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bala control

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    TTRANRANSSFFERERFFUNUNCCTTIIONON OFOF DDCC SESERVRVOO MMOTOROTOR

    EEXPXPT.T.NNOO ::

    AAIIMM::

    To determine the transfer function of the DC servomotor

    APPAAPPARRAATTUUSS RREEQQUUIIRREEDD::

    S.No Name of the Equipment Range Type Quantity

    THEOTHEORYRY::

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

    Remain constant. Then,N 1

    By decreasing the flux speed can be increased and vice versa. Hence this method iscalled field control method. The flux of the DC shunt motor can be changed by changingfield current, Ish with the help of shunt field rheostat. Since the Ish relatively small, theshunt 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 interpolarmachine, a ratio of maximum to minimum speed of 6:1 which is fairly common.

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    J = 0.039 Kg2mB = 0.030 N / rpm

    FFOORMURMULLAA::

    AArrmmaatureture CCoontrntrooll DD..CC.. SerServvoo mmoottoorr::

    It is DC shunt motor designed to satisfy the requirements of the servomotor. The fieldexcited by a constant DC supply. If the field current is constant then speed is directly

    proportional to armature voltage and torque is directlyproportional to armature current.

    Transfer Function =Km

    S (1 + TmS)> Km = 1 / Avg Kb> Tm = JRa / Kb Kt> Kt = T / Ia> Eb = V-Ia Ra

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    FFiieelldd CoContntrrooll DD..CC.. SSerervvoo mmoottoor:r:

    It is DC shunt motor designed to satisfy the requirements of the servomotor. In thismotor 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.

    OBSERVATION TABLE FOR TRANSFER FUNCTION ARMATURE

    CONTROL DC SERVO MOTOR:

    TTaabbllee NNoo.. 11 FiFindndiinngg thethe vvalaluuee ooffKKbb

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

    Avg Kb

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    TTaabbllee NNoo.. 22 TToo ffiindnd RRaa

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

    Avg Ra =

    PRPREECAUCAUTTIIOONNS:S:

    At starting,The field rheostat should be kept in minimum resistance position

    PRPROOCCEEDURDUREE FFORORTTRANRANSSFFERERFFUNUNCCTTIIONON OFOF ARMARMAATTURUREE CCOONNTTRROLOL DDCC

    SESERVRVOOMMOTOOTORR

    FFiindndiinngg KKbb

    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.

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    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.

    FFiindndiinngg 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 toAA 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

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    PRPROOCCEEDURDUREE FFOORR TTRANRANSSFFERERFFUUNCNCTTIIONON OFOF FFIIEELLDD CCOONNTTRROLOL DD..CC.. SESERVRVOOMMOTOOTORR::

    FFiindndiinngg 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 FFrespectively.

    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.

    FFiindndiinngg ZZff

    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.

    FFiindndiinngg KKttll

    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 FFrespectively.

    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 Ktl

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    OBSEOBSERVARVATTIIONON TTAABLEBLE FFOROR TTRARANNSSFFERER FUFUNNCCTTIIONON OFOF ARMAARMATTUURREE CCOONNTTRROLOL

    DDCC SESERVRVOOMMOTOOTORR::

    TTaabbllee NNoo::33 ToTo ffiindnd RRff

    Sl.No If(amp) Vf(Volt) R f (ohm)

    Avg Rf=

    TTaabbllee NNoo::44 ToTo ffiindnd ZZff

    Sl.No If(amp)mA

    Vf(Volt) Zf= Vf / If

    AAvvgg ZZff ==

    TTaabbllee NNoo:: 55 ToTo ffiindnd KKttl

    Sl.No If Ia S1 S2 T( N m) N (rpm)

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    MMOODDELEL GGRAPRAPHH

    T

    FIELD CURRENT ARMATURE CURRENT

    MMOODDELEL CACALLCUCULLAATTIIOONN

    Result:

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    TTRANRANSSFFERERFFUNUNCCTTIIONON OFOF AACC SESERVRVOO MMOTOROTOR

    EEXPXPT.T.NNOO ::

    DADATETE ::

    AAIIMM::

    To determine the transfer function of the given AC servomotor

    APPAAPPARRAATTUUSS RREEQQUUIIRREEDD::

    S.No Name of the Equipment Range Type Quantity

    NAMNAMEE PPLLAATETE DDEETTAAIILS:LS:

    OUTPUT:VOLTAGE:CURRENT:SPEED:

    FUFUSESE RARATTIINNGS:GS:

    Blocked rotor test: 125% of rated current.

    THEOTHEORYRY::

    An servo motor is basically a two phase induction type except for certain special designfeatures. 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 (Inductivereactance / resistance) ratio is small which result in liner speed torque characteristics. The excitationvoltage applied to two stator winding should have a phase difference of 90o

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    WWOORRKKIINNGG PRPRIINNCCIIPPLELE OFOF AACC SESERRVVOOMMOTOROTOR

    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. Thedirection of rotation depends on the phase relationship of the two current (or voltage).

    The rotating magnetic field sweeps over the rotor conductor. The rotor conductorsexperience a change in flux and so voltage are induced in rotor conductors. This voltagecirculates 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) isdeveloped on the rotor and the rotor starts moving in the same direction as that of rotatingmagnetic field.

    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 createsrotor flux.

    Due to the interaction of stator and rotor flux, a mechanical force (or torque) isdeveloped on the rotor and the rotor starts moving in the same direction as that of rotatingmagnetic field.

    FFOORMURMULLAA::

    Transfer Function =Laplace Transform of output

    Laplace Transform of input

    (s) / Es(s) = K1 / sJ + K2 + B = Km / 1 + s ------ (1)

    Km = K1 / (K2 + B) ------------------------------- motor gain constant (2)

    m = J / (K2 + B) ---------------------------------- motor time constant (3)

    Torque (T) = 9.81 * r * s Nm

    S = applied load in KgR = radius of shaft in m = 0.068 m

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    CCoonnssttaantnt VaValluueess::

    J = 52 gm cm2 = 0.05kg cm2, B = 0.01875

    TTaabbllee NNoo:: 11

    OBSEOBSERVARVATTIIONON TTAABLEBLE FFORORDDETEETERMRMIINNIIGG MMOTOROTORCCOONNSTSTANANTT KK11::

    S.No Load(kg)

    ControlVoltage (Vc)

    Torque(Nm)

    TTaabbllee NNoo:: 22

    OBSEOBSERVRV

    AATTIIONON

    TTAABLEBLE

    FFOROR

    DDETEETERMRM

    IINNIINNGG

    MMOTOTOORR

    CCOONNSTSTAA

    NNTT KK22::

    11

    S.No Speed (N)rpm

    Load(kg)

    Torque (Nm)

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    PRPREECAUCAUTTIIOONNS:S:

    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 drumbefore starting the experiment .

    BLOCK DIAGRAM OF SERVOMOTOR

    PRPROOCCEEDURDURE:E:FFoorr deterdetermmiinniingng mmoottoorr ccoonnssttaantnt KK11

    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

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    FFoorr deterdetermmiinniingng mmoottoorr ccoonnssttaantnt KK22

    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.

    MMOODDELEL GGRAPRAPHH

    MMOTOROTORCCOONNSTSTANANTT K2K2 MMOTOROTORCCOONNSTSTANANTT K1K1

    TK2 = T / N TV

    N K1= T / V

    Speed in rpm

    MMOODDELELCACALLCUCULLAATTIIOO

    N:N:

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    Result:

    Speed in rpm

    EXPT NO. 3 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, = (ln MP) 2 / ( 2 + (ln MP) 2)

    Where MP is peak percent overshoot obtained from the response graph

    2. Undamped natural frequency, n = / tp (1 - 2)

    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 =1014

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    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 ofn and 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

    THEORY:

    The type number of the system is obtained from the number of poles located at origin in agiven system. Type 0 system means there is no pole at origin. Type 1 system means there is onepole located at the origin.

    The order of the system is obtained from the highest power of s in the denominator ofclosed loop transfer function of the system

    The first order system is characterized by one pole or a zero. Examples of first ordersystems are a pure integrator and a single time constant having transfer function of the form K/sand K/ (sT+1). The second order system is characterized by two poles and upto two zeros. Thestandard form of a second order system is C(s) /R(s) = n2 / (s2 + 2 ns + n2) where isdamping ratio and n is undamped natural frequency.

    BLOCK DIAGRAM:

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    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

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    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 nearsteady 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.

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    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 n and 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

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    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)

    Type equation here.3. To find the closed loop response of type 0 second order system

    S.

    N

    o.

    Gain,

    K

    Peak

    percent

    Overshoo

    t, %MP

    Rise

    time,tr

    (sec)

    Peak

    time,

    tp (sec)

    Settling

    time,ts

    (sec)

    Graphical Theoretical

    Dam

    ping

    ratio

    Undamped

    natural

    frequency,

    n

    (rad/sec)

    Dam

    ping

    ratio

    Undamped

    natural

    frequency,

    n(rad/sec)

    MODEL GRAPH:

    MODEL CALCULATION:

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    RESULT:

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

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    Ex. No: 4 Transfer Function of Separately Excited D.C. Generator

    Aim:

    To determine the transfer function of separately excited D.C. generator.

    Apparatus Required:

    Ammeter

    Voltmeter

    Rheostat

    DPST

    Name plate Details:

    D.C Shunt Motor D.C Generator

    1. K.W =3 1. K.W = 3

    2. R.P.M = 1500 2. R.P.M = 1500

    3. Voltage = 220 V 3. Voltage = 220 V

    4. A = 14 A 4. A = 14 A

    5. If = 0.014 A

    Formula Used:

    ff

    g

    f

    g

    SLR

    K

    SV

    sE

    +

    =

    )(

    )(

    (Without load)

    ( ) ( )aaffg

    f

    f

    SLRSLR

    K

    SV

    sV

    ++=

    .)(

    )(

    (With load)

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    Where

    La=Inductance of armature winding

    Lf= Inductance of field winding

    Kg=Generator rheostat

    Za,Zf= Total impendence due to armature and field winding

    Ra,Rf= Effective armature and field rheostat.

    Circuit Diagram:

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    Tabulation:

    To find Ra

    Va

    (V)

    Ia

    (A)

    Ra

    (

    )

    To find Rf

    Vf

    V

    If

    A

    Rf

    To find Za

    Va

    V

    Ia

    A

    Za

    To find Zf

    Vf

    V

    If

    A

    Zf

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    O.C.characteristics:

    Eg (V)

    If (A)

    Procedure:

    Determination of generator constant Kg

    1. The circuit connections were completed as in fig.

    2. The D.C motor was started using three point starter and made field rheostat.

    3. The voltage corresponding to zero field current was noted

    4. Then generator field rheostat was adjusted and opened circuit (voltage for various field

    current was tabulated)

    5. The above procedure was repeated until the voltmeter reads the rated voltage.

    6. A graph was drawn between Eg and If from linear portion of curve Kg was found out using

    formula

    f

    g

    gI

    EK

    =

    To find La, Ra, Za and Lf:

    1. The connections was completed as shown in fig,

    2. DPST switch was closed and rheostat or single phase Variac were varied

    3. The voltmeter and ammeter reading were tabulated. The average values are found out.

    Result:

    Thus the transfer function of the separately excited D.C. generator was found.

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    EX NO:TRANSFER FUNCTION OF ARMATURE CONTROLLED DC MOTOR

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

    Apparatus Required:Formula :

    1)

    (s) Kt

    =

    Va(s) (Ra + sLa)(B + sJ ) + Kt Kb

    (s) ( Kt / RaB)

    =

    Va (s) (1 + as) (1 + ms) + (Kt Kb/ Ra B)

    (s) ( Km)

    =

    Va(s) (1 +as) (1+ms) + (Kt Kb)

    Where,

    Kt -Torque constant

    Kb Back emf constant

    J - Moment of inertia

    B - Friction coefficient

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    Ra -Armature resistance

    La -Armature inductance

    Km - Motor gain constant = (Kt / Ra B)

    m - Mechanical time constant = (J/B)

    a - Armature circuit = (La / Ra)

    time constant

    2)

    1 Va.Kt

    B = - Kb . Kt N.m / rad / sec.Ra

    Where,

    - rated speed in rad / sec. Va - rated armature voltage in volts.3)

    (I1 + I2) (V1 + V2)

    P1 = watts

    2 2

    t2

    Pm = P1 watts

    t1 - t2

    2 2 dN

    also, Pm = x N x J x watts.60 dt

    where,

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    P1 Average power dissipated in load

    Pm Mechanical power developed

    N -Rated speed in rpm

    4) For armature controlled motor

    Kt = Kb

    Procedure:

    (a) Determination of back e.m.f constant Kb: [refer figure: (1.1)]1. Give the connections as per the circuit diagram.

    2. Keep the potential divider at the minimum potential position and the field rheostat

    at the minimum resistance position.

    3. Close the DPST switch and adjust the field rheostat to get a suitable field current say100% of the rated value.

    4. Vary the armature voltage by varying the potential divider.5. Note down the ammeter, voltmeter readings and speed.6. Repeat the steps (4) & (5) for various values of armature voltage.7. Tabulate the readings and plot the graph (Eb Vs w).8. Calculate the back e.m.f constant Kb.

    (b) Determination of moment of inertia J: [refer figure: (1.2)]

    1. Give the connections as per the circuit diagram.2. Keep the armature rheostat at maximum resistance position and field rheostat at

    minimum resistance position.3. With the DPDT switch at 1-1 position, close the DPST switch.4. Slowly vary the armature rheostat to start the motor smoothly.5. Adjust the speed of motor to rated value by varying field rheostat.6. Open the DPDT switch and note the time taken by motor for its speed to fall from rated

    value to1100rpm. Denote it as t1

    7. Open the DPST switch and repeat the steps (2) to (5).8. Close the DPDT switch from 1-1 position to 2-2 position.9. Note the time taken by motor for its speed to fall from its rated speed to 1100rpm.

    Denote it as t2.10.Note the Ammeter and Voltmeter readings at the instant of closing the DPDT to 2-2

    position. Denote them as I1 & V1 respectively.11.Note the ammeter and voltmeter readings when the speed is 1100rpm. Denote them as I2

    & V2 respectively.28

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    D

    P

    S

    12. Tabulate the readings and calculate the value of J.

    (c) Determination of armature resistance Ra: [refer figure: (1.3)]

    1. Give the connections as per the circuit diagram.

    2. Initially, the rheostat load should be in OFF position.

    3. Close the DPST switch.

    4. Apply the load in steps and note down the corresponding ammeter and voltmeterreadings.

    5. Tabulate the readings and calculate the armature resistance.

    (d) Determination of armature inductance La: [refer figure: (1.4)]

    1. Give the connections as per the circuit diagram.2. Keep the 1 autotransformer at minimum potential position.3. Close the DPST switch.4. Vary the autotransformer insteps and note down the corresponding ammeter & voltmeter

    readings.5. Tabulate the readings and calculate the armature inductance.

    Result:Thus the transfer function of armature controlled dc motor was determine and it is found

    to be,

    (s)

    =

    Va (s)

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    A

    V

    A

    31

    Determination of Back e.m.f ConstantK

    b:

    -

    Z

    ZZ

    A

    AA

    MFigure: (1.1)

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    Determination of Moment of Inertia J:

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    1 21 2 z

    zz

    AVA

    AA

    M

    33

    D

    P

    D

    T

    Figure: (1.2)

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    D

    P

    S

    T

    VAA

    AA

    AA

    M

    34

    Determination of armature inductance La:Figure: (1.3)

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    D

    P

    S

    T

    VAA

    AA

    AA

    M

    35

    Determination of Armature ResistanceR

    a:

    Rheostat LoadFigure: (1.4)

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    MODEL GRAPH :

    DETERMINATION OF BACK EMF CONSTSNT Kb :

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    DETERMINATION OF MOMENT OF INERTIA J

    DETERMINATION OF ARMATURE IMPEDANCE Za :

    Average Za =La =

    37

    S.No Va(Volts)

    Ia(Amps)

    N(rpm)

    Eb= (Va-IaRa)(Volts)

    = (2N / 60)(rad /sec)

    S.No Speed(rpm)

    Time(sec)

    Voltage(volts)

    Current(amps)

    Pm(watts)

    J(N Sec2 / m)

    N1 N2 T1 T2 V1 V2 I1 I2

    S.No Va(volts)

    Ia(amps)

    Za(ohms)

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    DETERMINATION OF ARMATURE RESISTANCE Ra :

    S.No Va(volts)

    Ia(amps)

    Ra(ohms)

    Average Ra =

    MODEL CALCULATION:

    RESULT:

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    EX NO:

    TRANSFER FUNCTION OF FIELD CONTROLLED DC MOTOR

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

    Apparatus Required:

    Formula :

    1) sVf s=Kt Rf + s LfB + s J2) sVf s= KtRf .B1 + f s1+ ms

    3) (s)Vf (s)=Km(1 + f s) (1+ ms)

    Where,

    Km - Field controlled motor gain constant

    f - Field circuit time constant = (Lf / Rf) m - Mechanical time constant = (J / B)

    Kt Torque constant of field controlled motor

    J Moment of inertia

    Rf Field resistance

    Lf Field inductanceB Viscous friction co-efficient

    2)

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    B = (Vf Kt 1/ . Rf) N.m / rad / sec.

    Where,

    - rated speed in rad / sec.Vf - rated field voltage in volts.

    3)

    where

    P1 Average power dissipated in load

    Pm Mechanical power developed

    N Rated speed in rpm

    Procedure:

    (a) Determination of Torque constant Kt: [refer figure: (3.1)]

    1. Give the connections as per the circuit diagram.

    2. Keep the potential divider at the minimum potential position and the field

    rheostat at the minimum resistance position.

    3. Close the DPST switch and adjust the field current to a suitable value say

    50% of the rated value by varying field rheostat.

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    4. Apply low voltage to armature circuit by varying the potential divider and note downthe armature current.

    5. Apply the load gradually till the motor comes to reset.Note: While loading the motor, the armature current should be maintained to

    the noted value by adjusting the potential divider.

    6. Note down the spring balance readings corresponding to the field current.

    7. Repeat the steps (4) to (6) for different values of field current.8. Tabulate the readings and plot the graph (T Vs If)9. Calculate the torque constant Kt.

    (a) Determination of moment of inertia J: [refer figure: (3.2)]1. Give the connections as per the circuit diagram.

    1. Keep the armature rheostat at maximum resistance position and field rheostat atminimum resistance position.

    2. With the DPDT switch at 1-1 position, close the DPST switch.3. Slowly vary the armature rheostat to start the motor smoothly.4. Adjust the speed of motor to rated value by varying field rheostat.6. Open the DPDT switch and note the time taken by motor for its speed to fall from ratedvalue to1100rpm. Denote it as t1

    7. Open the DPST switch and repeat the steps (2) to (5).8. Close the DPDT switch from 1-1 position to 2-2 position.

    9. Note the time taken by motor for its speed to fall from its rated speed to 1100rpm.Denote it as t2.

    10. Note the Ammeter and Voltmeter readings at the instant of closing the DPDT to 2-2position. Denote them as I1 & V1 respectively.

    11. Note the ammeter and voltmeter readings when the speed is 1100rpm. Denote them as I2& V2 respectively.

    12. Tabulate the readings and calculate the value of J.

    (c) Determination of field resistance Rf: [refer figure: (3.3)]

    1. Give the connections as per the circuit diagram.2. Initially, the rheostat load should be in OFF position.

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    3. Keeping the armature circuit open, close the DPST switch.4. Apply the load in steps and note down the corresponding ammeter and voltmeter

    readings.5. Tabulate the readings and calculate the armature resistance.

    (d) Determination of field inductance Lf: [refer figure: (3.4)]

    1. Give the connections as per the circuit diagram.2. Keep the 1 autotransformer at minimum potential position.3. Keeping the armature circuit open, close the DPST switch.4. Vary the autotransformer insteps and note down the corresponding ammeter & voltmeter

    readings.5. Tabulate the readings and calculate the field inductance.

    =

    Vf (s)

    Co

    42

    M

    Determination of Torque Constant Kt:

    S1

    -

    F A220V

    DC

    D

    P

    S

    T

    S

    V

    A

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    Tabulation:Radius of the Brake drum = -----------m

    Va = .Volts, Ia = Amps

    Model Graph:

    43

    S.No Vf

    (Volts)

    If

    (Amps)

    Load Torque =

    9.81 (S1 S2)

    (N.M)

    S1

    (Kg)

    S2

    (Kg)

    A

    V

    Figure: (3.1)

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    Determination of moment of inertia J :

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    1 21 2+D

    S

    P

    T

    F

    F

    AVA

    AA

    M

    45

    D

    P

    D

    T

    Figure. (3.2)

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    S.No Speed Time Voltage Current P m

    ()

    J

    (N.S2 /m)N1

    (rpm)

    N2

    (rpm)

    t1

    (sec)

    t2

    (sec)

    V1

    (V)

    V2

    (V)

    I1

    (A)

    I2

    (A)

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    220V

    DC

    Z

    ZZ

    AA

    D

    P

    S

    T

    VA

    47

    Determination of Field ResistanceR

    f:

    Figure 3.3

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    Tabulation :

    S.No Vf

    (Volts)

    If

    (Amps)

    Rf

    (Ohms)

    Average Rf=

    Result:

    Thus the transfer function of field controlled dc motor is determined.

    Expt. No: 6.a) STABILITY ANALYSIS OF LINEAR SYSTEMS(Bode Plot)

    AIMTo obtain the bode plot for the given system whose transfer function is given as

    G(S)= 242(s+5) s(s+1)(S2+5s+121)

    and to find out whether the system is stable or not.

    APPARATUS REQUIRED

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    MATLAB Software

    THEORY

    A Linear Time-Invariant Systems is stable if the following two notions of systemstability are satisfied

    When the system is excited by Bounded input, the output is also a Boundedoutput.

    In the absence of the input, the output tends towards zero, irrespective of theinitial conditions.

    The following observations are general considerations regarding system stability and are

    If all the roots of the characteristic equation have negative real parts, then the impulse

    response is bounded and eventually decreases to zero, then system is stable. If any root of the characteristic equation has a positive real part, then system is

    unstable. If the characteristic equation has repeated roots on the j-axis, then system is

    unstable. If one are more non-repeated roots of the characteristic equation on the j-axis,

    then system is unstable.

    BODE PLOT:

    Consider a Single-Input Single-Output system with transfer function

    C(s) R(s)= b0 Sm + b1Sm-1 + + bma0 Sn+ a1Sn-1 + +an

    Where m < n.

    Rule 1 A system is stable if the phase lag is less than 180 at the frequency forwhich the gain is unity (one).

    Rule 2 A system is stable if the gain is less than one (unity) at the frequency forwhich the phase lag is 180.The application of these rules to an actual process requires evaluation of the gain andphase shift of the system for all frequencies to see if rules 1 and 2 are satisfied. This isobtained by plotting the gain and phase versus frequency. This plot is called BODEPLOT. The gain obtained here is open loop gain.

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    The stability criteria given above represent Limits of Stability. It is well to design asystem with a margin of safety from such limits to allow for variation in components andother unknown factors. This consideration leads to the revised stability criteria, or more

    properly, a Margin of Safety provided to each condition. The exact terminology is interms of a Gain Margin and Phase Margin from the limiting values quoted.

    If the phase lag is less than 140 at the unity gain frequency, the system is stable.This then, is a 40 Phase Margin from the limiting values of 180.

    If the gain is 5dB below unity (or a gain of about 0.56) when the phase lag is180, the system is stable. This is 5dB Gain Margin.

    ALGORITHM

    1. Write a Program to (or using SIMULINK) obtain the Bode plot for the given system.2. Access the stability of given system using the plots obtained.

    PROGRAM

    %BODE PLOT OF THE SYSTEM%

    %Enter the numerator and denominator of the transfer function

    num=[0 0 0 242 1210];

    den=[1 6 126 121 0];

    sys=tf(num,den)

    %Specify the frequency range and enter the command

    w=logspace(-2,4,1000);

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    bode(sys,w)

    xlabel('Frequency')

    ylabel( ' Phase angle in degrees Magnitude of G(s)')

    title('Bode Plot of the system 242(s+5)/s(s+1)(s 2+5*s+121)')

    %To determine the Gain Margin,Phase Margin, Gain crossover frequency and

    %Phase cross over frequency

    [ Gm, Pm, Wcp, Wcg ]= margin (sys)

    PROCEDURE TO OBTAIN BODE PLOT

    1. Rewrite the sinusoidal transfer function in the time constant form by replacing s by j

    2. Identify the corner frequencies associated with each factor of the transfer function.

    3. knowing the corner frequencies draw the asymptotic magnitude plot. This plot

    consists of straight line segments with line slope changing at each corner frequency

    by +20db/decade for a zero and -20db/decade for a pole. For a complex conjugate

    zero or pole the slope changes by + 40db/decade.

    4. Draw a smooth curve through the corrected points such that it is asymptotic to the line

    segments. This gives the actual log-magnitude plot.

    5. Draw phase angle curve for each factor and add them algebraically to get the phaseplot.

    MANUAL CALCULATIONS

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    i)The sinusoidal transfer function G (j) is obtained by replacing s by j in the given s domaintransfer function

    G(j)= 242(j +5)j (j +1)( j 2+5 j +121)

    On comparing the quadratic factor of G(s) with standard form of quadratic

    factor , and n can be evaluated.

    s2+5s+121 = s2+2ns + n2

    On comparison

    n2 = 121 2n= 5

    n =11 rad/sec = 0.227

    G(j)= 10(1+0.2j)j (1+j)( 1+0.4 j -0.0083 2)

    ii)CORNER FREQUENCIES

    The corner frequencies are c1=1rad/sec c2= 5 rad/sec and c3=11rad/sec

    Choose a low frequency l such that l< c1 and choose a high frequency h> c3.

    Let l=0.5 rad/sec and h=100 rad/sec

    Term Corner Frequency

    rad/sec

    Slope db/dec Change in slopedb/dec

    10

    j

    __ -20 __

    1 1 -20 -20-20= - 40

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    (1+j)

    (1+0.2j) 5 20 -40-20 = -20

    1

    ( 1+0.4 j -0.0083 2)

    11 -40 -40-20 = -60

    iii)MAGNITUDE PLOTS

    Calculate A at l, c1, c2, c3, and h

    Let A= | G(j)| in db

    At = l A= 20 log(10/0.5)=26.03db

    = c1 , A=20l og(10/1)=20db

    = c2 A= -40log(5/1)+20=-7.96 db

    = c3 A = -20log(11/5) - 7.96 = -14.80 db

    = h A = -60log(100/11)-14.80 = - 72.3 db

    These values are plotted in the semilog graph sheet taking frequency along the logarithmic scaleand magnitude in db along the linear scale

    iv)PHASE PLOT

    The phase angle of G(j) as a function of is given by

    = G(j) = tan-1 0.2 -90 tan-1 tan-1 0.04/(1 0.00832)

    tan-1 0.2 tan-1 tan-1 {0.04/

    (1 0.00832)}

    = G(j)

    1 11.3 45 2.31 -126.01

    10 63.43 84.29 63.44 -174.3

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    20 75.96 87.14 180-19.98=160 -261.18

    100 87014 89.43 180-2.9=177.1 -269.3

    These values are plotted in the semilog graph sheet taking the same frequency as before alongthe logarithmic scale and phase angle in degrees along the linear scale.

    OUTPUT (from simulation)

    242 s + 1210S4+6S3+126S2+121S

    Gm = 2.0273

    Pm = 41.8270

    Wcp = 10.0961

    Wcg = 3.6322

    OUTPUT (from graph)

    gc= gc=3.1rad/sec

    Phase margin =180+ gc = 180-134 = 46 degrees

    Gain Margin = 12 db

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    pc = 10.1 rad/sec

    BODE PLOT

    RESULT:i)The Bode plot is drawn for the given transfer function using MATLAB and verified

    manually

    ii) The system is stable

    Expt. No: 6. b) STABILITY ANALYSIS OF LINEAR SYSTEMS

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    (Root Locus Plot)

    AIM

    To obtain the Root locus plot for the given system whose transfer function is given as

    G(S)= Ks(s+3)(S2+3s+11.25)

    APPARATUS REQUIRED

    Mat lab Software

    THEORY

    ROOT LOCUS PLOT :

    The characteristic of the transient response of a closed-loop system is related to the location ofthe closed loop poles. If the system has a variable loop gain, then the location of the closed-looppoles depend on the value of the loop gain chosen. A simple technique known as Root LocusTechnique used for studying linear control systems in the investigation of the trajectories of theroots of the characteristic equation.

    This technique provides a graphical method of plotting the locus of the roots in the s-plane as agiven system parameter is varied over the complete range of values(may be from zero toinfinity). The roots corresponding to a particular value of the system parameter can then belocated on the locus or the value of the parameter for a desired root location can be determinedform the locus. The root locus is a powerful technique as it brings into focus the completedynamic response of the system . The root locus also provides a measure of sensitivity of rootsto the variation in the parameter being considered. This technique is applicable to both single aswell as multiple-loop systems.

    PROCEDURE:

    1. Write a Program to (or using SIMULINK) obtain the Root locus plot for the givensystem.

    2. Access the stability of given system using the plots obtained.

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    PROGRAM

    %ROOT LOCUS OF THE SYSTEM%

    num=[0 0 0 0 1]

    den=[1 6 20.25 33.75 0]

    sys=tf(num,den)

    rlocus(sys)

    v=[-10,10,-8,8];

    axis(v)

    xlabel('Real Axis')

    ylabel('Imaginary Axis')

    title('Root Locus of the system ')

    title('Root Locus Plot of the system K/s(s+3)( s2+3s+11.25))')

    MANUAL CALCULATIONS

    1. Number of poles =4, zeros = 0, number of root locus branches =4. Starting points s=0, -3

    & 1.5+ j3.

    2. Pole zero plot is as follows

    Section between 0 and -3 is part of root locus. One breakway point is between s=0 and

    s=-3.

    3. Angle of asymptotes are 45,135,225 and 315 degrees

    4. Centroid = -1.5

    5. Three Breakway points are -1.5,-1.5 + j 1.8371

    6. Intersection with imaginary axis s= + j2.37.

    7. Angle of departure -90, +90.57

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    8. Root locus is plotted.

    9. Stability for 0< K82.26 system is unstable

    OUTPUT

    num =

    0 0 0 0 1

    den =

    1.0000 6.0000 20.2500 33.7500 0

    Transfer function:

    1S4+6S3+20.25S2+33.75S

    GRAPH(from Simulation)

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    RESULT:

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    Expt. No: 6 c) STABILITY ANALYSIS OF LINEAR SYSTEMS

    (Nyquist Plot)

    AIM

    To obtain the Nyquist plot for the given system whose transfer function is given as

    G(S)= 50 (s+4)(s2+3s+3)

    and to find out whether the system is stable or not.

    APPARATUS REQUIRED

    Mat lab Software

    THEORY

    POLAR PLOTS OR NYQUIST PLOTS:

    The sinusoidal transfer function G(j) is a complex function is given by

    G(j) = Re[ G(j)] + j Im[G(j)] or

    G(j) = G(j) G(j) = M -----------(1)

    From equation (1), it is seen that G(j) may be represented as a phasor of magnitude M andphase angle . As the input frequency varies from 0 to , the magnitude M and phase angle changes and hence the tip of the phasor G(j) traces a locus in the complex plane. The locus thusobtained is known as

    POLAR PLOT.

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    The major advantage of the polar plot lies in stability study of systems. Nyquist related the stabilityof a system to the form of these plots. Polar plots are referred as NYQUIST PLOTS.

    NYQUIST stability criterion of determining the stability of a closed loop system by investigatingthe properties of the frequency domain plot of the loop transfer function G(s) H(s).

    Nyquist stability criterion provides the information on the absolute stability of a control system assimilar to Routh- Hurwitz criterion. Not only giving the absolute stability, but indicates Degree

    of Stability i.e Relative Stability of a stable system and the degree of instability of an unstablesystem and indicates how the system stability can be improved. The Nyquist stability citerion isbased on a Cauchys Residue Theorem of complex variables which is referred to as the principleof argument.

    Let Q(s) be a single valued function that has a finite number of poles in the s-plane. Supposethat an arbitrary closed path q is chosen in the s-plane so that the path does not go throughany one of the poles or zeros of Q(s); the corresponding q locus mapped in the Q(s) planewill encircle the origin as many times as the difference between the number of the zeros andthe number of poles of Q(s) that are encircled by the s-plane locus q.

    The principle of argument is given by

    N= Z - P

    Where N number of encirclemnts of the origin made by the Q(s) plane locus q.

    Z number of zeros of Q(s) encircled by the s-plane locus q in the s-plane.

    P - number of poles of Q(s) encircled by the s-plane locus q in the s-plane.

    ALGORITHM

    1. Write a Program to (or using SIMULINK) obtain the Nyquist plot for the given system.2. Access the stability of given system using the plots obtained.

    PROGRAM

    %NYQUIST PLOT

    %Enter the numerator and denominator of the transfer function

    num=[0 0 0 50]61

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    den=[1 7 12 12]

    sys=tf(num,den)

    %Specify the frequency range and enter the command

    nyquist(sys)

    v=[-3 5 -7 7]

    axis(v)

    xlabel('Real Axis');

    ylabel('Imaginary Axis');

    title('Nyquist Plot of the sytem 50/(s+4)(s^2+3s+3)')

    %To determine the Gain Margin,Phase Margin, Gain crossover frequency and

    %phase cross over frequency

    [Gm,Pm,Wcp,Wcg]=margin (sys)

    v =

    -3 5 -7 7

    Gm =

    1.4402

    Pm =

    11.1642

    Wcp =

    3.4643

    Wcg =

    2.9533

    MANUAL CALCULATIONS:

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    NYQUIST PLOT

    RESULT:

    Thus the Nyquist plot is drawn for the given transfer function using matlab and verified manually

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    Expt. No: 7a) DC POSITION CONTROL SYST M

    To study the characteristics of a dc position control system.

    APPARATUS REQUIRED:

    i) DC position control kit and Motor unit

    ii) MultimeterTHEORY:

    A pair of potentiometers acts as error-measuring device. They convert theinput and output positions into proportional electric signals. The desired position is set on the inputpotentiometer and the actual position is fed to feedback potentiometer. The difference between thetwo angular positions generates an error signal, which is amplified and fed to armature circuit ofthe DC motor. If an error exists , the motor develops a torque to rotate the output in such a way asto reduce the error to zero. The rotation of the motor steps when the error signal is zero, i.e., whenthe desired position is reached.

    PROCEDURE:

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    1. The input or ref potentiometer is adjusted nearer to zero initially.2. The command switch is kept in continuous mode and some value of forward gain Ka is

    selected.3. For various positions of input potentiometer (r) the positions of the response

    potentiometer (0) is noted. Simultaneously the reference voltage measured between Vr &E and the output voltage measured between VO & E are noted.

    4. A graph is plotted with0 along y-axis andralong x-axis.

    Tabular Column

    S.`No r

    degree

    O

    degree

    Vr in Volts VO in Volts

    MODEL GRAPH:

    Output

    Position

    (Deg)

    Input Position (Deg)

    RESULT

    Thus the dc position control system characteristics are studied and corresponding graphs are drawn

    Ex no 7 (b) AC POSITION CONTROL SYSTEM

    AIM: Study Of Synchro Transmitter and receiver

    APPARATUS REQUIRED

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    S.NoName of the Equipment

    Quantity1 Synchro transmitter and receiver unit 12 Multimeter (Digital / Analog ) 13 Patch cords As required

    THEORY:

    A synchro is an electromagnetic transducer commonly used toconvert an angular position of a shaft into an electric signal. It is commercially known as aselsyn or an autosyn. The basic synchro unit is usually called a synchro transmitter. Itsconstruction is similar to that of three phase alternator. The stator is of laminated silicon steeland is slotted to accommodate a balanced three phase winding which is usually of concentriccoil type and star connected. The rotor is dumb bell construction and its wound with aconcentric coil.

    AC voltage is applied to rotor winding through slip rings. Let and

    AC voltage

    Vr (t) = Vr sin ct be applied to the rotor of the synchro transmitter.

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    The voltage causes a flow of magnetizing current in rotor coil which produces asinusoidally time varying flux directed along its axis and distributed nearly sinusoidallyin the air gap along the stator periphery. Because of transformer action, voltage is inducedin each of the stator coil. As the air gap flux sinusoidally distributed the flux linking with anystator 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 thesynchro 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,is the angle made by rotor axis with the stator coil S2.

    The various stator voltages are

    Vs1n = KVrsin ct cos ( + 120o)

    Vs2n = KVrsin ct cos

    Vs1n = KVrsin ct cos ( + 240o)

    The terminal voltages of the stator are

    Vs1s2Vs1n Vs2n

    3 KVr

    Sin(2400sin ct)

    Vs2s3Vs2n Vs3n

    3 KVr

    Sin(1200sin ct)

    Vs3s1 Vs3n Vs1n 3 KVr Sin sin c t

    When = 0, Vs1s2 and Vs2 s3 have the maximum voltage and while Vs3s1 has zerovoltage. This position of rotor is defined as ht electrical zero of the transmitter and is used asreference for specifying the angular position of the rotor.

    Thus it is seen that the input to the synchro transmitter is the angular position ofits rotor shaft and the output is a set of three signal phase voltages. The magnitudes of thisvoltage are function of the shift position. The output of the synchro transmitter is applied tostator winding of synchro control transformer.

    The control transmitter is similar in construction to a synchro transmitter except forthe fact that rotor of the control transformer in made cylindrical in shape so that the air gap ispractically uniform. The system (transmitter and control transformer pair) acts an error

    detector, circulating current to the same phase but of different magnitudes flow through twostator coils. The result is establishment ofan indentical flux pattern in the air gap of the control transformer as the voltage drops in

    resistance and lockage reactances of two sets of stator coils are usuallysmall.

    OBSEOBSERVARVATTIIONON TTAABLEBLE

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    S.NoTransmitter

    (Degree)

    Receiver

    (Degree)Vs1 Vs2 Vs2 Vs3 Vs3 Vs1 Error

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

    E (t) = KVr cos sinr tThe synchro transmitter and control transformer thus act as an error detector giving a voltagesignal at the rotor terminals of the control transformer proportional to the angular differencebetween the transmitter control transformer shaft positions.PRPROOCCEEDURDURE:E:

    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:

    EX NO STEPPER MOTOR CONTROL SYSTEM

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    DATE

    AIM

    To control the stepper motor using microprocessor using 8085.

    APPARATUS REQUIRED

    (1) 8085 Micro Processor Kit(2) Power chord(3) stepper motor(4)stepper motor interface board

    ALGORITHM

    Step 1: Load the count value and data into memory locations

    Step 2: Get the first data and move it to interface board

    Step 3: Call delay routine

    Step 4: Decrement the count

    Step 5: If count is equal to zero, then go to next step else go to step 2

    Step 6 : Stop

    PROGRAM

    Memory address Label mnemonics opcode

    4100 START LXI H,LOOL UP 21 1A 41

    4103 MVI B,04 06 04

    4105 REPT MOV A,M 7E

    4106 OUT OC0H D3 CO4108 LXI D, 0303H 11 03 03

    410B DELAY NOP 00

    410C DCX D 1 B

    410D MOV A,E 7 B

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    410E ORA D B2

    410F JNL DELAY C2 0B 41

    4112 INX H 23

    4113 DCR B 05

    4114 JNZ PEPT C2 05 41

    4117 JMP START C3 00 41

    411A LOOK UP DB 09 05 06 0A 09 05 06 0A

    RESULT

    Expt. No9: DIGITAL SIMULATION OF FIRST ORDER SYSTEMS

    AIMTo digitally simulate the time response characteristics of a linear system without non

    linearities and to verify it manually.

    APPARATUS REQUIRED

    PC with MATLAB package

    THEORY

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

    The desired performance characteristics of a system of any order may be specified in termsof transient response to a unit step input signal. The transient response characteristics of a controlsystem 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

    PROCEDURE:

    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.

    4. From the sink block in the library drag thescope.

    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.

    9. From the step response obtained note down the rise time, peak time, peak overshoot and

    settling time.

    10. For the same transfer function write a matlab program to obtain the step response and verify

    both the results.

    PROGRAM

    %This is a MATLAB program to find the step response

    num=[0 25];

    den=[1 5];

    sys = tf (num,den);71

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    step (sys);

    PLOT

    RESULT

    The time response characteristics of a linear system without non linearities is simulated digitallyand verified manually

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    Expt. No 10: DIGITAL SIMULATION OF SECOND ORDER SYSTEMS

    DATE:

    AIM

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

    APPARATUS REQUIRED

    A PC with MATLAB package

    THEORY

    The time characteristics of control systems are specified in terms of time domainspecifications. Systems with energy storage elements cannot respond instantaneously and willexhibit 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 controlsystem 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

    PROCEDURE:

    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.

    4. From the sink block in the library drag thescope.

    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 respectiveblock.

    8. Start simulation and observe the results in scope.

    9. From the step response obtained note down the rise time, peak time, peak overshoot and

    settling time.

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    10. For the same transfer function write a matlab program to obtain the step response and verify

    both the results.

    PROGRAM

    num=[0 0 25];

    den=[1 6 25];

    sys = tf (num,den);

    step (sys);

    Model graph

    RESULT

    The time response characteristics of a linear system without non linearities is simulated digitallyand verified manually.


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