196444650 ee-2257-control-system-lab-manual

Control Systems Laboratory Manual
Page 1
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DDEEPARPARTTEEMMEENNTT OFOF ELEELECCTTRRIICACALL ANANDD ELEELECCTTRROONNIICCSS
EENNGGIINNEEEERRIINNGG

Page 2
EE2257 CONTROL SYSTEM LABORATORY
MMANUALANUAL
PRPREEPARPAREDED BYBY
VV.B.BAALLAAJIJI,, MM.Tech,.Tech, ((PPh.h.DD),), MM..II.S.T.E,.S.T.E, MM..II..AA.E.ENNG,G, MM..II.O..O.JJ.E.E
AASSSSIITTANANTT PRPROOFFESSOESSORR//EEEEEE DDEEPARPARTTMMEENNTT
DDHHANAANALLAAKSHKSHMMII CCOLLEGEOLLEGE OFOF EENNGGIINNEEEERRIINNGG
CCHEHENNANNAII

LLIISTST OFOF EEXPXPEERIMRIMEENNTSTS
1. Determination of transfer function of DC Servomotor
2. Determination of transfer function of AC Servomotor.
3. Analog simulation of Type - 0 and Type – 1 systems
4. Determination of transfer function of DC Generator
5. Determination of transfer function of DC Motor
6. Stability analysis of linear systems
7. DC and AC position control systems
8. Stepper motor control system
9. Digital simulation of first systems
10. Digital simulation of second systems

TTRANRANSSFFERER FFUNUNCCTTIIONON OFOF DDCC SESERVRVOO MMOTOROTOR
EEXPXPT.T.NNOO ::
DADATETE ::
AAIIMM::
To determine the transfer function of the DC servomotor

APPAAPPARRAATTUUSS RREEQQUUIIRREEDD::
S.No Name of the Equipment Range Type Quantity
THEOTHEORYRY::
Speed can be controlled by varying (i) flux per pole (ii) resistance of armature circuit and
(iii) applied voltage.
It is known that N Eb. If applied voltage is kept, Eb = V – IaRa will
Remain constant. Then, N 1
By decreasing the flux speed can be increased and vice versa. Hence this
method is called field control method. The flux of the DC shunt motor can be
changed by changing field current, Ish with the help of shunt field rheostat. Since
the Ish relatively small, the shunt filed rheostat has to carry only a small current,
which means Ish
2
R loss is small. This method is very efficient. In non-interpolar
machines, speed can be increased by this methods up to the ratio 2: 1. In interpolar
machine, a ratio of maximum to minimum speed of 6:1 which is fairly common.
FFOORMURMULLAA::
AArrmmaatureture CCoontrntrooll DD..CC.. SerServvoo mmoottoorr::
It is DC shunt motor designed to satisfy the requirements of the servomotor.
The field excited by a constant DC supply. If the field current is constant then
speed is directly proportional to armature voltage and torque is directly
proportional to armature current.

J = 0.039 Kg
2
m
B = 0.030 N / rpm
Transfer Function = Km
S (1 + TmS)
Km = 1 / Avg Kb
Tm = JRa / Kb Kt
Kt = T / Ia
Eb = V-Ia Ra
Constant Values

FFiieelldd CoContntrrooll DD..CC.. SSerervvoo mmoottoor:r:
It is DC shunt motor designed to satisfy the requirements of the servomotor.
In this motor the armature is supplied with constant current or voltage. Torque is
directly proportional to field flux controlling the field current controls the torque of
Sl.No If Ia S1 S2 N V T Eb Kb = Eb /
the motor.
Transfer Function =
K
Js
2
(1 + s)
K = Kt / Rf
2
= Lf / Rf = V Zf
2
– Rf
= 2 N / 60
/ 2 f / Rf
T = r ( S1 – S2 ) * 9.81 N-m and r = .075m
OBSERVATION TABLE FOR TRANSFER FUNCTION ARMATURE
CONTROL DC SERVO MOTOR:
TTaabbllee NNoo.. 11 FiFindndiinngg thethe vvalaluuee ooff KKbb

TTaabbllee NNoo.. 22 TToo ffiindnd RRaa
Avg Kb
Sl.No Volt Va Current Ia Ra = Va / Ia
Avg Ra =
PRPREECAUCAUTTIIOONNS:S:
At starting,
The field rheostat should be kept in minimum resistance position
PRPROOCCEEDURDUREE FFOROR TTRANRANSSFFERER FFUNUNCCTTIIONON OFOF ARMARMAATTURUREE CCOONNTTRROLOL
DDCC SESERVRVOOMMOTOOTORR::

FFiindndiinngg KKbb
1. Keep all switches in OFF position.
2. Initially keep voltage adjustment POT in minimum potential position.
3. Initially keep armature and field voltage adjustment POT in minimum
position.
4. Connect the module armature output A and AA to motor armature terminal
A and AA respectively, and field F and FF to motor field terminal F and FF
respectively.
5. Switch ON the power switch, S1, S2.
6. Set the field voltage 50% of the rated value.
7. Set the field current 50% of the rated value.
8. Tight the belt an take down the necessary readings for the table – 1 to find
the value of Kb.
9. Plot the graph Torque as Armature current to find Kt.
FFiindndiinngg RRaa
2. Initially keep voltage adjustment POT in minimum position.
potential position.
4. Connect module armature output A and AA to motor armature terminal A to
AA respectively.
5. Switch ON the power switch and S1.
6. Now armature voltage and armature current are taken by varying the
armature POT with in the rated armature current value.
7. The average resistance value in the table -2 gives the armature resistance.
PRPROOCCEEDURDUREE FFOORR TTRANRANSSFFERER FFUUNCNCTTIIONON OFOF FFIIEELLDD CCOONNTTRROLOL DD..CC..
SESERVRVOOMMOTOOTORR::
FFiindndiinngg RRff
2. Keep armature field voltage POT in minimum potential position.
potential position.
4. Connect module filed output F and FF to motor filed terminal F and FF
respectively.

5. Switch ON the power, S1 and S2.
6. Now filed voltage and filed current are taken by varying the armature POT
with in the rated armature current value.
7. Tabulate the value in the table no – 3 average resistance values give the fied
resistance.
FFiindndiinngg ZZff
2. Keep armature and field voltage POT in minimum position.
position.
4. Connect module varaic output P and N to motor filed terminal F and FF
respectively.
5. Switch on the power note down reading for the various AC supply by
adjusting varaic for the table no – 4.
FFiindndiinngg KKtt
ll
1. Keep all switches OFF position.
2. Initially keep voltage adjustment POT in minimum potential position.
position.
4. Connect the module armature output A and AA to motor armature terminal
and AA respectively, and field F and FF to motor field terminal F and FF
respectively.
5. Switch ON the power switch, S1 and S2.
6. Set the filed voltage at rated value (48V).
7. Adjust the armature voltage using POT on the armature side till it reaches
the 1100 rpm.
8. Tight the belt and take down the necessary reading for the table – 5 Kt
l
9. Plot the graph Torque as Field current to find Kt
l
OBSEOBSERVARVATTIIONON TTAABLEBLE FFOROR TTRARANNSSFFERER FUFUNNCCTTIIONON OFOF ARMAARMATTUURREE
CCOONNTTRROLOL DDCC SESERVRVOO MMOTOOTORR::
TTaabbllee NNoo::33 ToTo ffiindnd RRff
Sl.No If (amp) Vf (Volt) Rf (ohm)

Avg Rf =
TTaabbllee NNoo::44 ToTo ffiindnd ZZff
Sl.No
If (amp)
mA
Vf (Volt) Zf = Vf / If
TTaabbllee NNoo:: 55 ToTo ffiindnd KKtt
l
AAvvgg ZZff ==
Sl.No If Ia S1 S2 T( N – m) N (rpm)
MMOODDELEL GGRAPRAPH:H:
T T Kt
l
= T / If T T Kt = T / Ia
PREPARED BIfY V.BALAJI ,M.Tech, (Ph.D), AP/EEE, DCE
Ia
Page 9

Field Current Armature Current

PREPARED BY V.BALAJI ,M.Tech, (Ph.D), AP/EEE, DCE Page
1010
VIVA QUESTIONS:
1. What are the main parts of a DC servo motor?
2. What are the two types of servo motor?
3. What are the advantages and disadvantages of a DC servo motor?
4. Give the applications of DC servomotor?
5. What do you mean by servo mechanism?
6. What do you mean by field controlled DC servo motor?
MMOODDELEL CACALLCUCULLAATTIIOONN::

1111
Result:
TTRANRANSSFFERER FFUNUNCCTTIIONON OFOF AACC SESERVRVOO MMOTOROTOR
EEXPXPT.T.NNOO ::
DADATETE ::
AAIIMM::

1212
To determine the transfer function of the given AC servomotor
S.No Name of the Equipment Range Type Quantity
NAMNAMEE PPLLAATETE DDEETTAAIILS:LS:
OUTPUT :
VOLTAGE :
CURRENT :
SPEED :
FUFUSESE RARATTIINNGS:GS:
Blocked rotor test: 125% of rated current.
THEOTHEORYRY::
An servo motor is basically a two – phase induction type except for certain
special design features. A two – phase servomotor differ 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
(Inductive reactance / resistance) ratio is small which result in liner speed – torque
characteristics. The excitation voltage applied to two – stator winding should have
a phase difference of 90
o
WWOORRKKIINNGG PRPRIINNCCIIPPLELE OFOF AACC SESERRVVOOMMOTOROTOR
Voltages of equal rms magnitude and 90
o
phase difference excite the stator
winding. These results in exciting current i1 and i2 that are phase displaced by 90
o
and have equal rms value. These current are rise to a rotating magnetic field of
constant magnitude. The direction of rotation depends on the phase relationship of
the two current (or voltage).

1313
The rotating magnetic field sweeps over the rotor conductor. The rotor
conductors experience a change in flux and so voltage are induced in rotor
conductors. This voltage circulates current in the short circuited rotor conductors
and the current creates rotor flux.
Due to the interaction of stator and rotor flux, a mechanical force (or torque)
is developed on the rotor and the rotor starts moving in the same direction as that
of rotating magnetic field.
FFOORMURMULLAA::
Transfer Function =
Laplace Transform of output
Laplace Transform of input
(s) / Es(s) = K1 / sJ + K2 + B = Km / 1 + s ------ (1)
Km = K1 / (K2 + B) ------------------------------- motor gain constant (2)
m = J / (K2 + B) ---------------------------------- motor time constant (3)
Torque (T) = 9.81 * r * s Nm
S = applied load in Kg
R = radius of shaft in m = 0.068 m
CCoonnssttaantnt VaValluueess::
J = 52 gmcm
2
= 0.05kg cm
2
, B = 0.01875
TTaabbllee NNoo:: 11
OBSEOBSERVARVATTIIONON TTAABLEBLE FFOROR DDETEETERMRMIINNIIGG MMOTOROTOR CCOONNSTSTANANTT
KK11::
S.No
Load
(kg)
Control
Voltage (Vc)
Torque
(Nm)

TTaabbllee NNoo:: 22
OBSEOBSERVARVATTIIONON TTAABLEBLE FFOROR DDETEETERMRMIINNIINNGG MMOTOTOORR CCOONNSTSTANANTT
KK22::
S.No
Speed (N)
rpm
Load
(kg)
Torque (Nm)
PRPREECAUCAUTTIIOONNS:S:
i. Initially DPST switch should be in open condition.
ii. Keep the autotransformer in minimum potential position.
iii. In blocked rotor test, block the rotor by tightening the belt around the the
brake drum before starting the experiment.
BLOCK DIAGRAM OF SERVOMOTOR

PRPROOCCEEDURDURE:E:
FFoorr deterdetermmiinniingng mmoottoorr ccoonnssttaantnt KK11
1. Keep variac in minimum potential position.
2. Connect banana connectors “Pout to Pin” and “Nout to Nin”.
3. Connect 9pin D connector from the motor feed back to the input of module
VPET – 302.
4. Switch ON the 230V AC supply of the motor setup.
5. Switch ON the power switch.
6. Switch ON the S2 (main winding) and S1 (control winding) switches.
7. Set the rated voltage (230V) to control phase using VARIAC.
8. Apply load to the motor step by step until it reaching 0 rpm.
9. Take necessary readings for the table -1.
10.To calculate K1 plot the graph torque vs control winding.
FFoorr deterdetermmiinniingng mmoottoorr ccoonnssttaantnt KK22
1. Keep variac in minimum potential position.
2. Connect banana connectors “Pout to Pin” and “Nout to Nin”.
3. Connect 9pin D connector from the motor feed back to the input of module
VPET – 302.
4. Switch ON the 230V AC supply of the motor setup.
5. Switch ON the power switch.
6. Switch ON the S2 (main winding) and S1 (control winding) switches.

7. Set the rated voltage (230V) to control phase using VARIAC.
8. Apply load to the motor step by step until it reaches 0 rpm.
9. Take necessary readings for the table -2.
10.To calculate K2 plot speed vs torque curve.
MMOODDELEL GGRAPRAPHH
MMOTOROTOR CCOONNSTSTANANTT K2K2 MMOTOROTOR CCOONNSTSTANANTT K1K1
TK2 = T / NT
V
N
K1= T / V
Speed in rpm Speed in rpm
MMOODDELEL CACALLCUCULLAATTIIOONN::

VIVA QUESTIONS:
1. Define transfer function?
2. What is A.C servo motor? What are the main parts?
3. What is servo mechanism?
4. Is this a closed loop or open loop system .Explain?
5. What is back EMF?
Result:
ANALOG SIMULATION OF TYPE – 0 and TYPE – 1 SYSTEM
AIM:
To study the time response of first and second order type –0 and type- 1 systems.
APPARATUS REQUIRED:
1. Linear system simulator kit
2. CRO
FORMULAE USED:
PREPARED BY V.BALAJI ,M.Tech, (Ph.D), AP/EEE, DCE Page 17

1. Damping ratio, = (ln MP)
2
/ (
2
+ (ln MP)
2
)
Where MP is peak percent overshoot obtained from the response graph
2. Undamped natural frequency, n = / tp (1 -
2
)
Where tp is peak time obtained from the response graph
3. Closed loop transfer function of type-0 second order system is
C(s) / R(s) = G(s) / 1+G(s)
Where G(s) = K K2 K3 / [(1+sT1) (1 + sT2)]
K is the gain
K2 is the gain of the time constant – 1 block =10
T1 is the time constant time constant – 1 block = 1 ms
T2 is the time constant time constant – 2 block = 1 ms
4. Closed loop transfer function of type-1 second order system is
C(s) / R(s) = G(s) / 1+G(s)
Where G(s) = K K1 K2 /[s (1 + sT1)]
K is the gain
K1 is the gain of Integrator = 9.6
T1 is the time constant of time constant – 1 block = 1 ms
Theoretical Values of n and can be obtained by comparing the co-efficients of
the denominator of the closed loop transfer function of the second order system
with the standard format of the second order system where the standard format is
C(s) /R(s) = n
2
/ s
2
+ 2 ns + n
2
THEORY:
The type number of the system is obtained from the number of poles located at origin in a
given system. Type – 0 system means there is no pole at origin. Type – 1 system means there is
one pole located at the origin.
The order of the system is obtained from the highest power of s in the denominator of
closed loop transfer function of the system
The first order system is characterized by one pole or a zero. Examples of first order
systems are a pure integrator and a single time constant having transfer function of the form K/s
and K/ (sT+1). The second order system is characterized by two poles and upto two zeros. The
standard form of a second order system is C(s) /R(s) = n
2
/ (s
2
+ 2 ns + n
2
) where is
damping ratio and n is undamped natural frequency.
BLOCK DIAGRAM:
1. To find steady state error of type- 1 system

1919
2. To find steady state error of type- 0 system
3. To find the closed loop response of Type-1 second order system
4. To find the closed loop response of Type-0 second order system

2020
PROCEDURE:
1. To find the steady state error of type – 1 first order system
1. The blocks are connected using the patch cords in the simulator kit.
2. The input triangular wave is set to 1 V peak to peak in the CRO and this is applied
to the REF terminal of error detector block. The input is also connected to the X-
channel of CRO.
3. The output from the system is connected to the Y- channel of CRO.
4. The experiment should be conducted at the lowest frequency so keep the frequency
knob in minimum position to allow enough time for the step response to reach near
steady state.
5. The CRO is kept in X-Y mode and the steady state error is obtained as the vertical
displacement between the two curves.
6. The gain K is varied and different values of steady state errors are noted.
2. The input square wave is set to 1 V peak to peak in the CRO and this is applied
channel of CRO.
4. The CRO is kept in X-Y mode and the steady state error is obtained as the vertical
displacement between the two curves.
5. The gain K is varied and different values of steady state errors are noted.
3. To find the closed loop response of type – 0 and type- 1 second order system
2. The input square wave is set to 1 V peak to peak in the CRO and this is applied
channel of CRO.
4. The output waveform is obtained in the CRO and it is traced on a graph
sheet. From the waveform the peak percent overshoot, settling time, rise time,
peak time are measured. Using these values n and are calculated.

2121
5. The above procedure is repeated for different values of gain K and the values are
compared with the theoretical values.
TABULAR COLUMN:
S.No. Gain ,K Steady state error ess (V)
S.No. Gain ,K Steady state error ess (V)
3. To find the closed loop response of type – 0 second order system
S.No. Ga
in,
K
Peak
percent
Overshoot,
%MP
Rise
time,
tr
(sec)
Peak
time,
tp
(sec)
Settling
time,ts
(sec)
Graphical Theoretical
Dam
ping
ratio
Undamped
natural
frequency,
n (rad/sec)
Dam
ping
ratio
Undamped
natural
frequency,
n(rad/sec)

2222
4. To find the closed loop response of type – 1 second order system
S.No. Gain,
K
Peak
percent
Overshoot,
%MP
Rise
time,
tr
(sec)
Peak
time,
tp
(sec)
Settling
time,ts
(sec)
Graphical Theoretical
Dam
ping
ratio
Undamped
natural
frequency,
n
(rad/sec)
Damping
ratio
Undamped
natural
frequency,
n(rad/sec)
MODEL GRAPH:
MODEL CALCULATION:
RESULT

2323
STSTAABBIILLIITYTY ANAANALLYYSSIISS OFOF LLIINNEEAARR SSYYSTEMSTEM

2424
EEXPXPT.T.NNOO ::
DADATETE ::
AAIIMM::
(i) To obtain the bode plot, Nyquist plot and root locus of the given
transfer function.
(ii) To analysis the stability of given linear system using MATLAB.
APPARATUS REQUIRED:
System with MATLAB
THEOTHEORYRY::
FFrequenrequenccyy RReessppoonnsse:e:
The frequency response is the steady state response of a system when the
input to the system is a sinusoidal signal.
Frequency response analysis of control system can be carried either
analytically or graphically. The various graphical techniques available for
frequency response analysis are
1. Bode Plot
2. Polar plot (Nyquist plot)
3. Nichols plot
4. M and N circles
5. Nichols chart
BBoodede pplolott::
The bode plot is a frequency response plot of the transfer function of a
system. A bode plot consists of two graphs. One is plot of the magnitude of a
sinusoidal transfer function versus log . The other is plot of the phase angle of a
sinusoidal transfer function versus log .
The main advantage of the bode plot is that multiplication of magnitude can
be converted into addition. Also a simple method for sketching an approximate log
magnitude curve is available.

PPoollaarr pplloot:t:
The polar plot of a sinusoidal transfer function G (j ) on polar coordinates
as is varied from zero to infinity. Thus the polar plot is the locus of vectors
G (j ) G (j ) as is varied from zero to infinity. The polar plot is also called
Nyquist plot.
NNyyququiisstt SSttaabbililiityty CCrriiteterriioon:n:
If G(s)H(s) contour in the G(s)H(s) plane corresponding to Nyquist contour
in s-plane encircles the point – 1+j0 in the anti – clockwise direction as many times
as the number of right half s-plain of G(s)H(s). Then the closed loop system is
stable.
RRoooott LLooccuuss::
The root locus technique is a powerful tool for adjusting the location of
closed loop poles to achieve the desired system performance by varying one or
more system parameters.
The path taken by the roots of the characteristics equation when open loop
gain K is varied from 0 to are called root loci (or the path taken by a root of
characteristic equation when open loop gain K is varied from 0 to is called root
locus.)
FFrequenrequenccyy DDoommaiainn SpecSpeciiffiiccaattioionnss::
The performance and characteristics of a system in frequency domain are
measured in term of frequency domain specifications. The requirements of a
system to be designed are usually specified in terms of these specifications.
The frequency domain specifications are
1. Resonant peak, Mr
2. Resonant Frequency, r.
3. Bandwidth.
4. Cut – off rate
5. Gain margin
6. Phase margin

RReesosonnaantnt PPeeaakk,, MMrr
The maximum value of the magnitude of closed loop transfer function is
called the resonant peak, Mr. A large resonant peak corresponds to a large over
shoot in transient response.
RReesosonnaantnt FFreqrequuenenccyy,, rr
The bandwidth is the range of frequency for which the system gain is more
than -3db. The frequency at which the gain is -3db is called cut off frequency.
Bandwidth is usually defined for closed loop system and it transmits the signals
whose frequencies are less than cut-off frequency. The bandwidth is a measured of
the ability of a feedback system to produce the input signal, noise rejection
characteristics and rise time. A large bandwidth corresponds to a small rise time or
fast response.
CCut-Offut-Off RRaate:te:
The slope of the log-magnitude curve near the cut off frequency is called
cut-off rate. The cut-off rate indicates the ability of the system to distinguish the
signal from noise.
GGaaiinn MMaarrggiin,n, KKgg
The gain margin, Kg is defined as the reciprocal of the magnitude of open
loop transfer function at phase cross over frequency. The frequency at witch the
phase of open loop transfer function is 180 is called the phase cross over
frequency, pc.
PPhhasasee MMaarrgigin,n,
The phase margin , is that amount of additional phase lag at the gain cross
over frequency required to bring the system to the verge of instability, the gain
cross over frequency gc is the frequency at which the magnitude of open loop
transfer function is unity (or it is the frequency at which the db magnitude is zero).

1. Enter the command window of the MATLAB.
2. Create a new M – file by selecting File – New – M – File.
3. Type and save the program.
4. Execute the program by either pressing F5 or Debug – Run.
5. View the results.
6. Analysis the stability of the system for various values of gain.
PPrroobbllemem 11
Obtain the bode diagram for the following system
 x1   0 1   x1   1 1

y 1 
         
 x 2   25 4   x 2

 0 1  y 2 
 y 1   0 1   x1 
     
 y 2   25 4   x 2 
MAMATLTLAABB PPrrogogrraamm
a = [0 1 ; -25 -4]
b = [1 1 ; 0 1]
c = [1 1 ; 1 1]
d = [0 0 ; 0 0]
bode (a, b, c, d)
grid
title (‘BODE DIAGRAM’)
PPrroobbllemem 22
1

PREPARED BY V.BALAJI ,M.Tech, (Ph.D), AP/EsE(Es,
DCE1 )
Page 27

Draw the Nyquist plot for G(s) =
num = [0 0 0]
den = [1 1 0]
nyquist (num,den)
v = [-2,2,-5,5]
axis (v)
grid
title (‘Nyquist Plot’)
PPrroobbllemem 22
Obtain the root focus plot of the given open loop T.F is
G(s) H (s) =
s (
s
K
0 . 5 )( s
2
0 . 6
s
10 )
num = [0 0 0 0 1]
den = [11.1 10.3 5 0]
rlocus (num,den)
grid
title [‘Root Locus Plot’]

Result:
CCLOSEDLOSED LOOPLOOP SSPPEEDEED CCOONNTTRROLOL SSYYSTEMSTEM

EEXPXPT.T.NNOO ::
DADATETE ::
3030
AAIIMM::
To study the behavior of closed loop speed control system using
PID controller
(i) PID controller with motor
(ii) CRO
THEOTHEORYRY::
Closed loop system
Control system which the output has an effect upon the input quantity in such a
manner as to maintain the desired output value is called closed loop systems.
The open loop system can be modified as closed loop system by providing a
feedback. The provision of feedback automatically corrects the change in output
due disturbances. Hence the closed loop system is also called closed loop system.
The general block diagram of an automatic control system is given below. In
consists of an error detector, a controller, plant (open loop system) and feedback
path element.
The reference signal (or input signal) corresponds to desired output. The
feedback path elements sample the output and convert it to a signal of same type as
that of reference detector. The error signal generated by the error detector is the
difference between reference signal and feedback signal. The controller modifies
and amplifies the error signal to produce better control action. The modified error
signal is fed to the plant to correct its output.

1. Make the connections as per the circuit diagram.
2. Set the speed of the motor using set position.
3. Vary the gain values of P,I, and D controller until to get the set speed to
current speed.
4. Repeat the above procedure for different values of set speed.
Result:
STSTUDUDYY OFOF AACC SSYYNCNCHHRROO TTRANRANSSMMIITTERTTER ANANDD RREECCEEIIVVERER

EEXPXPT.T.NNOO ::
DADATETE ::
3232
AAIIMM::
To study the operation of AC synchro transmitter and receiver
S.No Name of the Equipment Quantity
1 Synchro transmitter and receiver unit 1 Nos
2 Multimeter (Digital / Analog ) 1 Nos
3 Patch cords As required
THEOTHEORYRY::
A synchro is an electromagnetic transducer commonly used to
convert an angular position of a shaft into an electric signal. It is commercially
known as a selsyn or an autosyn. The basic synchro unit is usually called a synchro
transmitter. Its construction is similar to that of three phase alternator. The stator is
of laminated silicon steel and is slotted to accommodate a balanced three phase
winding which is usually of concentric coil type and star connected. The rotor is
dumb bell construction and its wound with a concentric coil.
AC voltage is applied to rotor winding through slip rings. Let and
AC voltage
Vr (t) = Vr sin ct be applied to the rotor of the synchro transmitter.
The voltage causes a flow of magnetizing current in rotor coil which produces a
sinusoidally time varying flux directed along its axis and distributed nearly
sinusoidally in the air gap along the stator periphery. Because of transformer
action, voltage is induced in each of the stator coil. As the air gap flux sinusoidally
distributed the flux linking with any stator coil is proportional to the cosine of the
angle between the axes of rotor and stator coil. This flux voltage in each stator coil.
Voltages are in time.

phase with each other. Thus the synchro transmitter acts a like a single-phase
transformer in which the rotor coil is the primary and the stator coil is the
secondary.
Let Vs1n, Vs2n, Vs3n, be the voltage induced in the stator coils, S1, S2,
S3 respectively with respect to the neutral. Then for a rotor position of the synchro
transmitter, is the angle made by rotor axis with the stator coil S2.
The various stator voltages are
Vs1n = KVr sin ct cos ( + 120
o
)
Vs2n = KVr sin ct cos
Vs1n = KVr sin ct cos ( + 240
o
)
The terminal voltages of the stator are
Vs1s2
Vs2s3
Vs1n
Vs2n
Vs2n
Vs3n
3 KVr sin(
3 KVr sin(
240
o
120
o
sin c t
) sin c t
Vs3s1 Vs3n Vs1
n
3 KVr sin sin c t
When = 0, Vs1s2 and Vs2 s3 have the maximum voltage and while Vs3s1 has
zero voltage. This position of rotor is defined as ht electrical zero of the transmitter
and is used as reference for specifying the angular position of the rotor.
Thus it is seen that the input to the synchro transmitter is the angular
position of its rotor shaft and the output is a set of three signal phase voltages. The
magnitudes of this voltage are function of the shift position. The output of the
synchro transmitter is applied to stator winding of synchro control transformer.
The control transmitter is similar in construction to a synchro transmitter
except for the fact that rotor of the control transformer in made cylindrical in shape
so that the air gap is practically uniform. The system (transmitter and control
transformer pair) acts an error detector, circulating current to the same phase but of
different magnitudes flow through two stator coils. The result is establishment of
an indentical flux pattern in the air gap of the control transformer as the voltage
drops in resistance and lockage reactance’s of two sets of stator coils are usually
small.

3434
OBSEOBSERVARVATTIIONON TTAABLE:BLE:
S.No
Transmitter
(Degree)
Receiver
(Degree)
Vs1 – Vs2 Vs2 – Vs3 Vs3 – Vs1 Error
the synchro transmitter rotor, the voltage induced the control transformer rotor is
proportional to the cosine of the angle between the two rotors given by
E (t) = KVr cos sin r t

The synchro transmitter and control transformer thus act as an error detector giving
a voltage signal at the rotor terminals of the control transformer proportional to the
angular difference between the transmitter control transformer shaft positions.
1. Make the connections as per the patching diagram.
2. Switch ON the supply.
3. Vary the shaft position of the transmitter and observe the corresponding
changes in the shaft position of the receiver.
4. Repeat the above steps for different angles of the transmitter.
5. Tabulated the different voltage at the test points of S1 S2, S3S2, and S3S1.
Result:
CYCLE – 2
7. (a) Lag Compensator.

3636
7. (b) Lead Compensator.
8. Digital Simulation of Non-Liner System.
9. Digital Simulation of Liner System.
10. Digital Simulation of Type 0 and Type 1 System.
DDIIGGIITTAALL SSIIMUMULLAATTIIONON OFOF NNOONN-L-LININEEAARR SSYYSTEMSTEM
EEXPXPT.T.NNOO ::

DADATETE ::
AAIIMM::
To simulate the time response characteristic of liner system with simple
non-linearities like saturation and dead zone.
System with MATLAB 6.5
THEOTHEORYRY::
NNoon-Ln-Liineneaarr SSyysstetemmss::
The non linear system are system witch do not obey the principle of
superposition.
In practical engineering systems, there will be always some non linearity due
to friction, inertia, stiffness, backslash, hysteresis, saturation and dead – zone. The
effect of the non linear components can be avoided by restricting the operation of
the component over a narrow limited range.
CCllaassssiiffiiccaattiioonn ooff nnoonn lilinneeaarriittiieess::
The non linearities can be classified as incidental and intentional.
The incidental non linearities are those which are inherently present in
the system. Common examples of incidental non linearities are saturation, dead –
zone, coulomb friction, stiction, backlash, etc.
The intentional non linearities are those which are deliberately
inserted in the system to modify system characteristics. The most common
example of this type of non linearity is a relay.
SSaatuturraattioionn::

In this type of non linearity the output proportional to input for limited
range of input signals. When the input exceeds this range, the output tends to
become nearly constant.
All devices when driven by sufficient large signals, exhibit the
phenomenon of saturation due to limitations of their physical capabilities.
Saturation in the output of electronic, rotating and flow amplifiers, speed and
torque saturation in electric and hydraulic motors, saturation in the output of
sensors for measuring position, velocity, temperature, etc. are the well known
examples.
DDeeaadd ZZoone:ne:
The dead zone is the region in witch the output is zero for given input.
Many physical devices do not respond to small signals, i.e., if the input amplitude
is less than some small value, there will be no output. The region in which the
output is zero is called dead zone. When the input is increased beyond this dead
zone value, the output will be linear.
1. Double click on MATLAB 6.5 icon on desktop command window opens.
2. From File Tab, select New Model file.
3. A Simulink model screen opens a “untitled”.
4. From Simulink library – select necessary blocks and place in new model
screen.
Block
Constant - Simulink-Continuous
Simulator - Simulink-Math operator
Transfer function - Simulink-Continuous

Scope - Simulink –sink
Dead Zone, Saturation - Simulink-Non-linear
5. Select properties for each item and connect them as shown in diagrams.
6. Select simulation Tab and configuration parameters and select ode23tb
model.
7. Save file under ‘work’ directory.
8. Simulated the system with step and sine inputs with and without dead zone,
saturation non – linearities.
9. Name the signals as mentioned in diagram and observe signal names on
scope by right clicking on response curve and by opening axes.

Result:
EEXPXPT.T.NNOO :
DADATETE :
DDIIGGIITTAALL SSIIMUMULLAATTIIONON OFOF LLIINNEEAARR SSYYSTEMSTEM

AAIIMM::
To simulate the time response characteristic of higher-order Multi-
input multi output (MIMO) liner system using state variable formulation.
MATLAB 6.5
THEOTHEORYRY::
Time Domain Specification
The desired performance characteristics of control systems are specified in
terms of time domain specification. System with energy storage elements
cannot respond instantaneously and will exhibit transient responses, whenever
they are subjected to inputs or disturbances.
The desired performance characteristics of a system of any order may be
specified in terms of the transient response to a units step input signal.
The transient response of a system to a unit step input depends on the initial
conditions. Therefore to compare the time response of various systems it is
necessary to start with standard initial conditions. The most practical standard is
to start with the system at rest and output and all time derivatives there of zero.
The transient response of a practical control system often exhibits damped
oscillation before reaching steady state.
The transient response characteristics of a control system to a unit step input
are specified in terms of the following time domain specifications.
1. Delay time, td
2. Rise time, tr
3. Peak time, tp
4. Maximum overshoot, Mp

2
1 2
5. Setting time, ts
FFOORMURMULLAA::
Risetime
d
where tan 1 1
Damped frequency of oscillation, d n
8. Create a new workspace by selecting new file.
9. Complete your model.
10.Run the model by either pressing F5 or start simulation.
11.View the results.
12.Analysis the stability of the system for various values of gain.
PRPROBLEOBLEMM::
Obtain the step response of series RLC circuit with R = 1.3K , L = 26mH and
C=3.3 f using MATLAB M – File.
MAMATLTLAABB PRPROGOGRRAAMM FFOROR UUNNIITT IIMMPUPULSELSE PRPRSSPPOONNSSEE::
PRPROGOGRAMRAM::
num = [ 0 0 1 ]
den = [ 1 0.2 1 ]

impulse (num, den)
grid
title (‘ unit impulse response plot’)
MAMATLTLAABB PRPROGOGRRAAMM FFOROR UUNNIITT STEPSTEP PRPRSSPPOONNSE:SE:
PRPROGOGRAMRAM::
Format long e
num = [ 0 0 1.6e10 ]
den = [ 1 50000 1.6e10 ]
step (num, den)
grid on
title (‘step response of series RLC circuit’)
Result:
DIGITAL SIMULATION OF TYPE 0 AND TYPE 1 SYSTEM
EXPT.NO :
DATE :

AIM:
To simulate the time response characteristics of first order second
order, type 0 and type 1 system using MATLAB.
APPARATUS REQUIRED:
System employed with MATLAB 6.5
THEORY:
The desired performance characteristics of control system are specified in
terms of time domain specification. Systems with energy storage elements cannot
respond instantaneously and will exhibit transient responses, whenever they are
subjected to inputs or disturbances.
The desired performance characteristics of a system pf any order may be
specified in terms of the transient response to a unit step input signal.
The transient response of a system to unit step input depends on the initial
conditions. Therefore to compare the time response of various systems it is
necessary to start with standard initial conditions. The most practical standard is to
start with the system at rest and output and all time derivatives there of zero. The
transient response of a practical control system often exhibits damped oscillations
before reaching steady state.
The transient response characteristics of a control system to a unit step input
are specified in terms of the following time domain specifications.
1. Delay time, td
2. Rise time, tr
3. Peak time, tp
4. Maximum overshoot, Mp
5. Settling time, ts

The time domain specification is defined as follows.
1. Delay Time:
It is the taken for response to reach 50% of the final value, for the very first
time.
2. Rise Time:
It is the time taken for response to raise from 0 to 100% for the very first
time. For under damped system, the rise time is calculated from 0 to 100%. But for
over damped system it is the time taken by the response to raise from 10% to 90%.
For critically damped system, it is the time taken for response to raise from 5% to
95%.
Risetime
d
Where tan
1
1 s
2
/ s
Damped frequency of oscillation, d
3. Peak Time:
n 1 s
2
It is the time taken for the response to reach the peak value for the very first
time. (or) It is the taken for the response to reach the peak overshoot, tp.
Peak time = / d
4. Peak Overshoot (Mp):
It is defined as the ration of the maximum peak value measured from final
value to the final value.
Let final value = c (e)
Maximum vale = c (tp)
c ( t p
) c ( e )
Peak Overshoot, Mp =
c ( e )

1
t s
1
t
t
s
s
% M p e
s
3
x100
5. Settling Time:
It is defined as the time taken by the response to reach and stay within a
specified error. It is usually expressed as % of final value. The usual tolerable error
is 2% or 5% of the final value.
4
t s
for
n
3
for
n
2 % erroe
5 % erroe
FORMULA:
% M p e
s
3
x100
4
for
n
3
for
n
2 % erroe
5 % erroe

PROCEDURE:
Closed loop response of first order system:
4. Run the model by either pressing F5 or start simulation.
5. Analysis the stability of the system for various values of gain
Closed loop response of second order system:
4. Run the model by either pressing F5 or start simulation.
6. Analysis the stability of the system for various values of gain.
General MATLAB coding for closed loop response for type 0 and type1
system:
PROGRAM:
clear all

close all
clc
T1 = tf (2.25, [1 0.5 2.25 ])
p=pole (T1)
pre=abs (real (p(1))) pim=abs
(imag (p(1)))
wn=sqrt(pre*pre*+pim*pim)
damping _ratio=(pre/wn)
os=(exp(-1*pre*pi/pim))*100
tp=pi/pim
ts=4/pre
step(T1)
t=[ 0.1:0.1:25]
for x=1:length (t)
c (x)=1-1.01418*(cos (1.47902*t(x)-(9.59*pi/180))*exp(-25*t(x)))
end
figer
plot(t,c)
Result:
EEXPXPT.T.NNOO ::
DADATETE ::
LLAAGG CCOOMPMPEENNSSAATORTOR

System employed with MATLAB 6.5
THEOTHEORYRY::
The control systems are designed to perform specific taskes. When
performance specification are given for single input. Single output linear time
invariant systems. Then the system can be designed by using root locus or
frequency response plots.
The first step in design is the adjustment of gain to meet the desired
specifications. In practical system. Adjustment of gain alone will not be sufficient
to meet the given specifications. In many cases, increasing the gain may result poor
stability or instability. In such case, it is necessary to introduce additional devices
or component in the system to alter the behavior and to meet the desired
specifications. Such a redesign or addition of a suitable device is called
compensations. A device inserted into the system for the purpose or satisfying the
specifications is called compensator. The compensator behavior introduces pole &
zero in open loop transfer function to modify the performance of the system.
The different types of electrical or electronic compensators used are lead
compensator and lag compensator.
In control systems compensation required in the following situations.
1. When the system is absolutely unstable then compensation is required
to stabilize the system and to meet the desired performance.
2. When the system is stable. Compensation is provided to obtain the
desired performance.
LAG COMPENSATOR:
A compensator having the characteristics of a lag network is called a lag
compensator. If a sinusoidal signal is applied to a lag network, then in steady state
the output will have a phase lag with respect input.

1
1
Lag compensation result in a improvement in steady state performance but
result in slower response due to reduced bandwidth. The attenuation due to the lag
compensator will shift the gain crossover frequency to a lower frequency point
where the phase margin is acceptable. Thus the lag compensator will reduce the
bandwidth of the system and will result in slower transient response.
Lag compensator is essentially a low pass filter and high frequency noise
signals are attenuated. If the pore introduce by compensator is cancelled by a zero
in the system, then lag compensator increase the order of the system by one.
FORMULA:
Gain
B
A
y 0
x 0
20 log( B / A )
Phase sin ( x0
/ A)
sin ( y
0
/ B )
PROCEDURE:
With out compensator:
1. Make the connection as per the circuit diagram.
2. Apply the 2V p-p sin wave input and observe the waveform.
3. Very the frequency of the sin wave input and tabulate the values of xo and yo
4. Calculated gain and phase angle.
5. Draw the bode plot.
With lag compensator:
1. From the bode plot find the new gain crossover frequency.
2. Find out values and writ the frequency function. G(s).

3. From the transfer function calculated R1, R2 and C.
4. Set the amplifier gain at unity.
5.
6.
Insert the lag compensator with the help of passive
determine the phase margin of the plant.
Observe the step response of the compensated system.
components and
MATLAB coding with Compensator:
PROGRAM:
num = [ 0 0 100 5 ];
den = [ 400 202 1 0 ];
sys = (sys)
margin (sys)
[ gm, ph, wpc, wgc ] = margin (sys)
title (‘BODE PLOT OF COMPENSATED SYSTEM’)
MATLAB coding with out lag Compensator:
PROGRAM:
num = [ 0 0 5 ];
den = [ 2 1 0 ];
sys = tf (num, den)
bode (sys)
Margin (sys)
[ gm, ph, wpc, wgc ] = margin (sys).
title (‘BODE PLOT OF UNCOMPENSATED SYSTEM’);

Result:
LEAD COMPENSATOR:
A compensator having the characteristics of a lead network is called a lead
compensator. If sinusoidal signal is applied to a lead network, then in steady state
the output will have a phase lead with respect to input.

1
1
The lead compensator increase the bandwidth, which improves the speed of
response and also reduces the amount of overshoot. Lead compensation
appreciably improves the transient response, whereas there is a small change in
steady state accuracy. Generally lead compensation is provided to make an
unstable system as a stable system. A lead compensator is basically a high pass
filter and so it amplifies high frequency noise signals. If the pole is introduced by
the compensator is not cancelled by a zero in the system, then lead compensator
increases order of the system by one.
FORMULA:
Gain
B
A
y 0
x 0
20 log( B / A )
Phase sin ( x0
/ A)
sin ( y
0
/ B )
PROCEDUR:
1. Enter the command window of MATLAB.
2. Create a New M-File by selecting file New M-File.
4. Execute the program by pressing F5 or Debug Run.
6. Analyze the Results.
With lead compensator:
2. Create a new M – file by selecting File – New –M-File.

4. Execute the program by either pressing F5 or Debug – Run.
6. Analysis the result.
MATLAB coding with out Compensator for loop system
PROGRAM:
den=[ 1 0.739 0.921 0 ];
pitch=tf(num, den);
sys_cl=feedback (pitch,1);
de=0.2;
t=0:0.01:10;
figure
step(de*sys_cl, t)
sys_cl=feedback (pitch,10);
de=0.2;
t=0:0.01:10;
bode(sys_cl, t)
grid on
title ( 'BODE PLOT FOR CLOSED LOOP SYSTEM WITHOUT
COMPENSATOR')
MATLAB coding with Compensator for loop system
PROGRAM:

num=[1 151 0.1774 ];
den=[1 0.739 0.921 0 ];
pitch=tf(num, den);
alead=200;
Tlead=0.0025;
K=0.1;
lead=tf(K*[alead*Tlead 1], [Tlead 1]);
bode(lead*pitch)
sys_cl=feedback(lead*pitch,10);
de=0.2;
t=0:0.01:10;
figure
step (de*sys_cl, t)
title('BODE PLOT FOR CLOSED LOOP SYSTEM WITH
COMPENSATOR')
Result:

196444650 ee-2257-control-system-lab-manual

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196444650 ee-2257-control-system-lab-manual