Control System Lab Manual

Control Systems Laboratory Manual / II EEE, IV SEM

VELS UNIVERSITY

SCHOOL OF ENGINEERING

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

LABORATORY MANUAL

CLASS : II YEAR EEE

SEMESTER : IV (DEC 2010 – MAY 2011)

SUBJECT : CONTROL SYSTEMS LABORATORY

STAFF IN-CHARGE : V.Arivumani Assistant Professor

EEE Department

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

Prepared by V.Arivumani, AP/EEE, Vels University, Chennai 1


CONTROL SYSTEMS LABORATORY MANUAL

NAME :

CLASS :

SEMESTER :

ROLL NUMBER :

REGISTER NUMBER :



INDEX

S. No. Date Title of Experiment Page

No. Marks Signature



SYLLABUS

EE2257 CONTROL SYSTEM LABORATORY 0 0 3 2

1. Determination of transfer function of DC Servomotor2. Determination of transfer function of AC Servomotor.3. Analog simulation of Type - 0 and Type – 1 systems4. Determination of transfer function of DC Generator5. Determination of transfer function of DC Motor6. Stability analysis of linear systems7. DC and AC position control systems8. Stepper motor control system9. Digital simulation of first order systems10.Digital simulation of second order systems

P = 45 Total = 45DETAILED SYLLABUS

1. Determination of Transfer Function Parameters of a DC Servo Motor

AimTo derive the transfer function of the given D.C Servomotor and experimentally determine the transfer function parameters

Exercise1. Derive the transfer function from basic principles for a separately excited DC

motor.2. Determine the armature and field parameters by conducting suitable experiments.3. Determine the mechanical parameter by conducting suitable experiments.4. Plot the frequency response.

Equipment 1. DC servo motor : field separately excited – loading facility – variable voltage source - 1 No2. Tachometer : 1 No3. Multimeter : 2 Nos4. Stop watch : 1 No

2. Determination of Transfer Function Parameters of AC Servo Motor

Aim



To derive the transfer function of the given A.C Servo Motor and experimentally determine the transfer function parameters

Exercise1. Derive the transfer function of the AC Servo Motor from basic Principles.2. Obtain the D.C gain by operating at rated speed.3. Determine the time constant (mechanical)4. Plot the frequency response

Equipment 1. AC Servo Motor : Minimum of 100w – necessary sources for main winding and control winding – 1 No 2. Tachometer : 1 No 3. Stopwatch : 1 No 4. Voltmeter : 1 No

3. Analog Simulation of Type-0 And Type-1 System

AimTo simulate the time response characteristics of I order and II order, type 0 and type-1 systems.

Exercise1. Obtain the time response characteristics of type – 0 and type-1, I order and II

order systems mathematically.2. Simulate practically the time response characteristics using analog rigged up

modules.3. Identify the real time system with similar characteristics.

Equipment1. Rigged up models of type-0 and type-1 system using analog components.2. Variable frequency square wave generator and a normal CRO - 1 No (or) DC source and storage Oscilloscope - 1 No

4. Determination of Transfer function of DC Generator

AimTo determine the transfer function of DC generator

Exercise1. Obtain the transfer function of DC generator by calculating and gain

Equipment1. DC Generator2. Tachometer



3. Various meters 4. Stop watch

5. Determination of Transfer function of DC Motor

AimTo determine the transfer function of DC motor

Exercise1. Obtain the transfer function of DC motor by calculating and gain

Equipment1. DC Motor2. Tachometer 3. Various meters 4. Stop watch

6. Stability Analysis of Linear Systems

AimTo analyse the stability of linear systems using Bode / Root locus / Nyquist plot

Exercise1. Write a program to obtain the Bode plot / Root locus / Nyquist plot for the given

system2. Access the stability of the given system using the plots obtained3. Compare the usage of various plots in assessing stability

Equipment 1. System with MATLAB / MATHCAD / equivalent software - 3 user license

7. DC and AC position Control Systems

AimTo study the AC and DC position control system and draw the error characteristics between set point and error.

Exercise1. To study various position control systems and calculate the error between set point and output position 2. To measure outputs at various points (between stages)

Equipment 1. AC and DC position control kit with DC servo motor.

2. Power transistor3. Adder



8. Stepper Motor Control System

AimTo study the working of stepper motor

Exercise1. To verify the working of the stepper motor rotation using microprocessor.

Equipment 1. Stepping motor

2. Microprocessor kit 3. Interfacing card 4. Power supply

9. Digital Simulation of First Order System

AimTo digitally simulate the time response characteristics of first -order system

Exercise1. Write a program or build the block diagram model using the given software.2. Obtain the impulse, step and sinusoidal response characteristics.3. Identify real time systems with similar characteristics.

Equipment1. System with MATLAB / MATHCAD (or) equivalent software - minimum 3

user license.

10. Digital Simulation of Second Order Systems

AimTo digitally simulate the time response characteristics of second -order system

Exercise1. Write a program or build the block diagram model using the given software.2. Obtain the impulse, step and sinusoidal response characteristics.3. Identify real time systems with similar characteristics.

EquipmentSystem with MATLAB / MATHCAD (or) equivalent software - minimum 3 user license.



LIST OF EXPERIMENTS FIRST CYCLE:

1. Determination of transfer function of armature controlled DC servomotor.

2. Determination of transfer function of field controlled DC servomotor.

3. Determination of transfer function of AC servomotor.

4. Determination of transfer function of separately excited DC generator.

5. Determination of transfer function of DC motor.

6. DC position control system.

SECOND CYCLE:

7. Analog simulation of Type-0 and Type-1 systems.

8. Digital simulation of first order systems.

9. Digital simulation of second order systems

10. Stability analysis of linear systems.

11. Stepper motor control system.

12. AC position control system.



Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTOR

AIM: To determine the transfer function of armature controlled DC servo motor.

APPARATUS / INSTRUMENTS REQUIRED:

S. No Description Range Type Quantity1. DC servo motor trainer kit - 12. DC servo motor 13. Rheostat 500Ω/1A 1

4. Ammeter (0-1)A MC 1(0-100) mA MI 1

5. Voltmeter (0–300) V MC 1 (0–75) V MI 1

6. Stopwatch - 17. Patch cords - As required

THEORY:

In servo applications a DC motor is required to produce rapid accelerations from standstill. Therefore the physical requirements of such a motor are low inertia and high starting torque. Low inertia is attained with reduced armature diameter with a consequent increase in the armature length such that the desired power output is achieved. Thus, except for minor differences in constructional features a DC servomotor is essentially an ordinary DC motor. A DC servomotor is a torque transducer which converts electrical energy into mechanical energy. It is basically a separately excited type DC motor. The torque developed on the motor shaft is directly proportional to the field flux and armature current, Tm = Km Φ Ia. The back emf developed by the motor is Eb = Kb Φ ωm.. In an armature controlled DC Servo motor, the field winding is supplied with constant current hence the flux remains constant. Therefore these motors are also called as constant magnetic flux motors. Armature control scheme is suitable for large size motors.

ARMATURE CONTROLLED DC SERVOMOTOR:



FORMULAE USED:

Transfer function of the armature controlled DC servomotor is given as

θ(s) / Va(s) = Km / [s (1+sτa)(1+sτm ) + (Kb Kt /RaB)]where

Motor gain constant, Km = (Kt/RaB)

Motor torque constant, Kt = T / Ia

Torque, T in Nm = 9.55 Eb Ia Back emf, Eb in volts = Va – Ia Ra Va = Excitation voltage in volts Back emf constant, Kb = Va / ω

Angular velocity in rad/ sec = 2πN / 60

Armature time constant, τa = La / Ra

Armature Inductance, La in H= XLa / 2f XLa in =(Za

2 – Ra2)Za in = Va2 / Ia2

Armature resistance,Ra in = Va1 / Ia1

Mechanical time constant, τm = J / B

Moment of inertia, J in Kg m2 / rad = W x (60 / 2 ) 2 x dt/dN N

Stray loss, W in Watts = W’ x [ t2 / (t1-t2) ]Power absorbed, W’ in watts = Va Ia

t2 is time taken on load in secst1 is time taken on no load in secsdt is change in time on no load in secsdN is change in speed on no load is rpmN is rated speed in rpm



Frictional co-efficient, B in N-m / (rad / sec ) = W’’ / (2N / 60 )2

W’’ = 30 % of Constant lossConstant loss = No load i/p – Copper lossNo load I/P = V ( Ia + If )Copper loss = Ia

2 Ra

N is rated speed in rpm

PROCEDURE:

1. To determine the motor torque constant Kt and Back emf constant Kb:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Press the reset button to reset the over speed. Patch the circuit as per the patching diagram. Put the selection button of the trainer kit in the armature control mode. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220 V to the armature of the servomotor. Note the values of the armature current Ia, armature voltage Va, and speed N. Find the motor torque constant Kt and Back emf constant Kb using the above values.

Note:If the voltmeter and ammeter in the trainer kit is found not working external metersof suitable range can be used.

OBSERVATIONS:

S. No. Armature Voltage,Va

(V)Armature Current,Ia

(A)Speed,N

(rpm)

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE THE MOTOR TORQUE CONSTANT Kt AND BACK EMF CONSTANT Kb



PROCEDURE:

2. To determine armature resistance Ra:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode. The field terminal is left opened. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220 V to the armature of the servomotor. Note the values of the armature current Ia, armature voltage Va. Find the value of armature resistance Ra using the above values


OBSERVATIONS:

S. No. Armature Voltage, Va1

(V)Armature Current, Ia1

(A)Armature resistance,

Ra ()

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE ARMATURE RESISTANCE Ra



PROCEDURE:

3. To find armature inductance, La

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode. The field terminal is left opened. Switch ON the MCB. Note the values of the armature current Ia, armature voltage Va. Find the value of armature inductance La.using the above values


OBSERVATIONS:



(mA)Armature impedance

Za ()

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE ARMATURE INDUCTANCE, La



PROCEDURE:

4. To determine moment of inertia J and frictional co-efficient B:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode and the DPDT

switch in power circuit position. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and adjust the motor to run at rated speed. Note the values of armature current Ia, armature voltage Va, field current If, Speed N. Change the DPDT switch position from power circuit side to load side,

simultaneously noting the time taken t1 of the motor to come to rest from rated speed, using a stop watch.

Set the potentiometer to minimum position and change the DPDT switch to power circuit side

Connect a load of 500 Ohms in the load position Vary the pot and adjust the motor to run at rated speed Change the DPDT switch position from power circuit side to load side,


Find the values of moment of inertia J and frictional co-efficient B using the above values

OBSERVATIONS:

S. No.Armature

Voltage, Va

(V)

Armature Current,

Ia

(A)

Field Current, If

(A)

Speed, N

(rpm)

t1

(secs)

t2

(secs)

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE MOMENT OF INERTIA J , FRICTIONAL CO-EFFICIENT B: ( t1 – No load)



DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE MOMENT OF INERTIA J , FRICTIONAL CO-EFFICIENT B: ( t2 – load)



CALCULATIONS:

RESULT:

The transfer function of armature controlled DC servomotor is determined as

VIVA-VOCE QUESTIONS:

1. Define transfer function.2. What is DC servo motor? State the main parts.3. What is servo mechanism?4. Is this a closed loop or open loop system .Explain.5. What is back EMF?



Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF FIELD CONTROLLED DC SERVO MOTOR

AIM: To determine the transfer function of field controlled DC servo motor.


S. No Description Range Type Quantity1. DC servo motor trainer kit - 12. DC servo motor 13. Rheostat 500Ω/1A 1

4. Ammeter (0-1)A MC 1(0-100) mA MI 1


6. Stopwatch - 17. Patch cords - As required

THEORY:

In a field controlled DC Servo motor, the electrical signal is externally applied to the field winding. The armature current is kept constant. In a control system, a controller generates the error signal by comparing the actual o/p with the reference i/p. Such an error signal is no enough to drive the DC motor. Hence it is amplified by the servo amplifier and applied to the field winding. With the help of constant current source, the armature current is maintained constant. When there is change in voltage applied to the field winding, the current through the field winding changes. This changes the flux produced by field winding. This motor has large Lf / Rf ratio, so time constant of this motor is high and it can’t give rapid responses to the quick changing control signals.

FIELD CONTROLLED MOTOR:

FORMULAE USED:



Transfer function of field controlled DC servo motor is given as,

(s) / Vf (s) = Km / s (1+sTf) (1+sTm)where

Motor gain constant Km = Ktf / Rf B

Motor torque constant Ktf in N-m / A = T / If

Torque T in N-m = 9.55 Eb Ia / N Back EMF Eb in volts = Va – Ia Ra

Va = Excitation voltage in voltsArmature resistance,Ra in = Va1 / Ia1

Field resistance,Rf in = Vf1 / If1 Field time constant Tf = Lf / Rf

Field Inductance,Lf in H= XLf / 2f XLf in = (Zf

2 – Rf2)

Zf in = Vf2 / If2

Mechanical time constant Tm = J / B

Moment of inertia J in Kg m2 / rad = W x (60 / 2 ) 2 x dt/dN N

Stray loss, W in watts = W’ x [ t2 / (t1-t2) ]Power absorbed, W’ in Watts = Va Ia

t2 is time taken on load in secst1 is time taken on no load in secsdt is change in time on no load in secsdN is change in speed on no load is rpmN is rated speed in rpm

Frictional co-efficient, B in N-m / (rad / sec ) = W’’ / (2N / 60 )2

W’’ = 30 % of Constant lossConstant loss = No load i/p – Copper lossNo load I/P = V ( Ia + If )Copper loss = Ia

2 Ra

N is rated speed in rpm

PROCEDURE:

1. To determine the motor torque constant Ktf :



Check whether the MCB is in OFF position in the DC servomotor trainer kit Press the reset button to reset the over speed. Patch the circuit as per the patching diagram. Put the selection button of the trainer kit in the field control mode. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220V to the armature of the servomotor. Note the values of the armature current Ia, armature voltage Va, and speed N. Find the motor torque constant Kt f using the above values.


OBSERVATIONS:

S. No. Armature Voltage,Va

(V)Armature Current,Ia

(A)Speed,N

(rpm)

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF FIELD CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE THE MOTOR TORQUE CONSTANT Ktf



PROCEDURE:

2. To determine armature resistance Ra:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode. The field terminal is left opened. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220V to the armature of the servomotor. Note the values of the armature current Ia, armature voltage Va. Find the value of armature resistance Ra using the above values


OBSERVATIONS:



(A)Armature Resistance,

Ra ()

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF FIELD CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE ARMATURE RESISTANCE Ra



PROCEDURE:

3. To determine field resistance Rf:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the field control mode. The armature terminal is left opened. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220V to the field of the servomotor. Note the values of the field current If, field voltage Vf. Find the value of field resistance Rf using the above values


OBSERVATIONS:

S. No. Field Voltage, Va1

(V)Field Current, Ia1

(A)Field Resistance,

Rf ()

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF FIELD CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE FIELD RESISTANCE RF



PROCEDURE:

4. To determine Field Inductance, Lf

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the field control mode. The armature terminal is left opened. Switch ON the MCB. Note the values of the field current If2, field voltageVf2. Find the value of field inductance Lf.using the above values


OBSERVATIONS:

S. No. Field Voltage, Vf2

(V)Field Current, If2

(mA)Field Impedance

Zf ()

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF FIELD CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE FIELD INDUCTANCE, LF



PROCEDURE:

5. To determine moment of inertia J and frictional co-efficient B:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode and the DPDT

switch in power circuit position. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and adjust the motor to run at rated speed. Note the values of armature current Ia, armature voltage Va, field current If, Speed N. Change the DPDT switch position from power circuit side to load side,


Set the potentiometer to minimum position and change the DPDT switch to power circuit side

Connect a load of 500 Ohms in the load position Vary the pot and adjust the motor to run at rated speed Change the DPDT switch position from power circuit side to load side,


Find the values of moment of inertia J and frictional co-efficient B using the above values

OBSERVATIONS:

S. NoArmature

Voltage, Va

(V)

Armature Current,

Ia

(A)

Field Current, If

(A)

Speed, N

(rpm)

t1

(secs)

t2

(secs)

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF FIELD CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE MOMENT OF INERTIA J , FRICTIONAL CO-EFFICIENT B: ( t1 – No load)



DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTORPATCHING DIAGRAM TO DETERMINE MOMENT OF INERTIA J , FRICTIONAL CO-EFFICIENT B: ( t2 – load)



CALCULATIONS:

RESULT:

The transfer function of field controlled DC servomotor is determined as


1. What are the main parts of a DC servo motor?2. Name the two types of servo motor.3. State the advantages and disadvantages of a DC servo motor.4. Give the applications of DC servomotor.5. What is servo mechanism?6. What do you mean by field controlled DC servo motor?



Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OF AC SERVO MOTOR

AIM:To derive the transfer function of the given AC Servomotor.


S. No Description Range Type Quantity1. AC servo motor trainer kit - 12. AC servo motor 1

3. Ammeter (0-1) A MC 1(0-100) mA MI 1


5. Patch cords - As required

THEORY:

An AC servo motor is basically a two phase induction motor with some special design features. The stator consists of two pole pairs (A-B and C-D) mounted on the inner periphery of the stator, such that their axes are at an angle of 90o in space. Each pole pair carries a winding, one winding is called reference winding and other is called a control winding. The exciting current in the winding should have a phase displacement of 90o. The supply used to drive the motor is single phase and so a phase advancing capacitor is connected to one of the phase to produce a phase difference of 90o.The rotor construction is usually squirrel cage or drag-cup type. The rotor bars are placed on the slots and short-circuited at both ends by end rings. The diameter of the rotor is kept small in order to reduce inertia and to obtain good accelerating characteristics. The drag cup construction is employed for very low inertia applications. In this type of construction the rotor will be in the form of hollow cylinder made of aluminium. The aluminium cylinder itself acts as short-circuited rotor conductors. Electrically both the types of rotor are identical.

WORKING PRINCIPLE :

The stator windings are excited by voltages of equal magnitude and 90o phase difference. These results in exciting currents i1 and i2 that are phase displaced by 90o and have equal values. These currents give rise to a rotating magnetic field of constant magnitude. The direction of rotation depends on the phase relationship of the two currents (or voltages). This rotating magnetic field sweeps over the rotor conductors. The rotor conductor experience a change in flux and so voltages are induced rotor conductors. This voltage circulates currents in the short-circuited rotor conductors and currents create rotor flux. Due to the interaction of stator & rotor flux, a mechanical force (or torque) is developed on the rotor and so the rotor starts moving in the same direction as that of rotating magnetic field.



GENERAL SCHEMATIC OF AC SERVOMOTOR:

FORMULAE USED:

Transfer function, Gm (s) = Km / (1+ sm)

Where

Motor gain constant, Km = K / FO + F

K is T / CFO is T / NTorque, T is 9.81 X R (S1 S2)R is radius of the rotor in m

Frictional co-efficient, F = W / (2N / 60)2

Frictional loss, W is 30 % of constant loss in WattsConstant loss in watts = No load input – Copper loss No load i/p = V (IR+IC)V is supply voltage, VIR is current through reference winding, AIC is current through control winding, ACopper loss in watts = IC

2 RC

RC = 174 N is rated speed in rpm

Motor time constant, m = J / FO + FMoment of inertia J is d4 L R ρ / 32d is diameter of the rotor in m ( Given d =39.5 mm)LR is length of the rotor in m (Given L R =76 mm)ρ is density = 7.8 X 102 gm / m



PROCEDURE:

1. DETERMINATION OF FRICTIONAL CO-EFFICIENT, F

1. Check whether the MCB is in OFF position.2. Patch the circuit using the patching diagram.3. Switch ON the MCB4. Vary the control pot to apply rated supply voltage5. Note the control winding current, reference winding current, supply voltage and

speed.6. Find the frictional co-efficient using the above values

OBSERVATIONS:

S. No.Supply Voltage

V(V)

Control winding Current Ic

(A)

Reference Winding Current Ir

(A)

Speed N

(rpm)

CALCULATIONS:



DETERMINATION OF TRANSFER FUNCTION OF AC SERVO MOTOR PATCHING DIAGRAM TO DETERMINE FRICTIONAL CO-EFFICIENT F:



PROCEDURE:

2. To determine the motor gain constant Km

DETERMINATION OF FO FROM TORQUE - SPEED CHARACTERISTICS:1. Check whether the MCB is in OFF position.2. Patch the circuit using the patching diagram.3. Set the control pot in minimum position.4. Check whether the motor is in no load condition5. Switch ON the MCB6. Vary the control pot and apply rated voltage to the reference phase winding and

control phase winding. Note down the no load speed.7. Apply load in steps. For each load applied note down the speed and spring balance

readings. ( Take 3 or 4 sets of readings)8. Reduce the load fully and allow the motor to run at rated speed.9. Repeat steps 7 and 8 for 75 % control winding voltage.10. Draw the graph between speed and torque, the slope of the graph gives FO.

OBSERVATIONS:

S. No

Control voltage Vc1 = Control voltage Vc2 =Speed

N

(rpm)

Spring Balance values Torque

T(Nm)

SpeedN

(rpm)

Spring Balance values

Torque T

(Nm)S1

(kg)S2

(kg)S1

(kg)S2

(kg)

MODEL GRAPH: TORQUE - SPEED CHARACTERISTICS



DETERMINATION OF K FROM TORQUE - CONTROL VOLTAGE CHARACTERISTICS:

1. Check whether the MCB is in OFF position.2. Patch the circuit using the patching diagram.3. Set the control pot in minimum position.4. Check whether the motor is in no load condition5. Switch ON the MCB

6. Vary the control pot and apply rated voltage to the reference phase winding and control phase winding. Note down the no load speed.

7. Load the motor gradually; the speed of the motor will decrease. Vary the control pot and increase the control winding voltage till the speed obtained at no load is

reached. Note down control voltage and spring balance readings. 8. Repeat step 7 for various speeds and tabulate. (for 1000 rpm)

9. Plot the graph between torque and control winding voltage. The slope of the graph gives the value of K.

OBSERVATIONS:

S. No

Speed N1 = Speed N2 =ControlVoltage

Vc

(V)


Torque T

Nm

Speed

rpm


ControlVoltage

Vc

VS1

(kgS2

kgS1

Kg

S2

Kg

MODEL GRAPH: TORQUE - CONTROL VOLTAGE CHARACTERISTICS


Control Systems Laboratory Manual / II EEE, IV SEM DETERMINATION OF TRANSFER FUNCTION OF AC SERVO MOTOR

PATCHING DIAGRAM TO DETERMINE MOTOR GAIN CONSTANT KM:



CALCULATIONS:

RESULT:

The transfer function of AC servomotor is determined as


1. What are the main parts of an AC servomotor?2. State the advantages and disadvantages of an AC servo motor.3. Give the applications of AC servomotor.4. What do you mean by servo mechanism?5. What are the characteristics of an AC servomotor?



Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OF SEPARATELY EXCITED DC GENERATOR

AIM:

To obtain the transfer function of separately excited DC generator on no load and loaded condition.

APPARATUS / INSTRUMENTS REQUIRED:S. No Description Range Type Quantity

THEORY:

Derivation of transfer function of separately excited DC generator is as follows,

Applying KVL to the field side,

ef = Rf if + Lf (dif / dt) … (1)

Applying KVL to the armature side,

eg = Ra ia + La (dia / dt) + RL ia … (2)

VL = RL ia … (3)

Also since eg α if , let eg = Kg if … (4)

Taking Laplace transform of equation (1) we get

Ef (s) = Rf If(s) + sLf If(s)

Ef (s) = If (s) [Rf + sLf]

If (s) = Ef (s) / [Rf + sLf] … (5)



Taking Laplace transform of equation (2) we getEg (s) = Ra Ia(s) + sLa Ia(s) + RL Ia(s)Eg (s) = Ia(s) [Ra + sLa + RL] … (6)

Taking Laplace transform of equations (3) and (4) we getVL(s) = RL Ia( s) Therefore, Ia( s) = VL(s) / RL … (7)Eg(s) = Kg If(s) … (8)

Substituting. equations (7) and (8) in equation (6) we getKg If(s) = [Ra + sLa + RL] [VL(s) / RL] … (9)

Substituting the value of If (s) in the above equation we getKg Ef (s) / [Rf + sLf] = [Ra + sLa + RL] [ VL(s) / RL]

Hence transfer function, VL(s) / Ef (s) = Kg RL / [Rf + sLf] [Ra + sLa + RL] …(10)

For unloaded condition, Ia = 0Therefore transfer function VL(s) / Ef (s) = Kg / [Rf + sLf] … (11)

For loaded conditionLf = √ (Zf

2 – Rf2) / 2πf

La = √ (Za2 – Ra

2) / 2πf

Transfer function VL(s) / Ef (s) = Kg RL / [Rf (Ra + RL) (1+sτf) (1 + sτa)] … (12)

where τf = Lf / Rf and τa = La / (Ra + RL)

FORMULAE USED:

Transfer function of DC generator,On no load condition: VL(s) / Ef (s) = Kg / [Rf + sLf]where Kg is gain constant

Rf is field resistance in OhmsLf is field inductance in Henry

On loaded condition: VL(s) / Ef (s) = Kg RL / [Rf (Ra + RL) (1+sτf) (1 + sτa)] where Kg is gain constant

Field time constant τf = Lf / Rf Rf is field resistance in OhmsLf is field inductance in Henry

Armature time constant τa = La / (Ra + RL)Ra is armature resistance in OhmsLa is armature inductance in Henry



PROCEDURE:1. To determine the gain constant Kg :

No load or open circuit characteristics:

1. Connections are made as shown in the circuit diagram2. The motor field rheostat should be in minimum resistance position and the generator

field rheostat should be in maximum resistance position or minimum potential position while switching ON and switching OFF the supply side DPST switch.

3. Ensure that the DPST switch on the load side is open.4. Switch ON the supply DPST switch.5. Using the 3- point starter the DC motor is started and it is brought to rated speed by

adjusting the motor field rheostat.6. Keeping the DPST switch on the load side open, the generated voltage Eg and field

current If of generator is noted down by varying the generator field rheostat.7. The above step is repeated till 125 % of rated voltage is reached.8. A graph is plotted between Eg and If taking If along x- axis. A tangent to the linear

portion of the curve is drawn from the origin and slope of this line gives Kg. OBSERVATIONS:

MODEL GRAPH:

Prepared by V.Arivumani, AP/EEE, Vels University, Chennai

S. No. Field current, If

(A)Induced Voltage, Eg

(V)

47


CIRCUIT DIAGRAM:

To determine gain constant, Kg:

CALCULATIONS:



Load characteristics:

1. Connections are made as shown in the circuit diagram2. The motor field rheostat should be in minimum resistance position and the generator

field rheostat should be in maximum resistance position or minimum potential position while switching ON and switching OFF the supply side DPST switch.

3. Ensure that the DPST switch on the load side is open.4. Switch ON the supply DPST switch5. The generator is brought to its rated voltage by varying the generator field rheostat. 6. The DPST switch on the load side is closed, and the load is varied for convenient

steps of load current up to 120 % of its rated capacity and the voltmeter VL and ammeter Ia readings are observed. On each loading the speed should be maintained at rated speed.

7. A graph is plotted between VL and IL taking IL on x- axis. The slope of the graph gives Kg.

OBSERVATIONS:

MODEL GRAPH:

Prepared by V.Arivumani, AP/EEE, Vels University, Chennai

S. No. Terminal Voltage, VL

(V)Load Current, IL

(A)

49


PROCEDURE:

2. To determine field Inductance Lf :

1. Connections are made as per the circuit diagram.2. Auto transformer is varied in steps for different voltages and corresponding voltmeter

and ammeter readings are noted down.3. Field impedance Zf is calculated as V/I and the average value of Zf is obtained.4. Field resistance (Rf) is measured using multimeter.5. Field inductance (Lf) can be calculated using formula

Lf = √ (Zf2 – Rf

2) / 2πf

CIRCUIT DIAGRAM:

OBSERVATIONS:

S. No Field Voltage, V (V)

Field Current, I

(A)Field Impedence, Zf

(Ohms)

CALCULATIONS:



PROCEDURE:

3. Determination of armature inductance La :


and ammeter readings are noted down.3. Armature impedance Za is calculated as V/I and the average value of Za is obtained.4. Armature resistance Ra is measured using multimeter.5. Armature inductance La can be calculated using formula,

La = √ (Za2 – Ra

2) / 2πf

CIRCUIT DIAGRAM:

OBSERVATIONS:

S. No Armature Voltage, V (V)

Armature Current, I

(A)

Armature Impedence, Za

(Ohms)

CALCULATIONS:



CALCULATIONS:

RESULT:

The transfer function of separately excited DC generator is determined as



Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OF DC MOTOR

AIM:

To obtain the transfer function of field controlled DC motor.

APPARATUS / INSTRUMENTS REQUIRED:S. No Description Range Type Quantity

THEORY:



FIELD CONTROLLED MOTOR:

FORMULAE USED:

Transfer function of field controlled DC motor,

(s) / Vf (s) = Km / [s (1+sτf) (1 + sτm)] where

Motor gain constant, Km = Ktf / (BRf)Ktf is motor torque constant

Torque, T is 9.81 X R (S1 S2) R is radius of the brake drum in m

R = circumference of the brake drum/ (2 П)B is viscous co-efficient of frictionRf is field resistance in Ohms

Field time constant τf = Lf / Rf Rf is field resistance in OhmsLf is field inductance in HenryLf = √ (Zf

2 – Rf2) / 2πf

Zf is field impedence in OhmsRf is field resistance in Ohms

Mechanical time constant τm = J/BMoment of inertia J = Pav / [2П2(N1

2 – N22 )((1/t1)-(1/t2))]

Average power delivered to the load, Pav= (V1I1 + I12 Ra + V2I2 + I2

2 Ra) / 2Ra is armature resistance in OhmsLa is armature inductance in Henryt2 is time taken on load in secst1 is time taken on no load in secs

Viscous Co-efficient of friction, B = Pstray / (N1 + N2)2

Stray loss, Pstray = [2П2(N12 – N2

2 )] J / t1



PROCEDURE:

1. To determine motor torque constant, Ktf :

1. Connections are made as shown in the circuit diagram2. The armature current Ia of the motor is set to some value by adjusting the armature

circuit resistance. This value of Ia is maintained constant throughout the experiment.3. The field current If is varied in steps by adjusting the field rheostat and for each value

of If the brake drum is adjusted such that it just fails to rotate. The corresponding readings of ammeter and spring balances are noted.

4. The value of torque for each value of If is calculated5. A graph is plotted between torque T and field current IF taking IF along x-axis. The

slope of the graph gives the value of Ktf

OBSERVATIONS:

S. No.Armature current

Ia

(A)

Field current If

(A)

Spring balance readings Torque T

(Nm)S1

(kg)S2

(kg)

MODEL GRAPH:



CIRCUIT DIAGRAM:

CALCULATIONS:

PROCEDURE



2. To determine field Inductance Lf :


and ammeter readings are noted down.3. Field impedance Zf is calculated as V/I and the average value of Zf is obtained.4. Field resistance (Rf) is measured using multimeter.5. Field inductance (Lf) can be calculated using formula

Lf = √ (Zf2 – Rf

2) / 2πf

CIRCUIT DIAGRAM:

OBSERVATIONS:

S. No. Field Voltage, V (V)

Field Current, I

(A)Field Impedence, Zf

(Ω)

CALCULATIONS:



PROCEDURE:

3. To determine moment of inertia J and Viscous friction Co-efficient B:

1. Connections are made as shown in the circuit diagram2. The field current of the motor is set to some value by adjusting the field resistance.3. DPDT switch is thrown to position (1,11) and the motor is made to run at a speed N1

(1700 rpm) by adjusting the armature rheostat.4. DPDT switch is opened from position (1,11) and the stop watch is started

simultaneously. The time taken t1 for the speed to drop from N1(1700 rpm) to N2

( 1300 rpm) is noted.5. Again the DPDT switch is thrown to position (1,11) and the motor is made to run at a

speed greater than N1 (1700 rpm) by adjusting the armature rheostat.6. DPDT switch is thrown to position (2,21) and the stop watch is started when the motor

speed reaches N1 (1700 rpm). The time taken t2 for the speed to drop from N1 (1700 rpm) to N2( 1300 rpm) is noted. Simultaneously the readings of the ammeter and voltmeter corresponding to N1 and N2 are noted.

OBSERVATIONS:

S. No. N1

(rpm)t1

(Sec)V1

(V)I1

(A)N2

(rpm)T2

(Sec)V2

(V)I2

(A)

CALCULATIONS:



CIRCUIT DIAGRAM:

CALCULATIONS:



CALCULATIONS:

RESULT:

The transfer function of field controlled DC motor is determined as

Expt. No: Date:



DC POSITION CONTROL SYSTEM

AIM:

To study the characteristics of a DC position control system.


i) DC position control kit and Motor unit

ii) Multimeter

THEORY:

A DC position control system is a closed loop control system in which the position of the mechanical load is controlled with the position of the reference shaft. A pair of potentiometers acts as error-measuring device. They convert the input and output positions into proportional electric signals. The desired position is set on the input potentiometer and the actual position is fed to feedback potentiometer. The difference between the two angular positions generates an error signal, which is amplified and fed to armature circuit of the DC motor. The tachogenerator attached to the motor shaft produces a voltage proportional to the speed which is used for feedback. If an error exists, the motor develops a torque to rotate the output in such a way as to reduce the error to zero. The rotation of the motor stops when the error signal is zero, i.e., when the desired position is reached.

PROCEDURE:

1. The input or reference potentiometer is adjusted nearer to zero initially(R).2. The command switch is kept in continuous mode and some value of forward gain KA

is selected.3. For various positions of input potentiometer (R) the positions of the response

potentiometer (0) is noted. Simultaneously the reference voltage (VR) measured between the terminals VR & E and the output voltage (VO) measured between the terminals VO & E are noted.

4. A graph is plotted with 0 along y-axis and R along x-axis.



OBSERVATIONS:

S. No

Reference angular position,

θR

(degrees)

Output angular position, θO

(degrees)

Reference Voltage, Vr

(V)

Output VoltageVO

(V)

KA = KA = KA = KA = KA = KA = KA = KA =

MODEL GRAPH:



DC POSITION CONTROL SYSTEM



RESULT:

The DC position control system characteristics are studied and corresponding graphs are drawn.



Expt. No. Date:

ANALOG SIMULATION OF TYPE – 0 and TYPE – 1 SYSTEMS

AIM:To study the time response of first and second order type –0 and type- 1 systems.

APPARATUS / INSTRUMENTS REQUIRED: 1. Linear system simulator kit 2. CRO 3. Patch cords

FORMULAE USED:

Damping ratio, z= (ln MP)2 / (2 +(ln MP)2)Where MP is peak percent overshoot obtained from the time response graph

Undamped natural frequency, n = / [tp (1 - z2)]where tp is the peak time obtained from the time response graph

Closed loop transfer function of the type – 0 second order system is

C(s)/R(s) = G(s) / [1 + G(s) H(s)]where

H(s) = 1G(s) = K K2 K3 / (1+sT1) (1 + sT2) where K is the gain

K2 is the gain of the time constant – 1 block =10K3 is the gain of the time constant – 2 block =10T1 is the time constant of time constant – 1 block = 1 msT2 is the time constant of time constant – 2 block = 1 ms

Closed loop transfer function of the type – 1-second order system is C(s)/R(s) = G(s) / [1 + G(s) H(s)]

where H(s) = 1

G(s) = K K1 K2 / s (1 + sT1) where K is the gain K1 is the gain of Integrator = 9.6 K2 is the gain of the time constant – 1 block =10 T1 is the time constant of time constant – 1 block = 1 ms

THEORY:

The type number of the system is obtained from the number of poles located at origin in a given system. Type – 0 system means there is no pole at origin. Type – 1 system means there is one pole located at the origin. The order of the system is obtained from the highest power of s in the denominator of closed loop transfer function of the system. The first order system is



characterized by one pole or a zero. Examples of first order systems are a pure integrator and a single time constant having transfer function of the form K/s and K/(sT+1). The second order system is characterized by two poles and up to two zeros. The standard form of a second order system is G(s) = n

2 / (s2 + 2zns + n

2) where z is damping ratio and n is undamped natural frequency.

PROCEDURE:

1. To find the steady state error of type – 0 first order system

1. Connections are made in the simulator kit as shown in the block diagram.2. The input square wave is set to 2 Vpp in the CRO and this is applied to the REF

terminal of error detector block. The input is also connected to the X- channel of CRO.3. The output from the simulator kit is connected to the Y- channel of CRO.4. The CRO is kept in X-Y mode and the steady state error is obtained as the vertical displacement between the two curves.

5. The gain K is varied and different values of steady state errors are noted.

Block diagram of Type-0 first order system

OBSERVATIONS:

S. No. Gain, K Steady state error, ess 123

TRACES FROM CRO:Prepared by V.Arivumani, AP/EEE, Vels University, Chennai 66


For Gain, K =

For Gain, K =

For Gain, K=



LINEAR SYSTEM SIMULATORPATCHING DIAGRAM TO OBTAIN THE STEADY STATE ERROR OF TYPE – 0 FIRST ORDER SYSTEM



2. To find the steady state error of type – 1 first order system

1. The blocks are Connected using the patch chords in the simulator kit.2. The input triangular wave is set to 2 Vpp in the CRO and this applied o the REF

terminal of error detector block. The input is also connected to the X- channel of CRO.3. The output from the system is connected to the Y- channel of CRO.4. The experiment should be conducted at the lowest frequency to allow enough

time for the step response to reach near steady state.5. The CRO is kept in X-Y mode and the steady state error is obtained as the vertical

displacement between the two curves. 6. The gain K is varied and different values of steady state errors are noted. 7. The steady state error is also calculated theoretically and the two values are compared.

Block diagram of Type- 1 First order system

OBSERVATIONS:

S. No. Gain, K Steady state error, ess 123



TRACES FROM CRO:

For Gain, K =

For Gain, K =

For Gain, K =



LINEAR SYSTEM SIMULATORPATCHING DIAGRAM TO OBTAIN THE STEADY STATE ERROR OF TYPE – 1 FIRST ORDER SYSTEM



3. To find the closed loop response of type– 0 and type- 1 second order system

1. The blocks are connected using the patch chords in the simulator kit.2. The input square wave is set to 2 Vpp in the CRO and this applied to the REF terminal

of error detector block. The input is also connected to the X- channel of CRO.3. The output from the system is connected to the Y- channel of CRO.4. The output waveform is obtained in the CRO and it is traced on a graph sheet. From

the waveform the peak percent overshoot, settling time,rise time, peak time are measured. Using these values n and x are calculated.

5. The above procedure is repeated for different values of gain K and the values are compared with the theoretical values.

Block diagram to obtain closed loop response of Type-0 second order system

OBSERVATIONS:

S. No. GainK

Peak percent

Overshoot

%MP

Rise time

tr

(sec)

PeakTime

tp

(sec)

Settling time

ts

(sec)

Damping ratio

z

UndampedNatural

frequencyn

(rad/sec)

1

2

TRACES FROM CRO:

For Gain, K = For Gain, K =



Block diagram to obtain closed loop response of Type-1 second order system

OBSERVATIONS:

S. No. GainK

Peak percent

Overshoot

%MP

Rise time

tr

(sec)

PeakTime

tp

(sec)

Settling time

ts

(sec)

Damping ratio

z

UndampedNatural

frequencyn

(rad/sec)

1

2

TRACES FROM CRO:

For Gain, K = For Gain, K =



LINEAR SYSTEM SIMULATORPATCHING DIAGRAM TO OBTAIN THE CLOSED LOOP RESPONSE OF TYPE – 0 SECOND ORDER SYSTEM



LINEAR SYSTEM SIMULATORPATCHING DIAGRAM TO OBTAIN THE CLOSED LOOP RESPONSE OF TYPE – 1 SECOND ORDER SYSTEM



CALCULATIONS:

RESULT:

The time response of first and second order type-0 and type-1 systems are studied.


1. Define order and type number.2. What are dominant poles?3. What is a closed loop system?4. What is the effect of negative feedback?5. What are poles and zeros of a system?6. Define transfer function.



Expt. No. Date:

DIGITAL SIMULATION OF FIRST ORDER SYSTEMS AIM:

To digitally simulate the time response characteristics of a linear system without non- linearities and to verify it manually.

APPARATUS REQUIRED:

A PC with MATLAB package

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

Delay time td

Rise time tr Peak time tp

Maximum peak overshoot Mp

Settling time ts

STUDY OF BASIC MATLAB COMMANDS:

The name MATLAB stands for MATRIX LABORATORY. MATLAB was originally written to provide easy access to matrix software developed by the LINPACK and EISPACK projects. Today, MATLAB engines incorporate the LAPACK and BLAS libraries, embedding the state of the art in software for matrix computation. It has evolved over a period of years with input from many users. In university environments, it is the standard instructional tool for introductory and advanced courses in MATHEMATICS, ENGINEERING, AND SCIENCE. In industry, MATLAB is the tool of choice for high-productivity research, development, and analysis.

MATLAB is a high-performance language for technical computing. It integrates computation, visualization, and programming in an easy-to-use environment where problems and solutions are expressed in familiar mathematical notation. Typical uses include,

Math and computation Algorithm development Data acquisition Modeling, simulation, and prototyping Data analysis, exploration, and visualization Scientific and engineering graphics



Application development, including graphical user interface building

It is an interactive system whose basic data element is an array that does not require dimensioning. This allows you to solve many technical computing problems, especially those with matrix and vector formulations, in a fraction of the time it would take to write a program in a scalar non-interactive language such as C or Fortran. It also features a family of add-on application-specific solutions called toolboxes. Very important to most users of MATLAB, toolboxes allow you to learn and apply specialized technology. Toolboxes are comprehensive collections of MATLAB functions (M-files) that extend the MATLAB environment to solve particular classes of problems. Areas in which toolboxes are available include SIGNAL PROCESSING, CONTROL SYSTEMS, NEURAL NETWORKS, FUZZY LOGIC, WAVELETS, SIMULATION, AND MANY OTHERS.

Some practical examples of first order systems are RL and RC circuits.

PROCEDURE:

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.3. To build a SIMULINK model to obtain step response / sine response of a first order

system, the following procedure is followed:1. In MATLAB software open a new model in SIMULINK library browser.2. From the continuous block in the library drag the transfer function block.3. From the source block in the library drag the step input/ sine input.4. From the sink block in the library drag the scope.5. From the math operations block in the library drag the summing point.6. Connect all to form a system and give unity feedback to the system.7. For changing the parameters of the blocks connected double click the

respective block. 8. Start simulation and observe the results in scope. (Use a mux from the signal

routing block to view more than one graph in the scope) 9. Compare the simulated and theoretical results.

BLOCK DIAGRAM:

Step response of a first order system:



Sine response of a first order system:

2. MATLAB (m-file) program to obtain the step response and impulse response

% MATLAB program to find the step response

num=[ ];den=[ ];sys = tf (num,den);step (sys);grid

OUTPUT: (Paste the graph obtained from PC)

% MATLAB program to find the impulse response

num=[ ];den=[ ];sys = tf (num,den);impulse (sys);grid




CALCULATIONS:

Unit step response of the given RL series circuit:



Unit Impulse response of the given RLC series circuit:

RESULT:

The time response characteristics of a first order system is simulated digitally and verified manually.


1. What is MATLAB?2. What is the use of MATLAB Package?3. What are the toolboxes available in MATLAB?4. What is the use of a simulation?5. Differentiate real time systems and simulated systems.6. Give two examples for first order system.7. Name the standard test signals used in control system.8. What is time response?



Expt. No: Date:

DIGITAL SIMULATION OF SECOND ORDER SYSTEMSAIM:

To digitally simulate the time response characteristics of a second order system and verify manually.

APPARATUS REQUIRED

A PC with MATLAB Software

THEORY

The time 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:

Delay time td

Rise time tr Peak time tp

Maximum overshoot Mp

Settling time ts

PROCEDURE:

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.3. To build a SIMULINK model to obtain step response / sine response of a second

order system, the following procedure is followed:1. In MATLAB software open a new model in SIMULINK library browser.2. From the continuous block in the library drag the transfer function block.3. From the source block in the library drag the step input/ sine input.4. From the sink block in the library drag the scope.5. From the math operations block in the library drag the summing point.6. Connect all to form a system and give unity feedback to the system.7. For changing the parameters of the blocks connected double click the

respective block. 8. Start simulation and observe the results in scope. (Use a mux from the signal

routing block to view more than one graph in the scope)9. From the step response obtained note down the rise time, peak time, peak

overshoot and settling time.10. Compare the simulated and theoretical results.



BLOCK DIAGRAM:

Step response of a second order system:

Sine response of a second order system:

2. MATLAB program to obtain the step response and impulse response of second order system.

% MATLAB program to find the step responsenum=[ ];den=[ ];sys = tf (num,den);step (sys);




% MATLAB program to find the impulse response

num=[ ];den=[ ];sys = tf (num,den);impulse (sys);


CALCULATIONS:

Unit step response of the given RLC series circuit:



Unit impulse response of the given RLC series circuit:



RESULT:

The time response characteristics of the given second order system is simulated digitally and verified manually.


1. What is MATLAB?2. What is the use of MATLAB Package?3. What are the toolboxes available in MATLAB?4. What is the use of a simulation?5. Differentiate real time systems and simulated systems.6. Give two examples for second order system.7. Name the standard test signals used in control system.8. What is time response?



Expt. No: Date:

STABILITY ANALYSIS OF LINEAR SYSTEMS

a. USING BODE PLOT

AIM:

To obtain the bode plot and check for stability of the system with open loop transfer function, G(S) =

APPARATUS REQUIRED:


THEORY:

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

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

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,

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.

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

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

If one 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.

PROGRAM

%BODE PLOT OF THE SYSTEM

%Enter the numerator and denominator of the transfer function

num=[ ];

den=[ ];

sys=tf(num,den)

%Specify the frequency range and enter the command

w=logspace(-2,4,1000);

bode(sys,w)

xlabel('Frequency')

ylabel( ' Phase angle in degrees Magnitude of G(s) in decibels')

title('Bode Plot of the system ')

%To determine the Gain Margin, Phase Margin, Gain crossover frequency and

%Phase cross over frequency

margin(sys)

[ Gm, Pm, Wpc, Wgc ]= margin (sys)



MANUAL CALCULATIONS:



OUTPUT (from manual calculation):

OUTPUT (from program):

RESULT:

The Bode plot is drawn for the given transfer function using MATLAB and verified manually. From the plot obtained, the system is found to be ______________.


1. Define stability of Linear Time Invariant System.2. Give the stability conditions of system using Pole-Zero plot.3. Define Bode Plot.4. What is the use of Bode Plot?5. What the conditions of stability are in Bode plot?6. Define Stability criteria.7. Define Limits of stability.8. Define safe regions in stability criteria.9. Define Phase margin and Gain margin.



b. Using Root Locus

AIM:

To obtain the Root locus plot and to verify the stability of the system with transfer function,G(s) =

APPARATUS REQUIRED:


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:

1. Write a program to obtain the root locus plot for the given system.2. Assess the stability of given system using the plot obtained.

PROGRAM:

%ROOT LOCUS OF THE SYSTEM%

num=[ ]

den=[ ]

sys=tf(num,den)

rlocus(sys)

v=[-10,10,-8,8];

axis(v)



xlabel('Real Axis')

ylabel('Imaginary Axis')

title('Root Locus of the system')

title('Root Locus Plot of the system ')




OUTPUT (from manual calculation)

OUTPUT (from program):

RESULT:

The Root locus plot is drawn for the given transfer function, G(s)= ___________________using MATLAB and the range of gain K for stability is______________.


1. Define root locus technique.2. What are the conditions of stability in root locus criteria?3. What is the advantage of root locus technique?4. Which method of stability analysis is more advantageous?5. How the stability of unstable is improved?6. What are the methods to improve the stability?7. What is the use of compensators?8. What do you mean by Root-Loci?9. What is complementary Root Loci?10. What are contours?11. State the basic properties of Root Locus.12. How would you find the number of branches of Root Loci?13. How are the break away points of the root locus determined?14. How is the point of intersection of the asymptotes with real axis found out.



c. USING NYQUIST PLOT

AIM:

To obtain the Nyquist plot and check the stability of the system using Nyquist Stability Criterion for the given unity feedback system with transfer function

G(s)H(s) =

APPARATUS REQUIRED


THEORY:

NYQUIST STABILITY CRITERION:

POLAR PLOTS / 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.

PROCEDURE:

1. Write a program to obtain the Nyquist plot for the given system.2. Assess the stability of given system using the plot obtained.



PROGRAM

%NYQUIST PLOT%Enter the numerator and denominator of the transfer functionnum=[ ]den=[ ]sys=tf(num,den)

%Specify the frequency range and enter the commandnyquist(sys)v=[ ]axis(v)xlabel('Real Axis');ylabel('Imaginary Axis');title('Nyquist Plot of the system ‘)




OUTPUT ( from Manual calculation)

OUTPUT (from program)



RESULT:

The Nyquist plot is drawn for the given transfer function, G(s) = ______________________

using MATLAB and the system is found to be ______________________.


1. What is polar plot?2. What is Nyquist plot?3. Define the conditions of stability in polar plot.4. What is the use and advantage of polar plot?5. State Nyquist stability criterion.



Expt. No: Date:

STEPPER MOTOR CONTROL SYSTEM


Control System Lab Manual

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