07e509 Control Engineering Lab Manual Mpm,Spd

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07E509 CONTROL ENGINEERING LABORATORY 0 0 3 2

1. Determination of transfer function of armature controlled DC motor

2. Determination of transfer function of field controlled DC motor

3. Determination of transfer function of AC servo motor

4. Analog and Digital simulation of Type-0 and Type-1 system

5. Time response analysis simulation

6. a) Characteristics of synchros

b) Study of stepper motor 7. Digital simulation of systems with nonlinearity 8. Frequency response analysis simulation

9. Performance evaluation of P, PI and PID controllers

10. Stability analysis of linear systems simulation

11. Compensator design simulation

12. Mini-project


Experiment No: 1 Date:

TRANSFER FUNCTION OF ARMATURE CONTROLLED DC MOTOR

OBJECTIVE:

To find the Transfer function of DC Motor with the control signal applied to

Armature.

THEORY:

A DC machine can run as a motor, when a DC supply is given to its field winding

to produce magnetic flux while the same DC source is used to supply current to the

armature. Now the armature becomes a current carrying conductor and as it is kept in a

magnetic field, it develops mechanical force. The direction of the force is given by

Fleming’s Left Hand Rule. DC motor in control applications is used for delivering

mechanical power to control elements while taking electrical control signal as input.

Electrical input to DC motor is called control signal and that can be applied in two

ways. In one method, the control signal is applied to the field winding while fixed voltage

is applied to armature winding. This method is called Field controlled motor. In another

method, control signal is applied to armature winding and constant voltage is applied to

field winding. This method is called Armature controlled motor. Here the transfer

function of armature controlled motor is to be found out.

DERIVATION OF TRANSFER FUNCTION:

According to Kirchhoff’s Voltage Law (KVL),

eLRiV ba

aaa dtdi

a (1) The expression for back emf is n/A Z P (t)be Where n = speed of the shaft in revolutions per second

Hence )( (t)be t


dtd)( (t)b KKe bb t (2)

Substituting eqn.(2) for eb in eqn.(1)

dtd

dtd

KiLRiV ba

aaa

a s)((s) s]ss)[( VKLRI abaa a (3)

Torque in a motor t)((t) T(t) i a Here control signal is applied to the armature and field is constant excited

Hence t)(i a T(t) t)(i Ka a T(t) (4) This torque rotates the armature with inertia J and Friction Coefficient B , hence

T(t)dtdB

dt

2dJ

(5) Multiplying eqn.(3) by Ka, we get

(s)VK(s) sKK]sL[R (s)IK a a b aaaa a (6) Substituting eqn.(5) for KaIa in eqn.(6)

(s)VK(s) sKK)sL(R (s) Bs)2(Js a a b aaa

(s)VK(s) s]KK)sLBs)(R[(Js a a b aaa2


sKK)sLBs)(R(Js

K(s)V(s)

b aaa2a

a

]

BRKKs)

RLs)(1

BJBs[(1R

K

a

b a

a

a a

a

]BRKK)sT)(1sTs[(1

BRK

a

b aam

a

a

where La/Ra = Armature Time Constant, Ta

J/B = Mechanical Time Constant , Tm

PROCEDURE:

1. Determination of Ka

Power developed in d.c motor

TNIe ab )

602(

Ttt IK ab )()(

Since for armature controlled motor, torque T = Ka Ia

IKtIK aaab t )()(

Hence KK ba

The connections are made as shown in the circuit diagram 1 for the determination

of back emf of the d.c motor for different speeds. Initially 100 ohms resistance is

connected in series with the armature. Field winding rheostat is kept at its

minimum. When supply for the motor is given, there will be tendency for the


armature current to increase to a high value as back emf is zero at starting. But the

current is limited by the 100 ohms armature rheostat. When the motor has started

the armature resistance is gradually cut down completely to increase the motor

speed. Further to bring the motor speed to rated value, resistance in field rheostat

is increased so that motor reaches rated speed. This excitation is called rated or

normal excitation and it is not disturbed.

Once again the armature resistance is included to the maximum 100 ohms

while keeping the field excitation undisturbed. Now the slow speed at full

armature resistance is noted and correspondingly the voltage across the armature

is noted down. The speed is increased in steps by cutting the armature resistance

each time noting the speed and voltage across the armature. By measuring the

armature resistance back emf for each speed can be calculated. The readings are

tabulated as in Table 1.

Graph: Back emf , eb vs angular speed, w rad/sec

2. Measurement of Ra

Connections are made as per the circuit diagram 2 where armature of the d.c

motor is connected in series with low resistive load having current capacity equal to

full load current of the armature. A voltmeter is connected across the armature and an

ammeter is connected in series with the armature. First keeping the maximum

resistance in the armature supply is given. The voltmeter reading and ammeter

reading are noted down. Then armature resistance is decreased so that the armature

current becomes half the rated value. The readings are taken. Again the armature

resistance is further decreased so that the armature current is rated value. Again the

readings are taken and tabulated.

Table 1

Sl.No Va Ia Ra = Va/ Ia


Table 2

Sl.No Voltage Across

armature , Va

Armature

Current, Ia

Speed in

Rpm, N

w, rad/sec

Eb= V - IaRa

3. Measurement of Time constant of the motor, Tm The connections are made as shown in the circuit diagram 1. The experiment is no

load test on the motor. The armature of the d.c motor is kept at maximum resistance by

connecting a 100 ohms rheostat in series with the armature. The field rheostat is kept in

its minimum position. The motor is started and armature resistance is cut slowly so that

the motor is allowed to pick up speed. After fully cutting the armature resistance, motor

field resistance is increased to a position so that motor reaches rated speed. The field

current of the motor at this position is called rated excitation. This field rheostat position

is maintained for rated excitation.

Before starting the experiment, once again armature resistance of the motor is

brought to the maximum value of 100 ohms without disturbing the rated field excitation.

Now the supply for the motor is given and sufficient time is allowed so that the motor

reaches a steady speed, Ns whose value is noted. From this steady speed, 0.632 times of

the steady speed is calculated which is equal to 0.632 Ns. Let the mechanical Time

constant of the motor with 100 ohms armature resistance be Tmh. The Time constant Tmh

is the time taken for the motor to reach 0.632 Ns from starting. To find out Tmh the

supply is disconnected and the motor is restarted with switching on a stop watch. The

time taken by the motor to reach 0.632 Ns is Tmh. The actual motor time constant Tm


without the external resistance of 100 ohms will be much lower and it is calculated as

follows.

RT am α

RRT examh α

RR

RTT

exa

a

mh

m

)(RR

RTTexa

amhm

(7)

The actual Time constant, Tm of the motor will be a fraction of a second and difficult

to measure. Hence the higher Time constant with 100 ohms included in the armature, Tmh

is measured . From which Tm is found out by equation (7)

4. Determination of La

The connections are made similar to the circuit diagram 2 except 230V ac is applied

instead of dc. The voltmeter and ammeter are ac meters of appropriate range. The

experiment is repeated as was done for measurement of armature resistance.

Armature time constant , Ta = La/Ra

1. What is meant by Time constant?

2. Define armature Time constant.

3. Define Mechanical Time constant.

4. Prove Ka = Kb


5. How inductance is encountered in d.c motor?

6. What is the significance of back emf?

7. How do you determine armature Time constant?

8. How do you determine Mechanical Time constant?

9. Why motor Time constant is measured by including large resistance in the

armature?

TRANSFER FUNCTION OF FIELD CONTROLLED DC MOTOR

OBJECTIVE:

To find the Transfer function of DC Motor with the control signal applied to

Field.

THEORY:

A DC machine can run as a motor, when a DC supply is given to its field winding

to produce magnetic flux while the same DC source is used to supply current to the

armature. Now the armature becomes a current carrying conductor and as it is kept in a

magnetic field, it develops mechanical force and the direction of the force is given by

Fleming’s Left Hand Rule. DC motor in control applications is used for delivering

mechanical power to control elements for electrical control signal as input.

Electrical input to DC motor is called control signal and that can be applied in two

ways. In one method, the control signal is applied to the field winding while fixed voltage

is applied to armature winding and hence mechanical power is developed and that

depends totally on the magnitude of the control signal applied to field. This method is

called Field controlled motor. In another method, control signal is applied to armature

winding and constant voltage is applied to field winding. The mechanical power


developed here depends on the control signal that goes to armature. This method is called

Armature controlled motor.

DERIVATION OF TRANSFER FUNCTION:

According to Kirchoff Voltage Law(KVL)

t)(dtdi t)( VLIR f

ffff

(s)V]sL(s)[RI ffff (1) (t) T(t) if

(t) T(t) iK ff (s) T(s) iK ff (2)

T(t)dtdB2dt

2dJ

T(s) (s) Bs)2(Js (3) Multiplying eqn.(1) by Kf

s)(]s[ (s) VKLRiK ffffff (4) substituting (s) Bs)2(Js(s)IK f f in eqn.(4),

(s)VK]sL[R (s) Bs)2(Js f fff


ff

f

f sLRBs)2(Js

K(s)V(s)

ff

f

sLRBJssK

s))

RL(s)(1)

BJ(s(1

BRK

f

f f

f

)sT)(1sTs(1

Kfm

m

where Km = Motor Gain Constant

Tf = Field Time Constant

Tm = Mechanical Time Constant

PROCEDURE:

1. Measurement of Time constant of the motor, Tm

The connections are made as shown in the circuit diagram. The

experiment is no load test on the motor. The armature of the d.c motor is kept at

maximum resistance by connecting a 100 ohms rheostat in series with the

armature. The field rheostat is kept in its minimum position. The motor is started

and armature resistance is cut slowly so that the motor is allowed to pick up

speed. After fully cutting the armature resistance, motor field resistance is

increased to a position so that motor reaches rated speed. The field current of the

motor at this position is called rated excitation. This field rheostat position is

maintained for rated excitation.


Before starting the experiment, once again armature resistance of the

motor is brought to the maximum value of 100 ohms without disturbing the rated

field excitation. Now the supply for the motor is given and sufficient time is

allowed so that the motor reaches a steady speed, Ns whose value is noted. From

this steady speed, 0.632 times of the steady speed is calculated which is equal to

0.632 Ns. Let the mechanical Time constant of the motor with 100 ohms armature

resistance be Tmh. The Time constant Tmh is the time taken for the motor to reach

0.632 Ns from starting. To find out Tmh the supply is disconnected and the motor

is restarted with switching on a stop watch. The time taken by the motor to reach

0.632 Ns is Tmh. The actual motor time constant Tm without the external

resistance of 100 ohms will be much lower and it is calculated as follows.

RT am α

RRT examh α

RR

RTT

exa

a

mh

m

)(RR

RTTexa

amhm

(5)

The actual Time constant, Tm of the motor will be a fraction of a second

and difficult to measure. Hence the higher Time constant with 100 ohms included

in the armature, Tmh is measured . From which Tm is found out by equation (5)

2. Determination Of Field Resistance of the Motor, Rf

The motor field resistance is measured by passing known current through the field

winding. The voltage across the field winding is measured using a appropriate voltmeter.

The connections are made as shown in circuit diagram 2. Normally resistance of the field

winding of d.c machine will be around 200 ohms. Hence an external variable resistance


of 350 ohms/1.1 amps rheostat shall be connected in series with the field winding of the

motor. The rated 220 volts d.c shall be applied to the field winding of the motor in series

with the 350 ohms/1.1 amps rheostat kept at its maximum resistance. An ammeter is

connected in series with the field winding to measure the current and a voltmeter across

the winding to measure the voltage drop in the winding.

Keeping the resistance of the 350 ohms/1.1 amps rheostat in its maximum value,

the ammeter and voltmeter readings are noted. The resistance of the rheostat is reduced in

steps each time noting the ammeter and voltmeter readings. Three sets of readings are

taken and the average of the three resistances measured is the motor field resistance.

Sl.no Voltmeter

Reading, V

Ammeter

Reading ,I

Rf = V/I

1.

2.

3.

Average field resistance, Rf

3. Determination of Field winding inductance, Lf

The connections are made as shown in circuit 3 which is almost similar to circuit2

except a.c voltage is applied instead of d.c voltage. The ammeter and voltmeters are also

a.c meters of appropriate range. Three sets of readings are taken . The ratio of voltage to

current gives the impedance, Zf. The average of the impedances measured gives the

impedance of the field winding, Zf.

Sl.No Voltmeter

Reading, Vac

Ammeter

Reading, Iac

Field impedance,

Z

Average field impedance


Z2 = Rf2 + Xf

2

Xf = (Z2 - Rf

2)1/2

Lf = (Z2 - Rf2)1/2 / 2Π

4. Determination of Kf

T(t) = Ka Ia(t)

= Kf If (t)

Kb Ia = Kf If

Torque = Kf If = Ka Ia (rated)

Kf = Ka Ia (rated) / If(rated)

1. What is meant by Time constant of a device?

2. Define field Time constant of the motor.

3. Define Mechanical Time constant of the motor.

4. How field torque constant Kf is determined?

5. How inductance is encountered in d.c motor?

6. How Mechanical Time constant is determined?

7. Why high resistance is included in the armature to determine the mechanical Time

constant?



DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF

AC SERVO MOTOR AIM:

To derive the transfer function of the given AC servo motor and experimentally determine the transfer function parameters of field controlled DC motor. APPARATUS / INSTRUMENTS USED:

S.NO ITEM QUANTITY

1 Two phase AC Servo Motor Speed Control and Transfer Function Study Trainer Kit 1

2 Patch Chards As required 3 AC Voltmeter (0-20)V 1

THEORY: An AC Servomotor is basically a two phase induction motor except for certain special design features. It differs in the following two ways from a normal induction motor.

The rotor of the servomotor is built with high resistance, so that its X/R ratio is small which results in linear speed torque characteristics where for normal induction motors the X/R ratio is large.


The excitation voltage applied to two stator windings should have a phase difference of 90o.

Construction: The stator consists of two pole pairs mounted on the inner periphery of the stator, such that their axes are at angle of 90o in space. Each pole pair carries one winding. One is called reference winding and the other is called a control winding. The exciting current in the winding should have phase displacement of 90o. The supply used to drive the motor is single phase and so a phase advancing capacitor is connected to one of the phase to produce a phase difference of 90o. The rotor construction is usually a squirrel cage or drag-cup type. The squirrel cage rotor is made up of laminations. The rotor bars are placed on slots and short circuited at both ends by end rings. The diameter of the rotor is kept small in order to reduce inertia and to obtain good accelerating characteristics. The drag-cup type construction is employed for very low inertia applications. The rotor will be in the form of hollow cylinder made of aluminium, which acts as short circuited rotor conductors. Working Principle: The stator winding are excited by voltages of equal rms magnitude and 90o phase difference. This results in exciting currents i1 and i2 that are displaced by 90o and have equal rms values. These current give rises to a rotating magnetic field of constant magnitude. The direction of rotation depends on the phase relationship of the two currents (or voltages). The rotating magnetic field sweeps over the rotor conductors. The rotor conductors experience a change in flux and so voltages are induced in rotor conductors.Hence a circulating current is produced in the short circuited rotor and the currents create rotor flux. Due to the interaction of stator and rotor flux, a mechanical force (torque) developed on the rotor and it starts moving in the same direction as that of rotating magnetic field. The block schematic of an AC servo motor with control system components is shown in Fig 1. The reference winding is excited by a constant voltage source with frequency 50 Hz. The speed of the motor is controlled by controlling the control voltage. The error output of error detector is fed to PI controller. Due to the error, controller takes control action by giving control voltage to the firing circuit. The firing circuit generates the pulses to rotate the motor at the required speed. The characteristics of AC servo motor are similar to three phase induction motor. At no load, induction motor will run nearly at synchronous speed. When the motor gets loaded, the torque will increase to supply the load with the proportional decrease of speed and the torque reaches the maximum value at a particular speed. When the speed falls beyond the limit, both the speed and torque fall as shown in Fig 1.

Speed (N) in rpm

Torq

ue (T

) in

Nm

Synchronous speed


Fig 1. Speed - Torque Characteristics of 3- Induction Motor This part of the characteristics is not suitable for control applications. The

required characteristics for control applications is when the torque increases, the speed of the motor should fall from synchronous speed to zero speed. Practically this characteristic curve can be obtained by including more resistance in the rotor of the induction motor. The same is true for two phase induction motor. When more resistance is included in the rotor of the two phase induction motor, the characteristic curve becomes linear as shown in Fig. 2.

Fig 2. Speed - Torque Characteristics of AC Servo Motor

Two phase induction motor has two windings main winding (or) running winding and starting winding or control winding. Besides these, two phase AC servo motor has large

resistance in the rotor. The control voltage is applied to the control winding and fixed AC voltage having a phase difference of 90o with respect to control winding voltage is applied to the running winding. The control voltage results in the development of the motor torque. Unless the control winding is applied with AC voltage, the motor will not develop torque. The torque speed characteristic curve shown in Fig. 2 is a straight line and the torque applied by the motor will be increased with supply voltage.

Hence the torque of the AC servo motor can be expressed as

T(t) = f(Vc,ω) (1)

By Taylor’s series expansion

T = ω(t)ωT(t)V

VT

Vcωcc

(2)

V1 Torq

ue (T

) in

Nm

V3

V2

Speed (ω rad/sec)


Where Vc = AC voltage signal applied to control winding in Volts ω = Angular speed of the motor in rad/sec T = Torque developed by the motor in Nm

The partial derivative cV

T is denoted by Kc and

ωT is denoted by Km. The

factors Kc and Km can be determined by simple load test. Derivation of Transfer Function: As we know,

T(t) = ω(t)ωT(t)V

VT

Vcωcc

(3)

T(t) = KcVc(t) + Km ω(t) (4) The developed torque used to rotate the load on the shaft with inertia J and friction coefficient B, with an angular speed of )(t is given by

T(t) = (t)Bdt

)(dJ

t (5)

Taking Laplace Transformation for equations (4) and (5) and equating them KcVc(s) + Km )(s = (Js + B) )(s The input to the motor is control signal Vc(s) and the resulting mechanical output is )(s . Re-arranging the above equation

(Js + B - Km) )(s = KcVc(s) Transfer function =

m

c

K-BJsK

)(ω(s)

sV c

(6)

In the transfer function (B - Km) should be always positive otherwise the system

with AC servomotor becomes unstable. Fortunately the drooping characteristic gives negative value for Km and hence B - Km is always positive.

sT)(1)K-/(BK

)(ω(s) mc

sV c

where Mechanical Time Constant T =

)K-(BJ

m

If friction is assumed zero the transfer function becomes

)1()(

)()(

sTK

ss K

Vm

c

PROCEDURE: The AC servo motor needs two phase supply for its operation. In a three phase

supply, the voltage between terminals R and N will be in quadrature with voltage between Y and B as per the vector diagram. Thus, two phase ac voltages obtained.


1. G1K1 of pulse isolation output is connected to G1K1 of SCR1. 2. G2K2 of pulse isolation output is connected to G2K2 of SCR1. 3. Switch S1 is used to choose the control mode of operation. 4. 9-pin D-Connector is connected to backside of the trainer. 5. Speed indicator switch is kept in P.V mode. 6. The voltmeter is connected across the control winding C1 & C2. 7. 230 V AC Supply is switched on at which the running winding is applied

with rated voltage. 8. Pulse ON / OFF Switch S2 is switched on. 9. Rated voltage is applied (12 V) to the control phase winding by using

control voltage (C.V) pot. 10. The motor is loaded with known weights in steps. 11. For each step, the speed N in rpm is observed and angular speed

in rad/sec = 2**N/60 is calculated. 12. The procedure is continued till the motor reaches 0 rpm. 13. All the loads are removed and the control voltage pot is adjusted to 3/4th of

the rated voltage. 14. The same procedure is repeated and the readings are observed and

tabulated. 15. For each load, torque is calculated using the formula

T = 9.81*r*S*10-3 (in Nm) Where r is the radius of the shaft in m

S is the applied voltage in kg 16. The graph Angular Speed, (rad/sec) Vs Torque, T (Nm) is plotted for the

two control voltages.

17. The slope of any one of the characteristics ωT gives the motor constant

Km. 18. Kc can be found from the characteristics as follows:

A vertical line to represent constant is drawn from axis to meet the characteristics of both control voltages.

From the intersecting points of the characteristics by the vertical axis for two known control voltages, a horizontal line towards the torque axis is drawn.

The ratio of change in torque to the change in control voltage cV

T

is calculated which gives the value of Kc.

INFERENCE: RESULT: REVIEW QUESTIONS:


10. What type of motor is used as AC servomotor? Why? 11. What improvements are made in the servomotor characteristics? 12. How voltage constant Kc and speed constant Km are determined? 13. Does the negative value of speed constant affect stability of the element? 14. How two phase AC voltage is obtained from three phase AC? 15. Does the motor run when control winding voltage is zero? 16. Plot the frequency response of the above system and comment on stability. 17. Obtain Bode plot of the above system using MATLAB and give your comments. 18. A closed loop control system uses a two phase AC motor. The error signal is

amplified by an amplifier and its output is connected to the motor. The amplifier gain is 20 V/V and tachogenerator used in the feedback path has a gain constant Km = 0.2 V/(rad/sec). The moment of inertia and the coefficient of viscous friction referred to motor shaft side are respectively 1.5 x 10-4 kg-m2 and 1 x 10-4 Nm/(red/sec). The motor has a stall torque of 0.12 Nm at 100 V input and the torque decreases by 50 % when the speed increases to 2000 rpm. Draw the block diagram and determine the transfer function of the system.

TABULATION: Radius r = 0.0175 m

Control Voltage VC1 = V

Sl.No. Speed N in rpm

Angular Velocity

in rad/sec

Load S in grams

Torque T = 9.81*r*S*10-3 (Nm)

Control Voltage VC2 = V

Sl.No. Speed N in rpm

Angular Velocity

in rad/sec

Load S in grams

Torque T = 9.81*r*S*10-3 (Nm)


MODEL CALCULATIONS: Moment of Inertia J = kg/cm2 Friction Coefficient B = Nm/rad/sec For Control Voltage VC1 = V : Speed N = rpm Angular Velocity = 2**N/60 = rad/sec Load S = grams Torque T = 9.81*r*S*10-3 = Nm From the charactersitics,

KC = cV

T = Nm/rad/sec

Motor Constant Km = ωT = Nm/rad/sec

Therefore, Transfer function of AC servo motor

m

c

K-BJsK

)(ω(s)

sV c

=


Mechancial Time Constant T = )K-(B

Jm

= seconds

MODEL GRAPH: Experiment No: 4 Date:

STUDY OF SYNCHROS

AIM:

To study the characteristics of synchros as error detector. APPARATUS / INSTRUMENTS USED:

S.NO ITEM QUANTITY 1 Synchronous Transmitter and Receiver 1 Set 2 Patch Chards As required 3 Multimeter 1

THEORY: A synchro is an electromagnetic transducer commonly used to convert converting angular position difference between output shaft and input command dial into a proportional AC voltage.

The basic synchro is usually called a synchro transmitter. Its construction is similar to that of a three phase alternator. The stator (stationary member) is of laminated silicon steel and is slotted to accommodate a balanced three phase winding which is usually of concentric coil type (three identical coils are placed in the stator with their axis

Torq

ue (T

) in

Nm

VC2

VC1

Speed (ω rad/sec)


120o apart) and is star connected. The rotor is a dumb bell construction and wound with a concentric coil. The rotor winding is applied with an AC supply voltage through slip rings and this rotor is held fixed in a desired angular position say θr (i.e. input or reference). Let the AC voltage applied to the rotor of the transmitter be

Ei(t) = Em sin (2Π f t) (1)

This voltage causes a flow of magnetizing current in the rotor coil which produces a sinusoidally time varying flux directed along its axis and distributed nearly sinusoidally in the air gap along stator periphery. Because of transformer action, voltages are induced in each of the stator coils. As the air gap flux is sinusoidally distributed, the flux linking any stator coil is proportional to the cosine of the angle between rotor and stator coil axis and so is the voltage induced in each stator coil. The stator coil voltages are of course in time phase with each other. Thus the synchro transmitter acts like a single phase transformer in which rotor coil is the primary and the stator coil form three secondaries. Let e1n, e2n and e3n be the voltages induced in the stator coils S1, S2 and S3 respectively with respect to neutral. Then for the rotor position of the synchro transmitter shown in figure, the rotor axis makes an angle 0 with the axis of the stator coil S1. Thus for θr = 0o, the corresponding voltage induced by transformer section across the stator winding 1n is given by, e1n = K Em sin (2Π f t) cos θr = K Em sin (2Π f t) cos 0 = K Em sin (2Π f t) (2) where K is the constant of proportionality. As the stator windings 2n and 3n are 240º and 120º apart (angle measured as positive in anti-clockwise direction) with respect to the winding 1n, the voltages induced across them are, e2n = K Em sin (2Π f t) cos (θr + 240º) = K Em sin (2Π f t) cos (0 + 240º) = K Em sin (2Π f t) cos (240º) = -0.5 K Em sin (2Π f t) (3) e3n = K Em sin (2Π f t) cos(120º) = K Em sin (2Π f t) cos(θr + 120º) = K Em sin (2Π f t) cos(120º) = -0.5 K Em sin (2Π f t) (4) Equations (2), (3) and (4) indicate that three voltages e1n, e2n and e3n are single phase voltages of same frequency and have same phase but their magnitudes depend on rotor position.


Now, if the rotor of the synchro transmitter shifts in anti-clockwise direction through an angle θ, the voltages in the stator coils are, e1n = K Em sin (2Π f t) cos θr (5) e2n = K Em sin (2Π f t) cos(240º - θr) (6) e3n = K Em sin (2Π f t) cos(120º - θr) (7) Equations (5), (6) and (7) reveal that magnitudes of the voltages e1n, e2n and e3n vary sinusoidally with respect to θr as shown in figure . It is seen that when θr = 0o, the maximum voltage is induced in the stator coil S1. This position of rotor is defined as electrical zero of the transmitter and is used as reference for specifying the angular position of the rotor. Hence it is inferred that the synchro transmitter is the angular position of its rotor shaft and the output is a set of three single phase voltages. The magnitude of these voltages are functions of a shaft position.

The classical synchro system consists of two units: 1. Synchro Transmitter 2. Synchro Receiver

The synchro receiver is having almost the same constructional features. The two units are connected as shown in the figure 2. . The locations of transmitter and receiver can be away from each other. The rotor of synchro transmitter is salient pole type and that of synchro receiver is cylindrical type. The rotor of the synchro receiver is coupled to the output shaft of the control system. If the position of the output shaft is indicated as θo this results in an angular error θe = (θr - θo) between the positions of the input (reference) and the output shafts. Initially the winding S2 of the stator transmitter is positioned for maximum coupling with rotor winding. Suppose its voltage is V, the coupling between S1 and S2 of the stator and primary (rotor) winding is a cosine function. Therefore the effective voltages in these windings are proportional to 60o or they are V/2 each. So long as the rotors of the transmitters and receivers remain in this position, no current will flow between the windings because of voltage balance. When the rotor transmitter is moved to a new position, the voltage balance is distributed. Assume the rotor of the transmitter is moved through 30o, the stator winding voltages will be changed to zero, 0.866 V and 0.866 V respectively. Thus there is a voltage imbalance between the windings cause a currents to flow through the close circuit producing a torque that tends to rotate the rotor of the receiver to a new position where the voltage balance is again restored. This balance is restored only if the receiver turns through the same angle as the transmitter and also the direction of the rotation is the same as that of transmitter. The transmitter and receiver pair thus serves to transmit information regarding the angular position at one point to a remote point.

The magnitude of the output induced voltage eo developed across the rotor of the receiver is given by the following relation: eo = Kr sin (θr - θo) (8)


where Kr is a constant of proportionality. In control systems the angular error, i.e. (θr - θo) is usually small and expressed in radian, therefore, sin (θr - θo) ≡ (θr - θo) radian Hence, eo = Kr (θr - θo) (θr - θo) is expressed as θe (angular error) eo = Kr θe (radian) where Kr is expressed in V/rad. The above equation indicates that angular error is converted into a proportional voltage. Laplace transform of the above equation

Eo(s) = Kr θe(s) The Transfer function is given by

re

Eo(s) K(s)θ

(9)

Where Kr is known as the sensitivity or the gain of synchro-error detector. PRECAUTIONS:

The pointers for both rotors are to be handled gently. The pointers should not pulled out The rotor and stator terminals should not be shorted.

PROCEDURE:

To obtain the Input – Output characteristics of Synchro Transmitter:

19. The main supply to the system is connected with the help of cable provided (the interconnections S1 , S2 and S3 to S1

’, S2’ and S3

’ should not be made).

20. The mains supply to the unit is switched ON. 21. The supply to the rotor transmitter is also switched ON. The magnitude of

the AC voltage supplied to the rotor is fixed as 40 V. 22. The magnitude of the AC voltage of between each of the stator terminals

VS1S2, VS2S3 and VS3S1 are measured and tabulated for different rotor positions at equal intervals.

23. The graph, angular position Vs rotor voltages for all the three phases are plotted as shown in figure 3.

To obtain the Input – Output characteristics of Synchro Transmitter & Receiver

pair:


1. The mains supply cable is connected. 2. The terminals of transmitter S1, S2 and S3 are connected to S1

’, S2’ and S3

’ of receiver using patch cords respectively.

3. The mains supply is switched ON. 4. The rotor supply of both transmitter and receiver are also switched ON. 5. The rotor of receiver is tightly held at 0o. 6. The rotor position of synchro transmitter (pointer) is moved in steps of 10o

and the new rotor position of receiver is observed. 7. The output voltage is also observed and it varies sinusoidally with angular

error. 8. The plot input angular position in degrees Vs output angular position in

degrees is plotted as shown in figure 4. INFERENCE: RESULT: REVIEW QUESTIONS:

1. What is the principle of synchro? 2. How synchro is a Zero order device? 3. What is synchro torque transducer? 4. How synchro is used as error detector? 5. Why rotor of synchro receiver is cylindrical in shape?

TABULATION: Rotor Voltage = 40 V

Sl.No. Angular Displacement of Rotor in degrees

VS1S2 in Volts

VS2S3 in Volts

VS3S1 in Volts

Sl.No. Angular Displacement of Rotor in degrees Output Voltage

in Volts t r c


Study of DC stepper Motor Aim:

By conducting the suitable experiment study the construction and working principle of a DC stepper motor Theory: The stepper motor is a special type of synchronous motor which is designed to rotate through a specific angle for each electrical pulse received from its control unit In recent years, the growth of computer industry induced a wide-spread demand for stepper motor can be controlled directly by computers, microprocessors and programmable controllers. The stepping motor ideally suited for precision positioning of an object or precision control of speed without using a closed loop feed back. The shaft of the stepper motor rotates in a series of discrete angular intervals or steps, one step being taken each time a command pulse is received. When a definite number of pulses are supplied, the shaft turns through a definite known angle this makes the motor well suited for open loop position control because no feedback is required from the shaft. The only moving part in a stepping motor its rotor which has no winding, commutator or brushes. STEP ANGLE The angle through which the motor shaft rotates for each command pulse is called step angle β. Smaller the step angle, greater the number of steps / revolution and higher the resolution or accuracy of positioning obtained. The step angle can be as small as 0.72 ْ ◌ degrees or as large as 90 ْ ◌. But most common step sizes are 1.8 ْ ◌ , 2.5 ْ ◌ , 7.5 ْ ◌ & 15 ْ ◌ . The value of step angle can be expressed either in terms of a rotor and stator poles (teeth) Nr and Ns respectively or in terms of the number of stator phases (m) and number of rotor teeth (Nr).


360N NN~ Nβ

r

r

S

S

OR

Teethsrotor of No. PhasesStator No.of360

Nm360β

r

Resolution is the number of steps needed to complete one revolution.

Resolution = Revolution

Pulses of No.β

360Revolution

Steps of No.

The stepper motor has the extraordinary ability to operate at very high stepping rate of 20,000 steps per second in some motors. Operation at very high speed is called ‘slewing’. When in the slewing range, the motor generally emits an audible whine having a fundamental frequency equal to stepping rate .if f is the stepping frequency (or) pulse rate in pulse per second and β is step angle then the motor shaft speed is given by

360

fβn Rps. Where

Second360

revolution pulse of No. Resolution

and Second

Revolutionn ,Second

pulse of No.f

Stepping motors are designed to operate for long periods with the rotor held in a fixed position and with the rated current flowing in the stator winding for other motors this will result in collapse of back emf and very high current which can lead to quick burn out the stator winding. Permanent-Magnet Stepping Motor

Construction: It has a wound stator poles and the rotor is made of permanent magnet material like magnetically ‘hard’ ferrite. As shown in the Fig.1, the rotor has projecting poles but the rotor is cylindrical and has radially magnetized permanent magnets. The operating principle of such a motor can be understood with the help of Fig.1 (a) where the rotor has two poles and the stator has four poles. Since two stator poles are energized by one winding, the motor has two windings or phases marked A and B. The step angle of this motor β = 360°/mNr = 360° / 2 x 2 = 90° or β =(4 - 2) x 360° /2 x 4 = 90°. Working: When a particular stator phase is energized, the rotor magnetic poles move into alignment with the excited stator poles. The stator windings A and B can be excited with either polarity current. Fig.1 (a) shows the condition when phase A is excited with positive current. Here, θ = 0°. If excitation is now switched to Phase B as in Fig.1 (b), the rotor rotates by full step of 90° in the clockwise direction. Next, when phase A is excited with negative current, the rotor turns through another 90° in CW direction as shown in Fig. 1 (c). Similarly, excitation of phase B with negative current further turns the rotor through another 90° in the same direction as shown in Fig. 1 (d). After this, excitation of


phase A with positive current makes, the rotor turn through one complete revolution of 360°. It will be noted that in a permanent-magnet stepper motor, the direction of rotation depends on the polarity of the phase currents.

A A B B΄ Θ 1 0 0 0 0° 0 0 1 0 90° 0 1 0 0 180° 0 0 0 1 270° 1 0 0 0 0°

Truth tables for three possible current sequences for producing clockwise rotation are given in Fig.2. Table No.1 applies when only one phase is energized at a time in I-phase-ON mode giving step size of 90°. Table No.2 represents 2-phase-ON mode when two phases are energized simultaneously. The resulting steps are of the same size but the effective rotor pole positions are mid way between the two adjacent full-step positions. Table No.3 represents half-stepping when 1-phase-ON and2-phase-ON modes are used alternately. In this case, the step size becomes half of the normal step or one-fourth of the pole-pitch (i.e. 90° / 2 = 45° or 180°/4 = 45°). Micro step can also be employed which will give further reduced step size there by increasing the resolution The specification of the Stepper Motor:

1. Permanent magnet stepper motor 2. Number of phases =2 3. Step per revolution = 200 4. Step angle = 1.8° ± 0.1° Non-cumulative. 5. Holding Torque = 2.8 Kg. cm. to 84 Kg. cm. covered in different frame size

(1 Kg. cm. =0.1 N.m) Procedure:

1. Select the stepping rate by patching on the top panel to either 100 Hz or 10 Hz or 1 Hz pulse output

2. Set the direction control toggle switch to CCW rotation. 3. Set the CON/OSC toggle switch to the CON position. 4. Connect the motor by plugging the socket on to the controller unit. 5. Switch on the AC main supply then the stepper motor will start stepping in the

CCW direction on the continuous basis. 6. Repeat the above procedure at different stepping rate selectable by the patch card

connection.

A A B B΄ θ 1 0 0 0 0° 1 0 1 0 45° 0 0 1 0 90° 0 1 1 0 135° 0 1 0 0 180° 0 1 0 1 225° 0 0 0 1 270° 1 0 0 1 315° 1 0 0 0 0°

A A B B΄ θ 1 0 1 0 45° 0 1 1 0 135° 0 1 0 1 225° 1 0 0 1 315° 1 0 1 0 45°

1-Phase-On Mode Table No. 1

2-Phase-On Mode Table No. 2

1-Phase-On Mode & 2-Phase-On Mode

Table No. 3

Fig. 2


7. Repeat the above procedure for CW direction of rotation. 8. Repeat the above procedure for OSC mode of operation. 9. Repeat the above procedure for single step operation by linking the step rate

socket to ground and single pulse command given through the push switch marked MONO.

10. Observe the wave forms of the control logic and driver stage circuitry on a high input impedance oscilloscope (10 M Ω or more) and the plot the wave forms.

11. Measure the frequency of pulse output at socket marked 100 Hz and measures the rotation speed of the shaft with a non contact digital tachometer. The rotational speed related to the pulse frequency as per the following expression.

Speed R.P.M. 360

60f1.8n

Inference:


PERFORMANCE EVALUATION OF P, PI AND PID

CONTROLLERS

AIM:

To study the tuning process of P, PI & PID controller using first order system. APPARATUS / INSTRUMENTS USED:

S.NO ITEM QUANTITY 1 Level Trainer 1 2 PC 1

THEORY PROPORTIONAL CONTROLLER In proportional mode, a smooth linear relationship exists between the controller output and the error. Each value of error has a unique value of controller output and has


one to one correspondence. The range of error covering 0% to 100% controller output is called proportional band. P = Kp + Po

Kp = Proportional Gain , Po = Controller Gain

[Controller output P = Error * 100/ proportional Band]

If error is zero, the output is constant and equal to Po. For every 1% of error a convection of Kp % is added or subtracted from P. INTEGRAL CONTROLLER In this mode, there will be a infinite signal change in the rate of controller output with infinitesimal change in the error. This mode is often referred to as reset action. dp/dt = Ki ep

dp/dt = rate of controller output change

Ki = constant relating the rate of the error

The inverse of Ki is called the integral time Ti(=1/Ki). If we integrate the above equation, we can find the actual controller output at any time as where Po is the controller output at t = 0. This equation shows that the present controller output P(t) decreases on the history of error from the time when observation has started at t=0. From the equation di/dt =Ki ep, if the error doubles, the rate of controller output change also doubles. If the error is zero, the output stays fixed at a value. It is equal to what it was when the error is zero. If the error is not zero, the output will begin to increase or decrease at a rate of Ki % per second for every 1% of error. DERIVATIVE CONTROLLER In this mode, the controller output depends on the rate of change of error. This mode is also known as rate of anticipating control. This mode cannot be used alone because when the error is zero or constant, the controller has no output. The expression for the derivative controller is given by P = Kd dep /dt

Kd = derivative gain constant

dep /dt = rate of change of error

PROPORTIONAL –INTEGRAL CONTROLLER The expression for the controller output is given by


In this mode, the one-to-one correspondence of proportional mode is available and the integral mode eliminates the inherent offset. The proportional gain by design changes the net integration mode gain but Ki can be independently adjusted. PROPORTIONAL –INTEGRAL-DERIVATIVE CONTROLLER One the most powerful but complex controller mode operation combines the proportional, integral and derivative modes. This system can be used for virtually any process conditions. The analytical expression is This mode eliminates the offset of the proportional mode and still provides fast response. PROCEDURE

1. Connections are given as per circuit diagram. 2. Before the controller can be used, the parameters must be programmed. The

program mode is selected with the key operated switch. 3. DPC 5010 displace the prompt and the currently set values of each parameter.

Press ENTER if you don’t wish to change the present values. 4. Enter the new value, most significant digit as follows. 5. Press , digits starts increasing. 6. Press , for entering new value. 7. If all the digits have been entered, go to step 10. 8. Press selected digits are shifted to left. Repeat from step 6 for the next digit. 9. Press shift +Enter to enter the value in program memory. 10. Vary the parameters and analyze the system performance.

INFERENCE RESULT OBSERVATION

S.No Controller

Constant Settling Time

Maximum Peak Over Shoot

Steady State error

Rise Time (sec) Kp Kd Ki

P


PI

PID

BLOCK DIAGRAM OF LEVEL CONTROLLER


PR

OC

ESS

TAN

K

TR

AN

SMIT

TER

C

ON

TRO

LLER

A

DC

DA

C

POW

ER

SUPP

LY

SCR

C

ON

TRO

L U

NIT

PU

MP

W

ATE

R

RES

ERV

OIR



DIGITAL SIMULATION OF NON-LINEAR SYSTEMS

AIM:

To digitally simulate the time response characteristics of a linear system with simple non-linearities like saturation and dead zone.

APPARATUS / INSTRUMENTS USED:

S.NO ITEM QUANTITY 1 PC with MATLAB Software 1

PROCEDURE:

1. Derivation of transfer function of the given SISO linear system:

Apply Kirchhoff’s voltage law or Kirchhoff’s current law to form the differential equations describing electrical circuits comprising of resistors, capacitors, and inductors.

Apply Newton’s Laws to form the equations of motion for lumped

parameter mechanical systems (Translational and rotational) comprising of masses, springs, and dampers.

Form transfer functions from the describing differential

equations.

2. Digital simulation of time response characteristics of the above system:

Write MATLAB script and build the block diagram for the above system using SIMULINK blocks.

Plot the time response characteristics by simulating the system

for step, impulse and sinusoidal inputs.

Determine the time domain specifications analytically and verify with the simulated results.

Identify real systems with similar characteristics.

3. Digital simulation of time response characteristics of the above system

with simple non-linearity:


In the SIMULINK Block of the above linear system, include

simple non-linearities like saturation and dead zone. Plot the time response characteristics by simulating the system

for step, impulse and sinusoidal inputs.

Discuss the effect of non-linearity. EXAMPLE:

INFERENCE: RESULT: REVIEW QUESTIONS:

19. What are non-linear systems? 20. How to analyse non-linear systems? 21. Give some examples of non-linearities? 22. What is linearization of non-linear systems? 23. List the methods of analyzing non-liner systems. 24. What are limit cycles?

(Important Note: The simulated results are to be neatly pasted in the Observation and Record Note Books)



DIGITAL SIMULATION OF FIRST AND SECOND ORDER SYSTEM

AIM To digitally simulate the time response characteristics of first and second order linear system for different input signals APPARATUS / INSTRUMENTS USED:


THEORY:

First Order System From the general equation of transfer function for a first order system, substitute the value of time constant(T) and obtain the time response characteristics by manually and through simulation process for different type of input signals. Second Order System

Undamped System Assign damping factor, =0 in the general transfer function of the second order system. Perform the simulation process and obtain the time response characteristics for different input signals and verify the same by manually. Underdamped System Assign 1 in the general transfer function of the second order system. Perform the simulation process and obtain the time response characteristics for different input signals and verify the same by manually. Critically damped System Assign = 1 in the general transfer function of the second order system. Perform the simulation process and obtain the time response characteristics for different input signals and verify the same by manually.


Over damped System Assign 1 in the general transfer function of the second order system. Perform the simulation process and obtain the time response characteristics for different input signals and verify the same by manually.

PROCEDURE: Digital simulation of time response characteristics of the given system:

Write MATLAB script and build the block diagram for the above system using SIMULINK blocks.

Plot the time response characteristics by simulating the system for step,

impulse and sinusoidal inputs.

Determine the time domain specifications analytically and verify with the simulated results.

INFERENCE: RESULT: Experiment No: 10 Date:

STABILITY ANALYSIS USING MATLAB/SIMULINK


AIM: To digitally simulate the frequency response characteristics of the given system and analyse the stability of the system by

(i) Routh Hurwitz Method (ii) Root Locus Method (iii) Drawing Bode Plot (iv) Drawing Nichols Plot



THEORY: Routh Hurwitz Criterion This criterion represents a method of determining the location of poles of a characteristics equation with respect to the left and right half of the s-plane without actually solving the equation. By the location of the poles, the stability of the system can be analysed whether it is stable or unstable.

Bode Plot Magnitude and phase of the frequency response can be plotted directly on linear

scale. It consists of two parts of graph. The magnitude is expressed in gain as decibels against logarithmic values of frequency called Magnitude plot. Phase angle in degrees against logarithmic values of frequency called phase plot. Root Locus Method This is the graphically method in which movement of poles in ‘S’ plane are sketched when a particular parameter of a system is varied from zero to infinity. But for root locus method, gain is assumed to be a parameter which is to be varied from zero to infinity. Nichols Plot The Nichols chart consists of curves that are maps of constant M and N circles on a new coordinate system. The coordinate system plots gain in dB versus phase angle in degrees. When the transfer function parameters are transformed to M and N circles to log magnitude and phase angle coordinates, the resultant plot is Nichols Plot. It is very useful in determining closed loop response from open loop response. It gives more information by graphically relating open loop and closed loop information on single plot. It consists


of M and N circles that repeat every 3600. Note down that both M and N circles are symmetry about 1800. PROCEDURE:

Write the Matlab script for the given system. Plot the root locus by simulating the system. Plot Bode and Nichols plot for the given transfer function by simulating the

system. Find the frequency domain specifications by all the methods mentioned above

and analyze the system either stable or unstable. INFERENCE:

RESULT: Experiment No: 12 Date:


FREQUENCY RESPONSE ANALYSIS SIMULATION

AIM:

To determine the frequency response of a system with a given specifications using lead network



THEORY:

If the performance of a control system is not up to expectations as per desired

specifications then it is evident that some change in the system is needed to obtain the

desired performance. The change can be obtained by inserting a compensating device in

the control system.

Introducing compensation network in a control system after the error signal

results in reshaping the original transfer function and adjustment of time constant of

compensation network enables to obtain the system performance as per desired

specification.

A phase lead network gives the output voltage that leads the input voltage for

sinusoidal inputs. The transfer function of the lead network is given by

E0/Es = α(1+js)/(1+j αs) α<1

The pole-zero configuration of the lead network reveals that the zero is nearer to

the origin as compared to the pole; hence the effect of zero is dominant. Therefore the

lead network when introduced in series with forward path of a transfer function the phase

shift is increased. It also allows passing high frequencies and low frequencies are

attenuated. As the gain is reduced at low frequencies, additional gain is needed in the

system to account for the reduction in gain.


The phase lead compensation network in a control system shifts the gain

crossover frequency point to a higher value and therefore, the bandwidth is increased thus

improving the speed of response and overshoot is reduced but the state error does not

show much improvement.

PROCEDURE:

Write a M-file to draw the Bode plot.

Calculate the Phase margin and Gain margin values.

If the specifications are not matched with the desired value, a suitable

compensation network should be designed based on the required value.

Find the transfer function of the compensating network.

Draw the Bode plot for the compensated network and verify the result with the

desired value.

INFERENCE:

RESULT:

Documents

07e509 Control Engineering Lab Manual Mpm,Spd