78541246 131451 Control Systems Lab Manual

7/31/2019 78541246 131451 Control Systems Lab Manual

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SSN COLLEGE OF ENGINEERING

KALAVAKKAM- 603 110

DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING

131451 - CONTROL SYSTEMS

LAB MANUALDec 2011-April 2012

Name: _______________________________________

Reg. No.: _____________________________________

Year: II Sem: 4 Sec: A/B Dept: EEE


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DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING

131451- CONTROL SYSTEMS LABORATORY

NAME OF THE STUDENT : ____________________________

REGISTER NUMBER : ____________________________

CLASS : II- EEE- A / B

ACADEMIC YEAR : Dec 2011- Apr 2012

TOTAL MARKS : -------- / 10

SIGNATURE OF THE STAFF:


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LIST OF EXPERIMENTS

1. Determination of transfer function parameters of Armature controlledDC (servo) motor.

2. Determination of transfer function parameters of Field controlled DC(servo) motor.

3. Determination of transfer function parameters of an AC servomotor.4. Analog simulation of type-0 and type-1 systems5. Digital simulation of first order systems6. Digital simulation of second order systems.7. Stability analysis of linear systems.8. DC and AC position control systems.9. Stepper motor control system10. Determination of transfer function parameters of DC generators.11. Study of synchros12. Design and implementation of compensators.13. Design of P, PI and PID controllers.

.

P = 45, TOTAL = 45


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INDEX

Expt.

NO.DATE

Title MARKS

(10)

SIGN. OF

THE

STAFF

1

2

3

4

5

6

7

8

9

10

11

12

13

Total Marks

Signature of the faculty:


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EXPT. NO.:

DATE:

TRANSFER FUNCTION OF ARMATURE CONTROLLED

DC SERVOMOTORAIM:

To determine the transfer function of an armature controlled dc servomotor.

APPARATUS REQUIRED:

THEORY:

Transfer function is defined as the ratio of Laplace transform of the outputvariable to the Laplace transform of input variable at zero initial conditions.

Armature controlled DC shunt motor

In this system, Ra = Resistance of armature in La= Inductance of armature windings in HIa = Armature current in A

If= Field current in Ae = Applied armature voltage in V

eb = back emf in VTm = Torque developed by the motor in Nm


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J = Equivalent moment of inertia of motor and load referred tomotor shaft in kgm

2

B= Equivalent viscous friction coefficient of inertia of motorand load referred to motor shaft in Nm/(rad/s)

In Servo applications, DC motors are generally used in the linear range of the

magnetization curve. Therefore, the air gap flux is proportional to the field current. If

= KfIf ,where Kf is a constant. ----------------------------- (1)

The torque Tm developed by the motor is proportional to the product of the armaturecurrent and air gap flux.

Tm Ia

Tm =Ki Ia = Ki Kf IfIa , where Ki is aconstant --------------(2)

In the armature controlled DC motor, the field current is kept constant. So the above

equation can be written asTm = Kt Ia , Where Kt is known as motor torque constant.------ (3)

The motor back emf being proportional to speed is given by

eb d/dt,

eb = Kb d/dt, where Kb is the back emf constant.----------------(4)

The differential equation of the armature circuit ise = IaRa + La dIa/dt + eb ----------------------------------------- (5)

The torque equation is

Tm = Jd2/dt2 + B d/dt ------------------------------------------ (6)

Equating equations (3) and (6)

Jd2/dt2 + B d/dt = Kt Ia ---------------------------------------(7)

Taking Laplace transforms for the equations (4) to (7), we get

Eb(s) = Kb s (s) -------------------------------------------- (8)

(s La + Ra ) Ia(s) = E(s) Eb(s). ------------------------------- (9)

( J s2+ B s) (s) = Tm (s) = Kt Ia(s) ---------------------------- (10)

From equations (8) to (10) , the transfer function of the system is obtained as


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Block diagram

Using the above equations, the block diagram for the armature controlled DC motor isgiven below:

E(s)+ (s)(s)

- Eb(s)

1. Circuit diagram to determine Kt and Kb

1/[Ra+sLa] Kt 1/s[Js+B]

s Kb


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2. Circuit diagram to determine Ra:

3. Circuit diagram to determine La:

PROCEDURE:

i)Load test to determine Kt

1. Initially keep all the switches in the off position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch on the power and the SPST switches S1 and S2.5. Adjust the field voltage to the rated value.6. Apply the armature voltage until the motor runs at the rated speed.7. Apply load and note the armature voltage, current and spring balance readings.8. Calculate torque and plot the graph between torque and armature current.9. Determine Kt from graph.


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ii)No-Load test to determine Kb.

1. Initially keep all the switches in the off position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch on the power and the SPST switches S1 and S2.5. Set the field voltage to the rated value.6. Adjust the armature voltage and note the armature voltage, current and speed.7. Calculate the back emf eb and plot the graph between back emf and 8. Determine Kb from graph.

iii) To determine Ra:

1. Initially keep all the switches in the off position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch on the power and the SPST switches S1.5. Note the armature current for various armature voltages.6. Calculate Ra.

iv) To determine La:

1. Initially keep all the switches in the off position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch on the power .5. Apply ac voltage to armature winding6. Note down the current for various input ac voltage .7. Calculate Ra.

Tabulation to determine Kt

S.No. Field

Current

If (A)

Armature

Voltage

Va (V)

Armature

Current

Ia (A)

Spring

Balance

Readings

(kg)

Speed(N)

rpm

Torque

T =

9.81(S1-

S2)r(Nm)S1 S2

Where r is the radius of the brake drum. r = _____________m


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Tabulation to determine Kb

S.No. Armature

Voltage Va

(V)

Armature

Current Ia

(A)

Speed N

(rpm)

Eb= Va-Ia Ra

(V)

= 2N/60(rad/sec)

-Tabulation to determine Ra

S.No. Armature

Voltage

Va (V)

Armature

Current

Ia (A)

Ra

()

Calculation by least square method

Ra = [V1I1 +V2I2 +V3I3+V4I4 ] / (I12+I2

2+I3

2+I4

2)

Tabulation to determine Za

S.No. ArmatureVoltage

Va (V)

ArmatureCurrent

Ia (A)

Za

()

Average Za = ________


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MODEL GRAPH

To find Kt To find Kb

Torque Eb( V)(Nm)

Ia(A) (rad/sec)

MODEL CALCULATION

Ra = ..OhmsZa = .. OhmsLa = (Za

2Ra

2) / 2f = . H

f = 50 Hz J = 0.074 kg/m2, B = 0.001Nm/rad/sec From Graph,

Kt = Torque constant = T / Ia = Nm / A

Kb = Back emf constant = Eb / = . V/(rad/s)

RESULT:

INFERENCE:


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EXPT. NO.:

DATE:

TRANSFER FUNCTION OF FIELD CONTROLLED DC

SERVOMOTOR

AIM:

To determine the transfer function of a field controlled dc servomotor.

APPARATUS REQUIRED:

THEORY:

The transfer function is defined as the ratio of Laplace transform of the outputvariable to the Laplace transform of input variable at zero initial conditions.

Armature controlled DC shunt motor

In this system, Rf= Resistance of the field winding in Lf= Inductance of the field windings in HIa = Armature current in AIf= Field current in Ae = Applied armature voltage in Veb = back emf in Vef= Field voltage in V


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Tm = Torque developed by the motor in NmJ = Equivalent moment of inertia of motor and load referred to

motor shaft in kgm2B= Equivalent viscous friction coefficient of inertia of motor

and load referred to motor shaft in Nm/(rad/s)

In Servo applications, the DC motors are generally used in the linear range of themagnetization curve. Therefore the air gap flux is proportional to the field current.

If

= KfIf ,where Kfis a constant. -------------------------------- (1)

The torque Tm developed by the motor is proportional to the product of the armaturecurrent and air gap flux.

Tm Ia

Tm =K Ia = K KfIf Ia = Km Kf If , where Ki is aconstant ----(2)

Appling Kirchhoffs voltage law to the field circuit, we haveLfdIf/dt + RIf= ef ------------------------------------------------- (3)

Now the shaft torque Tm is used for driving the load against the inertia and frictionaltorque. Hence,

Tm = Jd2/dt2 + B d/dt ------------------------------------------- (4)

Taking Laplace transforms of equations (2) to (4), we get

Tm(s) = KmKfIf(s) ----------------------------------------------- (5)

Ef(s) = (s Lf+ Rf) If(s) -------------------------------------------- (6)

Tm(s) = (J s2+ B s) (s) ------------------------------------------- (7)

Solving equations (5) to (7), we get the transfer function of the system as


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1. Circuit diagram to determine KmKf

2. Circuit diagram to determine Rf

3. Circuit diagram to determine Lf


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PROCEDURE:

i)Load test to determine KmKf

1. Initially keep all the switches in the OFF position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch ON the power and the SPST switches S1 and S2.5. Apply 50% of the rated field voltage.6. Apply the 50% of the rated armature voltage.7. Apply load and note the field current and spring balance readings.8. Vary the field voltage and repeat the previous step.9. Calculate torque and plot the graph between torque and field current.10.Determine KmKffrom graph.

ii) To determine Rf

1. Initially keep all the switches in the OFF position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch ON the power and the SPST switch S2.5. Note the field currents for various field voltages.6. Calculate Rf.

iii) To determine Lf

1. Initially keep all the switches in the OFF position.2. Keep all the voltage adjustment knobs in the minimum position.3. Give connections.4. Switch ON the power .5. Apply AC voltage to field windings6. Note the currents for various input AC voltages.7. Calculate Lf.


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MODEL GRAPH

To find KmKf

Torque

(Nm)

If(A)

MODEL CALCULATIONS

1. Rf = ..Ohms2. Zf= .. Ohms3. Lf= (Zf2 Rf2 ) / 2f = . H4. f = 50 Hz5. J = 0.074 kg/m2, B = 0.001Nm/rad/s6. From Graph, KmKf = T /If = Nm / A

RESULT:

INFERENCE:


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EXPT. NO:DATE :

DETERMINATION OF TRANSFER FUNCTION PARAMETERS

OF AC SERVO MOTOR

AIM:

To derive the transfer function of the given AC servomotor andexperimentally determine the transfer function parameters.

APPRATUS REQUIRED:

FORMULA:

1. Motor transfer function(s) Km

=Eo (s) s (1+s.m)

2. Motor gain constant Km = K1K2 + B

3. Motor time constant m= JK2 + B

Where K1 = slope of torque - control phase voltage characteristicsK2= slope of torque -speed characteristicsJ = Moment of inertia of load and the rotorB= viscous frictional coefficient of load and the rotor

THEORY

When the objective of a system is to control the position of an object, then thesystem is called a servomechanism. The motors that are used in automatic controlsystems are called servomotors.

Servomotors are used to convert an electrical signal (control voltage) into anangular displacement of the shaft. In general, servomotors have the followingfeatures.

1. Linear relationship between speed and electrical control signal2. Steady state stability3. Wide range of speed control


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4. Linearity of mechanical characteristics throughout the entire speed range5. Low mechanical and electrical inertia6. Fast response

Derivation of Transfer Function:

Let Tm = Torque developed by the servomotor = angular displacement of the rotor = d / dt = angular speedTL = torque required by the loadJ = Moment of inertia of the load and the rotorB = Viscous frictional coefficient of the load and the rotorK1 = slope of the control phase voltage and torque characteristics.K2 = slope of the speed and torque characteristics.

The transfer function of the AC servomotor can be obtained by torque equation. Themotor developed torque is given by

Tm = K1 e c K2 d(1)dt

The rotating part of the motor and the load can be modeled by

TL = J d2 + B.d . (2)

dt2 dtAt equilibrium, the motor torque is equal to load torque. Hence,

K1 e c K2 d /dt = J d2 + B d ....(3)dt dt

Taking Laplace Transform

K1 Ec (s) K2 s (s) = J s2

(s) + B s (s)(4)

(s) K1 KmT.F = = =

Ec (s) s (K2+s J+B) s (1 +s m)

K1Where motor gain constant Km =

B + K2

Jand motor time constant m =

B + K2


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PROCEDURE:

I. DETERMINATION OF TORQUE SPEED CHARACTERISTICS

1. Give the connections.2. Connect voltmeter or a digital Multimeter across the control winding.3. Apply rated voltage to the reference phase winding and control phase

winding.4. Note the no load speed.5. Apply load in steps. For each load, note the speed.6. Repeat steps 4,5 for various control voltage levels and tabulate the readings.

II. DETERMINATION OF TORQUE CONTROL VOLTAGE CHARACTERISTICS

1. Make connections.2. Connect voltmeter or a digital Multimeter across the control phase winding3. Apply rated Voltage to Reference phase winding.4. Apply a certain voltage to the control phase winding and make the motor

run at low speed. Note the voltage and the no load speed.

5. Apply load to motor. Motor speed will decrease. Increase the controlvoltage until the motor runs at same speed as on no-load. Note the controlvoltage and load.

6. Repeat steps 5 for various loads7. Repeat 4-6 for various speeds and tabulate.

Torque Speed Characteristics

Radius of brake drum =________

Vc = Vc = Vc =

Load

g

N

rpm

Torque

N-m

Load

g

N

rpm

Torque

N-m

Load

g

N

rpm

Torque

N-m

Model Graph

Torque

N-m

Speed (rpm)


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Torque Control Voltage Characteristics

N1 = N2 = N3 =

Load

g

Vc

V

Torque

N-m

Load

g

Vc

V

Torque

N-m

Load

g

Vc

V

Torque

N-m

Model Graph

Torque

N-m

Control voltage (Volts)

From Graph , K1 =

K2 =

Given , B =

J =

From Calculations, Km =

m =


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RESULT:

INFERENCE:


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EXPT. NO:

DATE :

ANALOG SIMULATION OF TYPE-0 AND TYPE-1 SYSTEM

AIM:

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

APPARATUS REQUIRED:

THEORY:

Order of the system:

The order of the system is given by the order of the differential equationgoverning the system. The input-output relationship of a system can be expressed bytransfer function. Transfer function of a system is obtained by taking Laplacetransform of the differential equation governing the system and rearranging them asratio of output and input polynomials in s. The order is given by the maximumpower of s in denominator polynomial Q(s)

T(s) = P(s) / Q(s)

P(s) --- Numerator polynomialQ(s) --- Denominator polynomial

Q(s) =ao sn + a1sn-1 + a2 sn-2 + .+ an-1 s + an

If n=0, then system is Zero-Order system.

If n=1, then system is First-Order system.If n=0, then system is Second-Order system.

Type of the system

Type of the system is given by the number of poles of the loop transfer function at theorigin.

G(s)H(s) = K P(s) / Q(s)

(s+z1) (s+z2) (s+z3) ..=

sN

(s+p1) (s+p2) (s+p3) ..

If N=0, the system is a Type Zero system.If N=1, the system is a Type One system.If N=0, the system is a Type Two system.


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First Order Type 0 system

The generalized transfer function for first order Type 0 system is

T(s) = C(s) / R(s) = 1/(1+s) --------------------------------------------------------(1)

C(s) ---- Output of the systemR(s) ----- Reference input to the system.

If input is a Step input

R(s) = 1/s ----------------------------------------------------- (2)From eqn (1)

1C(s) = R(s) ---------------------------------------(3)

(1+s)

substituting for R(s),

1 1C(s) = ------------------------------------(4)

s (1+s)To find C(t) , Take Inverse Laplace Transform of eqn (4),

------------------(5)

PROCEDURE:

1. Give the connections as per the block diagram in the process control simulatorusing the front panel diagram .

2. Set the Input (set point) value using the set value knob.3. Observe the Output (process value or PV) using CRO and plot it in the graph.4. Tabulate the reading and calculate the % error.5. Repeat the procedure in closed loop condition.

C(t) = 1 e-t/


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TABULATION FOR FIRST ORDER SYSTEM:

(a)Type Zero system

Loop type Set Point

SP

(V)

Process

variable

PV(V)

Settling

Time

(s)

Error

SP-PV

(V)

% Error

SP-PV x 100%

SP

Open Loop

Closed Loop

(b)Type One System

Loop type Set

Point

SP

(V)

Process

variable

PV

(V)

Settling

Time

(s)

Error

SP-PV

(V)

% Error

SP-PV x 100%

SP

Open loop

Closed loop

TABULATION FOR SECOND ORDER SYSTEM

(a)Type Zero system

Loop type Set Point

SP

(V)

Process variable

PV

(V)

Settling

Time

(s)

Error

SP-PV

(V)

% Error

SP-PV x 100%

SP

Open loop

Closed loop

(b)Type One system

Loop type Set Point

SP

(V)

Process

variable

PV

(V)

Settling

Time

(s)

Error

SP-PV

(V)

% Error

SP-PV x 100%

SP

Open loop

Closed loop


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RESULT:

INFERENCE:


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EXPT. NO:

DATE :

DIGITAL SIMULATION OF FIRST ORDER SYSTEMS

(i) Digital Simulation of first order Linear and Non Linear SISO SystemsAIM:

To digitally simulate the time response characteristics of Linear and NonLinear SISO systems using state variable formulation.

APPARATU REQUIRED:

A PC with MATLAB package.THEORY:

SISO linear systems can be easily defined with transfer function analysis. Thetransfer function approach can be linked easily with the state variable approach.The state model of a linear-time invariant system is given by the followingequations:

X(t) = A X(t) + B U(t) State equationY(t) = C X(t) + D U(t) Output equation

Where A = n x n system matrix,B = n x m input matrix,C= p x n output matrix andD = p x m transmission matrix,

PROGRAM/ SIMULINK MODEL:


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PROGRAM:


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RESULT:

INFERENCE:


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(ii) Digital Simulation of Multi-Input Multi-Output Linear Systems

AIM:

To digitally simulate the time response characteristics of MIMO Linear system

using state-variable formulation.

APPARATUS REQUIRED:

PC MATLAB Package.

THEORY:

State Variable approach is a more general mathematical representation of asystem, which, along with the output, yields information about the state of the systemvariables at some predetermined points along the flow of signals. It is a direct time-

domain approach, which provides a basis for modern control theory and systemoptimization.

u1(t) y1(t)

u2(t) y2(t) U Y. .. .

um(t) yp(t). . . . . . . . X

x1(t) x

2(t) x

n(t)

.X(t) = A X(t) + B U(t) State equationY(t) = C X(t) + D U(t) Output equation

The state vector X determines a point (called state point) in an n - dimensional space,called state space. The state and output equations constitute the state model of thesystem.

Controlled systemState variables (n)

Controlledsystem


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PROGRAM:


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RESULT:

INFERENCE:


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Expt. No.:

Date:

DIGITAL SIMULATION OF SECOND ORDER SYSTEMS

AIM:

To digitally simulate the time response characteristics of second order linearand non-linear system with saturation and dead zone.

APPARATU REQUIRED:

A PC with MATLAB package.

PROGRAM / SIMULINK MODEL:


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SIMULINK MODEL:

RESULT:

INFERENCE:


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Expt. No.:

Date:

STABILITY ANALYSIS OF LINEAR SYSTEMS

AIM:

To analyze the stability of linear system using Bode plot/ Root Locus / NyquistPlot.

APPARATUS REQUIRED:

A PC with MATLAB package.

THEORY:

A Linear Time-Invariant Systems is stable if the following two conditions ofsystem stability are satisfied

When the system is excited by a bounded input, the output is alsobounded.

In the absence of the input, the output tends towards zero, irrespectiveof the initial conditions.

PROCEDURE:

1. Write a Program to obtain the Bode plot / Root locus / Nyquist plot for thegiven system.

2. Determine the stability of given system using the plots obtained.

PROGRAM:


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RESULT:

INFERENCE:


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EXPT. NO:

DATE :

CLOSED LOOP DC POSITION CONTROL SYSTEM

AIM:

To study the operation of closed loop position control system (DCServomotor) with a PI controller.

APPARATUS REQUIRED:

THEORY:

A pair of potentiometers is used to convert the input and output positions intoproportional electrical signals. The desired position is set on the input potentiometerand the actual position is fed to feedback potentiometer. The difference between thetwo angular positions generates an error signal, which is amplified and fed toarmature circuit of the DC motor. If an error exists , the motor develops a torque torotate the output in such a way as to reduce the error to zero. The rotation of the motorstops when the error signal is zero, i.e., when the desired position is reached.

Fig. 1 Block Diagram

+

-


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Fig.(2.).Front Panel

PROCEDURE:

1. Switch on the system. Keep the pulse release switch in OFF position.2. Vary the set point with the pulse release switch in the ON position and

note the output position.

3. Note SP voltage , PV voltage, P voltage and PI output voltage.4. Calculate KP using the formula KP = P/(SP-PV).


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TABULATION

RESULT:

INFERENCE:

S.NO. POSITION (degrees) Error (set output)in degrees

set output


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EXPT. NO:DATE :

CLOSED LOOP AC POSITION CONTROL SYSTEM

AIM:

To study the closed loop operation of AC position control system (ACServomotor) with PI controller.

APPARATUS REQUIRED:

THEORY:

CONSTRUCTIONAL DETAILS

The AC servomotor is a two-phase induction motor with some special designfeatures. The stator consists of two pole pairs (A-B and C-D) mounted on the innerperiphery of the stator, such that their axes are at an angle of 90

oin space. Each pole

pair carries a winding, one winding is called the reference winding and other windingis called the control winding. The exciting currents in the two windings should have aphase displacement of 90o. The supply used to drive the motor is single-phase andhence a phase advancing capacitor is connected to one of the phases to produce aphase difference of 90

o. The stator constructional features of AC servomotor are

shown in fig.1.The rotor construction is usually of squirrel cage or drag-cup type. The

squirrel cage rotor is made of laminations. The rotor bars are placed on the slots andshort-circuited at both ends by end rings. The diameter of the rotor is kept small inorder to reduce inertia and to obtain good accelerating characteristics. Drag cupconstruction is employed for very low inertia applications. In this type ofconstruction, the rotor will be in the form of hollow cylinder made of aluminium. Thealuminium cylinder itself acts as short-circuited rotor conductors.

WORKING PRINCIPLES

The stator windings are excited by voltages of equal rms magnitude and 90

o

phase difference. This results in exciting currents i1 and i2 displaced in phase by 90o

and having identical rms values. These currents give rise to a rotating magnetic fieldof constant magnitude. The direction of rotation depends on the phase relationship ofthe two currents (or voltages). The exciting current shown in fig.2 produces aclockwise rotating magnetic field. When i1 is shifted by 180

o, an anticlockwise

rotating magnetic field is produced. This rotating magnetic field sweeps over the rotorconductors. The rotor conductor experience a change in flux and so voltages are


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induced in rotor conductors. This results in circulating currents in the short-circuitedrotor conductors resulting in rotor flux.

Due to the interaction of stator & rotor flux, a mechanical force (or Torque) isdeveloped in the rotor and the rotor starts moving in the same direction as that ofrotating magnetic field.

Fig 1 Stator Construction of AC Servomotor

Fig 2.Waveforms of Stator & Rotor Excitation Current

Fig.3 Basic Block Diagram of AC Position Control System


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Fig.4. Block Diagram

PROCEDURE:

1. Switch ON the system. Keep the pulse release switch in the OFF position.2. Vary the set point with the pulse release switch in the ON and note the output

position.

3. Note the SP voltage, PV voltage, P voltage and PI output voltage.4. Calculate KP using the formula KP = P/(SP-PV).


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TABULATION

S.NO. SET

POSITION

(degrees)

OUTPUT

POSITION

(degrees)

ERROR=(set position-output

position)

(degrees)

RESULT:

INFERENCE:


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Ex. No:

Date:STEPPER MOTOR

Aim:To study the Stepper motor

Theory:

Stepper motors are highly accurate pulse-driven motors that change theirangular position in steps, in response to input pulses from digitally controlledsystems.

A stepper or stepping motor converts electronic pulses into proportionatemechanical movement. Each revolution of the stepper motor's shaft is made up ofa series of discrete individual steps. A step is defined as the angular rotation

produced by the output shaft each time the motor receives a step pulse. Thesetypes of motors are very popular in digital control circuits, such as robotics,

because they are ideally suited for receiving digital pulses for step control.

Each step causes the shaft to rotate a certain number of degrees.A step angle represents the rotation of the output shaft caused by each step,

measured in degrees.Figure.1. illustrates a simple application for a stepper motor. Each time the

controller receives an input signal, the paper is driven a certain incremental

distance.

Fig.1

In addition to the paper drive mechanism in a printer, stepper motors are also popularin machine tools, process control systems, tape and disk drive systems, andprogrammable controllers.


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The Common Features of stepper motors are

Brushless Stepper motors are brushless. The commentator and brushes ofconventional motors are some of the most failure-prone components, and theycreate electrical arcs that are undesirable or dangerous in some environments.

Load Independent Stepper motors will turn at a set speed regardless of loadas long as the load does not exceed the torque rating for the motor.

Open Loop Positioning Stepper motors move in quantified increments orsteps. As long as the motor runs within its torque specification, the position ofthe shaft is known at all times without the need for a feedback mechanism.

Holding Torque Stepper motors are able to hold the shaft stationary. Excellent response to start-up, stopping and reverse.

Types of Stepper Motor

1. Permanent-magnet stepper motor

The permanent-magnet stepper motor operates on the reaction between apermanent-magnet rotor and an electromagnetic field.Figure shows a basic two-pole PM stepper motor.The rotor shown in Figure (a) has a permanent magnet mounted at each end.The stator is illustrated in Figure (b). Both the stator and rotor are shown as

having teeth

Fig.2

The teeth on the rotor surface and the stator pole faces are offset so that

there will be only a limited number of rotor teeth aligning themselves with anenergized stator pole. The number of teeth on the rotor and stator determine the

step angle that will occur each time the polarity of the winding is reversed.The greater the number of teeth, the smaller the step angle.


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Fig.3

The holding torque is defined as the amount of torque required to move the rotorone full step with the stator energized.

An important characteristic of the PM stepper motor is that it can maintain theholding torque indefinitely when the rotor is stopped.Figure (a) shows a permanent magnet stepper motor with four stator windings.By pulsing the stator coils in a desired sequence, it is possible to control the speedand direction of the motor.Figure (b) shows the timing diagram for the pulses required to rotate the PM

stepper motor.

2.Variable-reluctance (VR) stepper motor

The variable-reluctance (VR) stepper motor differs from the PM stepper inthat it has no permanent-magnet rotor and no residual torque to hold the rotor atone position when turned off.

When the stator coils are energized, the rotor teeth will align with theenergized stator poles. This type of motor operates on the principle of minimizingthe reluctance along the path of the applied magnetic field. By alternating thewindings that are energized in the stator, the stator field changes, and the rotor is

moved to a new position.The stator of a variable-reluctance stepper motor has a magnetic core

constructed with a stack of steel laminations. The rotor is made of unmagnetized

soft steel with teeth and slots.

The relationship among step angle, rotor teeth, and stator teeth is expressed usingthe following equation:


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------(1)

In this circuit, the rotor is shown with fewer teeth than the stator. This ensures that

only one set of stator and rotor teeth will align at any given instant.The stator coils are energized in groups referred to asphases.

According to above Eq., the rotor will turn 30 each time a pulse is applied.

Figure (a) shows the position of the rotor when phase A is energized. As long asphase A is energized, the rotor will be held stationary.

Fig.4

When phase A is switched off and phase B is energized, the rotor will turn 30until two poles of the rotor are aligned under the north and south polesestablished by phase B.

= 360rs

rs

NN

NN


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By repeating this pattern, the motor will rotate in a clockwise direction. Thedirection of the motor is changed by reversing the pattern of turning ON andOFF each phase.

The disadvantage of this design for a stepper motor is that the steps aregenerally quite large (above 15).

Multistack stepper motors can produce smaller step sizes because the motoris divided along its axial length into magnetically isolated sections, or stacks.

Result:


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EXPT. NO.:

DATE:

DETERMINATION OF TRANSFER FUNCTION OF SEPRATELY

EXITED DC GENERATOR

AIM:

To determine the transfer function of separately exited generator.

APPARATUS REQUIRED:

Ammeter MC (0-1A), (0-10A)Ammeter MI (0-5A),(0-50mA)Voltmeter MC (0-300V)Voltmeter MI (0-300V)Rheostat 1000 / 1ARheostat 50 / 5AAuto Transformer 1 230V/270V

THEORY:

The transfer function of a separately excited generator can be represented in blockdiagram format as shown below

The transfer function is

IL(s)/Vf(s) = Kg /(Rf+sLf)(Rl+sLa)

WhereVf(s)- Excitation Voltage

Rf, Lf - Field resistance & InductanceIf(s) - Field Current

Kg Induced emf constant in V/AmpLl Total load InductanceRl Total load resistance

Kg can be obtained by conducting open circuit testRf, Lf, Ra, Lf can be found out by voltmeter- Ammeter method


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DC GENERATOR

Circuit Diagram

DETERMINATION OF Ra :

Circuit Diagram

Tabulation

S.No Va (V) Ifa (A) Ra =Va/Ia( )

1

2

3

4

5

6

7

8

9

10

Mean value of Ra =


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DETERMINATION OF Rf :

Circuit Diagram

S.No Vf (V) If (A) Rf =Vf/If(

)1

2

3

4

5

6

7

8

Determination of Lf

Circuit Diagram


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Tabulation

S.No Vf (V) If (A) Zf ( )=Vf/If

1

2

3

4

56

7

8

9

10

Mean value of Zf =

Determination of La

Circuit Diagram


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Tabulation

S.No Va (V) Ia (A) Za ( )=Va/Ia

1

2

3

45

6

7

8

9

10

Mean value of Za =

Open circuit characteristics


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PROCEDURE:

Determination of Kg:

1. The connections are made as shown in fig.2. DPST switch is closed3. The motor is started with help of starter4. The motor is brought to the rated speed by adjusting the motor field rheostat.

The drives the generated at rated speed.5. Note down the field current If and the open circuit voltage Eo.6. By adjusting the Rf, the field current is increased in convenient steps up to the

rated field current.7. In each step the readings of Eo and If are noted. Throughout the experiment

the speed is maintained at constant8. A plot of Eo Vs If is drawn by taking If on X axis and Eo on Y-axis. 9. A tangent to the linear portion of the curve is dran through the origin. The

slope of this line ,Eo Vs If gives Kg.

V-A method to obtain Ra, Rf, La & Lf

1. Give the connection as shown in fig to measure Ra & Rf and note downthe V & I

2. To measure La &Lf give the connection as shown in fig.3. Apply an AC voltage & measure the field reactance Zf & armature

reactance Za.4. Calculate Lf= Sqrt(Zf2 Rf2) /2f5. Calculate La = Sqrt(Za2 Ra2) /2f

Where f= supply frequency (50Hz)

RESULT:


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EXPT. NO.:

DATE:

STUDY OF SYNCHROS

AIM:

To study the characteristics of Synchros.

APPARATUS REQUIRED:

THEORY:

A Synchro is an electro-magnetic transducer used to convert an angularposition of a shaft into an electrical signal. It is commercially known as a Selsyn or an

Autosyn. The basic element of a synchro is a synchro transmitter whose constructionis very similar to that of a 3 phase Alternator. The stator is of concentric coil type, inwhich three identical coils are placed with their axis 120 apart, and is star connected.The rotor is of dumb bell shaped construction and is wound with a concentric coil. ACvoltage is applied to the rotor winding through slip rings.

Fig.1 Constructional Features of Synchro Transmitter

The constructional features and schematic diagram of a synchro transmitter andreceiver is shown in fig.1. Let an AC voltage Vc(t) = Vr Sin t be applied to the rotorof the synchro transmitter. The applied voltage causes a flow of a magnetizing currentin the rotor coil, which produces a sinusoidal time varying flux directed along its axisand distributed nearly sinusoidally in the air gap along the stator periphery. Becauseof transformer action, voltages are induced in each of the stator coils. As the air gapflux is sinusoidally distributed, the flux linking any stator coil is proportional to thecosine of the angle between the rotor and stator coil axis, and so is the voltageinduced in the stator coil. Thus, we see that synchro transmitter acts like a single-


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phase transformer in which the rotor coil is the primary and stator coil is thesecondary.

Fig .2 Schematic Diagram of Synchro Transmitter

Let Vs1, Vs2 & Vs3 be the voltage induced in the stator coils S1,S2 andS3 with respectto the neutral. Then, for the rotor position of the synchro transmitter shown in the fig.2 where the rotor axis makes an angle with the axis of the stator coil S2

Vs1 = K Vr Sint Cos ( + 120 ) ---------------------------- (1)

Vs2 = K Vr Sint Cos ( ) ----------------------------------(2)

Vs3 =K Vr Sint Cos ( + 240 ) ----------------------------(3)

The three terminal voltages of stator are

Vs1s2 = Vs1 - Vs2 =3 KVr Sin( + 240 ) Sin t -----------------------(4)

Vs2s3 = Vs2 - Vs3 = 3 KVr Sin( + 120 ) Sin t ----------------------(5)

Vs3s1 = Vs3 - Vs1 =3 KVr Sin() Sin t -----------------------(6)

When = 0, from equations (1), (2) and (3), it is seen that the maximum voltage isinduced in the stator coil S2 , while it follows from the equation (6) from that theterminal voltage Vs3s1 is zero . This position of the rotor is defined as the electricalzero of the transmitter and is used as reference for specifying the angular position ofthe rotor. The input to the synchro transmitter is the angular position of its rotor shaftand the output is a set of 3 single-phase voltages given by equations (4) to (6). The

magnitude of these voltages is function of the shaft position.

The outputs of the synchro transmitter are applied to the stator windings of a synchrocontrol transformer. The rotor of the control transformer is cylindrical in shape sothat the air gap is practically uniform. The system acts as an error detector.Circulating currents of the same phase but of different magnitude flow through thetwo sets of stator coils. This results in the establishment of an identical flux pattern inthe gap at the control transformer as the voltage drop in resistances and leakagereactances of the two sets of stator coils are usually small. The voltage induced in the


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control transformer rotor is proportional to the cosine of the angle between the tworotors () and is given by

E(t) = K1 Vr Cos Sin t

When =900, the voltage induced in the control transformer is zero. This position is

known as electrical zero position of the control transformer.

Fig. 3 Synchro Error Detector

PROCEDURE:

Tabulation 1:

1. Give connections as given in the circuit diagram.2. Vary the input position and note the output position.3. Plot the variation in output position with respect to the input position.

Tabulation 2:

1. Give excitation to the rotor winding.

2. Measure the output voltage across S1-S2, S2-S3 and S3-S1 of stator

windings for different rotor positions.

3. Plot the voltage Vs. angle characteristics.


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TABULATION: I

Sl.No Input

position

(degrees)

Output

position

(degrees)

Error

(degrees)

1 0

2 30

3 60

4 90

5 120

6 150

7 180

8 210

9 240

10 270

11 300

12 330

TABULATION : II

S.No Input angle

(degree)

Vs1 - Vs2

(V)

Vs2 - Vs3(V)

Vs3 - Vs1

(V)

1 0

2 303 60

4 90

5 120

6 150

7 180

8 210

9 240

10 270

11 30012 330

RESULT:

INFERENCE:


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EXPT. NO:

DATE :

DESIGN OF COMPENSATOR NETWORKS

AIM:

To design a compensator network for the process given in the Process Control

Simulator.

APPARATUS REQUIRED:

THEORY:

Practical feedback control systems are often required to satisfy designspecification in the transient as well as steady state regions. This is not possible byselecting good quality components alone (due to basic limitations and characteristicsof these components). Cascade compensation is most commonly used for this purposeand design of compensation networks figures prominently in any course in automaticcontrol systems.

In general, there are two situations in which compensation is required. In thefirst case the system is absolutely unstable and the compensation is required tostabilize it as well as to achieve a specified performance. In the second case thesystem is stable but the compensation is required to obtain the desired performance.The systems which are of type 2 or higher are usually unstable. For these systems,lead compensator is required, because the lead compensator increases the margin ofstability. For type 1 and type 0 systems stable operation is always possible. If the gainis sufficiently reduced, in such cases, any of three components viz. Lag, Lead, Lag Lead must be used to obtain the desired performance. The simulation of this behaviorof the Lead Lag Compensator can be done with the module (VLLN OI).

An electronic Lead - lag network using Operational amplifiers is givenfigure 1.

C2

C1 R2 R4

- -R1 + R3 +

Fig.1 LEAD -LAG NETWORK USING OPERATIONAL -AMPLIFIER


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The transfer function for this circuit can be obtained as follows :

Let Z1 = R1 C1

The second op-amp acts as a sign inverter with a variable gain to compensate for

the magnitude. The transfer function of the entire system is given by

G(j) = ( R4 R2/ R3 R1 ) (1+R1C1s) / (1+R2 C2 s)G(j) = ( R4 R2/ R3 R1 ) ( 1+T1

22) / ( 1+T222 )

where T1 = R1 C1 ; T2 = R2 C2

= angle G(j) = - tan-1(T1) tan-1

(T2).

Thus steady state output is

For an input =X sint,

Yss(t) =X (R4 R2/ R3 R1) (( 1+T122) / ( 1+T2

22 ))sin(wt tan-1 T1 tan-1

T2

)

From this expression, we find that if T1 > T2, then tan-1 T1 tan-1 T2 > 0.

Thus if T1 > T2 , then the network is a LEAD NETWORK.

If T1 < T2 , then the network is a LAG NETWORK.

DETERMINATION OF VALUES FOR ANGLE COMPENSATION:

Frequency of sine wave = 20 Hz

Angle to be compensated = 70

= tan-1 (2 f *T1) tan-1

(2 f *T2)

T1 = 10, then substituting in above equation

70 - tan-1

(2 * * 20 * 10) tan-1 (2 * * 20 *T2)

solving for T2

T2 = 0.003 .

Hence, the values of T1 and T2 are chosen from which values of R1 ,C1 , R2 and C2

can be determined .

For example, T1 =R1 C1 = 10 ; If C1= 1F, then R1 = 10 M.

T2 = 0.003 = R2 C2 then C2 =1 F, and hence R2 = 3 M.

These values produce a phase lead of 70 which is the desired compensation angle.

Nominal Value for R1 =1 M C1 = 0.1 F

R2 =20 K C2 = 0.01 F


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PROCEDURE:

1. Switch ON the power to the instrument.2. Connect the individual blocks using patch chords.3. Give a sinusoidal input as the set value .4. Measure the amplitude and frequency of the input signal.5. Measure the amplitude and phase shift of the output signal with respect to

the input sine wave using CRO.

6. Draw the magnitude versus frequency plot and phase versus frequencyplot.

7. Using the technique explained previously, calculate the values of R1, C1,R2 and C2 to compensate for the phase shift of the output signal.

8. Connect the components at the points provided.9. Now include the compensation block in the forward path before the

process using patch chords.

10. Now measure the phase shift of the output signal with the input andverify for compensation.

11. Draw the magnitude versus frequency plot and phase versus frequencyplot for the designed compensator.

Table 1 (for the process without compensation):

S.No. Input

Freq (Hz)

Output

Voltage(V)

Gain (dB)

20 log(Vo/Vin)

Phase shift

()


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Table I1 (for the process with compensation): Vin = V

S.No. Input

Freq (Hz)

Output

Voltage(V)

Gain (dB)

20 log(Vo/Vin)

P Phase shift

()

A Amplitude of input sine wave (V)

F Frequency of the input sine wave (Hz)

Phase shift (Degrees)

RESULT :

INFERENCE:


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EXPT. NO.:

DATE:

STUDY OF P, PI, PID CONTROLLERSAIM:

To study the P, PI, PID controller using MATLAB software .

APPARATUS REQUIRED:

THEORY:

The transient response of a practical control system often exhibits dampedoscillation before reaching steady state value. In specifying the transient response

characteristics of control systems to unit step input, it is common to specify thefollowing

i) Delay Time(Td)ii) Rise time(Tr)iii) Peal time( Tp)iv) Max. overshoot (Mp)v) Settling time( Ts)

Proportional control:

The output of the controller is proportional to inputU(t) = Kp e(t)

E(t) = error signalU(t) controller out[putKp = proportional constant

It amplifies the error signal and increases loop gain. Hence steady statetracking accuracy , disturbance signal rejection and relative stability areimproved.

Its drawbacks are low sensitivity to parameter variation and it producesconstant steady state error.

Proportional + Integral Control:

The output of the PI controller is given byt

U(t) = Kp [ e(t) + (1/Ti )e(t) dt ]0

where Kp is the proportionality constant and Ti is called the integral time.

This controller is also called RESET controller. It introduces a zero in the system and increases the order by 1. The type number of open loop system is increased by 1


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It eliminates steady state error. Damping ratio remains same. Increase in order decreases the stability of system.

Proportional + Integral Control + Differential Control:

The output of a PID controller is given by

t

U(t) = Kp [ e(t) + (1/Ti )e(t) dt + Td de(t)/dt ]0

The PID controller introduces a zero in the system and increases the damping.This reduces peak overshoot and reduces rise time. Due to increase in damping,ultimately peak overshoot reduces.

The stability of the system improves.In PID controller, all effects are combined. Proportional control stabilizes gain

but produces steady state error. Integral control eliminates error. Derivative controllerreduces rate of change of error.

TUNING OF PID CONTROLLERS

Proportional-integral-differential (PID) controllers are commonly employed inprocess control industries. Hence we shall present various techniques of tuning PIDcontrollers to achieve certain performance index for systems dynamic response. Thetechnique to be adopted for determining the proportional, integral and derivativeconstants of the controller depends upon the dynamic response of the plant.

In presenting the various tuning techniques we shall assume the basic controlconfiguration, wherein the controller input is the error between the desired output andthe actual output. This error is manipulated by the controller (PID) to produce acommand signal for the plant according to the relationship.

U(s)=Kp (1+(1/si)+sd)Where Kp= proportional gain constant

I= integral time constant.d= Derivative time constant.

PROCEDURE:

1. Give the step input to the system selected and obtain the response usingCRO.

2. For the obtained response (S-shaped curve), draw a tangent at theinflection point and find its intersection with the time axis and the linecorresponding to the steady-state value of the output.

3. Find the dead time L where the tangent cutting X- axis, and the timeconstant T which is specified in model graph.

4. From the value of L and T, find the value of Kp, I and d settings by usingthe following formulas: Kp = 1.2(T/L) , I = 2L and d = 0.5L.

5. Connect the unknown system in closed loop with the help of a PIDcontroller and substitute all those values obtained in the previous step.


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6. Simulate the system with a step input and view the response using CRO.7. Comment on the response obtained using controller.

(I)General Block Diagram

(II)Block Diagram for P Controller

(III)OP Amp P Controller Using Inverting Amplifier


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(IV) Block Diagram For PI Controller

(V)PI Controller Using Op-Amp

(VI) Block Diagram Of PID Controller


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Block Diagram Of Closed Loop Control Using PID Controller

RESULT:

INFERENCE:

PID(KP,Ti,Td

M(s)

Transfer Function

C(s)

C(s)

R(s) E(s)


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Documents

78541246 131451 Control Systems Lab Manual