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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