Control Systems Lab Manual

VIGNAN’S INSTITUTE OF ENGINEERING
FOR WOMEN
Approved by AICTE, New Delhi, Affiliated to JNTUK, Kakinada
Kapujaggaraju Peta, VSEZ(post), Visakhapatnam, Andhra Pradesh-530049.
CONTROL SYSTEMS
LABORATORY
MANUAL
JNTUK-R19 REGULATIONS
Prepared by
KAVURU SRINIVASARAO
Assistant Professor
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY: KAKINADA
KAKINADA 533 003, Andhra Pradesh, India
DEPARTMENT OF ELECTRICAL AND ELECTRONICS
ENGINEERING
COURSE STRUCTURE-R19
III Year I SEMESTER
L T P C
0 0 2 1
CONTROL SYSTEMS LABORATORY
Learning Objectives:
• To impart hands on experience to understand the performance of basic control
system components such as magnetic amplifiers, D.C. servo motors, A.C. Servo
motors and Synchros.
• To understand time and frequency responses of control system with and without
controllers and compensators.
Any 10 of the following experiments are to be conducted:
1. Time response of Second order system
2. Characteristics of Synchros
3. Effect of P, PD, PI, PID Controller on a second order systems
4. Design of Lag and lead compensation Magnitude and phase plot
5. Transfer function of DC motor
6. Bode Plot, Root locus, Nyquist Plots for the transfer functions of systems up to
5th order using MATLAB.
7. Controllability and Observability Test using MAT LAB.
8. Temperature controller using PID
9. Characteristics of magnetic amplifiers
10. Characteristics of AC servo motor
11. Characteristics of DC servo motor
12. Block Diagram Representation of Field Controlled DC servo Motor Using
Simulink.
Learning Outcomes:
After the completion of the course the student should be able to:
• Analyse the performance and working Magnetic amplifier, D.C and A.C. servo
motors and synchros.
• Design P,PI,PD and PID controllers
• Design lag, lead and lag lead compensators
• Control the temperature using PID controller
• Determine the transfer function of D.C Motor
• Control the performance of D.C and A.C Servo Motor.
• Test the controllability and observability.
• Judge the stability in time and frequency domain.

Table of Contents
1. Time Response Analysis of Second Order System................................................1
2. Characteristics of Synchro’s ..................................................................................9
3. Effect of P, PD, PI, PID Controller on a Second Order Systems ........................19
4. Design of Lag and Lead Compensation Magnitude and Phase Plot....................25
5. Transfer Function of a DC Motor........................................................................33
6. Controllability and Observability Test using MATLAB.....................................41
7. Bode Plot, Root Locus, Nyquist Plots for the Transfer Functions of Systems up
to 5th Order using MATLAB. .....................................................................................49
8. Temperature Controller using PID.......................................................................73
9. Characteristics of Magnetic Amplifiers...............................................................81
10. Characteristics of AC Servo Motor......................................................................89
11. Characteristics of DC Servo Motor......................................................................97
12. Block Diagram Representation of Field Controlled DC Servo Motor Using
Simulink.....................................................................................................................103
13. Add-On 01. Design Problem by Construction of Root Locus...........................111
14. Add-On 02. Design of Position Control System using Bode Plot (Frequency
Response Analysis)....................................................................................................119
15. Add-On 03. Design of Lag Compensator using Frequency Response Method 127
16. Add-On 04. Design of Lead Compensator using Frequency Response Method ....
.................................................................................................................137
17. Add-On 05: Design of Lag-Lead Compensator using Frequency Response
Method .................................................................................................................145
•

Time Response Analysis of Second Order System
Vignan’s Institute of Engineering for Women, Visakhapatnam 1
1. Time Response Analysis of
Second Order System

Circuit Diagram
Figure 1.1: Second Order System (RLC Network)
Figure 1.2: Response of RLC Network (Second Order System)

TIME RESPONSE ANALYSIS
OF SECOND ORDER SYSTEM
Aim:
To find the time response of a second order system.
Apparatus:
1. Time Response Kit
2. Multimeter
3. Digital CRO
4. Patch Cords
Theory:
Second order control systems are characterized by two poles and two zeros. For
the purpose of transient response studies usually zeros are not considered primarily
because of simplicity in calculation and also because the zeros do not affect the internal
nodes of the system. Overall transfer function of the second order system in standard
form is given by
2
2 2
( )
( )
( ) 2
n
n n
C S
T S
R S S S

 
= =
+ +
When n
 is called damping ratio n
 is the un-damped natural frequency. In order to
analyze the transient behaviour of the control systems, the first step is to obtain the
mathematical model of the system these dynamic behaviour is analysed under the
application of standard test signals is generally used for testing as it can be easily
generated. The time response performance is measured by computing time response
performance is indicated below.
1. Delay time ( d
t ): It is the time required for the response to read 50% of the final
value in first attempt.
Delay time,
1 0.7
d
n
t


+
=

Observation Tables:
Inductance of inductor coil, L=2H
Internal resistance of inductor coil, Rint=30Ω
Capacitance, C=0.32µF
Tabular Form for Under Damped System
Resistance Value, R =300 Ω
Damping Factor
300 30 0.32
0.066
2 2 2
R C
L


+
= = =
Delay Time.
td
Raise Time
tr
Peak Time,
tP
% Peak Overshoot
MP (%)
Settling Time,
TS
Theoretical 0.833ms 1.3ms 2.5ms 32.79 0.053 s
Practical 1.1ms 2ms 0.8ms 6.66ms 0.59

2. Rise time ( r
t ): It is the time required by the response to rise from 10% to 90% of
final value for over-damped systems and 0 to 100% of the final value for under-
damped system.
Rise time,
2
1
2
1
tan
1
r
n
t



 
− −
−
=
−
3. Peak time ( p
t ): It is the time required for the response to reach the peak of time
response or peak overshoot.
Peak time,
2
1
p
n
t

 
=
−
4. Peak overshoot (MP): It indicates the normalized difference between the time
response peak and steady state output and is defined as,
Peak overshoot,
2
1
100
P
M e


−
−
= 
5. Settling time (ts): It is the time required for the response to reach and stay with in
a specified tolerance level (usually 2 % or 5% of its final value)
Settling time,
1
for 2% error
3
for 5% error
S
n
n
t


=
=
Procedure:
1. Switch On the main supply to the unit observe signal source output by selecting
square wave (or) step input and by vary amplitude potentiometer. Make sure signal
source is correct before connecting the input of second order system.
2. Trace the input square wave signal.
3. Trace the output of the second order RLC circuit for R=300Ω, R=5000Ω and
R=10kΩ. Measure the R value using digital multi meter between Input and
terminals marked “R”.
4. Find the time response specifications for the various values of R.

Theoretical Calculations:
Calculations for Underdamped System:
Inductance of inductor coil, L=2H
Internal resistance of inductor coil, Rint=30Ω
Capacitance, C=0.32µF
Resistance Value, R =300 Ω
Damping Factor
300 30 0.32
0.066
2 2 2
R C
L


+
= = =
Natural Frequency of oscillations,
1 1
1250 / sec
2 0.32
n rad
LC


= =

2 2
1 1250 1 0.066 1247.27 / sec
d n rad
  
= − = − =
Delay time,
1 0.7 1 0.7 0.066
2.51
1250
d
n
t ms


+ + 
= = =
Rise time,
2 2
1 1
2 2
1 1 0.06
tan tan
0.06 1.3
1 1250 1 0.06
r
n
t ms

 

 
− −
− −
− −
= = =
− −
Peak time,
2 2
2.5
1 1250 1 0.06
p
n
t ms
 
 
= = =
− −
% Peak overshoot,
2 2
0.06
1 1 0.06
100 100 32.79%
P
M e e
 


− −
− −
=  =  =
Settling time,
4
for 2% error
4
= 0.053sec
0.066 1250
S
n
t

=
=


Precautions:
Handle the instruments gently and avoid errors while observing the waveforms.
Result:
Time domine specifications of the under damped second order system are
calculated and compared with practical readings. Influence of damping factor on the
time response is observed.

Characteristics of Synchro’s
2. Characteristics of Synchro’s

Figure 2.1:Torque Transmission Through the Synchro’s

CHARACTERISTICS OF SYNCHROS
Aim:
To study the characteristics of synchro’s.
Apparatus:
1. Synchro transmitter
2. Synchro receiver
3. Multi meter
4. Connecting wires.
Theory:
A Synchro is an electromagnetic transducer commonly used to convert an angular
position of a shaft into an electric signal.
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 120 degree
apart) and is Y connected. The rotor is a dumb bell construction and wound with a
concentric coil.
The classical synchro systems consists of two units.
1. Synchro transmitter (Tx)
2. Synchro receiver (Tr)
The synchro receiver is having almost the same constructional features. The two
units are connected as shown in figure no.2. Initially the winding S2 of the stator of
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 winding are proportional to
cos 60 degrees 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 windings because of
voltage balance.

Observations:
Part-A:
Rotor Shaft
Position in
Degrees
S1-S2
V
(Volts)
S2-S3
V
(Volts)
S3-S1
V
(Volts)
0 21.9 93.7 112.2
30 75.4 43.2 119.2
60 110.2 12.2 97.5
90 120 70.5 48.9
120 96 110.8 14.4
150 46 119.6 72.2
180 12.8 96.9 110.3
210 63.7 48.0 119.5
240 104.2 15.4 95.4
270 119.3 74.2 44.4
300 93.7 112.2 18.1
330 43.6 118.5 75.3
Part-B:
Transmitter: Rotor
Position in Degrees
Receiver: Rotor
Position in Degrees
Error in Degrees
0 0 0
30 30 2
60 58 2
90 88 5
120 116 4
150 146 4
180 177 1
210 209 0
240 240 0
270 270 0
300 300 0
330 330 0

When the rotor of Tx is moved to a new position, the voltage balance is
disturbed. Assume that the rotor of Tx is moved through 30 degrees, the stator winding
voltages will be changed to zero, 0.866V and 0.866V respectively. Thus there is a
voltage imbalance between the windings causes currents to flow through the close
circuit producing 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 Tx.
The Tx Tr pair thus serves to transmit information regarding angular position
at one point to a remote point.
Procedure:
Part-A:
1. Switch ON the power supply to synchro’s study kit and connect the voltmeter
across transmitter terminals.
2. Measure voltages between S1 – S2 ( 1- 2
S S
V )for various shaft positions from
0⁰ - 360⁰. (These voltages which have positive sign has to be decided from the
knowledge of voltage waveform). Similarly measure voltages between S2–S3
( 2- 3
S S
V )and S3 – S1 ( 3- 1
S S
V )for various shaft positions and tabulate readings.
3. Plot the graph of rotor position in degree Vs voltage across stator terminal voltages.
Part-B:
1. Connect the system to main supply.
2. With the help of patch cards establish connections between corresponding
terminals of Tx and Tr stators i.e., connect S1 to S1, S2 to S2 and S3 to S3 of Tx
and Tr respectively.
3. Switch on SW1 and SW2.
4. Move the pointer of the Tx and observe the new rotor position of Tr. Enter input
angular position and output angular position and plot the graph. This is the study
the phenomenon of transmission distance by electrical means.

Model Graphs:
Figure 2.2: Characteristics of Synchros

Precautions:
1. Handle the pointers for both the rotors in a gentle manner.
2. Do not attempt to pull out the pointers.
3. Do not short rotor or stator terminals.
Result:
Synchro’s characteristics are obtained, and the maximum error obtained is 5 degrees.

Front Panel Connections-PART-A:
Figure 2.3: Font End Panel Connections for Part-A Procedure

Front Panel Connections-PART-B:
Figure 2.4: Font End Panel Connections for Part-B Procedure

Effect of P, PD, PI, PID Controller on a Second Order Systems
3. Effect of P, PD, PI, PID
Controller on a Second Order
Systems

Block Diagram of Closed Loop System
Figure 3.1: Block Diagram of Closed Loop System

EFFECT OF P, PD, PI, PID CONTROLLER ON A
SECOND ORDER SYSTEMS
Aim:
To study the effect of P, PD, PI, PID controller on a second order system.
Apparatus:
• PID Controller Trainer Kit
• Patch Cards
Theory:
To meet two independent specifications a second order systems, need
modifications. This modification is termed as compensation should allow for high open
loop gains to specification steady state accuracy and yet preserve a satisfactory dynamic
performance. Some of the practical modification schemes are discussed below.
Derivative Error Compensation:
A system is said to possess derivative error compensation when the generation of
its output depends in some way on the rate of change of actuating signal. A controller
producing such a signal is called a proportional plus derivative controller or PD
controller. The advantage here is that as the damping increases due to compensation with
remaining fixed, the system setting time reduces.
Integral Error Compensation:
In this scheme, the output response depends is some manner up on the integral of the
actuating signal. This type of compensation is introduced by using a controller, which
produces an output signal consisting of two terms, one proportional to the actuating signal
and other proportional to its integral. Such a controller is called proportional plus Integral
or PI controller. This scheme is used to have high accuracy requirements.
Proportional Plus Integral Plus Derivative Controller:
Error integration in the forward path eliminates steady state velocity error but it increases
systems order making it more susceptible to instability. It also introduces a zero into the
forward path so that peak overshoot to step i/p cannot be easily eliminated. To increase
the damping factor of the dominant poles of a PI controlled system, we take advantage of
adding with derivative system.

Observation Tables:
Proportional Controller:
Proportional Controller Gain, Kp=5
Delay Time, td 5.6 ms
Rise time tr 26.40 ms
Peak time, tp -
Settling time, ts 26.40
Peak Overshot, Mp (%) -
Steady Stare Error ess 0.44V
Proportional-Integral Controller:
Controllers Gain, Kp=5, and KI=3
Delay Time, td 6 ms
Rise time tr 47.6ms
Peak time, tp -
Settling time, ts 47.6 ms
Steady Stare Error ess 0
Proportional-Differential Controller:
Controllers Gain, Kp=5, and KD=4
Rise time tr 28 ms
Peak time, tp -
Settling time, ts 28 m
Steady Stare Error ess 0.44V
Proportional-Differential Integral Controller:
Controllers Gain, Kp=5, KD=4, and KI=3
Rise time tr 35.60 ms
Peak time, tp -
Settling time, ts 35.60 ms
Steady Stare Error ess 0

Tuning Rules for PID Controllers:
If a mathematical model of a plant can be derived, then it is possible to apply
various design techniques for determining parameters of the controller that will meet
the transient and steady state specifications of the closed loop system. However, if the
plant is so complicated that its mathematical model cannot be easily obtained, then an
analytical approach to the design of a PID controller is not possible. Then we must
resist to experimental approaches to the running of PID controllers.
The process of selecting the controller parameters to meet given performance
specifications is known as controller tuning.
Procedure:
1. Switch ON the main supply to unit, observe input square wave amplitude in CRO.
Set the peak-to-peak voltage of square wave input is equal to 3V by varying the
potentiometer.
2. Give the connections as given in the diagram.
3. Connect the CRO at the output terminals.
4. Keep the controllers switches in ON position based on the type of controller.
5. Vary the gain knobs of controllers and take readings of delay time, raise time,
peak time, raise time, peak overshoot, settling time, and steady state error.
6. After taking the readings from CRO, turn OFF the supply.
Precautions:
• Handle the instruments gently
Result:
From the observation tables, we can conclude that,
• Proportional controller has steady state error, due to it increases the gain of closed
loop system (Note: PD controller also has steady state error).
• For PI controller, steady stare error is zero, because PI controller increases the
type number by one (Note: for PID controller also, steady state error is zero)
• For PD controller, transient performance is improved, i.e., rise time, delay time
reduced.
• For PID controller, both steady state and transient performance of the system are
improved.

Front Panel Connections:
Figure 3.2: Front Panel Connections of PID Trainer Kit

Design of Lag and Lead Compensation Magnitude and Phase Plot
4. Design of Lag and Lead
Compensation Magnitude and
Phase Plot

Circuit Diagram
Figure 4.1: Lag Compensation Network
Figure 4.2: Lead Compensation Network

DESIGN OF LAG AND LEAD COMPENSATION
MAGNITUDE AND PHASE PLOT
Aim:
To plot the response of land and lead compensators – magnitude and phase plot.
Apparatus:
1. Lag – Lead Compensation Kit
2. Cathode Ray Oscilloscope
3. Patch Cords
Theory:
All the control systems are designed to achieve specific objectives. A good
control system has less error, good accuracy, good speed of response, good relative
stability, good damping which will not cause under overshoots etc. For satisfactory
performance of the system, gain is adjusted first. In practice, adjustment of gain alone
cannot provide satisfactory results. This is because when gain is increased, steady state
behavior of the system improves but results into poor transient response, in some cases
may even instability. In such cases it is necessary to redesign the entire system.
Practically the design specifications are provided in terms of precise numerical values
according to which the system is designed.
The set of such specifications include peak overshoot, peak time, Wn , kp , kv ,
ka , G.M and P.M. etc. In practice, if a system is to be redesigned so as to meet the
required specifications, it is necessary to alter the system by adding an external device
to it. Such a redesign or alteration of system using an additional suitable device is called
compensation of a control system. While an external device, which is used to alter the
behavior of the system so, as to achieve given specification is called compensator.
Compensating Networks:
The compensator is a physical device. It may be an electrical network, mechanical
unit, pneumatic, hydraulic or combination of various types of devices.
The commonly used electrical compensating networks are
1. Lead network or lead compensator
2. Lag network or lag compensator

Observations:
Table 4-1: Observation Table for Lag Compensator
Frequency (Hz)
Phase Angle
(degree)
Output, Vo
(Volts)
Gain= o
i
V
20log
V
84 2 2.79 -0.630
101 3 2.77 -0.692
133 4 2.76 -0.724
164 5 2.76 -0.724
200 6 2.75 -0.755
261 8 2.73 -0.819
327 10 2.71 -0.880
554 16 2.62 -1.176
812 23 2.49 -1.618
1041 28 2.34 -2.150
1655 29 1.96 -3.690
2057 44 1.74 -4.730
Table 4-2: Observation Table for Lead Compensator
Frequency (Hz)
Phase Angle
(degree)
Output, Vo
(Volts)
Gain= o
i
V
20log
V
84 72 0.14 -26.61
100 73 0.17 -24.93
128 74 0.19 -23.96
139 76 0.23 -22.30
196 77 0.33 -19.17
362 75 0.60 -13.98
474 72 0.78 -11.70
859 62 1.31 -7.196
919 61 1.38 -6.74
1071 57 1.54 -5.79
2053 41 2.22 -2.61

Procedure:
Lead Compensator
1. Connections are made as per the Circuit diagram shown in Fig 1 so as to form a
phase lead network by selecting R1 = 10kΩ and C = 0.22µF
2. Switch ON the supply.
3. Using the CRO in X-Y mode, give input of the network to X – input of CRO and
output of the network to Y – input of CRO.
4. Set the sine wave amplitude to 3V.
5. Now vary the frequency from 25 Hz to 2000 Hz in steps and note down the
readings in the Table given below:
6. Calculate the gain and phase difference from the readings.
7. Now calculate the theoretical values of gain and phase difference.
8. Draw the Bode plot for both the theoretical & Practical Values.
Lag Compensator
1. Connections are made as per the Circuit diagram shown in Fig 2 so as to form a
phase lag network by selecting R1 = 10kΩ, R2 = 10kΩ and C = 0.22 µF.
2. Switch ON the supply.
3. Using the CRO in X – Y mode, give input of the network to X – input of CRO
and output of the network to Y – input of CRO.
4. Set the sine wave amplitude to 3V.
5. Now vary the frequency from 25 Hz to 2000 Hz in steps and note down the
readings in the Table given below:
6. Calculate the gain and phase difference from the readings.
7. Now calculate the theoretical values of gain and phase difference.
8. Draw the Bode plot for both the theoretical & Practical Values.
Result:
Frequency response of lag and lead compensators is studied, and magnitude and phase
angle pots are drawn. From the output,
• In lag compensator as the frequency increases, output voltage decreases and
goes out of phase.
• In lead compensator, at higher frequency output voltage is high and output
come to in-phase with input.

Lag Compensator Network
Figure 4.3: Front Panel Connections of Lag Compensator

Lead Compensator Network
Figure 4.4:Front Panel Connections of Lag Compensator

Transfer Function of a DC Motor
5. Transfer Function of a DC
Motor

Circuit Diagram
Figure 5.1: Circuit Diagram for Retardation Test to Determine the Constants of DC Motor Transfer Function

TRANSFER FUNCTION OF A DC MOTOR
Aim:
To determine the transfer function of the given dc motor after determining various
constants.
Name Plate Details of DC Motor:
Voltage 220V Speed 1500 RPM
Arm. Current 12A Power 3 HP
Field Current 0.6A Winding Shunt
Apparatus:
S.
No.
Name of the item Type Range Quantity
1 Ammeter PMMC (0-5) A 1
2 Voltmeter PMMC (0-300) V 1
3 Rheostat WW
300Ω/1.7A
300Ω/2A
50Ω/5A
1
1
1
4 Tachometer Digital (0-9999) rpm 1
5 Switch DPDT TCC 1
6 Stopwatch Digital - 1
7 Connecting Wires - - As per required
Theory:
Transfer function of a DC motor is designed as follows:
( )
( ) ( ) ( )
( ) (N in rpm)
=
2
( )
a b a a a
b
b b
di t
V t e t L i t R
dt
P
e t ZN
A
P d
Z
A dt
d
e t k
dt





= + +
 
=  
 
 
 
 
 
=
( ) ( )
b b
E t k S S

=   (1)

Observations:
To find out Ra:
Rated Voltage
a
V
Rated Current
a
I
Armature Resistance
a
a
a
V
R
I
=
220V 12A 2.7Ω
To find one-time t1:
Change in Speed (rpm) Time Taken (t1)
1500 31.89
To find t2 (when load is connected to armature):
Voltage (V) Current (I) Fall in Speed (rpm) Time Taken (t2)
206 0.6 1500 14.14 sec
Calculations:
3 2
( )
( ) ( ) ( )
t
a a a a a t b
k
S
V S J S L J R B L S B R k k S

=
  +  +  +  +
Case-I:
S
P =rate of change of kinetic energy
1
2
60
S
dN
P J N
dt

 
=   
 
 
(I)
Case-II:
S r
P P
+ =rate of change o kinetic energy
2
2
60
S r
dN
P P J N
dt

 
+ =   
 
 
(II)From (I) and (II),
1 1 2
1
2 2
S
S r
dN N
P dt t t
dN N
P P t
dt t
= = =
+

1
here, emf constant
2
b
P
k Z
A


 
=   =
 
 
( )
( ) ( ) ( )
a b a a a
di t
V t e t L i t R
dt
= + +
By applying Laplace transform,
( ) ( ) ( ) ( )
a b a a a a
V S E S SL I S I S R
= + +
 
( ) ( )
( ) a b
a
a a
V S E S
I S
SL R
−
=
+
(2)
1
Torque T(t)= ( )
2
= k ( )
1
here k Torque constant
2
a
t a
t
P
ZI t
A
I t
P
Z
A




 
 
 

 
= =
 
 
( ) ( )
t a
T S k I S
=  (3)
and also.
2
2
T(t)=J
d d
B
dt dt
 
+
( ) ( ) ( )
b b
S S k S E S

  =
2
( ) ( ) ( )
T S S J S S B S
 
=   +  
2
( )
( )
T S
S
S J S B
 =
 + 
(4)
From equations, 1-4, we can build DC motor block diagram as:
Figure 5.2: Block Diagram of Armature Controlled DC Shunt Motor
Transfer function of DC shunt motor is:
3 2
( )
( ) ( ) ( )
t
a a a a a t b
k
S

=
  +  +  +  +

1 2
( )
S S r
P t P P t
 = + 
2
1 2
206 0.6 14.14
98.46
31.89 14.14
r
S
P t
P W
t t
  
= = =
− −
98.46
S
P W
=
from (I)
2
2 1500 0
98.46 1500
60 31.89
0.12
J
J kg m
 −
   
=   
   
   

 =
220
| | | |
12
= 18.33
a
a a a
a
V
Z R jX
I
= + = =

2 2
2 2
18.33
18.33 2.7 18.13
2 18.13
18.13
2 50
= 0.057H
a a
a
a
a
R X
X
fL
L


 + =
= − = 
 =
=

Calculation of Kb,
a a
b
V I R
k

− 
=
220 12 2.7
2 1500
60
b
k

− 
=

1.194
b t
k k
= =
Transfer function:
3 2
( )
( ) ( ) ( )
t
a a a a a t b
k
S

=
  +  +  +  +
3 2 2
( )
( ) 0.5 0.057 (2.7 0.12 0.026 0.057) (0.026 2.7 1.194 )
t
a
k
S
V S S S S

=
  +  +  +  +
3 2
( ) 1.194
( ) 0.0342 0.325 1.495
a
S
V S S S S

=
 +  + 

Procedure:
1. Connections were made as per the circuit diagram.
2. Keep the motor field rheostat in the appropriate position. adjust the speed to the rated
value.
3. Give the supply by connecting DPST.
4. Vary the field rheostat value until the motor run above 10% rated speed.
5. Now open the DPST and DPDT switched simultaneously and start the stopwatch.
Note down time taken t1, here entire energy utilized to meet mechanical
losses(Pm) as the field is open.
6. Now repeat the steps 2, 3 and 4 again, now remove DPDT only, so losses in
kinetic energy meets both (Pi+Pm)
7. Now repeat the steps 2, 3,and 4 again. Now close DPDT switch 2-2’, note down
the Values of voltmeter, ammeter. Now armature acts as source as a result.
8. The decreasing in kinetic energy is to meet the (Pi+Pm+Pr) is due to flow of
current through 300Ω rheostat.
Result:
Transfer function of DC shunt motor is: 3 2
( ) 1.194
( ) 0.0342 0.325 1.495
a
S
V S S S S

=
 +  + 

Controllability and Observability Test using MATLAB.
6. Controllability and
Observability Test using
MATLAB.

MATLAB Code:
% CONTROLLABILITY AND OBSERVABILITY TEST USING MATLAB %
% 2
% SystemTransfer Function= --------------------------
% s^3 + 6 s^2 + 11 s + 6
Numerator_Coefficients=[2];
Denominator_Coefficients=[1 6 11 6];
Tansfer_Function_of_System=tf(Numerator_Coefficients,Denominator_Coefficients)
;
[A, B, C,D]=tf2ss(Numerator_Coefficients,Denominator_Coefficients)
%Controllability Test%
Controllability_Matrix=ctrb(A,B)
Rank_Cntrl=rank(Controllability_Matrix)
if Rank_Cntrl==length(A)
disp('Given System is Controllable')
else
disp('Given System is not Controllable')
end
%Observability Test%
Observability_Matrix=obsv(A,C)
Rank_Obsrv=rank(Observability_Matrix)
if Rank_Cntrl==length(A)
disp('Given System is Observable')
else
disp('Given System is not Observable')
end

CONTROLLABILITY AND OBSERVABILITY TEST
USING MATLAB
Aim:
To test the controllability and observability for the transfer function
3 2
( ) 2
( ) 6 11 6
y s
u s s s s
=
+ + +
Software: MATLAB
Theory:
Transfer function and graphical approaches such as bode, root locus and Nyquist plots
are called as conventional approaches. Basically, in all these methods we model the
given system in the form of the transfer function. The transfer function has the
following limitations
1. The initial conditions in the system are considered as zero.
2. Applicable only to the single input single output systems that are linear time-
invariant.
3. The analysis of the system output for specified input is obtained. No information
regarding the initial state of the system.
4. It gives an analysis of the system for some specific types of input like step, ramp
etc.
The drawback in the conventional approaches for modelling is overcome by another
method of modelling known as state-variable approach. In control systems, state space
representation is a mathematical model of a physical system as a set of input and output
and state variables related by first-order differential equations. The state space
representation also is known as ‘Time Domain Approach’ provides a convenient and
compact way to model and analyses system with multiple inputs and outputs.

Calculations:
Given the system transfer function
3 2
( ) 2
( ) 6 11 6
y s
u s s s s
=
+ + +
( ) ( ) ( ) ( )
3 2
6 11 6 2 ( )
s y s s y s sy s y s u s
+ + + =
 (1)
This is an nth
order system; n state variables are required. Let the first state variable be
equal to the x1 equal to the output y. Let second variable x2 be equal to the derivative
of the output variable and so on. This is,
1
2
3
x y
x y
x y
=
=
=
Equation-1 can be written in differential equation form
6 11 6 2 ( )
6 11 6 2 ( )
y y y y u s
y y y y u s
+ + + =
= − − − +
.
1
1
.
2 2
.
3
3
0 1 1 0
0 0 1 0
6 11 6 2
x
x
x x U
x
x
 
       
       
= +
       
       
− − −
     
 
 
   
1
2
3
10 0
x
y x
x
 
 
=  
 
 
Controllability:
Controllability matrix QC= [B AB A2
B]
0 1 1 0
0 0 1 0
6 11 6 2
A B
   
   
= =
   
   
− − −
   
0 0 2
0 2 12 , 0,
2 12 50
C C
Q Q
 
 
= − 
 
 
− −
 

Controllability and observability represent two major concepts of modern control
system theory. These originally theoretical concepts, introduced by R. Kalman in 1960,
are particularly important for practical implementations. They can be roughly defined
as follows.
Controllability: In order to be able to do whatever we want with the given dynamic
system under control input, the system must be controllable.
Observability: In order to see what is going on inside the system under observation,
the system must be observable.
Hence controllability matrix is the full rank of matrix ‘3’, hence the system is
controllable.
Procedure:
1. Open MATLAB by anyone of the following ways
• Select MATLAB icon.
• Call MATLAB from MATLAB command prompt.
2. Create new script file and enter the program
3. Execute the program by clicking on Run button
4. Observe the results in the command window.

Observability:
Observability matrix ( )
2
......... .....
T T T T T
O
Q C A C A C
 
 
=
 
1 0 0
0 1 0 1, 0
0 0 1
O O
Q Q
 
 
= = 
 
 
 
, hence given system is observable.

Result:
State space model for classic transfer function is verified both theoretically and
with MATLAB. Then given system is said to be controllable and observable.

Bode Plot, Root Locus, Nyquist Plots for the Transfer Functions of Systems up to 5th
Order using MATLAB.
7. Bode Plot, Root Locus,
Nyquist Plots for the Transfer
Functions of Systems up to
5th Order using MATLAB.

Order using MATLAB.

Order using MATLAB.
I. BODE PLOT
FOR THE TRANSFER FUNCTIONS OF SYSTEMS

Order using MATLAB.
Problem:
• Sketch bode plot for the following transfer function and determine the system gain
K for the cross over frequency to be 5 rad/sec
2
( )
(1 0.2 )(1 0.02 )
Ks
G s
s s
=
+ +
Solution:
The sinusoidal transfer function of G(jꞷ) is obtained by replacing s by jꞷ in the given
transfer function.
2
( )
( )
(1 0.2 )(1 0.02 )
K j
G j
j j


 
=
+ +
Let K=1,
2
( )
( )
(1 0.2 )(1 0.02 )
j
G j
j j


 
 =
+ +
Magnitude Plot:
The corner frequencies are 1
1
5 / sec
0.2
c rad
 = = and 2
1
50 / sec
0.02
c rad
 = =
The various terms of G(jꞷ) are listed in Table in the increasing order their corner
frequency. Also, the table shows the slope contributed by each term and the change in
the slope at the corner frequency.
Term
Corner Frequency
rad/sec
Slope
db/dec
Change in the
slope
db/dec
2
( )
j - +40
1
1 0.2 j
+ 1
1
5
0.2
c
 = = -20 40-20=20
1
1 0.02 j
+ 2
1
50
0.02
c
 = = 20 20-20=0
Choose a low frequency l
 such that 1
l c
 
 and choose a high frequency h
 such
that 2
h c
 
 .
Let l
 = rad/sec and h
 =100 rad/sec. Let, | ( ) |
A G j
= in db

Order using MATLAB.
I. BODE PLOT
Aim:
To plot the bode plot for given transfer function.
Software: MATLAB
Theory:
Bode diagrams are graphical representations of the frequency responses and are
used in solving design problems. Magnitude and phase functions are shown on separate
graphs using logarithmic frequency scale along the x-axis. Logarithm of the frequency to
base 10 is used for the x-axis of a graph. Zero frequency will correspond to negative infinity
on the logarithmic scale and will not show on the plots. The x-axis is graduated in log10 ω
and so every decade of frequency (e.g., … 0.001, 0.01, 0.1, 1, 10, 100, …) is equally spaced
on the x-axis.
The gain magnitude, represented by decibels defined as 20 log10 [A(ω)], is plotted
on the y-axis of magnitude plot. Since A(ω) dB can be both positive and negative, the y-
axis has both positive and negative values. Zero dB corresponds to a magnitude function
of unity. The y-axis for the phase function uses a linear scale in radians or degrees. Semi-
log graph paper makes it convenient to sketch Bode plots.
Bode plots are easily sketched by making asymptotic approximations first. The
frequency response function H(jω) is a rational function, and the numerator and
denominator are factorized into first-order terms and second-order terms with complex
roots.

Order using MATLAB.
Let us calculate A at 1 2
, , , and
l c c h
   
At 2 2 2
, 20log | ( ) | 20log 20log(0.5) 12
l A j db
   
= = = = = −
2 2 2
1, 20log | ( ) | 20log 20log(5) 28
c A j db
   
= = = = =
( )
1
2
2 1 2
1
50
, Slope from to log =20 log 28 48
5
c
c
c c c at
c
A A db
 

   
 =
 
= =  +  + =
 
 
( )
2
2
2
50
, Slope from to log =0 log 48 48
5
c
h
h c h at
c
A A db
 

   
 =
 
= =  +  + =
 
 
Let the points a, b, c and d be the points corresponding to frequencies
1 2
, , , and
l c c h
    respectively on the magnitude plot.
Phase Plot:
The phase angle of G(jꞷ) as the function of ꞷ is given by ,
1 1
( ) 180 tan (0.2 ) tan (0.02 )
G j
   
− −
=  = − −
ꞷ
rad/sec
( )
G j
 
= 
deg
0.5 174
1 168
130
10 106
50 50
100 30
Calculation of K:
Given that gain crossover frequency id 5 rad/sec. At ꞷ=5 rad/sec the gain is 28
db. If gain cross over frequency is 5 rad/sec, then at that frequency the db gain should
be zero. Hence to every point of magnitude plot a db gain of -28db should be added.
The addition of -28db shifts the plot downwards. The corrected magnitude plot is
obtained by shifting the plot with K=1 by -28db downwards. The magnitude correction
is independent of frequency. Hence the magnitude of -28db is contributed by the term
K. The value of K is calculated by equating 20 log K to -28db.
20log 28
28
log
20
K db
K
 = −
−
=
28
20
K=10 0.0398
−
 =

Order using MATLAB.
Procedure:
Program
% PLOTING THE BODE PLOT %
Numerator_Coefficients=[1 0 0];
Denominator_Coefficients=[0.004 0.22 1];
Tansfer_Function_of_System=tf(Numerator_Coefficients,Denominator_Coefficients);
bode(Tansfer_Function_of_System)
Gain_Margin=margin(Tansfer_Function_of_System)
Result:
Bode plot for given system transfer function
2
( )
(1 0.2 )(1 0.02 )
Ks
G s
s s
=
+ +
, and
system gain K for the gain cross-over frequency 5 rad/ sec is 0.0398.

Order using MATLAB.
Figure 7.1: Bode Plot of Given System

Order using MATLAB.
II. ROOT LOCUS

Order using MATLAB.
Problem:
A unity feedback control system has an open loop transfer function,
2
( )
( 4 13)
K
G s
s s s
=
+ +
Sketch the root locus and find the limiting value of K for
stability of the system.
Solution:
Step-01: To Locate the Poles and Zeros
The poles of open loop transfer function are roots of 2
( 4 13) 0
s s s
+ + =
The roots of the quadratic are,
2
4 4 4 13
2 3
2
s j
−  − 
= = − 
⸫ the poles are lying at s=0, 2 3
j
− + and 2 3
j
− −
Let us denote the poles as P1, P2 and P3.
Here P1=0. P2= 2 3
j
− + and P3= 2 3
j
− −
Step-02: To Find the Root Locus on the Real Axis
There is only one pole on the real axis at the origin. Hence if we choose any test
point on the negative real axis then to the right of that point the number of real poles
and zeros is one, which is odd number. Hence negative real axis will be part of root
locus.
Step-03: To Find the Angles of the Asymptotes and Centroid
Since there are three poles, the number of the root locus branches are three. There
is no finite zero . Hence all the three root locus branches end at zeros at infinity. The
number of the asymptotes required are three.
Angles of the asymptotes=
180(2 1)
q
n m
 +
−
; q=0,1,2……..n
Here n=3 and m=0. ⸫q=0,1,2,3
When q=0, Angles=
180
60
3

= 
When q=1, Angles=
180 3
180
3
 
= 
When q=2, Angles=
180 5
300 60
3
 
=  =

Order using MATLAB.
II. ROOT LOCUS
Aim:
To plot the root locus for given transfer function.
Software : MATLAB
Theory:
The root locus of a feedback system is the graphical representation in the
complexs-plane of the possible locations of its closed-loop poles for varying values of
a certain system parameter. The points that are part of the root locus satisfy the angle
condition. The value of the parameter for a certain point of the root locus can be
obtained using the magnitude condition. Suppose there is a feedback system with input
signal X(s) and output signal Y(s). The forward path transfer function G(s) is the
feedback path transfer function is H(s).
Thus, the closed-loop poles of the closed-loop transfer function are the roots of
the characteristic equation 1+G(s) H(s) =0. The roots of this equation may be found
wherever G(s)*H(s) = -1. The factoring of K and the use of simple monomials means
the evaluation of the rational polynomial can be done with vector techniques that add
or subtract angles and multiply or divide magnitudes. The vector formulation arises
from the fact that each monomial term (S-a) in the factored G(s)H(s) represents the
vector from (a) to (s) in the s-plane. The polynomial can be evaluated by considering
the magnitudes and angles of each of these vectors.
According to vector mathematics, the angle of the result of the rational
polynomial is the sum of zeros need be considered. This is known as the angle
condition. Similarly, the magnitude of the result of the rational polynomial is the
product of all the magnitudes in the numerator divided by the product of all the
magnitudes in the denominator. It turns out that the calculation of the magnitude is not
needed to determine if a point in the s- plane is part of the root locus because K varies
and can take an arbitrary real value. For each point of the root locus a value of K can
be calculated. This is known as the magnitude condition. The root locus only gives the
location of closed loop poles as the gain K is varied. The value of K does not affect the
location of the zeros. The open-loop zeros are the same as the closed- loop zeros.

Order using MATLAB.
When q=3, Angles=
180 7
420 60
3
 
=  =
Centroid=
Sum of poles - sum of zeros 0 2 3 2 3 0
1.33
3
j j
n m
− + − − −
= = −
−
Step-04: To find the Breakaway and Break-in Points
Closed loop Transfer Function=
2
2
2
( ) ( ) ( 4 13)
( ) 1 ( ) ( 4 13)
1
( 4 13)
K
C s G s K
s s s
K
R s G s s s s K
s s s
+ +
= = =
+ + + +
+
+ +
Characteristics equation 2
( 4 13)
s s s K
+ + + =0
⸫ 3 2
4 13
s s s K
+ + + =0 → 3 2
4 13
K s s s
= − − −
On differentiating the equation of K with respect to s, we get,
2
(3 8 13)
dK
s s
ds
= − + +
Put 0
dK
ds
= ; 2
(3 8 13) 0
s s
+ + =
2
8 8 4 13 3
1.33 1.6
2 3
s j
−  −  
= = − 

Check for K; When s= 1.33 1.6
j
− + , the value of the K is,
( ) ( ) ( )
3 2
3 2
( 4 13 ) 1.33 1.6 4 1.33 1.6 13 1.33 1.6
positive and real
K s s s j j j
 
= − + + = − − + + − + + − +
 

Since the values of K for 1.33 1.6
j
−  , are not real and positive, these points are not an
actual breakaway or break in points. The root locus has neither breakaway nor break in
points.
Step-05: To Find the Angle of Departure
Angle of the departure at the pole P2 = ( )
180 180 tan (3/ 2) 90 33.7
I O
−
 
− − + = −
 
Angle of the departure at the complex pole P3 is negative of angle departure of pole
P2.
Angle of the departure at the complex pole P3 = 33.7O
+

Order using MATLAB.
Procedure:
• Select MATLAB Icon.
• Call MATLAB from MATLAB Command Prompt.
3. Execute the program by clicking on Run Button
Program
% PLOTING THE ROOT LOCUS %
;
rlocus(Tansfer_Function_of_System)

Order using MATLAB.
Step-06: To Find the Crossing Point on Imaginary Axis
The characteristic equation is given by
3 2
4 13 0
s s s K
+ + + =
Put s=jꞷ,
3 2 3 2
( ) 4( ) 13( ) 0 - 4 13 0
j j j K j j K
     
+ + + =  − + + =
On equating imaginary part to zero, we
get,
3
3
2
13 0
13
13 13 3.6
 
 
 
− + =
− = −
=  =  = 
On equating real part to zero, we get,
2
2
4 0
4
4 13 52
K
K


− + =
=
=  =
The crossing point of root locus is 3.6
 . The value of K at this crossing point is
K=52.

Order using MATLAB.
Result:
The gain margin for the system of transfer function 2
( )
( 4 13)
K
G s
s s s
=
+ +
is
K=52 (this is the limiting value of K for the stability of the system).

Order using MATLAB.
Figure 7.2: Root Locus Plot of Given System

Order using MATLAB.
III. NYQUIST PLOT

Order using MATLAB.
Problem:
Draw the Nyquist plot for the system whose open loop transfer function is,
( ) ( )
( 2)( 10)
K
G s H s
s s s
=
+ +
. Determine the range of K for which closed loop system is
stable
Solution:
Given that,
0.05
( ) ( )
( 2)( 10) (1 0.5 )(1 0.1 )
2 1 10 1
2 10
K K K
G s H s
s s
s s s s s s
s
= = =
+ + + +
   
 +  +
   
   
The open loop transfer function has a pole at origin. Hence choose the Nyquist
contour on s-plane enclosing entire half plane except the origin.
The Nyquist contour has four sections C1, C2, C3 and C4. The mapping of each
section is performed separately, and the overall Nyquist plot is obtained by combing
the individual sections.
Mapping of Sections C1:
In the section C1, varies from 0 to +∞. The mapping of section C1 is given by the
locus of G(jꞷ)H(jꞷ) as ꞷ is varied from 0 to ∞. This locus is the polar plot of
G(jꞷ)H(jꞷ).
0.05
( ) ( )
(1 0.5 )(1 0.1 )
K
G s H s
s s s
=
+ +
Let s=jꞷ,
2 2 2
0.05 0.05 0.05
( ) ( )
(1 0.5 )(1 0.1 ) (1 0.6 0.05 ) 0.6 (1 0.05 )
K K K
G s H s
j j j j j j
        
 = = =
+ + + − − + −
When the locus of G(jꞷ)H(jꞷ) crosses real axis the imaginary term will be zero and
the corresponding frequency is the phase cross over frequency, ꞷpc.
2 2 1
At = , (1 0.05 ) 0 1 0.05 0 4.472 / sec
0.05
pc pc pc pc pc rad
     
 − =  − =  = =
2 2
0.05 0.05
At = 4.472 / sec, G(j )H(j )= 0.00417
0.6 0.6 4.472
pc
K K
rad K
   

= = = −
− − 

Order using MATLAB.
III. NYQUIST PLOT
Aim:
To draw and find out the gain margin for given system by using Nyquist plot.
Software: MATLAB
Theory:
In control theory and stability theory, the Nyquist stability criterion or Strecker–
Nyquist stability criterion, independently discovered by the German electrical engineer
Felix Strecker [de] at Siemens in 1930and the Swedish-American electrical engineer
Harry Nyquist at Bell Telephone Laboratories in 1932, is a graphical technique for
determining the stability of a dynamical system. Because it only looks at the Nyquist
plot of the open loop systems, it can be applied without explicitly computing the poles
and zeros of either the closed-loop or open-loop system (although the number of each
type of right-half-plane singularities must be known). As a result, it can be applied to
systems defined by non-rational functions, such as systems with delays. In contrast to
Bode plots, it can handle transfer functions with right half-plane singularities. In
addition, there is a natural generalization to more complex systems with multiple inputs
and multiple outputs, such as control systems for airplanes.
The Nyquist criterion is widely used in electronics and control system
engineering, as well as other fields, for designing and analyzing systems with feedback.
While Nyquist is one of the most general stability tests, it is still restricted to linear
time-invariant (LTI) systems. Non-linear systems must use more complex stability
criteria, such as Lyapunov or the circle criterion. While Nyquist is a graphical
technique, it only provides a limited amount of intuition for why a system is stable or
unstable, or how to modify an unstable system to be stable. Techniques like Bode plots,
while less general, are sometimes a more useful design tool.

Order using MATLAB.
The open loop system is type-1 and third order system. Also, it is a minimum phase
system with all poles. Hence the polar plot of G(jꞷ)H(jꞷ) starts at -90⁰ axis at infinity,
crosses real axis at -0.00417K and ends at origin in second quadrant
The mapping of section C2 from s-plane to G(s)H(s)-plane is obtained by
letting Rej
R
s Lt 
→
= in G(s)H(s) and varying  from +  /2 to -  /2. Since Rej
s 
→
and R → , the G(s)H(s) can be approximated as shown below,
3
0.05 0.05
( ) ( )
(1 0.5 )(1 0.1 ) 0.5 0.1
K K K
G s H s
s s s s s s s
=  =
+ +  
Let, Rej
R
s Lt 
→
=
( )
3
3 3
Re
Re
( ) ( ) 0
Re
j
j
R
j
j
R
R
s Lt
s Lt
K K
G s H s e
s Lt




−
→
→
→
=
=
= = =
When,
3
2
, G(s)H(s)=0
2
j
e



−
= (1)
When,
3
2
, G(s)H(s)=0
2
j
e



+
= − (2)
From the equations (1) and (2) we can say section C2 is in s-plane is mapped
as circular arc of zero radius around origin in G(s)H(s)-plane with argument varying
from -3 /2 to +3 /2
In the section C3, ꞷ varies from - to 0. The mapping of section C3 is given
by the locus of G(jꞷ)H(jꞷ).
The inverse polar plot is given by the mirror image of polar plot with
respect to real axis.
The mapping of section C4 from s-plane to G(s)H(s)-plane obtained by
letting
0
Rej
R
s Lt 
→
= in G(s)H(s) and varying  from - /2 to + /2. Since Rej
s 
→
and 0
R → , the G(s)H(s) can be approximated as shown below,

Order using MATLAB.
Procedure:
Program
% PLOTING THE NYQUIST PLOT %
Tansfer_Function_of_System=tf(Numerator_Coefficients,Denominator_Coefficients);
nyquist(Tansfer_Function_of_System)

Order using MATLAB.
0.05 0.05 0.05
( ) ( )
(1 0.5 )(1 0.1 ) 1 1
K K K
G s H s
s s s s s
=  =
+ +  
Let,
0
Rej
R
s Lt 
→
=
( )
0
0
0
Re
Re
0.05 0.05
( ) ( )
Re
j
j
R
j
j
R
R
s Lt
s Lt
K K
G s H s e
s Lt




−
→
→
→
=
=
= = = 
When, 2
, G(s)H(s)=
2
j
e



−
= −  (3)
When, 2
, G(s)H(s)=
2
j
e



+
=  (4)
From the equations (3) and (4) we can say that section C4 in s-plane is mapped
as a circular arc of infinite radius with argument (phase) varying from + /2 to- /2
Complete Nyquist Plot:
Stability Analysis:
When, -0.00417K=-1, the contour passes through (-1+j0) point and corresponding
value of K is the limiting value of K for stability.
1
Limiting value of K= 240
0.00417
 =

Order using MATLAB.
Result:
The gain margin for the system of transfer function ( )
( 2)( 10)
K
G s
s s s
=
+ +
is
K=240 .

Order using MATLAB.
Figure 7.3: Nyquist Plot of Given System

Temperature Controller using PID
8. Temperature Controller using
PID

Schematic Diagram
Figure 8.1: Internal Schematic Diagram of Temperature Control with PID Controller Kit

TEMPERATURE CONTROL USING PID
CONTROLLER
Aim:
To Study the temperature control using PID controller
Apparatus:
• Temperature Control - PID Controller kit
• Oven
Theory:
A proportional–integral–derivative controller (PID controller or three-term
controller) is a control loop mechanism employing feedback that is widely used in industrial
control systems and a variety of other applications requiring continuously modulated
control. A PID controller continuously calculates an error value e(t) as the difference
between a desired setpoint (SP) and a measured process variable (PV) and applies a
correction based on proportional, integral, and derivative terms (denoted P, I, and D
respectively), hence the name.
In practical terms, it automatically applies an accurate and responsive correction to
a control function. An everyday example is the cruise control on a car, where ascending a
hill would lower speed if only constant engine power were applied. The controller's PID
algorithm restores the measured speed to the desired speed with minimal delay and
overshoot by increasing the power output of the engine in a controlled manner.
In a temperature controller system, the controller accepts a temperature sensor as
input, such as an RTD or thermocouple and compares the actual temperature with the
rewired control temperature (or) setpoint. The error signal is provided to control element.
Atypical PID temperature controller application could be continuously vary a regulator
which can alter a process temperature. This may be pulsed switching device for electrical
heaters or by opening and closing a gas valve. A heat only PID temperature controller uses
a reverse output action i.e. more power applied when the temperature is below the setpoint
and less power when above. PID control for injecting and extension applications often
employ additional cooling control output and usually rewire multiple controllers.

Observation Table: Set Temperature =50O
C
P Alone is ON
P gain is max.
P+I are ON
P, I gains are max
P+D are ON
P, D gains are max
P+I+D are ON
P, I & D gains are max
Time in Sec
Run
Temperature
Time in Sec
Run
Temperature
Time in Sec
Run
Temperature
Time in Sec
Run
Temperature
0 36 0 36 0 36 0 36
5 36.3 5 36.9 5 38.6 5 36.6
10 38.5 10 39.6 10 40.8 10 38.8
15 40.4 15 41.3 15 42.9 15 40.6
20 42.3 20 43.0 20 44.5 20 41.9
25 44.9 25 44.5 25 45.9 25 43.4
30 46.2 30 45.5 30 47.5 30 44.7
35 47.4 35 47.1 35 48.3 35 46.3
40 48.2 40 48.1 40 48.7 40 47.3
45 48.6 45 49.1 45 49.0 45 48.4
50 48.9 50 49.9 50 49.2 50 49.1
55 49.1 55 50.2 55 49.4 55 50.2
60 49.2 60 50.3 60 49.5 60 50.3
65 49.4 65 50.3 65 49.6 65 50.3
70 49.5 70 50.3 70 49.6 70 50.3
75 49.6 75 50.3 75 49.6 75 50.3
80 49.5 80 50.3 80 49.6 80 50.3
85 49.5 85 50.1 85 49.6 85 50.2
90 49.5 90 50.0 90 49.6 90 50.2
95 49.5 95 50.0 95 49.6 95 50.0
100 49.5 100 50.0 100 49.6 100 50.0
105 49.6 105 50.0 105 49.6 105 50.0
110 49.6 110 50.0 110 49.6 110 50.0
115 49.5 115 50.0 115 49.6 115 50.0

Procedure:
1. Connect the temperature sensor from oven to the socket provided at the PID
controller.
2. Switch ON the power supply to PID unit.
3. Set the required (5V) say 50 o
C using potentiometer. This is temperature is displays
in digital temperature indicator, by selecting the toggle switch ON set position near
temperature indicator.
4. Keep the P switch ‘ON’, I&D switches at OFF position. Keep gain at maximum
position process temperature to set for past running of the process.
5. Switch ON output AC to the TRIAC and oven circuit.
6. Switch ON the fan.
7. Record the temperature of the oven by selecting the toggle switch at run position
for ever0.5 minute till the system reach steady state.
8. Cool the system till 30 or 35 o
C by switching OFF AC ON switch provided.
9. Do the experiment for PI, PD and PID controllers individually.
10. Plot the graph of the time vs. temperature.
Precautions:
• Handle the instruments gently

Model Graphs:
Figure 8.2: Time Response of Temperature Controller

Result:
From the observation tables, we can conclude that,
• Proportional controller has steady state error, due to it increases the gain of closed
loop system (Note: PD controller also has steady state error).
• For PI controller, steady stare error is zero, because PI controller increases the
type number by one (Note: for PID controller also, steady state error is zero)
• For PD controller, transient performance is improved.
• For PID controller, both steady state and transient performance of the system are
improved.

Characteristics of Magnetic Amplifiers
9. Characteristics of Magnetic
Amplifiers

Circuit Diagram
(i). Series Magnetic Amplifier:
Figure 9.1: Series Magnetic Amplifier
(ii). Parallel Magnetic Amplifier:
Figure 9.2: Parallel Magnetic Amplifier

CHARACTERISTICS OF MAGNETIC AMPLIFIER
Aim:
To plot the load characteristics of 1. Series Connected Magnetic Amplifier
2. Parallel Connected Magnetic Amplifier
Apparatus:
1. Magnetic Amplifier Kit
2. Patch Cards
3. Load
4. External DC Supply
5. Ammeter (0-100) mA-MC
6. Ammeter (0-1) A-MC
Theory:
The magnetic amplifier (colloquially known as a "mag amp") is an electromagnetic
device for amplifying electrical signals. The magnetic amplifier was invented early in the
20th century and was used as an alternative to vacuum tube amplifiers where robustness
and high current capacity were required.
Visually a mag amp device may resemble a transformer, but the operating
principle is quite different from a transformer – essentially the mag amp is a saturable
reactor. It makes use of magnetic saturation of the core, a non-linear property of a
certain class of transformer cores. For controlled saturation characteristics, the
magnetic amplifier employs core materials that have been designed to have a specific
B-H curve shape that is highly rectangular, in contrast to the slowly tapering B-H curve
of softly saturating core materials that are often used in normal transformers.
The typical magnetic amplifier consists of two physically separate but similar
transformer magnetic cores, each of which has two windings: a control winding and an
AC winding. Another common design uses a single core shaped like the number "8"
with one control winding and two AC windings as shown in the photo above. A small
DC current from a low-impedance source is fed into the control winding. The AC
windings may be connected either in series or in parallel, the configurations resulting
in different types of mag amps. The amount of control current fed into the control
winding sets the point in the AC winding waveform at which either core will saturate.
In saturation, the AC winding on the saturated core will go from a high-impedance state
("off") into a very low-impedance state ("on") – that is, the control current controls the
point at which voltage the mag amp switches "on".

Observation Tables:
(i) Series Magnetic Amplifier:
S. No.
Control Current,
IC (mA)
Load Current,
IL (mA)
1 0.010 0.108
2 0.022 0.227
3 0.030 0.303
4 0.040 0.384
5 0.050 0.435
6 0.062 0.492
7 0.070 0.522
8 0.085 0.522
(ii) Parallel Magnetic Amplifier:
S. No.
Control Current,
IC (mA)
Load Current,
IL (mA)
1 0.010 0.289
2 0.022 0.459
3 0.03 0.568
4 0.04 0.625
5 0.05 0.652
6 0.06 0.669
7 0.07 0.677
8 0.085 0.682

Procedure:
1. Make the connections as per the circuit diagram.
2. Keep control current setting knob at its extreme left position, which ensures zero
control current at starting.
3. Switch ON the unit.
4. Now gradually increase control current by rotating control current setting knob
clockwise in steps and noted down control current and corresponding load
current.
5. Plot the graph of load current v/s control current.
Precautions:
1. Handle the pointers for both the rotors in a gentle manner.
2. Do not attempt to pull out the pointer.
3. Apply the voltage slowly to control winding as coil may get heated up and
down.
4. Take the readings carefully and accurately

Model Graphs:
Figure 9.3: Model Graphs for Input-Output Characteristics of Series and
Parallel Connected Magnetic Amplifier

Result:
Input-Output characteristics of magnetic amplifier are obtained for series and
parallel connection. Series connected amplifier has the linear input -output
characterises and parallel magnetic amplifier has the exponential input-output
characteristics.

Characteristics of AC Servo Motor
10. Characteristics of AC Servo
Motor

Internal Schematic Diagram
Figure 10.1: Internal Schematic Diagram of AC Servo Motor Testing Kit

SPEED-TORQUE CHARACTERISTICS OF
AC SERVO MOTOR
Aim:
To Plot Speed – Torque Characteristics of A.C. Servo Motor.
Apparatus:
1. AC Servo Motor Testing Kit
2. Multi-Meter
3. Patch Cards
Theory:
An A.C. servomotor is basically a two-phase induction motor except for certain
special design features. A two-phase induction motor consisting of two stator windings
oriented 90 degrees electrically apart in space and excited by A.C. voltages which differ
in time phase by 90⁰. Generally, voltages of equal magnitude and 90⁰ phase difference
are applied to the two-stator phases thus making their respective fields 90⁰ apart in both
time and space, at synchronous speed. As the field sweeps over the rotor, voltages are
induced in it producing current in the short-circuited rotor. The rotating magnetic field
interacts with these currents producing a torque on the rotor in the direction of field
rotation.
The shape of the characteristics depends upon ratio of the rotor reactance (X) to
the rotor resistance (R). In normal induction motor X/R ratio is generally kept high so
as to obtain the maximum torque close to the operating region, which is usually around
5% slip. A two-phase servomotor differs in two ways from normal induction motor.
1. The rotor of the servomotor is built with high resistances so that its X/R ratio is
small and the torque speed characteristics are as shown in the figure (2).Curve (3)
is nearly linear in contrast to highly non-linear characteristics with large X/R ratio
is used for servo applications, then because of the positive slope for part of the
characteristics, the system using such a motor becomes unstable. The motor
construction is usually squirrel cage or drag cup type. The diameter of the rotor is
kept small in order to reduce inertia and thus to obtain good accelerating
characteristics. Drag cup construction is used for very low inertia operations.

Observations:
Table 10-1: Speed-Back Emf Characteristics
S. No. R. P. M Eb (Volts)
1 230 0.95
2 406 1.57
3 603 2.33
4 830 3.24
5 1012 3.90
6 1104 4.25
7 1346 5.13
8 1640 6.24
Table 10-2: Speed-Toque Characteristics of AC Servo Motor at C
V =180
S.
No.
Ia
(mA)
N
(RPM)
Eb
(Volts)
P
(Watts)
T= Toque (gm-cm)
3
P×1.019×10 ×60
2πN
1 0.053 1564 3.87 0.205 1.2754
2 0.07 1502 3.38 0.2366 1.5328
3 0.079 1482 3.07 0.2425 1.5920
4 0.112 1380 2.08 0.2329 1.6422
5 0.141 1288 1.31 0.1847 1.3953
6 0.163 1230 0.65 0.1059 0.8377
Table 10-3: Speed-Toque Characteristics of AC Servo Motor at C
V =120
S.
No.
Ia
(mA)
N
(RPM)
Eb
(Volts)
P
(Watts)
T= Toque (gm-cm)
3
P×1.019×10 ×60
2πN
1 0.05 1335 3.15 0.1575 1.148
2 0.077 1254 2.25 0.1732 1.344
3 0.100 1180 1.52 0.152 1.253
4 0.121 1102 0.93 0.1125 0.993
5 0.148 1023 0.02 0.0296 0.281

2. In servo applications, the voltages applied to the two-stator windings are seldom
balanced. One of the phases known as the reference phase is excited by constant
voltage and the other phase known as the control phase with respect to the voltage
supplied to the reference windings and it has a variable magnitude and polarity.
The control winding voltage is supplied from a servo amplifier. For low power
applications, A.C. servomotors are preferred because they are light weight, rugged
and there are no brush constant to maintain. The reference winding of the motor is
excited from a fixed voltage of 100 volts. The control winding voltage is obtained
through an R-C combination. The voltage available to control winding is varied by
the control of resistance R. The capacitance is used to generate a phase shift. By
varying the magnitude of the control voltage, it is possible to vary the speed of the
A.C. servomotor. The secondary of the transformer T1 provided the reference
winding and control winding voltage. Torque
Procedure:
1. Study all the controls carefully on the front panel.
2. Keep the switch SW3 in upward position, switch SW2 should also be in OFF
position.
3. Ensure P1 and P2 are in fully anticlockwise position.
4. Now, switch on SW1 and also switch on SW2. You can observe that A.C. servo
motor will start rotating and the speed will be indicated by the meter M1 on the
front panel.
5. With SW3 in OFF condition, vary the speed of the A.C. servomotor by moving P1
in clockwise direction and note the E.M.F. generated by the D.C. machine (Now
working as D.C. generator or tacho). Enter the results in Table no. 1 (use a D.C.
voltmeter in the range 0 to 2 volts or so).
6. Now switch SW3 in OFF condition, switch ON SW2 and keep the pot P1 in
minimum position. You can observe that the A.C. servomotor starts moving with
speed being indicted by the RPM indicator and control winding voltage (which is
variable by P1). Note the speed of A.C. servo motor. Now switch on SW3 and start
loading A.C. servo motor by controlling pot P2 in a slow fashion. Note down
corresponding values on Ia and N. Enter these values in Table 2.

Model Graphs:
Figure 10.2:Speed vs. Back EMF Graph
Figure 10.3: Speed Torque Characteristics of AC Servo Motor

8. Now you may set control-winding voltage to a new value of 30 volts after
switching off SW3. Again, repeat the process as indicated in step No. 5 i.e. Table
2 for a new value of control winding voltage.
9. Plot, the speed torque characteristics for various values of control winding
voltages. Study their nature.
Precautions:
1. Remove any load before turning ON the AC servo motor
2. Never overload the motor
Result:
Speed-Torque characteristics of AC servo motor are obtained at control
voltage levels of 150V and 180V.

Characteristics of DC Servo Motor
11. Characteristics of DC Servo
Motor

Block Diagram of DC Servo Motor Speed-Torque Characteristics Kit
Figure 11.1: Block Diagram of DC Servo Motor: Speed-Torque Characteristics Setup

CHARACTERISTICS OF DC SERVO MOTOR
Aim:
To plot Speed-Torque characteristics of D.C. Servomotor.
Apparatus:
1. DC Servo Motor Testing Kit
2. Multi-meter
3. Patch Cords
Theory:
There are many types of D.C. motors used in industries. D.C. motors that are used
in control systems are called D.C. servomotors. In D.C. servomotors the rotor inertia have
been made very small, with the result that motors with very high torque to inertia ratio are
commercially available. Some D.C. motors with relatively small power rating are used in
instruments and computer related instruments. D.C. Series, shunt and permanent magnet
D.C. servo motors with medium and large power ratings are used in robot systems and CNC
machines. And normally they are available with built in tacho generator for speed sensing
in closed loop control applications. Most important among the characteristics of the D.C.
servomotor is the maximum acceleration obtainable. For a given available torque the rotor
moment of inertia must be maximum. Since the servomotor operates under continuously
varying conditions, acceleration of the rotor occur from time to time. The servo motor must
be able to absorb the mechanical energy as well as to generate it. The performance of
servomotor when used as a brake is satisfactory.
Torque Measurement:
In order to measure the torque produced by the D.C. servomotor we have an
arrangement to produce the variable load on the D.C. servomotor. The D.C. servomotor
is fitted to the front panel and its shaft is extended outside the front panel. A small
aluminium break drum is fitted to the shaft for loading. A small leather belt and two
spring balances are fitted to the belt on both the sides and fitted to an MS channel. A
wheel is fitted to the one side of the shaft or varying the speed.

Observations:
Rated Voltage=24V
Break drum radius, r=2.5 cm
(i) Observation table for 60% of rated voltage=14.4V
S.
No.
Rated
Voltage,
V (Volts)
Speed
(rpm)
Current, I
(A)
W1 W2 W= W1-W2
Torque,
W×r
1 14.4 2533 0.121 0 0 0 0
2 14.4 2411 0.196 10 30 20 50
3 14.4 2209 0.356 25 65 40 100
4 14.4 1922 0.593 30 95 75 187.5
5 14.4 1889 0.529 40 125 85 212.5
(ii) Observation table for 40% of rated voltage=9.6V
S.
No.
Rated
Voltage,
V (Volts)
Speed
(rpm)
Current, I
(A)
W1 W2 W= W1-W2
Torque
W×r
1 9.6 1672 0.114 0 0 0 0
2 9.6 1550 0.186 10 30 20 50
3 9.6 1421 0.257 20 50 30 75
4 9.6 1278 0.339 30 80 50 125
5 9.6 1125 0.434 40 110 70 175
6 9.6 1040 0.502 50 160 110 205
Model Graphs:
Figure 11.2: Speed-Torque Characteristics of DC Servo Motor

Procedure:
1. Remove the load in no load condition. Switch ON the module.
2. Note down the no load current and no-load speed.
3. Adjust the potentiometer for the rated voltage of 24V.
4. Adjust the load in steps to a maximum of 200 gm – cm and current of 0.8 Amps
(do not exceed 0.8 amps)
5. At each load note down the speed.
6. Calculate the corresponding torque and plot the torque speed characteristics.
7. Repeat the procedure for 60% and 40% of the rated voltage
• Rated voltage
• 60% rated voltage
• 40% rated voltage
Precautions:
3. Remove any load before turning ON the AC servo motor
4. Never overload the motor
Result:
Speed-Torque characteristics of DC servomotor at the applied voltage levels
of 14.4V and 9.6V are obtained.
• Speed droop at 60% of rated voltage from no load to full load is 644 rpm.
• Speed droop at 40% of rated voltage from no load to full load is 632 rpm

Block Diagram Representation of Field Controlled DC Servo Motor Using Simulink
12. Block Diagram
Representation of Field
Controlled DC Servo Motor
Using Simulink

Derivation of Transfer Function of Field Controlled DC Servo Motor:
A field controlled DC servo motor is shown in below
Figure 12.1: Field Controlled DC Servo Motor
In this system,
f
R =field winding resistance (Ω)
f
L =field winding inductance (H)
f
e =field control voltage (V)
f
i =field current
M
T =Torque developed by the motor
J=equivalent moment of inertia of motor and load referred to motor shaft (kg-m2
)
B=equivalent viscous friction coefficient of motor and load referred to motor shaft
N m
rad s
−
−
 =angular displacement of motor shaft (rad)
In the field-controlled motor, the armature current is fed from a constant current
source. Therefore,
1
M f f a T f
T K K i i K i

= = Where, K is a constant

BLOCK DIAGRAM REPRESENTATION OF FIELD
CONTROLLED DC SERVO MOTOR USING SIMULINK.
Aim:
To represent the transfer function of field-controlled DC servo motor in MATLAB
Simulink and obtain step response1
.
Software: MATLAB/Simulink
Theory:
There are many types of D.C. motors used in industries. D.C. motors that are used
in control systems are called D.C. servomotors. In D.C. servomotors the rotor inertia have
been made very small, with the result that motors with very high torque to inertia ratio are
commercially available. Some D.C. motors with relatively small power rating are used in
instruments and computer related instruments. D.C. Series, shunt and permanent magnet
D.C. servo motors with medium and large power ratings are used in robot systems and CNC
machines. And normally they are available with built in tacho generator for speed sensing
in closed loop control applications. Most important among the characteristics of the D.C.
servomotor is the maximum acceleration obtainable. For a given available torque the rotor
moment of inertia must be maximum. Since the servomotor operates under continuously
varying conditions, acceleration of the rotor occur from time to time. The servo motor must
be able to absorb the mechanical energy as well as to generate it. The performance of
servomotor when used as a brake is satisfactory.
1
Refer the Website for DC Servo Motor Constants:
https://ctms.engin.umich.edu/CTMS/index.php?example=MotorSpeed&section=SystemModeling

The equation for the field circuit is
f
f f f f
di
e L R i
dt
= + (1)
Torque equation is
2
2 M T f
d d
J B T K i
dt dt
 

+ = = (2)
Taking the Laplace transform of equations 1 & 2, we get
( )
f f f f
L s R i E
+ =
( )
2
( ) ( ) ( )
M T f
Js Bs s T s K I s
 
+ = =
From the above equations, the transfer function of the motor is obtained as
( )( )
( )( )
( )
( )
( )
( ) 1 1
T
f f f
m
f f me
K
s
E s s L s R Js B
K
s
E s s s s


 

=
+ +
=
+ +
Where T
m
f
K
K
R B

= motor gain constant.
f
f
f
L
R
 = time constant of field circuit
me
J
B
 = mechanical time constant
Block diagram of the field-controlled DC servo motor is shown in diagram below
Simulink Block Diagram:

Procedure:
• Select MATLAB Icon.
• Call MATLAB from MATLAB Command Prompt.
10. Open the Simulink and create the new Simulink model.
11. Browse for elements in Simulink library
12. Then add and connect them as shown in the diagram
13. Give the values of Rf, Lf, KT’, J and B values in transfer function blocks
14. Run the model and record the results obtained in the Scope.

Result Graphs:
Figure 12.2: Step Response of Field Controlled DC Servo Motor (Rotor Angle)

Result:
Transfer function of field-controlled DC servo motor is derived and modelled
in the MATLAB-Simulink. And step response of system is observed.

Add-On 01. Design Problem by Construction of Root Locus
13. Add-On 01. Design Problem by
Construction of Root Locus

Root Locus by Hand Calculations:
The system control scheme can be redrawn as
Figure 13.1: Reduced Block Diagram of Servo Motor Control Scheme
The forward path transfer function can be written as
𝐺(𝑆) = 𝐾𝐴 ×
(𝑆 + 10)(𝑆 + 50)
𝑆
×
0.066
2.32 𝑆(0.0022 𝑆 + 0.316)
𝐺(𝑆) =
𝐾𝐴(𝑆 + 10)(𝑆 + 50)
𝑆2( 𝑆 + 144)
𝐻𝑒𝑟𝑒, 𝐾 =
𝐾𝐴 × 0.066
2.32 × 0.0022
Open loop poles: 0, 0, -144, and open loop zeros: -10, -50.
Step-1: To Locate the Poles and Zeros in S-plane
Poles of open loop transfer function are roots of equation S2
(S+144) =0, therefore the
poles are lying at S=0, 0 and -144. The zeros are lying at S=-10 and -50. Let denote the
poles as P1, P2 & P3 and zeros as Z1 & Z2. Here P1=0, P2=0 & P3=-144 and Z1=-10 &
Z2=-50. The poles are marked by the X (cross) and zeros by O in the graph.
Step-2: To Find Out the Root Locus on the Real Axis
All the poles and zero are lies on the real axis. In order to determine the parts of
root locus on the real axis, take test point on the real axis. If the total number of the
poles and zeros those left to the test point on the real axis is an odd number, then the
test point is in root locus. By the observation, on the real axis, root locus exists between
S=-10 to S=-50 and S=-144 to infinity.

DESIGN PROBLEM
BY CONSTRUCTION OF ROOT LOCUS
Aim:
To construct the root locus to calculate the amplifier gain of the servo motor, of
Autonomous Guided Vehicle (AGV).
Description of Root Locus Problem:
An Autonomous Guided Vehicle (AGV) is used to carry the payload to the
various destinations by an onboard computer. The vehicle having four wheels powered
by the two DC servo motors of identical rating the two back wheels being free. Each
motor independently controlled by a PID controller. The block diagram of the control
scheme is given in the figure. Where in the motor is provided with an internal feedback
loop.
Figure 13.2: Block Diagram of the Control Scheme
Calculate the value of the amplifier gain through root locus method such that dominant
pole pair have damping co-efficient ξ=1. Here α=10 and β=50.
Software Tool to be used:
• MATLAB
Theory:
The root locus method/ technique provides a graphical method of plotting the
locus of the roots in the S plane as the given system passes is varies over the complete
range of values. The roots corresponding to the particular value of the system parameter
can then located on the locus or value of the parameter for the desired root location can
be determined from the root locus.

Step-3: To Find the Angles of Asymptote and Centroid
Since there are three poles, the number of roots locus branches is three. One branch
starts at S=-144 and terminates at infinity. And there are two finite zeros, so remaining
two root locus branches stars at origin and will end at finite zeros. Hence the number
of asymptotes is one and angle of asymptotes is -180o
.
Step-4: To Find Break-in Point
Characteristic equation of the closed loop system is,
K (S2
+ 60S +500) + (S3
+144S2
) =0;
Then, K=
−(S3+144S2 )
( S2+ 60S +500)
;
On the differentiating K with respect to S, We get
𝑑𝐾
𝑑𝑆
=
( S2
+ 60S + 500)(−2S2
− 288S) − (−(S3
+ 144S2 ))(2S + 60)
( S2 + 60S + 500)2
For
𝑑𝐾
𝑑𝑆
=1, the numerator should be zero
60S3
+9640S2
+144000=0
By solving the above equation, roots are S=0, -144, -16.667. Only S=-16.667 on the
region between S=-10 to S=-50, so break-in point is 16.6667.
Root locus starts at origin and end at the zeros (S=-10 &S=-50), which is drawn in the
graph.
Step-5: Check for Amplifier Gain (KA):
At break-in point the dominant pole have ξ=1;
Break-in point is a d=16.667 distance from the origin, at this point,
𝐾 =
𝑑2|144−𝑑)
|10−𝑑| |50−𝑑|
=160.1
Amplifier gain at break-in point (ξ=1) is KA=
160 ×2.32 ×0.0022
0.066
=12.37

The root locus is a powerful technique as it brings into focus the complete
dynamic response of the system. An important advantage of the root locus method is
that the roots of the characteristic equation of the system can be directly obtained.
This results in the complete and accurate results of the transient and steady state
response of the controlled variables.
Procedure:
Math Script:
% PLOTING THE ROOT LOCUS %
Numerator_Coefficients=[1 60 500];
rlocus(Tansfer_Function_of_System)

Calculations:
Break in point from the MATLAB root locus=-17.2
Gain value at this point (from the output) = 159
Amplifier gain KA=
158.844 ×2.32 ×0.0022
0.066
=12.296
Tabular Form
Calculated MATLAB
Break-in Point -16.667 -17.2
Gain at Break-in Point 160.1 159
Amplifier Gain 12.37 12.296

Results and Conclusions:
The root locus is plotted by hand calculation and by using MATLAB and then
results are compared. Amplifier gain (at ξ=1) of AGV servo motor is KA=12.296 (from
MATLAB) and KA =12.37 (by hand calculations).

Figure 13.3:Root Locus Plot

Add-On 02.
Design of Position Control System using Bode Plot (Frequency Response Analysis)
14. Add-On 02.
Design of Position Control
System using Bode Plot
(Frequency Response
Analysis)

Add-On 02.
Calculations
Step-1: Initial Assumptions and Calculations
Let us take the amplifier to gain K=3.6, and convert the given system open loop transfer
function to time constant form.
G(S)=
360
𝑆(𝑆+100) (𝑆+36)
=
0.1
𝑆 (1+
S
36
) (1+0.01S)
The sinusoidal transfer function G(jω) is obtained by replacing S by jω
G(jω) =
0.1
jω (1+
jω
36
) (1+j0.01ω)
Step-2: Bode Plot for Amplifier Gain K=3.6
Magnitude Plot:
The corner frequencies are ωc1=36 rad/sec and ωc2=100 rad/sec. The various terms of
G(jω) are listed in the table below in increasing order of their frequencies. Also, the
table shows the slope contributed by each term and the change in the slope at the corner
frequency.
Table 14-1: Asymptotic Approximation Table for Construction of the Bode Plot
Term
Corner Frequency
(rad/sec)
Slope
(db/dec)
Change in Slope
(db/dec)
0.1
(jω)
-20
1
(1 +
jω
36
)
ωc1=36 -20 -20-20=-40
1
(1 + j0.01ω)
ωc2=100 -20 -40-20=-40
Choose low frequency ωl such that ωl< ωc1 and choose high frequency ωh such that ωh>
ωc2.Let, ωl=0.1 rad/sec, ωh=1000 rad/sec and A= |G(jω) | in db.
Let us calculate A at ωl, ωc1, ωc2, ωc3 and ωh
At ω= ωl, A=20 log |
0.1
0.1
|=0
At ω= ωc1, A=20 log |
0.1
36
|=-51.1261

Add-On 02.
DESIGN OF POSITION CONTROL SYSTEM USING
BODE PLOT (FREQUENCY RESPONSE ANALYSIS)
Aim:
To construct the bode plot to calculate the amplifier gain of the given system
and verify the results with simulation results.
Description of Problem:
For the position control system shown in below figure, find the gain of the pre-amplifier
gain, K, to yield a 9.5 overshoot in the transient response for the step input.
Figure 14.1: Block Diagram of the Position Control System and its Control
Scheme
Software Tool to be used:
• MATLAB
Theory:
Bode plot is a frequency response plot of the sinusoidal transfer function of a system.
Abode plot consists of two graphs. One is a plot of the magnitude of a sinusoidal
transfer function versus log ω. The other is a plot of the phase angle of the sinusoidal
transfer function versus log ω.
Bode can draw for both open loops and closed loop system. Usually, the bode
plot is drawn for an open loop system. The standard representation of the logarithmic
magnitude of open loop transfer function G(jω) is 20 log | G(jω) |, where the base of
the logarithm is 10. The unit used in this representation of the magnitude decibel usually
abbreviated db. The curves are drawn on a semi-log graph, using the log scale for
frequency and a linear scale for either magnitude or phase angle.

Add-On 02.
At ω= ωc2, A= [slope from ωc1 to ωc2 ×log
ω𝑐2
ω𝑐1
] + Aωc1 =[-40× log (100/36)] -51.12
=-71.55
At ω= ωh, A= [slope from ωc2 to ωh ×log
ωℎ
ω𝑐2
] + Aωc2 =[-60× log (1000/100)] – 71.55
=-111.55
Let the points a, b, c and d corresponding to the frequencies ωl, ωc1, ωc2 and ωh
respectively on the magnitude plot. In a semi-log graph, sheet chooses a scale of 1 unit
=50db on the y-axis. The frequencies are marked in decades from the 0.1 to 1000
rad/sec on the logarithm scale in the x-axis. Fix the points a, b, c and d on the graph.
Join the points by a straight line.
Phase Plot:
The phase angle of the G(jω) as the function of ω is given by
ϕ = - 90 –tan-1
(ω/36)–tan-1
(ω/100)
The phase angle of the G(jω) is calculated for various values of ω and listed in the
below table.
Table 14-2:Phase Angle of G(jω) at Different Frequencies
ω
(rad/sec)
ϕ ω
(rad/sec)
ϕ ω
(rad/sec)
ϕ
0.1 -90.21 20 -130 500 -254.57
1 -92.16 30 -146.50 1000 -262.22
2.5 -95.40 50 -170.81
4 -98.6 80 -194.43
10 -111.23 100 -205.20
On the same semi-log graph sheet choose a scale of 1unit=100 degrees on the y-axis on
the right side of the semi-log graph sheet. Mark the calculated phase angle on the graph
sheet. Join the points by a smooth curve.
Step-3: Calculation of Amplifier Gain Value
Amplifier gain of the system has to be chosen in such a way that the peak overshoot
should be 9.5%. The damping ratio of the closed-loop system to get 9.5% overshoot
can be calculated by using equation

Add-On 02.
Procedure:
Math Script:
numg=[1];
deng=[1 136 3600 0];
G=tf(numg,deng)
bode(G)
pos=9.5
Z=(-log(pos/100))/(sqrt(pi^2+log(pos/100)^2));
Pm=atan(2*Z/(sqrt(-2*Z^2+sqrt(1+4*Z^4))))*(180/pi);
w=0.01:0.01:1000;
[M,P]=bode(G,w);
Ph=-180+Pm;
K=1;
for i=1:1:length(P);
Diff(i)=(P(i)-Ph);
if Diff(i)<=0;
m=M(i);
Frq=w(i);
K=(1/m)
break
end
end

Add-On 02.
ξ=
−ln(%𝑂𝑆/100)
√𝜋2+𝑙𝑛(%𝑂𝑆/100)2
=
−ln(9.5/100)
√𝜋2+𝑙𝑛(9.5/100)2
=0.6
Phase margin required for this damping ratio can be calculated by using the equation
γmreq=tan−1 2ξ
√−2ξ2+√1+4ξ4
=tan−1 2×0.6
√−2×0.62+√1+4×0.64
=59.2o
We know, phase margin γ=180+ϕgc, here the ϕgc is a phase of the open loop transfer
function. The phase of the open loop transfer function for the required phase margin
(γmreq) is
ϕgc=180- γmreq
=180-59.2=120.8o
From the drawn bode plot, the gain at this phase angle (i.e. 120.8) is 44 dB. This gain
should be made zero to have a phase margin of 59.2o
. The amplifier gain to increase the
magnitude of the curve by 44 dB is 10
44
20
⁄
=158.48. Since the initial gain value is 3.6,
the obtained gain value has to multiply with 3.6 for calculation of the amplifier gain.
Amplifier gain K=3.6*158.48=570.56.
Below diagram shows the system control scheme block diagram with obtained gain
value.
Figure 14.2: Modified Block Diagram of Control Scheme of System
Tabular Form
Calculated MATLAB
Amplifier Gain 570.56 584.179

Add-On 02.
Results and Conclusions:
The Bode plot is plotted by hand calculation and by using MATLAB. amplifier
gain to get 9.5% overshoot is obtained, K=570.56 (by calculations) and K=584.17 (from
MATLAB).

Add-On 02.

Add-On 03.
Design of Lag Compensator using Frequency Response Method
15. Add-On 03.
Design of Lag Compensator
using Frequency Response
Method

Control Systems Lab Manual

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