe the motion of rotation are torque, angular displacement θ, angular velocity (ω) and angular acceleration(α) The three basic rotational mechanical components are 1) Moment of inertia J 2 ) Torsional spring 3) Viscous friction. Moment of inertia J is considered as an indication of the property of an element, which stores the kinetic energy of rotational motion. The moment of inertia of a given element depends on geometric composition about the axis of rotation and its density. When a body is rotating a reactive torque is produced which is equal to the product of its moment of inertia (J) and angular acceleration and is given by T= Jα = J d2 θ d t2 A well known example of a torsional spring is a shaft which gets twisted when a torque is applied to it. Ts = Kθ, θ is angle of twist and K is torsional stiffness.

There is viscous friction whenever a body rotates in viscous contact with another body. This torque acts in opposite direction so that angular velocity is ω given by

T = f ω = f d2 θ

Where ω = relative angular velocity between two bodies.

2

f = co efficient of viscous friction.

dt

Newton’s II law of motion states Σ T = J d2 θ. d t2 Gear wheel In almost every control system which involves rotational motion gears are necessary. It is often necessary to match the motor to the load it is driving. A motor which usually runs at high speed and low torque output may be required to drive a load at low speed and high torque.

Driving wheel N1

N2

Driven wheel

Analogous Systems Consider the mechanical system shown in fig A and the electrical system shown in fig B The differential equation for mechanical system is

d2x M +

dt2

dx + B

dt

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

The differential equation for electrical system is d2q

d2q

q

dt2

dt2

c

L +

+R

+

= e ---------- 2

Comparing equations (1) and (2) we see that for the two systems the differential equations are of identical form such systems are called “ analogous systems and the terms which occupy the corresponding positions in differential equations are analogous quantities” The analogy is here is called force voltage analogy

Table for conversion for force voltage analogy Mechanical System

Electrical System

Force (torque)

Voltage

Mass (Moment of inertia)

Inductance

Viscous friction coefficient

Resistance

Spring constant

Capacitance

Displacement

Charge

Velocity

Current.

Force – Current Analogy Another useful analogy between electrical systems and mechanical systems is based on force – current analogy. Consider electrical and mechanical systems shown in fig.

For mechanical system the differential equation is given by

d2x

M +

2

dt

dx +B

dt

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

For electrical system C

d2x 2

dt

+

1

R

+

dΦ

+

2

dt

Φ

= I(t)

L

Comparing equations (1) and (2) we find that the two systems are analogous systems. The analogy here is called force – current analogy. The analogous quantities are listed. Table of conversion for force – current analogy

Mechanical System

Electrical System

Force( torque)

Current

Mass( Moment of inertia)

Capacitance

Viscous friction coefficient

Conductance

Spring constant

Inductance

Displacement ( angular)

Flux

Velocity (angular)

Voltage

Although it is equally easy to write the equations for either form of system and thus equations for either form of system and thus there is no need to consider analogs, to simplify the analysis, there are significant advantages to the use of electrical analogos mechanical systems. For example it is not particularly convenient to setup a mechanical spring mass dashpot system and test its response in the laboratory because such components are not available in a wide variety of sizes, and are inconvenient to work with in any event. Since electrical components, as are current and voltage signals in a variety of forms for test inputs and since currents and voltages are accurately measured

with ease, it is often convenient to study the response equivalent to the mechanical system of interest, adjusting component values as required to provide the desired results.

CONTROL SYSTEMS Resource person: S. RAGHAVENDRA Selection Grade Lecturer E&EE Dept. SJCE, Mysore. REVIEW QUESTIONS ( Sessions 1 to 5 from 21-3-2005 to 29-3-2005) Chapter 1 Modeling of Physical Systems 1. Name three applications of control systems. 2. Name three reasons for using feedback control systems and at least one reason for not using them. 3. Give three examples of open- loop systems. 4. Functionally, how do closed – loop systems differ from open loop systems. 5. State one condition under which the error signal of a feedback control system would not be the difference between the input and output. 6. Name two advantages of having a computer in the loop. 7. Name the three major design criteria for control systems. 8. Name the two parts of a system’s response. 9. Physically, what happens to a system that is unstable? 10. Instability is attributable to what part of the total response. 11. What mathematical model permits easy interconnection of physical systems? 12. To what classification of systems can the transfer function be best applied? 13. What transformation turns the solution of differential equations into algebraic manipulations ? 14. Define the transfer function. 15. What assumption is made concerning initial conditions when dealing with transfer functions? 16. What do we call the mechanical equations written in order to evaluate the transfer function ? 17. Why do transfer functions for mechanical networks look identical to transfer functions for electrical networks?

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

Problems ( Sessions 1 to 5 from 21-3-2005 to 29-3-2005) 1.Write the differential equation relating to motion X of the mass M to the force input u(t)

K1

X

K2

(output)

M

U(t) (input)

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

X

K

X1

(output)

B2 M

F(t) (input)

B1

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

X1 K1

X2

f12 M1

M2

f(t)

f1

f2

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

Xi

K

M

X

M force f(t)

Xo

B

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

K Xi

Xo

Xi

Xo

Yk F 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.

b

f(t)

K

a

M B

X

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

X2 K1

Force f

a

b M2

M1 B1

K2

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

B1 J1 T(t) θ 9. Write the force equation for the mechanical system shown in figure.

T(t)

J1 θ1

K

J2 θ2

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

θ1 K 1 J1

Torque T

θ2 K 2 J2 B2

B1

θ3 K 3 J3 B3

11. Torque T(t) is applied to a small cylinder with moment of inertia J1 which rotates with in a larger cylinder with moment of inertia J2. The two cylinders are coupled by viscous friction B1. The outer cylinder has viscous friction B2 between it and the reference frame and is restrained by a torsion spring k. write the describing differential equations.

J2

K

J1 Torque T1, θ1

B2 B1

12. The polarized relay shown exerts a force f(t) = Ki. i(t) upon the pivoted bar. Assume the relay coil has constant inductance L. The left end of the pivot bar is connected to the reference frame through a viscous damper B1 to retard rapid motion of the bar. Assume the bar has negligible mass and moment of inertia and also that all displacements are small. Write the describing differential equations. Note that the relay coil is not free to move.

13. Figure shows a control scheme for controlling the azimuth angle of an armature controlled dc. Motion with dc generator used as an amplifier. Determine transfer function θL (s)

. The parameters of the plant are given below.

u (s) Motor torque constant Motor back emf constant Generator gain constant Motor to load gear ratio

= KT in N.M /amp = KB in V/ rad / Sec = KG in v/ amp = N2

N1 Resistance of the circuit = R in ohms. Inductance of the circuit = L in Henry Moment of inertia of motor = J Viscous friction coefficient = B Field resistance = Rf Field inductance = Lf

14. The schematic diagram of a dc motor control system is shown in figure where Ks is error detector gain in volt/rad, k is the amplifier gain, Kb back emf constant, Kt is torque constant, n is the gear train ratio = θ2 θ1

=

Tm

Bm = motion friction constant

T2

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

15. Obtain a transfer function C(s) /R(s) for the positional servomechanism shown in figure. Assume that the input to the system is the reference shaft position (R) and the system output is the output shaft position ( C ). Assume the following constants. Gain of the potentiometer (error detector ) K1 in V/rad

Amplifier gain ‘ Kp ’ in V / V Motor torque constant ‘ KT ’ in V/ rad Gear ratio N1 N2 Moment of inertia of load ‘J’ Viscous friction coefficient ‘f’

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

I

E0 C2

input

R

Output

K.Puttaswamy

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 1.1 shows an element of the block diagram. The arrow head pointing towards the block indicates the input and the arrow head away from the block represents the output. Such arrows are entered as signals. X(s)

G(s

Y(s)

Fig 1.1 The advantages of the block diagram representation of a system lie in the fact that it is easy to form the over all block diagram for the entire system by merely connecting the blocks of the components according to the signal flow and thus it is possible to evaluate the contribution of each component to the overall performance of the system. A block diagram contains information concerning dynamic behavior but does not contain any information concerning the physical construction of the system. Thus many dissimilar and unrelated system can be represented by the same block diagram.

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 fig1.2 +

R(s)

-

C(s) C(s)

Fig1.2 Note that a circle with a cross is the symbol which indicates a summing operation. The plus or minus sign at each arrow head indicates whether the signal is to be added or subtracted. Note that the quantities to be added or subtracted should have the same dimensions and the same units.

Block diagram of a closed loop system . Fig1.3 shows an example of a block diagram of a closed system

Summing point R(s)

Branch point C(s)

+

G(s)

-

Fig. 1.3 Block diagram of a closed loop system. The output C(s) is fed back to the summing point, where it is compared with reference input R(s). The closed loop nature is indicated in fig1.3. Any linear system may be represented by a block diagram consisting of blocks, summing points and branch points. A branch is the point from which the output signal from a block diagram goes concurrently to other blocks or summing points. When the output is fed back to the summing point for comparison with the input, it is necessary to convert the form of output signal to that of he input signal. This conversion is followed by the feed back element whose transfer function is H(s) as shown in fig 1.4. Another important role of the feed back element is to modify the output before it is compared with the input. B(s) R(s)

C(s) G(s

+

-

C(s)

B(s) H(s Fig 1.4

The ratio of the feed back signal B(s) to the actuating error signal E(s) is called the open loop transfer function. open loop transfer function = B(s)/E(s) = G(s)H(s) The ratio of the output C(s) to the actuating error signal E(s) is called the feed forward transfer function .

Feed forward transfer function = C(s)/E(s) = G(s) If the feed back transfer function is unity, then the open loop and feed forward transfer function are the same. For the system shown in Fig1.4, the output C(s) and input R(s) are related as follows. C(s) = G(s) E(s) E(s) = R(s) - B(s) = R(s) - H(s)C(s)

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

Eliminating E(s) from these equations C(s) = G(s)[R(s) - H(s)C(s)] C(s) + G(s)[H(s)C(s)] = G(s)R(s) C(s)[1 + G(s)H(s)] = G(s)R(s) C(s)

G(s) =

R(s) 1 + G(s)H(s) C(s)/R(s) is called the closed loop transfer function. The output of the closed loop system clearly depends on both the closed loop transfer function and the nature of the input. If the feed back signal is positive, then C(s)

G(s) =

R(s)

1 - G(s)H(s)

Closed loop system subjected to a disturbance Fig1.5 shows a closed loop system subjected to a disturbance. When two inputs are present in a linear system, each input can be treated independently of the other and the outputs corresponding to each input alone can be added to give the complete output. The way in which each input is introduced into the system is shown at the summing point by either a plus or minus sign.

Disturbance N(s) R(s)

+ +

-

G1(s)

+

G2(s)

C(s)

H(s Fig1.5 Fig1.5 closed loop system subjected to a disturbance. Consider the system shown in fig 1.5. We assume that the system is at rest initially with zero error. Calculate the response CN(s) to the disturbance only. Response is CN(s)

G2(s)

=

R(s)

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

On the other hand, in considering the response to the reference input R(s), we may assume that the disturbance is zero. Then the response CR(s) to the reference input R(s)is CR(s) = R(s)

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

The response C(s) due to the simultaneous application of the reference input R(s) and the disturbance N(s) is given by C(s) = CR(s) + CN(s) G2(s)

C(s) =

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

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

Procedure for drawing block diagram : To draw the block diagram for a system, first write the equation which describe 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 fig1.6(a). The equations for the circuit shown are R

ei

i

eo C

Fig. 1.6a

ei = iR + 1/c∫ idt

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

And eo = 1/c∫ idt

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

Equation (1) becomes ei = iR + eo ei - eo

=i

R

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

Laplace transforms of equations (2) & (3) are Eo(s) = 1/CsI(s)

-----------(4)

Ei(s) - Eo(s) = I(s)

--------(5)

R Equation(5) represents a summing operation and the corresponding diagram is shown in fig1.6(b). Equation (4) represents the block as shown in fig1.6(c). Assembling these two elements, the overall block diagram for the system shown in fig1.6(d) is obtained. Fig1.6(b) I(s) Ei(s)

+ _ Eo(s)

1/R

I(s)

Fig1.6(c) Eo(s) +

Fig1.6(b)

Eo(S)

1/C

1/R

I(s)

1/C

_

Fig1.6(d) REFERENCE: 1. MODERN CONTROL ENGINEERING BY OGATA 3. AUTOMATIC CONTROL SYSTEMS BY B.C. KHO

SIGNAL FLOW GRAPH

Eo(s)

The block diagram is useful for graphically representing control systems. For a complicated system, however the block diagram reduction process becomes time consuming . An alternate approach for finding the relationships among the system variables of complicated control system is the signal flow graph approach due to Mason. Signal flow graph is a diagram which represents a set of simultaneous linear algebraic equation .When applying the signal flow graph method to control system we must first transform linear differential equation into algebraic equation in ‘s’ Signal flow graph consist of a network, in which nodes are connected by directed branches. Each node represents a system variable and each branch connected between two nodes acts as a signal multiplier. Note that signal flows only in one direction .The direction of the signal flow is indicated by an arrow placed on the branch and the multiplication factor is indicated along the branch. The signal flow graph depicts the flow of signals from one point of a system to another and gives the relationship among the signals. Signal flow graph contains essentially the same information as a block diagram. The advantage of using signal flow graph is to represent a control system is that a gain formula or Mason’s gain formula is available which gives the relationships among the system variables without requiring a reduction of the graph

TERMINOLOGY Node: A node is a point representing a variable or signals. Transmittance : is a gain between the two nodes. Branch : is a directed line segment joining two nodes. The gain of the branch is a transmittance. Input node or source : is a node which has only outgoing branches. This corresponds to an independent variable . Output node or Sink :

An output node or sink is a node which has only incoming

branches. This corresponds to a dependent variable. Mixed node : is a node which has both incoming and outgoing branches. Path : is a traversal of connected branches in the direction of the branch arrows. Loop : is a closed path. Loop gain : is the product of the branch transmittance of a loop.

Non touching loops : Loops are non-touching if they do not posses any common nodes. Forward path : A forward path is a path from an input node to an output node which does not cross any nodes more then once. Forward path gain : is the product of the branch transmittances of a forward path.

Mixed mode Input node x1

a

x2

x4 input node

b

x3

1

x3 Output node

C Properties of signal flow graph 1. A branch indicates the functional dependence of one signal upon another. A signal passes through only in the direction specified by the arrow of the branch . 2. A node adds the signals of all incoming branches and transmits this sum to all outgoing branches. 3. A mixed node which has both incoming and outgoing branches may be treated as an output node by adding an outgoing branch of unity transmittance. 4. For a given system, signal flow graph is not unique. Many different signal flow graphs can be drawn for a given sytem by writing the system equations differently. Signal flow graph Algebra The independent and dependent variables of the equations become the input nodes and output nodes respectively. The branch transmittance can be obtained from the coefficients of the equations. To determine the input output relationship we may use Mason’s formula or we may reduce the signal flow graph to a graph containing only input and output nodes we use the following rules 1. The value of a node with one incoming branch as shown in fig. 1 x2 = ax1 2. The total transmittance of cascaded branches is equal to the product of all the branch transmittances. Cascaded branches can be combined into single branch the multiplying the transmittance as shown in fig. 2

3. Parallel branches may be combined by adding transmittances as shown in fig. 3 4. A mixed node may be eliminated as shown in fig. 4 5. A loop may be eliminated as shown in fig. 5 a

a

b

ab =

x1 x2 x2 = ax1 fig.(1)

x1

x2

x3

x1 x3 = abx1

Fig. (2)

x3

a x1

x2

a+b =

x1

b

Fig. (3)

x1

x1

a c b

(ac)x1

x4

x3

x2

ac

=

bc

x4

x2

(bc)x2 x1

x2 = (a + b) s1

x2

x4

=

x4

=

Fig. (4) a

x2

b

x3

x1

ab

x3

=

ab =

c

bc Fig. (5)

x3 = bx 2 x2 = ax1 + cx3

x3 = abx1 + bcx3 x3 = abx1 + b2cx2 = abx1 + b2c (ax1 + cx3) =

abx1 + b2cax1 + b2c2x3

x3 (1 – b2c2 ) = abx1 (1 + bc) x3 =

abx1 (1 + bc) 1 – b2c2

x1

1 - bc

x3

ab (1 + bc ) = ( 1 + bc) (1 - bc)

X1

ab x3 =

x1 1 - bc

Signal flow graph representaion of linear systems Here the graph can be drawn from the system equations or can be drawn by inspection of the physical system. Consider system defined by the following set of equations x1

=

a11x1 + a12x2 + a13x3 + b1u1

-

(1)

x2

=

a21x1 + a22x2 + a23x3 + b2u2

-

(2)

x3

=

a31x1 + a32x2 + a33x3

-

(3)

a11

a22 x2 x3

b1 u1

x1

x2

x3

x1

a21

a12

a23

b2

a13

u2

a31 a33 x1

x2

a32

x3

u1, u2 input variables x1x2 & x3 are output variables Final signal flow graph is drawn by combining these three equations of simultanious equations.

a31

b1

u1

x3

a11

a22 a21

a33

a32 x2

x1

a12

b2

1

x3

a23

u2 a13

Mason’s Gain formula for signal flow graph We want to determine the relationship between n an input and output variable of the signal flow graph. The transmittance between an input node and an output node is the overall gain or transmittance between the two nodes.

∑k pk

Mason’s gain formula p =

k

= path gain or transmittance of kth forward path

Pk

= determinant of the graph = 1 – (Sum of all different loop gains) + (sum of gain products of all possible combinations of two nontouching loops) – (Sum of the gain products of all possible combinations of three nontouching loops) =1–

La + a

k

LbL c -

b,c

LdLeLf + …………

d,e,f

= Cofactor of kth forward path, determinant of the graph with the loops touching the

kth forward path removed Or value of OR

for all loops except for those which touch the forward path ‘K’

Value of above

by eliminating all loop gains & associated products which are th

touching to the k forward path. BLOCK DIAGRAM

R(s)

R(s)

G(s)

+

C(s)

G(s)

R(s)

G(s)

C(s)

C(s)

R(s)

E(s)

1

C(s) H(s)

REFERENCE: 1. 2. 3.

-H(s)

MODERN CONTROL ENGINEERING BY OGATA AUTOMATIC CONTROL SYSTEMS B.C. KHO CONTROL SYSTEM BY V.A. BAKSHI & U.A. BAKSHI

G(s)

PROBLEM 1 : H2

+

+ -

+

G1

+

-

G2

G3

R(s)

C(s)

H1

Signal flow graph for the above Block Diagram is

(2)

1GG32 1 R(s) C(s) (1) H1 -1

(3) In this system there is only one forward path between the input R(s) and output C(s). The forward path gain is

P 1 = G1 G2 G3 In this figure there are three individual loops. The gains of these loops are

L1 = G1 G2 H1 L2 = - G2 G3 H2 L3 = - G1 G2 G3 Since all the three loops have a common branch, there are no nontouching loops. Hence determinant is given by = 1 - [L1+L2+L3] = 1 - [G1G2H1 – G2G3H2 – G1G2G3] The cofactor of the determinant along the forward path connecting the input node & output node by removing the loops that touches this path. Since the path p1 touches all the three loops.

We get Overall given

1

=1

c(s) R(s)

c(s) R(s)

p1

=

1

G1G2G3 = 1 - G1G2H1 + G2G3H2 + G1G2G3

2.

Use the signal flow graph method determine the gain c/R for the block diagram shown.

1 R

2 3

4

5 C

G1

+ G21 + 3 + -

H2

H1

Signal flow graph for the fig. Shown ∑k Pk

k

G4 R

1

G1

G2

G3

loop (3) 1

C

loop (1) H2 loop (2) - H1

Mason’s formula

No. of forward path

P=

p1 = G1G2G3

P2 = G1G2G4

There are three feed back loops L1 = G1G2H2 L2 = - G1G2G3H1 L3 = - G1G2G4H1 = 1 – (L1 + L2 + L3) = 1 – (G2G3H2 – G1G2G3H1 – G1G2G4H1) L1 touches L3, L2 touches L3 and L1 touches L2 No combinations of non touching loops can exists for the signal flow graph given, only first two terms exsists. 1

is obtained from 1=1 2= 1

p

=

p1

1 + p2

2

G1G2G3 + G1G2G4

=

1 - [G1G2G3H2 – G1G2G3H1 – G1G2G4H1] c/R = p

=

G1G2G3 + G1G2G4 1 - G1G2H2 + G1G2G3H1 + G1G2G4H1

3.

The following equation describes a control system. Construct the signal flow graph for it and obtain transfer function . and

Y2 U1

for u2 = 0

Y2 U2

for u1 = 0

Where Y2 = out put node and u1 & u2 are the inputs Y1 = a11 Y1 + a12 Y2 + b1 U1 ------ (1) Y2 = a21 x a22 Y2 + b2 U2 --------- (2) Solution The different node variables are Y1 & Y2 ,U1 & U2 are the inputs. Consider the equation signal flow graph is U1

b1

a11 Y1

y2 a12 a22 Similarly considering the equation 2 Signal flow graph is

Y1

a21

Y2 b2 u2

Combining these two signal the graphs, the total signal graph is

u1

aa2211

b1a21 1 YY2 1

output

Y2 ba212

U1

U2 = o

U2 = 0 u2 Q11 b1 Transfer function

Y2 u1

Assuming

u2 = 0

U1 Y1 u2 = 0

Q22 Q22

1 Y2

Signal flow graph is

a11

b1

a22

a21

U1

1

Y1

Y2 a12

Using Mason’s gain formula

T.F =

∑ k Pk k

k=1 k = No of forward paths = 1 P1 = b1 a21 L1 = a12 a21 L2 = a11 L3 = a22 L2 & L3 are non touching loops ∴

= 1 – [∑ all individual feed back loop gains] +

[∑ gain products of all possible combinations of two non

touching loops]

= 1 – [ L1 + L2 + L3 ] + L2 L3 1

= cofactor of the determinant eliminating all loop gains which are touching the

first forward path. 1

=1 Y2 U1

P1

=

b1a21

1

1– a12a21 –a11 – a22 + a11a22

Assuming u1 = o Then the graph is

a11

a22 a21

Y1

Y2 a12

b2 u2

T.F =

Y2 U2

U1 =0

K = No of forward path = 1 P1 = b2 Individual loops are L1 = a21 a12 L2 = a11 L3 = a22 nontouching loops are L2 & L3 = 1 – [ L1 + L2 + L3 ] + L2 L3 = 1 – a21 a12 – a11 – a22 + a11 a22 1=

Y2 U2 4.

1 – a11 p1 =

1

=

b2 (1-a11) 1-a21a12 – a11 – a22 + a11 a22

Construct the signal flow graph for the following set of system equations

Y2 = G1Y1 + G3Y3

--- --- (1)

Y3 = G4Y1 + G2Y2 + G5Y3

--- --- (2)

Y4 = G6Y2 + G7Y3

--- --- (3)

Find the transfer function

Y4 Y1

Solution Consider the equation 1 Signal flow graph is Y1

G1

Y2

Y3 G3

G4 G5

Consider the equation 2, signal flow graph is

Y1

Y2

Y3

G2 G6

Consider the equation 3, signal flow graph is Y2

Y4

Y3

G7

G4

Combining all the three, complete signal flow graph is

G5 G1

Y2

G2

Y1

No of forward path = K = 4 Transfer function = ∑ 4 K =1

G7 Y3

G3 pk Ak G6

Y4

=

p1

1 +

p2

2

+ p3

3

+ p4

4

P1 = G1G2G7, P2 = G4G7, P3 = G1G6 & P4 = G4 G3 G6 Individual loops are L1 = G2 G3 L2 = G5 (Self loop ) = 1 – (L1 + L2 ) = 1 – G2G3 –G5 There are no nontouching loop combinations 1 2 3 4

=1 = 1 = 1– G5

Booth loops are touching Booth loops are touching G5 is non touching the path p3

=1 Y4 Y1

5.

=

G1G2G7 + G4G7 + G1G6 (1 – G5 ) + G4G3G6 1 – G2G3 – G5

From the given block diagram, draw the signal flow the graph and find C(s) R(s)

1

2

R(s)

G4

6 5+ 7 +GG3 21 34

H2 H1

Signal Flow graph is

C(s)

R(s)

1

1

2

1

3

G1

4

G2

G4

-H2

-H1

-1 No of forward paths = K = 2 Forward path gains are P1 = G1 G2 G3 P2 = G4

Feed back loops are L1 = - G1G2H1

L4 = - G4

5

G3

6

7

1

C(s)

L2 = - G2G3H2

L5 = G2G4H1H2

L3 = - G1G2G3 There are no non touching loops = 1 – [ L1 + L2 + L3 + L 4 + L5 ] = 1 + G1G2H1 + G2G3H2 + G1G2G3 + G4 – G2G4 H1H2 considering the forward path G1G2G3 all loops touching ∴

1=1

Considering the forward path G4 , all the loops are touching the path G4 , C(S) =

p1

1+

p2

2

=1

2

R (s) =

1+G1G2H1 + G2G3H2 + G1G2G3 + G4 – G2G4

H1 H2 6.

G1G2G3 + G4

Given the Electrical Network 1. Find out the laplace transform of the given network and re draw the network in

‘S’ domain 2. Work down the equations for different branch current and node voltages.

3. Simulate each equation by drawing the corresponding signal flow graph. 4. Combine all the signal flow graphs to get the total signal flow graph. 5. Use Mason’s gain formula to derive the transfer function of the given network. 6. Formal the transfer function for the given network. R1

Vi

C

R2

L

Vo

Laplace transform of the given network is as shown R1

I1(s)

Vi(s)

R2

V1(s)

1

I2(s)

SC

Equations are I1(s)

= Vi (s) – V1 (s)

V1(s)

=

LS

Vo (s)

……

(1)

…….

(2)

R1

[ I1(s) – I2 (s)] CS

V1(s) - V0 (s) I2 (s)

=

………….

(3)

R2

V0(s) = I2(s) LS Signal flow graph for the equation (1)

Vi

1 R1

I1

V1

-

1 R1

Signal flow graph for equation 2

I1

1 Sc

V1

I2

-1 Sc Signal flow graph for equation 3 1 R2

V1

I2

Vo

-1 R2 Signal flow graph for equation 4

SL

I2

Vo

Total signal flow graph for the network is

Vi

I1

V1

11 Sc R R21

11 -R Sc21

Using Mason’s gain formula =

Vo Vi

∑ PK

=

No of forward paths = 1 V0 Vi

=

P1

1

k

I2

SL

Vo

L Forward path gain Ti = Feedback loops are

R1R2C

1 L1 = -

1 L2 = -

SR1C

SL L3 = -

SR2C

R2

Non touching Loops are L1 & L3 = 1 – [ L1 + L2 + L3 ] + L1L3 1 =1+

1 +

SL +

SR1C SR2C

L +

R2

R1R2C

All the loops are touching the forward path

V0

L R1R2C

L R1R2C

=

Vi SL

1 1+

1 +

SR1C

SL = SL +

SR2C

L

SR1R2C + R2 + R1 + S2LR1C +

+ R2

R1R2C

6. Find the transfer function for the given network using Mason’s gain formula C

Vi

Laplace transformed network is 1 CS

1 CS

R1

1 1 +SR1C

Vo

R2

R1 1 + SR1C Z

Z=

R1 + 1 /Sc

R1 V V I(s) RVi(s) V (s) io2(s) o (s)

R1 x 1/Sc

I (S ) = Vi(s) – Vo(s) Z

Z=

R1Sc + 1 = Vi(s) – Vo(s)

1 + R1Sc R1

Vo (s) = I (s) R2

From the equation (1) signal flow graph is

Vi

1+ R1SC R1

1

Vo

-

1 + R1Sc R1

From the equation (2) signal flow graph is

R1

I(s)

R2

Vo(s)

Then combined signal flow graph is

Vi

1 + R1Sc R1

1

R2

Vo

-

1+R1Sc R1

Using Mason’s gain formula, the number of forward path = 1 ∴ forward pathg gain P1 = R2 Loop gown L1 = - R2

1 + R1 Sc R1

1 + R1 Sc R1

Determinant of the graph

= 1 – L1 = 1 + R2 ( 1 + R1Sc ) R1 R1 + R2 ( 1 + R1Sc ) = R1

As L1 is touching the forward path 1=1 then Vo (s) P1 1 R2 ( 1 + R1SC ) 1 T(S) = = = = Vi(s) R1 [ R1 + R2 (1 + R1SC)] R1

8.

R2 (1 + R1 Sc) R1 + R2 (1 + R1 Sc )

BLOCK DIAGRAM FOR THE SIGNAL FLOW GRAPH

Procedure : 1. 2.

To obtain the block diagram from the given signal flow graph, it is necessary to write set of system equations representing the given signal flow graph. Assume suitable node variables, write equations for every node.

3.

(1)

while writing equations remember that the value of the variable represented by the node is the algebraic sum of all the signals entering at that node. The outgoing branches have no effect on the value of the node variable.

Obtain the Block diagram from the signal flow graph, shown below S

2S

3

2

1

X

W ZU Y -3 4

-5 Write the equations for various node variables Y, Z, W and U, Input node is X. There are only outgoing branches. Y = (2S ) X – 4Z - 5U Z = 3Y – 3W W = 2Z U = 1 W + SZ The block digram simulation of various equations are Equation (1) X +

2S -

-

Y

4

From Z

5

From U

From equation (2) Y

Z

3

+

From W

3

For equation (3) 2

Z

W

For equation (4) W

From Z

+

+

S

Combining all the block diagrams, the complete block diagram is

U

4

X

2S

Y

-

+

-

3

+

S

Z

2

-

W

+

3

5

REFERENCE: 1. 2.

CONTROL SYSTEMS BY V.A. BAKSHI & U.A. BAKSHI CONTROL SYSTEMS BY DR.GANESH RAO

Frequency – Domain analysis of control systems Introduction : It was pointed out earlier that the performance of a feedback control system is more preferably measured by its time domain response characteristics. This is in contrast to the analysis & design of systems in the communication field, where the frequency response is of more importance, since in this case most of the signals to be processed are either sinusoidal or periodic in nature. However analytically, the time response of a control system is usually difficult to determine, especially in the case of high order systems. In the design aspects, there are no unified ways of arriving at a designed system given the time-domain specifications, such as peak overshoot, rise time , delay time & setting time. On the other hand, there is a wealth of graphical methods available in the frequencydomain analysis, all suitable for the analysis & design of linear feedback control systems once the analysis & design are carried out in the frequency domain, time domain behavior of the system can be interpreted based on the relationships that exist between the time-domain & the frequency-domain properties. Therefore, we may consider that the main purpose of conducting control systems analysis & design in frequency domain is merely to use the techniques as a convenient vehicle toward the same objectives as with time-domain methods.

U

The starting point in frequency-domain analysis is the transfer function. For a single loop feed back system, the closed loop transfer function is written C(s) G(s) M(s) = = R(s) 1+ G(s) H(s)

(1)

Under the sinusoidal steady-state, we get s = jω; then equation(1.0) becomes, C(jω) M(jω) =

G(jω) =

R(jω)

(1.1) 1+ G(jω) H(jω)

The sinusoidal steady-state transfer relation M(jω), which is a complex function of ω , may be expressed in terms of a real & an imaginary part; that is, M((jω) = Re [ M (jω)] + j Im [ M(jω)]

(1.2)

Or , M(jω) can be expressed in terms of its magnitude & phase as Where

M(jω) = M (ω) φm(ω)

(1.3)

G(jω) M((ω) =

(1.4) 1 + G(jω) H(jω)

And G(jω) φm(ω) = 1 + G(jω) H(jω) (1.5) =

G(jω) -

1 + G(jω) H(jω)

since the analysis is now in the frequency domain, some of the terminology used in communication system may be applied to the present control system characterization. For instance, M(ω) of Eq. (1.4) may be regarded as the magnification of the feed back control system is similar to the gain or amplification of an electronic amplifier. In an audio amplifier, for instance, an ideal design criterion is that the amplifier must have a flat gain for all frequencies. Of course, realistically, the design criterion becomes that of having a flat gain in the audio frequency range. In control system the ideal design criterion is similar. If it is desirable to keep the output C(jω) identical to the input R(jω) at all frequencies, M(jω) must be unity for all frequencies. However, from Eq. (1.1) it is apparent that M(jω) can be unity only when G(jω) is infinite, while H(jω) is finite & nonzero. An infinite magnitude for g(jω) is, of course, impossible to achieve in practice, nor would it be desirable, since most control system become unstable when its loop gain becomes very high. Further more, all control system are subject noise. Thus in addition to responding to the input signal, the system should be able to reject & suppress noise & unwanted signals. This mean that the frequency response of a control system should have a cutoff characteristic in general, & sometimes even a band-pass characteristic. The phase characteristic of the frequency response are also of importance. The ideal situation is that the phase must be a linear function of frequency within the frequency range of interest . Figure 1.1 shows the gain & phase characteristics of an ideal low-pass filter, which is impossible to realize physically. Typical gain & phase characteristics of a feedback control system are shown in Fig. 1.2. The fact is that the great majority of control systems have the characteristics of a low-pass filter, so the gain decreases as the frequency increases.

M(ω)

0

ω

1 φm(ω)

0

ωc ω Deg Fig. 1.1. Gain-phase characteristics of an ideal low-pass filter.

0 M(ω)

Mp

φm(ω)

1 0

Deg

ω

Fig.1.2. Typical gain & phase characteristics of a feedback control system.

ω

Frequency-Domain characteristics: If a control system is to be designed or analyzed using frequency-domain techniques, we need a set of specification to describe the system performance. The following frequency-domain specifications are often used in practice. Peak response Mp : The peak response Mp is defined as the maximum value of M(ω) that is given in Eq.(1.4). In general, the magnitude of Mp gives an indication of the relative stability of a feed back control system. Normally, a large Mp corresponds to a large peak overshoot in the step response. For most design problems it is generally accepted that an optimum value Mp of should be somewhere between 1.1 & 1.5. Resonant frequency ωp : The resonant frequency ωp is defined as the frequency at which the peak resonance Mp occurs. Bandwidth : The bandwidth , BW, is defined as the frequency at which the magnitude of M(jω), M(ω), drops at 70.7 percent of its zero-frequency level, or 3 dB down from the zerofrequency gain. In general, the bandwidth of a control system indicates the noise-filtering characteristics of the system. Also, bandwidth gives a measure of the transient response properties, in that a large bandwidth corresponds to a faster rise time, since higherfrequency signals are passed on to the outputs. Conversely, if the bandwidth is small, only signals of relatively low frequencies are passed, & the time response will generally be slow & sluggish. Cutoff rate : Often, bandwidth alone is inadequate in the indication of the characteristics of the system in distinguishing signals from noise. Sometimes it may be necessary to specify the cutoff rate of the frequency response at the higher frequencies. However, in general, a steep cutoff characteristics may be accompanied by a large Mp, which corresponds to a system with a low stability margin. The performance criteria defined above for the frequency-domain analysis are illustrated on the closed-loop frequency response, as shown in Fig. 1.3.

There are other criteria defined that may be used to specify the relative stability & performance of a feedback control system. These are defined in the ensuring sections of this chapter. M(ω) Mp 1

Bandwidth

ω

0 ωp

BW

Fig.1.3. Typical magnification curve of a feedback control system.

Mp, , ωp & the bandwidth of a second-order system: For a second-order feedback control system, the peak resonance Mp, the resonant frequency ωp, & the bandwidth are all uniquely related to the damping ratio ξ & the natural undamped frequency ωn of the system. Consider the second-order sinusoidal steady-state transfer function of a closed-loop system, C(jω) ω2n M(jω) = = (1.6) R(jω) (jω)2 + 2 ζ ωn (jω) + ω2n 1 =

1 + j 2 (ω/ωn) ζ - (ω/ωn)2

We may simplify the last expression by letter u = ω/ ωn. The Eq. (1.6) becomes

1 M(ju) =

1 + j2 u ζ - u2

( 1.7)

The magnitude & phase of M (jω) are 1 M(ju) = M(u) =

2 2

[( 1 – u ) + ( 2 ζ u)2] ½

And 2ζ u

(1.8)

M(ju) = φm(u) = - tan -1

1 – u2

(1.9)

The resonant frequency is determined first by taking the derivative of M(u) with respect to u & setting it equal to zero. Thus dM(u)

1

[( 1 – u2 )2 + ( 2ζ u)2] –3/2 ( 4 u3 – 4u + 8uζ2) = 0

=-

(

1.10) du

2

from which

4u3 – 4u + 8uζ2 = 0

(

1.11) The root of Eq. (1.11) are up = 0

(1.12) and

up = √ 1 - 2ζ2

1.13)

(

The solution Eq. (1.12) merely indicates that the slope of the M(ω) versus ω curve is zero at ω = 0; it is not true maximum. The solution of eq. (1.13) gives the resonant frequency, ωp = ωn √ 1 - 2ζ2

(1.14)

Since frequency is a real quantity, Eq. (1.14) is valid only for 1 ≥ 2ζ2 or ζ ≤ 0.707. This means simply that for all values of ξ greater than 0.707, the solution of ωp = 0 becomes the valid one, & Mp = 1. Substituting Eq. (1.13) into Eq. (1.8) & simplifying, we get 1 Mp =

2√ 1 -ζ

2

(1.15)

It is important to note that Mp is a function of ζ only, whereas ωp is a function of ζ & ωn

5 4 3 Mp 2 1

0 0.5

0.707 1.0 1.5 Damping ratio ζ

2.0 1

Fig.1.4 Mp versus-damping ratio for a second – order system, Mp =

2ζ √ 1 - ζ2

1.0 0.8 up = ωp /ωn 0.6 0.4 0.2 0 0.5 0.707 1.0 Damping ratio ζ Fig 1.5. Normalized resonant frequency- versus-damping ratio for a second order system, Up = √ 1 - ζ2 .

Fig.1.4 & 1.5 illustrate the relationship between Mp & ζ , & u = ωp / ωn & ζ, respectively.

Bandwidth : Bandwidth BW of a system is a frequency at which M(ω) drops to 70.7% of its zero frequency level or 3 dB down from the zero frequency gain. Equating the Eq. 1 M(u) =

2 2

2 ½

= 0.707

[( 1 –u ) + ( 2δu) ] =

[( 1 –u2)2 + ( 2δu)2] ½

1 = 0.707 =

√2.

Squaring both sides ( 1 –u2)2 + 4δ2u2 = 2

1 + u4 – 2u2 + 4δ2u2

= 2

Let u2 = x 1 + x2 – 2x + 4δ2 x

= 2

x2 – 2x + 4δ2 x = 1 2 2 x – x ( 2 - 4δ ) =1 x2 – x ( 2 - 4δ2 ) – 1 = 0 a = 1,

b = - ( 2 - 4δ2 ) , c = -1 -b±√ b2 – 4ac x= 2a (2 – 4δ2 ) ±√ (2 – 4δ2 )2 + 4 = 2 (2 – 4δ ) ±√ (4 + 16δ4 – 16δ 2 + 4 2

= 2 (2 – 4δ2 ) ±√ 16 δ4 – 16 δ2 + 8 = 2

2 (1 – 2δ2 ) ± √ 4 + 16δ4 – 16δ 2 + 4 = 2 2 (1 – 2δ2 ) ± 2√2 + 4δ4 – 4δ2 = 2

2 (1 – 2δ2 ) ± 2√2 + 4δ4 – 4δ2 = 2 2 [(1 – 2δ2 ) ± √2 + 4δ4 – 4δ2] = 2 u2 = x = ω / ωn = u =

(1 – 2δ2 ) ±√ 2 + 4δ4 – 4δ 2 [(1 – 2δ2 ) ±√ (2 + 4δ4 – 4δ 2 ] 1/2

BW = ωn [ ( 1 – 2ζ2 ) + √4ζ4 - 4ζ2 +2]1/2

For the second order system under consideration, we easily establish some simple relationship between the time – domain response & the frequency-domain response of the system. 1. The maximum over shoot of the unit step response in the time domain depends upon ζ only. 2. The response peak of the closed - loop frequency response Mp depends upon ζ only. 3. The rise time increases with ζ , & the bandwidth decreases with the increase of ζ , for a fixed ωn, therefore, bandwidth & rise time are inversely proportional to each other. 4. Bandwidth is directly proportional to ωn. 5. Higher bandwidth corresponds to larger Mp.

Stability Analysis Every System, for small amount of time has to pass through a transient period, whether the 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 anything 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 whether the given system can reach steady state after passing through the transients successfully is called the stability analysis of the system. In this chapter, we will steady

1. 2. 3. 4. 5. 6.

The stability & the factor on which system stability depends. Stability analysis & location of closed loop poles. Stability analysis using Hurwitz method. Stability analysis using Routh-Hurwitz method. Special cases of Routh’s array. Applications of Routh-Hurwitz method.

Concept of stability: Consider a system i.e. a deep container with an object placed inside it as shown in fig(1) Force ‘F’

(a)

fig(1)

(b)

Now, if we apply a force to take out the object, as the depth of container is more, it will oscillate & settle down again at original position. Assume that force required to take out the object tends to infinity i.e. always object will oscillate when force is applied & will settle down but will not come out such a system is called absolutely stable system. No change in parameters, disturbances, changes the output. Now consider a container which is pointed one, on which we try to keep a circular object. In this object will fall down without any external application of force. Such system is called unstable system.

(a)

fig(2)

(b)

While in certain cases the container is shallow then there exists a critical value of force for which the object will come out of the container. F F F

Fig(3)

F
F>Fcritical

As long as FFcritical object will come out. Stability depends on certain conditions of the system, hence system is called conditionally stable system. Pendulum where system keeps on oscillating when certain force is applied. Such systems are neither stable nor unstable & hence called critically stable or marginally stable systems

Stability of control systems: The stability of a linear closed loop system can be determined from the locations of closed loop poles in the S-plane. If the system has closed loop T.F. C(s) 10 R(s) (S+2) (S+4) Output response for unit step input R(s) =

1

S 10 S(S+2) (S+4)

C(s)

A S

B S+2

C S+4

Find out partial fractions 1 C(s) = 8 S = C(s) =

1 4 S+2

1 8 S+4

10

1.25 2.5 1.25 S S+2 S+4 1.25 – 2.5e-2t+1.25e-4t

=Css + Ct(t) If the closed loop poles are located in left half of s-plane, Output response contains exponential terms with negative indices will approach zero & output will be the steady state output. I.e. Ct (t) = 0 t ∞ Transient output = 0 Such a system is called absolutely stable systems. Now let us have a system with one closed loop pole located in right half of s- plane C(s) 10 R(s) S(S-2)(s+4) A S

=

+

B S-2 2t

+

C 10 S+4 –4t

C(t) = - 1.25 + 0.833e + 0.416e Here there is one exponential term with positive in transient output Therefore Css = - 1.25 t 0

C(t) 0

1 2 4 ∞

+ 4.91 + 44.23 +2481.88 ∞

From the above table, it is clear that output response instead of approaching to steady state value as t ∞ due to exponential term with positive index, transients go on increasing in amplitude. So such system is said to be unstable. In such system output is uncontrollable & unbounded one. Output response of such system is as shown in fig(4). C(t) C(t) ∞ ∞ OR Steady state output t (a)

fig(4)

t (b) For such unstable systems, if input is removed output may not return to zero. And if the input power is turned on, output tends to ∞ . If no saturation takes place in system & no mechanical stop is provided then system may get damaged. If all the closed loop poles or roots of the characteristic equation lies in left of s-plane, then in the output response contains steady state terms & transient terms. Such transient terms approach to zero as time advances eventually output reaches to equilibrium & attains steady state value. Transient terms in such system may give oscillation but the amplitude of such oscillation will be decreasing with time & finally will vanish. So output response of such system is shown in fig5 (a) & (b). C(t)

C(t)

Steady state -----------------------

Damped oscillations

--------------------------OR Steady state output

t (a)

t fig 5

(b)

BIBO Stability : This is bounded input bounded output stability. Definition of stable system: A linear time invariant system is said to be stable if following conditions are satisfied. 1. When system is excited by a bounded input, output is also bounded & controllable.

2. In the absence of input, output must tend to zero irrespective of the initial conditions. Unstable system: A linear time invariant system is said to be unstable if, 1. for a bounded input it produces unbounded output. 2. In the absence of input, output may not be returning to zero. It shows certain output without input. Besides these two cases, if one or more pairs simple non repeated roots are located on the imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such systems are said to be critically or marginally stable systems. Critically or Marginally stable systems: A linear time invariant system is said to be critically or marginally stable if for a bounded input its output oscillates with constant frequency & Amplitude. Such oscillation of output are called Undamped or Sustained oscillations. For such system one or more pairs of non repeated roots are located on the imaginary axis as shown in fig6(a). Output response of such systems is as shown in fig6(b). C(t) Constant Amplitude & frequency oscillations

X Jω2

-------------------------

--------------------------------------- steady state output X

Jω1

-----------------------------------σ

X - Jω1 X - Jω2

Fig 6(a) non repeated poles on Jω axis.

t

If there are repeated poles located purely on imaginary axis system is said to be unstable. C(t) Jω1 x x σ

Jω1

----------------------------------------steady state output

s-plane x x t

Conditionally Stable: A linear time invariant system is said to be conditionally stable, if for a certain condition if a particular parameter of the system, its output is bounded one. Otherwise if that condition is violated output becomes unbounded system becomes unstable. i.e. Stability of the system depends the on condition of the parameter of the system. Such system is called conditionally stable system. S-plane can be divided into three zones from stability point of view. Jω axis

Left half of s-plane

Right half of s-plane

Real Stable

Unstable S-Plane

(repeated) unstable Sl. No 1.

Nature of closed loop poles. Real negative i.e in LHS of splane

(non repeated roots) Marginally stable

Location of closed loop poles in s-plane Jω

Step response C(t)

Stability condition

--------------------x

x

Absolutely stable

σ

-02 -01

t Real positive in RHS of s-plane

Jω C(t)

2. x a1

∞ ------------------------

σ

Unstable

t

increasing towards ∞

3.

Complex conjugate with negative real part

C(t) Jω Jω 1

x

σ

-a1

Absolutely stable.

x

t

-Jω 2 Damped oscillation 4.

Complex conjugate with positive real part

Ct Jω1

x

-Jω 1

-x

σ

Unstable. t oscillations with increasing amplitude

5.

Non repeated pair on imaginary axis

C(t) Jω x Jω1 x -Jω2

Marginally or critically stable σ

t

C(t)

OR Jω

t

X Jω2 x Jω1

σ

x –Jω1

Marginally or critically stable Sustained oscillations with two frequencies ω1 & ω2

x – Jω2

6.

Repeated pair on imaginary axis

Two non repeated pairs on imaginary axis. Jω C(t) x x Jω1

-----------------------σ

x x -Jω1

Unstable t oscillation of increasing amplitude

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 Jω axis, settling time will be large. As the roots move away from Jω 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. C(t)

Jω

Stable for P1 x

x

σ

----------------------------------

P2 P1 Relatively more stable for P2 t

Jω C(t) x

stable for ∝1

x

∝2

∝1

x

x

σ --------------------------------------------------------------

more stable for ∝2 Jω axis t Relative stability improves

s-plane

Routh – Hurwitz Criterion :

This represents a method of determining the location of poles of a characteristics equation with the respect to the left half & right half of the s-plane without actually solving the equation. The T.F.of any linear closed loop system can be represented as, C(s) =

b0 sm + b1 sm-1 +….+ bm n

n-1

R(s) a0 s + a1 s + …. + an Where ‘a’ & ‘b’ are constants. To find the closed loop poles we equate F(s) =0. This equation is called as Characteristic Equation of the system. n n-1 n-2 F(s) = a0 s + a1 s + a2 s + ….. + an = 0. Thus the roots of the characteristic equation are the closed loop poles of the system which decide the stability of the system.

Necessary Condition to have all closed loop poles in L.H.S. of s-plane. In order that the above characteristic equation has no root in right of s-plane, it is necessary but not sufficient that, 1. All the coefficients off the polynomial have the same sign. 2. Non of the coefficient vanishes i.e. all powers of ‘s’ must be present in descending order from ‘n’ to zero. These conditions are not sufficient.

Hurwitz’s Criterion : The sufficient condition for having all roots of characteristics equation in left half of splane is given by Hurwitz. It is referred as Hurwitz criterion. It states that: The necessary & sufficient condition to have all roots of characteristic equation in left half of s-plane is that the sub-determinants DK, K = 1, 2,………n obtained from Hurwitz determinant ‘H’ must all be positive.

Method of forming Hurwitz determinant: a1 a0

a3 a2

a5 a4

…….. a2n-1 ..…… a2n-2

0

a1

a3

.……. a2n-3

0

a0

a2

…….. a2n-4

0

0

a1

……... a2n-5

-

-

-

-

H=

0

-

-

-

……...

-

……...

-

…….

an

The order is n*n where n = order of characteristic equation. In Hurwitz determinant all coefficients with suffices greater than ‘n’ or negative suffices must all be replaced by zeros. From Hurwitz determinant subdeterminants, DK, K= 1, 2, ….n must be formed as follows:

D1 =

a1

D2 =

a1 a0

a3 a2

D3 =

a1 a0 0

a3 a2 a1

a5 a4 a3

DK =

H

For the system to be stable, all above determinants must be positive. Determine the stability of the given characteristics equation by Hurwitz,s method. Ex 1: F(s)= s3 + s2 + s1 + 4 = 0 is characteristic equation. a0 = 1, a1 = 1, a2 = 1, a3 = 4, n = 3 a3 a2 a1

a1 a0 0

H=

D1 = 1

=1

1 4 1 1

D2 = 1

4

= -3 0

a5 a4 a3

=

1 4 0 1 1 0 0 1 4

D3 =

1 1 0 0 1 4

= 4 –16 = -12.

As D2 & D3 are negative, given system is unstable.

Disadvantages of Hurwitz’s method : 1.

For higher order system, to solve the determinants of higher order is very complicated & time consuming. 2. Number of roots located in right half of s-plane for unstable system cannot be judged by this method. 3. Difficult to predict marginal stability of the system. Due to these limitations, a new method is suggested by the scientist Routh called Routh’s method. It is also called Routh-Hurwitz method.

Routh’s Stability Criterion: It is also called Routh’s array method or Routh-Hurwitz’s method Routh suggested a method of tabulating the coefficients of characteristic equation in a particular way. Tabulation of coefficients gives an array called Routh’s array. Consider the general characteristic equation as, n n-1 n-2 F(s) = a0 s + a1 s + a2 s + ….. + an = 0.

Method of forming an array : Sn

a0

a2

a4

a6

Sn-1

a1

a3

a5

a7

Sn-2

b1

b2

b3

Sn-3

c1

c2

c3

-

-

-

-

-

-

-

-

S0

……….

an

Coefficients of first two rows are written directly from characteristics equation. From these two rows next rows can be obtained as follows.

b1 =

a1 a2 – a0 a3 a1

,

b2 =

a1 a4 – a0 a5 a1

,

b3 =

a1 a6 – a0 a7 a1

From 2nd & 3rd row , 4th row can be obtained as C1 =

b1 a3 – a1 b2

,

C2 =

b1

b1 a5 – a1 b3 b1 0

This process is to be continued till the coefficient for s is obtained which will be an. From this array stability of system can be predicted.

Routh’s criterion : The necessary & sufficient condition for system to be stable is “ All the terms in the first column of Routh’s array must have same sign. There should not be any sign change in first column of Routh’s array”. If there are sign changes existing then, 1. System is unstable. 2. The number of sign changes equals the number of roots lying in the right half of the s-plane. Examine the stability of given equation using Routh’s method : Ex.2:

s3+6s2 + 11s + 6 =0

Sol:

a0 = 1,

a1 = 6,

a2 =11,

S3

1

11

S2

6

6

S1 S0

11 * 6 – 6 6 6

=10

a3 = 6,

0

As there is no sign change in the first column, system is stable.

n=3

Ex. 3 Sol:

s3 + 4s2 + s + 16 = 0 a0 =1, a1 = 4, a2 = 1,

S3

1

1

S2

+4

16

S1

4 - 16 = 4 +16

S0

-3

a3 = 16

0

As there are two sign changes, system is unstable. Number of roots located in the right half of s-plane = number of sign changes = 2.

Special Cases of Routh’s criterion : Special case 1 : First element of any of the rows of Routh’s array is zero & same remaining rows contains at least one non-zero element. Effect : The terms in the new row become infinite & Routh’s test fails. e.g. : s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 S5

1

3

2

S4

2

6

1

S3

0

1.5

0

S2

∞

….

…

Special case 1 Routh’s array failed

Following two methods are used to remove above said difficulty. First method : first

Substitute a small positive number ‘ε’ in place of a zero occurred as a element in the row. Complete the array with this number ‘ε’. Then examine lim

Sign change by taking ε

. Consider above Example. 0

S5

1

3

2

S4

2

6

1

S3

ε

1.5

0

S2

6ε - 3 ε

1

0

S1

1.5(6ε - 3)

0 -ε

ε (6ε - 3) ε S0

1

To examine sign change, Lim Lim = 6ε - 3 = 6 3 ε 0 ε ε 0 ε =6-∞ = - ∞ sign is negative. Lim 1.5(6ε – 3) - ε2 = Lim 9ε - 4.5 - ε2 ε 0 6ε -3 ε 0 6ε - 3 = 0 – 4.5 – 0 0 -3 =

+ 1.5

sign is positive

Routh’s array is, S5

1

3

2

S4

2

6

1

S3

+ε

1.5

0

S2

-

∞

1

0

S1

+1.5

0

0

S0

1

0

0

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 ‘1/Z’ in original equation. Taking L.C.M. rearrange characteristic equation in descending powers of ‘Z’. Then complete the Routh’s array with this new equation in ‘Z’ & examine the stability with this array. F(s) = s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 s=1/Z

Consider Put

∴1 + 2 + 3 +6 +2 + 1=0 Z5 Z4 Z3 Z2 Z Z5 + 2Z4+ 6Z3+3Z2+2Z+ 1 = 0 Z5

1

6

2

Z4

2

3

1

Z3

4.5

1.5

0

Z2

2.33

1

0

Z1

- 0.429

0

Z0

1

As there are two sign changes, system is unstable.

Special case 2: All the elements of a row in a Routh’s array are zero.

Effect : The terms of the next row can not be determined & Routh’s test fails. S5

a

b

c

S4

d

e

f

S3

0

0

0

Row of zeros, special case 2

This indicates no availability of coefficient in that row. Procedure to eliminate this difficulty: 1. Form an equation by using the coefficients of row which is just above the row of zeros. Such an equation is called an Auxiliary equation denoted as A(s). For above case such an equation is, A(s) = ds4 + es2 + f Note that the coefficients of any row are corresponding to alternate powers of ‘s’ starting from the power indicated against it.

So‘d’ is coefficient corresponding to s4 so first term is ds4 of A(s). Next coefficient ‘e’ is corresponding to alternate power of ‘s’ from 4 i.e. s2 Hence the term es 2 & so on. 2. Take the derivative of an auxiliary equation with respect to ‘s’. i.e. dA(s) = 4d s3 + 2e s ds

3. Replace row of zeros by the coefficients of dA(s) ds S5

a

b

c

S4

d

e

f

S3

4d 2e 0

4. Complete the array of zeros by the coefficients.

Importance of auxillary equation : Auxillary equation is always the part of original characteristic equation. This means the roots of the auxillary equation are some of the roots of original characteristics equation. Not only this but roots of auxillary equation are the most dominant roots of the original

characteristic equation, from the stability point of view. The stability can be predicted from the roots of A(s)=0 rather than the roots of characteristic equation as the roots of A(s) = 0 are the most dominant from the stability point of view. The remaining roots of the characteristic equation are always in the left half & they do not play any significant role in the stability analysis. e.g. Let F(s) = 0 is the original characteristic equation of say order n = 5. Let A(s) = 0 be the auxillary equation for the system due to occurrence of special case 2 of the order m = 2. Then out of 5 roots of F(s) = 0, the 2 roots which are most dominant (dominant means very close to imaginary axis or on the imaginary axis or in the right half of s-plane) from the stability point of view are the 2roots of A(s) = 0. The remaining 5 –2 = 3 roots are not significant from stability point of view as they will be far away from the imaginary axis in the left half of s-plane. The roots of auxillary equation may be, 1. A pair of real roots of opposite sign i.e.as shown in the fig. 8.10 (a). jω

jω x

x

x

σ σ

σ x

Fig. 8. 10 (b)

Fig 8. 10(a)

2. A pair of roots located on the imaginary axis as shown in the fig. 8.10(b). 3. The non-repeated pairs of roots located on the imaginary axis as shown in the fig.8.10 (c). jω jω x x

xx σ

x x

Fig. 8.10(c)

σ xx

Fig. 8.10(d).

4. The repeated pairs of roots located on the imaginary axis as shown in the Fig.8.10 (d). Hence total stability can be determined from the roots of A(s) = 0, which can be out of four types shown above.

Change in criterion of stability in special case 2:

After replacing a row of zeros by the coefficients of dA(s) , complete the Routh’s array. ds But now, the criterion that, no sign in 1st column of array for stability, no longer remains sufficient but becomes a necessary. This is because though A(s) is a part of original characteristic equation, dA(s) is not, which is in fact used to complete the array. ds So if sign change occurs in first column, system is unstable with number of sign changes equal to number of roots of characteristics equation located in right half of s-plane. But there is no sign changes, system cannot be predicted as stable . And in such case stability is to be determined by actually solving A(s) = 0 for its roots. And from the location of roots of A(s) = 0 in the s-plane the system stability must be determined. Because roots A(s) = 0 are always dominant roots of characteristic equation.

Application of Routh’s of criterion : Relative stability analysis : If it is required to find relative stability of system about a line s = - σ . i.e. how many roots are located in right half of this line s = - σ, the Routh’s method can be used effectively. To determine this from Routh’s array, shift the axis of s – plane & then apply Routh’s array i.e. substitute s = s 1 - σ, (σ = constant) in characteristic equation. Write polynomial in terms of s1. Complete array from this new equation. The number of sign changes in first column is equal to number of roots those are located to right of the vertical line s = - σ. Imaginary jω

σ

-σ

0

Determining range of values of K : In practical system, an amplifier of variable gain K is introduced . The closed loop transfer function is C(s) KG(s) = R(s) 1+ KG(s) H(s)

Hence the characteristic equation is F(s) = 1+ KG(s) H(s) = 0 So the roots of above equation are dependent on the proper selection of value of ‘K’. So unknown ‘K’ appears in the characteristic equation. In such case Routh’s array is to be constructed in terms of K & then the range of values of K can be obtained in such away that it will not produce any sign change in first column of the Routh’s array. Hence it is possible to obtain the range of values of K for absolute stability of the system using Routh’s criterion. Such a system where stability depends on the condition of parameter K, is called conditionally stable system.

Advantages of Routh’s criterion : Advantages of routh’s array method are : 1. Stability of the system can be judged without actually solving the characteristic equation. 2. No evaluation of determinants, which saves calculation time. 3. For unstable system it gives number of roots of characteristic equation having positive real part. 4. Relative stability of the system can be easily judged. 5. 6. 7.

By using the criterion, critical value of system gain can be determined hence frequency of sustained oscillations can be determined. It helps in finding out range of values of K for system stability. It helps in finding out intersection points of roots locus with imaginary axis.

Limitation of Routh’s criterion: 1. It is valid only for real coefficients of the characteristic equation. 2. It does not provide exact locations of the closed loop poles in left or right half of s-plane. 3. It does not suggest methods of stabilizing an unstable system. 4. Applicable only to linear system. Ex.1. s6 + 4s5 +3s4 – 16s2- 64s – 48 = 0 Find the number of roots of this equation with positive real part, zero real part & negative real part Sol:

S6

1

3

-16

-48

S5

4

0

-64

0

S4

3

0

-48

S3

0

0

0

0

A(s) = 3S4 – 48 = 0

Lim

0

= 12S3

S6

1

3

-16

-48

S5

4

0

-64

0

S4

3

0

-48

0

S3

12

0

0

0

S2

(ε ) 0

-48

S1

576 ε -48

0

S0

ε

dA ds

576 ε

0 0

0 0

+∞

Therefore one sign change & system is unstable. Thus there is one root in R.H.S of the s – plane i.e. with positive real part. Now Solve A(s) = 0 for the dominant roots A(s) = 3s4 – 48 =0 Put S2 = Y ∴

3Y2 = 48 S2 = + 4 S

= ±2

∴ Y2 =16,

∴ Y = ±√16 = ± 4

S2 = -4 S = ± 2j

So, S = ± 2j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S. indicated by a sign change is S = ± 2 as obtained by solving A(s) = 0. Total there are 6 roots as n = 6. Roots with Positive real part = 1 Roots with zero real part =2 Roots with negative real part = 6 –2 – 1 = 3

Ex.2:

For unity feed back system, k G(s) =

, Find range of values of K, marginal value of

K S (1 + 0.4s) ( 1 + 0.25 s)

Sol: Characteristic equation,

& frequency of sustained oscillations.

1 + G (s) H (s) = 0 & H(s) = 1

K ∴1+

=0 s(1 + 0.4s) ( 1 + 0.25s)

s [ 1 + 0.65s + 0.1s2} + K = 0 ∴ 0.1s3 + 0.65s2 +s + K = 0

S3

0.1

1

From s0,

K>0

S2

0.65

K

from s1,

S1

0.65 – 0.1K 0.65 K

0

0.65 – 0.1K > 0 ∴ 0.65 > 0.1 K ∴ 6.5 > K

S0

∴ Range of values of K, 0 < K < 6.5

Now marginal value of ‘K’ is that value of ‘K’ for which system becomes marginally stable. For a marginal stable system there must be row of zeros occurring in Routh’s array. So value of ‘K’ which makes any row of Routh array as row of zeros is called marginal value of ‘K’. Now K = 0 makes row of s0 as row of zeros but K = 0 can not be marginal value because for K = 0, constant term in characteristic equation becomes zeros ie one coefficient for s0 vanishes which makes system unstable instead of marginally stable. Hence marginal value of ‘K’ is a value which makes any row other than s0 as row of zeros. ∴ 0.65 – 0.1 K mar = 0 ∴ K mar = 6.5 To find frequency, find out roots of auxiliary equation at marginal value of ‘K’ A(s) = 0.65 s2 + K = 0 ;

∴ 0.65 s2 + 6.5 = 0 Because K = 6.5 s2 = -10 s = ± j 3.162 comparing with s = ± jω ω = frequency of oscillations = 3.162 rad/ sec. Ex : 3 For a system with characteristic equation F(s) = s5 + s4 + 2s3 + 2s2 + 3s +15 = ∞, examine the stability Solution :

S5

1

2

3

S4

1

2

15

S3

0

-12

0

S5

1

2

3

S4

1

2

15

S3

ε

-12

0

S2 S1 S0

S2 S1

S0

(2ε + 12) 15 ε (2ε + 12)( -12 ) – 15ε ε 2ε + 12 ε 15

0

0

Lim ε 0

2ε + 12

12 =2+∞ = +∞

=2 + ε

ε

Lim ε 0

(2ε + 12)( -12 ) – 15ε ε 2ε + 12 ε

=

Lim ε 0

-24 ε - 144 – 15 ε2 2ε + 12

0 – 144 - 0 =

= - 12 0 + 12

S5

1

2

3 15

S4

1

2

S3

ε

-12

0 There are two sign changes,

so system is unstable.

2

+∞

15

S1

- 12

0

S0

15

S

0

Ex : 4 Using Routh Criterion, investigate the stability of a unity feedback system whose open loop transfer function is e -sT G(s) = s(s+1) Sol :

The characteristic equation is 1 + G(s) H(s) = 0 e -sT

∴

1 +

= 0 s(s+1)

∴

s2 + s + e –sT = 0

Now e – sT can be Expressed in the series form as

e

–sT

s2 T2 = 1 – sT +

+ …… 2!

Truncating the series & considering only first two terms we get esT = 1 – sT ∴ s2 + s + 1 – sT = 0 ∴ s2 + s ( 1- T ) + 1 = 0 So Routh’s array is S2

1

1

S

1-T

0

S0

1

∴ 1 – T > 0 for stability ∴ T<1 This is the required condition for stability of the system.

Ex : 5

Determine the location of roots with respect to s = -2 given that F(s) = s4 + 10 s3 + 36s2 + 70s + 75

Sol :

shift the origen with respect to s = -2 s = s1 – 2

(s′ – 2 ) 4 + 10 (s′ – 2)3 + 36(s′ – 2 )2 + 70 ( s′ –2) + 75 = 0 s′4 + 2s′3 + 0s′2 + 14s′ + 15 = 0 S′4

1

0

15

S′3

2

14

0

15

0

S′2

-7

S′1

18.28

0

0

S′ 0

15

Two sign change, there are two roots to the right of s = -2 & remaining ‘2’ are to the left of the line s = -2. Hence the system is unstable.

Date : 2 – 5 -2005 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 7. The stability & the factor on which system stability depends. 8. Stability analysis & location of closed loop poles. 9. Stability analysis using Hurwitz method. 10. Stability analysis using Routh-Hurwitz method. 11. Special cases of Routh’s array. 12. Applications of Routh-Hurwitz method. Concept of stability: Consider a system i.e a deep container with an object placed inside it as shown in fig(1) force ‘F’

(a)

fig(1)

(b)

Now, if we apply a force to take out the object, as the depth of container is more, it will oscillate & settle down again at original position.

Assume that force required to take out the object tends to infinity i.e always object will oscillate when force is applied & will settle down but will not come out such a system is called absolutely stable system. No change in parameters, disturbances, changes the output. Now consider a container which is pointed one, on which we try to keep a circular object. In this object will fall down without any external application of force. Such system is called Unstable system.

(a) fig(2) (b) While in certain cases the container is shallow then there exsists a critical value of force for which the object will come out of the container. F F F

Fig(3)

F
F>Fcritical

As long as FFcritical object will come out. Stability depends on certain conditions of the system, hence system is called conditionally stable system. Pendulum where system keeps on oscillating when certain force is applied. Such systems are neither stable nor unstable & hence called critically stable or marginally stable systems Stability of control systems: The stability of a linear closed loop system can be determined from the locations of closed loop poles in the S-plane. If the system has closed loop T.F. C(s) 10 R(s) (S+2) (S+4) Output response for unit step input R(s) = C(s) Find out partial fractions

1

S 10 S(S+2) (S+4)

A S

B S+2

C S+4

1 C(s) = 8 S = C(s) =

1 4 S+2

1 8 S+4

10

1.25 2.5 1.25 S S+2 S+4 -2t -4t 1.25 – 2.5e +1.25e

=Css + Ct(t) If the closed loop poles are located in left half of s-plane, Output response contains exponential terms with negative indices will approach zero & output will be the steady state output. i.e. Ct (t) = 0 t ∞ Transient output = 0 Such system are called absolutely stable systems.

Now let us have a system with one closed loop pole located in right half of s- plane C(s) 10 R(s) S(S-2)(s+4) A S

=

+

B S-2 2t

+

C 10 S+4 –4t

C(t) = - 1.25 + 0.833e + 0.416e Here there is one exponential term with positive in transient output Therefore Css = - 1.25 t 0 1 2 4 ∞

C(t) 0 + 4.91 + 44.23 +2481.88 ∞

From the above table, it is clear that output response instead of approaching to steady state value as t ∞ due to exponential term with positive index, transients go on increasing in amplitude. So such system is said to be unstable. In such system output is uncontrollable & unbounded one. Output response of such system is as shown in fig(4).

C(t)

∞

C(t) ∞ OR Steady state output t (a)

fig(4)

t (b) For such unstable systems, if input is removed output may not return to zero. And if the input power is turned on, output tends to ∞ . If no saturation takes place in system & no mechanical stop is provided then system may get damaged. If all the closed loop poles or roots of the characteristic equation lies in left of s-plane, then in the output response contains steady state terms & transient terms. Such transient terms approach to zero as time advances eventually output reaches to equilibrium & attains steady state value. Transient terms in such system may give oscillation but the amplitude of such oscillation will be decreasing with time & finally will vanish. So output response of such system is shown in fig5 (a) & (b). C(t)

C(t)

Steady state -----------------------

Damped oscillations

--------------------------OR Steady state output

t (a)

t fig 5

(b)

BIBO Stability : This is bounded input bounded output stability. Definition of stable system: A linear time invariant system is said to be stable if following conditions are satisfied. 3. When system is excited by a bounded input, output is also bounded & controllable. 4. 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, 3. for a bounded input it produces unbounded output. 4. In the absence of input, output may not be returning to zero. It shows certain output without input.

Besides these two cases, if one or more pairs simple non repeated roots are located on the imaginary axis of the s-plane, but there are no roots in the right half of s-plane, the output response will be undamped sinusoidal oscillations of constant frequency & amplitude. Such systems are said to be critically or marginally stable systems. Critically or Marginally stable systems: A linear time invariant system is said to be critically or marginally stable if for a bounded input its output oscillates with constant frequency & Amplitude. Such oscillation of output are called Undamped or Sustained oscillations. For such system one or more pairs of non repeated roots are located on the imaginary axis as shown in fig6(a). Output response of such systems is as shown in fig6(b). C(t) Constant Amplitude & frequency oscillations

X Jω2

-------------------------

--------------------------------------- steady state output X

Jω1

-----------------------------------σ

X - Jω1 X - Jω2

Fig 6(a) non repeated poles on Jω axis.

t

If there are repeated poles located purely on imaginary axis system is said to be unstable. C(t) Jω1 x x σ

Jω1

----------------------------------------steady state output

s-plane x x t

Conditionally Stable: A linear time invariant system is said to be conditionally stable, if for a certain condition if a particular parameter of the system, its output is bounded one. Otherwise if that condition is violated output becomes unbounded system becomes unstable. i.e. Stability of the system depends the on condition of the parameter of the system. Such system is called conditionally stable system.

S-plane can be divided into three zones from stability point of view. Jω axis

Left half of s-plane

Right half of s-plane

Real Stable

Unstable S-Plane

(repeated) unstable Sl.No 1.

Nature of closed loop poles. Real negative i.e in LHS of splane

(non repeated roots) Marginally stable

Location of closed loop poles in s-plane Jω

Step response

Stability condition

C(t) ---------------------

x

x

Absolutely stable

σ

-02 -01

t

2.

Real positive in RHS of splane

Jω

C(t)

x a1

σ

∞ ------------------------

Unstable t

increasing towards ∞

3.

Complex conjugate with negative real part

C(t) Jω Jω 1

x

σ

-a1

Absolutely stable.

x

t

-Jω 2 Damped oscillation 4.

Complex conjugate with positive real part

Ct Jω1

x

-Jω 1

-x

σ

Unstable. t oscillations with increasing amplitude

5.

Non repeated pair on imaginary axis

C(t) Jω x Jω1

Marginally or critically stable σ

x -Jω2

OR

t C(t)

Jω X Jω2 x Jω1

t Marginally or

σ

critically stable Sustained oscillations with two frequencies ω1 &

x –Jω1

ω2

x – Jω2 Two non repeated pairs on imaginary axis.

6.

Repeated pair on imaginary axis

Jω C(t) x x Jω1

-----------------------σ

Unstable t

x x -Jω1

oscillation of increasing amplitude

Reference : 1. Modern Control Engineering by Ogata. 2. Automatic control system by B.C.Kho. 3. Control Engineering by U.A. Bakshi.

4 – 5-2005. 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 Jω axis, settling time will be large. As the roots move away from Jω 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. Jω

C(t) Stable for P1

x

x

σ

----------------------------------

P2 P1 Relatively more stable for P2 t

Jω C(t) x

stable for ∝1

x

∝2

∝1

x

x

σ --------------------------------------------------------------

more stable for ∝2 Jω axis t Relative stability improves

s-plane

Routh – Hurwitz Criterion : This represents a method of determining the location of poles of a characteristics equation with the respect to the left half & right half of the s-plane without actually solving the equation.

The T.F.of any linear closed loop system can be represented as, C(s) =

b0 sm + b1 sm-1 +….+ bm n

n-1

a0 s + a1 s + …. + an R(s) 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. n n-1 n-2 F(s) = a0 s + a1 s + a2 s + ….. + an = 0. Thus the roots of the characteristic equation are the closed loop poles of the system which decide the stability of the system. Necessary Condition to have all closed loop poles in L.H.S. of s-plane. In order that the above characteristic equation has no root in right of s-plane, it is necessary but not sufficient that, 3. All the coefficients off the polynomial have the same sign. 4. 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 splane is given by Hurwitz. It is referred as Hurwitz criterion. It states that: The necessary & sufficient condition to have all roots of characteristic equation in left half of s-plane is that the sub-determinants DK, K = 1, 2,………n obtained from Hurwitz determinant ‘H’ must all be positive.

Method of forming Hurwitz determinant: a1 a0

H=

a3 a2

a5 a4

…….. a2n-1 ..…… a2n-2

0

a1

a3

.……. a2n-3

0

a0

a2

…….. a2n-4

0

0

-

-

-

-

0

a1

……... a2n-5

-

-

-

……...

-

……...

-

…….

an

The order is n*n where n = order of characteristic equation. In Hurwitz determinant all coefficients with suffices greater than ‘n’ or negative suffices must all be replaced by zeros. From Hurwitz determinant subdeterminants, DK, K= 1, 2, ….n must be formed as follows:

D1 =

a1

D2 =

a1 a0

a3 a2

D3 =

a1 a0 0

a3 a2 a1

a5 a4 a3

DK =

H

For the system to be stable, all above determinants must be positive. Determine the stability of the given characteristics equation by Hurwitz,s method. Ex 1: F(s)= s3 + s2 + s1 + 4 = 0 is characteristic equation. a0 = 1, a1 = 1, a2 = 1, a3 = 4, n = 3 a1 a0 0

H=

D1 = 1

D2 =

D3 =

a3 a2 a1

a5 a4 a3

=

1 4 0 1 1 0 0 1 4

=1

1 4 1 1

1 4 0 1 1 0 0 1 4

= -3

= 4 –16 = -12.

As D2 & D3 are negative, given system is unstable.

Disadvantages of Hurwitz’s method :

4.

For higher order system, to solve the determinants of higher order is very complicated & time consuming. 5. Number of roots located in right half of s-plane for unstable system cannot be judged by this method. 6. 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, n n-1 n-2 F(s) = a0 s + a1 s + a2 s + ….. + an = 0.

Method of forming an array : Sn

a0

a2

a4

a6

Sn-1

a1

a3

a5

a7

Sn-2

b1

b2

b3

Sn-3

c1

c2

c3

-

-

-

-

-

-

-

-

S0

……….

an

Coefficients of first two rows are written directly from characteristics equation. From these two rows next rows can be obtained as follows.

b1 =

a1 a2 – a0 a3 a1

,

b2 =

a1 a4 – a0 a5 a1

,

b3 =

a1 a6 – a0 a7 a1

From 2nd & 3rd row , 4th row can be obtained as C1 =

b1 a3 – a1 b2

,

C2 =

b1

b1 a5 – a1 b3 b1 0

This process is to be continued till the coefficient for s is obtained which will be an. From this array stability of system can be predicted.

Routh’s criterion : The necessary & sufficient condition for system to be stable is “ All the terms in the first column of Routh’s array must have same sign. There should not be any sign change in first column of Routh’s array”. If there are sign changes existing then, 3. System is unstable. 4. The number of sign changes equals the number of roots lying in the right half of the s-plane. Examine the stability of given equation using Routh’s method : Ex.2:

s3+6s2 + 11s + 6 =0

Sol:

a0 = 1,

a1 = 6,

a2 =11,

S3

1

11

S2

6

6

S1

11 * 6 – 6 6 6

S0

=10

a3 = 6,

0

As there is no sign change in the first column, system is stable. Ex. 3 Sol:

s3 + 4s2 + s + 16 = 0 a0 =1, a1 = 4, a2 = 1,

a3 = 16

n=3

S3

1

1

S2

+4

16

S1

4 - 16 = 4 +16

S0

-3

0

As there are two sign changes, system is unstable. Number of roots located in the right half of s-plane = number of sign changes = 2. Reference : 1. modern control Engineering by Ogato 2.Automatic control System by B.C.Kuo. 3. Control engineering by U.A.Bakshi.

Date : 5 – 5 –2005. Special Cases of Routh’s criterion : Special case 1 : First element of any of the rows of Routh’s array is zero & same remaining rows contains at least one non-zero element. Effect : The terms in the new row become infinite & Routh’s test fails. e.g. : s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 S5

1

3

2

S4

2

6

1

S3

0

1.5

0

S2

∞

….

…

Special case 1 Routh’s array failed

Following two methods are used to remove above said difficulty. First method : first

Substitute a small positive number ‘ε’ in place of a zero occurred as a element in the row. Complete the array with this number ‘ε’. Then

examine lim Sign change by taking ε

. Consider above Example. 0

S5

1

3

2

S4

2

6

1

S3

ε

1.5

0

S2

6ε - 3 ε

1

0

S1

1.5(6ε - 3)

0 -ε

ε (6ε - 3) ε S0

1

To examine sign change, Lim Lim = 6ε - 3 = 6 3 ε 0 ε ε 0 ε =6-∞ = - ∞ sign is negative. 2 Lim 1.5(6ε – 3) - ε = Lim 9ε - 4.5 - ε2 ε 0 6ε -3 ε 0 6ε - 3 = 0 – 4.5 – 0 0 -3 = Routh’s array is,

+ 1.5

sign is positive

S5

1

3

2

S4

2

6

1

S3

+ε

1.5

0

S2

-

∞

1

0

S1

+1.5

0

0

S0

1

0

0

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 ‘1/Z’ in original equation. Taking L.C.M. rearrange characteristic equation in descending powers of ‘Z’. Then complete the Routh’s array with this new equation in ‘Z’ & examine the stability with this array. F(s) = s5 + 2s4 + 3s3 + 6s2 + 2s + 1 = 0 s=1/Z

Consider Put

∴1 + 2 + 3 +6 +2 + 1=0 Z Z5 Z4 Z3 Z2 Z5 + 2Z4+ 6Z3+3Z2+2Z+ 1 = 0 Z5

1

6

2

Z4

2

3

1

Z3

4.5

1.5

0

Z2

2.33

1

0

Z1

- 0.429

0

Z0

1

As there are two sign changes, system is unstable.

Special case 2 : All the elements of a row in a Routh’s array are zero. Effect : The terms of the next row can not be determined & Routh’s test fails. S5

a

b

c

S4

d

e

f

S3

0

0

0

Row of zeros, special case 2

This indicates no availability of coefficient in that row. Procedure to eliminate this difficulty : 5. Form an equation by using the coefficients of row which is just above the row of zeros. Such an equation is called an Auxillary equation denoted as A(s). For above case such an equation is, A(s) = ds4 + es2 + f Note that the coefficients of any row are corresponding to alternate powers of ‘s’ starting from the power indicated against it.

So ‘d’ is coefficient corresponding to s4 so first term is ds4 of A(s). Next coefficient ‘e’ is corresponding to alternate power of ‘s’ from 4 i.e. s2 Hence the term es 2 & so on. 6. Take the derivative of an auxillary equation with respect to ‘s’. i.e. dA(s) = 4d s3 + 2e s ds

7. Replace row of zeros by the coefficients of dA(s) ds S5

a

b

c

S4

d

e

f

S3

4d 2e 0

8. Complete the array of zeros by the coefficients.

Importance of auxillary equation :

Auxillary equation is always the part of original characteristic equation. This means the roots of the auxillary equation are some of the roots of original characteristics equation. Not only this but roots of auxillary equation are the most dominant roots of the original characteristic equation, from the stability point of view. The stability can be predicted from the roots of A(s)=0 rather than the roots of characteristic equation as the roots of A(s) = 0 are the most dominant from the stability point of view. The remaining roots of the characteristic equation are always in the left half & they do not play any significant role in the stability analysis. e.g. Let F(s) = 0 is the original characteristic equation of say order n = 5. Let A(s) = 0 be the auxillary equation for the system due to occurrence of special case 2 of the order m = 2. Then out of 5 roots of F(s) = 0, the 2 roots which are most dominant (dominant means very close to imaginary axis or on the imaginary axis or in the right half of s-plane) from the stability point of view are the 2roots of A(s) = 0. The remaining 5 –2 = 3 roots are not significant from stability point of view as they will be far away from the imaginary axis in the left half of s-plane. The roots of auxillary equation may be, 5. A pair of real roots of opposite sign i.e.as shown in the fig. 8.10 (a). jω

jω x

x

x

σ σ

σ x

Fig. 8. 10 (b)

Fig 8. 10(a)

6. A pair of roots located on the imaginary axis as shown in the fig. 8.10(b). 7. The non-repeated pairs of roots located on the imaginary axis as shown in the fig.8.10 ©.

jω

jω x x

xx σ

x

σ

x

xx

Fig. 8.10©

Fig. 8.10(d).

8. The repeated pairs of roots located on the imaginary axis as shown in the Fig.8.10 (d). Hence total stability can be determined from the roots of A(s) = 0, which can be out of four types shown above.

Change in criterion of stability in special case 2 : After replacing a row of zeros by the coefficients of dA(s) , complete the Routh’s array. ds st But now, the criterion that, no sign in 1 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 = - σ . i.e. how many roots are located in right half of this line s = - σ, the Routh’s method can be used effectively. To determine this from Routh’s array, shift the axis of s – plane & then apply Routh’s array i.e. substitute s = s 1 - σ, (σ = constant) in characteristic equation. Write polynomial in terms of s1. Complete array from this new equation. The number of sign changes in first column is equal to number of roots those are located to right of the vertical line s = - σ.

Imaginary jω

σ

-σ

0

Determining range of values of K : In practical system, an amplifier of variable gain K is introduced . The closed loop transfer function is C(s) KG(s) = R(s) 1+ KG(s) H(s) Hence the characteristic equation is F(s) = 1+ KG(s) H(s) = 0 So the roots of above equation are dependent on the proper selection of value of ‘K’. So unknown ‘K’ appears in the characteristic equation. In such case Routh’s array is to be constructed in terms of K & then the range of values of K can be obtained in such away that it will not produce any sign change in first column of the Routh’s array. Hence it is possible to obtain the range of values of K for absolute stability of the system using Routh’s criterion. Such a system where stability depends on the condition of parameter K, is called conditionally stable system.

Advantages of Routh’s criterion : Advantages of routh’s array method are : 8. Stability of the system can be judged without actually solving the characteristic equation. 9. No evaluation of determinants, which saves calculation time. 10. For unstable system it gives number of roots of characteristic equation having positive real part. 11. Relative stability of the system can be easily judged. 12. By using the criterion, critical value of system gain can be determined hence frequency of sustained oscillations can be determined. 13. It helps in finding out range of values of K for system stability. 14. It helps in finding out intersection points of roots locus with imaginary axis.

Limitation of Routh’s criterion : 5. 6.

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. 7. It does not suggest methods of stabilizing an unstable system. 8. Applicable only to linear system.

Ex.1. s6 + 4s5 +3s4 – 16s2- 64s – 48 = 0 Find the number of roots of this equation with positive real part, zero real part & negative real part Sol:

S6

1

3

-16

-48

S5

4

0

-64

0

S4

3

0

-48

S3

0

0

0 dA

4

A(s) = 3S – 48 = 0

0

= 12s3

ds S6

1

3

-16

-48

S5

4

0

-64

0

S4

3

0

-48

0

S3

12

0

0

0

S2

(ε )0

-48

0

S1

576 ε -48

0

S0 Lim ε

576 0 ε

0

0 0

+∞

Therefore One sign change & system is unstable. Thus there is one root in R.H.S of the s – plane i.e. with positive real part. Now solve A(s) = 0 for the dominant roots A(s) = 3s4 – 48 =0 Put S2 = Y ∴

3Y2 = 48 S2 = + 4

∴ Y2 =16, S2 = -4

∴ Y = ±√16 = ± 4

S

= ±2

S = ± 2j

So S = ± 2j are the two parts on imaginary axis i.e. with zero real part. Root in R.H.S. indicated by a sign change is S = ± 2 as obtained by solving A(s) = 0. Total there are 6 roots as n = 6. Roots with Positive real part = 1 Roots with zero real part =2 Roots with negative real part = 6 –2 – 1 = 3 Reference : 1. Modren control Engineering byOgato 2. Automatic control system by B.C.kuo 3. Control engineering by U.A. Bakshi.

Date : 9 –5 – 2005. Ex.2 :

For unity feed back system, k G(s) =

, Find range of values of K, marginal value of

K S(1 + 0.4s) ( 1 + 0.25 s)

Sol : Characteristic equation,

1 + G (s) H (s) = 0 & H(s) = 1

K ∴1+

& frequency of sustained oscillations.

=0 s(1 + 0.4s) ( 1 + 0.25s)

s [ 1 + 0.65s + 0.1s2} + K = 0

∴ 0.1s3 + 0.65s2 +s + K = 0

S3

0.1

1

From s0,

K>0

S2

0.65

K

from s1,

S1

0.65 – 0.1K 0.65 K

0

0.65 – 0.1K > 0 ∴ 0.65 > 0.1 K ∴ 6.5 > K

S0

∴ Range of values of K, 0 < K < 6.5

Now marginal value of ‘K’ is that value of ‘K’ for which system becomes marginally stable. For a marginal stable system there must be row of zeros occurring in Routh’s array. So value of ‘K’ which makes any row of Routh array as row of zeros is called marginal value of ‘K’. Now K = 0 makes row of s0 as row of zeros but K = 0 can not be marginal value because for K = 0, constant term in characteristic equation becomes zeros ie one coefficient for s0 vanishes which makes system unstable instead of marginally stable. Hence marginal value of ‘K’ is a value which makes any row other than s0 as row of zeros. ∴ 0.65 – 0.1 K mar = 0 ∴ K mar = 6.5 To find frequency, find out roots of auxiliary equation at marginal value of ‘K’ A(s) = 0.65 s2 + K = 0 ; ∴ 0.65 s2 + 6.5 = 0 Because K = 6.5 2 s = -10 s = ± j 3.162 comparing with s = ± jω ω = frequency of oscillations = 3.162 rad/ sec.

Ex : 3 For a system with characteristic equation F(s) = s5 + s4 + 2s3 + 2s2 + 3s +15 = ∞, examine the stability Solution :

S5

1

2

3

S4

1

2

15

S3

0

-12

0

S5

1

2

3

S4

1

2

15

S3

ε

-12

0

S2 S1 S0

S2 S1

S0

Lim ε 0

(2ε + 12) 15 ε (2ε + 12)( -12 ) – 15ε ε 2ε + 12 ε 15

2ε + 12

0

12 =2+∞ = +∞

=2 + ε

0

ε

Lim ε 0

(2ε + 12)( -12 ) – 15ε ε 2ε + 12 ε

=

Lim ε 0

-24 ε - 144 – 15 ε2 2ε + 12

0 – 144 - 0 =

= - 12 0 + 12

S5

1

2

3 15

S4

1

2

S3

ε

-12

0 There are two sign changes,

so system is unstable.

S2

+∞

15

S1

- 12

0

S0

15

0

Ex : 4 Using Routh Criterion, investigate the stability of a unity feedback system whose open loop transfer function is e -sT G(s) = s(s+1) Sol :

The characteristic equation is 1 + G(s) H(s) = 0

e -sT ∴

1 +

= 0 s(s+1) s2 + s + e –sT = 0

∴

Now e – sT can be Expressed in the series form as

e

–sT

s2 T2 = 1 – sT +

+ …… 2!

Trancating the series & considering only first two terms we get esT = 1 – sT ∴ s2 + s + 1 – sT = 0 ∴ s2 + s ( 1- T ) + 1 = 0 So routh’s array is S2

1

1

S

1-T

0

S0

1

∴ 1 – T > 0 for stability ∴ T<1 This is the required condition for stability of the system.

Ex : 5

Determine the location of roots with respect to s = -2 given that F(s) = s4 + 10 s3 + 36s2 + 70s + 75

Sol :

shift the origen with respect to s = -2

s = s1 – 2 (s′ – 2 ) 4 + 10 (s′ – 2)3 + 36(s′ – 2 )2 + 70 ( s′ –2) + 75 = 0 s′4 + 2s′3 + 0s′2 + 14s′ + 15 = 0

S′4

1

0

15

S′3

2

14

0

15

0

S′2

-7

S′1

18.28

S′ 0

15

0

0

Two sign change, there are two roots to the right of s = -2 & remaining ‘2’ are to the left of the line s = -2. Hence the system is unstable.

12 – 5 - 2005.

Frequency – Domain analysis of control systems: Introduction: It was pointed out earlier that the performance of a feedback control system is more preferably measured by its time domain response characteristics. This is in contrast to the analysis & design of systems in the communication field, where the frequency response is of more importance, since in this case most of the signals to be processed are either sinusoidal or periodic in nature. However analytically, the time response of a control system is usually difficult to determine, especially in the case of high order systems. In the design aspects, there are no unified ways of arriving at a designed system given the time-domain specifications, such as peak overshoot, rise time , delay time & setting time. On the other hand, there is a wealth of graphical methods available in the frequencydomain analysis, all suitable for the analysis & design of linear feedback control systems once the analysis & design are carried out in the frequency domain, time domain behavior of the system can be interpreted based on the relationships that exist between the time-domain & the frequency-domain properties. Therefore, we may consider that the main purpose of conducting control systems analysis & design in frequency domain is merely to use the techniques as a convenient vehicle toward the same objectives as with time-domain methods.

The starting point in frequency-domain analysis is the transfer function. For a single loop feed back system, the closed loop transfer function is written C(s) G(s) M(s) = = R(s) 1+ G(s) H(s)

(1)

Under the sinusoidal steady-state, we get s = jω; then equation(1.0) becomes, C(jω) M(jω) =

G(jω) =

R(jω)

(1.1) 1+ G(jω) H(jω)

The sinusoidal steady-state transfer relation M(jω), which is a complex function of ω , may be expressed in terms of a real & an imaginary part; that is, M((jω) = Re [ M (jω)] + j Im [ M(jω)]

(1.2)

Or , M(jω) can be expressed in terms of its magnitude & phase as Where

M(jω) = M (ω) φm(ω)

(1.3)

G(jω) M((ω) =

(1.4) 1 + G(jω) H(jω)

And G(jω) φm(ω) = 1 + G(jω) H(jω) (1.5) =

G(jω) -

1 + G(jω) H(jω)

since the analysis is now in the frequency domain, some of the terminology used in communication system may be applied to the present control system characterization. For instance, M(ω) of Eq. (1.4) may be regarded as the magnification of the feed back control system is similar to the gain or amplification of an electronic amplifier. In an audio amplifier, for instance, an ideal design criterion is that the amplifier must have a flat gain for all frequencies. Of course, realistically, the design criterion becomes that of having a flat gain in the audio frequency range. In control system the ideal design criterion is similar. If it is desirable to keep the output C(jω) identical to the input R(jω) at all frequencies, M(jω) must be unity for all frequencies. However, from Eq. (1.1) it is apparent that M(jω) can be unity only when G(jω) is infinite, while H(jω) is finite & nonzero. An infinite magnitude for g(jω) is, of course, impossible to achieve in practice, nor would it be desirable, since most control system become unstable when its loop gain becomes very high. Further more, all control system are subject noise. Thus in addition to

responding to the input signal, the system should be able to reject & suppress noise & unwanted signals. This means that the frequency response of a control system should have a cutoff characteristic in general, & sometimes even a band-pass characteristic. The phase characteristics of the frequency response are also of importance. The ideal situation is that the phase must be a linear function of frequency within the frequency range of interest. Figure 1.1 shows the gain & phase characteristics of an ideal low-pass filter, which is impossible to realize physically. Typical gain & phase characteristics of a feedback control system are shown in Fig. 1.2. The fact is that the great majority of control systems have the characteristics of a low-pass filter, so the gain decreases as the frequency increases.

M(ω)

0

ω

1 φm(ω)

0

ωc ω Deg Fig. 1.1. Gain-phase characteristics of an ideal low-pass filter.

0 M(ω)

Mp

φm(ω)

1 0

Deg

ω

Fig.1.2. Typical gain & phase characteristics of a feedback control system.

ω

Frequency-Domain characteristics: If a control system is to be designed or analyzed using frequency-domain techniques, we need a set of specification to describe the system performance. The following frequency-domain specifications are often used in practice. Peak response Mp: The peak response Mp is defined as the maximum value of M (ω) that is given in Eq.(1.4). In general, the magnitude of Mp gives an indication of the relative stability of a feed back control system. Normally, a large Mp corresponds to a large peak overshoot in the step response. For most design problems it is generally accepted that an optimum value Mp of should be somewhere between 1.1 & 1.5. Resonant frequency ωp : The resonant frequency ωp is defined as the frequency at which the peak resonance Mp occurs. Bandwidth: The bandwidth, BW, is defined as the frequency at which the magnitude of M (jω), M(ω), drops at 70.7 percent of its zero-frequency level, or 3 dB down from the zerofrequency gain. In general, the bandwidth of a control system indicates the noise-filtering characteristics of the system. Also, bandwidth gives a measure of the transient response properties, in that a large bandwidth corresponds to a faster rise time, since higherfrequency signals are passed on to the outputs. Conversely, if the bandwidth is small, only signals of relatively low frequencies are passed, & the time response will generally be slow & sluggish. Cutoff rate: Often, bandwidth alone is inadequate in the indication of the characteristics of the system in distinguishing signals from noise. Sometimes it may be necessary to specify the cutoff rate of the frequency response at the higher frequencies. However, in general, a steep cutoff characteristic may be accompanied by a large Mp, which corresponds to a system with a low stability margin. The performance criteria defined above for the frequency-domain analysis are illustrated on the closed-loop frequency response, as shown in Fig. 1.3. There are other criteria defined that may be used to specify the relative stability & performance of a feedback control system. These are defined in the ensuring sections of this chapter. M (ω) Mp

1

Bandwidth

ω

0 ωp

BW

Fig.1.3. Typical magnification curve of a feedback control system.

Mp, ωp & the bandwidth of a second-order system: For a second-order feedback control system, the peak resonance Mp, the resonant frequency ωp, & the bandwidth are all uniquely related to the damping ratio ξ & the natural undamped frequency ωn of the system. Consider the second-order sinusoidal steady-state transfer function of a closed-loop system, C(jω) ω2n M(jω) = = (1.6) 2 2 R(jω) (jω) + 2 ζ ωn (jω) + ω n 1 =

1 + j 2 (ω/ωn) ζ - (ω/ωn)2

We may simplify the last expression by letter u = ω/ ωn. The Eq. (1.6) becomes

1 M(ju) = 1 + j2 u ζ - u

2

( 1.7)

The magnitude & phase of M (jω) are 1 M (ju) = M (u) =

[( 1 – u2 )2 + ( 2 ζ u)2] ½

(1.8)

And M(ju) = φm(u) = - tan

-1

2ζ u (1.9)

1 – u2 The resonant frequency is determined first by taking the derivative of M(u) with respect to u & setting it equal to zero. Thus dM(u)

1 =-

[( 1 – u2 )2 + ( 2ζ u)2] –3/2 ( 4 u3 – 4u + 8uζ2) = 0

(

1.10) du

2

from which

4u3 – 4u + 8uζ2 = 0

(

1.11) The root of Eq. (1.11) are up = 0

(1.12) and

up = √ 1 - 2ζ2

1.13)

(

The solution Eq. (1.12) merely indicates that the slope of the M(ω) versus ω curve is zero at ω = 0; it is not true maximum. The solution of eq. (1.13) gives the resonant frequency, ωp = ωn √ 1 - 2ζ2

(1.14)

Since frequency is a real quantity, Eq. (1.14) is valid only for 1 ≥ 2ζ2 or ζ ≤ 0.707. This means simply that for all values of ξ greater than 0.707, the solution of ωp = 0 becomes the valid one, & Mp = 1. Substituting Eq. (1.13) into Eq. (1.8) & simplifying, we get 1 Mp =

2 √ 1 - ζ2

(1.15)

It is important to note that Mp is a function of ζ only, whereas ωp is a function of ζ & ωn

5

4 3 Mp 2 1

0 0.5

0.707 1.0 1.5 Damping ratio ζ

2.0 1

Fig.1.4 Mp versus-damping ratio for a second – order system, Mp =

1.0 0.8 up = ωp /ωn 0.6

2ζ √ 1 - ζ2

0.4 0.2 0 0.5 0.707 1.0 Damping ratio ζ Fig 1.5. Normalized resonant frequency- versus-damping ratio for a second order system, Up = √ 1 - ζ2 .

Fig.1.4 & 1.5 illustrate the relationship between Mp & ζ , & u = ωp / ωn & ζ, respectively.

Bandwidth : Bandwidth BW of a system is a frequency at which M(ω) drops to 70.7% of its zero frequency level or 3 dB down from the zero frequency gain. Equating the Eq. 1 M(u) =

=

[( 1 –u2)2 + ( 2δu)2] ½ 2 2

2 ½

[( 1 –u ) + ( 2δu) ]

= 0.707 1 = 0.707 =

√2.

Squaring both sides ( 1 –u2)2 + 4δ2u2 = 2 1 + u4 – 2u2 + 4δ2u2 Let u2 = x 1 + x2 – 2x + 4δ2 x

= 2

x2 – 2x + 4δ2 x

= 1

= 2

x2 – x ( 2 - 4δ2 ) =1 x2 – x ( 2 - 4δ2 ) – 1 = 0 a = 1,

b = - ( 2 - 4δ2 ) , c = -1 -b±√ b2 – 4ac x= 2a (2 – 4δ2 ) ±√ (2 – 4δ2 )2 + 4 = 2 (2 – 4δ ) ±√ (4 + 16δ4 – 16δ 2 + 4 2

= 2 (2 – 4δ2 ) ±√ 16 δ4 – 16 δ2 + 8 = 2

2 (1 – 2δ2 ) ± √ 4 + 16δ4 – 16δ 2 + 4 = 2 2 (1 – 2δ2 ) ± 2√2 + 4δ4 – 4δ2 = 2 2 (1 – 2δ2 ) ± 2√2 + 4δ4 – 4δ2 = 2 2 [(1 – 2δ ) ± √2 + 4δ4 – 4δ2] 2

= 2 u2 = x = ω / ωn = u =

(1 – 2δ2 ) ±√ 2 + 4δ4 – 4δ 2 [(1 – 2δ2 ) ±√ (2 + 4δ4 – 4δ 2 ] 1/2

BW = ωn [ ( 1 – 2ζ2 ) + √4ζ4 - 4ζ2 +2]1/2

For the second order system under consideration, we easily establish some simple relationship between the time – domain response & the frequency-domain response of the system. 6. The maximum over shoot of the unit step response in the time domain depends upon ζ only. 7. The response peak of the closed - loop frequency response Mp depends upon ζ only. 8. The rise time increases with ζ , & the bandwidth decreases with the increase of ζ , for a fixed ωn, therefore, bandwidth & rise time are inversely proportional to each other. 9. Bandwidth is directly proportional to ωn. 10. Higher bandwidth corresponds to larger Mp.

Reference: 1. Modern control engineering by Ogata 2. Automatic control system by B.C Kuo.

USHA

Time response analysis of control systems: Introduction: Time is used as an independent variable in most of the control systems. It is important to analyse the response given by the system for the applied excitation, which is function of time. Analysis of response means to see the variation of out put with respect to time. The output behavior with respect to time should be within these specified limits to have satisfactory performance of the systems. The stability analysis lies in the time response analysis that is when the system is stable out put is finite The system stability, system accuracy and complete evaluation is based on the time response analysis on corresponding results.

DEFINITION AND CLASSIFICATION OF TIME RESPONSE Time Response: The response given by the system which is function of the time, to the applied excitation is called time response of a control system.

Practically, output of the system takes some finite time to reach to its final value. This time varies from system to system and is dependent on different factors. The factors like friction mass or inertia of moving elements some nonlenierities present etc. Example: Measuring instruments like Voltmeter, Ammeter. Classification: The time response of a control system is divided into two parts. 1 Transient response ct(t) 2 Steady state response css(t) . . . c(t)=ct(t) +cSS(t) Where c(t)= Time Response Total Response=Zero State Response +Zero Input Response Transient Response: It is defined as the part of the response that goes to zero as time becomes very large. i,e, Lim ct(t)=0 t Æ∞ A system in which the transient response do not decay as time progresses is an Unstable system. C(t) Css(t) Ct(t) Step

ess = study state error

The transient response may be experimental or oscillatory in nature.

2. Steady State Response: O It is defined the part of theTime response which remains after complete transient Transient time Study state response vanishes from the system output. Time Fig-01 c(t) exponentional . isi,e, Lim ct(t)=css(t) t Æ∞ The time domain analysis essentially involves the evaluation of the transient and Steady state response of the control system.

Standard Test Input Signals For the analysis point of view, the signals, which are most commonly used as reference inputs, are defined as standard test inputs. • • •

The performance of a system can be evaluated with respect to these test signals. Based on the information obtained the design of control system is carried out. The commonly used test signals are 1. Step Input signals. 2. Ramp Input Signals. 3. Parabolic Input Signals.

4. Impulse input signal. Details of standard test signals 1. Step input signal (position function) It is the sudden application of the input at a specified time as usual in the figure or instant any us change in the reference input Example :a. If the input is an angular position of a mechanical shaft a step input represent the sudden rotation of a shaft. b. Switching on a constant voltage in an electrical circuit. c. Sudden opening or closing a valve. r(t) A

t

O

When, A = 1, r(t) = u(t) = 1 The step is a signal who’s value changes from 1 value (usually 0) to another level A in Zero time. In the Laplace Transform form R(s) = A / S Mathematically r(t) = u(t) = 1 for t > 0 = 0 for t < 0

2. Ramp Input Signal (Velocity Functions): It is constant rate of change in input that is gradual application of input as shown in fig (2 b). r(t) Ex:- Altitude Control of a Missile Slope = A

t O The ramp is a signal, which starts at a value of zero and increases linearly with time. Mathematically r (t) = At for t ≥ 0 = 0 for t≤ 0. In LT form R(S) = A S2 If A=1, it is called Unit Ramp Input

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

{

t for t ≥ 0 0 for t ≤ 0

In LT form R(S) = A = 1 S2 S2

3. Parabolic Input Signal (Acceleration function): • The input which is one degree faster than a ramp type of input as shown in fig (2 c) or it is an integral of a ramp . • Mathematically a parabolic signal of magnitude A is given by r(t) = A t2 u(t) 2 2 At for t ≥ 0 = 2 0 for t ≤ 0

•

Slope = At

r(t) t

In LT form R(S) = A S3 If A = 1, a unit parabolic function is defined as r(t) = t2 u(t) 2 ie., r(t) =

{

t2 for t ≥ 0 2 0 for t ≤ 0

In LT for R(S) = 1 S3 4. Impulse Input Signal : It is the input applied instantaneously (for short duration of time ) of very high amplitude as shown in fig 2(d) Eg: Sudden shocks i e, HV due lightening or short circuit. It is the pulse whose magnitude is infinite while its width tends to zero. r(t) 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 δ(t) Mathematically it can be expressed as r(t) = A for t = 0 = 0 for t ≠ 0 In LT form R(S) = 1 if A = 1

Standard test Input Signals and its Laplace Transforms. r(t) Unit Step Unit ramp Unit Parabolic Unit Impulse

R(S) 1/S 1/S2 1/S3 1

First order system:The 1st order system is represent by the differential Eq:- a1dc(t )+aoc (t) = bor(t)------ (1) dt Where, e (t) = out put , r(t) = input, a0, a1 & b0 are constants. Dividing Eq:- (1) by

a0, then a1. d c(t ) + c(t) = bo.r (t) a0 dt ao T . d c(t ) + c(t) = Kr (t) ---------------------- (2) dt

Where, T=time const, has the dimensions of time = a1 & K= static sensitivity = b0 a0 a0 Taking for L.T. for the above Eq:-

[ TS+1] C(S) = K.R(S)

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

K . 1+TS

If K=1, Then G(S) =

[ It’s a dimensionless T.F.] ……I

This system represent RC ckt. A simplified bloc diagram is as shown.; R(S)+

1 TS

C(S)

Unit step response of 1st order system:Let a unit step i\p u(t) be applied to a 1st order system, Then, r (t)=u (t) & R(S) = 1 . ---------------(1) S W.K.T. C(S) = G(S). R(S) C(S) = 1 . 1 . = 1 . T . ----------------- (2) 1+TS S S TS+1

Taking inverse L.T. for the above Eq:then, C(t)=u (t) – e –t/T ; t.>0.------------- (3)

slope = 1 . T

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

c (t)

The smaller the time const. T. the faster the system response. The slope of the tangent line at at t= 0 is 1/T.

0.632

Since dc = 1 .e -

dt

T

t/T

1 – e –t/T

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

t T

From Eq:- (4) , We see that the slope of the response curve c(t) decreases monotonically from 1 . at t=0 to zero. At t= ∞ T

Second order system:The 2nd order system is defined as,

a2 d2 c(t) + a1 dc(t) + a0 c(t) = b0 .r(t)-----------------(1) 2 dt dt Where c(t) = o/p & r(t) = I/p -- ing (1) by a0, 2

a2 d c(t) + a1 . 2 a0 dt a0 2

a2 d c(t) + 2 a0 dt

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

2a1 . √ a2 . 2√a0 √ a0 . √ a2

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

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

K . where, S(1+ST) T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so that (a) The peak overshoot of unity step response of the system is reduced from 75% to 25% (b) The damping ratio increases from 0.1 to 0.6. Solution: G(S) =

K . S(1+ST)

Let the value of damping ratio is, when peak overshoot is 75% & when peak overshoot is 25% ξ .

Mp =

∏

⎯e √ 1- ξ2 ξ .

ln 0. 75 =

√ 1- ξ

∏

⇒

ξ .

0.0916 =

√ 1- ξ2

2

(0.0084) (1-ξ2) = ξ2 (1.0084 ξ2) = 0.0084 ξ = 0.091

ξ1 = 0.091 ξ2 = 0.4037

k . S+S T . 1+ K . = 2 S+S T 2

w.k.t. T.F. =

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

T.F. =

=

K . 2 S + S T+K

K/T . S2 + S + K . T T

Comparing with std Eq :Wn =

, 2 ξWn = 1 . T

K . T

Let the value of K = K1 When ξ= ξ1 & K = K2 When ξ = ξ2. Since 2 ξWn = 1 . , ξ = 1 . = 1 . T

2TWn

2√ KT

1 .

∴

ξ1 . = 2 √ K 1T ξ2 1

=

2√ K2T

K2 . K1

K2 . ⇒ K1

0.091 = 0.4037

K2 . = 0.0508 K1

K2 = 0.0508 K1 1 . = 20 0.0508

a) The amplitude K has to be reduced by a factor = b) Let ξ = 0.1 Where gain is K1 and ξ = 0.6 Where gain is K2

∴ 0.1 = 0.6

K2 . = 0.027 ⇒ K2 = 0.027 K1 K1

K2 . K1

The amplitude gain should be reduced by

1 . = 36 0.027

4) Find all the time domain specification for a unity feed back control system whose open loop T.F. is given by G(S) =

25 . S(S+6)

Solution: 25 . G(S) = =

25 . S(S+6)

∴

Wd = Wn √ 1- ξ2 =

=

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

25 . S2 + ( 6S+25 )

W2n = 25 , ⇒ Wn = 5,

tr

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

2 ξWn = 6 ⇒ ξ = 6 . = 0.6 2x5

= 5 √ 1- (0.6)2 = 4

∏ -β , β = tan-1 Wd Wd σ

β = tan-1 ( 4/3 ) = 0.927 rad.

σ = ξWn = 0.6 x 5 = 3

tp = ∏ . = 3.14 = 0.785 sec. Wd

4 ξ .

MP =

=

∏

⎯e √ 1- ξ

⎯e √1-

2

ts = 4 .

0.6 .

for 2% =

ξWn

x3.4 = 9.5%

0.62

4 . = 1.3 ………3sec. 0.6 x 5

5) The closed loop T.F. of a unity feed back control system is given by Determine (1) Damping ratio ξ (2) Natural undamped response frequency Wn. (3) Percent peak over shoot Mp (4) Expression for error resoponse.

5 . C(S) = 2 R(S) S + 4S +5

Solution: 5 . , Wn2 = 5 ⇒ Wn = √5 = 2.236 C(S) = R(S) S2 + 4S +5 2ξWn = 4 ⇒ ξ = MP =

4 . = 0.894. Wd = 1.0018 2 x 2.236 ξ .

⎯e √ 1- ξ

⎯e √ 1-(0.894)

2

W. K.T. C(t) = e-ξWnt

0.894 .

∏ =

Cos Wdtr +

= e-0.894x2.236t

ξ

√ 1-ξ

X 3.14 = 0.19% 2

. sin wdtr 2

Cos 1.0018t + 0.894 . sin 1.0018t

√ 1-(0.894)2

6) A servo mechanism is represent by the Eq:d2θ + 10 dθ = 150E , E = R-θ is the actuating signal calculate the dt value of damping ratio, undamped and damped dt2 frequency of ascillation.

Soutions:-

d2θ + 10 dθ = 15 ( r - θ ) , 2

dt

dt

= 150r – 150θ.

Taking L.T., [S2 + 10S + 150] θ (S) = 150 R (S). θ (S) = 150 . R(S) S2 + 10S + 15O Wn2 = 150 ⇒ Wn = 12.25. ………………………….rad sec .1 2ξWn = 10 ⇒ ξ =

10 2 x 12.25

. = 0.408.

Wd = Wn √ 1 - ξ 2 = 12.25 √ 1- (0.408)2 = 11.18. rad 1sec. 7) Fig shows a mechanical system and the response when 10N of force is applied to the system. Determine the values of M, F, K,. K

x(t)inmt f(t)

0.00193

The T.F. of the mechanical system is , X(S) = 1 . 2 F(S) MS + FS = K f(t) = Md2X + F dX + KX dt2 dt F(S) = (MS2 + FS + K) x (S)

0.02

M F

x 1

2

3

4

5

Given :- F(S) = 10 S.

∴

X(S) =

10 . S(MS2 + FS + K) SX (S) = 10 . MS2 + FS + K

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

10 S

O

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

= 500.) MP = 0.00193 = 0.0965 = 9.62%

(K

0.02 MP = ⎯e

ξ

. Π ⇒ ln 0.0965 = √ 1 - ξ2

ξ . Π

√ 1 - ξ2

0.744 = ξ . ⇒ 0.5539 = ξ2 . √ 1 - ξ2 √ 1 - ξ2 0.5539 – 0.5539 ξ2 = ξ2 ξ = 0.597 = 0.6 tp = Π = Wd Π

3 =

Sx(S) =

Π . Wn√ 1 – ξ2 . ⇒ Wn = 1.31…… rad / Sec. Wn√ 1 – (0.6)2

10/ M . (S2 + F S + K ) M M

Comparing with the std. 2nd order Eq :-, then, Wn2 = K

⇒ Wn =

M F = 2ξWn

K

(1.31)2 = 500 .

M

M = 291.36 kg.

M

F = 2 x 0.6 x 291 x 1.31

M

F = 458.7 N/M/ Sec.

8) Measurements conducted on sever me mechanism show the system response to be c(t) = 1+0.2e-60t – 1.2e-10t , When subjected to a unit step i/p. Obtain the expression for closed loop T.F the damping ratio and undamped natural frequency of oscillation . Solution:

C(t) = 1+0.2e-60t –1.2e-10t Taking L.T., C(S) = 1 . + 0.2 . – 1.2 . S S+60 S+10 C(S) . =

600 / S . S + 70S + + 600 2

∴ R(S) =

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

600 / S . S + 70S + + 600 2

Comparing, Wn2 = 600,

24.4 …..rad / Sec

2 ξWn = 70, ⇒ ξ =

70 . = 1.428 2 x 24.4

9) The C.S. shown in the fig employs proportional plus error rate control. Determine the value of error rate const. Ke, so the damping ratio is 0.6 . Determine the value of settling time, max overshoot and steady state error, if the i/p is unit ramp, what will be the value of steady state error without error rate control.

θR θ(S)

__

10) A feed back system employing o/p damping is as shown in fig. 1) Find the value of K1 & K2 so that closed loop system resembles a 2nd order system with ξ = 0.5 & frequency of damped oscillation 9.5 rad / Sec. 2) With the above value of K1 & K2 find the % overshoot when i/p is step i/p 3) What is the % overshoot when i/p is step i/p, the settling time for 2% tolerance?

R

K1

+ __

1 S(1+S)

K2S

. C

+

K1 C . = . 2 R S + ( 1 + K2 ) S + K1 Wn2 = K1 ⇒ Wn = 2ξWn = 1 + K2 ⇒ ξ = Wd rad/Sec

Wn √ 1 - ξ2

=

√ K1 1 + K2 2√ K1 ⇒

∴ Wn =

9.5

.

10.96

√ 1 – 0.52

K1 = (10.96)2 = 120.34 2ξWn = 1 + K2 , K2 = 9.97 ξ

MP =

⎯e ts =

. Π

= 16.3%

√ 1 - ξ2 4 . = 4 . = 0.729 sec ξWn 0.5 x 10.97

Steady state Error :Steady state errors constitute an extremely important aspect of system performance. The state error is a measure of system accuracy. These errors arise from the nature of i/p’s type of system and from non-linearties of the system components. The steady state performance of a stable control system are generally judged by its steady state error to step, ramp and parabolic i/p.

Consider the system shown in the fig.

G(S) R(S)

E(S)

C(S) H(S)

C(S) = G(S) . …………………………(1) R(S) 1+G(S) . H(S) The closed loop T.F is given by (1). The T.F. b/w the actuating error signal e(t) and the i/p signal r(t) is, E(S) = R(S) – C(S) H(S) = 1 – C(S) . H(S) R(S) R(S) R(S)

= 1 – G(S) . H(S) . 1 + G(S) . H(S) =

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

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

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

∴ E(S)

=

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

.R(S) ……………………….(1)

The steady state error ess may be found by the use of final value theorem and is as follows;

ess

= lt

t→∞

Substituting (1),

e(t) = lt SE(S) S→ O

ess

= lt

S→O

S.R(S) . ……………….(2) 1+G(S) . H(S)

Eq :- (2) Shows that the steady state error depends upon the i/p R(S) and the forward T.F. G(S) and loop T.F G(S) . H(S). The expression for steady state errors for various types of standard test signals are derived below; 1) Steady state error due to step i/p or position error constant (Kp):The steady state error for the step i/p is

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

ess

From Eq:- (2),

= lt

S→O

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

= 1 + lt

S→O

1 G(S). H(S)

.

lt G(S) . H(S) = Kp (S → O )

Where Kp = proportional error constant or position error const.

∴ ess

=

1 . 1 + Kp

(1 + Kp)

ess

⇒

=1

Note :- ess

Kp = 1 - ess

ess

=

R

. for non-unit step i/p

1 + Kp

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

ess of the system with a unit ramp i/p or unit velocity i/p is

r ( t) = t. u(t) , Traking L -T, R(S) = 1/S2 Substituting this to ess Eq:-

∴ ess

= lt S →O

lt

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

S →O

ess

= lt

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

1 .

∴ ess

= 1 .

S + Kv

S →O

Kv

Velocity error is not an error in velocity , but it is an error in position error due to a ramp i/p 3) Steady state error due to parabolic i/p or static acceleration co-efficient (Ka) :The steady state actuating error of the system with a unit parabolic i/p (acceleration i/p) which is defined by r(t) + 1 . t2 Taking L.T. R(S)= 1 . 2 S3

ess

= lt S →O

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

lt

S →O

∴ ess

S →O

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

2

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

= lt S →O

Note :-

lt

ess =

1 . S + Ka 2

=

1 .

Ka

R . for non unit parabolic. Ka

Types of feed back control system :The open loop T.F. of a unity feed back system can be written in two std, forms; 1) Time constant form

and

2) Pole Zero form,

∴ G(S) = K(TaS +1) (TbS +1)………………….. Sn(T1 S+1) (T2S + 1)……………….

Where K = open loop gain. Above Eq:- involves the term Sn in denominator which corresponds to no, of integrations in the system. A system is called Type O, Type1, Type2,……….. if n = 0, 1, 2, ………….. Respectively. The Type no., determines the value of error co-efficients. As the type no., is increased, accuracy is improved; however increasing the type no., aggregates the stability error. A term in the denominator represents the poles at the origin in complex S plane. Hence Index n denotes the multiplicity of the poles at the origin. The steady state errors co-efficient for a given type have definite values. This is illustration as follows.

1) Type – O system :- If, n = 0, the system is called type – 0, system. The steady state error are as follows;

. .

[ .

Let, G(S) = K . S+1

ess

(Position) =

. . . Kp = lt

H(s) = 1]

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

G(S) . H(S) = lt

S →O

ess

S →O

(Velocity) = 1 . = Kv

ess

S+1

(acceleration) = 1 . = Ka

1 .= O

Ka = lt S2 G(S) . H(S) = lt S→O

S→ O

K . = O. S+1

(Position) = 1 . = O 1+∞ Kp = lt

G(S) . H(S) = lt S →O

S→ O

Kv =

lt S→ O

ess

S2

∞

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

2) Type 1 –System :-

ess

S K . = O.

S→ O

S→O

= K

∞

1 .= O

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

K . S+1

1 . 1 + Kp

S

K . S(S+1)

(Velocity) = 1 . K

= K

K . S( S + 1)

= ∞

K . S (S + 1)

ess

(acceleration) = 1 . = 1 . = O O Ka = lt S2 S→O

K

∞

. = O.

S (S + 1)

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

K .= S (S + 1) 2

S→O

. . .

ess

(Position) = 1 . = O

Kv = lt

S

S→O

.

ess

. .

. .

ess

2

(Velocity) = 1 . = O S

S→O

∞ .= ∞

K S (S + 1)

2

Ka = lt

.

∞

∞ K . = K. S (S + 1) 2

(acceleration) = 1 . K

3) Type 3 –System :- Gives Kp = Kv = Ka = ∞ (Onwards)

& ess = O.

K . S (S + 1) 2

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 coefficient while maintaining the transient response within an acceptable limit. PROBLEMS; 1. The unit step response of a system is given by C (t) = 5/2 +5t – 5/2 e-2t. Find the T. F of the system. T/P = r(t) = U (t). Taking L.T, R(s) = 1/S. Response C(t) = 5/2+5t-5/2 e-2t Taking L.T, C(s) = 5 =

5

2

1

+

S

5 1 S2

- 5

1 + 2 S S2

1

2 (S+2)

1 S+2

2

S2+2S+2S+4— 2

-

C(s) = 5 S(S+2)+2(S+2)-S2

= 5 2

S2(S+2)

=

2 10 (S+1)

S2 (S+2) T.F = C (S) = 10 (S+1) S

=

10

(S+1) R (S)

S2(S+2)

S(S+2)

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

S (S+10)

system

is

Find the static error constant and the steady state error of the system when subjected to an i/p given by the polynomial R(t)

= Po +

p1t + P2 t2 2

G(s) =

100

position error co-efficient

S (S+10) KP = lt S

0

S

Similarly KV = lt S

0

0

S

0

S

Given :- r(t)

ess

R1

state

R2

S (S+10)

=

R1

0 Ka = lt S2 G(S)

R2 +

1+Kp

=

R3 +

= 10

Po+P1t +P2 t2

=

steady

0

= 100

S (S+10)

2 Therefore ess

100 x s

100 x s2 0

100

=:

SG(s) = lt

lt S

S (S+10)

G(s) = lt

+

R3 +

Kv

error

Ka

P0 P2

P1 +

+

Ess = 0+0.1 P1 + ∞ = ∞ 3. Determine the error co-efficeint and static G(s) = 1 And H(s) = (S+2)

error

S(S+1) (S+10)

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

for

Kp = lt lt

G(S) H(S) =

(0+2) =∞ 0(0+1) (0+10)

Kv = lt lt

G(S) H(S) =

(0+2) = 1/5 = 0.2 0(0+1) (0+10)

Ka = 0

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

1

1 =

Kv

=5 0.2

Unit parabolic i/p = 1/0 = ∞ 50 4. A feed back C.S is described as S (S+2) (S+5) G(S) = H(S)=1/s.For unit step i/p,cal steady state error constant and errors.

Kp = lt lt

G(S) H(S) =

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

Kv = lt lt

G(S) H(S) =

Ka = lt lt

G(S) H(S) =

=∞

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

=∞

S2 x 50 =

50 =

Lt

S2 (S+2) (S+5)

5

The steady error

state

Ess = lt S 0

S. 1/S 1+50

S 0

2

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

= 0/50 = 0 K 5. A certain feed back C.S is described by following C.S G(S) =

H(S) = 1 S2 (S+20) (S+30)

Determine steady state error co-efficient and also determine the value of K to limit the steady to 10 units due to i/p r(t) = 1 + 10 + t 20/2 t2.

Kp = lt lt

G(S) H(S) =

50 2

S (S+20) (S+30)

=∞

Kv = lt S 0

Ka = lt S

S K 2 S (S+20) (S+30)

0

=∞

S2 K K 2 S (S+20) (S+30)600

Steady state error:-

1

1

+ 1+Kp

Error due to unit step

=0 1+∞

10 Kv

20 Error due to para i/p, , r (t) =

10 +

Error due o r(t) ramp

=0 ∞

40 =

Ka

i/p

20 x 600 =

2Ka

K

i/p

12000 = K

0+0 12000 = 10 = K = 1200 K

First order system:The 1st order system is represent by the differential Eq:- a1dc(t )+aoc (t) = bor(t)------ (1) dt Where, e (t) = out put , r(t) = input, a0, a1 & b0 are constants. Dividing Eq:- (1) by

a0, then a1. d c(t ) + c(t) = bo.r (t) a0 dt ao T . d c(t ) + c(t) = Kr (t) ---------------------- (2) dt

Where, T=time const, has the dimensions of time = a1 & K= static sensitivity = b0 a0 a0 Taking for L.T. for the above Eq:-

[ TS+1] C(S) = K.R(S)

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

K . 1+TS

1 . [ It’s a dimensionless T.F.] 1+TS ……I

If K=1, Then G(S) =

This system represent RC ckt. A simplified bloc diagram is as shown.; R(S)+

1 TS

C(S)

Unit step response of 1st order system:Let a unit step i\p u(t) be applied to a 1st order system, Then, r (t)=u (t) & R(S) = 1 . ---------------(1) S W.K.T. C(S) = G(S). R(S) 1 . T . ----------------- (2) C(S) = 1 . 1 . = 1+TS S S TS+1

Taking inverse L.T. for the above Eq:then, C(t)=u (t) – e –t/T ; t.>0.------------- (3)

slope = 1 . T

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

c (t)

The smaller the time const. T. the faster the system response. The slope of the tangent line at at t= 0 is 1/T.

0.632

Since dc = 1 .e -

dt

T

t/T

1 – e –t/T

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

t T

From Eq:- (4) , We see that the slope of the response curve c(t) decreases monotonically from 1 . at t=0 to zero. At t= ∞ T

Second order system:The 2nd order system is defined as, a2 d2 c(t) + a1 dc(t) + a0 c(t) = b0 .r(t)-----------------(1) 2 dt dt Where c(t) = o/p & r(t) = I/p -- ing (1) by a0, 2

a2 d c(t) + a1 . 2 a0 a0 dt 2

a2 d c(t) + 2 a0 dt

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

2a1 . √ a2 . 2√a0 √ a0 . √ a2

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

Step response of 2nd order system: The T.F. = C(s) = Wn2 Based on value R(s) 32+2 ξWnS+ Wn2 The system may be, 2) Under damped system (0< ξ<1) 3) Critically damped system (ξ=1) 4) Over damped system (ξ>1)

1) Under damped system :- (0< ξ<1) In this case C(s) can be written as R(s) C(s) = R(s)

Wn2 (S+ ξwn + jwd ) (S+ ξwn - jwd )

Where wd = wn 1- ξ2

The Freq. wd is called damped natural frequency

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

W n2 . X R (S) = W n2 . = (S2+2 ξ Wn S+ Wn2) (S2+2 ξ WnS+ Wn2)

C(S) = 1 . S =

1 . S

C(S) =

1 . S

1 . S

S+2ξWn . 2 S +2 ξWnS+ Wn 2

S+ξWn . 2 2 (S+ ξ Wn) + Wd

__ ξ Wn . --------------- (5) 2 2 (S+ ξ Wn + Wd )

S+ξWn . __ ξ . 2 2 2 √ 1- ξ (S+ ξWn) + Wd

Taking ILT, C(t) = 1-e-ξWnt

COS Wdt +

Wd . (S+ ξ Wn) 2 + Wd2

ξ . Sin Wdt ------------ (6) √ 1- ξ2

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

∴

e(t) = r(t) − c(t) . = 1− c(t) =

At t =

e-ξWnt COS Wdt +

ξ . Sin Wdt √ 1- ξ2

--------------------- (7) t > o.

∞, error exists b/w the i/p & o/p.

If the damping ratio ξ= O, the response becomes undamped & oscillations continues indefinitely. The response C(t) for the zero damping case is , c(t) =1-1(COS wnt) =1- COS wnt ; t > O --------------------- (8) From Eq:- (8) , we see that the Wn represents the undamped natural frequency of the system. If the linear system has any amount of damping the undamped natural frequency cannot be observed experimentally. The frequency, which may be observed, is the damped natural frequency.

Wd =wn√ 1− ξ2 This frequency is always lower than the undamped natural frequency. An increase in ξ would reduce the damped natural frequency Wd . If ξ is increased beyond unity, the response over damped & will not oscillate.

Critically damped case:- ( ξ=1). If the two poles of C(S) are nearly equal, the system may be approximated by a R(S) Critically damped one. For a step I/p R(S) = 1/S

∴ C(S)

=

W n2 . 2 2 S +2 ξ Wn S+ Wn

=

1 . S

1 . (S + Wn)

=

1 . S

Wn 2 . ( S + Wn )2S

Taking I.L.T.,

•

1 . S Wn . ( S+ Wn)2

C(t) = 1 – e -Wnt (1+wnt)

Over damped system :- (ξ > 1) If this case, the two poles of C(S) are negative, real and unequal. R(S) For a unit step I/p R(S) = 1/S , then, C(S) =

(S+ ξ Wn + Wn √ ξ - 1

W n2

) ( S+ξ Wn - Wn √ ξ – 1)

2

.

2

Taking ILT, C(t) = 1+

1

.

S√ ξ2 – 1 (ξ+ √ ξ2 – 1) 1 S√ ξ2 – 1 (ξ+ √ ξ2 – 1)

.

⎯e (ξ + √ ξ2 – 1) Wn t.

⎯e (ξ + √ ξ2 – 1) Wn t.

C(t) = 1+

Wn

S √ ξ2 – 1

.

e-S1t . -

S1

e-S2t .

; t>O

S2

Where S1 = (ξ + √ ξ2 – 1) Wn S 2 = ( ξ - √ ξ 2 – 1 ) Wn

Time response (Transient ) Specification (Time domain) Performance :The performance characteristics of a controlled system are specified in terms of the transient response to a unit step i/p since it is easy to generate & is sufficiently drastic. MP

The transient response of a practical C.S often exhibits damped oscillations before reaching steady state. In specifying the transient response characteristic of a C.S to unit step i/p, it is common to specify the following terms. 1) Delay time (td) 2) Rise time (tr)

Response curve 3) Peak time (tp) 4) Max over shoot (Mp) 5) Settling time (ts)

1) Delay time :- (td) It is the time required for the response to reach 50% of its final value for the 1st time.

2) Rise time :- (tr) It is the time required for the response to rise from 10% and 90% or 0% to 100% of its final value. For under damped system, second order system the 0 to 100% rise time is commonly used. For over damped system, the 10 to 90% rise time is commonly used.

3) Peak time :- (tp)

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

4) Maximum over shoot :- (MP) It is the maximum peak value of the response curve measured from unity. The amount of max over shoot directly indicates the relative stability of the system.

5) Settling time :- (ts) It is the time required for the response curve to reach & stay with in a range about the final value of size specified by absolute percentage of the final value (usually 5% to 2%). The settling time is related to the largest time const., of C.S.

Transient response specifications of second order system :W. K.T.

for the second order system,

T.F. = C(S) = Wn 2 . ------------------------------(1) 2 2 R(S) S +2 ξ WnS+ Wn Assuming the system is to be underdamped (ξ< 1) Rise time tr W. K.T. C(tr) = 1- e-ξWnt Cos Wdtr + ξ

. sin wdtr

√ 1-ξ2

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

Cos wdtr + ξ

√ 1-ξ

Cos wdtr + ξ . sin wdtr = tan wdtr = - √ 1-ξ2

√ 1-ξ

2

. sin wdtr 2

=

wd .

ξ

σ jW

Thus, the rise time tr is , 1 . tan-1 - w d = Π- β secs Wd σ wd When β must be in radians. tr =

jWd Wn √ 1-ξ2

β

-σ

Wn σ

ξWn S- Plane

Peak time :- (tp) Peak time can be obtained by differentiating C(t) W.r.t. t and equating that derivative to zero. dc

= O = Sin Wdtp

Wn

. e-ξWntp

dt t = t p √ 1-ξ2 Since the peak time corresponds to the 1st peak over shoot.

∴ Wdtp = Π = tp =

Π . Wd

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

Maximum overshoot :- (MP) The max over shoot occurs at the peak time. i.e. At t = tp = Π . Wd Mp = e –(σ/ Wd) Π or e –(ξ / 1-ξ2) Π

Settling time :- (ts) An approximate value of ts can be obtained for the system O < ξ <1 by using the envelope of the damped sinusoidal waveform. Time constant of a system = T =

1 . ξWn Setting time ts = 4x Time constant. = 4x 1 . for a tolerance band of +/- 2% steady state. ξWn

Delay time :- (td) The easier way to find the delay time is to plot Wn td VS ξ. Then approximate the curve for the range O<ξ< 1 , then the Eq. becomes, Wn td = 1+0.7 ξ

∴ td

PROBLEMS:

= 1+0.7 ξ Wn

(1) Consider the 2nd order control system, where ξ = 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 :- ξ = 0.6, Wn = 5rad /sec, tr = ?, tp = ?, Mp = ?, ts = ? Wd = Wn σ

√ 1- ξ2

= 5 √ 1-(0.6) 2 = 4

= ξWn = 0.6 x 5 = 3.

tr =

∏

-β, β = tan-1 Wd

=

tan-1

Wd

σ

4

= 0.927 rad

3

tr = 3.14 – 0.927 = 0.55sec. 4 tp = ∏ = 3.14 = 0.785 sec. Wd 4 ξ .

MP = MP = ⎯e

ts :-

⎯e √ 1- ξ2 (3/4) x 3.14

∏ =

⎯e

σ / Wd

∏

= 0.094 x 100 = 9.4%

For the 2% criteria., ts = 4 . = 4 . = 1.33 sec. ξWn 0.6x5 For the 5% criteria., ts = 3 = 3 = 1 sec σ 3

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

K . S ( S+10) Determine the value of K so that the system has a damping factors of 0.5 For this value of K determine settling time, peak over shoot & time for peak over shoot for unit step i/p

LCS The error co-efficient Kp, Kv, & Ka describes the ability of the system to eliminate the steady state error therefore they are indicative of steady

state performance. It is generally described to increase the error coefficient while maintaining the transient response within an acceptable limit. PROBLEMS; 2. The unit step response of a system is given by C (t) = 5/2 +5t – 5/2 e-2t. Find the T. F of the system. T/P = r(t) = U (t). Taking L.T, R(s) = 1/S. Response C(t) = 5/2+5t-5/2 e-2t Taking L.T, C(s) = 5 =

1

+

5 1

- 5

1 + 2 S S2

1

-

1 S+2

5 2

S

S2

2 (S+2)

2

S2+2S+2S+4— 2

C(s) = 5 S(S+2)+2(S+2)-S2

= 5 3

S2(S+2)

=

2

S2 (S+2)

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

(S+1)

R (S)

S2(S+2)

=

10

S(S+2)

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

S (S+10)

system

is

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 +

2

p1t + P2 t2

G(s) =

100

position error co-efficient

S (S+10) KP = lt S

0

S

Similarly KV = lt S

0

0

S

0

S

Given :- r(t)

ess

R1 R3

state

R2 +

= 10

S (S+10)

=

0 Ka = lt S2 G(S)

Po+P1t +P2 t2

=

steady

0

= 100

S (S+10)

2 Therefore ess

100 x s

100 x s2 0

100

=:

SG(s) = lt

lt S

S (S+10)

G(s) = lt

R1

R2 +

1+Kp

= +

R3 +

Kv

error

Ka

P0 P2

P1 +

+

Ess = 0+0.1 P1 + ∞ = ∞ 3. Determine the error co-efficeint and static G(s) = 1 And H(s) = (S+2)

error

S(S+1) (S+10)

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

for

Kp = lt lt

G(S) H(S) =

(0+2) =∞ 0(0+1) (0+10)

Kv = lt lt

G(S) H(S) =

(0+2) = 1/5 = 0.2 0(0+1) (0+10)

Ka = 0

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

1

1 =

Kv

=5 0.2

Unit parabolic i/p = 1/0 = ∞ 50 4. A feed back C.S is described as S (S+2) (S+5) G(S) = H(S)=1/s.For unit step i/p,cal steady state error constant and errors.

Kp = lt lt

G(S) H(S) =

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

=∞

Kv = lt lt

G(S) H(S) =

Ka = lt lt

G(S) H(S) =

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

=∞

S2 x 50 =

50 =

Lt

S2 (S+2) (S+5)

5

The steady error

state

Ess = lt S 0

S. 1/S 1+50

S 0

2

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

= 0/50 = 0 K 5. A certain feed back C.S is described by following C.S G(S) =

H(S) = 1 S (S+20) (S+30) 2

Determine steady state error co-efficient and also determine the value of K to limit the steady to 10 units due to i/p r(t) = 1 + 10 + t 20/2 t2.

Kp = lt lt

Kv = lt S 0

G(S) H(S) =

50 S2 (S+20) (S+30)

S K 2 S (S+20) (S+30)

=∞

=∞

Ka = lt S

0

S2 K K S2 (S+20) (S+30)600

Steady state error:-

1

1

+ 1+Kp

Error due to unit step

=0 1+∞

10 Kv

20 Error due to para i/p, , r (t) =

10 +

Error due o r(t) ramp

=0 ∞

40 =

Ka

i/p

20 x 600 =

2Ka

K

i/p

12000 = K

0+0 12000 = 10 = K = 1200 K

K . where, S(1+ST) T&K are constants having + Ve values.By what factor (1) the amplitude gain be reduced so that (a) The peak overshoot of unity step response of the system is reduced from 75% to 25% (b) The damping ratio increases from 0.1 to 0.6.

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

Solution: G(S) =

K . S(1+ST)

Let the value of damping ratio is, when peak overshoot is 75% & when peak overshoot is 25% Mp =

ξ .

∏

⎯e √ 1- ξ2 ξ .

ln 0. 75 =

√ 1- ξ

∏

⇒

ξ .

0.0916 =

√ 1- ξ2

2

(0.0084) (1-ξ2) = ξ2 (1.0084 ξ2) = 0.0084 ξ = 0.091

ξ1 = 0.091 ξ2 = 0.4037

k . S+S T . 1+ K . = 2 S+S T 2

w.k.t. T.F. =

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

T.F. =

=

K . 2 S + S T+K

K/T . S2 + S + K . T T

Comparing with std Eq :Wn =

, 2 ξWn = 1 . T

K . T

Let the value of K = K1 When ξ= ξ1 & K = K2 When ξ = ξ2. Since 2 ξWn = 1 . , ξ = 1 . = 1 . T

2TWn

2√ KT

1 .

∴

ξ1 . = 2 √ K 1T ξ2 1 2√ K2T

0.091 = 0.4037

=

K2 . K1

K2 . ⇒ K1

K2 . = 0.0508 K1

K2 = 0.0508 K1 a) The amplitude K has to be reduced by a factor = b) Let ξ = 0.1 Where gain is K1 and

1 . = 20 0.0508

ξ

= 0.6 Where gain is K2

∴ 0.1 =

K2 . = 0.027 ⇒ K2 = 0.027 K1 K1

K2 . K1

0.6

The amplitude gain should be reduced by

1 . = 36 0.027

4) Find all the time domain specification for a unity feed back control system whose open loop T.F. is given by G(S) =

25 . S(S+6)

Solution: 25 . G(S) = =

∴

25 . S(S+6)

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

25 . S + ( 6S+25 )

Wd = Wn √ 1- ξ2 =

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

2

W2n = 25 , ⇒ Wn = 5,

tr

=

2 ξWn = 6 ⇒ ξ = 6 . = 0.6 2x5

= 5 √ 1- (0.6)2 = 4

∏ -β , β = tan-1 Wd Wd σ

σ = ξWn = 0.6 x 5 = 3

β = tan-1 ( 4/3 ) = 0.927 rad.

tp = ∏ . = 3.14 = 0.785 sec. Wd

4 ξ .

MP =

=

∏

⎯e √ 1- ξ

⎯e √1-

2

ts = 4 . ξWn

for 2% =

0.6 .

x3.4 = 9.5%

0.62

4 . = 1.3 ………3sec. 0.6 x 5

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

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

Solution: C(S) = 5 . , Wn2 = 5 ⇒ Wn = √5 = 2.236 R(S) S2 + 4S +5 2ξWn = 4 ⇒ ξ = MP =

4 . = 0.894. Wd = 1.0018 2 x 2.236 ξ .

0.894 .

∏ =

⎯e √ 1- ξ

⎯e √ 1-(0.894)

2

W. K.T. C(t) = e-ξWnt

Cos Wdtr +

= e-0.894x2.236t

ξ

√ 1-ξ

X 3.14 = 0.19% 2

. sin wdtr 2

Cos 1.0018t + 0.894 . sin 1.0018t

√ 1-(0.894)2

6) A servo mechanism is represent by the Eq:d2θ + 10 dθ = 150E , E = R-θ is the actuating signal calculate the dt2 dt value of damping ratio, undamped and damped frequency of ascillation.

Soutions:-

d2θ + 10 dθ = 15 ( r - θ ) , 2

dt

= 150r – 150θ.

dt

Taking L.T., [S2 + 10S + 150] θ (S) = 150 R (S). θ (S) = 150 . 2 R(S) S + 10S + 15O Wn2 = 150 ⇒ Wn = 12.25. ………………………….rad sec .1 2ξWn = 10 ⇒ ξ =

10 2 x 12.25

. = 0.408.

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

7) Fig shows a mechanical system and the response when 10N of force is applied to the system. Determine the values of M, F, K,. K

x(t)inmt f(t)

0.00193

The T.F. of the mechanical system is , X(S) = 1 . F(S) MS2 + FS = K f(t) = Md2X + F dX + KX dt2 dt 2 F(S) = (MS + FS + K) x (S)

0.02

M F

x 1

2

3

4

5

Given :- F(S) = 10 S.

∴

X(S) =

10 . S(MS + FS + K) SX (S) = 10 . 2 MS + FS + K 2

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

10 S

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

O

= 500.) MP = 0.00193 = 0.0965 = 9.62% 0.02 MP = ⎯e

ξ

. Π ⇒ ln 0.0965 = √ 1 - ξ2

ξ . Π

√ 1 - ξ2

0.744 = ξ . ⇒ 0.5539 = ξ2 . √ 1 - ξ2 √ 1 - ξ2 0.5539 – 0.5539 ξ2 = ξ2 ξ = 0.597 = 0.6 tp = Π = Wd 3 =

Sx(S) =

Π

10/ M

Π . Wn√ 1 – ξ2 . ⇒ Wn = 1.31…… rad / Sec. Wn√ 1 – (0.6)2 .

(K

(S2 + F S + K ) M M Comparing with the std. 2nd order Eq :-, then, Wn2 = K

⇒ Wn =

(1.31)2 = 500 .

K

M F = 2ξWn M

M

M = 291.36 kg.

M

F = 2 x 0.6 x 291 x 1.31 F = 458.7 N/M/ Sec.

9) Measurements conducted on sever me mechanism show the system response to be c(t) = 1+0.2e-60t – 1.2e-10t , When subjected to a unit step i/p. Obtain the expression for closed loop T.F the damping ratio and undamped natural frequency of oscillation . Solution:

C(t) = 1+0.2e-60t –1.2e-10t Taking L.T., C(S) = 1 . + 0.2 . – 1.2 . S S+60 S+10 C(S) . =

600 / S . S2 + 70S + + 600

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

∴ R(S) = 1 . S

600 / S . S2 + 70S + + 600

Comparing, Wn2 = 600,

24.4 …..rad / Sec

2 ξWn = 70, ⇒ ξ =

70 . = 1.428 2 x 24.4

9) The C.S. shown in the fig employs proportional plus error rate control. Determine the value of error rate const. Ke, so the damping ratio is 0.6 . Determine the value of settling time, max overshoot and steady state error, if the i/p is unit ramp, what will be the value of steady state error without error rate control.

θR θ(S)

__

10) A feed back system employing o/p damping is as shown in fig. 4) Find the value of K1 & K2 so that closed loop system resembles a 2nd order system with ξ = 0.5 & frequency of damped oscillation 9.5 rad / Sec. 5) With the above value of K1 & K2 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 2% tolerance?

R

1

K1

+

. C

S(1+S)

__

K2S

C . = K1 . 2 R S + ( 1 + K2 ) S + K1 Wn2 = K1 ⇒ Wn = 2ξWn = 1 + K2 ⇒ ξ = Wd rad/Sec

√ K1 1 + K2 2√ K1

Wn √ 1 - ξ2

=

⇒

∴ Wn =

9.5

.

√ 1 – 0.52

K1 = (10.96)2 = 120.34 2ξWn = 1 + K2 , K2 = 9.97 ξ

MP =

⎯e ts =

. Π

= 16.3%

√ 1 - ξ2 4

. =

4

.

= 0.729 sec

10.96

+

ξWn

0.5 x 10.97

Steady state Error :Steady state errors constitute an extremely important aspect of system performance. The state error is a measure of system accuracy. These errors arise from the nature of i/p’s type of system and from non-linearties of the system components. The steady state performance of a stable control system are generally judged by its steady state error to step, ramp and parabolic i/p.

Consider the system shown in the fig.

G(S) R(S)

E(S)

C(S) H(S)

C(S) = G(S) . …………………………(1) R(S) 1+G(S) . H(S) The closed loop T.F is given by (1). The T.F. b/w the actuating error signal e(t) and the i/p signal r(t) is, E(S) = R(S) – C(S) H(S) = 1 – C(S) . H(S) R(S) R(S) R(S)

= 1 – G(S) . H(S) . 1 + G(S) . H(S) =

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

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

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

∴ E(S)

=

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

.R(S) ……………………….(1)

The steady state error ess may be found by the use of final value theorem and is as follows;

ess

= lt

e(t) = lt SE(S)

t→∞

S→ O

ess

Substituting (1),

= lt

S→O

S.R(S) . ……………….(2) 1+G(S) . H(S)

Eq :- (2) Shows that the steady state error depends upon the i/p R(S) and the forward T.F. G(S) and loop T.F G(S) . H(S). The expression for steady state errors for various types of standard test signals are derived below; 2) Steady state error due to step i/p or position error constant (Kp):The steady state error for the step i/p is I/P r(t) = u(t). Taking L.T., R(S) = 1/S.

ess

From Eq:- (2),

= lt

S→O

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

= 1 + lt

S→O

1 G(S). H(S)

lt G(S) . H(S) = Kp (S → O )

Where Kp = proportional error constant or position error const.

∴ ess

=

1 . 1 + Kp

(1 + Kp)

ess

⇒

=1

Note :- ess

Kp = 1 - ess

ess

=

R 1 + Kp

. for non-unit step i/p

.

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

ess of the system with a unit ramp i/p or unit velocity i/p is

r ( t) = t. u(t) , Traking L -T, R(S) = 1/S2 Substituting this to ess Eq:-

∴ ess

= lt S →O

lt

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

S →O

ess

= lt

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

∴ ess

1 . S + Kv

S →O

= 1 . Kv

Velocity error is not an error in velocity , but it is an error in position error due to a ramp i/p 3) Steady state error due to parabolic i/p or static acceleration co-efficient (Ka) :The steady state actuating error of the system with a unit parabolic i/p (acceleration i/p) which is defined by r(t) + 1 . t2 Taking L.T. R(S)= 1 . 3 S3

ess

= lt S →O

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

lt

S →O

∴ ess

S →O

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

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

= lt S →O

Note :-

lt

ess =

1 . S + Ka 2

=

1 .

Ka

R . for non unit parabolic. Ka

2

2

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

1) Time constant form

2) Pole Zero form,

∴ G(S) = K(TaS +1) (TbS +1)………………….. Sn(T1 S+1) (T2S + 1)……………….

Where K = open loop gain. Above Eq:- involves the term Sn in denominator which corresponds to no, of integrations in the system. A system is called Type O, Type1, Type2,……….. if n = 0, 1, 2, ………….. Respectively. The Type no., determines the value of error co-efficients. As the type no., is increased, accuracy is improved; however increasing the type no., aggregates the stability error. A term in the denominator represents the poles at the origin in complex S plane. Hence Index n denotes the multiplicity of the poles at the origin. The steady state errors co-efficient for a given type have definite values. This is illustration as follows. 2) Type – O system :- If, n = 0, the system is called type – 0, system. The steady state error are as follows;

. .

[ .

Let, G(S) = K . S+1

ess

(Position) =

. . . Kp = lt

H(s) = 1]

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

G(S) . H(S) = lt

S →O

ess

S →O

(Velocity) = 1 . = Kv

1 .= O

Kv = lt G(S) . H(S) = lt S→O

S→ O

∞

S K . = O. S+1

K . S+1

1 . 1 + Kp = K

ess

(acceleration) = 1 . = Ka

Ka = lt S2 G(S) . H(S) = lt S→O

S2

K . = O. S+1

S→ O

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

2) Type 1 –System :-

ess

∞

1 .= O

K . S (S + 1)

(Position) = 1 . = O 1+∞ Kp = lt

G(S) . H(S) = lt S →O

S→ O

Kv =

lt

S

S→ O

(Velocity) = 1 . K

ess

(acceleration) = 1 . = 1 . = O O S→O

K

= ∞

= K

K . S(S+1)

ess

Ka = lt S2

K . S( S + 1)

∞

. = O.

S (S + 1)

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

K .= S2 (S + 1)

S→O

. . .

ess

(Position) = 1 . = O

Kv = lt

S

S→O

.

ess

. . Ka = lt

S→O

∞

∞ .= ∞

K S (S + 1) 2

(Velocity) = 1 . = O 2

S

∞ K . = K. S2 (S + 1)

K . S (S + 1) 2

. . .

ess

(acceleration) = 1 . K

3) Type 3 –System :- Gives Kp = Kv = Ka = ∞ (Onwards)

& ess = O.