Ee2257 Control System Lab Manual

Control Systems Laboratory Manual

PREPARED BY V.BALAJI ,M.Tech, (Ph.D), AP/EEE, DCE Page 1

DDEEPPAARRTTEEMM EENNTT OOFF EELL EECCTTRRII CCAALL AANNDD EELL EECCTTRROONNII CCSS

EENNGGII NNEEEERRII NNGG

EE2257 CONTROL SYSTEM LABORATORY

MM AANNUUAALL

PPRREEPPAARREEDD BBYY

VV..BBAALL AAJJII ,, MM ..TTeecchh,, ((PPhh..DD)),, MM ..II ..SS..TT..EE,, MM ..II ..AA..EENNGG,, MM ..II ..OO..JJ..EE

AASSSSII TTAANNTT PPRROOFFEESSSSOORR//EEEEEE DDEEPPAARRTTMM EENNTT

DDHHAANNAALL AAKK SSHHMM II CCOOLL LL EEGGEE OOFF EENNGGII NNEEEERRII NNGG

CCHHEENNNNAAII



LL II SSTT OOFF EEXXPPEERRII MM EENNTTSS

1. Determination of transfer function of DC Servomotor 2. Determination of transfer function of AC Servomotor. 3. Analog simulation of Type - 0 and Type – 1 systems 4. Determination of transfer function of DC Generator 5. Determination of transfer function of DC Motor 6. Stability analysis of linear systems 7. DC and AC position control systems 8. Stepper motor control system 9. Digital simulation of first systems 10. Digital simulation of second systems



TTRRAANNSSFFEERR FFUUNNCCTTII OONN OOFF DDCC SSEERRVVOO MM OOTTOORR

EEXXPPTT..NNOO ::

DDAATTEE ::

AAII MM ::

To determine the transfer function of the DC servomotor

AAPPPPAARRAATTUUSS RREEQQUUII RREEDD::

S.No Name of the Equipment Range Type Quantity



TTHHEEOORRYY::

Speed can be controlled by varying (i) flux per pole (ii) resistance of armature circuit and (iii) applied voltage. It is known that N ∝ Eb. If applied voltage is kept, Eb = V – IaRa will

φ Remain constant. Then, N ∝ 1

φ By decreasing the flux speed can be increased and vice versa. Hence this method is called field control method. The flux of the DC shunt motor can be changed by changing field current, Ish with the help of shunt field rheostat. Since the Ish relatively small, the shunt filed rheostat has to carry only a small current, which means Ish

2 R loss is small. This method is very efficient. In non-interpolar machines, speed can be increased by this methods up to the ratio 2: 1. In interpolar machine, a ratio of maximum to minimum speed of 6:1 which is fairly common. FFOORRMM UULL AA::

AArr mmaattuurr ee CCoonnttrr ooll DD..CC.. SSeerr vvoo mmoottoorr ::

It is DC shunt motor designed to satisfy the requirements of the servomotor. The field excited by a constant DC supply. If the field current is constant then speed is directly proportional to armature voltage and torque is directly proportional to armature current. Km Transfer Function = S (1 + TmS)

Km = 1 / Avg Kb Tm = JRa / Kb Kt Kt = ∆T / ∆Ia Eb = V-Ia Ra

Constant Values J = 0.039 Kg2m B = 0.030 N / rpm



FFiieelldd CCoonnttrr ooll DD..CC.. SSeerr vvoo mmoottoorr ::

It is DC shunt motor designed to satisfy the requirements of the servomotor. In this motor the armature is supplied with constant current or voltage. Torque is directly proportional to field flux controlling the field current controls the torque of

the motor. K Transfer Function = Js2 (1 + ζs)

K = Kt / Rf ζ = Lf / Rf = V Zf

2 – Rf2 / 2πf / Rf

ω = 2πN / 60 T = r ( S1 – S2 ) * 9.81 N-m and r = .075m

OBSERVATION TABLE FOR TRANSFER FUNCTION ARMATURE

CONTROL DC SERVO MOTOR:

TTaabbllee NNoo.. 11 FFiinnddiinngg tthhee vvaalluuee ooff KK bb

Sl.No If Ia S1 S2 N V T Eb ωωωω Kb = Eb / ωωωω



Avg Kb

TTaabbllee NNoo.. 22 TToo ff iinndd RRaa

Avg Ra = PPRREECCAAUUTTII OONNSS:: At starting,

• The field rheostat should be kept in minimum resistance position

PPRROOCCEEDDUURREE FFOORR TTRRAANNSSFFEERR FFUUNNCCTTII OONN OOFF AARRMM AATTUURREE CCOONNTTRROOLL DDCC SSEERRVVOOMM OOTTOORR::

Sl.No Volt Va Current Ia Ra = Va / Ia



FFiinnddiinngg KK bb

1. Keep all switches in OFF position. 2. Initially keep voltage adjustment POT in minimum potential position. 3. Initially keep armature and field voltage adjustment POT in minimum

position. 4. Connect the module armature output A and AA to motor armature terminal

A and AA respectively, and field F and FF to motor field terminal F and FF respectively.

5. Switch ON the power switch, S1, S2. 6. Set the field voltage 50% of the rated value. 7. Set the field current 50% of the rated value. 8. Tight the belt an take down the necessary readings for the table – 1 to find

the value of Kb. 9. Plot the graph Torque as Armature current to find Kt.

FFiinnddiinngg RRaa

1. Keep all switches in OFF position. 2. Initially keep voltage adjustment POT in minimum position. 3. Initially keep armature and field voltage adjustment POT in minimum

potential position. 4. Connect module armature output A and AA to motor armature terminal A to

AA respectively. 5. Switch ON the power switch and S1. 6. Now armature voltage and armature current are taken by varying the

armature POT with in the rated armature current value. 7. The average resistance value in the table -2 gives the armature resistance.

PPRROOCCEEDDUURREE FFOORR TTRRAANNSSFFEERR FFUUNNCCTTII OONN OOFF FFII EELL DD CCOONNTTRROOLL DD..CC.. SSEERRVVOOMM OOTTOORR:: FFiinnddiinngg RRff

1. Keep all switches in OFF position. 2. Keep armature field voltage POT in minimum potential position. 3. Initially keep armature and field voltage adjustment POT in minimum

potential position. 4. Connect module filed output F and FF to motor filed terminal F and FF

respectively.



5. Switch ON the power, S1 and S2. 6. Now filed voltage and filed current are taken by varying the armature POT

with in the rated armature current value. 7. Tabulate the value in the table no – 3 average resistance values give the fied

resistance. FFiinnddiinngg ZZ ff

1. Keep all switches in OFF position. 2. Keep armature and field voltage POT in minimum position. 3. Initially keep armature and field voltage adjustment POT in minimum

position. 4. Connect module varaic output P and N to motor filed terminal F and FF

respectively. 5. Switch on the power note down reading for the various AC supply by

adjusting varaic for the table no – 4.

FFiinnddiinngg KK ttll

1. Keep all switches OFF position. 2. Initially keep voltage adjustment POT in minimum potential position. 3. Initially keep armature and field voltage adjustment POT in minimum

position. 4. Connect the module armature output A and AA to motor armature terminal

and AA respectively, and field F and FF to motor field terminal F and FF respectively.

5. Switch ON the power switch, S1 and S2. 6. Set the filed voltage at rated value (48V). 7. Adjust the armature voltage using POT on the armature side till it reaches

the 1100 rpm. 8. Tight the belt and take down the necessary reading for the table – 5 Kt

l 9. Plot the graph Torque as Field current to find Kt

l

OOBBSSEERRVVAATTII OONN TTAABBLL EE FFOORR TTRRAANNSSFFEERR FFUUNNCCTTII OONN OOFF AARRMM AATTUURREE CCOONNTTRROOLL DDCC SSEERRVVOO MM OOTTOORR:: TTaabbllee NNoo::33 TToo ff iinndd RRff

Sl.No I f (amp) Vf (Volt) Rf (ohm)



Avg Rf =

TTaabbllee NNoo::44 TToo ff iinndd ZZ ff

Sl.No I f (amp)

mA Vf (Volt) Zf = Vf / If

AAvvgg ZZff == TTaabbllee NNoo:: 55 TToo ff iinndd KK tt

l

Sl.No If Ia S1 S2 T( N – m) N (rpm)

MM OODDEELL GGRRAAPPHH::

∆ If

∆ T ∆Ktl = ∆T / ∆If

T

Field Current

T

∆ Ia

∆ T ∆Kt = ∆T / ∆Ia

Armature Current



VIVA QUESTIONS:

1. What are the main parts of a DC servo motor? 2. What are the two types of servo motor?

3. What are the advantages and disadvantages of a DC servo motor? 4. Give the applications of DC servomotor? 5. What do you mean by servo mechanism? 6. What do you mean by field controlled DC servo motor?

MM OODDEELL CCAALL CCUULL AATTII OONN::



Result:

TTRRAANNSSFFEERR FFUUNNCCTTII OONN OOFF AACC SSEERRVVOO MM OOTTOORR

EEXXPPTT..NNOO ::

DDAATTEE ::

AAII MM ::



To determine the transfer function of the given AC servomotor


S.No Name of the Equipment Range Type Quantity

NNAAMM EE PPLL AATTEE DDEETTAAII LL SS::

OUTPUT : VOLTAGE : CURRENT : SPEED :

FFUUSSEE RRAATTII NNGGSS::

Blocked rotor test: 125% of rated current.

TTHHEEOORRYY::

An servo motor is basically a two – phase induction type except for certain special design features. A two – phase servomotor differ in the following two ways from a normal induction motor. The rotor of the servomotor is built with high resistance. So that its X / R (Inductive reactance / resistance) ratio is small which result in liner speed – torque characteristics. The excitation voltage applied to two – stator winding should have a phase difference of 90o

WWOORRKK II NNGG PPRRII NNCCII PPLL EE OOFF AACC SSEERRVVOOMM OOTTOORR Voltages of equal rms magnitude and 90o phase difference excite the stator winding. These results in exciting current i1 and i2 that are phase displaced by 90o and have equal rms value. These current are rise to a rotating magnetic field of constant magnitude. The direction of rotation depends on the phase relationship of the two current (or voltage).



The rotating magnetic field sweeps over the rotor conductor. The rotor conductors experience a change in flux and so voltage are induced in rotor conductors. This voltage circulates current in the short circuited rotor conductors and the current creates rotor flux. Due to the interaction of stator and rotor flux, a mechanical force (or torque) is developed on the rotor and the rotor starts moving in the same direction as that of rotating magnetic field. FFOORRMM UULL AA:: Laplace Transform of output

Transfer Function = Laplace Transform of input ω(s) / Es(s) = K1 / sJ + K2 + B = Km / 1 + sζ ------ (1) Km = K1 / (K2 + B) ------------------------------- motor gain constant (2) ζm = J / (K2 + B) ---------------------------------- motor time constant (3)

Torque (T) = 9.81 * r * s Nm

• S = applied load in Kg • R = radius of shaft in m = 0.068 m

CCoonnssttaanntt VVaalluueess:: J = 52 gmcm2 = 0.05kg cm2, B = 0.01875

TTaabbllee NNoo:: 11 OOBBSSEERRVVAATTII OONN TTAABBLL EE FFOORR DDEETTEERRMM II NNII GG MM OOTTOORR CCOONNSSTTAANNTT KK 11::

S.No Load (kg)

Control Voltage (Vc)

Torque (Nm)



TTaabbllee NNoo:: 22 OOBBSSEERRVVAATTII OONN TTAABBLL EE FFOORR DDEETTEERRMM II NNII NNGG MM OOTTOORR CCOONNSSTTAANNTT KK 22::

S.No Speed (N)

rpm Load (kg)

Torque (Nm)

PPRREECCAAUUTTII OONNSS::

i. Initially DPST switch should be in open condition. ii. Keep the autotransformer in minimum potential position. iii. In blocked rotor test, block the rotor by tightening the belt around the the

brake drum before starting the experiment. BLOCK DIAGRAM OF SERVOMOTOR



PPRROOCCEEDDUURREE:: FFoorr ddeetteerr mmiinniinngg mmoottoorr ccoonnssttaanntt KK 11

1. Keep variac in minimum potential position. 2. Connect banana connectors “Pout to Pin” and “Nout to Nin”. 3. Connect 9pin D connector from the motor feed back to the input of module

VPET – 302. 4. Switch ON the 230V AC supply of the motor setup. 5. Switch ON the power switch. 6. Switch ON the S2 (main winding) and S1 (control winding) switches. 7. Set the rated voltage (230V) to control phase using VARIAC. 8. Apply load to the motor step by step until it reaching 0 rpm. 9. Take necessary readings for the table -1. 10. To calculate K1 plot the graph torque vs control winding.

FFoorr ddeetteerr mmiinniinngg mmoottoorr ccoonnssttaanntt KK 22 1. Keep variac in minimum potential position. 2. Connect banana connectors “Pout to Pin” and “Nout to Nin”. 3. Connect 9pin D connector from the motor feed back to the input of module

VPET – 302. 4. Switch ON the 230V AC supply of the motor setup. 5. Switch ON the power switch. 6. Switch ON the S2 (main winding) and S1 (control winding) switches.



7. Set the rated voltage (230V) to control phase using VARIAC. 8. Apply load to the motor step by step until it reaches 0 rpm. 9. Take necessary readings for the table -2. 10. To calculate K2 plot speed vs torque curve.

MM OODDEELL GGRRAAPPHH

MM OOTTOORR CCOONNSSTTAANNTT KK 22 MM OOTTOORR CCOONNSSTTAANNTT KK 11

MM OODDEELL CCAALL CCUULL AATTII OONN::

∆T

∆N

K2 = ∆T / ∆N

Speed in rpm

∆T

∆V

K1= ∆T / ∆V

Speed in rpm



VIVA QUESTIONS:

1. Define transfer function?

2. What is A.C servo motor? What are the main parts?

3. What is servo mechanism?

4. Is this a closed loop or open loop system .Explain?

5. What is back EMF?

Result:

ANALOG SIMULATION OF TYPE – 0 and TYPE – 1 SYSTEM AIM:

To study the time response of first and second order type –0 and type- 1 systems. APPARATUS REQUIRED: 1. Linear system simulator kit 2. CRO FORMULAE USED:



1. Damping ratio, ξ= √ (ln MP) 2 / (π2 + (ln MP)

2) Where MP is peak percent overshoot obtained from the response graph 2. Undamped natural frequency, ωn =π / tp √ (1 - ξ2) Where tp is peak time obtained from the response graph 3. Closed loop transfer function of type-0 second order system is C(s) / R(s) = G(s) / 1+G(s)

Where G(s) = K K2 K3 / [(1+sT1) (1 + sT2)] K is the gain K2 is the gain of the time constant – 1 block =10 K3 is the gain of the time constant – 2 block =10 T1 is the time constant time constant – 1 block = 1 ms T2 is the time constant time constant – 2 block = 1 ms 4. Closed loop transfer function of type-1 second order system is C(s) / R(s) = G(s) / 1+G(s)

Where G(s) = K K1 K2 /[s (1 + sT1)] 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 Theoretical Values of ωn and ξ can be obtained by comparing the co-efficients of the denominator of the closed loop transfer function of the second order system with the standard format of the second order system where the standard format is C(s) /R(s) = ωn

2 / s2 + 2ξωns +ωn 2

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 upto two zeros. The standard form of a second order system is C(s) /R(s) = ωn

2 / (s2 + 2ξωns + ωn

2) where ξ is damping ratio and ωn is undamped natural frequency. BLOCK DIAGRAM: 1. To find steady state error of type- 1 system



2. To find steady state error of type- 0 system

3. To find the closed loop response of Type-1 second order system

4. To find the closed loop response of Type-0 second order system



PROCEDURE: 1. To find the steady state error of type – 1 first order system 1. The blocks are connected using the patch cords in the simulator kit. 2. The input triangular wave is set to 1 V peak to peak 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 system is connected to the Y- channel of CRO. 4. The experiment should be conducted at the lowest frequency so keep the frequency knob in minimum position 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. 2. To find the steady state error of type – 0 first order system 1. The blocks are connected using the patch cords in the simulator kit. 2. The input square wave is set to 1 V peak to peak 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 system 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.

3. To find the closed loop response of type – 0 and type- 1 second order system 1. The blocks are connected using the patch cords in the simulator kit. 2. The input square wave is set to 1 V peak to peak 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 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 ωn and ξ are calculated.



5. The above procedure is repeated for different values of gain K and the values are compared with the theoretical values. TABULAR COLUMN: 1. To find the steady state error of type – 1 first order system

S.No. Gain ,K Steady state error ess (V)

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

S.No. Gain ,K Steady state error ess (V)

3. To find the closed loop response of type – 0 second order system S.No. Ga

in, K

Peak percent Overshoot, %M P

Rise time, tr

(sec)

Peak time, tp (sec)

Settling time,ts (sec)

Graphical Theoretical

Dam ping ratioξξξξ

Undamped natural frequency, ωωωωn (rad/sec)

Damping ratioξξξξ

Undamped natural frequency,ωωωωn(rad/sec)



4. To find the closed loop response of type – 1 second order system

MODEL GRAPH: MODEL CALCULATION: RESULT

S.No. Gain, K

Peak percent Overshoot, %M P

Rise time, tr

(sec)

Peak time, tp (sec)

Settling time,ts (sec)

Graphical Theoretical Dam ping ratio ξξξξ

Undamped natural frequency, ωωωωn (rad/sec)

Damping ratio ξξξξ

Undamped natural frequency, ωωωωn(rad/sec)



SSTTAABBII LL II TTYY AANNAALL YYSSII SS OOFF LL II NNEEAARR SSYYSSTTEEMM



EEXXPPTT..NNOO ::

DDAATTEE :: AAII MM ::

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

transfer function.

(ii) To analysis the stability of given linear system using MATLAB.

APPARATUS REQUIRED: System with MATLAB TTHHEEOORRYY:: FFrr eeqquueennccyy RReessppoonnssee:: The frequency response is the steady state response of a system when the input to the system is a 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 Plot 2. Polar plot (Nyquist plot) 3. Nichols plot 4. M and N circles 5. Nichols chart

BBooddee pplloott:: The bode plot is a frequency response plot of the transfer function of a system. A bode plot consists of two graphs. One is plot of the magnitude of a

sinusoidal transfer function versus log ω. The other is plot of the phase angle of a

sinusoidal transfer function versus logω. The main advantage of the bode plot is that multiplication of magnitude can be converted into addition. Also a simple method for sketching an approximate log magnitude curve is available.



PPoollaarr pplloott :: The polar plot of 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 vectors

[G (jω) ] ∠ G (jω) as ω is varied from zero to infinity. The polar plot is also called Nyquist plot.

NNyyqquuiisstt SSttaabbii ll ii ttyy CCrr ii tteerr iioonn:: If G(s)H(s) contour in the G(s)H(s) plane corresponding to Nyquist contour in s-plane encircles the point – 1+j0 in the anti – clockwise direction as many times as the number of right half s-plain of G(s)H(s). Then the closed loop system is stable.

RRoooott LL ooccuuss:: 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 roots of the characteristics equation when open loop

gain K is varied from 0 to ∞ are called root loci (or the path taken by a root of

characteristic equation when open loop gain K is varied from 0 to ∞ is called root locus.)

FFrr eeqquueennccyy DDoommaaiinn SSppeeccii ff iiccaatt iioonnss:: The performance and characteristics of a system in frequency domain are measured in term of frequency domain specifications. The requirements of a system to be designed are usually specified in terms of these specifications. The frequency domain specifications are

1. Resonant peak, Mr 2. Resonant Frequency, ωr. 3. Bandwidth. 4. Cut – off rate 5. Gain margin 6. Phase margin



RReessoonnaanntt PPeeaakk,, MM rr

The maximum value of the magnitude of closed loop transfer function is called the resonant peak, Mr. A large resonant peak corresponds to a large over shoot in transient response.

RReessoonnaanntt FFrr eeqquueennccyy,, ωωωωωωωωrr

The bandwidth is the range of frequency for which the system gain is more than -3db. The frequency at which the gain is -3db is called cut off frequency. Bandwidth is usually defined for closed loop system and it transmits the signals whose frequencies are less than cut-off frequency. The bandwidth is a measured of the ability of a feedback system to produce the input signal, noise rejection characteristics and rise time. A large bandwidth corresponds to a small rise time or fast response.

CCuutt--OOff ff RRaattee:: The slope of the log-magnitude curve near the cut off frequency is called cut-off rate. The cut-off rate indicates the ability of the system to distinguish the signal from noise.

GGaaiinn MM aarr ggiinn,, KK gg The gain margin, Kg is defined as the reciprocal of the magnitude of open loop transfer function at phase cross over frequency. The frequency at witch the phase of open loop transfer function is 180 is called the phase cross over

frequency, ωpc.

PPhhaassee MM aarr ggiinn,, γγγγγγγγ

The phase marginγ, is that amount of additional phase lag at the gain cross over frequency required to bring the system to the verge of instability, the gain

cross over frequency ωgc is the frequency at which the magnitude of open loop transfer function is unity (or it is the frequency at which the db magnitude is zero).

PPRROOCCEEDDUURREE::



1. Enter the command window of the MATLAB. 2. Create a new M – file by selecting File – New – M – File. 3. Type and save the program. 4. Execute the program by either pressing F5 or Debug – Run. 5. View the results. 6. Analysis the stability of the system for various values of gain.

PPrr oobblleemm 11 Obtain the bode diagram for the following system MM AATTLL AABB PPrr ooggrr aamm a = [0 1 ; -25 -4] b = [1 1 ; 0 1] c = [1 1 ; 1 1] d = [0 0 ; 0 0] bode (a, b, c, d) grid title (‘BODE DIAGRAM’) PPrr oobblleemm 22

−−=

+

−−=

2

1

2

1

2

1

2

1

2

1

425

10

10

11

425

10

x

x

y

y

y

y

x

x

x

x

)1(

1

+ss



Draw the Nyquist plot for G(s) = MM AATTLL AABB PPrr ooggrr aamm num = [0 0 0] den = [1 1 0] nyquist (num,den) v = [-2,2,-5,5] axis (v) grid title (‘Nyquist Plot’) PPrr oobblleemm 22 Obtain the root focus plot of the given open loop T.F is G(s) H (s) = MM AATTLL AABB PPrr ooggrr aamm num = [0 0 0 0 1] den = [11.1 10.3 5 0] rlocus (num,den) grid title [‘Root Locus Plot’]

)106.0)(5.0( 2 +++ ssss

K



Result:

CCLL OOSSEEDD LL OOOOPP SSPPEEEEDD CCOONNTTRROOLL SSYYSSTTEEMM



EEXXPPTT..NNOO ::

DDAATTEE ::

AAII MM ::

To study the behavior of closed loop speed control system using

PID controller


(i) PID controller with motor (ii) CRO

TTHHEEOORRYY:: Closed loop system Control system which the output has an effect upon the input quantity in such a manner as to maintain the desired output value is called closed loop systems. The open loop system can be modified as closed loop system by providing a feedback. The provision of feedback automatically corrects the change in output due disturbances. Hence the closed loop system is also called closed loop system. The general block diagram of an automatic control system is given below. In consists of an error detector, a controller, plant (open loop system) and feedback path element. The reference signal (or input signal) corresponds to desired output. The feedback path elements sample the output and convert it to a signal of same type as that of reference detector. The error signal generated by the error detector is the difference between reference signal and feedback signal. The controller modifies and amplifies the error signal to produce better control action. The modified error signal is fed to the plant to correct its output. PPRROOCCEEDDUURREE::



1. Make the connections as per the circuit diagram. 2. Set the speed of the motor using set position. 3. Vary the gain values of P,I, and D controller until to get the set speed to

current speed. 4. Repeat the above procedure for different values of set speed.

Result:

SSTTUUDDYY OOFF AACC SSYYNNCCHHRROO TTRRAANNSSMM II TTTTEERR AANNDD RREECCEEII VVEERR



EEXXPPTT..NNOO ::

DDAATTEE ::

AAII MM ::

To study the operation of AC synchro transmitter and receiver


S.No Name of the Equipment Quantity

1 Synchro transmitter and receiver unit 1 Nos

2 Multimeter (Digital / Analog ) 1 Nos

3 Patch cords As required

TTHHEEOORRYY:: A synchro is an electromagnetic transducer commonly used to convert an angular position of a shaft into an electric signal. It is commercially known as a selsyn or an autosyn. The basic synchro unit is usually called a synchro transmitter. Its construction is similar to that of three phase alternator. The stator is of laminated silicon steel and is slotted to accommodate a balanced three phase winding which is usually of concentric coil type and star connected. The rotor is dumb bell construction and its wound with a concentric coil. AC voltage is applied to rotor winding through slip rings. Let and AC voltage

Vr (t) = Vr sin ωct be applied to the rotor of the synchro transmitter. The voltage causes a flow of magnetizing current in rotor coil which produces a sinusoidally time varying flux directed along its axis and distributed nearly sinusoidally in the air gap along the stator periphery. Because of transformer action, voltage is induced in each of the stator coil. As the air gap flux sinusoidally distributed the flux linking with any stator coil is proportional to the cosine of the angle between the axes of rotor and stator coil. This flux voltage in each stator coil. Voltages are in time.



phase with each other. Thus the synchro transmitter acts a like a single-phase transformer in which the rotor coil is the primary and the stator coil is the secondary. Let Vs1n, Vs2n, Vs3n, be the voltage induced in the stator coils, S1, S2, S3 respectively with respect to the neutral. Then for a rotor position of the synchro

transmitter, θ is the angle made by rotor axis with the stator coil S2. The various stator voltages are

Vs1n = KVr sinωct cos (θ + 120o)

Vs2n = KVr sinωct cosθ

Vs1n = KVr sinωct cos (θ + 240o) The terminal voltages of the stator are

When θ = 0, Vs1s2 and Vs2 s3 have the maximum voltage and while Vs3s1 has zero voltage. This position of rotor is defined as ht electrical zero of the transmitter and is used as reference for specifying the angular position of the rotor. Thus it is seen that the input to the synchro transmitter is the angular position of its rotor shaft and the output is a set of three signal phase voltages. The magnitudes of this voltage are function of the shift position. The output of the synchro transmitter is applied to stator winding of synchro control transformer. The control transmitter is similar in construction to a synchro transmitter except for the fact that rotor of the control transformer in made cylindrical in shape so that the air gap is practically uniform. The system (transmitter and control transformer pair) acts an error detector, circulating current to the same phase but of different magnitudes flow through two stator coils. The result is establishment of an indentical flux pattern in the air gap of the control transformer as the voltage drops in resistance and lockage reactance’s of two sets of stator coils are usually small.

tKVnVsnVssVs

tKVnVsnVssVs

tKVnVsnVssVs

cr

co

r

co

r

ωθ

ωθ

ωθ

sinsin31313

sin)120sin(33232

sin240sin(32121

=−=

+=−=

+=−=



OOBBSSEERRVVAATTII OONN TTAABBLL EE::

S.No Transmitter

(Degree) Receiver (Degree)

Vs1 – Vs2 Vs2 – Vs3 Vs3 – Vs1 Error

the synchro transmitter rotor, the voltage induced the control transformer rotor is proportional to the cosine of the angle between the two rotors given by

E (t) = KVr cos θ sin ωr t



The synchro transmitter and control transformer thus act as an error detector giving a voltage signal at the rotor terminals of the control transformer proportional to the angular difference between the transmitter control transformer shaft positions.

PPRROOCCEEDDUURREE:: 1. Make the connections as per the patching diagram. 2. Switch ON the supply. 3. Vary the shaft position of the transmitter and observe the corresponding

changes in the shaft position of the receiver. 4. Repeat the above steps for different angles of the transmitter. 5. Tabulated the different voltage at the test points of S1 S2, S3S2, and S3S1.

Result:

CYCLE – 2

7. (a) Lag Compensator.



7. (b) Lead Compensator.

8. Digital Simulation of Non-Liner System.

9. Digital Simulation of Liner System.

10. Digital Simulation of Type 0 and Type 1 System.

DDII GGII TTAALL SSII MM UULL AATTII OONN OOFF NNOONN--LL II NNEEAARR SSYYSSTTEEMM

EEXXPPTT..NNOO ::



DDAATTEE ::

AAII MM ::

To simulate the time response characteristic of liner system with simple

non-linearities like saturation and dead zone.


System with MATLAB 6.5

TTHHEEOORRYY::

NNoonn--LL iinneeaarr SSyysstteemmss::

The non linear system are system witch do not obey the principle of

superposition.

In practical engineering systems, there will be always some non linearity due

to friction, inertia, stiffness, backslash, hysteresis, saturation and dead – zone. The

effect of the non linear components can be avoided by restricting the operation of

the component over a narrow limited range.

CCllaassssii ff iiccaatt iioonn ooff nnoonn ll iinneeaarr ii tt iieess::

The non linearities can be classified as incidental and intentional.

The incidental non linearities are those which are inherently present in

the system. Common examples of incidental non linearities are saturation, dead –

zone, coulomb friction, stiction, backlash, etc.

The intentional non linearities are those which are deliberately

inserted in the system to modify system characteristics. The most common

example of this type of non linearity is a relay.

SSaattuurr aatt iioonn::



In this type of non linearity the output proportional to input for limited

range of input signals. When the input exceeds this range, the output tends to

become nearly constant.

All devices when driven by sufficient large signals, exhibit the

phenomenon of saturation due to limitations of their physical capabilities.

Saturation in the output of electronic, rotating and flow amplifiers, speed and

torque saturation in electric and hydraulic motors, saturation in the output of

sensors for measuring position, velocity, temperature, etc. are the well known

examples.

DDeeaadd ZZoonnee::

The dead zone is the region in witch the output is zero for given input.

Many physical devices do not respond to small signals, i.e., if the input amplitude

is less than some small value, there will be no output. The region in which the

output is zero is called dead zone. When the input is increased beyond this dead

zone value, the output will be linear.


1. Double click on MATLAB 6.5 icon on desktop command window opens.

2. From File Tab, select New Model file.

3. A Simulink model screen opens a “untitled”.

4. From Simulink library – select necessary blocks and place in new model

screen.

Block

Constant - Simulink-Continuous

Simulator - Simulink-Math operator

Transfer function - Simulink-Continuous



Scope - Simulink –sink

Dead Zone, Saturation - Simulink-Non-linear

5. Select properties for each item and connect them as shown in diagrams.

6. Select simulation Tab and configuration parameters and select ode23tb

model.

7. Save file under ‘work’ directory.

8. Simulated the system with step and sine inputs with and without dead zone,

saturation non – linearities.

9. Name the signals as mentioned in diagram and observe signal names on

scope by right clicking on response curve and by opening axes.







Result:

DDII GGII TTAALL SSII MM UULL AATTII OONN OOFF LL II NNEEAARR SSYYSSTTEEMM

EEXXPPTT..NNOO :

DDAATTEE :



AAII MM ::

To simulate the time response characteristic of higher-order Multi-

input multi output (MIMO) liner system using state variable formulation.


MATLAB 6.5

TTHHEEOORRYY::

Time Domain Specification

The desired performance characteristics of control systems are specified in

terms of time domain specification. System 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 the transient response to a units step input signal.

The transient response of a system to a unit step input depends on the initial

conditions. Therefore to compare the time response of various systems it is

necessary to start with standard initial conditions. The most practical standard is

to start with the system at rest and output and all time derivatives there of zero.

The transient response of a practical control system often exhibits damped

oscillation before reaching steady state.

The transient response characteristics of a control system to a unit step input

are specified in terms of the following time domain specifications.

1. Delay time, td

2. Rise time, tr

3. Peak time, tp

4. Maximum overshoot, Mp



5. Setting time, ts

FFOORRMM UULL AA::

Damped frequency of oscillation,


7. Enter the command window of the MATLAB.

8. Create a new workspace by selecting new file.

9. Complete your model.

10. Run the model by either pressing F5 or start simulation.

11. View the results.

12. Analysis the stability of the system for various values of gain.

PPRROOBBLL EEMM ::

Obtain the step response of series RLC circuit with R = 1.3KΩ, L = 26mH and

C=3.3µf using MATLAB M – File.

MM AATTLL AABB PPRROOGGRRAAMM FFOORR UUNNII TT II MM PPUULL SSEE PPRRSSPPOONNSSEE::

PPRROOGGRRAAMM ::

num = [ 0 0 1 ]

den = [ 1 0.2 1 ]

d

Risetimeω

θπ −=

ζζθ

21 1

tan−= −where

21 ζωω −= nd



impulse (num, den)

grid

title (‘ unit impulse response plot’)

MM AATTLL AABB PPRROOGGRRAAMM FFOORR UUNNII TT SSTTEEPP PPRRSSPPOONNSSEE::

PPRROOGGRRAAMM ::

Format long e

num = [ 0 0 1.6e10 ]

den = [ 1 50000 1.6e10 ]

step (num, den)

grid on

title (‘step response of series RLC circuit’)

Result:

DIGITAL SIMULATION OF TYPE 0 AND TYPE 1 SYSTEM

EXPT.NO :

DATE :



AIM:

To simulate the time response characteristics of first order second

order, type 0 and type 1 system using MATLAB.

APPARATUS REQUIRED:

System employed with MATLAB 6.5

THEORY:

The desired performance characteristics of control system are specified in

terms of time domain specification. 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 pf any order may be

specified in terms of the transient response to a unit step input signal.

The transient response of a system to unit step input depends on the initial

conditions. Therefore to compare the time response of various systems it is

necessary to start with standard initial conditions. The most practical standard is to

start with the system at rest and output and all time derivatives there of zero. The

transient response of a practical control system often exhibits damped oscillations

before reaching steady state.

The transient response characteristics of a control system to a unit step input

are specified in terms of the following time domain specifications.

1. Delay time, td

2. Rise time, tr

3. Peak time, tp

4. Maximum overshoot, Mp

5. Settling time, ts



The time domain specification is defined as follows.

1. Delay Time:

It is the taken for response to reach 50% of the final value, for the very first

time.

2. Rise Time:

It is the time taken for response to raise from 0 to 100% for the very first

time. For under damped system, the rise time is calculated from 0 to 100%. But for

over damped system it is the time taken by the response to raise from 10% to 90%.

For critically damped system, it is the time taken for response to raise from 5% to

95%.

Damped frequency of oscillation,

3. Peak Time:

It is the time taken for the response to reach the peak value for the very first

time. (or) It is the taken for the response to reach the peak overshoot, tp.

Peak time = µ / ωd

4. Peak Overshoot (Mp):

It is defined as the ration of the maximum peak value measured from final

value to the final value.

Let final value = c (e)

Maximum vale = c (tp)

Peak Overshoot, Mp =

d

Risetimeω

θπ −=

ssWhere /1tan 21 −= −θ21 snd −= ωω

)(

)()(

ec

ectc p −



erroefort

erroefort

ns

ns

%53

%24

ςω

ςω

=

=

erroefort

erroefort

ns

ns

%53

%24

ςω

ςω

=

=

5. Settling Time:

It is defined as the time taken by the response to reach and stay within a

specified error. It is usually expressed as % of final value. The usual tolerable error

is 2% or 5% of the final value.

FORMULA:

31%

ςς

−

−= seM p

100x

31%

ςς

−

−= seM p

100x



PROCEDURE:

Closed loop response of first order system:





5. Analysis the stability of the system for various values of gain

Closed loop response of second order system:






6. Analysis the stability of the system for various values of gain.

General MATLAB coding for closed loop response for type 0 and type1

system:

PROGRAM:

clear all



close all

clc

T1 = tf (2.25, [1 0.5 2.25 ])

p=pole (T1)

pre=abs (real (p(1)))

pim=abs (imag (p(1)))

wn=sqrt(pre*pre*+pim*pim)

damping _ratio=(pre/wn)

os=(exp(-1*pre*pi/pim))*100

tp=pi/pim

ts=4/pre

step(T1)

t=[ 0.1:0.1:25]

for x=1:length (t)

c (x)=1-1.01418*(cos (1.47902*t(x)-(9.59*pi/180))*exp(-25*t(x)))

end

figer

plot(t,c)

Result:

LL AAGG CCOOMM PPEENNSSAATTOORR

EEXXPPTT..NNOO ::

DDAATTEE ::




System employed with MATLAB 6.5

TTHHEEOORRYY::

The control systems are designed to perform specific taskes. When

performance specification are given for single input. Single output linear time

invariant systems. Then the system can be designed by using root locus or

frequency response plots.

The first step in design is the adjustment of gain to meet the desired

specifications. In practical system. Adjustment of gain alone will not be sufficient

to meet the given specifications. In many cases, increasing the gain may result poor

stability or instability. In such case, it is necessary to introduce additional devices

or component in the system to alter the behavior and to meet the desired

specifications. Such a redesign or addition of a suitable device is called

compensations. A device inserted into the system for the purpose or satisfying the

specifications is called compensator. The compensator behavior introduces pole &

zero in open loop transfer function to modify the performance of the system.

The different types of electrical or electronic compensators used are lead

compensator and lag compensator.

In control systems compensation required in the following situations.

1. When the system is absolutely unstable then compensation is required

to stabilize the system and to meet the desired performance.

2. When the system is stable. Compensation is provided to obtain the

desired performance.

LAG COMPENSATOR:

A compensator having the characteristics of a lag network is called a lag

compensator. If a sinusoidal signal is applied to a lag network, then in steady state

the output will have a phase lag with respect input.



Lag compensation result in a improvement in steady state performance but

result in slower response due to reduced bandwidth. The attenuation due to the lag

compensator will shift the gain crossover frequency to a lower frequency point

where the phase margin is acceptable. Thus the lag compensator will reduce the

bandwidth of the system and will result in slower transient response.

Lag compensator is essentially a low pass filter and high frequency noise

signals are attenuated. If the pore introduce by compensator is cancelled by a zero

in the system, then lag compensator increase the order of the system by one.

FORMULA:

PROCEDURE:

With out compensator:

1. Make the connection as per the circuit diagram.

2. Apply the 2V p-p sin wave input and observe the waveform.

3. Very the frequency of the sin wave input and tabulate the values of xo and yo

4. Calculated gain and phase angle.

5. Draw the bode plot.

With lag compensator:

1. From the bode plot find the new gain crossover frequency.

2. Find out β values and writ the frequency function. G(s).

0

0

x

y

A

BGain ==

)/log(20 AB=

)/(sin 01 AxPhase −−=θ

)/(sin 01 By−−=



3. From the transfer function calculated R1, R2 and C.

4. Set the amplifier gain at unity.

5. Insert the lag compensator with the help of passive components and

determine the phase margin of the plant.

6. Observe the step response of the compensated system.

MATLAB coding with Compensator:

PROGRAM:

num = [ 0 0 100 5 ];

den = [ 400 202 1 0 ];

sys = (sys)

margin (sys)

[ gm, ph, wpc, wgc ] = margin (sys)

title (‘BODE PLOT OF COMPENSATED SYSTEM’)

MATLAB coding with out lag Compensator:

PROGRAM:

num = [ 0 0 5 ];

den = [ 2 1 0 ];

sys = tf (num, den)

bode (sys)

Margin (sys)

[ gm, ph, wpc, wgc ] = margin (sys).

title (‘BODE PLOT OF UNCOMPENSATED SYSTEM’);



Result:

LEAD COMPENSATOR:

A compensator having the characteristics of a lead network is called a lead

compensator. If sinusoidal signal is applied to a lead network, then in steady state

the output will have a phase lead with respect to input.



The lead compensator increase the bandwidth, which improves the speed of

response and also reduces the amount of overshoot. Lead compensation

appreciably improves the transient response, whereas there is a small change in

steady state accuracy. Generally lead compensation is provided to make an

unstable system as a stable system. A lead compensator is basically a high pass

filter and so it amplifies high frequency noise signals. If the pole is introduced by

the compensator is not cancelled by a zero in the system, then lead compensator

increases order of the system by one.

FORMULA:

PROCEDUR:

1. Enter the command window of MATLAB.

2. Create a New M-File by selecting file New M-File.

3. Type and save the program.

4. Execute the program by pressing F5 or Debug Run.


6. Analyze the Results.

With lead compensator:


2. Create a new M – file by selecting File – New –M-File.

0

0

x

y

A

BGain ==

)/log(20 AB=

)/(sin 01 AxPhase −−=θ

)/(sin 01 By−−=



3. Type and save the program.

4. Execute the program by either pressing F5 or Debug – Run.


6. Analysis the result.

MATLAB coding with out Compensator for loop system

PROGRAM:

den=[ 1 0.739 0.921 0 ]; pitch=tf(num, den); sys_cl=feedback (pitch,1); de=0.2; t=0:0.01:10; figure step(de*sys_cl, t) sys_cl=feedback (pitch,10); de=0.2; t=0:0.01:10; bode(sys_cl, t) grid on title ( 'BODE PLOT FOR CLOSED LOOP SYSTEM WITHOUT COMPENSATOR')

MATLAB coding with Compensator for loop system

PROGRAM:



num=[1 151 0.1774 ]; den=[1 0.739 0.921 0 ]; pitch=tf(num, den); alead=200; Tlead=0.0025; K=0.1; lead=tf(K*[alead*Tlead 1], [Tlead 1]); bode(lead*pitch) sys_cl=feedback(lead*pitch,10); de=0.2; t=0:0.01:10; figure step (de*sys_cl, t) title('BODE PLOT FOR CLOSED LOOP SYSTEM WITH

COMPENSATOR')

Result:

Documents

Ee2257 Control System Lab Manual