CONTROL SYSTEMS AND SIMULATION LABORATORY MANUAL

CONTROL SYSTEMS AND SIMULATION LABORATORY MANUAL FOR YEAR 2016-2017 BY DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING DHULLAPALLY, KOMPALLY SE...
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CONTROL SYSTEMS AND SIMULATION LABORATORY MANUAL FOR YEAR 2016-2017 BY

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

DHULLAPALLY, KOMPALLY SECUNDERABAD-500014

St. MARTIN’S ENGINEERING COLLEGE DHULAPALLY, SECUNDERABAD Department of Electrical and Electronics Engineering

CONTROL SYSTEMS AND SIMULATION LAB LIST OF EXPERIMENTS: S.NO 1

NAME OF THE EXPERIMENT TIME RESPONSE OF SECOND ORDER SYSTEM

2

CHARACTERISTC OF SYNCHRO’S

3

PROGRAMMABLE LOGIC CONTROLLER-STUDY AND VERIFICATION OF TRUTH TABLES OF LOGIC GATES,SIMPLE BOOLEAN EXPRESSIONS ANSD APPLICATION OF SPEED CONTROL OF MOTOR

4

EFEECT OF FEEDBACK ON DC SEVRO MOTOR

5

TRANFER FUNCTION OF DC MOTOR

6

EFFECT OFP,PD,PI,PID CONTROLLER ON A SECOND ORDER SYSTEMS

7

LAG AND LEAD COMPENSATION-MAGNITUDE AND PHASE PLOT

8

TRANSFER FUNCTION OF DC GENERATOR

9

TEMPARATURE CONTROLLER USING PID

10

CHARACTRISTICS OF MAGNETIC AMPLIFIERS

11

CHARECTRISTICS OF AC SERVO MOTOR

12

PSPICE SIMULATION OF OP-AMP BASED INTEGRATOR AND DIFFERENTIATOR CIRCUITS

13

LINER SYSTEM ANALYSIS(TIME-DOMAIN ANALYSIS,ERROR – ANALYSIS)USING MATLAB

14

STABILITY ANALYSIS(BODE,ROOT LOCUS,NYQUIST)OF LINEAR TIME INVARIANT SYSTEM USING MATLAB

15

STATE SPACE MODEL FOR CLASSICAL TRANSFER FUNCTION USING MATLAB VERIFICATION

IN-CHARGE

HOD (EEE)

EXPERIMENT 1: TIME RESPONSE OF SECOND ORDER SYSTEM 1.1 OBJECTIVE: To compute the Time Response of a second order system (theoretically and practically). 1.2 RESOURCES: 1. 2. 3. 4.

Time Response of a second order system Trainer Kit C.R.O. Multi meter. Patch cords.

1.3 CIRCUIT DIAGRAM:

1.4 PROCEDURE: 1. Switch ON the Main supply and observe the signal source output by varying potentiometer 2. Apply Square wave or step input by varying amplitude potentiometer. 3. Make sure signal source is connected before the input of the second order system. 4. Now select square wave signal. Draw the input square wave signal. 5. Connect the output of square wave signal source to second order system using RLC. 6. Adjust the resistance value in the RLC circuit for different damping factors. 7. For different values of damping factor, observe second order response. 8. Verify time response specifications theoretically and practically.

Note:

Use 3 pin grounded main supply to the unit avoid line interference. Use proper CRO probes to see the output wave forms. For all these cases note down the time response specifications and compare them with theoretical values.

1.5 OBSERVATION TABLE: Sl No

R in Ohm

L in Henry

C in Micro Farad

1

500

2

0.32

2

1000

2

0.32

3

1500

2

0.32

4

2000

2

0.32

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

_

10000

2

0.32

Time domain specification Rise time (tr) Peak time (tp) Delay time (td) Setting time (ts) Peak over shoot (MP)

δ (Damping Factor)

Theoretical

td

tr

Mp

tp

ts

Practical

ess

1.6 MODEL GRAPH:

Fig: STEP RESPONSE OF AN UNDERDAMPED STSTEM.

CONNECTION DIAGRAM FOR SECOND ORDER SYSTEM USING RLC 1.7 RESULT:

1.8 PRE LAB QUESTIONS: 1. What is Time Response? 2. Define Delay Time, Rise Time, Peak Time, Peak Over Shoot, Settling Time? 3. Define type and order of a system? 4. Distinguish between Type and Order of a system? 1.9 POST LAB QUESTIONS 1. What is Steady State Error? 2. The damping ratio of system is 0.6 and the natural frequency of oscillation is 8 rad/ sec. Determine the rise time. 3. Define Positional Error Constant and Velocity Error Constant?

EXPERIMENT 2: CHARACTERISTICS OF SYNCHROS

2.1 OBJECTIVE:

To study

i) Synchro Transmitter characteristics. ii) Synchro Transmitter – Receiver Characteristics.

2.2 RESOURCES:

1. Synchro Transmitter – Receiver Kit. 2. Patch chords. 2.3 BLOCK DIAGRAM:

2.4 CIRCUIT DIAGRAM:

2.5 PROCEDURE:

Transmitter Characteristics:

1. Connect the mains supply to the system with the help of a cable provided. Do not connect any patch cords to terminals marked S1, S2 and S3, R1 and R2. 2. Switch ON mains of the unit. 3. Initially switch ON Sw1, starting from ZERO position, note down the voltage between stator winding terminals (i.e. VS1S2, VS2S3 and VS3S1) in a sequential manner. 4. Measure these voltages by using AC voltmeter provided in the trainer and note down the readings. 5. Plot a graph of angular position of rotor voltages for all three phases.

Transmitter-Receiver Characteristics:

1. Connect the mains supply cable. 2. Connect S1, S2 and S3 terminals of synchro transmitter to S1, S2 and S3 of synchro receiver by patch cords provided respectively. 3. Switch ON mains supply and also S1 and S2 on the kit. 4. Move the pointer i.e. rotor position of synchro transmitter Tx in steps of 30˚and observe the new rotor position in synchro receiver. 5. Observe that whenever Tx rotor is rotated, the Tr rotor follows if for both the directions of rotations and their positions are in good agreement. 6. Note down the input angular position and output angular position and plot the graph.

2.6 TABULAR COLUMN:

Transmitter Characteristics:

Sl. No.

Rotor Position

VS1S2

VS2S3

VS3S1

Transmitter-Receiver Characteristics: Sl. No.

Angular Position of Transmitter

Angular Position of Receiver

2.7 MODEL GRAPH: Synchro Transmitter Characteristics:

VS2S3

VS3S1

Amplitude (V)

VS1S2

Rotor Position

Synchro Transmitter-Receiver Characteristics:

θr

θS

2.8 PRE LAB QUESTIONS:

1. 2. 3. 4. 5.

Define the term "synchro."? Name the two general classifications of synchros? List the different synchro characteristics and give a brief explanation of each? Explain the operation of a basic synchro transmitter and receiver? Name the two types of synchro identification code?

2.9 LAB ASSIGNMENTS: 1. To draw the characteristics of receiver? 2. How it will works in navy systems? 3. Explain what happens when the rotor leads on Synchro transmitter and synchro receiver are reversed? 4. Draw the five standard schematic symbols for synchro and identify all connections?

2.10 POST LAB QUESTIONS:

1. 2. 3. 4. 5.

State the difference between a synchro transmitter and a synchro receiver? Explain the operation of a simple synchro transmission system? State the purposes of differential synchros? State the purposes and functions of multispeed synchro systems? List the basic components that compose a torque synchro system?

2.11 RESULT:

Hence, the synchro transmitter characteristics and synchro transmitter – Receiver characteristics are studied.

EXPERIMENT 3:PROGRAMMABLE LOGIC CONTROLLER

3.1 OBJECTIVE: To verify the truth tables of the logic gates using programmable logic controller.

3.2 RESOURCES: (i) PLC (ii) Dox mini – software.

3.3 Theory: The programmable logic controller, PLC is a solid state equipment design to perform the function of logical decision making for industrial control application.

3.4 General specifications:

Power supply

: 24 V DC

Standard I/O configuration Frequency

: up to 2 kHz

3.5 Addressing:

Input: 1000.0 to 1000.7 Output: 0000.0 to 0000.5 Flags: 2000.0 to 2031.F Ramword: 20000 to 20511 Table:40000 to 40099

: 8 inputs / 6 outputs

3.6 Instructions:

1.

-| |-

This is a single normally open.

2. 3. 4. 5.

-| /|This is a single normally close. –()This represents the coil of output. It is used for result display. – (/) This represents the inverter coil of output. –(S) This represents the latch type of coil. It is set to 1 and latched if the result of ladder programmed prior to it is 1 (true). 6. –(R) It resets the latched coil if the result of ladder programmed prior to it is 1 (true). 7. -| P|This instruction detects a +ve edge of the result just prior to this instruction. The output of this instruction remains ON for one scan period. 8. -| N|This instruction detects a –ve edge of the result just prior to this instruction. 9. – (MCR) -

Master control relay.

10. – (ME) -

Master end.

11. – (LBL) -

Label.

12. – (JMP) -

Jump.

13. – (FUN) -

To select the desired function from function block.

3.7 WRITING THE PROGRAM:

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

Dox- mini software works on only Mi-Dos mode. Restart in Dos mode. Open Dox-mini software menu. Press ‘Alt+F’ open the fine menu. Select ‘write to file’ option. Press enter key. Give the file name press enter key. Write the desired program in the blank space by using keys which are shown in window. 7. After writing the program press F5 to save the program.

3.8 Instructions to execute the program:

1. 2. 3. 4. 5.

Go to online menu by pressing ‘Alt+O’. Select download PC → PLC options press enter key. It displays ‘Down loading successful, put PLC in run mode”, press yes. Next it displays ‘PLC in RUN mode’ press OK. Execution is ended.

3.9 Precautions:

1. Apply ‘High’ or ‘Low’ to input terminals. 2. Don’t give any external power supply to the unit. 3. Don’t short make any interconnection between high (and24V) and low ground.

3.10 Example programs:

1. AND GATE:

2. OR GATE:

3. NOT GATE::

4. EX – NOR:

5. NAND GATE by using De-Morgan’s law:

6. NOR GATE by using De-Morgan’s law:

7. MULTIPLEXER:

In this program 1000.6 and 1000.7 are the selected lines to select the inputs from 1000.0 to 1000.3 we get the display as 0000.0.

8. Half Adder:

Note:

A (1000.0) & B (1000.1) are inputs Sum is output at 0000.1 carry is 0000.0

3.11 Result: All the truth tables of the logic gates have been verified using programmable logic controller.

EXPERIMENT 4: EFFECT OF FEEDBACK ON DC SERVOMOTOR 4.1 OBJECTIVE :

To study the performance characteristics of a dc motor angular position control system

4.2 RESOURCES:

1. DC Position control unit 2. Oscilloscope 4.3 DESCRIPTION OF THE EQUIPMENT:

The equipment consists of a DC Motor, connected to a load through gears. The angular position is sensed by a 3600 rotation potentiometer attached to it. A calibrated disk mounted on the potentiometer indicates its angular position in degrees. A small tachogenerator is also attached to provide rate feedback.

4.4 RATING OF THE MOTOR:

12V DC, 1.2 amps, 50 rpm Torque: 750 gm-cm

The main unit consists of the following systems:

COMMAND: Two operating modes are provided. Continuous command is given by the rotation of the potentiometer with a calibrated disc showing the angle. A step command can also be given which is equivalent to about 1500 through a switch.

ERROR DETECTOR: This is a 4-input, 1-output block. The inputs and outputs are as indicated in figure

GAIN BLOCKS: The forward path gain is adjustable from 0 to 10 and the tachogenerator channel has a gain, which can be varied from 0 to 1.

DRIVER: The driver circuit is a power amplifier suitable to run the motor in either direction.

WAVEFORM CAPTURE UNIT:

Since the response of the mechanical system is too slow for an oscilloscope to capture a waveform, capture/display unit is provided to store the wave form and then display it in ordinary oscilloscope.

4.5 PROCEDURE: Procedure for operation of waveform capture /display:

1. Switch on power/press reset. The unit goes into display mode. It displays X and Y axis. 2. Calibrate the time scale of the display. Feed the X-output to Y-input of the CRO and determine its time period and amplitude and find time/voltage. This gives the required calibrations. 3. Press mode switch. The unit goes into capture mode and is ready to record the response. At the end of the capture cycle the mode automatically shifts top display mode and the waveform is displayed on the scope.

Position control through continuous command:

1. Step switch should be in off position. 2. Move the command potentiometer in small steps and observe the rotation of the load potentiometer. Record σR, VR, σ02 and V0 for as few values of KA. 3. Calculate the error for different amplifier gains KA Position control through step command:

1. Ensure tachogenerator feedback is negative and the gain is zero. 2. Adjust reference potentiometer to get VR=0 and set KA=2 3. Connect a CRO. Connect the output to the y-plates and X-output to the X-plates. Press the switch to capture mode. 4. Apply the step input. Wait till the storage is complete and the mode returns to display mode. The waveform will be observed on the scope. Trace the waveform. 5. Compute Mp, Tp, ts and steady state error. Find σ and Vn. Step response with tachogenerator feedback:

1. Adjust KA=7 and KD=0.1. Obtain the response and find all the performance measures. 2. Repeat for different values of KD.

4.5 RESULT:

EXPERIMENT 5: TRANSFER FUNCTION OF DC MOTOR

5.1 OBJECTIVE:

To study the DC motor and DC generator characteristics DC motor speed-torque characteristics Step response of DC motor

5.2 CIRCUIT DIAGRAM: MEASUREMENT OF ARMATURE RESISTANCE 100Ω/2A

-

+ A

+ (0-24)V

V

DVM -

A M AA

5.3 PROCEDURE:

1. Connect the circuit as shown in the figure. Keeping field circuit open. 2. Motor shaft should not rotate. 3. Vary the input voltage from 0-100V from the controller and note dowr ammeter and 4. 5. 6.

voltmeter readings and enter in the tabular column. Calculate the Resistance = V/I Repeat the same for different input voltages. The average resistance value gives the armature resistance.

TABLE Va Volts

1a Amps

Ra = VJh Ohms

5.4 MEASUREMENT OF ARMATURE RESISTANCE

A

A M 0-230V AV AA

V

1. Make the connections as given in the circuit diagram. Keep field circuit open. 2. Vary the input AC Voltage from the controller and note down voltmeter and Ammeter readings and enter in the tabular column. 3. Calculate za, xa and La 4. Repeat the same for different input voltages 5. The average Value gives the La

f = 50Hz Ra – ohms

SI. No.

Va

la

Za = Va Ia

Xa =√Za2 Ra2

La= Xa 2πf

5.5 MEASERMENT OF FIELD RESISTANCE

+

A

F

0-230V AC

+ V

FF

Connect the circuit as shown in figure. Keep the armature winding open Vary the input DC Supply from the controller and note down voltmeter and ammeter readings. V / l ratio will give the field Resistance 4. Repeat the same for different input voltage and find out RF. 5. The average value gives RF. 1. 2. 3.

SLNo.

Vf Volts

If Amps

Rf=Vf/Ifohms

5.6 MEASERMENT OF FIELD INDUCTANCE

+

A

F

0-230V AC

+ V

FF

1. Connect the circuit as shown in the figure 2. Keep the armature winding openVary the input AC supply from the controller and note down voltmeter and ammeter readings and enter in the tabular column. 3. Calculate

Zf,Xf and Lf

4. Repeat the same for different input voltages.

TABLE

SI. No.

Vf

If

Zf = Vf If

X f=√Zf2 -R f2

L f=X f 2∏f

5.7 ARMATURE CONTROLLED DC MOTOR

Load Test on DC Motor:A F1

A

F2

M

0 – 220V

AA

1. Circuit connections are make as per the circuit diagram. 2. Connect 220V fixed DC supply to the field of DC motor and Brake drum belt should be loosened. 3. Start the motor by applying 0-220V variable DC supply from the controller till the motor rotates at its rated speed. 4. Note down meter readings which indicates no load reading. 5. Apply load in steps upto rated current of the motor and note down corresponding IL, N, F1and F2 readings. Switch OFF the armature DC supply using Armature supply ON/OFF switch and them Radius: R= 6.5 cm switch OFF the MCB.

T ∆T KT = ∆La Ia

TABLE

SLNo.

h

F2

N rpm

T=(F1~F2) 6.5X9.81N-cm

Speed Control by Armature Voltage Control

A

A

F

0-220V DC V

M

220V DC AA

FF

1. Circuit connections are mode as per the circuit diagram. 2. Connect 220V fixed DC supply to the motor field keep the armature control pot at its minimum position and switch at OFF position. 3. Switch ON the MCB, Switch ON the armature control switch. Vary the armature voltage and note down the speed and the corresponding meter readings. 4. Repeat the same for different annature voltages.

SI. No.

Ia

N

V

Eb = V - I a R a

Draw the graph of back emf V/S speed.

ω=2∏N 60

Eb ∆ Eb ∆ω ω

Transfer Function of Armature Controlled DC Motor = KT S[[Ra+ SLa] [SJm + fm] + KT Kb] Ra

=

Armature Resistance

La

=

Armature Inductance

la

=

Armature Current

Eb

=

Back emf

T

=

Torque developed

J

=

Moment of Inertia = 0.024 Kg-m2

B

=

Frictional Co-efficient - 0.8

Kb

=

Back emf constant

KT

=

Torque Constant

By KirchofFs law IaRa + La dla dt Since

flux

+

Eb = v

is

constant

Torque

is

proportional to la. Tm α Ia Tm α KTIa Also for Mechanical System J

dѲm2 dt 2

+ B dѲm dt

_ Tm

Also Back emf Eb d t o angular velocity of shaft Eb =Kb

dѲm dt

Dynamic Equation Tm = KT la Eb = Kb Wm

Laplace Equipment Tm (s) - KT la (s) Eb (S) = Kb Wm (s)

Va(s) - Eb(s) = Ra la (s) + S1a Ia(s) Tm(s) = JmS2 Ѳm (s) + Sfm Ѳm(s)

Va - Eb = Ra la + La dla dt Tm = Jm d2Ѳm + fm dѲm dt2 dt

By solving the Laplace equation we obtain the transfer function as Ѳ (s) V(s) Kj S ([Ra + SLa] [SJm + fm) + KT Kb]) 5.8 TRANSFER FUNCTION OF FIF.T.1) CONTROLLED OF DC MOTOR PROCEDURE: -

+ A A

Ia A M V 0-220V

AA

100-200V

1. Make the connections as given in the circuit diagram. 2. Motor field voltage should be maximum and belt is loosened 3. Switch ON the supply to the controller unit. Switch ON the armature supply. Vary the armature voltage till the motor speed comes to rated speed and note down meter readings. 4. Vary the field supply till we obtain 20% above rated speed of motor and note down the readings.

TABULAR COLUMN 1) Field Control Method: SI. No.

If amps

I* amps

N speed

V volts

T = I,Ia

5.9 TRANSFER FUNCTION DERIVATION Transfer function of a field control DC motor relates angular shift in the shaft and the field input voltage. The armature current Ia supplied is kept constant. The field current 1/produces a flux in the motor which in turn produces a torque at the motor shaft. The moment of inertia and coefficient of vises friction are Jm (kg/m2) and fm (N-m / rad/see) respectively. The angular shaft in the motor shaft being 9m radians or angular velocity 'W rad/sec. Dynamic Equations

Laplace Equations

Tm α I f Tm = Kf I f

Tm (s) = KfIf(s)

Vf = R f I f + L f d I f dt Tm = Im d2Ѳm + fm dѲm dt2 dt

Tm (s) = S2 Jm Ѳm(s) + S fm Ѳm(s)

or

or

Tm = Jm dωm + fm ωm dt

Tm (s) = S Im ωm (s) + Fm ωm (s)

Vf (s) = Rf If (s) + SLf If (s)

The overall Transfer function of field control D C motor is

Transfer function =

Kf

S(Rf+SLf)(SJm + fm) CALCULATION; From graph (K/) = Field resistance (Rf) = Inductance (Lf)

=

Transfer function

=

Kf

S(R f +SL f ) (SJm + fm) 5.10 TRANSFER FUNCTION OF DC - GENERATOR: (Separately excited and Self Excited) This setup consists of the following units to conduct the above experiment a. DC - Motor - Generator set - 0.5HP/220V / 1500 rpm. b. Controller unit suitable for the above motor - Generator set with Digital meters DC motor - Generator Set - 0.5HP/220V / 1500 rpm Field Resistance Rf - 376 ohms Field Inductance L f - 13.5H

TRANSFER FUNCTION OF DC GENERATOR (Separately Excited)

+

A M

V

Eg -

AA

0-220V DC

F

G

F

FF

FF

If 220V DC 100 - 220V DC 1. 2. 3. 4.

Make the connections as given in the circuit diagram Connect 220V fixed DC supply to the Motor field Connect 100-220V Variable DC supply to the Generator field Connect 0-220V Variable DC supply to the armature.

5. Switch on the MCB keeping armature voltage control pot at its minimum position & ON/OFF switch at OFF position and also variable field voltage pot at its maximum position 6. Now switch ON the Armature control switch and vary the armature control potentiometer till the motor rotates at its rated speed. 7. Note down If and Eg and entered in the tabular column. 8. Now vary the generator field supply and note down Eg for different Igs and entered in the tabular column. 9. Draw the graph of Eg volts v/s If. Transfer function = K E / Lf SfRf*/Lf

Eg Kg =

If

∆ Eg ∆ If

= Kg/Lf S + R f Lf

SI. No.

If Amps

Eg Volts

EXPERIMENT 6: EFFECT OF P, PI, PID CONTROLLER ON A SECOND ORDER SYSTEM

6.1 OBJECTIVE: To study the steady state performance of an analog P, PI & PID controller using simulated system. 6.2 RESOURCES:

3.

PID Controller

4.

Patch Chords

6.3 BLOCK DIAGRAM:

6.4 PROCEDURE:

1.

Make the connections as per the block diagram.

2.

Set input DC amplitude to 1V.

3.

Adjust I, D to Zero.

4.

For various values of P, measure Vf, Vi and Ve using meter provided on the kit and note down the readings.

6.5 TABULAR COLUMN:

P Controller

Sl. No.

P

(Variable Proportionate gain)

Vi

Vf

Ve

1. 2. 3. 4.

PI Controller

(Constant Proportionate gain and Variable Integral gain)

P=

Sl. No.

I

Vi

Vf

Ve

1. 2. 3. 4.

PID Controller

(Constant Proportionate gain & Integral gain and Variable Differential gain)

P=

Sl. No. 1. 2. 3. 4.

I=

D

Vi

Vf

Ve

6.6 PRELAB QUESTIONS:

1. What is a controller? 2. What is the difference between a compensator and controller? 3. Write a brief note about Proportional Controller? 4. Write a brief note about Derivative Controller? 5. Write a brief note about Integral Controller?

6.7 LAB ASSIGMNETS:

1. To observe open loop performance of building block and calibration of PID Controls? 2. To study P, PI and PID controller with type 0 system with delay? 3. To study P, PI and PID controller with type 1 system?

6.8 POSTLAB QUESTIONS:

1. Write a brief note about PID Controller? 2. Compare the performance of PI and PD controller? 3. Which controller is used for improving the transient response of the system? 4. Which controller is used for improving the steady state response of the system? 5. What is the purpose of PID controller?

6.10 RESULT:

Hence the steady state performance of an analog P, PI & PID controller has been studied using simulated system.

EXPERIMENT 7: STUDY OF LEAD – LAG COMPENSATION NETWORKS 7.1 OBJECTIVE: To study of Lead-Lag compensation networks. 7.2 RESOURCES: 1.

Lead-Lag network study unit.

7.3 CIRCUIT DIAGRAM: Lag Compensation Network:

Lead Compensation Network:

Lag-Lead Compensator Network:

7.4 PROCEDURE: 1.

Switch on the main supply to unit observe the sine wave signal by varying the frequency and amplitude potentiometer.

2.

Now make the network connections for Lag, Lead and Lead-Lag networks connect the sine wave output to the networks input.

3.

Note down the peak actuator input using digital voltmeter provided, now the meter will shows peak voltage.

4.

Set the amplitude of sine wave to some value ex: 3 Volts peak, 4Volts Peak etc.,

5.

Now vary the frequency and note down frequency, phase angle difference and output voltage peak for different frequencies and tabulate all the readings.

6.

Calculate the theoretical values of phase angle difference and gain compare these values with the practical values.

7.

Plot the graph of phase angle versus frequency (phase plot) and gain versus frequency (magnitude plot).

8.

Repeat the same for different values of R and C.

9.

Repeat the same for different sine wave amplitude.

10. Repeat the same experiment for lead and lag-lead networks. FORMULAE: Lag Network : Phase angle, Φ = tan -1ωRC

1 Gain =

RC

(

ω2 + 1

RC )

2

FORMULAE: Lead Network:

(

-1 Phase angle, Φ = tan 1

Gain GC=

( jω) =

Vout Vin

ωRC

)

jωRC 1 + jωRC

Vout R = Vin Z

Z=

(

R2 + 1

ωC )

2

Vin = V FORMULAE: Lag-Lead Network:

R1C1 = τ1 , R 2C 2 = τ 2 β=

R1 + R 2 >1 R2

β=

10k + 10k = 2.0 10k

α=

R2 0.5 < 1 R1 + R 2

The transfer function of such a compensator is given by

Ê s + 1 ˆÊ s + Á τ1 ˜Á ¯Ë G (s) = Ë Ê s + 1 ˆÊ s + Á βτ1 ˜Á Ë ¯Ë (lag)

1 ˆ˜ τ2 ¯ 1 ˆ˜ ατ 2 ¯

(lead)

CALCULATIONS:

R1C1 = τ1 =0.002, R 2C2 = τ 2 =0.002 ω0 = 1

τ1×τ2

The frequency at which phase angle zero is ω0 = 1

τ1×τ2

ω0= 2??=500 f =500/6.28 = 80Hz

7.5 TABULAR COLUMN : Lag Network : Sl. No.

Frequency (Hz)

Phase angle Ф in degrees

Output voltage VO (Volts)

Gain = Vo/Vin

Phase angle Ф in degrees

Output voltage VO (Volts)

Gain = Vo/Vin

Phase angle Ф in degrees

Output voltage VO (Volts)

Gain = Vo/Vin

TABULAR COLUMN : Lead Network : Sl. No.

Frequency (Hz)

TABULAR COLUMN : Lag-Lead Network: Sl. No.

Frequency (Hz)

7.6 PRELAB QUESTIONS: 1. Write a brief note about Lag Compensator. 2. Write a brief note about Lead Compensator. 3. Write a brief note about Lag Lead Compensator.

4. Which compensation is adopted for improving transient response of a negative unity feedback system?

7.7LAB ASSIGNMENTS: 1. To study the open loop response on compensator and

Close loop transient

response. 2. The max. phase shift provided for lead compensator with transfer function 3. G(s)=(1+6s)/(1+2s)

7.8 POSTLAB QUESTIONS: 1. Which compensation is adopted for improving steady response of a negative unity feedback system? 2. Which compensation is adopted for improving both steady state and transient response of a negative unity feedback system? 3. What happens to the gain crossover frequency when phase lag compensator is used? 4. What happens to the gain crossover frequency when phase lead compensator is used? 5. What is the effect of phase lag compensation on servo system performance?

7.9RESULT: Hence, the Lag, Lead and Lead-Lag compensation networks are studied.

EXPERIMENT 8: TRANSFER FUNCTION OF DC GENERATOR 8.1 OBJECTIVE: To determine the transfer function of separately excited DC generator using thyristor controller. 8.2 RESOURECES:

1. Thyristor controller. 2. MGSET (Motor Generator) 3. DC Ammeter 4. DC Voltmeter 5. AC Ammeter 6. AC Voltmeter

8.3 BLOCK DIAGRAM:

Vf

Vf

1/Rf+sLf

Kg/Rf+sLf

Kg

Eg

Eg

Vf(s)

(Kg/Lf) /

Measurement of Field Resistance:

Measurement of Field Inductance:

Eg

AC Voltage Controller:

8.4 PROCEDURE: 1. Make the connection as given in the circuit diagram. 2. Connect motor field supply to field adjust the motor field voltage to its rated voltage 220V. 3. Connect generator field supply to field of generator and keep the generator field potentiometer to its minimum position. 4. Now switch ON the armature thyristor controller till the motor runs at its rated speed -1500 r.p.m. 5. Now note down If and Eg and enter in the tabular column. 6. Now vary generator field controller and note down Eg for different If ‘s and enter in the tabular column. 7. Draw the graph of Eg volts VS If amps. Kg = ∆ Eg / ∆ Ig

Measurement of Field Resistance:

1. Make the connection as given in the circuit diagram. 2. Armature winding should be kept open. 3. Apply suitable DC voltage form GENERATOR FIELD CONTROLLER and note down V and I readings. 4. V/I ratio which given field resistance. 5. Repeat the step for various input DC voltage and find out field resistance. 6. Take the average value as the field resistance.

Measurement of Field Inductance:

1. Make the connection as given in the circuit diagram. 2. Armature winding should be kept open. 3. Connection thyristor controller as AC voltage controller as shown. 4. Vary the AC voltage applied and note down Vf and If and enter the readings in the tabular column and find out Zf , Xf and Lf . 5. Note down Vf and If for different AC voltage controller calculate Lf. 6. Take the average value as the field inductance.

8.5 TABULAR COLUMN:

1. Measurement of Field Resistance: Sl. No.

Vf (volts)

If (Amps)

Rf = Vf/If (Ω)

2. Measurement of Field Inductance: Sl. No.

Vf (volts)

If (Amps)

Zf = Vf/If (Ω)

Xf =√ Zf2 – Rf2

Lf = Xf/2∏f

3.

AC Voltage Controller:

Sl. No.

Ef (volts)

If (Amps)

Kg (Ω)

8.6 CALCULATIONS:

Field Resistance Rf =

Field Inductance Lf =

Kg = ∆ Eg / ∆ If

Transfer function of separately excited generator T(s) = Eg(s)/ Vf(s)

8.7 MODEL GRAPHS:

Eg

∆ ∆ If

8.8 PRE LAB QUESTIONS 1. What are the different types of generator? 2. Explain the characteristics of shunt generator? 3. Explain the characteristics of series generator? 4. Explain the characteristics of compound generator?

8.9 LAB ASSIGNMENTS:

1. To determine the transfer function of DC series generator using thyristor controller? 2. To determine the transfer function of DC shunt generator using thyristor controller? 3. To determine the transfer function of DC compound generator using thyristor controller? 4. To determine the transfer function of synchronous generator using thyristor controller?

8.11RESULT:

Hence the transfer function of separately excited DC generator using thyristor controller has been determined.

EXPERIMENT 9:TEMPERATURE CONTROL USING PID CONTROLLER

9.1 OBJECTIVE : To study the performance of PID controller used to control the temperature of an oven. 9.2 RESOURCES: 1. PID Controller 2. Patch chords Theory:

error e(t) Reference

Relay

Hys

Actuator

Plant (Oven)

Proportional

KP

Derivation

KD Integral

KD 0

10 m V/ C

Temperature Block Diagram Feedbackof the Temperature Controller

Output e(t) Temperature

CONTROLLER In temperature control systems, commonly used basic control actions are v ON-OFF or relay. v Proportional. v Proportional -- Integral. v Proportional -- Integral -- Derivative.

A brief description of these are give below : (a) ON - OFF or Relay type Controllers, it is a two position controllers, consist of a simple and inexpensive switch/relay. Therefore, it is used very commonly in both industrial and domestic control systems. Some applications include air conditioner, refrigerator, oven, healers with thermostat. Solenoid operated two position values are commonly used in hydraulic and pneumatic systems. The basic input-output behaviors of this controller is shown in Fig. 4. The two positions of the controller are M1 and M2, and H is the Hysteresis or differential gap. The Hysteresis enables the controller output to remain at its present value till the input or error has increased a little beyond zero. Hysteresis helps in avoiding too frequent switching of the control, although a large value results in greater errors. The response of a system with ON-OFF controller is shown in Fig. 5. Describing function technique is a standard method for the analysis of non-linear systems, for instance, one with an ON-OFF controller.

(b) Proportional Controller is simply an amplifier of gain Kp which amplifies the error signal and passes it to the actuator. The noise, drift and bias currents of this amplifier set the lower limit of the input signal which may be handled reliably and therefore decide the minimum possible value of the error between the input signal and output. Also the saturation characteristics of this amplifier sets the linear and non-linear regions of its operation. A typical proportional controller may have an input-output characteristics as in Fig. 6. Such controller gives non-zero steady state error to step input for a type-0 system as indicated earlier. The proportional (P) block in system consists of a variable gain amplifier having maximum value. Kpmax of 20.

e(t)

Fig. 6. Proportional Controller with

(c)

Proportional-Integral (Pi) Controller : Mathematical equation of such a controller is given by : t t m(t) = Kpe(t) + K1 ∫e(t) dt= Kpe(t) + 1/T1∫ e(t) dt o o

Kp e(t)

m(t)

+

and a block diagram representation is shown in Fig. 7. It may be easily seen

K1/s

that this controller introduces a -pole at the origin, i.e. increases the system type number by unity. The steady slate error of the system is therefore reduced or

Fig. 7. P

I

Controller

eliminated. Qualitatively, any small error signal e(t), present in the system, would get continuously integrated and generate actuator signal

mm forcing the plant output to exactly correspond to trie reference input so that the error is zero. In practical s> sterns, the error may not be exactly zero due to imperfections in an electronic integrator caused by bias current needed, noise and drift present and leakage of the integrator capacitor. The integrator (I) block in the present system is realized with a circuit shown in Fig. 8 an has a transfer function. Gr(s) = l/(41s) = K,/s

(2)

The integral gain is therefore adjustable in the range 0 to 0.024 (approx.). Due to the tolerance of large capacitance's, the value of K| is approximate.

(d) --- Proportional - Derivative (PD) Controller :

Kp

Mathematical equation governing the operation of this controller is given as :

e(t)

m(t)

+

m (ct) = Kp e(t) + Kp de(t) /dt

KD/s KD/s

and a block diagram representation is shown in Fig. 7 (b) = Kpc(t) + Td dc(t) /dt. Fig. 7(b).

P-D Controller.

The derivative (D) block in this system is realized with the circuit of Fig. (9) This has a transfer function of Cd(s) = 19-97(s) approx.

(3)

+ 1000µ

e(t)

+ 1000µ

__

82K

+ I

Fig. 8. Circuit for Integrator

The derivative gain is there for adjustable in the rage 0 to 20 approximately. Again, the appreciation is due to the higher tolerance in the value of large capacitances. Mathematical equations governing the operation of this controller is as (e)

Proportional-Integral- Derivative(P1D) Controller : Mathematical equations governing the operation of this controller is as 1 m(t) = Kpe(t) + K1 ∫e(t)dt = KDde(t) / dt o 1 = Kpe(t) + 1/T1 ∫e(t)dt = TDde(t) / dt 0 So that in the lap lace transform domain,

M(s) / E(s) = (Kp + TDS + 1T1s) A simple analysis would show that the derivative block essentially increases the damping ratio of the system and therefore improves the dynamic performance by reducing overshoot. The PID controller therefore, helps in redacting the steady state error with an improvement in the transient response. The derivative (D) block in this system is realized with the circuit of Fig. 9. This has a transfer function. GD(S) . = 19.97 s (approx. )

(4)

The derivative gain is therefore adjustable in the range 0 to 20 approximately. Again, the approximation is due to the higher tolerance in the values of large capacitance's. PID controller is one of the most widely used controller because of its simplicity. By adjusting is coefficients Kp. K D (or TD) the controller can be used with a variety of system the process of setting the controller coefficients to suit a given plant is known as tuning are man) methods of 'tuning' a PID controller. In the present experiment, the method of Ziegler-Nichol has been introduced which is suitable for the oven control system although better methods arc available and may be attempted. TEMPERATURE MEASUREMENT : The oven temperature can be sensed by a variety of transducer like thermostat, thermocouple, RTD and IC temperature sensors. In the present set-up, the maximum oven temperature is around 90 C which is well within the operating range of IC temperature sensor like LM35. further, these sensors are linear and have a good sensitivity, viz. 1µA/K. associated electronic circuits convert this output to 10m V/0C which may be easily measured by a DVM. The time constant of the sensor has however been neglected in the analysis since it is insignificant compared with the oven time constant.

EXPERIMENT AL WORK : A variety of experiments may be conducted with the help of this unit. The principal advantage of the mat is that all power sources and metering are built in and one needs only a watch to be able to note down the temperature readings at precise time instants. After each run the oven has to be cooled to early me room temperature, which may take about 20-30 minutes.

9.3PROCEDURE: Plant identification is the first step before an attempt can be obtained experimentally from its before mi attempt can be made to control it. In the present case the oven equations are obtained experimentally from its step response as out lined below : (1)

Keep Run/Wait switch to 'WAIT' S2, to Set/Meas. Switch to "SET" and open 'FEEDBACK' terminals, (refer panel drawing).

(2)

Connect P output to the actuator input and switch ON the unit.

(3)

Set P potentiometer to 0.5 which gives Kp=10. Adjust reference potentiometer to read 5.0 on the DVM. This provides an input of 0.5 V to the driver.

(4)

Put Set/Meas. Switch to the 'MEASURE’ position and note down the room temperature.

(5)

Put Run/Wai switch to 'RUN' position and note temperature reading every 15 sec, till be temperature becomes almost constant.

(6)

Plot temperature-lime curve on a graph paper. Referring to Fig. 1 0 , calculate T| and T2 and hence write the transfer function of the oven including its driver as G(s)

:

K e x p ( 0 -sT2) / (1+sT1), with T i n °C.

ON - OFF CONTROLLER : (7)

Keep Run/Wait switch to ' W A I T position and allow the oven to cool the room temperature. Short " F E E D BACK' terminals.

(8)

Keep Set/Meas. switch to the 'SET* position and adjust reference potentiometer to the desired output temperature, say 50(,C. by seeing on the digital display.

(9)

Connect R output to the driver input. Outputs of P, D, and I must be disconnected from driver input. Select "HI” or “LO” value of Hysteresis. (First keep the Hysteresis switch to L O ) .

(10)

Switch Set/Meas. switch to 'MEASURE' and Run/Wait switch to 'RUN' position. Read and record oven temperature every 15/30 sec., for about 20 minutes.

(11)

Plot a graph between temperature and time and observe the oscillations (Fig. 12) in the steady state. Note down the magnitude of oscillations.

(12)

Repeat above steps with the “J\HI” setting for Hysteresis and observe the rise time, steady-state error and percent overshoot.

PROPOTIONAL CONTROLLER : Ziegler and Nichols suggest the value of Kp for this condition as : Kp

= ( 1 / K ) x T 1/ T 2

(13)

Starting with a cool oven, keep switch Run/Wait switch to 'WAIT' position and connect P output to the driver input. Keep R. D and 1 outputs disconnection. Short 'FEEDBACK" terminals.

(14)

Set P potentiometer to the above calculated value o KP, keeping in mind that the maximum gain is 10.

(15)

Select and set the desired temperature to say 50 C.

(16)

Keep Run./Wait switch to "Run" position and record temperature readings as before.

(17)

Plot the observations on a linear graph paper and observe the rise time, steadystate error and percent overshoot (Fig. 13).

PROPORTIONAL-INTEGRAL CONTROLLER : Ziegler and Nichols suggest the value of KPand K1 for this conditions as : Kr = (0.9/K) x T1/T2; T1 = 1 / K 1 = 3.3 T2, giving K 1 = 1 / 3.3 T2. (18)

(19) (20) (21) (22)

Starting with a cool oven, keep Run/Wait switch to 'WAIT' position, connect P and I outputs to the actuator input and disconnect R & D output. Short "FEEDBACK" terminals. Set P & I potentiometers to the above value of Kp and K1 respectively, keeping in mind that the maximum value of Kp is 10 and that of K1 is 0.024. Select and set the desired temperature to say 50°C. Keep Run/Wait switch to 'RUN* position and record temperature readings as before. Plot the response on a graph paper and observe the steady-slate error and percent overshoot. (Fig. 14).

PROPORTIONAL - DERIVATIVE CONTROLLER : Ziegler and Nichols suggested the value of Kp and Kd for this condition as KP = ( 0.9 / k ) T1 T2 Kd = (T0 = 0.5T2 (23)

Starting with a cool oven, beep run / wait switch to 'WAIT' position and connect P and output to actuator input keep R and I output disconnected. Short feed back terminals.

(24)

Set P. and D potentiometer according to the above calculated values keeping in mind that the maximum values for there arc 20 and 20 respectively.

(25)

Select and set the desired temperature to say 50 C.

(26)

Keep Run / Wait switch to ' R U N * position and record temperature - time readings.

(27)

Plot the response on a linear graph paper and observe the rise time steady state errors etc.

PROPORTIONAL - INTEGRAL - DERIVATIVE CONTROLLER: Ziegler and Nichols suggest the value of Kp K D and K1 for this controller as : Kp = ( 1 . 2 K ) x T 1 / T 2 : : Tx = 2T2, giving K1 = 1 / 2 T2. Ko = TD = 0.5T2. (23) Starling with a cool oven, keep Run/Wail switch to ' W A I T , position and connect P, D & I outputs to actuator input. Keep K output disconnected. Short 'FEEDBACK' terminals. (24)

Set P, I & D potentiometers according to the above calculated value of Kp, K1 and KD keeping in mind that the maximum value for these are 20. 0.024 and 23.5 respectively.

(25)

Select and set the desired temperature to say 50 0C.

(26)

Keep Run/Wait Switch to "RUN", and record temperature-time readings.

(27)

Plot the response on a linear graph paper and observe t h e rise time, steadystale error and percent overshoot. (See Fig. 1 5 ) . Compare the results to the various controller options.

CALCULATIONS:

(1)

Open - Loop Measurement : The constant for oven pulse controller is given by : Final temperature Oven K =

Ambient Temp.

---------------------------------------------------------Input (Volts)

From the graph between temperature and time (Fig. 1 1 ) , the final oven temperature for input of 0.5 volt is 70.4°C. Hence. K =53.2 / 0.5

106.4 o C / V .

With reference to Fig. 10. T 1 and T2, as measured from the open-loop graph are : T1 = 2 sec.. T2 = 237 sec. (Note that these values may differ from unit to unit). (c)

Calculation for Kp, K|, K D :

The coefficient settings according to Ziegler and Nichols are different for different types of control. The calculations for them are illustrated below:

(i)

P Control : Kp = (l/K) x T1 / T2 With temperature

(1/158.8.4) x 337 / 21 - 0.071 V / ° C .

sensor sensitivity of 10mV/t and maximum gain of

P – Amplifier as 10KPmax = 0.1V/°C. Hence P-setting required for proportional control is 71%. The temperature Vs. time plot is shown in Fig. 1 3 . (ii)

P - l Control : Kp = (0.9 / K) x T1 / T2 - (0.9/158.8.4) x 237 /21 - 0.639 Hence. P-setting required

63.9%.

T 1 = 3.3 T2 = 3 . 3 x 2 1 = 6 9 . 3 ; K1 = 1/T, - 1/69.3 = 0.0144 sec. K I m a v = 1/ 4 1 = 0.0244, [ s e c F q . 2 } 1 - setting = (0.1 4 4 / 0.0244) x 100 = 0.59 x 1 0 0 - 59%. The temperature vs time plot is shown i n Fig. 1 4 . iii) PD Control : Kp = (0.9/ 158.8)237 21 =0.0639 Hence P - retting required 63.9%. T1 = 3.3. T2 =3.3x21 =69.3. Kd = Td = 0.5 T2 =0.5x21 = 10.5 Sec. D - Setting = (10.5 / 20) \ 100 = 52.5% The temperature vs time plot is shown in Fig. 14 A. (iv) P - I - D Control : KP = (1.2/K) x T1/T2 = (1.2/158.8) x 237/21 = 0.853 This gives a P-coefficient setting of 85.3% T1 = 2.0T2 = 2.0 x 21 = 42; K1 = 1 / 42 = 0.0239/sec. = 0.0238/see. I - setting = (0.0239 / 0.0244) x 100 = 0.979 x 100 = 97.8%. KD = TD = 0.5 T2 = 0.5X21 = 10.5 sec. K Dmax = 23.5 sec, [sec Eq. 3j D-setting = (10.5/23.5)x 100 = 0.447 x 100 = 44.7%. The temperature vs time plot is shown in Fig. 15.

Results:

EXPERIMENT : CHARACTERISTICS OF MAGNETIC AMPLIFIER

A=

-0.5000

-0.5000

1.0000

0

0 0

B=

-1.0000

0

0

1.0000 0

0 0

1.0000

0 0

1 0 0 0

C=

0

1.0000

1.5000

1.000

EXPERIMENT 10: CHARACTERISTICS OF MAGNETIC AMPLIFIER

10.1 OBJECTIVE:

To study the control characteristics of Magnetic Amplifier i) Series connection

ii) Parallel Connection

10.2 RESOURCES:

1. Magnetic Amplifier Demonstration kit. 2. A Rheostat of 50Ω and 5A. 3. Patch chords

10.3 CIRCUIT DIAGRAM:

Series connected Magnetic Amplifier:

Parallel connected Magnetic Amplifier:

10.4 PROCEDURE:

Series Connected Magnetic Amplifier:

1. Keep toggle switch in position D on front panel. 2. Keep control current setting knob at its extreme left position 3. (i.e., Rotate in anti clockwise direction) which ensures Zero control current at starting.

4. With the help of patch cords connect following terminals on the front panel a. Connect AC to A1 b. Connect B1 to A2 c. Connect B2 to L 5. Now connect variable rheostat (or bulb) on front panel. 6. Switch ON the unit.

6. Now increase control current by rotating the corresponding knob in steps and note down the Control Current and corresponding Load Current. 7. Plot the graphs of Load Current Vs Control Current. Parallel Connected Magnetic Amplifier:

1. Keep toggle switch in position D on front panel. 2. Keep control current setting knob at its extreme left position (i.e. Rotate in anti clockwise direction) which ensures Zero control current at starting. 3. With the help of patch cords connect following terminals on the front panel i) Connect AC to A1 ii) Connect A1 to A2 iii) Connect B1 to B2 iv) Connect B2 to L

4. Now connect variable rheostat (or bulb) on front panel. 5. Switch ON the unit. 6. Now increase control current by rotating the corresponding knob in steps and note down the readings of Control Current and corresponding Load Current. 7. Plot the graphs of Load Current Vs Control Current.

10.5 TABULAR COLUMN:

Series Connected Magnetic Amplifier:

Sl. No.

Control Current (IC)

Load Current (IL)

Parallel Connected Magnetic Amplifiers:

Sl. No.

Control Current (IC)

IL (mA)

10.6 MODEL GRAPH:

IC (mA)

Series connected Magnetic Amplifier

Load Current (IL)

IL (mA)

Parallel connected Magnetic Amplifier

IC (mA)

10.7 PRE LAB QUESTIONS: 1. What is a magnetic amplifier? 2. What are the applications of magnetic amplifier? 3. What are the advantages and disadvantages of magnetic amplifier? 4. Describe various methods of changing inductance?

10.8 LAB ASSIGNMENTS 1. In series connection how the characteristics will change when inductive load is connected? 2. In parallel connection how the characteristics will change when inductive load is connected? 3. Compare the input and output characteristics in both the modes?

10.9 POST LAB QUESTIONS: 1. Describe saturable core reactor? 2. Give the purpose of saturable reactor in magnetic amplifier? 3. Describe in detail the circuitry of magnetic amplifier?

10.10 RESULT:

Hence the series and parallel magnetic amplifiers are studied and corresponding graphs are plotted.

EXPERIMENT 11: CHARACTERISTICS OF AC SERVOMOTOR

11:1 OBJECTIVE:

To study the Speed-Torque and Speed-Back e.m.f. characteristics of AC Servomotor.

11:2 RESOURCES:

5. AC Servo Motor Kit 6. Patch Chords 7. Multimeter

11.3 BLOCK DIAGRAM:

11.4 PROCEDURE:

1. Speed, Back emf (Eb) Characteristics:

1. Study all the front panel controls and features carefully 2. Initially keep load switch and servomotor switch in OFF position. 3. Before switching ON the instrument, keep load control potentiometer P1 and speed control potentiometer P2 are in fully anti-clockwise (i.e. minimum ) position. 4. Now switch ON the instrument and also switch ON the servomotor. You can observe that AC Servomotor will start rotating and the speed will be indicated by the rpm meter on front panel. 5. With load switch OFF position, vary the speed of the AC servomotor by rotating speed control potentiometer P2 in clockwise direction and note down the back

emf (Eb) generated by the DC machine at Tp terminals at different speed. Use digital millimeter to measure back emf voltage. Tabulate these readings. 6. Plot the graph Back emf Speed Vs voltage (Eb) 2. Speed, Torque Characteristics:

1. Initially keep load switch and servomotor switch in OFF position. 2. Before switching ON the instrument, keep load control potentiometer P1 and speed control potentiometer P2 are in fully anti-clockwise (i.e. minimum ) position. 3. Now switch ON the instrument and also switch ON the servomotor. You can observe that AC Servomotor will start rotating and the speed will be indicated by the rpm meter on front panel. 4. By rotating P2, set control winding voltage (Vc) at 45V. Use digital millimeter to measure control winding voltage. Note down the speed of Ac servomotor in table. Now keep P1 at minimum position and switch ON the load switch to apply the load. Note down the back emf voltage (Eb) from Tp Terminals. Note down the Eb and Ia values. 5. Repeat above step for different values (at least 3) of load control potentiometer. Note down the corresponding value of Ia, Eb and speed. 6. Repeat the above two steps for Vc = 55V and 60V.

11.5 TABULAR COLUMN: 1. Speed, Back emf (Eb) Characteristics:

Sl. No.

Speed (N) in rpm

Back emf (Eb) in V

2. Speed, Torque (T) Characteristics:

Vc =

Sl. No.

Speed (N)

Back emf (Eb)

Current (Ia)

(rpm)

(Volts)

(Amps)

Power (P)

Torque (T)

(P=Eb x I a)

= (Px1.019x104 x60)/2N

(Watts)

(N-m)

11.6 MODEL GRAPHS:

11.7 PRE LAB QUESTIONS:

1. What are the main parts of an ac servo motor? 2. What are the advantages and disadvantages of an AC servo motor over dc servo motor? 3. Give the applications of Ac servomotor? 4. Define servo mechanism?

11.8

LAB ASSIGNMENTS:

1. Explain how the parameters changes when load changes 2. To study of A.C. motor position control through continuous command. 3. To study of error detector on A.C. motor position control through step command. 4. To study of A.C. position control through dynamic response.

11.9

POST LAB QUESTIONS:

1. What is the difference between regulator & servomechanism? 2. What are the components of AC position control? 3. How is position control achieved? 4. What are the applications of AC servomotors? 5. What is meant by the dynamic response of DC servomotor? 6. Why a revolver is used in Ac servo motor?

11.10 RESULT:

Hence, the speed-torque characteristics of an AC Servomotor are studied and the graphs are drawn.

EXPERIMENT 12: SIMULATION OF OP-AMP BASED INTEGRATOR & DIFFERENTIATOR CIRCUITS

12.1 OBJECTIVE:

To simulate op-amp based integrator and differentiator circuits using PSPICE.

12.2 PROBLEM STATEMENT:

a) Simulate the practical differentiator circuit given below(fig1) using PSPICE and plot the transient response of the output voltage for a duration of 0 to 40ms in steps of 50µsec the op-amp which is modeled by the circuit has Ri= 2mΩ, Ro=75Ω, R1=10kΩ and C1=1.5619µf, Ao=2×105.

b) Simulate the practical integrator circuit given below(fig2) using PSPICE and plot the transient response of the output voltage for a duration of 0 to 40ms in steps of 50µsec the op-amp which is modeled by the circuit has Ri= 2mΩ, R0=75Ω, R1=10kΩ and C1=1.5619µf, Ao=2×105.

12.3 CIRCUIT DIAGRAM:

12.4 PROCEDURE 1. Click on PSPICE icon. 2. From FILE menu click on NEW button and select Text file to open untitled window 3. Enter the following program in untitled window. (a) Program (Differentiator): VIN 1 0 PWL(0 0 1MS 1V 2MS 0V 3MS 1V 4MS 0V)

R1

1

2

100

RF

3

4

10K

RX

5

0

10K

RL

4

0

100K

C1

2

3

0.4UF

*SUBCKT CALL FOR OPAMP XA1

3

5

.SUBCKT RI

1

4

0

OPAMP

OPAMP 1 2

2

7

4

2 .0E6

*VOLTAGE CONTROLLED CURRENT SOURCE WITH GAIN OF 1m GB

4

3

R1

3

4

10K

4

1.5619UF

C1

3

1

2

0.1M

*VOLTAGE CONTROLLED VOLTAGE SOURCE WITH GAIN OF 2E5 EA

4

RO

5 5

3

7

4

2E5

75

.ENDS .TRAN

10US

.PRINT

4MS

TRAN

V(4,0)

V(1,0)

.PROBE .END (b) Program (Integrator): VIN

1

0PWL (0

1V) R1

1

RF

2

4

1MEG

RX

3

0

2.5K

0

100K

RL

4

2

2.5K

0

1NS -1V 1MS -1V 1.0001MS 1V 2MS +1V 2.0001MS -1V 3MS -1V 3.0001MS 1V 4MS

C1

2

4

0.1UF

*SUBCKT CALL FOR OPAMP XA1

2

3

.SUBCKT RI

1

4

0

OPAMP

OPAMP 1 2

2

7

4

2.0E6

*VOLTAGE CONTROLLED CURRENT SOURCE WITH GAIN OF 0.1M GB

4

3

1

R1

3

4

C1

3

4

2

0.1M

10K 1.5619UF

*VOLTAGE CONTROLLED VOLTAGE SOURCE WITH GAIN OF 2E5 EA

4

5

3

4

RO

5

7

75

2E5

.ENDS .TRAN

50US

4MS

.PRINT

TRAN

V(4,0)

V(1,0)

.PROBE .END 4. Save the above program by clicking on SAVE button from FILE menu (or) Ctrl+S 5. Run the program by clicking RUN button (or) from Simulation menu select RUN button and clear the errors (if any). 6. Observe the output from View menu select output file. 7. Observe the required plots at respective points by selecting Add Trace from Trace Menu.

12.5 MODEL GRAPH:

Differentiator:

Transient Response

Integrator:

Time (sec)

12.6 PRE LAB QUESTIONS:

1. What is an Operational amplifier? 2. What are the ideal characteristics of op amp? 3. What is Virtual Short? 4. What are the applications of op amp? 5. What are the advantages and disadvantages of op amp?

12.7 LAB ASSIGNMENTS:

1. What is the output voltage for the differentiator if the input is sinusoidal? 2. What is the output voltage for the integrator if the input is sinusoidal? 3. How the op amp acts as voltage multiplier?

12.8 POST LAB QUESTIONS:

1. Explain how the waveform is converted in differentiator? 2. Explain how the waveform is converted in integrator?

12.9 RESULT:

Hence, the PSPICE simulation of op-amp based integrator and differentiator circuits have been done.

EXPERIMENT NO – 13- LINEAR SYSTEM ANALYSIS USING MATLAB

13.1 OBJECTIVE: To write a program and simulate dynamical system of I/O model

13.2 RESOURCES: MATLAB Software

13.3 Program:

num=input('enter the numerator of transfer function:') den=input('enter the denominator of transfer function:') s=tf(num,den) step(s) title('the response of second order system is:') Xlabel('t in secs') Ylabel('v in volts') [nt,dt]=tfdata(s,'v') wn=sqrt(dt(3)) z=dt(2)/(2*wn) disp('the rise time is:') tr=pi-(atan^-1(sqrt(1-z^2)/z)/(wn*sqrt(1-z^2))) disp('the peak time is:') tp=pi/(wn*sqrt(1-z^2))

disp('the settling time is:') ts=4/(z*wn) disp('the peak over shoot is:') pos=exp((-z*pi)/sqrt(1-z^2))*100 end

OUTPUT:

enter the numerator of transfer function:100

num =

100

enter the denominator of transfer function:[1 5 100]

den =

1

5

100

Transfer function: 100 --------------s^2 + 5 s + 100

nt =

0

0

100

5

100

dt =

1

wn =

10

z=

0.2500

the rise time is:

tr = 2.8238

the peak time is:

tp =

0.3245

the settling time is:

ts =

1.6000

the peak over shoot is:

pos =

44.4344

the response of second order system is: 1.5

v in volts

1

0.5

0

0

0.5

1

1.5 t in secs (sec)

RESULT:

2

2.5

EXPERIMENT 14 : ROOT LOCUS, BODE PLOT AND NYQUIST PLOT USING MATLAB

14.1 OBJECTIVE:

To analyze frequency response of a system by plotting Root locus, Bode plot and Nyquist plot using MATLAB software.

14.2 RESOURECES:

1. MATLAB 7 Software 2. Personal Computer 14.3 PROCEDURE:

1. Click on MATLAB icon. 2. From FILE menu click on NEW button and select SCRIPT to open Untitled window 3. Enter the following program in untitled window. 14.4 PROGRAM:

For Root Locus Plot: %Root Locus Plot clear all; clc;

disp(‘Transfer Function of given system is : \n’); num = input (‘Enter Numerator of the Transfer Function:\ n’); den = input (‘Enter Denominator of the Transfer Function :\ n’); G = tf(num,den); figure(1); rlocus(G); For Bode Plot: %Bode Plot clear all; clc; disp(‘Transfer Function of given system is : \n’); num = input (‘Enter Numerator of the Transfer Function : \ n’); den = input (‘Enter Denominator of the Transfer Function : \ n’); G = tf(num,den); figure(2); bode(G); %margin(G); It can be used to get Gain Margin, Phase Margin etc [Gm,Pm,Wpc,Wgc] = margin(G); disp(‘Phase Cross Over frequency is : \n’); Wpc

disp(‘Gain Cross Over frequency is : \n’); Wgc disp(‘Phase Margin in degrees is : \n’); Pm disp(‘Gain Margin in db is : \n’); Gm = 20*log(Gm) Gm if (WgcWpc) disp(‘Closed loop system is unstable’) else disp(‘Closed loop system is Marginally stable’) end 4. Save the above program by clicking on SAVE button from FILE menu (or) Ctrl+S 5. Run the program by clicking RUN button (or) F5 and clear the errors (if any). 6. Observe the output on the MATLAB Command Window and plots from figure window.

14.5 MODEL GRAPHS:

Imaginary

Root Locus plot:

-1

Real Axis (Sec )

Phase (deg)

Magnitude (dB)

Bode Plot:

Frequency (rad/sec)

OUTPUT:

Phase Cross Over frequency is: Wpc =

Gain Cross Over frequency is : Wgc =

Phase Margin in degrees is : Pm = Gain Margin in db is : GM =

14.6 THEORETICAL CALCULATION:

1. Phase Margin:

1. For a given Transfer Function G(s), get G(jω) by placing s= jω. 2. Separate Magnitude and Phase terms from G(jω). 3. Equate magnitude of G(jω) to ONE and get ω value, this ω is called Gain Cross Over Frequency (ωgc) 4. Substitute ωgc in place of G(jω), get the phase angle (φ). 5. Now Phase margin (PM) = 180 + φ

2. Gain Margin: 1. For a given Transfer Function G(s), get G(jω) by placing s= jω 2. Separate Magnitude and Phase terms from G(jω). 3. Equate imaginary part to ZERO and get ω value, this ω is called Phase Cross Over Frequency (ωpc) 4. Substitute ωpc in real pat, get the corresponding gain (K). 5. Now Gain Margin (GM) = 20 log10 (1/K)

3. Maximum Allowable Gain:

1. For a given Transfer Function G(s), place K in the numerator and get the characteristic equation Q(s) = 1 + G(s). 2. Get Q(jω) by placing s = jω. 3. Separate imaginary and real terms from Q(jω). 4. Equate imaginary part to ZERO and get ω values, these values called Imaginary Cross Over points. 5. Substitute ωpc in real pat and equate real part of G(jω) to ZERO and get the corresponding gain (K). 6. This gain is called maximum Allowable Gain (Kmax) or Limiting value of the Gain for stability.

14.7 TABULAR COLUMN:

Specification

Gain Margin (GM) and

Phase Margin (PM) and

Gain Cross Over Frequency (Wgc)

Phase Cross Over Frequency (Wpc)

Maximum allowable gain (Kmax)

From MATLAB

Theoretical

14.8 PRE LAB QUESTIONS:

1. What is gain margin and phase margin? 2. What is gain cross over frequency and phase crossover frequency? 3. What are the different types of stability conditions? 4. What are the advantages and disadvantages of root locus, bode & nyquist plot? 5. What are the advantages of frequency response analysis?

14.9 LAB ASSIGNMENTS: 1. For the above function, if a pole is added, how the stability will be effected for all the plots? 2. For the above function, if a pole is removed, how the stability will be effected for all the plots? 3. For the above function, if a zero is added, how the stability will be effected for all the plots? 4. For th?e above function, if a zero is removed, how the stability will be effected for all the plots

14.10 POST LAB QUESTIONS:

1. What is complementary Root Loci? 2. What are contours? 3. How can you analyze the stability of system with bode, nyquist?

14.11 RESULT: Hence Root Locus Plot, Bode Plot and Nyquist plot of given transfer function has been plotted and verified with theoretical calculations.

EXPERIMENT 15: STATE SPACE MODEL FOR CLASSICAL TRANSFER FUNCTION USING MATLAB 15.1 OBJECTIVE:

To Transform a given Transfer Function to State Space Model and from State Space Model to Transfer Function using MATLAB.

15.2 RESOURCES:

1. MATLAB 7 Software 2. Personal Computer.

15.3 PROCEDURE:

1. Click on MATLAB icon. 2. From FILE menu click on NEW button and select SCRIPT to open Untitled window. 3. Enter the following program in untitled window. 15.4 PROGRAM:

For Transfer Function to State Space Model:

%Transfer Function to State Space Model Clear all;

clc; disp(‘Transfer Function of given system is : \n’); Num = [2 3 2]; Den = [2 1 1 2 0]; sys = tf(num,den); Disp(‘Corresponding State Space Model A,B,C,D are: \n’); [A,B,C,D] = tf2ss(num,den) A B C D

15.5 PROGRAM: For State Space Model to Transfer Function: %State Space Model to Transfer Function Clear all; clc; disp(‘A,B,C,D Matrices of given State Space Model are :: \n’); A = [1 2;3 4 ] B = [1;1] C = [1 0]

D = [0] [num,den] = ss2tf(A,B,C,D); Disp((‘And corresponding Transfer Function is : \n ‘); Sys = tf(num,den); Sys 7. Save the above program by clicking on SAVE button from FILE menu (or) Ctrl+S 8. Run the program by clicking RUN button (or) F5 and clear the errors (if any). 9. Observe the output from on the MATLAB Command Window. OUTPUT: Transfer Function to State Space Model: Transfer Function of given system is

Transfer Function:

2s^2 + 3s + 2 -----------------------------2s^4 + s^3 + s^2 + 2s

Corresponding State Space Model A, B, C, D are:

A=

-0.5000

-0.5000

1.0000

0

0 0

-1.0000

0

0

1.0000 0

0 0

1.0000

0 0

B=

1 0 0 0

C=

0

D=

0

1.0000

1.5000

OUTPUT: State Space Model to Transfer Function:

A,B,C,D Matrices of given State Space Model are :

A =

B =

1

2

3

4

1 1

C

=

1

D

=

0

0

1.000

and corresponding Transfer Function is:

s–2 ----------------s^2 – 5s -2

15.6 PRELAB QUESTIONS:

1. What are the advantages & disadvantages of state space analysis? 2. What are the disadvantages of transfer function? 3. What are the different functions in MATLAB? 4. What is workspace and command window?

15.7 LAB ASSIGNMENTS:

8s + 1 1.

------------------------------ formulate state space model? 9s^3 + s^2 + s + 2

s^4 +s^3+s^2+s+ 1 2.

---------------------------9s^3 + s^2 + s + 2

formulate state space model?

s 3.

------------------------------ formulate state space model? 9s^3 + s^2 + s + 2

15.8 POSTLAB QUESTIONS:

1. How to call MATLAB in batches? 2. Explain Handle graphics in MATLAB? 3. Explain the following commands: Acker, Bode, Ctrb, Dstep, Feedback, Impulse, Margin, Place, Rlocus, stairs

15.9 RESULT:

Hence, the given transfer function to state space model and state space model to transfer function is transformed by using MATLAB.