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90981041 control-system-lab-manual
1. SSN COLLEGE OF ENGINEERING
KALAVAKKAM- 603 110
DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING
EE 2257 - CONTROL SYSTEMS
LAB MANUAL
Dec 2009-April 2010
Name: _______________________________________
Reg. No.: _____________________________________
Year: II Sem: 4 Sec: A/B Dept: EEE
2. 2
DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING
EE 2257- CONTROL SYSTEMS LABORATORY
NAME OF THE STUDENT : ____________________________
REGISTER NUMBER : ____________________________
CLASS : II- EEE- A / B
ACADEMIC YEAR : Dec 2009- Apr 2010
TOTAL MARKS : -------- / 10
SIGNATURE OF THE STAFF:
3. 3
LIST OF EXPERIMENTS
1. Determination of transfer function parameters of Armature controlled
DC (servo) motor.
2. Determination of transfer function parameters of Field controlled DC
(servo) motor.
3. Determination of transfer function parameters of an AC servomotor.
4. Analog simulation of type-0 and type-1 systems
5. Digital simulation of first order systems
6. Digital simulation of second order systems.
7. Stability analysis of linear systems.
8. DC and AC position control systems.
9. Stepper motor control system
10. Determination of transfer function parameters of DC generators.
11. Study of synchros
12. Design and implementation of compensators.
13. Design of P, PI and PID controllers.
.
P = 45, TOTAL = 45
4. 4
INDEX
SIGN. OF
Expt. Title MARKS
DATE THE
NO. (10)
STAFF
1
2
3
4
5
6
7
8
9
10
11
12
13
Total Marks
Signature of the faculty:
5. 5
EXPT. NO.:
DATE:
TRANSFER FUNCTION OF ARMATURE CONTROLLED
DC SERVOMOTOR
AIM:
To determine the transfer function of an armature controlled dc servomotor.
APPARATUS REQUIRED:
THEORY:
Transfer function is defined as the ratio of Laplace transform of the output
variable to the Laplace transform of input variable at zero initial conditions.
Armature controlled DC shunt motor
In this system, Ra = Resistance of armature in Ω
La= Inductance of armature windings in H
Ia = Armature current in A
If = Field current in A
e = Applied armature voltage in V
eb = back emf in V
Tm = Torque developed by the motor in Nm
6. 6
J = Equivalent moment of inertia of motor and load referred to
motor shaft in kgm2
B= Equivalent viscous friction coefficient of inertia of motor
and load referred to motor shaft in Nm/(rad/s)
In Servo applications, DC motors are generally used in the linear range of the
magnetization curve. Therefore, the air gap flux φ is proportional to the field current.
φ α If
φ = Kf If ,where Kf is a constant. ----------------------------- (1)
The torque Tm developed by the motor is proportional to the product of the armature
current and air gap flux.
Tm α φ Ia
Tm =Ki φ Ia = Ki Kf If Ia , where Ki is a constant --------------(2)
In the armature controlled DC motor, the field current is kept constant. So the above
equation can be written as
Tm = Kt Ia , Where Kt is known as motor torque constant.------ (3)
The motor back emf being proportional to speed is given by
eb α dθ/dt,
eb = Kb dθ/dt, where Kb is the back emf constant.----------------(4)
The differential equation of the armature circuit is
e = IaRa + La dIa/dt + eb ----------------------------------------- (5)
The torque equation is
Tm = Jd2θ/dt2 + B dθ/dt ------------------------------------------ (6)
Equating equations (3) and (6)
Jd 2θ/dt2 + B dθ/dt = Kt Ia ---------------------------------------(7)
Taking Laplace transforms for the equations (4) to (7), we get
Eb(s) = Kb s θ(s) -------------------------------------------- (8)
(s La + Ra ) Ia(s) = E(s) – Eb(s). ------------------------------- (9)
( J s2+ B s) θ(s) = Tm (s) = Kt Ia(s) ---------------------------- (10)
From equations (8) to (10) , the transfer function of the system is obtained as
7. 7
Block diagram
Using the above equations, the block diagram for the armature controlled DC motor is
given below:
E(s)
+ θ (s)
ω(s) 1/[Ra+sLa] Kt 1/s[Js+B]
- Eb(s)
s Kb
1. Circuit diagram to determine Kt and Kb
8. 8
2. Circuit diagram to determine Ra:
3. Circuit diagram to determine La:
PROCEDURE:
i)Load test to determine Kt
1. Initially keep all the switches in the off position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch on the power and the SPST switches S1 and S2.
5. Adjust the field voltage to the rated value.
6. Apply the armature voltage until the motor runs at the rated speed.
7. Apply load and note the armature voltage, current and spring balance readings.
8. Calculate torque and plot the graph between torque and armature current.
9. Determine Kt from graph.
9. 9
ii)No-Load test to determine Kb.
1. Initially keep all the switches in the off position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch on the power and the SPST switches S1 and S2.
5. Set the field voltage to the rated value.
6. Adjust the armature voltage and note the armature voltage, current and speed.
7. Calculate the back emf eb and plot the graph between back emf and ω
8. Determine Kb from graph.
iii) To determine Ra:
1. Initially keep all the switches in the off position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch on the power and the SPST switches S1.
5. Note the armature current for various armature voltages.
6. Calculate Ra.
iv) To determine La:
1. Initially keep all the switches in the off position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch on the power .
5. Apply ac voltage to armature winding
6. Note down the current for various input ac voltage .
7. Calculate Ra.
Tabulation to determine Kt
S.No. Armature Armature Spring Balance Torque
Voltage Current Readings (kg) T = 9.81(S1-S2)r
Va (V) Ia (A) S1 S2 (Nm)
Where r is the radius of the brake drum. r = _____________m
10. 10
Tabulation to determine Kb
S.No. Armature Armature Speed N Eb= Va-Ia Ra ω = 2πN/60
π
Voltage Va Current Ia (rpm) (rad/sec)
(V) (A) (V)
-
Tabulation to determine Ra
S.No. Armature Armature Ra
Voltage Current (Ω)
Va (V) Ia (A)
Calculation by least square method
Ra = [V1I1 +V2I2 +V3I3+V4I4 ] / (I12+I22+I32+I42)
Tabulation to determine Za
S.No. Armature Armature Za
Voltage Current (Ω)
Va (V) Ia (A)
Average Za = ________Ω
11. 11
MODEL GRAPH
To find Kt To find Kb
Torque Eb( V)
(Nm)
Ia(A) ω(rad/sec)
MODEL CALCULATION
Ra = ……..Ohms
Za = …….. Ohms
La = √(Za2 –Ra2 ) / 2πf = ……. H
f = 50 Hz
J = 0.074 kg/m2, B = 0.001Nm/rad/sec
From Graph,
Kt = Torque constant = ∆T / ∆ Ia = ………… Nm / A
Kb = Back emf constant = ∆Eb / ∆ ω = ………. V/(rad/s)
RESULT:
INFERENCE:
12. 12
EXPT. NO.:
DATE:
TRANSFER FUNCTION OF FIELD CONTROLLED DC
SERVOMOTOR
AIM:
To determine the transfer function of a field controlled dc servomotor.
APPARATUS REQUIRED:
THEORY:
The transfer function is defined as the ratio of Laplace transform of the output
variable to the Laplace transform of input variable at zero initial conditions.
Armature controlled DC shunt motor
In this system, Rf = Resistance of the field winding in Ω
Lf= Inductance of the field windings in H
Ia = Armature current in A
If = Field current in A
e = Applied armature voltage in V
eb = back emf in V
ef = Field voltage in V
13. 13
Tm = Torque developed by the motor in Nm
J = Equivalent moment of inertia of motor and load referred to
motor shaft in kgm2
B= Equivalent viscous friction coefficient of inertia of motor
and load referred to motor shaft in Nm/(rad/s)
In Servo applications, the DC motors are generally used in the linear range of the
magnetization curve. Therefore the air gap flux φ is proportional to the field current.
φ α If
φ = Kf If ,where Kf is a constant. -------------------------------- (1)
The torque Tm developed by the motor is proportional to the product of the armature
current and air gap flux.
Tm α φ Ia
Tm =K′φ Ia = K′ Kf If Ia = Km Kf If , where Ki is a constant ----(2)
Appling Kirchhoff’s voltage law to the field circuit, we have
Lf dIf/dt + RIf = ef ------------------------------------------------- (3)
Now the shaft torque Tm is used for driving the load against the inertia and frictional
torque. Hence,
Tm = Jd2θ/dt2 + B dθ/dt ------------------------------------------- (4)
Taking Laplace transforms of equations (2) to (4), we get
Tm(s) = KmKf If (s) ----------------------------------------------- (5)
Ef(s) = (s Lf + Rf) If(s) -------------------------------------------- (6)
Tm(s) = (J s2+ B s) θ(s) ------------------------------------------- (7)
Solving equations (5) to (7), we get the transfer function of the system as
14. 14
1. Circuit diagram to determine KmKf
2. Circuit diagram to determine Rf
3. Circuit diagram to determine Lf
15. 15
PROCEDURE:
i)Load test to determine KmKf
1. Initially keep all the switches in the OFF position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch ON the power and the SPST switches S1 and S2.
5. Apply 50% of the rated field voltage.
6. Apply the 50% of the rated armature voltage.
7. Apply load and note the field current and spring balance readings.
8. Vary the field voltage and repeat the previous step.
9. Calculate torque and plot the graph between torque and field current.
10. Determine KmKf from graph.
ii) To determine Rf
1. Initially keep all the switches in the OFF position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch ON the power and the SPST switch S2.
5. Note the field currents for various field voltages.
6. Calculate Rf.
iii) To determine Lf
1. Initially keep all the switches in the OFF position.
2. Keep all the voltage adjustment knobs in the minimum position.
3. Give connections.
4. Switch ON the power .
5. Apply AC voltage to field windings
6. Note the currents for various input AC voltages.
7. Calculate Lf.
16. 16
Tabulation to determine KmKf
S.No Field Spring Balance Torque
Current Readings (kg) T = 9.81(S1-S2)r
If (A) S1 S2 (Nm)
Where r is the radius of the brake drum. r = _____________m
Tabulation to determine Rf
S.No. Field Field Rf
Voltage Current (Ω)
Ef (V) If (A)
Calculation by least square method
Rf = [V1I1 +V2I2 +V3I3+V4I4 ] / (I12+I22+I32+I42)
Tabulation to determine Zf
S.No. Field Field Zf
Voltage Current (Ω)
Ef (V) If (A)
Average Zf = ________Ω
17. 17
MODEL GRAPH
To find KmKf
Torque
(Nm)
If (A)
MODEL CALCULATIONS
1. Rf = ……..Ohms
2. Zf = …….. Ohms
3. Lf = √(Zf2 –Rf2 ) / 2πf = ……. H
4. f = 50 Hz
5. J = 0.074 kg/m2, B = 0.001Nm/rad/s
6. From Graph, KmKf = ∆T / ∆If = ………… Nm / A
RESULT:
INFERENCE:
18. 18
EXPT. NO:
DATE :
DETERMINATION OF TRANSFER FUNCTION PARAMETERS
OF AC SERVO MOTOR
AIM:
To derive the transfer function of the given AC servomotor and
experimentally determine the transfer function parameters.
APPRATUS REQUIRED:
FORMULA:
1. Motor transfer function
θ(s) Km
=
Eo (s) s (1+s.τm)
2. Motor gain constant Km = K1
K2 + B
3. Motor time constant τm= J
K2 + B
Where K1 = slope of torque - control phase voltage characteristics
K2= slope of torque -speed characteristics
J = Moment of inertia of load and the rotor
B= viscous frictional coefficient of load and the rotor
THEORY
When the objective of a system is to control the position of an object, then the
system is called a servomechanism. The motors that are used in automatic control
systems are called servomotors.
Servomotors are used to convert an electrical signal (control voltage) into an
angular displacement of the shaft. In general, servomotors have the following
features.
1. Linear relationship between speed and electrical control signal
2. Steady state stability
3. Wide range of speed control
19. 19
4. Linearity of mechanical characteristics throughout the entire speed range
5. Low mechanical and electrical inertia
6. Fast response
Derivation of Transfer Function:
Let Tm = Torque developed by the servomotor
θ = angular displacement of the rotor
ω = dθ / dt = angular speed
TL = torque required by the load
J = Moment of inertia of the load and the rotor
B = Viscous frictional coefficient of the load and the rotor
K1 = slope of the control phase voltage and torque characteristics.
K2 = slope of the speed and torque characteristics.
The transfer function of the AC servomotor can be obtained by torque equation. The
motor developed torque is given by
Tm = K1 e c – K2 dθ…………………………………(1)
dt
The rotating part of the motor and the load can be modeled by
TL = J d2θ + B.dθ …………………………. …………(2)
dt2 dt
At equilibrium, the motor torque is equal to load torque. Hence,
K1 e c – K2 dθ /dt = J d 2θ + B dθ …………....(3)
dt dt
Taking Laplace Transform
K1 Ec (s) – K2 s θ(s) = J s2 θ(s) + B s θ(s)………………(4)
θ (s) K1 Km
T.F = = =
Ec (s) s (K2+s J+B) s (1 +s τm)
K1
Where motor gain constant Km =
B + K2
J
and motor time constant τm =
B + K2
20. 20
PROCEDURE:
I. DETERMINATION OF TORQUE SPEED CHARACTERISTICS
1. Give the connections.
2. Connect voltmeter or a digital Multimeter across the control winding.
3. Apply rated voltage to the reference phase winding and control phase
winding.
4. Note the no load speed.
5. Apply load in steps. For each load, note the speed.
6. Repeat steps 4,5 for various control voltage levels and tabulate the readings.
II. DETERMINATION OF TORQUE – CONTROL VOLTAGE CHARACTERISTICS
1. Make connections.
2. Connect voltmeter or a digital Multimeter across the control phase winding
3. Apply rated Voltage to Reference phase winding.
4. Apply a certain voltage to the control phase winding and make the motor
run at low speed. Note the voltage and the no load speed.
5. Apply load to motor. Motor speed will decrease. Increase the control
voltage until the motor runs at same speed as on no-load. Note the control
voltage and load.
6. Repeat steps 5 for various loads
7. Repeat 4-6 for various speeds and tabulate.
Torque Speed Characteristics
Radius of brake drum =________
Vc = Vc = Vc =
Load N Torque Load N Torque Load N Torque
g rpm N-m g rpm N-m g rpm N-m
Model Graph
Torque
N-m
Speed (rpm)
21. 21
Torque –Control Voltage Characteristics
N1 = N2 = N3 =
Load Vc Torque Load Vc Torque Load Vc Torque
g V N-m g V N-m g V N-m
Model Graph
Torque
N-m
Control voltage (Volts)
From Graph , K1 =
K2 =
Given , B=
J=
From Calculations, Km =
τm =
23. 23
EXPT. NO:
DATE :
ANALOG SIMULATION OF TYPE-0 AND TYPE-1 SYSTEM
AIM:
To simulate the time response characteristics of I order and II order, type 0
and type-1 systems.
APPARATUS REQUIRED:
THEORY:
Order of the system:
The order of the system is given by the order of the differential equation
governing the system. The input-output relationship of a system can be expressed by
transfer function. Transfer function of a system is obtained by taking Laplace
transform of the differential equation governing the system and rearranging them as
ratio of output and input polynomials in ‘s’. The order is given by the maximum
power of ‘s’ in denominator polynomial Q(s)
T(s) = P(s) / Q(s)
P(s) --- Numerator polynomial
Q(s) --- Denominator polynomial
Q(s) =ao sn + a1sn-1 + a2 sn-2 + ………….+ an-1 s + a n
If n=0, then system is Zero-Order system.
If n=1, then system is First-Order system.
If n=0, then system is Second-Order system.
Type of the system
Type of the system is given by the number of poles of the loop transfer function at the
origin.
G(s)H(s) = K P(s) / Q(s)
(s+z1) (s+z2) (s+z3) ……..
=
sN (s+p1) (s+p2) (s+p3) ……..
If N=0, the system is a Type Zero system.
If N=1, the system is a Type One system.
If N=0, the system is a Type Two system.
24. 24
First –Order Type ‘0’ system
The generalized transfer function for first order Type –0 system is
T(s) = C(s) / R(s) = 1/(1+sτ) --------------------------------------------------------(1)
C(s) ---- Output of the system
R(s) ----- Reference input to the system.
If input is a Step input
R(s) = 1/s ----------------------------------------------------- (2)
From eqn (1)
1
C(s) = R(s) ---------------------------------------(3)
(1+sτ)
substituting for R(s),
1 1
C(s) = ------------------------------------(4)
s (1+sτ)
To find C(t) , Take Inverse Laplace Transform of eqn (4),
C(t) = 1 – e-t/τ ------------------(5)
PROCEDURE:
1. Give the connections as per the block diagram in the process control simulator
using the front panel diagram .
2. Set the Input (set point) value using the set value knob.
3. Observe the Output (process value or PV) using CRO and plot it in the graph.
4. Tabulate the reading and calculate the % error.
5. Repeat the procedure in closed loop condition.
25. 25
TABULATION FOR FIRST ORDER SYSTEM:
(a)Type Zero system
Loop type Set Point Process Settling Error % Error
SP variable Time SP-PV SP-PV x 100%
(V) PV (s) (V) SP
(V)
Open Loop
Closed Loop
(b)Type One System
Loop type Set Process Settling Error % Error
Point variable Time SP-PV SP-PV x 100%
SP PV (s) (V) SP
(V) (V)
Open loop
Closed loop
TABULATION FOR SECOND ORDER SYSTEM
(a)Type Zero system
Loop type Set Point Process variable Settling Error % Error
SP PV Time SP-PV SP-PV x 100%
(V) (V) (s) (V) SP
Open loop
Closed loop
(b)Type One system
Loop type Set Point Process Settling Error % Error
SP variable Time SP-PV SP-PV x 100%
(V) PV (s) (V) SP
(V)
Open loop
Closed loop
27. 27
EXPT. NO:
DATE :
DIGITAL SIMULATION OF FIRST ORDER SYSTEMS
(i) Digital Simulation of first order Linear and Non Linear SISO Systems
AIM:
To digitally simulate the time response characteristics of Linear and Non
Linear SISO systems using state variable formulation.
APPARATU REQUIRED:
A PC with MATLAB package.
THEORY:
SISO linear systems can be easily defined with transfer function analysis. The
transfer function approach can be linked easily with the state variable approach.
The state model of a linear-time invariant system is given by the following
equations:
X(t) = A X(t) + B U(t) State equation
Y(t) = C X(t) + D U(t) Output equation
Where A = n x n system matrix,
B = n x m input matrix,
C= p x n output matrix and
D = p x m transmission matrix,
PROGRAM/ SIMULINK MODEL:
30. 30
(ii) Digital Simulation of Multi-Input Multi-Output Linear Systems
AIM:
To digitally simulate the time response characteristics of MIMO Linear system
using state-variable formulation.
APPARATUS REQUIRED:
• PC
• MATLAB Package.
THEORY:
State Variable approach is a more general mathematical representation of a
system, which, along with the output, yields information about the state of the system
variables at some predetermined points along the flow of signals. It is a direct time-
domain approach, which provides a basis for modern control theory and system
optimization.
u1(t) y1(t)
u2(t) Controlled system y2(t) U Controlled Y
. State variables (n) . system
. .
um(t) yp(t)
. ....... X
x1(t) x2(t) xn(t)
.
X(t) = A X(t) + B U(t) State equation
Y(t) = C X(t) + D U(t) Output equation
The state vector X determines a point (called state point) in an n - dimensional space,
called state space. The state and output equations constitute the state model of the
system.
33. 33
Expt. No.:
Date:
DIGITAL SIMULATION OF SECOND ORDER SYSTEMS
AIM:
To digitally simulate the time response characteristics of second order linear
and non-linear system with saturation and dead zone.
APPARATU REQUIRED:
A PC with MATLAB package.
PROGRAM / SIMULINK MODEL:
35. 35
Expt. No.:
Date:
STABILITY ANALYSIS OF LINEAR SYSTEMS
AIM:
To analyze the stability of linear system using Bode plot/ Root Locus / Nyquist
Plot.
APPARATUS REQUIRED:
A PC with MATLAB package.
THEORY:
A Linear Time-Invariant Systems is stable if the following two conditions of
system stability are satisfied
When the system is excited by a bounded input, the output is also
bounded.
In the absence of the input, the output tends towards zero, irrespective
of the initial conditions.
PROCEDURE:
1. Write a Program to obtain the Bode plot / Root locus / Nyquist plot for the
given system.
2. Determine the stability of given system using the plots obtained.
PROGRAM:
37. 37
EXPT. NO:
DATE :
CLOSED LOOP DC POSITION CONTROL SYSTEM
AIM:
To study the operation of closed loop position control system (DC
Servomotor) with a PI controller.
APPARATUS REQUIRED:
THEORY:
A pair of potentiometers is used to convert the input and output positions into
proportional electrical 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
stops when the error signal is zero, i.e., when the desired position is reached.
1/T1
KP Chopper Motor Gear
+
-
Position
Sensor
Fig. 1 Block Diagram
38. 38
Fig.(2.).Front Panel
PROCEDURE:
1. Switch on the system. Keep the pulse release switch in OFF position.
2. Vary the set point with the pulse release switch in the ON position and
note the output position.
3. Note SP voltage , PV voltage, P voltage and PI output voltage.
4. Calculate KP using the formula KP = P/(SP-PV).
39. 39
TABULATION
S.NO. POSITION (degrees) Error (set – output)
in degrees
set output
RESULT:
INFERENCE:
40. 40
EXPT. NO:
DATE :
CLOSED LOOP AC POSITION CONTROL SYSTEM
AIM:
To study the closed loop operation of AC position control system (AC
Servomotor) with PI controller.
APPARATUS REQUIRED:
THEORY:
CONSTRUCTIONAL DETAILS
The AC servomotor is 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 an angle of 90o in space. Each pole
pair carries a winding, one winding is called the reference winding and other winding
is called the control winding. The exciting currents in the two windings should have a
phase displacement of 90 o. The supply used to drive the motor is single-phase and
hence a phase advancing capacitor is connected to one of the phases to produce a
phase difference of 90 o. The stator constructional features of AC servomotor are
shown in fig.1.
The rotor construction is usually of squirrel cage or drag-cup type. The
squirrel cage rotor is made of laminations. The rotor bars are placed on the slots 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. 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 short-circuited rotor conductors.
WORKING PRINCIPLES
The stator windings are excited by voltages of equal rms magnitude and 90 o
phase difference. This results in exciting currents i1 and i2 displaced in phase by 90 o
and having identical 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 shown in fig.2 produces a
clockwise rotating magnetic field. When i1 is shifted by 180 o, an anticlockwise
rotating magnetic field is produced. This rotating magnetic field sweeps over the rotor
conductors. The rotor conductor experience a change in flux and so voltages are
41. 41
induced in rotor conductors. This results in circulating currents in the short-circuited
rotor conductors resulting in rotor flux.
Due to the interaction of stator & rotor flux, a mechanical force (or Torque) is
developed in the rotor and the rotor starts moving in the same direction as that of
rotating magnetic field.
Fig 1 Stator Construction of AC Servomotor
Fig 2.Waveforms of Stator & Rotor Excitation Current
Fig.3 Basic Block Diagram of AC Position Control System
42. 42
PROCEDURE:
1. Switch ON the system. Keep the pulse release switch in the OFF position.
2. Vary the set point with the pulse release switch in the ON and note the output
position.
3. Note the SP voltage, PV voltage, P voltage and PI output voltage.
4. Calculate KP using the formula KP = P/(SP-PV).
TABULATION
S.NO. SET OUTPUT ERROR=(set position-output
POSITION POSITION position)
(degrees) (degrees) (degrees)
RESULT:
INFERENCE:
43. 43
Ex. No:
Date:
STEPPER MOTOR
Aim:
To study the Stepper motor
Theory:
Stepper motors are highly accurate pulse-driven motors that change their
angular position in steps, in response to input pulses from digitally controlled
systems.
A stepper or stepping motor converts electronic pulses into proportionate
mechanical movement. Each revolution of the stepper motor's shaft is made up of
a series of discrete individual steps. A step is defined as the angular rotation
produced by the output shaft each time the motor receives a step pulse. These
types of motors are very popular in digital control circuits, such as robotics,
because they are ideally suited for receiving digital pulses for step control.
Each step causes the shaft to rotate a certain number of degrees.
A step angle represents the rotation of the output shaft caused by each step,
measured in degrees.
Figure.1. illustrates a simple application for a stepper motor. Each time the
controller receives an input signal, the paper is driven a certain incremental
distance.
Fig.1
In addition to the paper drive mechanism in a printer, stepper motors are also popular
in machine tools, process control systems, tape and disk drive systems, and
programmable controllers.
The Common Features of stepper motors are
• Brushless – Stepper motors are brushless. The commentator and brushes of
conventional motors are some of the most failure-prone components, and they
create electrical arcs that are undesirable or dangerous in some environments.
44. 44
• Load Independent – Stepper motors will turn at a set speed regardless of load
as long as the load does not exceed the torque rating for the motor.
• Open Loop Positioning – Stepper motors move in quantified increments or
steps. As long as the motor runs within its torque specification, the position of
the shaft is known at all times without the need for a feedback mechanism.
• Holding Torque – Stepper motors are able to hold the shaft stationary.
• Excellent response to start-up, stopping and reverse.
Types of Stepper Motor
1. Permanent-magnet stepper motor
The permanent-magnet stepper motor operates on the reaction between a
permanent-magnet rotor and an electromagnetic field.
Figure shows a basic two-pole PM stepper motor.
The rotor shown in Figure (a) has a permanent magnet mounted at each end.
The stator is illustrated in Figure (b). Both the stator and rotor are shown as
having teeth
Fig.2
The teeth on the rotor surface and the stator pole faces are offset so that
there will be only a limited number of rotor teeth aligning themselves with an
energized stator pole. The number of teeth on the rotor and stator determine the
step angle that will occur each time the polarity of the winding is reversed.
The greater the number of teeth, the smaller the step angle.
45. 45
Fig.3
The holding torque is defined as the amount of torque required to move the rotor
one full step with the stator energized.
An important characteristic of the PM stepper motor is that it can maintain the
holding torque indefinitely when the rotor is stopped.
Figure (a) shows a permanent magnet stepper motor with four stator windings.
By pulsing the stator coils in a desired sequence, it is possible to control the speed
and direction of the motor.
Figure (b) shows the timing diagram for the pulses required to rotate the PM
stepper motor.
2.Variable-reluctance (VR) stepper motor
The variable-reluctance (VR) stepper motor differs from the PM stepper in
that it has no permanent-magnet rotor and no residual torque to hold the rotor at
one position when turned off.
When the stator coils are energized, the rotor teeth will align with the
energized stator poles. This type of motor operates on the principle of minimizing
the reluctance along the path of the applied magnetic field. By alternating the
windings that are energized in the stator, the stator field changes, and the rotor is
moved to a new position.
The stator of a variable-reluctance stepper motor has a magnetic core
constructed with a stack of steel laminations. The rotor is made of unmagnetized
soft steel with teeth and slots.
The relationship among step angle, rotor teeth, and stator teeth is expressed using
the following equation:
46. 46
N s − Nr
ψ= × 360° ------(1)
Ns × Nr
In this circuit, the rotor is shown with fewer teeth than the stator. This ensures that
only one set of stator and rotor teeth will align at any given instant.
The stator coils are energized in groups referred to as phases.
According to above Eq., the rotor will turn 30° each time a pulse is applied.
Figure (a) shows the position of the rotor when phase A is energized. As long as
phase A is energized, the rotor will be held stationary.
Fig.4
• When phase A is switched off and phase B is energized, the rotor will turn 30°
until two poles of the rotor are aligned under the north and south poles
established by phase B.
47. 47
• By repeating this pattern, the motor will rotate in a clockwise direction. The
direction of the motor is changed by reversing the pattern of turning ON and
OFF each phase.
• The disadvantage of this design for a stepper motor is that the steps are
generally quite large (above 15°).
• Multistack stepper motors can produce smaller step sizes because the motor
is divided along its axial length into magnetically isolated sections, or stacks.
Result:
48. 48
EXPT. NO.:
DATE:
DETERMINATION OF TRANSFER FUNCTION OF SEPRATELY
EXITED DC GENERATOR
AIM:
To determine the transfer function of separately exited generator.
APPARATUS REQUIRED:
Ammeter MC (0-1A), (0-10A)
Ammeter MI (0-5A),(0-50mA)
Voltmeter MC (0-300V)
Voltmeter MI (0-300V)
Rheostat 1000 / 1A
Rheostat 50 / 5A
Auto Transformer 1¢ 230V/270V
THEORY:
The transfer function of a separately excited generator can be represented in block
diagram format as shown below
The transfer function is
IL(s)/Vf(s) = Kg /(Rf+sLf)(Rl+sLa)
Where
Vf(s)- Excitation Voltage
Rf, Lf - Field resistance & Inductance
If(s) - Field Current
Kg – Induced emf constant in V/Amp
Ll – Total load Inductance
Rl – Total load resistance
Kg can be obtained by conducting open circuit test
Rf, Lf, Ra, Lf can be found out by voltmeter- Ammeter method
52. 52
PROCEDURE:
Determination of Kg:
1. The connections are made as shown in fig.
2. DPST switch is closed
3. The motor is started with help of starter
4. The motor is brought to the rated speed by adjusting the motor field rheostat.
The drives the generated at rated speed.
5. Note down the field current If and the open circuit voltage Eo.
6. By adjusting the Rf, the field current is increased in convenient steps up to the
rated field current.
7. In each step the readings of Eo and If are noted. Throughout the experiment
the speed is maintained at constant
8. A plot of Eo Vs If is drawn by taking If on X axis and Eo on Y-axis.
9. A tangent to the linear portion of the curve is dran through the origin. The
slope of this line ,Eo Vs If gives Kg.
V-A method to obtain Ra, Rf, La & Lf
1. Give the connection as shown in fig to measure Ra & Rf and note down
the V & I
2. To measure La &Lf give the connection as shown in fig.
3. Apply an AC voltage & measure the field reactance Zf & armature
reactance Za.
4. Calculate Lf= Sqrt(Zf2 – Rf2) /2πf
5. Calculate La = Sqrt(Za2 – Ra2) /2πf
Where f= supply frequency (50Hz)
RESULT:
53. 53
EXPT. NO.:
DATE:
STUDY OF SYNCHROS
AIM:
To study the characteristics of Synchros.
APPARATUS REQUIRED:
THEORY:
A Synchro is an electro-magnetic transducer used to convert an angular
position of a shaft into an electrical signal. It is commercially known as a Selsyn or an
Autosyn. The basic element of a synchro is a synchro transmitter whose construction
is very similar to that of a 3 phase Alternator. The stator is of concentric coil type, in
which three identical coils are placed with their axis 120˚ apart, and is star connected.
The rotor is of dumb bell shaped construction and is wound with a concentric coil. AC
voltage is applied to the rotor winding through slip rings.
Fig.1 Constructional Features of Synchro Transmitter
The constructional features and schematic diagram of a synchro transmitter and
receiver is shown in fig.1. Let an AC voltage Vc(t) = Vr Sin ωt be applied to the rotor
of the synchro transmitter. The applied voltage causes a flow of a magnetizing current
in the rotor coil, which produces a sinusoidal time varying flux directed along its axis
and distributed nearly sinusoidally in the air gap along the stator periphery. Because
of transformer action, voltages are induced in each of the stator coils. As the air gap
flux is sinusoidally distributed, the flux linking any stator coil is proportional to the
cosine of the angle between the rotor and stator coil axis, and so is the voltage
induced in the stator coil. Thus, we see that synchro transmitter acts like a single-
54. 54
phase transformer in which the rotor coil is the primary and stator coil is the
secondary.
Fig .2 Schematic Diagram of Synchro Transmitter
Let Vs1, Vs2 & Vs3 be the voltage induced in the stator coils S1,S2 and S3 with respect
to the neutral. Then, for the rotor position of the synchro transmitter shown in the fig.
2 where the rotor axis makes an angle θ with the axis of the stator coil S2
Vs1 = K Vr Sinωt Cos ( θ + 120 ) ---------------------------- (1)
Vs2 = K Vr Sinωt Cos ( θ ) ----------------------------------(2)
Vs3 =K Vr Sinωt Cos ( θ + 240 ) ----------------------------(3)
The three terminal voltages of stator are
Vs1s2 = Vs1 - Vs2 =√3 KVr Sin( θ + 240 ) Sin ωt -----------------------(4)
Vs2s3 = Vs2 - Vs3 = √3 KVr Sin( θ + 120 ) Sin ωt ----------------------(5)
Vs3s1 = Vs3 - Vs1 =√3 KVr Sin(θ) Sin ωt -----------------------(6)
When θ = 0, from equations (1), (2) and (3), it is seen that the maximum voltage is
induced in the stator coil S2 , while it follows from the equation (6) from that the
terminal voltage Vs3s1 is zero . This position of the rotor is defined as the “electrical
zero” of the transmitter and is used as reference for specifying the angular position of
the rotor. The input to the synchro transmitter is the angular position of its rotor shaft
and the output is a set of 3 single-phase voltages given by equations (4) to (6). The
magnitude of these voltages is function of the shaft position.
The outputs of the synchro transmitter are applied to the stator windings of a “synchro
control transformer”. The rotor of the control transformer is cylindrical in shape so
that the air gap is practically uniform. The system acts as an error detector.
Circulating currents of the same phase but of different magnitude flow through the
two sets of stator coils. This results in the establishment of an identical flux pattern in
the gap at the control transformer as the voltage drop in resistances and leakage
reactances of the two sets of stator coils are usually small. The voltage induced in the
55. 55
control transformer rotor is proportional to the cosine of the angle between the two
rotors (φ) and is given by
E(t) = K1 Vr Cos φ Sin ωt
When φ =900, the voltage induced in the control transformer is zero. This position is
known as electrical zero position of the control transformer.
Fig. 3 Synchro Error Detector
PROCEDURE:
Tabulation 1:
1. Give connections as given in the circuit diagram.
2. Vary the input position and note the output position.
3. Plot the variation in output position with respect to the input position.
Tabulation 2:
1. Give excitation to the rotor winding.
2. Measure the output voltage across S1-S2, S2-S3 and S3-S1 of stator
windings for different rotor positions.
3. Plot the voltage Vs. angle characteristics.
57. 57
EXPT. NO:
DATE :
DESIGN OF COMPENSATOR NETWORKS
AIM:
To design a compensator network for the process given in the Process Control
Simulator.
APPARATUS REQUIRED:
THEORY:
Practical feedback control systems are often required to satisfy design
specification in the transient as well as steady state regions. This is not possible by
selecting good quality components alone (due to basic limitations and characteristics
of these components). Cascade compensation is most commonly used for this purpose
and design of compensation networks figures prominently in any course in automatic
control systems.
In general, there are two situations in which compensation is required. In the
first case the system is absolutely unstable and the compensation is required to
stabilize it as well as to achieve a specified performance. In the second case the
system is stable but the compensation is required to obtain the desired performance.
The systems which are of type 2 or higher are usually unstable. For these systems,
lead compensator is required, because the lead compensator increases the margin of
stability. For type 1 and type 0 systems stable operation is always possible. If the gain
is sufficiently reduced, in such cases, any of three components viz. Lag, Lead, Lag –
Lead must be used to obtain the desired performance. The simulation of this behavior
of the Lead – Lag Compensator can be done with the module (VLLN – OI).
An electronic Lead - lag network using Operational amplifiers is given
figure 1.
C2
C1 R2 R4
- -
R1 + R3 +
Fig.1 LEAD -LAG NETWORK USING OPERATIONAL -AMPLIFIER
58. 58
The transfer function for this circuit can be obtained as follows :
Let Z1 = R1 C1
The second op-amp acts as a sign inverter with a variable gain to compensate for
the magnitude. The transfer function of the entire system is given by
G(jω) = ( R4 R2/ R3 R1 ) (1+R1C1s) / (1+R2 C2 s)
G(jω) = ( R4 R2/ R3 R1 ) ( √1+T12ω2) / ( √1+T22ω2 )
where T1 = R1 C1 ; T2 = R2 C2
φ = angle G(jω) = - tan-1(T1ω) – tan-1(T2ω).
Thus steady state output is
For an input =X sinωt,
Yss(t) =X (R4 R2/ R3 R1) (( √1+T12ω2) / ( √1+T22ω2 ))sin(wt – tan-1 T1ω –tan-1 T2ω
)
From this expression, we find that if T1 > T2, then tan-1 T1ω –tan-1 T2ω > 0.
Thus if T1 > T2 , then the network is a LEAD NETWORK.
If T1 < T2 , then the network is a LAG NETWORK.
DETERMINATION OF VALUES FOR ANGLE COMPENSATION:
Frequency of sine wave = 20 Hz
Angle to be compensated = 70º
φ= tan-1 (2π f * T1) –tan-1 (2π f * T2)
T1 = 10, then substituting in above equation
70 - tan-1 (2 * π * 20 * 10) –tan-1 (2 * π * 20 *T2)
solving for T2
T2 = 0.003 .
Hence, the values of T1 and T2 are chosen from which values of R1 ,C1 , R2 and C2
can be determined .
For example, T1 =R1 C1 = 10 ; If C1= 1µF, then R1 = 10 M .
T2 = 0.003 = R2 C2 then C2 =1 µF, and hence R2 = 3 M .
These values produce a phase lead of 70º which is the desired compensation angle.
Nominal Value for R1 =1 M C1 = 0.1 µF
R2 =20 K C2 = 0.01 µF
59. 59
PROCEDURE:
1. Switch ON the power to the instrument.
2. Connect the individual blocks using patch chords.
3. Give a sinusoidal input as the set value .
4. Measure the amplitude and frequency of the input signal.
5. Measure the amplitude and phase shift of the output signal with respect to
the input sine wave using CRO.
6. Draw the magnitude versus frequency plot and phase versus frequency
plot.
7. Using the technique explained previously, calculate the values of R1, C1,
R2 and C2 to compensate for the phase shift of the output signal.
8. Connect the components at the points provided.
9. Now include the compensation block in the forward path before the
process using patch chords.
10. Now measure the phase shift φ of the output signal with the input and
verify for compensation.
11. Draw the magnitude versus frequency plot and phase versus frequency
plot for the designed compensator.
Table – 1 (for the process without compensation):
S.No. Input Output Gain (dB) Phase shift φ
Freq (Hz) Voltage(V) 20 log(Vo/Vin) º
(º)
60. 60
Table – I1 (for the process with compensation): Vin = V
S.No. Input Output Gain (dB) P Phase shift φ
Freq (Hz) Voltage(V) 20 log(Vo/Vin) º
(º)
A – Amplitude of input sine wave (V)
F –Frequency of the input sine wave (Hz)
Φ – Phase shift (Degrees)
RESULT :
INFERENCE:
61. 61
EXPT. NO.:
DATE:
STUDY OF P, PI, PID CONTROLLERS
AIM:
To study the P, PI, PID controller using MATLAB software .
APPARATUS REQUIRED:
THEORY:
The transient response of a practical control system often exhibits damped
oscillation before reaching steady state value. In specifying the transient response
characteristics of control systems to unit step input, it is common to specify the
following
i) Delay Time(Td)
ii) Rise time(Tr)
iii) Peal time( Tp)
iv) Max. overshoot (Mp)
v) Settling time( Ts)
Proportional control:
The output of the controller is proportional to input
U(t) = Kp e(t)
E(t) = error signal
U(t) controller out[put
Kp = proportional constant
• It amplifies the error signal and increases loop gain. Hence steady state
tracking accuracy , disturbance signal rejection and relative stability are
improved.
• Its drawbacks are low sensitivity to parameter variation and it produces
constant steady state error.
Proportional + Integral Control:
The output of the PI controller is given by
t
U(t) = Kp [ e(t) + (1/Ti )∫ e(t) dt ]
0
where Kp is the proportionality constant and Ti is called the integral time.
• This controller is also called RESET controller.
• It introduces a zero in the system and increases the order by 1.
• The type number of open loop system is increased by 1
62. 62
• It eliminates steady state error. Damping ratio remains same.
• Increase in order decreases the stability of system.
Proportional + Integral Control + Differential Control:
The output of a PID controller is given by
t
U(t) = Kp [ e(t) + (1/Ti )∫ e(t) dt + Td de(t)/dt ]
0
The PID controller introduces a zero in the system and increases the damping.
This reduces peak overshoot and reduces rise time. Due to increase in damping,
ultimately peak overshoot reduces.
The stability of the system improves.
In PID controller, all effects are combined. Proportional control stabilizes gain
but produces steady state error. Integral control eliminates error. Derivative controller
reduces rate of change of error.
TUNING OF PID CONTROLLERS
Proportional-integral-differential (PID) controllers are commonly employed in
process control industries. Hence we shall present various techniques of tuning PID
controllers to achieve certain performance index for systems dynamic response. The
technique to be adopted for determining the proportional, integral and derivative
constants of the controller depends upon the dynamic response of the plant.
In presenting the various tuning techniques we shall assume the basic control
configuration, wherein the controller input is the error between the desired output and
the actual output. This error is manipulated by the controller (PID) to produce a
command signal for the plant according to the relationship.
U(s)=Kp (1+(1/sτi)+sτd)
Where Kp= proportional gain constant
τI= integral time constant.
τd= Derivative time constant.
PROCEDURE:
1. Give the step input to the system selected and obtain the response using
CRO.
2. For the obtained response (S-shaped curve), draw a tangent at the
inflection point and find its intersection with the time axis and the line
corresponding to the steady-state value of the output.
3. Find the dead time L where the tangent cutting X- axis, and the time
constant T which is specified in model graph.
4. From the value of L and T, find the value of Kp, τI and τd settings by using
the following formulas: Kp = 1.2(T/L) , τI = 2L and τd = 0.5L.
5. Connect the unknown system in closed loop with the help of a PID
controller and substitute all those values obtained in the previous step.
63. 63
6. Simulate the system with a step input and view the response using CRO.
7. Comment on the response obtained using controller.
(I)General Block Diagram
(II)Block Diagram for P Controller
(III)OP –Amp P Controller Using Inverting Amplifier
64. 64
(IV) Block Diagram For PI Controller
(V) PI Controller Using Op-Amp
(VI) Block Diagram Of PID Controller
65. 65
Block Diagram Of Closed Loop Control Using PID Controller
M(s)
R(s) E(s) C(s)
PID(KP,Ti,Td ) Transfer Function
C(s)
RESULT:
INFERENCE:
