1
DEPARTMENT OF EEE
CONTROL SYSTEMS
1. TIME RESPONSE OF SECOND ORDER SYSTEMAIM:To determine the time
response specifications of a second order system and verifying it
by using MATLAB.
APPARATUS:PC with MATLAB softwareTHEORY:
Second-order state determined systems are described in terms of
two state variables. Physical second-order system models contain
two independent energy storage elements which exchange stored
energy, and may contain additional dissipative elements; such
models are often used to represent the exchange of energy between
mass and stiffness elements in mechanical systems; between
capacitors and inductors in electrical systems, and between fluid
inheritance and capacitance elements in hydraulic systems. In
addition second-order system models are frequently used to rep-
resent the exchange of energy between two independent energy
storage elements in different energy domains coupled through a
two-port element, for example energy may be exchanged between a
mechanical mass and a fluid capacitance (tank) through a piston, or
between an electrical inductance and mechanical inertia as might
occur in an electric motor. Engineers often use second- order
system models in the preliminary stages of design in order to
establish the parameters of he energy storage and dissipation
elements required to achieve a satisfactory response. Second-order
systems have responses that depend on the dissipative elements in
the system. Some systems are oscillatory and are characterized by
decaying, growing, or continuous oscillations.
Other second order systems do not exhibit oscillations in their
responses. In this section we define a pair of parameters that are
commonly used to characterize second-order systems, and use them to
define the conditions that generate non-oscillatory, decaying or
continuous oscillatory, and growing (or unstable) responses.
When the resistance, inductance, and capacitance are connected
in series to the voltage source e and the voltage across the
capacitor is taken as output.
The mathematical equations are
e(t) = R i(t) +L di/dt+(1/C) ( i dt and eo =(1/C) ( i dt
Ei(s)/Eo(s) = (s2+(R/L) s+(1/LC))LC
Eo(s)/Ei(s) =1/(s2+(R/L) s+(1/LC))LC
Compare with characteristic equation s2+2(wns+wn2=0
wn = 1/(LC, ( = (R/2)* (C/L, = cos-1(()
Damping frequency = wd = wn(1-(2
TIME RESPONSE SPECIFICATIONS:
(i) Delay Time: It is the time taken to reach 50% of its final
value.
td = (1+0.7( )/ wn
(ii) Rise Time: It is the time taken to rise from 10% to 90% for
over damped system.
It is the time taken for the system response to rise from 0 to
100% for under damped system.
It is the time taken for the system response to rise from 5% to
95% for theoretically damped system.
td = [((-tan-1((1-(2/()]/wd
(iii) Peak Time: It is the time taken for the response to reach
peak value for the first attempt.
tp = (/wd
(iv) Settling Time: It is the time taken to reach and stay
within the tolerable limit (2-5%).
ts = 4/((wn)
(v) Peak Overshoot: It is the ratio of maximum peak value
measured to the final value.
Mp = e-((/((1-(2)THEORETICAL CALCULATIONS:
(- to be done by the student-)
PROGRAM:
NUM=[0 0 81];
DEN=[1 4 81];
SYS=TF(NUM,DEN)
STEP(SYS)
TITLE(STEP RESPONSE);
GRIDOBSERVATION TABLE:
TimeTheoretical valuesPractical Values
td (sec)
tr (sec)
tp (sec)
ts (sec)
Mp (%)
RESULT:
The time response specifications of second order system are
determined and verified using MATLAB.
VIVA VOCE:
1. What is the need to analyze the time response?
2. Define transient response.
3. Define steady state response.
4. Define steady state error.
2. DC SERVO MOTOR SPEED-TORQUE CHARACTERISTCSAIM:To study speed
torque characteristics dc servo motor.APPARATUS:
DMM 2 no s
Connecting wires
DC Servo Motor kit
CIRCUIT DIAGRAM & BLOCK DIAGRAM:
Fig.1 Circuit Diagram for DC Servomotor Speed-Torque
Characteristics
Fig2. Block Diagram for DC Servomotor Speed-Torque
CharacteristicsTHEORY:
The DC servo motors resemble a dc shunt motor turned inside out.
Dc servo motors feature permanent magnets, located on the rotor, or
a wound rotor excited by dc voltage through slip rings, requires
that the flux created by the current carrying conductors in the
stator rotate around the inside of the stator in order to achieve
servo motor action. The servo motor features a rotating field is
obtained by placing three stator windings around the interior of
the stator punching. The windings are then interconnected so that
introducing a three-phase excitation voltage to the three stator
windings (which are separated by 120 electrical degrees) produces a
rotating magnetic field. Brushless dc servo motor construction
speeds heat dissipation and reduces rotor inertia.
The DC servo motor features permanent magnet poles on the rotor,
which are attracted to the rotating poles of the opposite magnetic
polarity in the stator creating torque. As in the dc shunt motor,
the dc servo motor offers torque, which is proportional to the
strength of the permanent magnetic field and the field created by
the current carrying conductors. The magnetic field in the dc servo
motor stator rotates at a speed proportional to the frequency of
the applied voltage and the number of poles.
The rotor rotates in synchronism with the rotating field, thus
the name synchronous motor is often used to designate servo motors
of this design. More recently, this servo motor design has been
called an electrically commutated motor (ECM) due to its similarity
to the dc shunt motor. In the dc shunt motor, the flux generated by
the current carrying winding (rotor) is mechanically commutated to
stay in position with respect to the field flux. In the synchronous
dc servo motor, the flux of the current carrying winding rotates
with respect to the stator; but, like the dc motor, the current
carrying flux stays in position with respect to the field flux that
rotates with the rotor. The major difference is that the
synchronous dc servo motor maintains position by electrical
commutation, rather than mechanical commutation.
PROCEDURE:
FOR PLOTTING SPEED TORQUE CHARACTERISTICS OF DC SERVO MOTOR
1) Adjust spring balance so that there is minimum load on the
servo motor. Note that you have to pull the knob K in up ward
direction to apply load on the servo motor. You may make use of
holes to apply a fixed load in the system by using screw.
2) Ensure the pot P (speed control) is in maximum and
anticlockwise position.
3) Switch on the supply and slightly press the control knob in
anti clock wise direction so that self start relay is turned ON and
armature voltage is applied to the armature from zero onwards.
4) Connect the digital or analog millimeter across the terminal
marked armature voltage in the range o to 35 volts.5) Adjust P so
that Va = 10v and P2 so that Vf = 20v
6) Note down T1 ,T2 and speed and enter the result in the
table1
7) Keeping Va= 10v, adjust T1 up to 500 gm in suitable steps to
get a set of readings.
8) Now for Va= 15, 20v repeat step 6.
9) Plot speed torque characteristics.
10) You may repeat above steps for various values of field Vg by
controlling pot P2.OBSERVATION:
Table 1 ; Radius of pulley ; R=3.54 cms, VF=20 voltsArmature
voltage constant Va =10, 15, 20, 25 V, etc
A: VF= 20V and VA =10VPulley radius R=3.5cms S.NoT1 gmT2
gmT=(T1-T2)Torque
=T *3.5 gm-cmSpeed [RPM]Ia
[amps]
B: VF= 20V and VA =15V
S.NoT1 gmT2 gmT=(T1-T2)Torque
=T *3.5 gm-cmSpeed [RPM]Ia
[amps]
C: VF= 20V and VA =20VS.NoT1 gmT2 gmT=(T1-T2)Torque
=T *3.5 gm-cmSpeed [RPM]Ia
[amps]
EXPECTED GRAPH:
PRECAUTIONS:
1) The speed control knob should be always in the most anti
clock wise position before switching ON the equipment
2) In order to increase Va, rotate the knob in the clock wise
direction in a gentle fashion.
3) In order to increase the load on servo motor adjust the
spring balance in a care full fashion.RESULT:The speed-torque
characteristics of DC servo motor are drawn.VIVA VOCE:
1. What is meant by servo motor?
2. How it is different from DC motor?
3. Explain the advantages of DC servo motor.
4. Draw the characteristics of DC servo motor.3. TRANSFER
FUNCTION OF A DC MOTOR
AIM: To study the torque speed characteristics and determine the
transfer function of a DC motor.
APPARATUS:
1. Trainer kit of a DC motor
2. DMM meters 2 nos3. Connecting wires
CIRCUIT DIAGRAM:
THEORY:
DC motors are the most commonly used actuators in
electro-mechanical control systems or servomechanisms. Compared to
actuators like 2-phase ac motor and stepper motor has the advantage
of higher torque and simpler driving circuit. However, the presence
of a compensator and a set of brushes with the problems of sparking
make the DC motor somewhat less durable. This of course is not true
for a present day well designed DC servomotor.
The study of the dynamic characteristics of the DC motor is
important because the overall performance of the control system
depends on it. A standard analysis procedure is to model the
various subsystems and then combine these to develop the model of
the overall system.
This experiment is designed to obtain the torque-speed
characteristics, compute the various parameters, and finally
determine the transfer function of a DC motor.
PROCEDURE:
MOTOR AND GENERATOR CHARACTERISTICS:
1. Set Motor switch to ON set RESET switch to RESET set LOAD
switch to 0 position.
2. Vary Ea in small steps and take readings.
3. Plot N VS Ea and Eg VS N obtain the slopes and compute Km and
KG.
TORQUE- SPEED CHARACTERISTICS:
1. Set Motor switch to OFF set RESET switch to RESET set LOAD
switch to 0 position.
2. Connect Ea to the voltmeter and set Ea = 6V.3. Shift the
Motor switch to ON measure armature in put (Ea), motor current (Ia)
& motor speed in rpm record the readings.
4. Set the LOAD switch to 1, 2. . 5 and take readings as
above.
5. Complete the table motor voltage Ea = 6 volts; Ra = 28.
6. Plot torque VS speed cures on a graph paper.
7. Complete B from the slope of torque speed curve and average
Kb from the table.
8. Repeat above for Ea = 8v, 10v, 12v and record the average
values of motor parameters B and Kb.STEP RESPONSE:
1. Set Motor switch to OFF set RESET switch to RESET set LOAD
switch to 0 position.
2. Connect Ea to the volt meter and set it to 6V.
3. Switch ON the motor and measure Eg & the speed in rpm
these are the steady state generator voltage Eg and steady state
motor speed N respectively.
4. Set ES to 63.2% of Eg measure above this is the generator Vg
at which the counter will stop counting.
5. Switch OFF the motor set RESET switch to READY.
6. Now switch the motor ON record the counter reading as time
constant in mille seconds.
7. Repeat above with Ea = 8V, 9V and tabulate the results.
8. Substitute the values of Km and Tm in equation
Gm(s) = Km / (STm + 1) = w(s) / Ea(s).
9. Using the average values of Tm, B, Kb and Ra calculate the
motor inertia from equation I = Tm (B+Kb/Ra).OBSERVATION:
MOTOR AND GENERATOR CHARACTERISTICS:S.NoEa (volts)Ia (mA)N
(rpm)Eg (volts)
TORQUE SPEED CHARACTERISTICS:
S.NoLoad StepIa (mA)N (rpm)W= 2(N/60 (rad/sec)Eb = Ea-IaRa
(volts)Kb = Eb / wTm = KbIa (N-m)
STEP RESPONSE:
S.NoEa (volts)Eg (volts)N (rpm)Es = 0.632Eg (volts)Time constant
Tm msecGain constant Km = (N/ 30 Ea
EXPETED GRAPH:
FORMULAE USED:
Motor gain constant = Km = KT /(RaB+KTKb)Motor time constant =
Tm = Ra I/(RaB+KTKb)Steady state armature current, Ia = (Ea Eb)/ Ra
= (Ea/Ra) (KbW/Ra)
Steady state torque generated, Tm = KTIa
Tm = - KTKb / Ra(w) + KT / Ra (Ea)
Kb = Eb / W = (Ea - IaR) / W [volts/rad/sec]
Average Kb = 22.53x10-3 volts/rad/sec
B- Coefficient of viscous friction (N-m/rad/sec)
RESULT: The transfer function of a DC motor is determined and
speed-torque characteristics are drawn.VIVA VOCE:
1. Define transfer function.
2. How transfer function is different from voltage gain?
3. Explain the advantages of transfer function.
4. PID CONTROLLER
AIM:To study the performance characteristics of an analog PID
controller using simulated system.
APPARATUS:
1. PID Controller kit
2. Connecting wires
3. C R O
4. Digital voltmeter.CIRCUIT DIAGRAM:
THEORY:
A proportionalintegralderivative controller (PID controller) is
a generic control loop feedback mechanism (controller) widely used
in industrial control systems. A PID controller attempts to correct
the error between a measured process variable and a desired set
point by calculating and then outputting a corrective action that
can adjust the process accordingly.
The PID controller calculation (algorithm) involves three
separate parameters; the Proportional, the Integral and Derivative
values. The Proportional value determines the reaction to the
current error, the Integral value determines the reaction based on
the sum of recent errors, and the Derivative value determines the
reaction based on the rate at which the error has been changing.
The weighted sum of these three actions is used to adjust the
process via a control element such as the position of a control
valve or the power supply of a heating element.
By "tuning" the three constants in the PID controller algorithm,
the controller can provide control action designed for specific
process requirements. The response of the controller can be
described in terms of the responsiveness of the controller to an
error, the degree to which the controller overshoots the set point
and the degree of system oscillation. Note that the use of the PID
algorithm for control does not guarantee optimal control of the
system or system stability.
Some applications may require using only one or two modes to
provide the appropriate system control. This is achieved by setting
the gain of undesired control outputs to zero. A PID controller
will be called a PI, PD, P or I controller in the absence of the
respective control actions. PI controllers are particularly common,
since derivative action is very sensitive to measurement noise, and
the absence of an integral value may prevent the system from
reaching its target value due to the control action.
PROCEDURE:
Controller Response:
1. Apply a square wave signal of 100 mv, P-P at the in put of
the error detector connect P I and D o/p s to the summer and
display controller O/P on the CRO.
2. With P-potentiometer set to zero obtain maximum value of
P-P Square wave O/P P-P Square wave o/p
Kc = --------------------------------- =
-----------------------------
P-P square wave I/P 0.1
3. with I - potentiometer set to maximum and P, D potentiometer
to zero , a ramp will be seen on C R O .maximum value of K is then
given by
4 x f x (P-P) triangular curve O/P ramp in
K I (max) =
-------------------------------------------------------
P-P square wave amplitude in volts
Where f is the frequency of I/P
4. Set D - potentiometer to maximum and P and I potentiometers
to zero. A series of sharp pulses will be seen on C R O. this is
obviously not suitable for calibrating the D -potentiometer
applying a triangular wave at the I/P of the error detector a
square wave is seen on the C R O
P-P Square wave O/P
Kd(max) = -------------------------------------
4 x f x (P-P) Triangular wave I/P
5. Set all the three potentiometers = P, I and D to maximum
values and apply a square wave I/P of 100 mv (P-P). Observe and
trace the stop response of P I D controller, identify the effects
of P, I and D controls individually on the shape of this
response.
II. Proportional control:
1) Make connections as shown in the fig, with process made up of
time delay and time constant blocks. Notice that the C R O
operations in the X - Y mode ensures stable display even at low
frequencies.
2) Set input amplitude to 1v (P-P) and frequency to low
value.
3) For various values of Kc = 2-2, 2-4 . . . . . measure from
screen the value if peak over shoot and steady state error and
tabulate graph.
EXPETED GRAPH:
CALCULATIONS:
(a) P- control:
I/P = Square wave amp ----0.1v (p-p)
O/p = square wave amps2.0 (p-p)
O/p voltage (p-p)
Kc (max) 2.0/0.1 =20 = ---------------------------
I/p voltage (p-p)
(b) I - control:
I/p = square wave amplitude of 0.1v (p-p)
T=?
F = 1/T
O/P=Triangular wave of amplitude v (P-P)
Ki (max) = 4 x f x o/p voltage (p-p)
--------------------------------
I/P Voltage [P-P]
(c). D - Control:
Input Triangular wave of amplitude V (p-p)
Time =?
F =1/t
O/P Square wave of amplitude V(P-P)
O/P voltage (P-P)
K d (max) = ----------------------
4 x f x I/P voltage (P-P)
RESULTS:The performance characteristics of analog controller
using simulated system are drawn.
VIVA VOCE:
1. What is the need to add proportional control scheme in the
system?
2. Give the advantages of integral control over proportional
control.
3. Explain the advantages of derivative control scheme.
4. What is need have included PID controller in the system?
5. STATE SPACE MODEL FOR CLASSICAL TRANSFER FUNCTION USING
MATLAB VERIFICATION
(I) CONVERSION OF TRANSFER FUNCTIONS TO STATE SPACE MODEL:
AIM:To obtain the state space model for the given transfer
function and verifying it using MATLAB.
THEORY:
In control engineering, a state space representation is a
mathematical model of a physical system as a set of input, output
and state variables related by first-order differential equations.
To abstract from the number of inputs, outputs and states, the
variables are expressed as vectors and the differential and
algebraic equations are written in matrix form (the last one can be
done when the dynamical system is linear and time invariant). The
state space representation (also known as the "time-domain
approach") provides a convenient and compact way to model and
analyze systems with multiple inputs and outputs. With p inputs and
q outputs, we would otherwise have to write down laplace transforms
to encode all the information about a system. Unlike the frequency
domain approach, the use of the state space representation is not
limited to systems with linear components and zero initial
conditions. "State space" refers to the space whose axes are the
state variables. The state of the system can be represented as a
vector within that space.
The internal state variables are the smallest possible subset of
system variables that can represent the entire state of the system
at any given time. State variables must be linearly independent; a
state variable cannot be a linear combination of other state
variables. The minimum number of state variables required to
represent a given system, n, is usually equal to the order of the
system's defining differential equation. If the system is
represented in transfer function form, the minimum number of state
variables is equal to the order of the transfer function's
denominator after it has been reduced to a proper fraction. It is
important to understand that converting a state space realization
to a transfer function form may lose some internal information
about the system, and may provide a description of a system which
is stable, when the state-space realization is unstable at certain
points. In electric circuits, the number of state variables is
often, though not always, the same as the number of energy storage
elements in the circuit such as capacitors and
inductors.THEORITICAL CALCULATIONS:
(-to be done by the student-)
PROGRAM:
NUM = [1 3 3]
DEN = [1 2 3 1]
[A, B, C, D] = TF2SS(NUM, DEN)
OUTPUT: -2 -3 -1 1
A = 1 0 0 B = 0 C = 1 3 3 D = 0
0 1 0 0(II) CONVERSION OF STATE SPACE MODEL TO TRANSFER
FUNCTIONAIM:To obtain the transfer function for the given state
space model and Verifying it using MATLAB. -2 1 0 1
A = -3 0 1 B = 3 C = 1 0 0 D = 0
-1 0 0 3
THEORETICAL CALCULATIONS:
( - to be done by the student - )PROGRAM:
A = [-2 1 0; -3 0 1; -1 0 0]
B = [1; 3; 3]
C = [1 0 0]
D = [0]
[NUM, DEN] = SS2TF (A, B, C, D)
OUTPUT:
RESULT: The state space model of the given transfer function has
been verified using MATLAB and also verified for transfer function
the given state space model using MATLAB.VIVA VOCE:
1. What do you understand by state space model?
2. Explain the advantages of state space model over transfer
function approach.
3. Give the formula for transfer function in state space
model.
4. What do mean by state vector?
6. EFFECT OF FEEDBACK ON A GIVEN DC MOTOR
AIM:
To study the performance characteristics of a DC
motor.APPARATUS:
Trainer kit
Tachometer generator
Connecting wires
CIRCUIT DIAGRAM:
THEORY:
The direct current (DC) motor is one of the first machines
devised to convert electrical power into mechanical power.
Permanent magnet (PM) direct current converts electrical energy
into mechanical energy through the interaction of two magnetic
fields. One field is produced by a permanent magnet assembly; the
other field is produced by an electrical current flowing in the
motor windings. These two fields result in a torque which tends to
rotate the rotor. As the rotor turns, the current in the windings
is commutated to produce a continuous torque output.DC motors are
the most commonly used actuators in electro-mechanical control
systems or servomechanisms. Compared to actuators like 2-phase ac
motor and stepper motor has the advantage of higher torque and
simpler driving circuit. However, the presence of a compensator and
a set of brushes with the problems of sparking make the DC motor
somewhat less durable. This of course is not true for a present day
well designed DC servomotor.
PROCEDURE:
CLOSED LOOP PERFORMANCE:
1. Set VR = 1V and KA = 5
2. Record the speed N in rpm and the tachogenerator voltage VT
and steady state error ESS = VR - VT.
3. Repeat the above procedure for different values of KA.
4. Compare in each case, the steady state error computed using
the formula.TRANSFER FUNCTION OF MOTOR TACHO GENERATOR:
1. Set VR = 1V and KA = 3.
2. Record the speed N in rpm and the tachogenerator output
VT.
3. Repeat the same with VR = 1V and KA = 4, 5, 6, 7 . . . 10
& tabulate the measured motor voltage (VM = VRKA) steady state
motor speed N in rpm and tacho generator out put VT.
4. Plot N VS VM and VTVSN obtain KM from the linear regain of
the speed in rad/sec. WSS/motor voltage tacho generator gain
KT = VT, volt sec / Wss rad
5. Apply square wave signal and find the time constant using
formula given below.
6. Obtain the motor transfer using,
G(s) = Km / STm+1
OBSERVATION:
MOTOR AND TACHO GENERATOR CHARACTERISTICS:S.NoKA settingN
(rpm)VT (volts)Vm = VRKA (volts)Experimental
Ka = Vm / VR
CLOSED LOOP PERFORMANCE:
S.NoKA settingN (rpm)VT (volts) Ess = VR-VT (volts)
EXPECTED GRAPH:
THEORETICAL FORMULAE:
Keff = (KAKMKT) / (1+ KAKMKT)
Teff = 1 / (2f in [1-VT (p-p) / Vm (p-p)KMKT] )
Km = shaft speed (N) / motor voltage (Vm)
Where Km motor gain constant
And KT = VT / Wss voltage/rad
Where KT tacho generator gain
RESULT:
Effect of feed back on a given control system is studied.
VIVA VOCE:
1. What is meant by feed back?
2. Explain the advantages of negative feed back over the
positive feed back.
3. What happen when positive feedback is given a motor?
4. Give one example for closed and open loop control system.
7. ROOT LOCUS AND BODE PLOT FROM MATLAB(i) ROOT LOCUS
PLOT:AIM(i):To plot the Root locus for the given transfer function
and to verify it using MATLAB.
APPARATUS:PC with MATLAB software.
THEORY: ROOT LOCUS:In control theory, the root locus is the
locus of the poles and zeros of a transfer function as the system
gain K is varied on some interval. The root locus is a useful tool
for analyzing single input single output (SISO) linear dynamic
systems. A system is stable if all of its poles are in the
left-hand side of the s-plane (for continuous systems) or inside
the unit circle of the z-plane (for discrete systems).
In addition to determining the stability of the system, the root
locus can be used to identify the damping ratio and natural
frequency of a system. Where lines of constant damping ratio can be
drawn radially from the origin and lines of constant natural
frequency can be drawn as arcs whose center points coincide with
the origin. By selecting a point along the root locus that
coincides with a desired damping ratio and natural frequency a gain
can be calculated and implemented in the controller.
Suppose there is a motor with a transfer function expression
P(s), and a controller with both an adjustable gain K and a
transfer function expression C(s). A unity feedback loop is
constructed to complete this feedback system. For this system, the
overall transfer function is given by
Thus the closed-loop poles (roots of the characteristic
equation) of the transfer function are the solutions to the
equation 1+ KC(s)P(s) = 0. The principal feature of this equation
is that roots may be found wherever KCP = -1. The variability of K
(that's the gain you can choose for the controller) removes
amplitude from the equation, meaning the complex valued evaluation
of the polynomial in s K(s)C(s) needs to have net phase of 180 deg,
wherever there is a closed loop pole. We are solving a root
cracking problem using angles alone! So there is no computation
per-se, only geometry. The geometrical construction adds angle
contributions from the vectors extending from each of the poles of
KC to a prospective closed loop root (pole) and subtracts the angle
contributions from similar vectors extending from the zeros,
requiring the sum be 180. The vector formulation arises from the
fact that each polynomial term in the factored CP,(s-a) for
example, represents the vector from a which is one of the roots, to
s which is the prospective closed loop pole we are seeking. Thus
the entire polynomial is the product of these terms, and according
to vector mathematics the angles add (or subtract, for terms in the
denominator) and lengths multiply (or divide). So to test a point
for inclusion on the root locus, all you do is add the angles to
all the open loop poles and zeros. Indeed a form of protractor, the
"spirule" was once used to draw exact root loci.
From the function T(s), we can also see that the zeros of the
open loop system (CP) are also the zeros of the closed loop system.
It is important to note that the root locus only gives the location
of closed loop poles as the gain K is varied, given the open loop
transfer function. The zeros of a system can not be moved.
Using a few basic rules, the root locus method can plot the
overall shape of the path (locus) traversed by roots as the value
of K varies. The plot of the root locus then gives an idea of the
stability and dynamics of this feedback system for different values
of k.Roots of the transfer function move on the s-plane tracing a
particular path when gain is changed from 0 to (. This path is
called root locus.
Open loop transfer function =
Closed loop transfer function =
The characteristic equation is = 0
(
To make above equation true, ------(1)
------(2)
A plot satisfying (1) and (2) is the root locus. The constant
part in is called the Gain.ROOT LOCUS PLOT USING MATLAB:
The characteristic equation can be written as .
The command rlocus (num, den) gives the root locus plot.
If the system is defined in state space, root locus is obtained
by the command rlocus (A, B, C, D).
THEORETICAL CALCULATIONS:
(-to be done by the student-)
PROGRAM:
NUM = INPUT(ENTER NUMERATOR OF THE TF);
DEN = INPUT(ENTER DENOMINATOR OF THE TF);
SYS=TF (NUM, DEN)
RLOCUS (SYS)
GRIDOUTPUT:The Root locus for the given transfer function has
been obtained and verified it by using MATLAB.
(ii) BODE PLOT FROM MATLABAIM:
To obtain the Bode Plot for the given transfer function and to
verify it using MATLAB.
APPARATUS:
PC with MATLAB software.
THEORY:
A Bode magnitude plot is a graph of log magnitude versus
frequency, plotted with a log-frequency axis, to show the transfer
function or frequency response of a linear, time-invariant
system.
The Bode plot is named after Hendrik Wade Bode. It is usually a
combination of a Bode magnitude plot and Bode phase plot
The magnitude axis of the Bode plot is usually expressed as
decibels, that is, 20 times the common logarithm of the amplitude
gain. With the magnitude gain being logarithmic, Bode plots make
multiplication of magnitudes a simple matter of adding distances on
the graph (in decibels), since
A Bode phase plot is a graph of phase versus frequency, also
plotted on a log-frequency axis, usually used in conjunction with
the magnitude plot, to evaluate how much a frequency will be
phase-shifted. For example a signal described by: Asin(t) may be
attenuated but also phase-shifted. If the system attenuates it by a
factor x and phase shifts it by the signal out of the system will
be (A/x)sin(t). The phase shift is generally a function of
frequency.
Phase can also be added directly from the graphical values, a
fact that is mathematically clear when phase is seen as the
imaginary part of the complex logarithm of a complex gain.BODE PLOT
USING MATLAB:
A stable linear system subjected to a sinusoidal input gives
sinusoidal output of the same frequency after steady state
conditions are reached. However, the magnitude and phase angle
change. The output magnitude and phase depends on the input
frequency. Bode plot give this relation in a graphical way. It can
be proved that if s is replaced by jw, the transfer function gives
steady state response to sinusoidal inputs where w is the angular
frequency. The command bode (num, den) produces the bode plot.
The command (mag, phase, w) =bode (num, den, w) can be used for
specified frequency points contained in w-vector. Result is stored
in magnitude and phase matrices. The command mag dB=20*log (mag)
produces magnitude in dB.
The command log space (d1, d2) generates 50 points between 10d1
and 10d2, w=log space (1, 2) generates 50 points between 10-1 and
102 i.e., and 100 rad/sec. but if we have to generate 100 points
use the command, w=log space (-1, 2, 100).
THEORETICAL CALCULATIONS:
(- to be done by the student-)PROGRAM:
NUM = INPUT(ENTER NUMERATOR OF THE TF);
DEN = INPUT(ENTER DENOMINATOR OF THE TF);
SYS=TF (NUM, DEN);
BODE (SYS)
GRID
OUTPUT: The Bode plot for the given transfer function has been
obtained and verified it by using MATLAB.
RESULT:For the given transfer function, root locus plot and bode
plot are drawn and verified by using MATLAB.VIVA VOCE:1. Define
root locus plot.
2. Give the advantages of root locus.
3. Is root locus plot drawn on open loop or closed loop
system?
4. Give the advantages of bode plot over Nyquist plot.
5. Define gain cross over frequency.
6. Define phase cross over frequency.
7. Define gain margin and phase margin.
8. TEMPERATURE CONTROL SYSTEM
AIM:
To study the performance of various types of controller used to
control the temperature of an oven.APPARATUS:
Temperature control unit
Techno meter
- 1
Stop clock
- 1
CIRCUIT DIAGRAM:
THEORY:
Temperature control system is one of the most common industrial
control systems. Here the plant to be controlled is an electric
oven. As we know, characteristics and performance of many devices
change with a change in temperature making them difficult to use in
a particular operation. The change in temperature is caused by a
change in environment. To hold characteristics constant in a
changing environment we must supply or remove heat to compensate
for variations in ambient temperature. This is accomplished with
temperature controllers.
Most installations of temperature controllers supply heat, or
remove heat (chill), to hold the temperature at a constant point
somewhat above or below the ambient temperature. Electronic
temperature controllers are most often used to vary the supply of
an electric current through a resistance heater to accomplish this
when the controlled temperature is to be above ambient.The
controlled device or material can also be stabilized at some
temperature below environment by controlling the flow of a
refrigerant through a heat exchanger. Yet another type of low
temperature control system (called buck and boost) supplies cooling
to drop the temperature below the desired set point and then
controls the temperature by supplying heat via a controller to get
the exact temperature setting. This type of operation is needed
when the desired set point is close to the ambient temperature. In
an ideal world, once we set the temperature of an area or device,
the temperature would remain the same over any length of time.
Unfortunately we do not live in an ideal world. If one were to
observe the temperature of a controlled item over a period of time
it would be rare to always find that item at the exact target (set
point) temperature. Temperature would vary above and below the set
point most of the time. What we are concerned about, therefore, is
the amount of variation. One of the newer temperature controller
designs uses a sophisticated means of reducing this variation. This
controller is known as a PID controller.
In order to understand the operation of a PID
(Proportional-Integral-Differential) controller, we should review a
few basic definitions.
Derivative - is a value which expresses the rate of change of
another value. For instance, the derivative of distance is
speed.
Integral - is the opposite of a derivative. The integral of
acceleration is velocity and the integral of velocity is
distance.
Proportional - means a value varying relative to another value.
The output of a proportional controller is relative to (or a
function of) the difference between the temperature being
controlled and the set point. The controller will be full on at
some temperature which is well below the set point (or desired
temperature). It will be full off at some point above the set
point.
For a given constant power condition, heat loss through
insulation will cause the actual temperature to be slightly less
than it would be in a well insulated heated area. This difference
is the "I" in PID. It can be manually corrected by changing the
position of the proportional band center point (called offset) so
the result is the temperature you want to hold. The problem is that
if the heat loss conditions change and the system begins to lose
heat faster, then that changes the offset and you may not be there
to manually correct it. To compensate for this, we monitor the
change of that temperature point by watching the change in
temperature of the sensor. We then take the derivative of that
change (get a value for the rate of change in temperature - the "D"
in PID ) which is then added to the Integral value to make an
automatic correction. Basic control actions commonly used in
temperature control systems are:
a. on-off controller b. Proportional controller
c. Proportional-Integral controller
d. Proportional-Integral-Derivative controller
PROCEDURE:
I. OPEN LOOP TESTING:
1. Keep switch S1 to WAIT, S2 to SET and open FEED BACK.
2. Connect potentiometer (p) output to driver i/p and switch on
the unit.
3. Set potentiometer P to 0.5 which gives Kp = 10 adjust
reference potentiometer to read 5 on the dmm.
4. Put switch S2 to the measure position and note down the room
temperature.
5. Put switch S1 to run position and note down the room
temperature readings every 30 seconds till the temperature becomes
almost constant.
6. Plot temperature time curve on a graph paper calculate T1 and
T2 hence write the transfer function of the oven including its
driver as
G(s) = Ke (ST2) / (1+ST1) with T in 0C.
II. P CONTROLER:
Kp for P controller is a Kp = T1 / (K T2)
1. starting with cool oven, keep switch S1 to WAIT position
& connect P to output to the driver i/p keep R, D, and I o/ps
disconnected short FEED BACK terminals.
2. Set up potentiometer to the above calculated value of Kp
keeping in mind that maximum gain is 10.
3. Plot the observation on a linear graph paper and observe the
rise time, study state error and output overshoot.
III. P I CONTROLER:
1. Starting with cool oven, keep switch S1 to WAIT position
& connect P & I output to the driver i/p and disconnect R,
D. o/ps short FEED BACK terminals.
2. Set P and I potentiometer to the above values of KP and K
respectively select and set the desired temperature to say 60 keep
switch S2 to RUN position and record temperature plot the
observation on graph.
3. Starting with a cool oven, keep switch S1 to WAIT position
& connect P, I, and D o/ps to driver i/p keep R output
disconnected short feed back terminals.4. Set P, I & D
potentiometer according to calculated values.IV. P ID
CONTROLER:
1. Starting with cool oven, keep switch S1 to WAIT position
& connect P & I output to the driver i/p and disconnect R,
D. o/ps short FEED BACK terminals.
2. Set P, I, D according to the above calculated values of KP,
KI (or) KD keeping in mind that there is a maximum value are 20,
0.0245 and 23.5 respectively.
3. Select & set the desired temperature time readings.
4. Plot the response on a graph paper and observe Tr Steady
state error and percentage over shoots.
OBSERVATIONS:
P CONTROLER:S.NoTIMETEMPERATURE
P I CONTROLER:S.NoTIMETEMPERATURE
PID CONTROLER:S.NoTIMETEMPERATURE
EXPETED GRAPH:
RESULT:
The performance of various types of controllers i.e., P, PI and
PID controllers to control the temperature of an oven are
studied.
VIVA VOCE:1. Define control system.
2. Define open loop control system.
3. Define open loop control system.
4. Is temperature control system open loop are closed loop
control system?
9. CONVERSION OF STATE SPACE MODEL TO NYQUIST PLOTAIM: To obtain
the Nyquist plot from the given state model and to verify it using
MATLAB. 0 1 0
A = -3 -4 B = 1 C = 10 0 D = 0
THEORY:A Nyquist plot is used in automatic control and signal
processing for assessing the stability of a system with feedback.
It is represented by a graph in polar coordinates in which the gain
and phase of a frequency response are plotted. The plot of these
phasor quantities shows the phase as the angle and the magnitude as
the distance from the origin. This plot combines the two types of
Bode plot magnitude and phase on a single graph, with frequency as
a parameter along the curve. The Nyquist plot is named after Harry
Nyquist, a former engineer at Bell Laboratories. The high frequency
response is at the origin. The plot provides information on the
poles and zeros of the transfer function (eg. from the angle at
which the curve approaches the origin).
Assessment of the stability of a closed-loop negative feedback
system is done by applying the Nyquist stability criterion to the
Nyquist plot of the open-loop system (i.e. the same system without
its feedback loop). This method is easily applicable even for
systems with delays which may appear difficult to analyze by means
of other methods. Nyquist and related plots are classic methods of
assessing stability, but have been supplemented or supplanted by
computer-based mathematical tools in recent years. Such plots
remain a convenient method for an engineer to get an intuitive feel
for a circuit.
THEORETICAL CALCULATIONS:
( - to be done by the student - )
PROGRAM:
A = [0 1;-3 -4]
B = [0;1]
C = [10 0]
D = [0]
[NUM, DEN] = SS2TF (A, B, C, D)
NYQUIST (TF (NUM, DEN))
TITLE (NYQUIST PLOT);
GRID
RESULT:Nyquist plot for the given state model has been obtained
and verified it using MATLAB.VIVA VOCE:
1. Define the term state.
2. Define the term state variable.
3. Give the statement on Nyquist stability criterion?
4. Give advantage of Nyquist plot over bode plot.
10. STUDY OF SYNCHRO
AIM:
(1) To obtain the stator voltages corresponding to the given
rotor positions for the given Synchro transmitter.
(2) To connect the given Synchro transmitter pair as suggested
and study it as an error detector.
APPARATUS:
Synchro transmitter and receiver pair kit
1-phase power supply (0-100V)
Volt meter (0-100V)
Connecting leads
THEORY:
The most important unit in a modern transmission system is the
synchro. Synchros of different types transmit, receive, or combine
signals among stations which may be widely separated; for example,
they transmit gun order signals from a computer to the automatic
control equipment at a gun mount. The simplest types of synchro
units are the synchro transmitter (sometimes called synchro
generator) and the synchro receiver (sometimes called synchro
motor). The transmitter is a device that transmits an electrical
signal corresponding to the angle of rotation of its shaft. The
receiver is a device that, when it receives such a signal, causes
its own shaft (if not appreciably loaded) to rotate to an angle
corresponding to the signal.
The transmitter and receiver are identical in construction
except that the motor has a damper (not illustrated) -a device that
keeps it from running away when there are violent changes in its
electrical input.
To understand how a synchro functions, think of it for the
moment as a transformer in which the primary and secondary are
wound on separate cores. When a current flows in the primary, it
forms a magnetic field in its core. As the current changes and
reverses (which it does constantly, being an alternating current)
so does the magnetic field. The changes in the field induce current
in the secondary (whose circuit is closed through a load). The
currents in the secondary produce their own magnetic field. At any
instant, the induced or secondary field opposes in direction that
produced by the primary.
Now consider a synchro transmitter connected to a receiver as in
fig, so that the rotors are fed by the same AC line and the stator
coils of the receiver load the corresponding coils of the
transmitter. The currents induced in the transmitter stator
flowalso in the receiver, and produce the resultant stator fields
shown by the white arrows. Thus the receiver rotor, which produces
a magnetic field similar to that of the transmitter rotor (because
it is excited by the same AC line), always, because it is free to
rotate, assumes exactly the same angular position (relative to the
stator) as does the transmitter rotor. When the transmitter rotor
is turned-say 30 degrees, as in fig, the resultant field produced
by the stator turns too, as it did in fig, so does the receiver
stator field.
Fig 1.a Schematic diagram of synchro transmitter
Fig 1.b Stator voltages
Synchro control transformer has a wound rotor coil and three
stator coils, but the internal construction is different. The rotor
is round instead of bobbin-shaped (to keep it from tending to line
up with a magnetic field as a receiver rotor does) and is wound
with finer wire to increase electrical impedance and limit the
amount of current it will carry. The synchro control transformer
has 2 inputs, 1 mechanical (its rotor is driven by the mechanism or
load whose position it regulates) and the other electrical (the
synchro signal from the transmitter which is to control the load).
The electrical (synchro) 3-wire input goes into the control
transformers stator. The stators field acts as the primary of the
transformer; the rotor is its secondary. The output thus comes from
the rotor and varies with its position with respect to the stator.
This output is not a synchro signal; it is a voltage whose value
and polarity with respect to the AC supply depend on the position
of the control transformers rotor with respect to the stator.
Fig.3. Synchro pair as error detector
PROCEDURE (SYNCHRO TRANSMITTER):1) Connect the system to main
supply
2) Switch on the main switch and SW1 3) Do not connect any wires
between the stator winding of TX & TR.
4) Starting from zero position (i.e. Knob of synchro transmitter
at 0 degree) note- down the voltages between stator terminals i.e.
ES1S2, ES2S3, & ES3S1 in a sequential fashion. 5) Rotate the
knob by 30 degree and note down ES1S2, ES2S3, and ES3S1.6) Repeat
step no.5 for further rotation by 30 degree.OBSERVATION TABLE 1:
-S.No.Rotor angle inDegreesStator voltages
10Es1S2ES2S3ES3S1
230
360
4.
5
6.
7.
..
13360
PROCEDURE (SYNCHRO PAIR): 1) Connect the system to main
supply
2) With the help of patch card establish connections between
corresponding terminals of TX & TR stators. i.e., connect S1 to
S1 , S2to S2 and S3 to S3 of TX & TR stator respectively .
3) Switch on SW1 & SW2 as well as main of instrument.
4) Move the pointer (rotor position of TX in steps of 30 degrees
and observe the new position of TR. 5) Enter the input angular
position and output angular position in the tabular column and plot
the graph.OBSERAVTION TABLE 2: -S.No.Rotation of rotor of Synchro
Transmitter, (tRotation of rotor of Synchro Receiver,(r
10
230
.
.
.330
13.
PRECAUTIONS:
1. The connections should be tight and clean.
2. The multimeter should be used carefully.
3. The supply to the rotor of synchro transmitter should not
exceed 30Volts.
RESULT:
1) The graph of ES1S2, ES2S3, and ES3S1 has been plotted for
various positions of rotor.
It is observed that the voltages are displaced by 120 degrees.
2) The rotor of receiver follows the position of rotor of
transmitter.VIVA-VOCE:
1. What are the trade names for synchro?
2. What is electrical zero position of transformer?3. What is
null position of control transformer?4. Give one application for
synchro.11. COMPENSATION DESIGNOBJECTIVE:To study the effects of
different cascade compensation networks for a given
system.APPARATUS:
Trainer kit
Connecting wires
CROCIRCUIT DIAGRAM:
THEORY:
Practical feedback control systems are often required to satisfy
design specifications in the transient as well as steady state
regions. This is usually not possible by selecting good quality
components alone, due to basic physical limitations and
characteristics of these components. Cascade compensation is most
commonly used for this purpose and the design of compensation
networks figures prominently in any course on automatic control
systems. Due to the absence of any laboratory experience, however,
the concepts of compensation remain rather vague. This unit has
been designed to enable the students to go through the complete
design procedure and finally verify the performance improvements
provided by compensation.
A simulated second order system with variable gain is taken as
the unsatisfactory system. Simulated system has the advantage of
predictable performance which is necessary if the verification of
the results is to be meaningful. Built-in variable frequency square
wave and sine wave generators are provided for time domain and
frequency domain testing of the system. The frequency may be varied
in the range 25Hz 800Hz and its value read on a built-in frequency
meter on the panel. Although most practical control systems have
bandwidth up to a few Hz only, a higher bandwidth has been chosen
for the simulated system to facilitate viewing on a CRO. A
pre-wired amplifier makes the implementation of the compensation
network extremely simple. Only a few passive components need
plugging into the circuit. Lead and lag networks may be designed
and tested on the set-up using both frequency domain and s-plane
procedures.
The experimental set-up is accompanied by the supporting
literature which becomes of vital importance as a major part of the
experiment involves theoretical design of compensation networks.
Although a complete coverage of design philosophy is not feasible
in this document, all efforts have been made to describe the
salient features and design steps of the four problems listed
above. Also included is a typical design, explicitly covered with
compensation network parameter calculation and final results.1.
PROCEDURE:BODE PLOT OF THE PLANT:2. Disconnect the compensation
terminals and apply an input, say 1Vp-p, to the plant from the
built in sine wave source.3. From the low frequency end of the
magnitude plot, obtain the error coefficient and the steady state
error.4. Calculate the forward path gain K necessary to meet the
steady state error specifications.5. Set the above value of K,
short the compensation terminals and observe the step response of
the closed loop system. Compute the time domain performance
specifications.1. Shift the magnitude by 20 log10(K) and obtain the
value of phase margin. Compare with the given specifications of
phase margin.LAG NETWORK DESIGN:2. From the bode plot, find a
frequency where PMactual=PMspecified+a safety margin (50-100). This
is new gain cross over frequency wg,new.3. Measure gain at wg,new.
This must equal the high frequency attenuation of the lag network,
(20log(). Compute (.4. Choose Zc=1/T, at approx. 0.1 wg,new and Pc=
1/(T. accordingly.5. Write the transfer function Gc(s) and
calculate R1, R2 and C.6. Implement Gc(s) with the help of the few
passive components and and the amplifier provided for this purpose.
The gain of the amplifier must be set at unity.7. Insert the
compensator and determine experimentally the phase margin of the
plant.8. Observe the step response of the compensated system.
Obtain the values of Mp, tp, ess and (.
LEAD NETWORK DESIGN: 1. From the bode plot, calculate the
required phase lead as phase lead needed ((m ) = PMspecified -
PMavailable + safety margin (50-100).2. calculate ( for the lead
network from ( = (1 - sin(m) / (1 + sin(m)3. calculate new gain
cross over frequency wg,new such that |G|wg,new = 10 log (4. this
step ensures that maximum phase lead shall be added at the new gain
cross over frequency.
5. the corner frequencies are now calculated from 1/T = ( wm and
1/(T = wm/ (. 6. Implement Gc(s) with the help of the few passive
components and the amplifier provided for this purpose. The gain of
the amplifier is to be set equal to 1/ .7. Insert the compensator
and determine experimentally the phase margin of the plant with
compensator.
8. Observe the step response of the compensated system. Obtain
the values of Mp, tp, ess and (.OBSERVATION:
Frequency response measurements: fHz
ABx0y0Gain dBPhase in Degrees
Component values for implementationR1 = 19.64 k( ( 20 k(R2 =
9.09 k( (9.1 k(C1=1(FRESULT: The effects of different cascade
compensation networks for a given system are studied.VIVA VOCE:
1. What do mean by compensation?2. What is the purpose of lag
compensator?3. What is the purpose of lead compensator?
4. Discuss the comparison of lag, lead and lead-lag
networks.
TRR COLLEGE OG ENGINEERING
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