Lab Manual New CS

VELTECH UNIVERSITY U5EEA22 CONTROL SYSTEMS LAB MANUAL

EXP NO 1 DETERMINATION OF TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVO MOTOR

AIM: To determine the transfer function of armature controlled DC servo motor.

APPARATUS / INSTRUMENTS REQUIRED:

DC Servo Motor Trainer Kit and connecting wires

THEORY:

In servo applications a DC motor is required to produce rapid accelerations from standstill. Therefore the physical requirements of such a motor are low inertia and high starting torque. Low inertia is attained with reduced armature diameter with a consequent increase in the armature length such that the desired power output is achieved. Thus, except for minor differences in constructional features a DC servomotor is essentially an ordinary DC motor. A DC servomotor is a torque transducer which converts electrical energy into mechanical energy. It is basically a separately excited type DC motor. The torque developed on the motor shaft is directly proportional to the field flux and armature current, Tm = Km Φ Ia. The back emf developed by the motor is Eb = Kb Φ ωm.. In an armature controlled DC Servo motor, the field winding is supplied with constant current hence the flux remains constant. Therefore these motors are also called as constant magnetic flux motors. Armature control scheme is suitable for large size motors.

ARMATURE CONTROLLED DC SERVOMOTOR:

FORMULAE USED:

Transfer function of the armature controlled DC servomotor is given as

θ(s) / Va(s) = Km / [s (1+sτa)(1+sτm ) + (Kb Kt /RaB)]where

Motor gain constant, Km = (Kt/RaB)

Motor torque constant, Kt = T / Ia

Torque, T in Nm = 9.55 Eb Ia Back emf, Eb in volts = Va – Ia Ra Va = Excitation voltage in volts Back emf constant, Kb = Va / ω

1 | P a g e U 5 E E A 2 2 C O N T R O L S Y S T E M S L A B M A N U A L


Angular velocity in rad/ sec = 2πN / 60

Armature time constant, τa = La / Ra

Mechanical time constant, τm = J / B

Ra =10 ΩLa =0.8 mHJ=0.02gm=cm2

PROCEDURE:

1. To determine torque:

Remove the load and Switch on the module. Note down the no load current and no load speed Adjust the potentiometer for the rated voltage of 24 V Adjust the load in steps till the current of 0.8 amps is obtained At each load note down the speed. Calculate the corresponding torque and plot the torque speed characteristics. Repeat the procedure for 60% and 40% of the rated voltage

OBSERVATIONS:

S. No.

Armature Voltage,Va

(V)

Armature Current,Ia

(A)

Speed,N(rpm)

Load W (in gram)

TorqueT=WxR

gram-cm)

W=W2-W1

R=Radius of the pulley-2.5 Cm



SPEED Vs TORQUE CHARACTERISTICS

Speed in RPM

Rated Voltage (24 V) 60%Rated Voltage 40% Rated Voltage

Torque in gm-am

2. To Determine Kt and B

Switch on the mains of the supply unit and switch on the DC power supply to the motor.

Run the motor at some nominal speed , Progressively increasing loads upto 80% of the rated maximum current.

Determine the following in the process.

SI No Armature Current(A) Ia

Speed (N)in RPM

Angular velocityω=2πN/60

Torque T=W*R in gm-cm

Torque constant Kt=T/Ia

Friction Coefficient B=T/ ω in gm-cm/(rad/s)

To determine Kb

SI No Armature voltage Va

Armature Current(A) Ia

Speed (N)in RPM

Angular velocityω=2πN/60

Back EMF Eb=Va-Ia.Ra

Back EMF constant Kb=Eb/ω

Result



This the Transfer function of DC servo motor is obtained experimentally

VIVA-VOCE QUESTIONS:

1. Define transfer function. 2. What is DC servo motor? State the main parts. 3. What is servo mechanism? 4. Is this a closed loop or open loop system. Explain? 5. What is back EMF?

EXP NO 2 DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF



FIELD CONTROLLED DC SERVO MOTOR

AIM: To determine the transfer function of field controlled DC servo motor.


THEORY:

In a field controlled DC Servo motor, the electrical signal is externally applied to the field winding. The armature current is kept constant. In a control system, a controller generates the error signal by comparing the actual o/p with the reference i/p. Such an error signal is no enough to drive the DC motor. Hence it is amplified by the servo amplifier and applied to the field winding. With the help of constant current source, the armature current is maintained constant. When there is change in voltage applied to the field winding, the current through the field winding changes. This changes the flux produced by field winding. This motor has large Lf / Rf ratio, so time constant of this motor is high and it can’t give rapid responses to the quick changing control signals.

FIELD CONTROLLED MOTOR:

FORMULAE USED:

Transfer function of field controlled DC servo motor is given as,

(s) / Vf (s) = Km / s (1+sTf) (1+sTm)where

Motor gain constant Km = Ktf / Rf B

Motor torque constant Ktf in N-m / A = T / If

Torque T in N-m = 9.55 Eb Ia / N Back EMF Eb in volts = Va – Ia Ra

Va = Excitation voltage in voltsArmature resistance,Ra in = Va1 / Ia1

Field resistance,Rf in = Vf1 / If1

Field time constant Tf = Lf / Rf



Field Inductance,Lf in H= XLf / 2f XLf in = (Zf

2 – Rf2)

Zf in = Vf2 / If2

Mechanical time constant Tm = J / B

Moment of inertia J in Kg m2 / rad = W x (60 / 2 ) 2 x dt/dN N

Stray loss, W in watts = W’ x [ t2 / (t1-t2) ]Power absorbed, W’ in Watts = Va Ia

t2 is time taken on load in secst1 is time taken on no load in secsdt is change in time on no load in secsdN is change in speed on no load is rpmN is rated speed in rpm

Frictional co-efficient, B in N-m / (rad / sec ) = W’’ / (2N / 60 )2

W’’ = 30 % of Constant lossConstant loss = No load i/p – Copper lossNo load I/P = V ( Ia + If )Copper loss = Ia

2 Ra

N is rated speed in rpm

PROCEDURE:

1. To determine the motor torque constant Ktf :

Check whether the MCB is in OFF position in the DC servomotor trainer kit Press the reset button to reset the over speed. Patch the circuit as per the patching diagram. Put the selection button of the trainer kit in the field control mode. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220V to the armature of the servomotor. Note the values of the armature current Ia, armature voltage Va, and speed N. Find the motor torque constant Kt f using the above values.

Note:If the voltmeter and ammeter in the trainer kit is found not working external metersof suitable range can be used.

OBSERVATIONS:



S. No. Armature Voltage,Va

(V)Armature Current,Ia

(A)Speed,N

(rpm)

PROCEDURE:

2. To determine armature resistance Ra:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode. The field terminal is left opened. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220V to the armature of the servomotor. Note the values of the armature current Ia, armature voltage Va. Find the value of armature resistance Ra using the above values


OBSERVATIONS:

S. No. Armature Voltage, Va1

(V)Armature Current, Ia1

(A)Armature Resistance,

Ra ()

PROCEDURE:

3. To determine field resistance Rf:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the field control mode. The armature terminal is left opened. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and apply rated voltage of 220V to the field of the servomotor. Note the values of the field current If, field voltage Vf. Find the value of field resistance Rf using the above values




OBSERVATIONS:

S. No. Field Voltage, Va1

(V)Field Current, Ia1

(A)Field Resistance,

Rf ()

PROCEDURE:

4. To determine Field Inductance, Lf

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the field control mode. The armature terminal is left opened. Switch ON the MCB. Note the values of the field current If2, field voltageVf2. Find the value of field inductance Lf.using the above values


OBSERVATIONS:

S. No. Field Voltage, Vf2

(V)Field Current, If2

(mA)Field Impedance

Zf ()

PROCEDURE:

5. To determine moment of inertia J and frictional co-efficient B:

Check whether the MCB is in OFF position in the DC servomotor trainer kit Patch the circuit as per the patching diagram Put the selection button of the trainer kit in the armature control mode and the DPDT

switch in power circuit position. Check the position of the potentiometer; let it initially be in minimum position. Switch ON the MCB. Vary the pot and adjust the motor to run at rated speed. Note the values of armature current Ia, armature voltage Va, field current If, Speed N. Change the DPDT switch position from power circuit side to load side,

simultaneously noting the time taken t1 of the motor to come to rest from rated speed, using a stop watch.



Set the potentiometer to minimum position and change the DPDT switch to power circuit side

Connect a load of 500 Ohms in the load position Vary the pot and adjust the motor to run at rated speed Change the DPDT switch position from power circuit side to load side,

simultaneously noting the time taken t2 of the motor to come to rest from rated speed, using a stop watch.

Find the values of moment of inertia J and frictional co-efficient B using the above values

OBSERVATIONS:

S. NoArmature

Voltage, Va

(V)

Armature Current,

Ia

(A)

Field Current, If

(A)

Speed, N

(rpm)

t1

(secs)

t2

(secs)

RESULT:

The transfer function of field controlled DC servomotor is determined as


1. What are the main parts of a DC servo motor?2. Name the two types of servo motor.3. State the advantages and disadvantages of a DC servo motor.4. Give the applications of DC servomotor.5. What is servo mechanism?6. What do you mean by field controlled DC servo motor

EXP NO 3 DETERMINATION OF TRANSFER FUNCTION OF AC SERVO MOTORAIM:To derive the transfer function of the given AC Servomotor



Apparatus Required

AC Servo motor Trainer Kit , Patch chords

THEORY:

An AC Servomotor is basically a two phase induction motor except for certain spcial design features. A two phase induction motor consisting of teo stator windings oriented 90 degrees electrically apart in space excited by AC volatages wehich differ in time phase by 90 degrees. Generally voltages of equal magnitude and 90 degrees phase difference is applied to the two stator phases thus making their respective fields 90 degrees apart in both time and space at the synchronous speed. As there field sweeps over the rotor, voltages are induced in it producing current in the short circuited rotor. The rotating magnetic firld interacts with these currents producing a torque on the rotor in the direction of the field rotation.

The shape of the characteristics depends upon the ratio of the rotor reactance (X) to the rotor resistance(R) On normal Induction motors X/R ratio is generally kept high so as to obtain the maximum torque close to the opeearting region that is 5% slip.

GENERAL SCHEMATIC OF AC SERVOMOTOR:

FORMULA USED:Transfer Function of AC Servo motor is given by Θ(s)/Ec(s) =Km/s(τms+1)Where Km= Motor Gain Constant- K1/(K2 +B)τm= Motor Time Constant-J/(K2+B)

PROCEDURE: Plot the speed torque characteristics. From the graph determine K2. Plot Control winding voltage vs torque characteristics for constant speed. From the graph determine K1. Determine Km and τm by using values B and J given. Substitute Km and τm in the equation of the transfer function.

Determination of Motor Constant K2

Control Voltage = 230 V (rated Voltage)



Si No Speed (RPM) Torque (gm-cm)

Control Voltage = 200 V (rated Voltage)Si No Speed (RPM) Torque (gm-cm)

Graph of Torque Vs Speed

Torque(gm-cm) Slope = K2

Speed(RPM)

To Detrmine the motor Constant K1

Apply the rated voltage 230 V to the control winding. Apply load on the motor gradually till the motor will run say 1300 rpm (N1). Note down I and calculate Torque. Decrease the load slightly till the motor runs at say 1600 rpm. Reduce Vc slightly so that the motor comed to N1 rpm Note down I and calculate the torque. Repeat the procedure for N2 say 1000 rpm.

Graph of Torque Vs Vc

N1

Torque N2



Slope = K1

Result:

Thus the Transfer Function of AC servo motor is obtained Experimentally.


1. What are the main parts of an AC servomotor?2. State the advantages and disadvantages of an AC servo motor.3. Give the applications of AC servomotor.4. What do you mean by servo mechanism?5. What are the characteristics of an AC servomotor?

Exp No 4 DIGITAL SIMULATION OF FIRST ORDER SYSTEMS

AIM

To digitally simulate the time response characteristics of a linear system without non- linearities and to verify it manually.



APPARATUS REQUIRED:

A PC with MATLAB packageTHEORY::

The time response characteristics of control systems are specified in terms of time domain specifications. Systems with energy storage elements cannot respond instantaneously and will exhibit transient responses, whenever they are subjected to inputs or disturbances.

The desired performance characteristics of a system of any order may be specified in terms of transient response to a unit step input signal. The transient response characteristics of a control system to a unit step input is specified in terms of the following time domain specifications

Delay time td

Rise time tr Peak time tp

Maximum peak overshoot Mp

Settling time ts



PROGRAM:

num = [0 1]den = [1 1]Step (num, den)

FORMULA USED:

1. Rise time, tr = Π-θ/ωd

2. Peak time, tp = Π/ωd 3. Settling time, ts = 4/ξωn

SIMULINK MODEL:



PROCEDURE:

1. Derive the transfer function of a RL series circuit.2. Assume R= 1 Ohms L = 0. 1 H. Find the step response theoretically and plot it on a

graph sheet.3. To build a SIMULINK model to obtain step response / sine response of a first order

system, the following procedure is followed:1. In MATLAB software open a new model in SIMULINK library browser.2. From the continuous block in the library drag the transfer function block.3. From the source block in the library drag the step input/ sine input.4. From the sink block in the library drag the scope.5. From the math operations block in the library drag the summing point.6. Connect all to form a system and give unity feedback to the system.7. For changing the parameters of the blocks connected double click the respective

block. 8. Start simulation and observe the results in scope. (Use a mux from the signal

routing block to view more than one graph in the scope) 9. Compare the simulated and theoretical results.

MODEL GRAPH

RESULT:

The time response characteristics of a first order system is simulated digitally and verified manually.




1. What is MATLAB?2. What is the use of MATLAB Package?3. What are the toolboxes available in MATLAB?4. What is the use of a simulation?5. Differentiate real time systems and simulated systems.6. Give two examples for first order system.7. Name the standard test signals used in control system.8. What is time response?

Exp No 5 DIGITAL SIMULATION OF SECOND ORDER SYSTEMAIM:



To simulate the time response of second order system using MATLAB

APPARATUS REQUIRED

A PC with MATLAB Software

THEORY

The time characteristics of control systems are specified in terms of time domain specifications. Systems with energy storage elements cannot respond instantaneously and will exhibit transient responses, whenever they are subjected to inputs or disturbances. The desired performance characteristics of a system of any order may be specified in terms of transient response to a unit step input signal. The transient response characteristics of a control system to a unit step input is specified in terms of the following time domain specifications:

Delay time td

Rise time tr Peak time tp

Maximum overshoot Mp

Settling time ts

FORMULA USED1. Rise time, tr = Π-θ/ωd

2. Peak time, tp = Π/ωd 3. Settling time, ts = 4/ξωn

4. Maximum Peak Overshoot(%Mp) = e- ξπ/(1-ξ2)1/2

SIMULINK MODEL

PROGRAM

Num = [0 0 100]Den = [1 6 100]Step (num, den)

PROCEDURE

Derive the transfer function of a RLC series circuit.



Assume R= 1 Ohms, L = 0. 1 H and C = 1 micro Farad. Find the step response theoretically and plot it on a graph sheet

To build a SIMULINK model to obtain step response / sine response of a first order system, the following procedure is followed:

In MATLAB software open a new model in SIMULINK library browser. From the continuous block in the library drag the transfer function block. From the source block in the library drag the step input/ sine input. From the sink block in the library drag the scope. From the math operations block in the library drag the summing point. Connect all to form a system and give unity feedback to the system. For changing the parameters of the blocks connected double click the respective

block. Start simulation and observe the results in scope. (Use a mux from the signal routing

block to view more than one graph in the scope) Compare the simulated and theoretical results.

MODEL GRAPH

RESULT:

The time response characteristics of the given second order system is simulated digitally and verified manually


1. Give two examples for second order system.2. Name the standard test signals used in control system.3. What is time response?

Exp No 6 STABILITY ANALYSIS OF LINEAR SYSTEM

AIM



(i) To obtain the bode plot, Nyquist plot and the root locus of the given transfer function.

(ii) To analyse the stability of given linear system using MatlabApparatus Required System with MATLABTHEORYFrequency Response: The frequency response is the steady state response of a system when the input to the system is sinusoidal signal.Frequency response analysis of control system can be carried either analytically or graphically. The various graphical techniques available for frequency response analysis are

1. Bode plot2. Nyquist plot3. Nicholas plot4. Mand N circles5. Nicholas Chart

Bode PlotThe Bode plot is a frequency response plot of the gransfer function of a system. A bode plot consiss of two graphs. One is plot of the magnitude of a sinusoidal transfer fubction versus log ω. The other is plot of the phase angle of a sinusoidal transfer function versus log ω. The main advamtage of the bode plot is that the multiplication of mafnitude can be converted into addition. Also a simple methid for sketching an approximate log magnitude curve is available.Polar Plot:The polar plot a sinusoidal transfer function G(jω) on polar coordinates as ω is varied from zero to infinity. Thus the polar plot is the locus of vector [G(jω)] angle [G(jω)] as ω is varied from zero to infinity. The polar plot is also called nyquist plot.Nyquist stability Criterion:If G(s)H(s) contour in the G(s)H(s) plain corresponding to Nyquist contour in s-plane circles the point-1+j0 in the anti-clock wise direction many times as the number of right half s-plane of G(s)H(s). Then the closed loop system is stable.Root Locus:The Root Locus technique is a powerful tool for adjusting the location of closed loop poles to achieve the desired system performance by varying one or more system parameters.The path taken by the roos of the characteristics equation when open loop gain K is varied for zero to infinity are called the root loci (or the path taken bya root of characteristic equation when open loop fain K ios varied form zero to infiny is called root locus).Frequency Domaun Specifications:The performance and characteristics of a system in frequency domain are measured in term of frequemcy domain specification. The requirements of a system to be designed are ysually specified in terms if this specification.

1. Resonant Peak, Mr2. Resonant Frequency, ωr3. Bandwidth4. Cut-off rate5. Gain margin6. Phase Margin

PROCEDURE1. Enter the command window of the MATLAB2. Create a new M-file by selecting File-New-M-file



3. Type and save the program4. Execute the program by either pressing F5 or Debug-run.5. View the results.6. Analyse the stability of the system for various valuew of gain.

PROGRAM:

1. NYQUIST PLOT: Clear all Num = [0 0 1] Den = [1 0.8 1] Nyquist (num, den)

2. BODE PLOT: Num = [0 0 25] Den = [1 4 25] Bode (num, den)

3. ROOT LOCUS: Num = [0 0 0 18] Den = [15 20 16 0] Rlocus (num, den)

RESULT:The plots are drawn for the given transfer function using MATLAB and verified manually. From the plot obtained, the system is found to be ______________.


1. Define stability of Linear Time Invariant System. 2. Give the stability conditions of system using Pole-Zero plot. 3. Define Bode Plot. 4. What is the use of Bode Plot? 5. What the conditions of stability are in Bode plot? 6. Define Stability criteria. 7. Define Limits of stability. 8. Define safe regions in stability criteria. 9. Define Phase margin and Gain margin.

Exp No 7 DETERMINATION OF TRANSFER FUNCTION OF SELF EXCITED DC GENERATOR



AIM:To obtain the transfer function of self excited DC generator on no load and loaded

condition.APPARATUS / INSTRUMENTS REQUIRED:S. No Description Range Type Quantity

1. Ammeter (0-20) A/(0-2.5)A MC 1/12. Voltmeter (0-300)V MC 13. Rheostat 445Ώ/1A, 200

Ώ/1.4 AWire wound 1/1

4 Starter - 3 point 15 Load 3 KW Resistive 16 Connecting Wires - - Amount Required

THEORY:

Derivation of transfer function of separately excited DC generator is as follows,Applying KVL to the field side,Ef = Rf If + Lf (dIf / dt) ------(1)Applying KVL to the armature side,Eg = Ra Ia + La (dIa / dt) + RL Ia ----(2)VL = RL Ia ----(3)Also since Eg α if , let Eg = Kg If ---- (4)Taking Laplace transform of equation (1) we getEf (s) = Rf If(s) + sLf If(s)Ef (s) = If (s) [Rf + sLf]If (s) = Ef (s) / [Rf + sLf] ---(5)Taking Laplace transform of equation (2) we getEg (s) = Ra Ia(s) + sLa Ia(s) + RL Ia(s) ----(6)Eg (s) = Ia(s) [Ra + sLa + RL]Taking Laplace transform of equations (3) and (4) we getVL(s) = RL Ia( s) … (7)Therefore, Ia( s) = VL(s) / RL

Eg(s) = Kg If(s) … (8)Substituting. equations (7) and (8) in equation (6) we getKg If(s) = [Ra + sLa + RL] [VL(s) / RL] … (9)Substituting the value of If (s) in the above equation we getKg Ef (s) / [Rf + sLf] = [Ra + sLa + RL] [ VL(s) / RL]Hence transfer function,VL(s) / Ef (s) = Kg RL / [Rf + sLf] [Ra + sLa + RL] …(10)For unloaded condition, Ia = 0Therefore transfer function VL(s) / Ef (s) = Kg / [Rf + sLf] … (11)For loaded conditionLf = √ (Zf

2 – Rf2) / 2πf

La = √ (Za2 – Ra

2) / 2πfTransfer function VL(s) / Ef (s) = Kg RL / [Rf (Ra + RL) (1+sτf) (1 + sτa)] … (12)where τf = Lf / Rf and τa = La / (Ra + RL)



FORMULAE USED:

Transfer function of DC generator,

On no load condition: VL(s) / Ef (s) = Kg /[Rf + sLf] where Kg is gain constant

Rf is field resistance in Ohms Lf is field inductance in HenryOn loaded condition: VL(s) / Ef (s) = Kg RL / [Rf (Ra + RL) (1+sτf) (1 + sτa)]where Kg is gain constantField time constant τf = Lf / Rf

Rf is field resistance in OhmsLf is field inductance in Henry

Armature time constant τa = La / (Ra + RL)Ra is armature resistance in OhmsLa is armature inductance in Henry

PROCEDURE:

1. To determine the gain constant Kg :No load or open circuit characteristics:

1. Connections are made as shown in the circuit diagram 2. The motor field rheostat should be in minimum resistance position and the generator

field rheostat should be in maximum resistance position or minimum potential position while switching ON and switching OFF the supply side DPST switch.

3. Ensure that the DPST switch on the load side is open. 4. Switch ON the supply DPST switch. 5. Using the 3- point starter the DC motor is started and it is brought to rated speed by

adjusting the motor field rheostat. 6. Keeping the DPST switch on the load side open, the generated voltage Eg and field

current If of generator is noted down by varying the generator field rheostat. 7. The above step is repeated till 125 % of rated voltage is reached. 8. A graph is plotted between Eg and If taking If along x- axis. A tangent to the linear

portion of the curve is drawn from the origin and slope of this line gives Kg.

OBSERVATIONS:

Field current, If Induced Voltage, EgS. No.

(A) (V)



MODEL GRAPH:

LOAD CHARACTERISTICS:

1. Connections are made as shown in the circuit diagram 2. The motor field rheostat should be in minimum resistance position and the generator

field rheostat should be in maximum resistance position or minimum potential position while switching ON and switching OFF the supply side DPST switch.

3. Ensure that the DPST switch on the load side is open. 4. Switch ON the supply DPST switch 5. The generator is brought to its rated voltage by varying the generator field rheostat. 6. The DPST switch on the load side is closed, and the load is varied for convenient

steps of load current up to 120 % of its rated capacity and the voltmeter VL and ammeter Ia readings are observed. On each loading the speed should be maintained at rated speed.

7. A graph is plotted between VL and IL taking IL on x- axis. The slope of the graph gives Kg.

OBSERVATIONTerminal Voltage, VL Load Current, IL

S. No.(V) (A)



MODEL GRAPH:



RESULT:

The transfer function of separately excited DC generator is determined as

Exp No 8 DETERMINATION OF TRANSFER FUNCTION OF DC MOTORAIM:

To obtain the transfer functions of DC shunt motor.



APPARATUS REQUIRED:

S. No Description Range Type Quantity1 Ammeter (0-2.5) A MC 22. Voltmeter (0-300) V MC 13 Starter - 3 pt starter 14 Rheostat 200Ώ/1.4 A Wire wound 15 Connecting Wires - - Amount required

THEORY:

FIELD CONTROLLED MOTOR:

FORMULAE USED:Transfer function of field controlled DC motor,(s) / Vf (s) = Km / [s (1+sτf) (1 + sτm)]

where

Motor gain constant, Km = Ktf / (BRf)Ktf is motor torque constantTorque, T is 9.81 X R (S1-S2)R is radius of the brake drum in mR = circumference of the brake drum/ (2 П)B is viscous co-efficient of frictionRf is field resistance in Ohms

Field time constant τf = Lf / Rf

Rf is field resistance in Ohms Lf is field inductance in Henry

Lf = √ (Zf2 – Rf

2) / 2πfZf is field impedence in Ohms

Rf is field resistance in OhmsPROCEDURE:

1. To determine motor torque constant, Ktf :



1. Connections are made as shown in the circuit diagram 2. The armature current Ia of the motor is set to some value by adjusting the

armature circuit resistance. This value of Ia is maintained constant throughout the experiment.

3. The field current If is varied in steps by adjusting the field rheostat and for each values of If the brake drum is adjusted such that it just fails to rotate. The corresponding readings of ammeter and spring balances are noted.

4. The value of torque for each value of If is calculated 5. A graph is plotted between torque T and field current IF taking IF along x-axis. The

slope of the graph gives the value of Ktf

OBSERVATIONS:

Armature current Field current Spring balance readings Torque

S. No. Ia If TS1 S2

(A) (A) (kg) (kg) (Nm)

MODEL GRAPH:



CIRCUIT DIAGRAM

RESULT:

The transfer function of field controlled DC motor is determined as

Exp No 9 ANALOG SIMULATION OF TYPE 0 AND TYPE 1 SYSTEMAIM:

To study the time response of first and second order type –0 and type- 1 systems.



20AL AA F

-(0-2.5)A

MC 200 , (0-2.5)A1.4 A A MC

A1

-(0-300)V V F1

MMC -A2

20A F2

D

P

S

220V, D.C. Tsupply S


1. Linear system simulator kit 2. CRO 3. Patch cords

FORMULAE USED:

Damping ratio, z= (ln MP)2 / (2 +(ln MP)2)Where MP is peak percent overshoot obtained from the time response graph

Undamped natural frequency, n = / [tp (1 - z2)]where tp is the peak time obtained from the time response graph

Closed loop transfer function of the type – 0 second order system is

C(s)/R(s) = G(s) / [1 + G(s) H(s)]where

H(s) = 1G(s) = K K2 K3 / (1+sT1) (1 + sT2) where K is the gain

K2 is the gain of the time constant – 1 block =10K3 is the gain of the time constant – 2 block =10T1 is the time constant of time constant – 1 block = 1 msT2 is the time constant of time constant – 2 block = 1 ms

Closed loop transfer function of the type – 1-second order system is C(s)/R(s) = G(s) / [1 + G(s) H(s)]

where H(s) = 1

G(s) = K K1 K2 / s (1 + sT1) where K is the gain K1 is the gain of Integrator = 9.6 K2 is the gain of the time constant – 1 block =10 T1 is the time constant of time constant – 1 block = 1 ms

THEORY:

The type number of the system is obtained from the number of poles located at origin in a given system. Type – 0 system means there is no pole at origin. Type – 1 system means there is one pole located at the origin. The order of the system is obtained from the highest power of s in the denominator of closed loop transfer function of the system. The first order system is characterized by one pole or a zero. Examples of first order systems are a pure integrator and a single time constant having transfer function of the form K/s and K/(sT+1). The second order system is characterized by two poles and up to two zeros. The standard form of a second order system is G(s) = n

2 / (s2 + 2zns + n

2) where z is damping ratio and n is undamped natural frequency.

Block diagram of Type- 1 First order system



Block diagram to obtain closed loop response of Type-0 second order system

Block diagram to obtain closed loop response of Type-1 second order system

ProcedurePROCEDURE:

1. To find the steady state error of type – 0 first order system

1. Connections are made in the simulator kit as shown in the block diagram.2. The input square wave is set to 2 Vpp in the CRO and this is applied to the REF

terminal of error detector block. The input is also connected to the X- channel of CRO.

3. The output from the simulator kit is connected to the Y- channel of CRO.4. The CRO is kept in X-Y mode and the steady state error is obtained as the vertical displacement between the two curves.

5. The gain K is varied and different values of steady state errors are noted.

2. To find the steady state error of type – 1 first order system

1. The blocks are Connected using the patch chords in the simulator kit.2. The input triangular wave is set to 2 Vpp in the CRO and this applied o the REF




3. The output from the system is connected to the Y- channel of CRO.4. The experiment should be conducted at the lowest frequency to allow enough

time for the step response to reach near steady state.5. The CRO is kept in X-Y mode and the steady state error is obtained as the vertical

displacement between the two curves. 6. The gain K is varied and different values of steady state errors are noted. 7. The steady state error is also calculated theoretically and the two values are

compared.3. To find the closed loop response of type– 0 and type- 1 second order system

1. The blocks are connected using the patch chords in the simulator kit.2. The input square wave is set to 2 Vpp in the CRO and this applied to the REF


3. The output from the system is connected to the Y- channel of CRO.4. The output waveform is obtained in the CRO and it is traced on a graph sheet.

From the waveform the peak percent overshoot, settling time,rise time, peak time are measured. Using these values n and x are calculated.

5. The above procedure is repeated for different values of gain K and the values are compared with the theoretical values.

WIRING SEQUENCE:

FOR TYPE 0 SYSTEMS:

+12V-1, GND-2,-12V-3, A1-4, FG-A9, GND-48, A11-A13, A14-CIACH1, 6-A15, A16-CIACH0, GND-49.

FOR TYPE 1 SYSTEM:

+12V-1,GND-2,-12V-3,A1-4,6-12,14-38,FG-A9,FG GND-48,A11-A13,A14-CIA CH1,39-A15,A16-CIA CH0,CIA GND-49.

RESULT:

The time response of first and second order type-0 and type-1 systems are studied.


1. Define order and type number.2. What are dominant poles?3. What is a closed loop system?4. What is the effect of negative feedback?5. What are poles and zeros of a system?6. Define transfer function.

Exp No 10 DC POSITION CONTROL SYSTEM

AIM To study the behavior of closed loop speed control using PID control


Kp

1/T1

Position sensor

Chopper Motor Gear


APPARATUS REQUIRED PID controller kit with the motor, PID software, PCTHEORY

Control system which the output has an effect upon the input quantity in sucha manner as to maintain the desired output value is called a closed loop system.

The open loop system cvan be modified as closed loop system by providing a feedback. The provision of feed back automatically corrects the change in output due to disturbances. Hence the closed loop system is also called closed loop system. The general block diagram of an automatic control system is given below. It consists of an error detector, a controller, a plant(open loop system) and feedback path element

The reference signal(or input signal) corresponds to desired output. The feedback path elements sampled the output and convert it to a signal of same type as that to reference detector. The error signal generated by the error detector is the difference between reference signal and feedback signal. The controller modifies and amplifies error signal to produce better control action. The modified error signal is fed to the plant to correct its output.

A pair of potentiometers is used to convert the input and output position into proportional electrical signals. The desired is set on the input potentiometer and the actual position is fed to feedback potentiometer. The difference between the two angular position generates an error signal, which is amplified and fed to armature circuit of the DC motor. If an error exceeds, the motor develops a torque to rotate the output in such a way as to reduce the error to zero. The rotation of the motor stops when the error signal is zero,.i.e., when the desired position is reached.

-

PROCEDURE:

1. PROPORTIONAL CONTROL WITH STEP CHANGE IN SET POINT:

PB 20%, Ti = 64000sec, Td = 0 sec, Kd = 10, Ts = 0.1s, Output lower limit 0, Upper limit 100+reverse action, source for measured variable channel 0, source for set value channel 1.

WIRING SEQUENCE:

+12V-1, -12V-3, GND-2, FG(A17)-A13, A14-CIA CH1, CIA CH0-11, 32-4, CIA GND-36, CIA DAC O/P(5)-31.



2. PROPORTIONAL + INTEGRAL CONTROL WITH STEP CHANGE SET POINT:


WIRING SEQUENCE:


3. PROPORTIONAL+INTEGRAL+DERIVATIVE CONTROL WITH STEP CHANGE IN SET POINT:


WIRING SEQUENCE:


RESULT:

Thus the DC position control system using P, PI, and PID controller is done.

Exp No 11 STEPPER MOTOR CONTROL SYSTEM

AIM:

To write a simple program to interface a stepper motor with 8255 and to interface it.APPARATUS REQUIRED:



1. Switched mode power supply 2. IC port-8255 3. XPO ST/DC Kit 4. XPO85 Kit 5. Stepper motor

PROCEDURE:

For giving the input:

1. Press reset ‘R’. 2. Press sub ’S’. 3. Press ‘CR’ 2 times to enter data. 4. Enter 7A33 in the address then press CR. 5. Enter the respective data in the respective memory location..

To get the output:

1. Press reset ‘R’ 2. Press ‘G’ for getting the output. 3. Then press ‘G’ for getting the output then press ‘CR’ two times thus the stepper

motor. rotates according to the program.

To enter the missing code in between the program:

1. Press ‘CNT+CR’ at the same time. 2. The input code moves one step before. 3. Press ‘CR’ to enter the missing code.

PROGRAM:

ADDRESS OPCODE COMMENTS

7A33 31 00 21START:LXISP,2100

H7A36 3E 80 MVIA,80H,INIT 8255 PORTS



7A38 D3 0BOUT CR55,A,B,C AS O/P

PORTS

7A3A 16 20MVID,20H,INITIALISE DELAY COUNTER

7A3C 1E 10 MVIE,10H7A3E 3E 00 MVIA,00H7A40 D3 08 OUT PORT A7A42 0E 00 MVIC,00H7A44 21 62 7A7A47 06 04 MVIB,04H7A4A D3 08 OUTPORT A7A4C CD 15 06 CALL DELAY7A4F CD 15 06 CALL DELAY

7A52 0CIHR

C

7A53 79MOVA,

C7A54 FE 64 CPI 19H

7A56 CA 61 7AJ2

NEXT7A59 23 INX H7A5A 05 DCR B7A5B C2 49 7A JNZ BACK7A5E C3 44 7A JMP AGAIN7A61 CF NEXT:RST17A62 067A63 09 03 DB 0CH

ANTICLOCKWISE CLOCKWISESTEP A1 A2 B1 B2 DATA STEP A1 A2 B1 B2 DATA

1 1 0 0 1 AH 1 1 0 0 1 AH2 0 1 0 1 5H 2 0 1 1 0 6H3 0 1 1 0 6H 3 0 1 0 1 5H4 1 0 1 0 AH 4 1 0 0 1 AH

RESULT:

Thus the stepper motor was interfaced with the 8051 microcontroller and motor is rotated in forward and reverse direction.






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