Lab Manual_Control System Engineering

Gandhinagar Institute of Technology

Department of Electronics and Communication

Control System Engineering (2141004)

Gandhinagar Institute of Technology 2141004 CSE

LIST OF EXPERIMENTS

Sr.

No. Experiment Objective

Page

No. Date Sign

1. To study and perform about open loop control system.

2. To study and perform the basic feedback system.

3. To study and perform about test signal generator.

4. To study and perform about type “0” control system.



7. To digitally simulate the time response characteristics

of a linear system without non linearities and to verify

it manually using MATLAB/SCILAB software.

8. To obtain the bode plot for the given system whose

transfer function is given and to find out whether the

system is stable or not using MATLAB/SCILAB

software.

9. To obtain the Root locus plot for the given system

whose transfer function is given using

MATLAB/SCILAB software.

10. To obtain the Nyquist plot for the given system whose

transfer function is given using MATLAB/SCILAB

software.


Experiment: 1

AIM: To study and perform about open loop & close loop control system.

APPARATUS: Open loop control kit, Multimeter, Patch cords.

THEORY:

Open Loop Control System:

An open-loop controller, also called a non-feedback controller, is a type of controller that

computes its input into a system using only the current state and its model of the system. Fig. 1.1

shows a simple open loop control system. Its operation is very simple, when an input signal directs

the control element to respond, an output will be produced.

Numerous open-loop control system may be found in industry and in home, some example are all

the servomechanisms, most process control system, automatic hot water heaters and automatic

home heating systems, washing machines, light switches, gas ovens, etc.

An open-loop control system we can get control output but it is not controlled fully by itself means

for desired output we have to do manually operation.

Reference input r(t) is applied to the controller which generates the actuating signal u(t). u(t)

actuates the process to give controlled output c(t). The control action has nothing to do with status

of output c(t). Hence system is open loop.

A washing machine is an example of an open loop control system. Fig. 1.2 shows its block

diagram. The input and output of an open loop system are unrelated. An example is that the

operation of a washing machine does not depend on the cleanness of the clothes, but rather on the

preset time. Both the structure and the control process of an open loop control system are very

simple, but the result of the output depends on whether the input signal is appropriate or not.

Controller

Plant Controlled

Reference

Fig. 1.1 Open loop system

u(t)

http://en.wikipedia.org/wiki/Controller_%28control_theory%29

http://en.wikipedia.org/wiki/State_%28controls%29

http://en.wikipedia.org/wiki/Mathematical_model


Fig. 1.2 Block diagram of an open loop system - Washing Machine

More sophisticated example of an open loop control system is the burglar alarm system (Fig1.3).

The function of the sensor is to collect data regarding the concerned house. When the electronic

sensor is triggered off (for example, by the entry of an unauthorized person), it will send a signal

to the receiver. The receiver will then activate the alarm, which will in turn generate an alarm

signal. The alarm signal will not cease until the alarm is stopped manually.

Fig.1.3 Block diagram of an open loop system – Burglar Alarm

An another example is a conveyor system that is required to travel at a constant speed. For a

constant voltage, the conveyor will move at a different speed depending on the load on the motor

(represented here by the weight of objects on the conveyor). In order for the conveyor to run at a

constant speed, the voltage of the motor must be adjusted depending on the load. In this case, a

closed-loop control system would be necessary.

An irrigation sprinkler system, programmed to turn on at set times could be an example of an open-

loop system if it does not measure soil moisture as a form of feedback. Even if rain is pouring

down on the lawn, the sprinkler system would activate on schedule, wasting water.

Stepper motors are often used for open-loop control of position. A stepper motor rotates to one of

a number of fixed positions, according to its internal construction. Sending a stream of electrical

pulses to it causes it to rotate by exactly that many steps, hence the name. Such motors are often

used, together with a simple initial datum sensor (a switch that is activated at the machine's home

position), for the control of simple robotic machines or even the commonplace inkjet printer head.

The drawback of open-loop control of steppers is that if the machine load is too high, or the motor

attempts to move too quickly, then steps may be skipped. The controller has no means of detecting

this and so the machine continues to run slightly out of adjustment, until reset.

So the drawback of an open loop control system is that it is incapable of making automatic

adjustments. Even when the magnitude of the output is too big or too small, the system will not

make the appropriate adjustments. For this reason, an open loop control system is not suitable for

use as a complex control system. Sometimes it may even require monitoring and response from

http://en.wikipedia.org/wiki/Irrigation_sprinkler

http://en.wikipedia.org/wiki/Soil

http://en.wikipedia.org/wiki/Moisture

http://en.wikipedia.org/wiki/Stepper_motor

http://en.wikipedia.org/wiki/Inkjet_printer


the user. For example, when a washing machine finishes cleaning the clothes, the user will need

to check whether the clothes are clean or not; if they are not, they have to be put back into the

machine and washed again.

PROCEDURE :

Connect the jumper between I1 & I2 Connector

Keep both ports in fully anti-clockwise position

Plug in two-pin power chord to mains of 230, 50 CPS.

Put the mains switch in ‘ON’ position Red Neon will glow.

Set the voltage, by voltage adjustment port.

Connect the digital voltmeter at the controlled output.

Press start switch and observe control output. It will increase up to on time of timer.

Press reset switch for set zero control output.

Take the reading for different on time.

OBSERVATION TABLE:

Sr. No. Adjust Input Adjust Disturbance Controlled Output

1

2

3

4

CONCLUSION :

EXERCISES :

Q1 Draw the control block diagram for a house hold automatic coffee machine.


Experiment: 2

AIM: To study and perform the basic feedback system

APPARATUS: Basic feedback control kit, Multimeter, Patch cords.

THEORY:

Close Loop Control System:

“A system in which the control action is somehow dependent on the output is called as closed

loop system.”

Sometimes, we may use the output of the control system to adjust the input signal. This is called

feedback. Feedback is a special feature of a closed loop control system. A closed loop control

system compares the output with the expected result or command status, then it takes appropriate

control actions to adjust the input signal. Therefore, a closed loop system is always equipped with

a sensor, which is used to monitor the output and compare it with the expected result. Fig. 2.1

shows a simple closed loop system. The output signal is fed back to the input to produce a new

output. A well-designed feedback system can often increase the accuracy of the output.

The most elementary feedback control system has three components: a plant (the object to be

controlled, no matter what it is, is always called the plant), a sensor to measure the output of the

plant, and a controller to generate the plant’s input. Usually, actuators are lumped in with the plant.

We begin with the block diagram in Figure. Notice that each of the three components has two

inputs, one internal to the system and one coming from outside, and one output.

Fig. 2.1 Block diagram of a closed loop control system with feedback

These signals have the following interpretations:

r - Reference or command input y - Plant output and measured signal

v - Sensor output n - Sensor noise

u - Actuating signal, plant input d - External disturbance


The three signals coming from outside r, d, and n are called exogenous inputs. In what follows we

shall consider a variety of performance objectives, but they can be summarized by saying that y

should approximate some pr- specified function of r, and it should do so in the presence of the

disturbance d, sensor noise n, with uncertainty in the plant. We may also want to limit the size of

u. Frequently, it makes more sense to describe the performance objective in terms of the

measurement v rather than y, since often the only knowledge of y is obtained from v.

Feedback can be divided into positive feedback and negative feedback. Positive feedback causes

the new output to deviate from the present command status. For example, an amplifier is put next

to a microphone, so the input volume will keep increasing, resulting in a very high output volume.

Negative feedback directs the new output towards the present command status, so as to

allow more sophisticated control. For example, a driver has to steer continuously to keep his car

on the right track.

Most modern appliances and machinery are equipped with closed loop control systems. Examples

include air conditioners, refrigerators, automatic rice cookers, automatic ticketing machines, etc.

An air conditioner, for example, uses a thermostat to detect the temperature and

control the operation of its electrical parts to keep the room temperature at a preset constant. Fig.

2.2 shows the block diagram of the control system of an air conditioner.

Fig. 2.2 Block diagram of the control system of an air conditioner

One advantage of using the closed loop control system is that it is able to adjust its output

automatically by feeding the output signal back to the input. When the load changes, the error

signals generated by the system will adjust the output. However, closed loop control systems are

generally more complicated and thus more expensive to make.

A type of control system that automatically changes the output based on the difference between

the feedback signal to the input signal. In a closed-loop control system, a feedback controller

monitors the output (the vehicle's speed) and adjusts the control input (the throttle) as necessary to

keep the control error to a minimum (to maintain the desired speed). This feedback dynamically

compensates for disturbances to the system, such as changes in slope of the ground or wind speed.

So a feedback control systems is one which the output signal has a direct effect upon the control

system. That is closed loop control system are feedback control system. The actuating error signal


which is the difference between the input signals and the output signal, is fed to the controller so

as to reduce the error and bring the output of the system to a desired value. In other words, the

term closed loop implies the use of feedback action in order to reduce system error.

An advantage of the closed loop control system is that the use of the feedback makes the system

response relatively insensitive to external disturbances and internal variations in the system

parameter, it is thus possible to use relatively in accurate and inexpensive components to obtain

the accurate control of a given plant, whereas this is impossible in the open loop case. However,

on the other hand stability is a major problem in the closed control system, since it may tends to

over correct errors, which may cause oscillation of constant or changing amplitude.

PROCEDURE :

Connect the multimeter first with the reference signal and then with control output.

Connect the other multimeter to the error signal.

Tabulate the reading of the Multimeter.

OBSERVATION TABLE:-

Sr. No. Reference Signal (V) Control Output

Signal (V)

Error Signal

(mV)

1

2

3

4

CONCLUSION :

EXERCISES:

Q 1. Which three parts are commonly involved in a control system? How to categorize this

control system?

Q 2. What is the main difference between an open loop control system and a closed control

system? Use block diagram to elaborate.

Q 3. Explain the positive and negative feedback in a closed loop control system. What are the

differences between them and how they affect the control system?


Experiment: 3

AIM : To study and perform about standard test signal generator

APPARATUS: Test signal generator Kit, Patch cords, Digital multimeter

THEORY:

The control system is that by means of which any quality of interest in a machine, mechanism or

other equipment is maintain or altered in accordance in a desire manner. An automatic control or

regulating system is a feedback control system. In which the reference input or the desire output

is either constant or slowly varying with time and in which the primary task is to maintain the

actual output at the desire value at the presence of disturbances.

The characteristics of actual input signals are a sudden shock, a sudden change, a constant velocity,

and constant acceleration. The dynamic behavior of a system is therefore judged and compared

under application of standard test signals – an impulse, a step, a constant velocity, and constant

acceleration. Another standard signal of great importance is a sinusoidal signal.

There are many examples of automatic regulating system some of which are the automatic controls

of pressure, temperature, flow and of electric quantities such as voltage, current and frequency.

System Error is a major of the accuracy of a control system. Errors in a control system can be

attributed to many factors. Changes in the reference input will cause unavoidable errors during

transient periods and may also cause steady state errors.

Steady state errors are a major of the accuracy of a control system. The steady state performance

of a control system is generally judged by the steady state error due to step, ramp or acceleration

inputs, which are considered to be the standard test signals. Any physical control system inherently

suffered steady state error in response to certain types of inputs, but the same system may exhibit

non zero steady state error to a ramp input. Whether or not a given system will exhibit steady state

error for a given type is input depends upon the type of open loop transfer function of the system.

Impulse signal

The impulse signal imitates the sudden shock characteristic of actual input signal.

If A=1, the impulse signal is called unit impulse signal.

0 0

0 )(

t

tAt


Fig 3.1 Impulse Signal

Step signal

The step signal imitate the sudden change characteristic of actual input signal.

If A=1, the step signal is called unit step signal.

Fig 3.2 Step Signal

Ramp signal

The ramp signal imitates the constant velocity characteristic of actual input signal.

If A=1, the ramp signal is called unit ramp signal.

0 0

0 )(

t

tAtu

0 0

0 )(

t

tAttr


Fig 3.3 Ramp Signal

Fig 3.4 Ramp Signal with slope A Fig 3.5 Unit Ramp Signal

Parabolic signal

The parabolic signal imitates the constant acceleration characteristic of actual input signal.

If A=1, the parabolic signal is called unit parabolic signal.

Fig 3.6 Parabolic Signal

0 0

0 2)(

2

t

tAt

tp


Fig 3.7 Parabolic Signal with slope A Fig 3.8 Unit Parabolic Signal

Relation between standard Test Signals

Laplace Transform of Standard Test Signals

• Impulse

• Step

0 0

0 )(

t

tAt

AstL )()}({

0 0

0 )(

t

tAtu

S

AsUtuL )()}({


• Ramp

• Parabolic

PROCEDURE:

Plug in two pin power chord to main supply of 250volt, 50cps.

Keep the output selector switch in position ‘1’

Connect digital voltmeter across output terminal the DVM should be in DC volt range.

Adjust the step level Adjacent control the DC voltage should be variable from 0 to 10 volt

DC

Now keep the selector switch in position ‘Z’ ramp LED will glow

Keep the “DISCHARGEW RESET” switch in discharge position. Now bring the switch in

“RESTART MODE TABULATE” your results as below with respect to watch.

OBSERVATION TABLE:

(Step signal, Ramp signal and Parabolic signal)

Time (sec) Output voltage

V(Step)

Output voltage

V(Ramp)

Output voltage

V(Parabolic)

0

20

40

60

80

100

CONCLUSION:

0 0

0 )(

t

tAttr

2)()}({

s

AsRtrL

0 0

0 2)(

2

t

tAt

tp

3

2)()}({

S

AsPtpL


Experiment: 4

AIM: To study and perform about TYPE “0” control system APPARATUS: Test signal generator Kit, Type “0” control system Kit, Patch cords, Digital

multimeter

THEORY:

Classification of control system:

Control system may be classified according to the ability to follow step input, ramp parabolic

inputs etc. This is a reasonable classification scheme because actual inputs may frequently be

considered as combination of steady state error a due to these individual input are indicative of the

goodness of the system.

Consider the following open loop transfer function.

G(s)H(s) = 𝐾(𝑇𝑎𝑠+1)(𝑇𝑏𝑠+2)…..𝑇𝑚𝑠+1

𝑆(𝑇1𝑆+1)(𝑇2𝑆+2)….(𝑇𝑛𝑆+1)

It involves the terms S in the denominator, representing a pole of multiplicity N at the origin. The

present classification is based on the number of integrations indicated by the open loop transfer

function. A system is called type 0,type 1,type2 if N=0,N=1,N=2…. etc.

The number of poles at the origin of the loop gain transfer function (i.e. the number of

integrators) defines the system's type number.

Respectively note that this classification is different from that of the order of the system. As the

type no. is increased accuracy is improved, compromise between steady state accuracy and the

relative stability is necessary. In practice it is rather exceptional to have type 3 or higher systems

because we find it to generally difficult to design stable system having more than two integration

in the feed forward path.

Type ‘0’ Control System

The block of type 0 control system is engraved on the front panel.

Here G(s) = K, H(s) = 1

Hence open loop transfer function is

G(s) H(s) = K . 1 =K

As index of‘s’ term in denominator is zero. we can say this is a type ‘0’control system.

We will now study steady state response of the type ‘0’ control system.


Steady state error ‘ess’ of the control system is defined as follows:-

ess = lim𝑡→∞

∫ 𝑒(𝑡)∞

0

Or in Laplace form , it is defined as

ess = lim𝑠→∞

∫ 𝑒(𝑠)∞

0

Where e(t) and e(s) are error signals.

Steady State Errors

A unity feedback closed loop control system is shown in Fig.4.1

Fig 4.1 Unity feedback control system

The closed loop transfer function is

The transfer function between the error signal e(t) and input signal r(t) is

Which can be rearranged as

The final value theorem provides a convenient way to find the steady state error

Three types of static error constants are

1) Static position error constant Kp due to unit step input.

2) Static velocity error constant Kv due to unit ramp input.

3) Static acceleration error constant Ka due to unit parabolic input

which indicate the figures of merit of control systems.

Static Position Error Constant

The steady state error of the system for a unit step input is


The static position error constant Kp is defined by

Thus, the steady state error in terms of the static position error constant Kp is given by

For a type 0 system,

Type 0 system is incapable of following a ramp input in the steady state.

Static Velocity Error Constant

The steady state error of the system with a unit ramp input is given by

The static velocity error constant Kv is defined by

Thus, the steady state error in terms of the static velocity error constant Kv is

given by

The term velocity error is used here to express the steady state error for a ramp input. The

velocity error is an error in position due to ramp input.


Static Acceleration Error Constant

The steady state error of the system with a unit parabolic input is given by

The static acceleration error constant Ka is defined by


Thus, the steady state error in terms of the static acceleration error constant Ka is given by


Type 0 and systems are incapable of following a parabolic in the steady state.

PROCEDURE:

Connect test signal generator of type “0” system and then the output of the system to

multimeter

Apply each signal one by one to the “0” order kit and fine change in voltage with respect

to time.

Apply same procedure for each practical for all three test signal to ramp, step and

parabolic

OBSERVATION TABLE I:

Step Input

Sr. No. Input

R(s) Volt

Output

C(s) Volt

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 1

2 2

3 3

4 4

5 5

6 6

7 7

8 8


OBSERVATION TABLE II:

Ramp Input

Sr. No. Input

R(s) mV

Output

C(s) mV

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 25

2 50

3 75

4 100

5 125

6 150

7 175

8 200

OBSERVATION TABLE III:

Parabolic Input

Sr. No. Input

R(s) mV

Output

C(s) mV

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 100

2 200

3 300

4 400

5 500

6 600

7 700

8 800

CONCLUSION:

EXERCISES :

Q 1. For a negative unity feedback system determine the steady state error due to unit step,

unit ramp and unit parabolic input of the following system :


Experiment: 5

AIM: To study and perform about TYPE “1” control system

APPARATUS: Test signal generator Kit, Type “1” control system Kit, Patch cords, Digital

multimeter

THEORY:


The block of type ‘1’ control system is engraved on the front panel.

Here G(s) =1

𝑇𝑠, H(s) = 1


G(s) H(s) = 1

𝑇𝑠 . 1 =

1

𝑇𝑠

As index of ‘s’ term in denominator is zero, we can say this is a type ‘1’control system. We will

now study steady state response of the type ‘0’ control system.


ess = lim𝑡→∞

∫ 𝑒(𝑡)∞

0


ess = lim𝑠→∞

∫ 𝑒(𝑠)∞

0

where e(t) and e(s) are error signals.


For a type 1 system

If a zero steady state error for a step input is desired, the type of the system must be one or

higher.



For a type 1 system

Type 1 system with unity feedback can follow the ramp input with a finite error.


For a type 1 system

Type 1 systems are incapable of following a parabolic in the steady state.

PROCEDURE :

Connect generator kit of various input signals and various input no of type system board.

After giving input obtain output properly through multimeter.

Obtain output of different type of input signal


Step Input

Sr. No. Input

R(s) Volt

Output

C(s) Volt

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 1

2 2

3 3

4 4

5 5

6 6

7 7

8 8



Ramp Input

Sr. No. Input

R(s) mV

Output

C(s) mV

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 25

2 50

3 75

4 100

5 125

6 150

7 175

8 200


Parabolic Input

Sr. No. Input

R(s) mV

Output

C(s) mV

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 100

2 200

3 300

4 400

5 500

6 600

7 700

8 800

CONCLUSION:

EXERCISES :

Q 1. Given the control system shown in the figure below, find the value of K so that there is 10% error in

the steady state.


Experiment: 6

AIM: To study and perform about TYPE “2” control system

APPARATUS: Test signal generator Kit, Type “2” control system Kit, Patch cords,

Digital multimeter

THEORY:


The block of type 0 control system is engraved on the front panel.

Here G(s) = 1

𝑇1𝑆.

1

𝑇2𝑆, H(s)=1


G(s) H(s) = 1

𝑇1𝑆.

1

𝑇2𝑆 . 1 =

1

𝑇1𝑆.

1

𝑇2𝑆

As index of ‘s’ term in denominator is zero. we can say this is a type ‘2’control system.

We will now study steady state response of the type ‘0’ control system.


ess = lim𝑡→∞

∫ 𝑒(𝑡)∞

0


ess = lim𝑠→∞

∫ 𝑒(𝑠)∞

0

where e(t) and e(s) are error signals


For a type 2 system


If a zero steady state error for a step input is desired, the type of the system must be one or

higher.


For a type 2 system

Type 2 and higher order systems can follow a ramp input with zero steady state error.


For a type 2 system

Type 2 system with unity feedback can follow the parabolic input with a finite error.

Table 1 shows the summary of error constants and steady state errors for unit step, unit ramp and

unit parabolic inputs to a unity feedback loop.

Table 1: Summary of steady state errors for unity feedback systems


Fig 6.1 Steady state errors in closed loop systems

PROCEDURE :

Connect test signal generator output at 2 order systems input and then system kit output to

multimeter.

Apply step input and take reading at from multimeter with respect to time.

Apply each signal one by one and find change in output voltage from 2-order kit with

respect to time.



Step Input

Sr. No. Input

R(s) Volt

Output

C(s) Volt

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 1

2 2

3 3

4 4

5 5

6 6

7 7

8 8


Ramp Input

Sr. No. Input

R(s) mV

Output

C(s) mV

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 25

2 50

3 75

4 100

5 125

6 150

7 175

8 200


Parabolic Input

Sr. No. Input

R(s) mV

Output

C(s) mV

Error Signal

R(s) – C(s)

Error/step

[R(s) – C(s)]/R(s)

1 100

2 200

3 300

4 400

5 500

6 600

7 700

8 800


CONCLUSION :

EXERCISES :

Q 1. For a negative unity feedback system determine the steady state error due to unit step,

unit ramp and unit parabolic input of the following system :


Experiment: 7

AIM: To digitally simulate the time response characteristics of a linear system without

non- linearities and to verify it manually.

APPARATUS: A PC with MATLAB/SCILAB software

THEORY:

The time response characteristics of control systems are specified in terms of time domain

specifications. Systems with energy storage elements cannot respond instantaneously and will

exhibit transient responses, whenever they are subjected to inputs or disturbances.

The desired performance characteristics of a system of any order may be specified in terms of

transient response to a unit step input signal. The transient response characteristics of a control

system to a unit step input is specified in terms of the following time domain specifications:

1. Delay time td

2. Rise time tr

3. Peak time tp

4. Maximum peak overshoot Mp

5. Settling time ts

PROCEDURE:

For RL Circuit:

1. Derive the transfer function of a RL series circuit.

2. Assume R= 1 Ohms L = 0. 1 H. Find the step response theoretically and plot it on a

graph sheet.

For RLC Circuit:

1. Derive the transfer function of a RLC series circuit.

2. Assume R= 1 Ohms, L = 0. 1 H and C = 1 micro Farad. Find the step response

theoretically and plot it on a graph sheet.

MANUAL CALCULATIONS:

Unit step response of the given RL series circuit:


Unit step response of the given RLC series circuit:

MATLAB/SCILAB PROGRAM :

RESPONSE CURVE:


CONCLUSION :

EXERCISES :

Q 1. Differentiate real time systems and simulated systems.

Q 2. Give two examples for first order system.

Q 3. What is meant by Delay time ?

Q 4. What is meant by Rise time?

Q 5. What is meant by over damping?

Q 6. What is meant by peak time?

Q 7. What is meant by setting time?


Experiment: 8

AIM : To obtain the bode plot for the given system whose transfer function is given as

𝐺(𝑠) =242(𝑠+5)

𝑠(𝑠+1)(𝑠2+5𝑠+121)

and to find out whether the system is stable or not.


THEORY:

A Linear Time-Invariant Systems is stable if the following two notions of system stability are

satisfied –

1. When the system is excited by Bounded input, the output is also a Bounded output.

2. In the absence of the input, the output tends towards zero, irrespective of the initial

conditions.

The following observations are general considerations regarding system stability and are-

1. If all the roots of the characteristic equation have negative real parts, then the impulse

response is bounded and eventually decreases to zero, then system is stable.

2. If any root of the characteristic equation has a positive real part, then system is unstable.

3. If the characteristic equation has repeated roots on the jω-axis, then system is unstable.

4. If are more non-repeated roots of the characteristic equation on the jω-axis, then system

is unstable.

BODE PLOT :

Consider a Single-Input Single-Output system with transfer function

C(s) b0 sm + b1 sm-1 + ……+ bm

=

R(s) a0 sn + a1sn-1 + ……+an

Where m < n.

Rule 1 A system is stable if the phase lag is less than 180˚ at the frequency for

which the gain is unity (one).

Rule 2 A system is stable if the gain is less than one (unity) at the frequency for

which the phase lag is 180˚.


The application of these rules to an actual process requires evaluation of the gain and phase shift

of the system for all frequencies to see if rules 1 and 2 are satisfied. This is obtained by plotting

the gain and phase versus frequency. This plot is called BODE PLOT. The gain obtained here is

open loop gain. The exact terminology is in terms of a Gain Margin and Phase Margin from

the limiting values quoted.

If the phase lag is less than 140˚ at the unity gain frequency, the system is stable.

This then, is a 40˚ Phase Margin from the limiting values of 180˚.

If the gain is 5dB below unity (or a gain of about 0.56) when the phase lag is 180˚,

the system is stable. This is 5dB Gain Margin.

PROCEDURE:

Step 1: Write a program to obtain the Bode plot for the given system.

Step 2: Assess the stability of given system using the plot obtained.


1. The sinusoidal transfer function G (jω) is obtained by replacing s by jω in the given s

domain transfer function

𝐺(𝑗𝜔) =242(𝑗𝜔 + 5)

𝑗𝜔(𝑗𝜔 + 1)(𝑗𝜔2 + 5𝑗𝜔 + 121)

On comparing the quadratic factor of G(s) with standard form of quadratic factor, ζ and

ωn can be calculated.

2. Corner Frequency Table:

3. Magnitude Plots:


4. Phase Plots:


OUTPUT( FROM SIMULATION):

OUTPUT(FROM GRAPH):


CONCLUSION :

EXERCISES :

Q 1. What is polar plot?

Q 2. What is frequency response?

Q 3. Define stability of Linear Time Invariant System.

Q 4. Give the stability conditions of system using Pole-Zero plot.

Q 5. Define Limits of stability.

Q 6. Define Phase margin and Gain margin.


Experiment: 9

AIM : To obtain the Root locus plot for the given system whose transfer function is given as

𝐺(𝑆) =𝐾

𝑆2 + 3𝑆 + 11.25


THEORY:

ROOT LOCUS PLOT :

The characteristic of the transient response of a closed-loop system is related to the location of the

closed loop poles. If the system has a variable loop gain, then the location of the closed-loop poles

depend on the value of the loop gain chosen. A simple technique known as “Root Locus

Technique” used for studying linear control systems in the investigation of the trajectories of the

roots of the characteristic equation. This technique provides a graphical method of plotting the

locus of the roots in the s-plane as a given system parameter is varied over the complete range of

values(may be from zero to infinity). The roots corresponding to a particular value of the system

parameter can then be located on the locus or the value of the parameter for a desired root location

can be determined form the locus. The root locus is a powerful technique as it brings into focus

the complete dynamic response of the system . The root locus also provides a measure of sensitivity

of roots to the variation in the parameter being considered. This technique is applicable to both

single as well as multiple-loop systems.

PROCEDURE:

Step 1: Write a program to obtain the Root Locus plot for the given system.






OUTPUT(FROM GRAPH):


CONCLUSION :

EXERCISES :

Q 1. What do you mean by Root-Loci?

Q 2. Define root locus technique.

Q 3. Write the condition for magnitude & angle criterion?

Q 4. Define the break way point ?

Q 5. What is bandwidth?


Experiment: 10

AIM : To obtain the Nyquist plot for the given system whose transfer function is given as

𝐺(𝑆) =50

(𝑆 + 4)(𝑆2 + 3𝑆 + 3)


THEORY:

POLAR PLOTS OR NYQUIST PLOTS:

The sinusoidal transfer function G(jω) is a complex function is given by

G(jω) = Re[ G(jω)] + j Im[G(jω)] or

G(jω) = │G(jω) │ ∟G(jω) = M ∟Φ -----------(1)

From equation (1), it is seen that G(jω) may be represented as a phasor of magnitude M

and phase angle Φ. As the input frequency varies from 0 to ∞, the magnitude M and phase angle

Φ changes and hence the tip of the phasor G(jω) traces a locus in the complex plane. The locus

thus obtained is known as POLAR PLOT.

The major advantage of the polar plot lies in stability study of systems. Nyquist related

the stability of a system to the form of these plots. Polar plots are referred as NYQUIST PLOTS.

NYQUIST stability criterion of determining the stability of a closed loop system by investigating

the properties of the frequency domain plot of the loop transfer function G(s) H(s).

Nyquist stability criterion provides the information on the absolute stability of a control

system as similar to Routh- Hurwitz criterion. Not only giving the absolute stability, but indicates

“Degree of Stability” i.e “Relative Stability” of a stable system and the degree of instability of an

unstable system and indicates how the system stability can be improved. The Nyquist stability

citerion is based on a Cauchy’s Residue Theorem of complex variables which is referred to as the

“principle of argument”.

Let Q(s) be a single –valued function that has a finite number of poles in the s-plane.

Suppose that an arbitrary closed path Гq is chosen in the s-plane so that the path does not go

through any one of the poles or zeros of Q(s); the corresponding Гq locus mapped in the Q(s) plane

will encircle the origin as many times as the difference between the number of the zeros and the

number of poles of Q(s) that are encircled by the s-plane locus Гq. The principle of argument is

given by

N= Z - P

Where N – number of encirclemnts of the origin made by the Q(s) –plane locus Гq.


Z – number of zeros of Q(s) encircled by the s-plane locus Гq in the s-plane.

P - number of poles of Q(s) encircled by the s-plane locus Гq in the s-plane.

PROCEDURE:

Step 1: Write a program to obtain the Nyquist plot for the given system.






OUTPUT(FROM GRAPH):

CONCLUSION :

Documents

Lab Manual_Control System Engineering