Control Lab New

7/28/2019 Control Lab New

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EXPERIMENT No, 01

OBJECT:DC SPEED CONTROL SYSTEM

(A). To study D.C. Speed control system system on open loop.

(B). To study of transistor performance, another time signal is added at the input of control circuit.

(C). To study how eddy current breaking is being disturbance rejected by close and open loop.

APPARATUS REQUIRED: Experimental kit, Dual Trace CRO, and connecting leads.

THEORY: Speed control is a very commonly in many industrial applications such as rolling mills, spinnmills, paper factories etc. The present unit is low power dc motor speed control system designed a

laboratory experiments. The various components and subsystems have been carefully integrated, and t

experiments are designed to illustrate the important performance characteristics in a simply way. Figure

shows a schematic of the system, different blocks and parts of which are described below:

BLOCK DIAGRM:-

G(S)

EROOR MOTOR 600Mv/1Hz AMPLIFIER SPEED,W+ + +REF/0 To 2V+

- H(S)

TACHO

FORMULAES USED :

= Where KM is motor gain constant,

G(s) = = KA. .(1)

Where KA is the gain of amplifier. Again, the tachogenerarot transfer function (or gain) may be written as,

H(s)= 1

KA

0 TO 10


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This yields the closed loop transfer function of the complete system as

= = ...(2)

STEADY STATE ERROR:

Defining positional error coefficient, Kp as,

KP= G(S).H(S)=KA.KM.KT,

The steady state error, eSS, to step input R u (t), is given by,

=eSS= ...(3)TRANSIENT RESPONSE: For a step input VR (s)= R/s, Eq.2 yields

..(4)

(t)= (1-e- .t) ....(5)

Where effective time constant

Tdt=T/(KAKMKT+1)

The transient response has an exponential character similar to a capacitor charging through

resistor. Further, the effective time constant T eff decreases with increasing KA making t

motor response faster.

Consider a general first order, type-o transfer function of the form

C(s)/R(s) =K/s=1

e(t) =R.K{1-exp(1-)}

For a square wave of p-p value of R as input, it is easy to see that . ...(6)Where is the frequency of the square wave

T= ( ) for open loop system (7a)

Where VM (p-p)=KA - VS(p-p)


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T= for open loop system (7b)

Where K eff=

DISTURBANCE REJECTION: One of the important features of a feedback control system is its ability

reduce the effect of external disturbances. From Fig. 2, the disturbance transfer function for V R =0, may

written as

For a unit step disturbance, p(s)-D/s, the stead state output speed is given by

ss = D/KAKMKT+1(8)

Thus, the steady state speed change caused by an external disturbance should reduce as tgain KP is increased. Also, the performance should be much superior to the open loop case

e., with feedback disconnected (KT=0).

PROCEDURE: The experiments suggested in this section start with a study of the loop system and

subsystems. This is followed by the performance evaluation of the closed loop system for various operat

conditions like forward path gain and disturbance.

(A).SIGNAL AND REFERENCE:

Set KA =0. Connect DVM to measure the range of variation of reference VR. Switch ON the square wave signals VS and measures its amplitude and frequency using a calibrated C

The frequency of this signal is about 1Kz, which makes the CRO display a very inconvenient

measurements. It suggested that the amplitude may be measured with time-base switched OFF, and

frequency, simply count the number of pulses (as seen on CRO screen), in say 60 seconds, using

watch.

MOTOR AND TACHOGENERATOR:

Set VR=1V and KA=3. The motor may be running at a low speed. Record speed N in rpm and tachogenerator output V

T.

Repeat with VR =1 snf KA =1,510. And tabulate measured voltage VM (VRKA). Steady state motor speN in rpm (orSS=N2/60 in radiance/sec.) and tachogenerator output VT.

Plot N vs. VM, and VT vs. N. obtain KM and KT from the linear region of the curves (see Fig.5). Motor gain constant, KM = ,and


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Tachogenerator gain, KT =

To calculate motor time constant, with square wave wave signal VS ON, set VR KA sothat the peak-peak variation of VM lies between 3-8V. This would ensure a reasonably linear operation of the mot

Use Eq.7(a)to calculate the motor time constant T.(Caution: The CRO must be keeping in dc input mode for this measurement).

Obtain the motor transfer function usingG(s)=

DISTURBANCE

Set KP =5 and adjust the reference VR to get a speed reading close to 1200rpm. The brake setting shobe at i.e., no braking.

Record and tabulate the motor speed variation for different setting of the eddy current brake. Calculate percentage decrease in speed at each setting of the brake, starting from no braking.

(B).CLOSE LOOP PERFORMANCE:

Performance of the closed loop system is evaluated in terms of steady state error and disturbance rejectas functions of forward gain. The FEEDBACK terminals are

STEADY STATE ERROR:

Set V R =1 volt and K A =5 the motor may be running at a low speed. Measure and record speed N in rp

tachogenerator VT,and the steady state error ess (-VR-VT).

Repeat above for KA =5, 10, 15,20 100

Compare in each case the value of steady state error computed from Eq.(3) i.e.

e ss= TRANSIENT PERFORMANCE:

Set VR 0.5V and KA=5. Switch ON the square wave signal and measure peak-to-peak amplitude ofand VT use Eq.(7b) to calculate system time constant T effwith R=VS (p-p) and C=VT(p-p).The value

K may be obtained from(2) as

Keff=, after substituting VT =KT.

DISTURBANCE REJECTION:

With KA =5, FEEDBACK terminals shorted and the brake setting at 0, adjust reference VR to get a speclose to 1200rpm.

Record and tabulate the variation in speed for different setting of the eddy current brake. Calculpercentage decrease in speed at each setting of the brake.

Repeat above for KA =10.50, 100. Compare the percentage decrease in speed for at various brake settings for open loop, close loop w

KA=5, and closed loop with KA=10. Comment on the results.


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OBSERVATION TABLE :

(a) Motor and Tachogenerator CharacteristicsVR=1voltS.NO. KA

Setting

N rpm VT volt VM volt Experimental

KA=VM/VR

1 3

2 4

3 5

4 6

5 7

6 8

7 9

8 10

T= | |(b) Closed loop performance(i) Steady state errorVR=1 volt

S.NO. KA Setting N rpm VT volt Ess(VR-VT)Experimental ess=1/(1+KAKMKT )Theoretical1 5

2 10

3 15

4 20

5 25

6 50

7 75

8 100

(ii). System time constant

From equation 7(b)

Eeff= KAKMKT /KAKMKT+1

(iii). Disturbance Rejection

Speed =120rpm (approx.) at brake 0, i.e., no load

reak Setting0 1 2 3 4

pen Loop Speed, rpm1207 1198 1022 669 658

osed Loop (KA=5) 1197 1190 1149 1037 1035

osed Loop (KA=10) 1205 1200 1177 1106 1103

osed Loop (KA=50) 1201 1201 1195 1176 1175

osed Loop (KA=100) 1201 1200 1196 1185 1184

RESULTS: Typical results obtained on an experimental unit are given below for guidance.

PRECAUTIONS: 1. Make the circuit as per the circuit diagram.

2. Handle the kit properly.3. Take the observations carefully.


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EXPERIMENT No: 03

OBJECT:AC MOTOR POSITION CONTROL-(a). To study AC motor position control through continuous command.

(b). To study of error detector on A.C. motor position control through step command.

(c). To study of A.C. position control through dynamic response.

(a). To study AC motor position control through continuous command.

APPARATUS REQUIRED: Experiment kit, Dual Trace CRO, and Connecting Leads.

THEORY: The set of comprises two parts:

(1). The motor unit.

(2). The control unit.

A. THE MOTOR UNIT: It consists a two phase AC servomotor. It has the technical specifications as:

Operating voltage: 120V AC, maximum current 0.15Amp.

Rated shaft Speed: 2400rpm, Inertia 4.610-4

Kg/cm2.

Torque : 0.08510-2

kg/m.

The motor drive a potentiometric load through gear train. The gear ratio is 1:40 hence the load shaft rotatio =60 rpm.The angular displacement is sensed by a 360

0servo potentiometer. A graduated disc is mounted upon

potentiometer to indicate angular position with 10

resolutions. The complete unit is housed in see throu

cabinet. A cable is attached to the unit with 9 pin D type connection with the control unit.

B. THE CONTROL UNIT. It consists power supply, servo amplifier, error detector and commapotentiometer. There is facility given to record the transient period of position control system under st

signal. The description as follows:


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CIRCUIT DIAGRAM:

PROCEDURE: (A)

1. Connect motor unit with the control unit.2. Switch ON the power. Set KA = 4.3. Connect CRO with the Vr socket. Starting from one end say 300, measure the voltage amplitude A.4. Move the reference potentiometer to 900 and measure the output voltage here as B. the constan

related to command signal is

C volt /0 (about 8mVpp/0 approx)5. Connect CRO other channel with socket V0 Move command potentiometer in step of 300 approximat

and note the R (from command potentiometer),0 from feedback potentiometer VR and V0 frthe results R - 0 R - 0), for each step for each set of KA, where K = gain = selecgain KA 20.

6. Keep command potentiometer at 1800. Set KA = 4. Not the position of the feedback potentiometRotate slowly the command. Repeat the step for gain KA at 7.

From the observations it is found that motor follows the command signal with the finite error

low gain settings. Graphically it is shown below. At low gain K A = 420=80, the system exists so

delay (the motor follows command position after some displacement) and larger steady st

error, where at K=140 the system procedure low error with small delay hence faster response.

PROCEDURE: (B)


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1. Set gain KA = 4. Position reference pot to 1200.2. Connect CRO one channel at reference socket (Vr). Not the signal amplitude in VAC p-p.3. Connect other channel with feedback socket (V0) and measure the AC p-p voltage there as V0.4. Now apply step signal and note the rms signal value at Vr socket. After elapse of time when mo

move to previous position (11 second later) move the reference dial for same reading obtained w

step signal. Note the dial position and out the step signal actuation in degree.

For example start degree is 120 and newer is 200 than step signal actuation is equal to 200 120 = 8

5. Now remove CRO one channel Connected with feedback socket, and connect it with the error to 2- 120 = 80

0

.6. Apply step command and not the error output ( a noisy signal may appear in mV order) e ss 7. Set gain K A = 7 , apply same step and find out the e ss Connect CRO back with V0 is initial value and V

find value, where V0 is voltage at 1200

and V0 after applying step signal in steady state.

PROCEDURE: (C)

1. Select KA =4. Keep command potentiometer at 1200. Connect CRO X - Y output sockets with refereto ground in X - Y mode. Set Y at

and X at .A zig zag curve will be appear upon Cselect mode A.

2. Press capture key briefly, a spot will be appear upon screen. Press step key briefly, the motor will run. Wait capture time is complete (about 4 second). After completion of capture time the captu

waveform will be displayed upon scream.

RESULT:

PRECAUTIONS:

4. Make the circuit as per the circuit diagram.5. Handle the kit properly.6. Take the observations carefully.


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EXPERIMENT NO-6

Aim : The PID controller:

(A). To observe open loop performance of building blocks & Calibration of PID controls:

(B). To study P, PI and PID controller with type 0 system with delay.

(C).To study P, PI and PID controller with type 1 system

Theory:-This set up is designed to study performance of analog PID controller with simulated proce

The board has built is signal source, building blocks for simulated process and PID controller with bu

in regulated dc supply to operate the system.The PID controller has three adjustable parameters, as P, I,D; each has 10 turn potentiometers w

dial knobs which are subdivided for .02 resolutions. Three sockets are provided to add or out any

desired control P,I orD. At input of PID controller an adder is provided which sums the reference a

feedback signals. The input and output of PID controller has no phase shift.

EXPERIMENT NO-6(A)

(a). To observe open loop performance of building blocks

(b). Calibration of PID controls:

(a1). The process blocks: feed square wave 1Vpp 20HZsignl to the input of time constant bloConnect dual trace CRO as shown in fig.3 find out the time constant from the appearing wave shape.

(a2).Feed similar signal to the input of integrated input and measure output signal. The integrati

capacitor charges on a constant rate of 0.5/r, hence the triangular wave output is, Vpp=0.5/R( /2C

/4(1/RC )=Ki..=4Vpp/ (vin).Where, is the input signal time period .

(A3). Feed similar signal to the input of delay block. Find out the delay time L, and time constant T.

(a4). Feed similar signal to the input of uncommitted amplifier input and observe the output for ga

and phase.

(a5). Feed similar signal to the each input of adder block alternately and observe the output.

Write down the transfer function of each block.

(b) Calibration of PID controls:

Before commencing the experiment it is advised to calibrate the control positions. To calibrate t

control the procedure given below should be adopted.

(b1). To calibrate the P control: Adjust P control knob to maximum (10 on major scale) Connect CRO

socket w.r.t. ground. Apply a square wave signal at error input of 50mSec duration with amplitude

200mVpp.

The maximum gain value obtain at screen of CRO connected at PID output as,

Kp (max)=


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(b2).To calibrate I control set I control to maximum (10 on major scale) CRO to I socket and groun

Apply 200Vpp square wave signal of 50ms, at input of error amp and note the output on CRO. T

maximum value of ki, from step a2, will be,

Ki (max)=

(b3).To calibrates the D control set D control to maximum and CRO with D socket. Apply a triangu

pulses of 1.0 Vpp (50mSec =20Hz) at input of error amplifier and note the pp output of square wave

output as,

Kd (max)=

EXPERIMENT NO-6(B)

To study P, PI and PID controller with type 0 system with delay.

The P controller with type 0 system with delay.

1. Connect the circuit as shown in fig 4a. Feed 1vpp square wave input 20 Hz (50 mSec). TriggCRO with the input signal, Kept CRO in CAL mode.

2. Trace the process reaction curve paper, in regard with input as shown in fig 4b. Find out tterms T and L

3. Now close the loop, as shown in fig 5. Switchover CRO for XY mode. Adjust P control as giventable3, and note the X and Y values.

4. Find out the value of Kp at where the response observed oscillatory. It is the critical gain Kct.5. Increase the Kp more thus the system generate oscillation. Switchover CRO to normal mode a

measure the time period of oscillations as Pct.

6. Switchover again in XY mode and adjust P control to value (0.5 Kct).Find out the % os and steastate error.

7. Adjust P control to the value. The L and T are obtained from open loop studies step 0, 2, Fout the % os and ess.

The % os= Ypp Xpp /2.

Table 1:-The P controller with type 0 system.


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Kp Xpp Ypp overshoot ess

1.0 . ..

2.0

3.0

3.5

4.0

4.5

The PI controller with type 0 system with delay. Circuit As fig 6.1. From table 2, the Kp is set for 0.4

and Ki adjusted for different values as shown in table 4. The steps are similar to exp c. CRO in Xy mod

Table 2:- PI controller with with type 0system. Kp= 0.45 Kct=.

I dial Ki,s-1

Xpp Ypp overshoot ess

8.00 .. .. .. ..

8.50

9.00

9.50

2. Adjust I control to value in table 2,(0.833 pct).Find out the %os and steady state error.

3. Adjust p control to the value shown in table 1. Adjust I control to given value in table1. Fine out t

%os and ess.

The % overshoot= Ypp-Xpp/2

The steady state error ess= Square wave amplitude-Xpp.

The PID controller with type 0 system with delay.

1. Connect the circuit as fig 7. Set P control for 0.6 Kct and I control 0.5 Pct. Select Xy mode fCRO.

2. Proceed with XY with D control adjusted as table 5.3. Adjust D control to value shown in table 2,(0.125 Pct).Find out the %os and steady state errorTable 3:- PID controller with type 0 system. Kp= 0.6 Kct, ki= 0.5Pct.


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D dial Kd,s-1

Xpp Ypp Overshoot ess

0.00 0.00 . .. .. ..

0.20

0.40

0.60

0.80

1.00

1.20

1.40

EXPERIMENT NO-6(C)

To study P, PI and PID controller with type 1 system.

The P controller with type 1 system .fig 8.

The steps are similar to steps taken in experiment. Tabulate the results as shown in table no.1.

The PI controller with type 1 system .fig 9.

Set Kp= 7.5. Adjust Ki from dial setting at 5.00, to .8.00 and tabulate the %os and ess.

The PID controller with type 1 system .fig 10.

Sat Kp= 7.5 and Ki=7.00(dial). Adjust Kd from dial setting at 1.00, to5.00 and tabulate the %os and es

Disconnect the integral and derivative control leads. The system is only with P controller. Select C

for trigger mode (normal mode). Measure the tp, and Mp from the response curve.

Compute the transfer function of the system from the observation. Now put the values of I and D froPID table prepared from exp PID. Write the transfer function.

RESULT:

PRECAUTIONS: i. Make the circuit as per the circuit diagram

ii. Handle the kit properly.

iii. Take the observations carefully.


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EXPERIMENT No.-02

AIM: DC MOTOR POSITION CONTROL

(A)To study of potentiometer displacement constant on D.C. motor position control.(B)To study of D.C. position control through continuous command.(C)To study of D.C. position control through step command.(D)To study of D.C. position control through dynamic response.APPARATUS REQUIRED: Experiment kit, Dual Trace CRO & Connecting Leads.

THEORY: The set up comprises two parts The motor Unit and The Control Unit.

THE MOTOR UNIT: It consists permanent magnet armature controlled geared servo motor. It has t

technical specifications as

Operating Voltage : 12V DC, 5W

Rated Shaft speed : 50 RPM (reduced by gear train, otherwise 2400 hence n = 0.02)

Torque : 3.5 Kg/cm at load shaft.

The angular displacement is sensed by a 3600

servo potentiometer. A graduated disc is mounted upon t

potentiometer to indicate angular position with 10

resolutions. A small DC motor is driven by the servo mo

to generate the speed proportional voltage which are used as tacho output for velocity feedback. A miniatu

toggle switch is provided at rear side of the motor unit to change the polarity of these tacho voltages. T

complete unit is housed in see though cabinet. A cable is attached to the unit with 9 pin D-type connector

connection with the control unit.

THE CONTROL UNIT : This unit has reference servo potentiometer, voltage, source, error detect

amplifier, motor drive circuit, a RAM card and necessary regulated supplies for the circuits.

BLOCK DIAGRAM


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EXPERIMENT No. 08

OBJECT: LINEAR SYSTEM SIMULATOR

(a) Open loop response(i) Error detector with gain, (ii) Time constant, (iii) Integrator(b) Close loop system(I) First order system (II) Second order system (III) third order system

THEORY:

OPEN LOOP RESPONSE: As a first step, the open loop transfer function of all the blocks viz. integrator, ti

constant, uncommitted amplifier and error detector/adders are to be determines experimentally.

measurements are carried out with the help of a measuring oscilloscope and the signal source is the bu

square wave generator in each case. Further, to get a properly synchronized waveform, especially for sm

values of signal, it will be convenient to use the built- in trigger source while keeping the CRO in exter

triggering mode. A double beam CRO for the simultaneous viewing of input and output is recommended. Nthat the value of K1, K2, K3 and T1,T2 obtained experimentally may differ somewhat from their nominal valu

indicated in section 2.2 due to component tolerances and experimental errors.

SAMPLE CALCULATION FOR DATA AT S. NO. 2: The above observations are for the typ

second order configuration shown in Fig. 7. The calculations here would require some results obtain

earlier from open-loop measurements, which are given next.

(i) INTEGRATOR: Given a 1V p-p square wave input of time period , the integrating capacitor in tblock charges from a constant current of 0.5/R, and hence the p-p amplitude of the triangular outpu

given by

Vp-p = = = = K1, so that K1 =

Measured value of Vp-pfor = 25 msec is 60 mV. Hence K1 is 9.6.

(ii) TIME CONSTANT: The block has a transfer function of the form K2/(sT1+1). Open-lomeasurements give K2 = 10 and T1 = 1 msec.

Calculation for and n

: The maximum overshoot MP as a fraction of steady-value is related

damping coefficient by

MP = exp (-/ , so that = Thus for MP= 0.12, the value of damping coefficient by measurement is 0.559. Using relation between

and tp,


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n

= = = 861 rad / sec.

Calculation for #

and n#: Open-loop transfer function of the total system may be written as

==

, for K 7

Closed loop transfer function G (s) is written as

G(s) =

Comparing coefficients in the denominator of the expression for G(s),

n2

= 672 , or n# = 819 red / sec.2n = , or # = 0.609

(a)Typical results obtained for a closed loop 3

rd

order system (Fig. 8) are as shown in table - 4 and table5 for two types of wave form

Input 1 volt (p-p) square wave

Table 4

S.No. K MP tP , mess #

n#

.

rad/sec

1. 0.5 0.25 7 0.403 490

2. 0.9 0.50 5 0.215 643

3. 1.4 0.60 4 0.160 795

4. 1.6 1.00 2.5 0 1256

5. (System oscillates at n > 1256 rad/ sec

Input: Triangular wave, 1 volt p-p, r(t)=2Rt

Table 5

S.No. K R V/sec ess

V Ess#

V1. 10 80 0.16 0.167

2. 7 80 0.26 0.238

3. 5 80 0.38 0.333

Measured on CRO

# calculated from open loop transfer function determined in section 4.1.


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C (t)

C (t) = Kt

tFig. 2 (a) UNIT STEP RESPONSE OF FIRST ORDER TRANSFER FUNCTION.

Fig. 2 (b) SQUARE WAVE RESPONSE OF FIRST ORDER TRANSFER FUNCTION

(i) ERROR DETECTOR CUM VARIABLE GAIN Apply a 100 mV square wave signal to any of the three inputs. Set the gain setting potentiometer to 10.0. Measure the p-p output voltage and note its sign. Calculate the gain. This is maximum value

gain possible for this block.

Repeat for the other two inputs one by one. Write the equation of this block and verify by connecting the signal to all there input.

(ii) TIME CONSTANT

Apply a 100 mV p-p square wave of known frequency (measured by CRO). For texperiment, the frequency should be selected towards the lower end to ensure t

steady state is nearly reached.

The waveform on the CRO be traced on a tracing paper for this analysis. From the trace calculate the time = T at which the response reaches 63.2%. This is t

constant.

From the trace determine the steady state value of the response. The value of Kgiven by the ratio of p-p steady-state output to the p-p input amplitude.


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Write transfer function of the block as discussed in section 3.1.Appendix 2shows the actual output waveforms of different open loop system as obtain

through a digital storage CRO, Techtronics, TDS-210.

(iii) INTEGRATOR Apply a 1 volt. p-p square wave input of known frequency (frequency measur

by the CRO).

Measure the p-p output voltage of the integrator block, which is a triangular waand also note its phase.

Calculate the gain constant K of the integrator discussed in section 3.1, and wrthe transfer function of this block.

(b)CLOSE LOOP SYSTEM- FIRST ORDER SYSTEM: Two forms of first order closed loop system, as shown in Fig. 5 are possible. Ma

proper connections for the configuration chosen.

Apply a 1 volt p-p square wave input and trace the output waveform on a tracpaper for K = 0.5, 1.0, 1.5 calculate the time constant in each case and compa

with theoretical results of section 3.3.

Also calculate the steady state errors for the above cases and compare with theoretical results If the open loop transfers function in the chosen configuration was of type-1. The steady sterror above would be zero for a step input. To find steady state error for ramp input, apply a 1 volt

p triangular wave input. Keeping the CRO in X-Y mode, connect system input to the X input and t

system output to the Y input. A trace as shown in Fig. 6 will be seen on the CRO in which the verti

displacement between the two curves is the steady state error.

Repeat the measurement for steady state error for different values of K and compthe result with theoretical calculations.

Not: In case of first order type-1 system experiments must be conducted at the lowest frequency to all

enough time for the step response to reach near steady state. For still better results an external oscillat

may be used.

(II) SECOND ORDER SYSTEM: Choose and wire a suitable second system configuration from Fig. 7. Apply a 1 volt p-p square wave input and trace the output waveform on a tracing paper for differe

values of K. obtain peak percent overshoot, settling time, rise time and steady state errors form t

tracing, and calculate and n. Compare with theoretical results.

In the case of a type-1 system, obtain steady-state error to ramp input as suggestedin section 4.2.

(III) THIRD ORDER SYSTEM: Connect the third order system in Fig. 8 and apply a square wave input of 1V (p-p). Sketch the Root Locus diagram, and, for = 0.2, calculate the value of K and n. Repeat for

(sustained oscillations case).

Verify the above by setting the gain potentiometer at calculate value of K. and applying momentarilsquare wave input.


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Appendix-2 shows the actual output waveform of various closed loop system as obtained throug

Digital Storage CRO, Techtronic, TDS-210.

TYPICAL RESULTS:

Some typical result for closed loop system are given next.

(b) Typical results obtained for a closed loop 1st order type-0 system (second diagram of Fig. 5) are shoin table 2.

Input: Square wave, 1 volt p-p, max. Frequency (88Hz).

Table 2

S.NO. K C () p-p, volts ess = t-c(),

volts

1. 0.02 0.8 0.2 0.8

2. 0.04 0.6 0.3 0.7

3. 0.10 0.5 0.5 0.5

4. 0.24 0.3 0.7 0.3

5. 0.44 0.2 0.8 0.26. 1.0 negligible 0.9 0.1

(c) Typical results for a closed loop 2nd order type-1 system are as shown in table 3.Input: Square wave, 1 volt p-p

Time in msec and in rad/sec

Table 3

S.No. K MP tr tP ts

n

# n

#

1. 10 0.173 2.4 3.6 10 0.487 999 0.510 9792. 7 0.12 3.2 4.4 10 0.559 861 0.609 919

3. 5 0.06 4.4 6.0 10 0.667 703 0.721 693

Calculate from experimental values of MP and tP.

# Calculate from open transfer function determined 4.1.

Result

PRECAUTIONS: i.Make the circuit as per the circuit diagram.

ii. Handle the kit properly.

iii. Take the observations carefully.


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EXPERIMENT No.- 7

OBJECT: LEAD LAG COMPENSATOR

(a) To study the open loop response on compensator(b)Close loop transient response

THEORY: Compensation network oftenly used to made appreciable improvement in transient response a

small change in steady state accuracy. This set up has facilitate to study and design implemantion of su

networks. Three of such networks are given in the set up and performance of other designed networks can

implemented using few passive components.

Set up description: The set up is divided in three parts (a) the signal sources, (b) the uncompensated syst

and (c) compensators.

a. There are two signal sources provided in the set up.1. Sine wave : It is available through socket in marked frequencies spanning across two decades as, 10

to 1000 Hz. The signal has (8 Vp-p (approx) uncelebrated output with reference to ground. The freque

selection is made with coarse selector between 10 or 100 and multiplied with dial digit.

2. Square wave : It is available in fixed frequency, 40 Hz. The output is fixed about 1 Vp-p in amplitude.3. Phase angle meter: It is an analog meter provided to take readings of respective phase angle betwe

given signal input output, where its one input is connected internally with sine wave source. Tolera 5%, 1 count.b. Three are simulated systems of unknown dynamics.1. Process or plant : It consists of an active network simulation of second order system. The trans

function is, K / ( sT + 1 )2.

2. Gain 2 block: It is amplifier provided to compensate the gain KC. the gain setting is variable between, =0 10, by mean of a calibrated dial. The input/output is phase shifted by 1800.

3. Error detector cum gain : This block has two input ( e1, e2 ) and an amplifier has an output equal to eKA(e 1 + e 2), where KA is variable between 1 10 by a calibrated dial e1 is 1Vp-p sq wave. The input/output as in phas

4. Compensation circuit : There are three compensation circuits as Lag, Lead and lag-lead with transfunctions in form of Gc (s) =(Kc) (sT1 + 1/ sT2 + 1 ) and (Kc) ( sT1 + 1 / sT2 + 1 ) where (K c

implemented by gain compensation amplifier by gain 2 block.

PROCEDURE: 1

The experimental work is divided in two parts, (a e ) the open loop response and (A B ) the close lo

response. In open response in first step the magnitude / frequency and phase / frequency plots are perform

A dual trace CRO is essential for the experimental steps (optional, phase angle meter).

Exp. A. The open loop response: fig. 1, process.

a.1. Connect the sine source with process/ plant input. Connect CRO across input and output (dual trace mod

Measure input voltage A = ..Vpp.


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a.2. Start from the low frequency end (10 Hz), increase frequency in step of 10,20,30,40,50 Hz for linear mo

or 10,20,40,80,100,200 for octace mode and note the output voltage Vpp as B. Not the phase difference

each test frequency.

a.3. Prepare a table between input output volts, the gain magnitude in dB (20 log10 B/A) and phase angle

degree.

a.4. Sketch the phase / frequency (Bode plot) response for smooth curves upon semi log paper .From the l

frequency end obtain the error coefficient and required gain KA.

S.No. f Hz A Vp B Vpp Gain dB /

1 10 V V ..dB

n0

2 20

3 40

4 80

5 100

6 200

7 400

8 800

9 1000

Typical table (example) for plotting of response curve (Process or plant).

S.No. Freq A Vpp B Vpp Gain dB 0

1 10 V

2 203 40

4 80

5 100

6 200

7 400

8 800

9 1000

Exp:- fig. 2. Open loop response of compensators.

b.1. . Connect the sine source with process/ plant input. Connect CRO across input and output (dual tra

mode). Measure input voltage A = ..Vpp.

a.2. Start from the low frequency end (10 Hz) not the output voltage Vpp as B.

Note the phase difference for each test frequency as previous experiment.


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b.3. Prepare a table between input output volts, the gain magnitude in dB ( 20 log10 B/A)

and phase angle in degree.

Typical table (example) for plotting of response curve (Lag compensator).


1 10 V

2 20

3 40

4 80

5 100

6 200

7 400

8 800

9 1000

Typical table (example) for plotting of response curve (Lead ).


1 10 V

2 20

3 40

4 80

5 100

6 200

7 400

8 800

9 1000

Typical table (example) for plotting of response curve (Lag-Lead ).


1 10 V

2 20

3 40

4 80

5 100

6 200

7 400

8 800

9 1000

Exp.: - A close transient response. (Process or plant)

A.1. Connect the system as shown in fig. 4. The step (square wave) signal of 40 Hz, 1 Vpp at the one input of

error detector is available.


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A.2. Adjust K1 (gain 1) to the value found from the plot (K=3.2 approx, for less=0.) gain raised to 10dB/octa

approx. fine adjust K1 for required shape of the response. Sketch it upon the paper.

A.3. From the transient response measure maximum overshoot Mp, steady state error ess and the peak ti

tp.

n = DESIGN OF LEAD COMPENSATOR:

1. From the plot find out the gain crossover frequency (raised) and phase from 1800. It is equal to actphase a.

2. Select the specific phase s, desired. Add tolerance of 5 100 to it for compensation. Calculattenuation factor as,

== (1-sin m)

Find the new gain crossover frequency g new = 20 log 1/ dBFind out the new gain crossover frequency from the plot at this attenuation.

Calculate the corner frequencies as,

1/ T = .g new and 1 / T = g new / calculate the necessary gainKc required, while taking gain K 1 , as

Kc = K1/ Write the transfer function for the compensator as

Gc(s) = Kc

Compute the value of R1C and R2, with C from the transfer function as,

Time constant t = R1C = 1/ T,

and R2 / R1 + R2


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C

R1

INPUT R2 OUTPUT

DESIGN OF LAGCOMPENSATOR:

1. From the plot find out the gain crossover frequency (0 db ) and phase from 1800. It is equal to actphase a.

2.Select specific phase s, desired. Add tolerance of 5 100 to it for compensation.Find the new gain crossover frequency g new. Measure gain at this frequency, it must be equal to h

frequency attenuation of lag network 20 . Compute as

g new = 20 log 1/ dbcalculate the corner frequencies as, 1/T = at one decade or an octave below of wg new. The ot

corner frequency equal to Po = 10 /

calculate the necessary gain Kc required, while taking gain K1, as

Kc = K1 /

Write the transfer function for the compensator as

Gc(s) = Kc

Compute the value of R1 and R2, with C from the transfer function as,

Time constant t = R2C = 1/ T,

and = R2 + R1 / R2

R1 C

INPUT R2 OUTPUT


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DESIGN OF LAG LEAD COMPENSATOR:

1. From the plot find out the gain crossover frequency (raised) and phase from 1800. It is equal to actphases a at low and high frequency side.

2. Select the specific phase s, desired. Add tolerance of 5 100 to it for compensation. Calculattenuation factor as,

a = (1 sin m) for lead

find the new gain crossover frequency g new as,

g new = 20 log 1/ Find out the new gain crossover frequency from the plot at this attenuation.Calculate the corner frequency as,

1 / T = .g new and 1 / T = g new / Calculate the necessary gainKc = K1 /


Gc(s) = Kc for leadCompute the value of R1

*and R2

*. with C2 from the transfer function as,

Time constant t = R1C2 = 1/T,

and = R2 / R1 + R2

1. From the plot find out the gain crossover frequency (0 db) and phase from 1800. It is equal to actphase a.

2. Select the specific phase s, desired. Add tolerance of 5 100 to it for compensation.Find the new gain crossover frequency g new. Measure gain at this frequency, it must be equal

high frequency attenuation of lag network 20 Log . Compute as

g new = 20 log 1 / Calculate the corner frequencies as, 1 / T at one decoder or an octave below of wg new. The ot

corner frequency equal to Po = 10/ 1/T

Calculate the necessary gain Kc required, while taking gain K1, as

Kc = K1 /


Gc(s) = Kc for lag

Compute the value of R1*

and R2*

. with C1 from the transfer function as,

Time constant t = R2C2 = 1/T,


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and = R2 + R1 / R2

Combining both networks as givens for corner frequencies as, f1, f2 , f3 and f4 from lower to higher si

Combine both networks, *Keeping R1 lead = R1 lag and R2 lead = R2 lag.

C2

R2 C1

INPUT R1 OUTPUT

f1 = 1/

R1 C1

f2 = 1/ R1C1

f3 = 1/ R2C2

f4 = / R2C2

Control Lab New

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