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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Objectives Results

Inputs U

Outputs C

( a )

( b )

• 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

CONTROL SYSTEM

CONTROL SYSTEM

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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Speed of the Prime mover Induced Voltage

Output Inputs

Time Cleanliness of clothes

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.

Rotational Generator

Fig 1-2 Rotational Generator

EXAMPLE – 2 washing machine

Most ( but not all ) washing machines are operated in the following manner. After the clothes to be washed have been put into the machine, the soap or detergent, bleach and water are entered in proper amounts as specified by the manufacturer. The washing time is then set on a timer and the washer is energized. When the cycle is completed, the machine shuts itself off. In this example washing time forms input and cleanliness of the clothes is identified as output.

Washing Machine Fig 1-3 Washing Machine

EXAMPLE – 3 WATER TANK LEVEL CONTROL

To understand the concept further it is useful to consider an example let it be desired to maintain the actual water level 'c ' in the tank as close as possible to a desired level ' r '. The desired level will be called the system input, and the actual level the controlled variable or system output. Water flows from the tank via a valve Vo , and enters the tank from a supply via a control valve Vc. The control valve is adjustable manually.

Fig 1-4 b) Open loop control

WATER TANK

Desired Water level r

Valve VC

Valve VO

Water in

Water out

C

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Desired temp. ro c

Electronic thermostat

Controller Forward path element + C

Feed back path element

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)

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.

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

Controlled output C

Desired hot water. temp ro

c

Brain of operator (r-c)

Muscles and Valve

Actual Water temp Co C

+ C

Thermometer

+

Human operator

Thermometer

Hot water

Drain

Cold water

Steam

Steam

-

- +

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

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

Desired temp. ro c

Relay switch

Actual Temp. Co C Furnace Blower House +

Outdoor temp change (disturbance)

Electronic thermostat

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Session 4 -28.03.2005

REQUIREMENTS FOR THE CONTROL SYSTEM

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.

Steady State error

4

0

Floor

Input command

Time Fig. 1-8 Elevator input and output

Transient response

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

Example: 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.

Transfer function

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 + a1 + ………+ an-1 + an C = bo + b1

+ bm-1 + bm r

for n m

Taking Laplace transform on both sides and assuming zero initial conditions,

d2x dt2

+ dx dt

+

d2x

dt2

dx

dt

6 3 X = F

t2 t + + X = F

dnc dtn

dn-1c dtn-1

dm r dtm

dc dt

dm-1 r dtm-1

dr dt

>

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

C(s) bosm + b1sm-1 + …………+ bm

R(s) 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.

=

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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

• 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

a bApplied force F Piston

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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

It is also called ‘mechanical transformer’

Equating the moments of the force

F1 L1 = F2 L 2 F 2 = F1 L1 L2

Rotational mechanical system

The rotational motion of a body may be defined as motion about a fixed axis. The variables generally used to describe the motion of rotation are torque, angular displacement 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

L1 L2F2

F1 effort

Load

Fulcrum

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

T = f ω = f d2 θ Where ω = relative angular velocity between two bodies.

d t2 f = co efficient of viscous friction.

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.

Analogous Systems Consider the mechanical system shown in fig A and the electrical system shown in fig B The differential equation for mechanical system is M + + B + K X = f (t) ---------- 1 The differential equation for electrical system is

d2x

dt2

dx

dt

N1

N2

Driving wheel

Driven wheel

d2q

dt2

d2q

dt2 q

c

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

For mechanical system the differential equation is given by

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

For electrical system

C + + + = I ( t )

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

Table of conversion for force – current analogy Mechanical System Electrical System

Force( torque) Current

Mass( Moment of inertia) Capacitance

Viscous friction coefficient Conductance

Spring constant Inductance Displacement Flux ( angular)

Velocity (angular) Voltage

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

d2x

dt2 dx

dt

d2x

dt2

1

R dΦ

dt2 Φ

L

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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?

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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)

X (output)

U(t) (input) 2. Write the force equation for the mechanical system shown in figure

X (output)

X1

F(t) (input)

M K1 K2

M K B2

B1

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

3. Write the differential equations for the mechanical system shown in figure. 4. Write the modeling equations for the mechanical systems shown in figure. 5. For the systems shown in figure write the differential equations and obtain the transfer functions indicated.

f12 K1 M1

f1 f(t) M2

X2

f2

X1

M

Xi

Xo

M

B force f(t)

XK

Xi Xo Xi Xo

Yk

C

K F

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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

M

B

X K

f(t)

a

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

7. Write the equations of motion in terms of given mechanical quantities. 8. Write the force equations for the mechanical systems shown in figure.

T(t) θ 9. Write the force equation for the mechanical system shown in figure. 10. Write the force equation for the mechanical system shown in figure.

M1 M2 ba

X2K1

K2

Force f

X1 B1

B1

J1

K J1 J2

θ1 θ2

T(t)

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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. 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.

θ1 θ2

J1 J2 J3 K1 K2 K3

Torque T B1 B2 B3

θ3

K

B2

J1

Torque T1, θ1

B1

J2

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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 = KT in N.M /amp Motor back emf constant = KB in V/ rad / Sec Generator gain constant = KG in v/ amp Motor to load gear ratio = N2

N 1 Resistance of the circuit = R in ohms. Inductance of the circuit = L in Henry Moment of inertia of motor = J Viscous friction coefficient = B Field resistance = Rf Field inductance = Lf

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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 = Tm Bm = motion friction constant

θ1 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

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Amplifier gain ‘ Kp ’ in V / V

Motor torque constant ‘ KT ’ in V/ rad

Gear ratio N1 N2

Moment of inertia of load ‘J’

Viscous friction coefficient ‘f’

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

C1

I E0 C2 R Output input

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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) Y(s) G(s

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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

+

-

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Summing point Branch point R(s) C(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) C(s) B(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 .

G(s)

G(s

H(s

+ -

+ -

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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) C(s) G1(s) G2(s)+ -

++

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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) G1(s)G2(s) =

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

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

the disturbance N(s) is given by

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

G2(s) C(s) = [G1(s)R(s) + N(s)] 1 + G1(s)G2(s)H(s) Procedure for drawing block diagram :

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

Ceoei i

Fig. 1.6a

H(s

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

ei = iR + 1/c∫ idt -----------(1) And

eo = 1/c∫ idt ---------(2)

Equation (1) becomes

ei = iR + eo

ei - eo = i --------------(3)

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

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) Eo(S) Ei(s) + I(s) _ Fig1.6(c) Eo(s)

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

Fig1.6(b) _ Fig1.6(d) REFERENCE: 1. MODERN CONTROL ENGINEERING BY OGATA

3. AUTOMATIC CONTROL SYSTEMS BY B.C. KHO

SIGNAL FLOW GRAPH

1/R1/C

1/R 1/C

R

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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.

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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 x4 input node Input node x1 a x2 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

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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 =

x 1 x2 x 1 x2 x3 x1 x 3 x2 = ax1 x3 = abx1

fig.(1) Fig. (2) a x1 x2 a + b x2 = (a + b) s1 = x1 x2

b Fig. (3) x1 x1 a ac c x4 x4 b x3 = bc x2 x2 x4 = (ac)x1 x4 = (bc)x2

Fig. (4) x1 a x2 b x3 x1 ab x3 ab

= = x1 1 - bc x3

bc

Fig. (5) x3 = bx 2 x3 = abx1 + bcx3 x2 = ax1 + cx3 x3 = abx1 + b2cx2

= abx1 + b2c (ax1 + cx3)

= abx1 + b2cax1 + b2c2x3

x3 (1 – b2c2 ) = abx1 (1 + bc)

abx1 (1 + bc) x3 =

1 – b2c2

c

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

ab (1 + bc ) = X1

( 1 + bc) (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 b1 x2 u1 x1 x2 x3 x1 a21 x3 a12 b2 a23 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.

ab x3 = x1 1 - bc

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

a31 a22 a11 a33 b1 a21 a32 1 u1 x1 x2 x3 x3 a12 b2 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 k Mason’s gain formula p =

Pk = path gain or transmittance of kth forward path

= 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 + LbL c - LdLeLf + ………… a b,c d,e,f

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

kth forward path removed

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

OR

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Value of above by eliminating all loop gains & associated products which are

touching to the kth forward

path.

BLOCK DIAGRAM

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

R(s) + G(s) C(s) R(s) 1 E(s) G(s) C(s) - H(s) -H(s)

REFERENCE: 1. MODERN CONTROL ENGINEERING BY OGATA 2. AUTOMATIC CONTROL SYSTEMS B.C. KHO 3. CONTROL SYSTEM BY V.A. BAKSHI & U.A. BAKSHI

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

PROBLEM 1 :

Signal flow graph for the above Block Diagram is

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

The forward path gain is

P1 = G1 G2 G3

In this figure there are three individual loops. The gains of these loops are

R(s) C(s)

G1 G2 G3 + +

H1

+-

H2

- +

-1

H1

(1)

(3)

(2)

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

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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 1 = 1 Overall given c(s) p1 1 R(s)

G1G2G3 c(s) R(s)

1 - G1G2H1 + G2G3H2 + G1G2G3 2. Use the signal flow graph method determine the gain c/R for the block diagram shown.

1 2 R 3 4 5 C

=

=

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Signal flow graph for the fig. Shown

Mason’s formula

No. of forward path p1 = G1G2G3 P2 = G1G2G4

G1

G1 G2 G3

H2

H1

+ + + +

- +

R 1 G1 G2 G3 1 C

H2

- H1

loop (1)

loop (2)

loop (3) G4

∑k Pk k

P =

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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

p1 1 + p2 2

p = G1G2G3 + G1G2G4

= 1 - [G1G2G3H2 – G1G2G3H1 – G1G2G4H1]

G1G2G3 + G1G2G4

c/R = p = 1 - G1G2H2 + G1G2G3H1 + G1G2G4H1

3. The following equation describes a control system. Construct the signal flow

graph for it and obtain transfer function .

Y2 and Y2

U1 for u2 = 0 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 Y1

a11

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

y2

a22 Similarly considering the equation 2 Signal flow graph is Y1 b2 u2 Combining these two signal the graphs, the total signal graph is

a21 Y2

u1 output

b1

Y1 Y2

a22 a11

a21 1

a12

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Transfer function Y2

u1 u2 = 0 Assuming u2 = 0

Y2 U1 U2 = o U2 = 0 Q11 Q22 b1 Q22 1 U1 Y1 Y2

a12

u2

b2

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Signal flow graph is Using Mason’s gain formula 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]

b1 a21 1 U1

a11 a22

a12

Y1 Y2

∑ k Pk k T.F =

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

= 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 P1 1 b1a21 U1

Assuming u1 = o

Then the graph is

Y2 T.F = 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 = 1 – a11

Y2 p1 1 b2 (1-a11) = =

U2 1-a21a12 – a11 – a22 + a11 a22

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

a11 a22

Y1 Y2

a12

a21

u2

b2

= 1– a12a21 –a11 – a22 + a11a22

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Y1 G1 Y2 Y3 G3

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

Consider the equation 2, signal flow graph is

Consider the equation 3, signal flow graph is Y2 Y3 G7 Y4 Combining all the three, complete signal flow graph is

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

Y1 Y2 G2 Y3

G4 G5

G6

Y1

G1 G2

G4

G5

Y4Y3

G7

G3

G6

Y2

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

= 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 = 1 Booth loops are touching 2 = 1 Booth loops are touching 3 = 1– G5 G5 is non touching the path p3 4 = 1 Y4 G1G2G7 + G4G7 + G1G6 (1 – G5 ) + G4G3G6 = Y1 1 – G2G3 – G5 5. From the given block diagram, draw the signal flow the graph and find C(s) R(s)

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Signal Flow graph is

1 2 R(s) C(s)

G4

G1 G2 G3

H1 H2

4 3 5 6 7 + + + + +

- - -

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

No of forward paths = K = 2

Forward path gains are P1 = G1 G2 G3

P2 = G4

Feed back loops are

L1 = - G1G2H1 L4 = - G4

R(s) 1 1 2 1 3 G1 4 G2 5 G3 6 7 1 C(s)

-1

-H2 -H1

G4

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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 , 2 = 1

C(S) p1 1 + p2 2 = R (s) G1G2G3 + G4 = 1+G1G2H1 + G2G3H2 + G1G2G3 + G4 – G2G4 H1 H2 6. 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 R2 Vi C Vo

L

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Laplace transform of the given network is as shown

Vi(s) Equations are I1(s) = …… (1) R1

V1(s) = ……. (2)

CS

I2 (s) = …………. (3)

R2

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

1 Vi R1 I1 V1

- 1 R1

Signal flow graph for equation 2 I1 V1 I2 - 1 Sc Signal flow graph for equation 3

I2(s) Vo (s)

R1 R2V1(s)

1 SC

I1(s)

1 Sc

1 R2

LS

Vi (s) – V1 (s)

[ I1(s) – I2 (s)]

V1(s) - V0 (s)

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

Signal flow graph for equation 4 I2 Vo Total signal flow graph for the network is

Using Mason’s gain formula

∑ PK k = =

No of forward paths = 1

V0 P1 1 Vi =

- 1 R2

V1 I2 Vo

SL

Vi I1 V1 I2 SL Vo

1 R1 1 Sc 1 R2

1 R1 1 Sc 1 R2 - - -

Vo Vi

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

L Forward path gain Ti =

R1R2C Feedback loops are

1 1 SL L1 = - L2 = - L3 = - SR1C SR2C R2 Non touching Loops are L1 & L3

= 1 – [ L1 + L2 + L3 ] + L1L3

1 1 SL L = 1 + + + +

SR1C SR2C R2 R1R2C

All the loops are touching the forward path L R1R2C

L V0 R1R2C SL = = Vi 1 1 SL L SR1R2C + R2 + R1 + S2LR1C + SL 1 + + + + SR1C SR2C R2 R1R2C

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

C R1 Vi R2 Laplace transformed network is 1 1 CS 1 +SR1C

Vo

1 CS

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

R1 x 1/Sc

R1 + 1 /Sc I (S ) = Vi(s) – Vo(s)

Z Z = R1

R1Sc + 1 = Vi(s) – Vo(s) 1 + R1Sc R1 Vo (s) = I (s) R2 From the equation (1) signal flow graph is 1+ R1SC R1 Vi 1 Vo - 1 + R1Sc R1 From the equation (2) signal flow graph is

Z

Vi(s) R2

R1 1 + SR1C

Vo (s) Vi (s) R2

R1

I(s) Vo (s)

Z =

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

I(s) R2 Vo(s) Then combined signal flow graph is 1 + R1Sc R1 1 R2 Vo Vi 1+R1Sc - R1 Using Mason’s gain formula, the number of forward path = 1 ∴ forward pathg gain P1 = R2 1 + R1 Sc R1 Loop gown L1 = - R2 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 R2 (1 + R1 Sc) T(S) = = = = Vi(s) R1 [ R1 + R2 (1 + R1SC)] R1 + R2 (1 + R1 Sc )

R1

8. BLOCK DIAGRAM FOR THE SIGNAL FLOW GRAPH

Procedure :

1. 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.

2. Assume suitable node variables, write equations for every node.

=

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

3. 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.

(1) Obtain the Block diagram from the signal flow graph, shown below

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 3 2 1 X

Y Z W U

- 4 - 3

- 5

S

2S

4

Y

From Z

+-

-

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

From equation (2) Y

For equation (3) For equation (4) W From Z S Combining all the block diagrams, the complete block diagram is

5 From U

3 From W

+

-

Z

2 Z W

++

U

3

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

REFERENCE: 1. CONTROL SYSTEMS BY V.A. BAKSHI & U.A. BAKSHI 2. 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 frequency-domain analysis, all suitable for the analysis & design of linear feedback control systems once the analysis & design are carried out in the frequency domain, time domain behavior of the system can be interpreted based on the relationships that exist between the time-domain & the frequency-domain properties. Therefore, we may consider that the main purpose of conducting control systems analysis & design in frequency domain is merely to use the techniques as a convenient vehicle toward the same objectives as with time-domain methods.

4 S

3

5

2S 3 2 X U Y Z W +

+ +-

- -

www.jntuworld.com

www.jntuworld.com

www.jwjobs.net

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