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EE2257 CONTROL SYSTEM LABORATORY
032
0
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
P = 45 Total = 45
Detailed Syllabus
1.
Determination of Transfer Function Parameters of a DC Servo Motor
Aim
To derive the transfer function of the given D.C Servomotor and experimentally determine the
transfer function parameters
Exercise
1. Derive the transfer function from basic principles for a separately excited DC motor.
2. Determine the armature and field parameters by conducting suitable experiments.
3. Determine the mechanical parameter by conducting suitable experiments.
4. Plot the frequency response.
Equipment
1.
DC servo motor
2.
3.
4.
2.
Tachometer
Multimeter
Stop watch
: field separately excited – loading facility
– variable voltage source - 1 No
: 1 No
: 2 Nos
: 1 No
Determination of Transfer Function Parameters of AC Servo Motor
Aim
To derive the transfer function of the given A.C Servo Motor and experimentally
determine the transfer function parameters
Exercise
1.
Derive the transfer function of the AC Servo Motor from basic
Principles.
2.
Obtain the D.C gain by operating at rated speed.
3.
Determine the time constant (mechanical)
4.
Plot the frequency response
Equipment
1.
AC Servo Motor
2.
3.
4.
3.
Tachometer
Stopwatch
Voltmeter
: Minimum of 100w – necessary
sources for main winding and
control winding – 1 No
: 1 No
: 1 No
: 1 No
Analog Simulation Of Type-0 And Type-1 System
Aim
To simulate the time response characteristics of I order and II order, type 0 and
type-1 systems.
Exercise
1. Obtain the time response characteristics of type – 0 and type-1, I order and II order
systems mathematically.
2. Simulate practically the time response characteristics using analog rigged up modules.
3. Identify the real time system with similar characteristics.
Equipment
1. Rigged up models of type-0 and type-1 system using analog components.
2. Variable frequency square wave generator and a normal CRO - 1 No
(or)
DC source and storage Oscilloscope - 1 No
4.
Determination of Transfer function of DC Generator
Aim
To determine the transfer function of DC generator
Exercise
1. Obtain the transfer function of DC generator by calculating  and gain
Equipment
1.
DC Generator
2.
Tachometer
3.
Various meters
4.
Stop watch
5.
Determination of Transfer function of DC Motor
Aim
To determine the transfer function of DC motor
Exercise
1. Obtain the transfer function of DC motor by calculating  and gain
Equipment
1.
DC Motor
2.
Tachometer
3.
Various meters
4.
Stop watch
6.
Stability Analysis of Linear Systems
Aim
To analyse the stability of linear systems using Bode / Root locus / Nyquist plot
Exercise
1. Write a program to obtain the Bode plot / Root locus / Nyquist plot for the
given system
2. Access the stability of the given system using the plots obtained
3. Compare the usage of various plots in assessing stability
Equipment
1.
7.
System with MATLAB / MATHCAD / equivalent software - 3 user license
DC and AC position Control system
Aim
To study the AC and DC position control system and draw the error characteristics
between setpoint and error.
Exercise
1. To study various positions and calculate the error between
setpoint and output. position
2. To measure outputs at various points (between stages)
Equipment
1.
2.
3.
8.
AC and DC position control kit with DC servo motor.
Power transistor
Adder
Stepper Motor Control System
Aim
To study the working of stepper motor
Exercise
1. To verify the working of the stepper motor rotation using
microprocessor.
Equipment
1.
2.
3.
4.
9.
Stepping motor
Microprocessor kit
Interfacing card
Power supply
Digital Simulation of First order System
Aim
To digitally simulate the time response characteristics of first -order system
Exercise
1. Write a program or build the block diagram model using the given
software.
2. Obtain the impulse, step and sinusoidal response characteristics.
3. Identify real time systems with similar characteristics.
Equipment
1. System with MATLAB / MATHCAD (or) equivalent software - minimum 3 user license.
10.
Digital Simulation of Second order Systems
Aim
To digitally simulate the time response characteristics of second -order system
Exercise
1. Write a program or build the block diagram model using the given
software.
2. Obtain the impulse, step and sinusoidal response characteristics.
3. Identify real time systems with similar characteristics.
Equipment
System with MATLAB / MATHCAD (or) equivalent software - minimum 3 user license.
Expt. No: 1. a)DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF
FIELD CONTROLLED DC SERVO MOTOR
AIM: To determine the transfer function of field controlled DC servo motor
APPRATUS REQUIRED:
1. DC servo motor trainer kit
2. DC Servo motor
3. Digital Multi meter
FORMULE USED:
1. Field resistance,Rf in  = Vf1 / If1
2.Armature resistance,Ra in  = Va / Ia
3. Field Inductance,Lf in H= XLf / 2f
where XLf in  = (Zf2 – Rf2)
Zf in  = Vf2 / If2
4. Power absorbed, W’ in watts = Va Ia
5. Stray loss, W in watts = W’ x [ t2 / (t1-t2) ]
where W’ is Power absorbed in watts
t2 is time taken on load in secs
t1 is time taken on no load in secs
6. Moment of inertia J in Kg m2 / rad = W x (60 / 2)2 x dt/dN
N
Where W is stray loss in watts
dt is change in time on no load in secs
dN is change in speed on no load is rpm
N is rated speed in rpm
7. Frictional co-efficient, B in N-m / (rad / sec ) = W / (2N / 60 )2
where W is stray loss in watts
N is rated speed in rpm
8. Transfer function  (s) / Vf (s) = Km / s (1+sTf) (1+sTm)
where Motor gain constant Km = Ktf / Rf B
Torque constant Ktf in N-m / A = T / If
Torque T in N-m = 9.55 Eb Ia / N
Back EMF Eb in volts = V – Ia Ra
V = Excitation voltage in volts
Field time constant Tf = Lf / Rf
Mechanical time constant Tm = J / B
THEORY:
`DC Servo motor is basically a torque transducer which converts electrical energy into
mechanical energy It is basically a separately excited type DC motor. The torque developed on the
motor shaft is directly proportional to the field flux and armature current,Tm = Km  Ia. The back
emf developed by the motor is Eb = Kb  m
In a field controlled DC Servo motor, the electrical signal is externally applied to the field
winding. Hence current through field winding is controlled in turn controlling the flux. In a control
system, a controller generates the error signal by comparing the actual o/p with the reference i/p.
Such an error signal is no enough to drive the DC motor. Hence it is amplified by the servo
amplifier and applied to the field winding. With the help of constant current source, the armature
current is maintained constant.
When there is change in voltage applied to the field winding, the current through the field
winding changes. This changes the flux produced by field winding. This motor has large Lf / Rf
ratio, so time constant of this motor is high and it can’t give rapid responses to the quick changing
control signals.
CIRCUIT DIAGRAM
1. For field controlled motor
PROCEDURE:
1. To find Field Resistance, Rf
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Put the selector button in field mode.
4. Block the rotor with full load.
5. Leave the armature terminal in open.
6. Check the position of the potentiometer in minimum point.
7. Switch on the MCB, vary the pot and take voltage Vf1 and current If1 readings.
8. Calculate field resistance Rf = Vf1 / If1
2. To find Armature Resistance, Ra
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Put the selector button in armature mode.
4. Block the rotor with full load.
5. Leave the field terminal in open.
6. Check the position of the potentiometer in minimum point.
7. Switch on the MCB, vary the pot and take voltage Va and current Ia
readings.
8. Calculate armature resistance Ra = Va / Ia
3. To find Field Inductance, Lf
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Block the rotor with full load.
4. Switch on the MCB and take voltage Vf2 and current If2 readings.
5. Calculate field inductance Lf.
4. To find moment of inertia, and frictional co-efficient, B
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Put the selector button in armature mode and DPDT switch in power circuit
position.
4. Check the position of the potentiometer in minimum point.
5. Switch ON the MCB and vary the pot from min to max and adjust the motor to
run at rated speed.
6. Change the DPDT switch from power circuit side to load side.
7. Note down the time taken by the motor to come to rest. This value is t1 and set
the pot to min position.
8. Change the DPDT switch in power circuit position.
9. Connect 500 / 1A load in load position.
10. Vary the pot to run the motor at rated speed and change the DPDT switch
position from power circuit side to load side and note down the voltage Va and
current Ia at the instant of changing the switch. Also note down the time ast2
and from Va and Ia find average voltage and current.
5. To find the transfer function parameters
1. Check the MCB position in OFF condition.
2. Press the reset button to reset the over speed.
3. Patch the circuit as per the patching diagram.
4. Put the selection button in the field control mode.
5. Check the position of the potentiometer in minimum point.
6. Connect the armature of DC servo motor to fixed DC source.
7. Connect the field of DC servo motor across the voltmeter.
8. Switch on the MCB.
9. Vary the pot and in turn vary the speed.
10. Apply rated voltage of 220 V to armature and 150 V to field.
11. Note down the field current, field voltage and speed.
12. Find the transfer function (s) / Vf (s) = Km / s (1+sTf) (1+sTm).
Note:
If the voltmeter and ammeter in the trainer kit is found not working external meters of respective
range can be connected in that place.
TABULAR COLUMN:
1. For field resistance Rf
S.No
Vf1 (V)
If1 (A)
Rf ()
2. For armature resistance Ra
S.No
Va (V)
Ia (A)
Ra()
3. For field inductance
S.No
Vf2 (V)
If2 (A)
Zf ()
4. For transfer function parameters
S.No Vf
(V)
If
Ia
N
T
(A)
(A)
(rpm)
(N-m)
Tf
Tm
Ktf
MODEL CALCULATION:
REFERENCE:
1. NAGRATH & GOPAL, “Control Systems”.
RESULT:
The Transfer function of field controlled DC servomotor is determined as
Expt. No: 1.b) DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF
ARMATURE CONTROLLED DC SERVO MOTOR
AIM: To determine the transfer function of armature controlled DC servo motor
APPRATUS REQUIRED:
1. DC servo motor trainer kit
2. DC Servo motor
3.Digital Multi meter
FORMULA:
1.Armature resistance,Ra in  = Va1 / Ia1
2.Armature Inductance,La in H= XLa / 2f
where XLa in  = (Za2 – Ra2)
Za in  = Va2 / Ia2
3. Power absorbed, W’ in watts = Va Ia
4. Stray loss, W in watts = W’ x [ t2 / (t1-t2) ]
where W’ is Power absorbed in watts
t2 is time taken on load in secs
t1 is time taken on no load in secs
5.Moment of inertia J in Kg m2 / rad = W x (60 / 2 )2 x dt/dN
N
Where W is stray loss in watts
dt is change in time on no load in secs
dN is change in speed on no load is rpm
N is rated speed in rpm
6. Frictional co-efficient, B in N-m / (rad / sec) = W / (2N / 60)2
where W is stray loss in watts
N is rated speed in rpm
7. Transfer function  (s) / Va (s) = Kt
RaB / s {(1+sTa) (1+sTm ) +Kb Kt /RaB
where Torque consant Kt = T / Ia
Torque T in N-m = 9.55 Eb Ia / N
Back EMF Eb in volts = V – Ia Ra
V = Excitation voltage in volts (220 V)
Back emf constant Kb = Va / ω
Angular velocity  in rad/ sec = 2πN / 60
THEORY:
DC Servo motor is basically a torque transducer which converts electrical energy into
mechanical energy It is basically a separately excited type DC motor. The torque developed on the
motor shaft is directly proportional to the field flux and armature current,Tm = Km Φ Ia. The back
emf developed by the motor is Eb = Kb Φ ωm.
In an armature controlled DC Servo motor, the control signal available from the servo
amplifier is applied to the armature of the motor.This signal is based on the feedback information ,
supplied to the controller.Due to this armature current changes which in turn changes the torque
produced. The field winding is supplied with constant current hence the flux remains constant.
Therefore these motors are called as constant magnetic flux motors.
CIRCUIT DIAGRAM
1. For armature controlled DC Servomotor
PROCEDURE:
1. To find Armature Resistance, Ra
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Put the selector button in armature mode.
4. Block the rotor with full load.
5. Leave the field terminal in open.
6. Check the position of the potentiometer in minimum point.
7. Switch on the MCB, vary the pot and take voltage Va and current Ia readings.
8.Calculate armature resistance Ra = Va / Ia
2. To find armature inductance, La
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Block the rotor with full load.
4. Switch on the MCB and take voltage Va2 and current Ia2 readings.
5. Calculate armature inductance La.
3.To find moment of inertia, and frictional co-efficient, B
1. Check the MCB position in OFF condition.
2. Patch the circuit as per the patching diagram
3. Put the selector button in armature mode and DPDT switch in power circuit
position.
4. Check the position of the potentiometer in minimum point.
5. Switch ON the MCB and vary the pot from min to max and adjust the motor to
run at rated speed.
6. Change the DPDT switch from power circuit side to load side.
7. Note down the time taken by the motor to come to rest. This value is t1 and set
the pot to min position.
8. Change the DPDT switch in power circuit position.
ad position.
10. Vary the pot to run the motor at rated speed and change the DPDT switch
position from power circuit side to load side and note down the voltage Va and
current Ia at the instant of changing the switch. Also note down the time ast2
and from Va and Ia find average voltage and current.
4. To find the transfer function parameters
1. Check the MCB position in OFF condition.
2. Press the reset button to reset the over speed.
3. Patch the circuit as per the patching diagram.
4. Put the selection button in the armature control mode.
5. Check the position of the potentiometer in minimum point.
6. Connect the field of DC servomotor to fixed DC source.
7. Connect the armature of DC servomotor across the voltmeter.
8. Switch on the MCB.
9. Vary the pot and in turn vary the speed.
10. Apply rated voltage of 220 V to armature.
11. Note down the armature current, armature voltage and speed.
12. Find the transfer function (s) / Va(s) = KtRaB/ s{(1+sTa)(1+sTm ) +Kb Kt /RaB
Note:
If the voltmeter and ammeter in the trainer kit is found not working external meters of respective
range can be connected in that place.
TABULAR COLUMN:
1. For armature resistance Ra
S.No
Va (V)
Ia (A)
Ra()
2. For armature inductance La
S.No
Vf2 (V)
If2 (A)
Zf ()
3. For transfer function parameters
S.No Va
(V)
Ia
Eb
N
T

(A)
(V)
(rpm)
(N-m)
rad/sec
Kb
Kt
MODEL CALCULATION:
REFERENCE:
1. NAGRATH & GOPAL, “Control Systems”.
RESULT:
The Transfer function of armature controlled DC servomotor is determined as
Expt. No: 2
DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF AC
SERVO MOTOR
AIM:
To derive the transfer function of the given A.C Servo Motor and experimentally determine
the transfer function parameters from
a. Torque Speed Characteristics
b. Control Voltage characteristics
APPRATUS REQUIRED:
1.
2 AC servomotor speed control and transfer function trainer
2.
Speed sensor.
FORMULAE USED:
1.
Motor transfer function , Gm (s) = Km / (1+ sm)
Where Motor gain constant, Km = K / FO + F
Where K is T / C
FO is T / N
Torque, T is 9.81 X r (S1  S2)
R is radius of the rotor in m
Frictional co-efficient, F = W / (2N / 60)2
Frictional loss, W is 30 % of constant loss in watts
Constant loss in watts = No load i/p – Copper loss
No load i/p = V (IR+IC)
V is supply voltage, V
IR is current through reference winding, A
IC is current through control winding, A
Copper loss in watts = IC2 RC
RC = 174
N is rated speed in rpm
Where Motor time constant, m = J / FO + F
Where moment of inertia J is d4lR ρ / 32
d is diameter of the rotor = 0.0395m
lR is length of the rotor =0.076m
ρ is density = 0.78 Kgm/ m
THEORY
CONSTRUCTIONAL DETAILS
The AC servo motor is basically a two phase induction motor with some special
design features. The stator consists of two pole pairs (A-B and C-D) mounted on the inner
periphery of the stator, such that their axes are at angle of 90o in space. Each pole pair carries a
winding, One winding is called reference winding and other is called a control winding. The
exciting current in the winding should have a phase displacement of 90o. The supply used to drive
the motor is single phase and so a phase advancing capacitor is connected to one of the phase to
produce a phase difference of 90o. The stator constructional features of AC servo motor are shown
in fig.1.
The rotor construction is usually squirrel cage or drag-cup type. Rotor construction
of AC servomotor is shown in fig.2. The squirrel cage rotor is made of laminations. The rotor bars
are placed on the slots and short-circuited at both ends by end rings. The diameter of the rotor is
kept small in order to reduce inertia and to obtain good accelerating characteristics. The drag cup
construction is employed for very low inertia applications. In this type of construction the rotor will
be in the form of hollow cylinder made of aluminium. The aluminium cylinder itself acts as shortcircuited rotor conductors. Electrically both the types of rotor are identical.
WORKING PRINCIPLE AS AN ORDINARY INDUCTION MOTOR
The stator windings are excited by voltages of equal rms magnitude and 90o phase
difference. This result in exciting currents i1 and i2 that are phase displaced by 90o and have equal
rms values. These currents give rise to a rotating magnetic field of constant magnitude. The
direction of rotation depends on the phase relationship of the two currents (or voltages). The
exciting current produces a clockwise rotating magnetic field and phase shift of 180 o in i1 will
produce an anticlockwise rotating magnetic field. This rotating magnetic field sweeps over the rotor
conductors. The rotor conductor experience a change in flux and so voltages are induced rotor
conductors. This voltage circulates currents in the short-circuited rotor conductors and currents
create rotor flux. Due to the interaction of stator & rotor flux, a mechanical force (or Torque) is
developed on the rotor and so the rotor starts moving in the same direction as that of rotating
magnetic field.
ADVANTAGES OF AC SERVOMOTOR
1 Control of AC servomotor is so easier than induction motor, because of controlling only the
control phase winding voltage of magnitude 12V or 24V and not main winding voltage of
230V.
2 Direction of motor reversal is also obtained by interchanging the control phase winding
voltage.
DISADVANTAGES OF AC SERVOMOTOR
1 The characteristics are quite non-linear and are more difficult to control especially for
servomechanism
APPLICATIONS
2 AC servomotors are best suited for low power application such as instrument servo (e.g.
control of pen in x-y records) and computer related equipment (Disk, tape drives,
printers etc.)
BLOCK DIAGRAM OF SERVOMOTOR
PROCEDURE:
1.
DETERMINATION OF FRICTIONAL CO_EFFICIENT, F
1. Check the MCB position in OFF condition and patch the circuit using the
patching diagram.
2. Measure the control winding current, reference winding current and supply
voltage.
3. Find the frictional co-efficient, F = W / (2N / 60) 2
2
DETERMINATION OF TORQUE SPEED CHARACTERISTICS AND FO
1. Check the MCB position in OFF condition and patch the circuit using the patching
diagram.
2. Set the speed control pot in minimum position and load in free condition
3. Apply rated voltage to the reference phase winding and control phase winding.
4. Note down the no load speed.
5. Apply load in steps. For each step note down speed and load applied and calculate the
Torque as T = 9.81 X r X (S1 S2).
6. Reduce the load fully and allow the motor to run at rated speed.
7. Repeat steps 4, 5 for 75 % and 50 % of control winding voltage levels and tabulate
reading.
8. Draw the graph between speed and torque and from the graph find T and N and
calculate FO as T / N
3.
DETERMINATION OF TORQUE CONTROL VOLTAGE CHARACTERISTICS & K
1. Check the MCB position in OFF condition and patch the circuit using the patching
diagram.
2. Apply rated Voltage to Reference phase winding. Apply a certain voltage to the
Control phase winding and make the motor run at rated speed.
3. Load the motor gradually and the speed of the motor will decrease. Increase the
Control winding voltage till the speed obtained at no load is reached. Note down
control voltage and load readings.
4. Repeat the above procedure for various speeds and tabulate.
5. Calculate the Torque as T = 9.81 X r X(S1  S2) and plot the graph between
torque and control winding voltage. Find T and C and then find K.
TORQUE SPEED CHARACTERISTICS
N
Rpm
Vc1 =
S1
S2
Kg Kg
T
N-m
N
Rpm
Vc2 =
S1
S2
Kg
Kg
T
N-m
N
Rpm
Vc3 =
S1
S2
Kg Kg
MODEL GRAPH
TORQUE
N-m
SPEED (RPM)
T
N-m
TORQUE –CONTROL VOLTAGE CHARACETERISTICS
S1
Kg
S2
Kg
N1 =
Vc
V
T
N-m
S1
Kg
N2 =
S2
Vc
Kg
V
T
N-m
S1
Kg
N3 =
S2
Vc
Kg
V
T
N-m
MODEL GRAPH
TORQUE
N-m
Control voltage (Volts)
Given,
B = 0.01875 X 10-3 N-m / Rpm
J = 0.052 X 10-3 Kg m2
VIVA QUESTIONS
1. What are the main parts of an ac servo motor?
2. What are the advantages and disadvantages of an AC servo motor?
3. Give the applications of Ac servomotor?
4. What do you mean by servo mechanism?
REFERENCE
1. NAGRATH & GOPAL, “Control Systems”.
2. Lab Manual – Transfer Function Derivation of AC Servo motor System
RESULT:
From the Torque-Speed characteristics and Control-voltage Characteristics the transfer
function of AC servomotor is determined as
Expt No. 3
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:
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
K3 is the gain of the time constant – 2 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
K2 is the gain of the time constant – 1 block =10
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 / s2 + 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) = n2 / (s2 + 2ns + n2) where  is damping
ratio and n is undamped natural frequency.
BLOCK DIAGRAM:
1. To find steady state error of type- 1 system
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
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 Xchannel 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. To find the steady state error of type – 0 first order system
1. The blocks are connected using the patch cords in the simulator kit.
2. The input square 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 Xchannel of CRO.
3. The output from the system is connected to the Y- 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
1. The blocks are connected using the patch cords in the simulator kit.
2. The input square 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 Xchannel of CRO.
3. The output from the system is connected to the Y- 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.
5. The above procedure is repeated for different values of gain K and the values are
compared with the theoretical values.
TABULAR COLUMN:
1. To find the steady state error of type – 1 first order system
S.No. Gain ,K
Steady state error ess (V)
2. To find the steady state error of type – 0 first order system
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 Peak
in, percent
K Overshoot,
%MP
Rise
time,
tr
(sec)
Peak
time,
tp
(sec)
Settling
time,ts
(sec)
Graphical
Theoretical
Dam
ping
ratio

Dam
ping
ratio

Undamped
natural
frequency,
n (rad/sec)
Undamped
natural
frequency,
n(rad/sec)
4. To find the closed loop response of type – 1 second order system
S.No. Gain, Peak
K
percent
Overshoot,
%MP
Rise
time,
tr
(sec)
Peak Settling Graphical
time, time,ts Dam Undamped
tp
(sec)
ping
natural
(sec)
ratio frequency,
n
(rad/sec)
Theoretical
Damping Undamped
ratio
natural
frequency,

n(rad/sec)
MODEL GRAPH:
MODEL CALCULATION:
RESULT:
The time response of first and second order type- 0 and type- 1 systems are studied.
Expt. No: 6.a)
Plot)
STABILITY ANALYSIS OF LINEAR SYSTEMS(Bode
AIM
To obtain the bode plot for the given system whose transfer function is given as
G(S)=
242(s+5)
s(s+1)(s2+5s+121)
and to find out whether the system is stable or not.
APPARATUS REQUIRED
MATLAB Software
THEORY
A Linear Time-Invariant Systems is stable if the following two notions of system
stability are satisfied
 When the system is excited by Bounded input, the output is also a Bounded output.
 In the absence of the input, the output tends towards zero, irrespective of the initial
conditions.
The following observations are general considerations regarding system stability and are
If all the roots of the characteristic equation have negative real parts, then the impulse
response is bounded and eventually decreases to zero, then system is stable.



If any root of the characteristic equation has a positive real part, then system is
unstable.
If the characteristic equation has repeated roots on the jω-axis, then system is
unstable.
If one are more non-repeated roots of the characteristic equation on the jω-axis, then
system is unstable.
BODE PLOT :
Consider a Single-Input Single-Output system with transfer function
b0 sm + b1 sm-1 + ……+ bm
C(s)
=
R(s)
a0 sn + a1sn-1 + ……+an
Where m < n.
 Rule 1 A system is stable if the phase lag is less than 180˚ at the frequency for
which the gain is unity (one).
 Rule 2 A system is stable if the gain is less than one (unity) at the frequency for
which the phase lag is 180˚.
The application of these rules to an actual process requires evaluation of the gain and phase
shift of the system for all frequencies to see if rules 1 and 2 are satisfied. This is obtained
by plotting the gain and phase versus frequency. This plot is called BODE PLOT. The gain
obtained here is open loop gain.
The stability criteria given above represent Limits of Stability. It is well to design a system
with a margin of safety from such limits to allow for variation in components and other
unknown factors. This consideration leads to the revised stability criteria, or more properly,
a Margin of Safety provided to each condition. The exact terminology is in terms of a Gain
Margin and Phase Margin from the limiting values quoted.
 If the phase lag is less than 140˚ at the unity gain frequency, the system is stable.
This then, is a 40˚ Phase Margin from the limiting values of 180˚.
 If the gain is 5dB below unity (or a gain of about 0.56) when the phase lag is 180˚,
the system is stable. This is 5dB Gain Margin.
ALGORITHM
1. Write a Program to (or using SIMULINK) obtain the Bode plot for the given system.
2. Access the stability of given system using the plots obtained.
PROGRAM
%BODE PLOT OF THE SYSTEM%
%Enter the numerator and denominator of the transfer function
num=[0 0 0 242 1210];
den=[1 6 126 121 0];
sys=tf(num,den)
%Specify the frequency range and enter the command
w=logspace(-2,4,1000);
bode(sys,w)
xlabel('Frequency')
ylabel( ' Phase angle in degrees Magnitude of G(s)')
title('Bode Plot of the system 242(s+5)/s(s+1)(s^2+5*s+121)')
%To determine the Gain Margin,Phase Margin, Gain crossover frequency and
%Phase cross over frequency
[ Gm, Pm, Wcp, Wcg ]= margin (sys)
PROCEDURE TO OBTAIN BODE PLOT
1. Rewrite the sinusoidal transfer function in the time constant form by replacing s by jω
2. Identify the corner frequencies associated with each factor of the transfer function.
3. Knowing the corner frequencies draw the asymptotic magnitude plot. This plot
consists of straight line segments with line slope changing at each corner frequency
by +20db/decade for a zero and -20db/decade for a pole. For a complex conjugate
zero or pole the slope changes by + 40db/decade.
4. Draw a smooth curve through the corrected points such that it is asymptotic to the line
segments. This gives the actual log-magnitude plot.
5. Draw phase angle curve for each factor and add them algebraically to get the phase
plot.
MANUAL CALCULATIONS
i)The sinusoidal transfer function G (jω) is obtained by replacing s by jω in the given s domain
transfer function
G(jω)=
242(jω +5)
jω (jω +1)( jω 2+5 jω +121)
On comparing the quadratic factor of G(s) with standard form of quadratic
factor , ζ and ωn can be evaluated.
s2+5s+121 = s2+2ζωns + ωn2
On comparison
ωn2 = 121
2ζωn= 5
ωn =11 rad/sec
ζ = 0.227
G(jω)=
10(1+0.2jω)
jω (1+jω)( 1+0.4 jω -0.0083ω 2)
ii)CORNER FREQUENCIES
The corner frequencies are ωc1=1rad/sec ωc2= 5 rad/sec and ωc3=11rad/sec
Choose a low frequency ωl such that ωl< ωc1 and choose a high frequency ωh> ωc3.
Let ωl=0.5 rad/sec and ωh=100 rad/sec
Term
10
jω
1
Corner
Frequency
rad/sec
__
Slope db/dec
Change in slope
db/dec
-20
__
1
-20
-20-20= - 40
(1+jω)
(1+0.2jω)
1
( 1+0.4 jω -0.0083ω 2)
5
11
20
-40
-40-20 = -20
-40-20 = -60
iii)MAGNITUDE PLOTS
Calculate A at ωl, ωc1, ωc2, ωc3, and ωh
Let A= | G(jω)| in db
At ω= ωl A= 20 log(10/0.5)=26.03db
ω= ωc1 , A=20log(10/1)=20db
ω= ωc2 A= -40log(5/1)+20=-7.96 db
ω= ωc3 A = -20log(11/5) - 7.96 = -14.80 db
ω= ωh A = -60log(100/11)-14.80 = - 72.3 db
These values are plotted in the semilog graph sheet taking frequency along the logarithmic scale
and magnitude in db along the linear scale
iv)PHASE PLOT
The phase angle of G(jω) as a function of ω is given by
Φ = ‹G(jω) = tan-1 0.2ω -90 – tan-1 ω – tan-1 0.04ω/(1 – 0.0083ω2)
Ω
tan-1 0.2ω
tan-1 ω
tan-1 {0.04ω/
Φ = ‹G(jω)
(1 – 0.0083ω2)}
0.5
507
26.56
1.15
-112
1
11.3
45
2.31
-126.01
5
45
78.96
14.04
-138
10
63.43
84.29
63.44
-174.3
11
65.5
84.8
85.8
-195.4
20
75.96
87.14
180-19.98=160
-261.18
50
84.3
88.85
180-6=174
-268.55
100
87014
89.43
180-2.9=177.1
-269.3
These values are plotted in the semilog graph sheet taking the same frequency as before along
the logarithmic scale and phase angle in degrees along the linear scale.
OUTPUT (from simulation)
242 s + 1210
----------------------------s^4 + 6 s^3 + 126 s^2 + 121 s
Gm = 2.0273
Pm = 41.8270
Wcp = 10.0961
Wcg = 3.6322
OUTPUT (from graph)
ωgc= Φgc=3.1rad/sec
Phase margin γ=180+ Φgc = 180-134 = 46 degrees
Gain Margin = 12 db
ωpc = 10.1 rad/sec
BODE PLOT
RESULT:
i)The Bode plot is drawn for the given transfer function using MATLAB and verified
manually
ii) The system is stable
Expt. No: 6. b)
STABILITY ANALYSIS OF LINEAR SYSTEMS
(Root Locus Plot)
AIM
To obtain the Root locus plot for the given system whose transfer function is given as
G(S)=
K
s(s+3)(s2+3s+11.25)
APPARATUS REQUIRED
Mat lab Software
THEORY
ROOT LOCUS PLOT :
The characteristic of the transient response of a closed-loop system is related to the location of the
closed loop poles. If the system has a variable loop gain, then the location of the closed-loop poles
depend on the value of the loop gain chosen. A simple technique known as “Root Locus
Technique” used for studying linear control systems in the investigation of the trajectories of the
roots of the characteristic equation.
This technique provides a graphical method of plotting the locus of the roots in the s-plane as a
given system parameter is varied over the complete range of values(may be from zero to infinity).
The roots corresponding to a particular value of the system parameter can then be located on the
locus or the value of the parameter for a desired root location can be determined form the locus.
The root locus is a powerful technique as it brings into focus the complete dynamic response of the
system . The root locus also provides a measure of sensitivity of roots to the variation in the
parameter being considered. This technique is applicable to both single as well as multiple-loop
systems.
PROCEDURE:
1. Write a Program to (or using SIMULINK) obtain the Root locus plot for the given system.
2. Access the stability of given system using the plots obtained.
PROGRAM
%ROOT LOCUS OF THE SYSTEM%
num=[0 0 0 0 1]
den=[1 6 20.25 33.75 0]
sys=tf(num,den)
rlocus(sys)
v=[-10,10,-8,8];
axis(v)
xlabel('Real Axis')
ylabel('Imaginary Axis')
title('Root Locus of the sytem ')
title('Root Locus Plot of the system K/s(s+3)(s^2+3s+11.25))')
MANUAL CALCULATIONS
1. Number of poles =4, zeros = 0, number of root locus branches =4. Starting points s=0, -3 &
1.5+ j3.
2. Pole – zero plot is as follows
Section between 0 and -3 is part of root locus. One breakway point is between s=0 and s=3.
3. Angle of asymptotes are 45,135,225 and 315 degrees
4. Centroid = -1.5
5. Three Breakway points are -1.5,-1.5 + j 1.8371
6. Intersection with imaginary axis s= + j2.37.
7. Angle of departure -90, +90.
8. Root locus is plotted.
9. Stability for 0< K<82.26 system is stable.
K=82.26 system is marginally stable.
K>82.26 system is unstable
OUTPUT
num =
0 0 0 0 1
den =
1.0000 6.0000 20.2500 33.7500
Transfer function:
1
--------------------------------s^4 + 6 s^3 + 20.25 s^2 + 33.75 s
0
GRAPH(from Simulation)
RESULT:
Thus the Root Locus plot is drawn for the given transfer function using matlab and verified
manually.
Expt. No: 6 c)
STABILITY ANALYSIS OF LINEAR SYSTEMS
(Nyquist Plot)
AIM
To obtain the Nyquist plot for the given system whose transfer function is given as
G(S)=
50
(s+4)(s2+3s+3)
and to find out whether the system is stable or not.
APPARATUS REQUIRED
Mat lab Software
THEORY
POLAR PLOTS OR NYQUIST PLOTS:
The sinusoidal transfer function G(jω) is a complex function is given by
G(jω) = Re[ G(jω)] + j Im[G(jω)]
or
G(jω) = │G(jω) │ ∟G(jω) = M ∟Φ -----------(1)
From equation (1), it is seen that G(jω) may be represented as a phasor of magnitude M and
phase angle Φ. As the input frequency varies from 0 to ∞, the magnitude M and phase angle Φ
changes and hence the tip of the phasor G(jω) traces a locus in the complex plane. The locus thus
obtained is known as
POLAR PLOT.
The major advantage of the polar plot lies in stability study of systems. Nyquist related the stability
of a system to the form of these plots. Polar plots are referred as NYQUIST PLOTS.
NYQUIST stability criterion of determining the stability of a closed loop system by investigating
the properties of the frequency domain plot of the loop transfer function G(s) H(s).
Nyquist stability criterion provides the information on the absolute stability of a control system as
similar to Routh- Hurwitz criterion. Not only giving the absolute stability, but indicates “Degree
of Stability” i.e “Relative Stability” of a stable system and the degree of instability of an unstable
system and indicates how the system stability can be improved. The Nyquist stability citerion is
based on a Cauchy’s Residue Theorem of complex variables which is referred to as the “principle
of argument”.
Let Q(s) be a single –valued function that has a finite number of poles in the s-plane. Suppose that
an arbitrary closed path Гq is chosen in the s-plane so that the path does not go through any one of
the poles or zeros of Q(s); the corresponding Гq locus mapped in the Q(s) plane will encircle the
origin as many times as the difference between the number of the zeros and the number of poles of
Q(s) that are encircled by the s-plane locus Гq.
The principle of argument is given by
N= Z - P
Where N – number of encirclemnts of the origin made by the Q(s) –plane locus Гq.
Z – number of zeros of Q(s) encircled by the s-plane locus Гq in the s-plane.
P - number of poles of Q(s) encircled by the s-plane locus Гq in the s-plane.
ALGORITHM
1. Write a Program to (or using SIMULINK) obtain the Nyquist plot for the given system.
2. Access the stability of given system using the plots obtained.
PROGRAM
%NYQUIST PLOT
%Enter the numerator and denominator of the transfer function
num=[0 0 0 50]
den=[1 7 12 12]
sys=tf(num,den)
%Specify the frequency range and enter the command
nyquist(sys)
v=[-3 5 -7 7]
axis(v)
xlabel('Real Axis');
ylabel('Imaginary Axis');
title('Nyquist Plot of the sytem 50/(s+4)(s^2+3s+3)')
%To determine the Gain Margin,Phase Margin, Gain crossover frequency and
%phase cross over frequency
[Gm,Pm,Wcp,Wcg]=margin (sys)
v=
-3 5 -7
Gm =
1.4402
Pm =
11.1642
Wcp =
3.4643
Wcg =
2.9533
7
MANUAL CALCULATIONS:
NYQUIST
PLOT
VIVA QUESTIONS
1. What is polar plot?
2. What is Nyquist plot?
3. Define the conditions of stability in polar plot.
4. What is the use and advantage of polar plot.
REFERENCE
1. NAGRATH & GOPAL,” Control Systems”.
2. MATLAB User Manual .
3. Control Systems by Nagoor gani
RESULT:
Thus the Nyquist plot is drawn for the given transfer function using matlab and verified manually
Expt. No: 8
DC POSITION CONTROL SYSTE M
To study the characteristics of a dc position control system.
APPARATUS REQUIRED:
i)
DC position control kit and Motor unit
ii)
Multimeter
THEORY:
A pair of potentiometers acts as error-measuring device. They convert the
input and output positions into proportional electric signals. The desired position is set on the input
potentiometer and the actual position is fed to feedback potentiometer. The difference between the
two angular positions generates an error signal, which is amplified and fed to armature circuit of
the DC motor. If an error exists , the motor develops a torque to rotate the output in such a way as
to reduce the error to zero. The rotation of the motor steps when the error signal is zero, i.e., when
the desired position is reached.
PROCEDURE:
1. The input or ref potentiometer is adjusted nearer to zero initially.
2. The command switch is kept in continuous mode and some value of forward gain Ka is
selected.
3. For various positions of input potentiometer (r) the positions of the response potentiometer
(0) is noted. Simultaneously the reference voltage measured between Vr & E and the
output voltage measured between VO & E are noted.
4. A graph is plotted with 0 along y-axis and r along x-axis.
Tabular Column
S.`No
θr
θO
degree
degree
Vr in Volts
VO in Volts
MODEL GRAPH:
Output
Position
(Deg)
Input Position (Deg)
RESULT
Thus the dc position control system characteristics are studied and corresponding graphs are drawn.
Expt. No9:
DIGITAL SIMULATION OF FIRST ORDER SYSTEMS
AIM
To digitally simulate the time response characteristics of a linear system without non
linearities and to verify it manually.
APPARATUS REQUIRED
A PC with MATLAB package
THEORY
The time characteristics of control systems are specified in terms of time domain
specifications. 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 of any order may be specified in terms
of transient response to a unit step input signal. The transient response characteristics of a control
system to a unit step input is specified in terms of the following time domain specifications
 Delay time td
 Rise time tr
 Peak time tp
 Maximum overshoot Mp
 Settling time ts
PROCEDURE:
1. In MATLAB software open a new model in simulink library browser.
2. From the continuous block in the library drag the transfer function block.
3. From the source block in the library drag the step input.
4. From the sink block in the library drag thescope.
5. From the math operations block in the library drag the summing point.
6. Connect all to form a system and give unity feedback to the system.
7. For changing the parameters of the blocks connected double click the respective
block.
8. Start simulation and observe the results in scope.
9. From the step response obtained note down the rise time, peak time, peak overshoot and
settling time.
10. For the same transfer function write a matlab program to obtain the step response and verify
both the results.
PROGRAM
%This is a MATLAB program to find the step response
num=[0 0 25];
den=[1 6 25];
sys = tf (num,den);
step (sys);
PLOT
RESULT
The time response characteristics of a linear system without non linearities is simulated digitally
and verified manually.