SSN COLLEGE OF ENGINEERING KALAVAKKAM- 603 110DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING EE 2257 - CONTROL SYSTEMS LAB MANUAL Dec 2009-April 2010Name: _______________________________________Reg. No.: _____________________________________Year: II Sem: 4 Sec: A/B Dept: EEE
2DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING EE 2257- CONTROL SYSTEMS LABORATORYNAME OF THE STUDENT : ____________________________REGISTER NUMBER : ____________________________ CLASS : II- EEE- A / BACADEMIC YEAR : Dec 2009- Apr 2010TOTAL MARKS : -------- / 10SIGNATURE OF THE STAFF:
3 LIST OF EXPERIMENTS1. Determination of transfer function parameters of Armature controlled DC (servo) motor.2. Determination of transfer function parameters of Field controlled DC (servo) motor.3. Determination of transfer function parameters of an AC servomotor.4. Analog simulation of type-0 and type-1 systems5. Digital simulation of first order systems6. Digital simulation of second order systems.7. Stability analysis of linear systems.8. DC and AC position control systems.9. Stepper motor control system10. Determination of transfer function parameters of DC generators.11. Study of synchros12. Design and implementation of compensators.13. Design of P, PI and PID controllers. . P = 45, TOTAL = 45
4 INDEX SIGN. OFExpt. Title MARKS DATE THENO. (10) STAFF 1 2 3 4 5 6 7 8 9 10 11 12 13 Total Marks Signature of the faculty:
5EXPT. NO.:DATE: TRANSFER FUNCTION OF ARMATURE CONTROLLED DC SERVOMOTORAIM: To determine the transfer function of an armature controlled dc servomotor.APPARATUS REQUIRED:THEORY: Transfer function is defined as the ratio of Laplace transform of the outputvariable to the Laplace transform of input variable at zero initial conditions.Armature controlled DC shunt motorIn this system, Ra = Resistance of armature in Ω La= Inductance of armature windings in H Ia = Armature current in A If = Field current in A e = Applied armature voltage in V eb = back emf in V Tm = Torque developed by the motor in Nm
6 J = Equivalent moment of inertia of motor and load referred to motor shaft in kgm2 B= Equivalent viscous friction coefficient of inertia of motor and load referred to motor shaft in Nm/(rad/s)In Servo applications, DC motors are generally used in the linear range of themagnetization curve. Therefore, the air gap flux φ is proportional to the field current. φ α If φ = Kf If ,where Kf is a constant. ----------------------------- (1)The torque Tm developed by the motor is proportional to the product of the armaturecurrent and air gap flux. Tm α φ Ia Tm =Ki φ Ia = Ki Kf If Ia , where Ki is a constant --------------(2)In the armature controlled DC motor, the field current is kept constant. So the aboveequation can be written as Tm = Kt Ia , Where Kt is known as motor torque constant.------ (3)The motor back emf being proportional to speed is given by eb α dθ/dt, eb = Kb dθ/dt, where Kb is the back emf constant.----------------(4)The differential equation of the armature circuit is e = IaRa + La dIa/dt + eb ----------------------------------------- (5)The torque equation is Tm = Jd2θ/dt2 + B dθ/dt ------------------------------------------ (6)Equating equations (3) and (6) Jd 2θ/dt2 + B dθ/dt = Kt Ia ---------------------------------------(7)Taking Laplace transforms for the equations (4) to (7), we get Eb(s) = Kb s θ(s) -------------------------------------------- (8) (s La + Ra ) Ia(s) = E(s) – Eb(s). ------------------------------- (9) ( J s2+ B s) θ(s) = Tm (s) = Kt Ia(s) ---------------------------- (10) From equations (8) to (10) , the transfer function of the system is obtained as
7Block diagramUsing the above equations, the block diagram for the armature controlled DC motor isgiven below:E(s)+ θ (s)ω(s) 1/[Ra+sLa] Kt 1/s[Js+B] - Eb(s) s Kb 1. Circuit diagram to determine Kt and Kb
82. Circuit diagram to determine Ra:3. Circuit diagram to determine La:PROCEDURE: i)Load test to determine Kt 1. Initially keep all the switches in the off position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch on the power and the SPST switches S1 and S2. 5. Adjust the field voltage to the rated value. 6. Apply the armature voltage until the motor runs at the rated speed. 7. Apply load and note the armature voltage, current and spring balance readings. 8. Calculate torque and plot the graph between torque and armature current. 9. Determine Kt from graph.
9 ii)No-Load test to determine Kb. 1. Initially keep all the switches in the off position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch on the power and the SPST switches S1 and S2. 5. Set the field voltage to the rated value. 6. Adjust the armature voltage and note the armature voltage, current and speed. 7. Calculate the back emf eb and plot the graph between back emf and ω 8. Determine Kb from graph. iii) To determine Ra: 1. Initially keep all the switches in the off position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch on the power and the SPST switches S1. 5. Note the armature current for various armature voltages. 6. Calculate Ra. iv) To determine La: 1. Initially keep all the switches in the off position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch on the power . 5. Apply ac voltage to armature winding 6. Note down the current for various input ac voltage . 7. Calculate Ra.Tabulation to determine KtS.No. Armature Armature Spring Balance Torque Voltage Current Readings (kg) T = 9.81(S1-S2)r Va (V) Ia (A) S1 S2 (Nm) Where r is the radius of the brake drum. r = _____________m
10Tabulation to determine KbS.No. Armature Armature Speed N Eb= Va-Ia Ra ω = 2πN/60 π Voltage Va Current Ia (rpm) (rad/sec) (V) (A) (V) -Tabulation to determine RaS.No. Armature Armature Ra Voltage Current (Ω) Va (V) Ia (A)Calculation by least square methodRa = [V1I1 +V2I2 +V3I3+V4I4 ] / (I12+I22+I32+I42)Tabulation to determine ZaS.No. Armature Armature Za Voltage Current (Ω) Va (V) Ia (A)Average Za = ________Ω
11 MODEL GRAPH To find Kt To find KbTorque Eb( V)(Nm) Ia(A) ω(rad/sec) MODEL CALCULATION Ra = ……..Ohms Za = …….. Ohms La = √(Za2 –Ra2 ) / 2πf = ……. H f = 50 Hz J = 0.074 kg/m2, B = 0.001Nm/rad/sec From Graph, Kt = Torque constant = ∆T / ∆ Ia = ………… Nm / A Kb = Back emf constant = ∆Eb / ∆ ω = ………. V/(rad/s)RESULT:INFERENCE:
12EXPT. NO.:DATE: TRANSFER FUNCTION OF FIELD CONTROLLED DC SERVOMOTORAIM: To determine the transfer function of a field controlled dc servomotor.APPARATUS REQUIRED:THEORY: The transfer function is defined as the ratio of Laplace transform of the outputvariable to the Laplace transform of input variable at zero initial conditions.Armature controlled DC shunt motorIn this system, Rf = Resistance of the field winding in Ω Lf= Inductance of the field windings in H Ia = Armature current in A If = Field current in A e = Applied armature voltage in V eb = back emf in V ef = Field voltage in V
13 Tm = Torque developed by the motor in Nm J = Equivalent moment of inertia of motor and load referred to motor shaft in kgm2 B= Equivalent viscous friction coefficient of inertia of motor and load referred to motor shaft in Nm/(rad/s)In Servo applications, the DC motors are generally used in the linear range of themagnetization curve. Therefore the air gap flux φ is proportional to the field current. φ α If φ = Kf If ,where Kf is a constant. -------------------------------- (1)The torque Tm developed by the motor is proportional to the product of the armaturecurrent and air gap flux. Tm α φ Ia Tm =K′φ Ia = K′ Kf If Ia = Km Kf If , where Ki is a constant ----(2)Appling Kirchhoff’s voltage law to the field circuit, we have Lf dIf/dt + RIf = ef ------------------------------------------------- (3)Now the shaft torque Tm is used for driving the load against the inertia and frictionaltorque. Hence, Tm = Jd2θ/dt2 + B dθ/dt ------------------------------------------- (4)Taking Laplace transforms of equations (2) to (4), we get Tm(s) = KmKf If (s) ----------------------------------------------- (5) Ef(s) = (s Lf + Rf) If(s) -------------------------------------------- (6) Tm(s) = (J s2+ B s) θ(s) ------------------------------------------- (7) Solving equations (5) to (7), we get the transfer function of the system as
141. Circuit diagram to determine KmKf2. Circuit diagram to determine Rf3. Circuit diagram to determine Lf
15PROCEDURE: i)Load test to determine KmKf 1. Initially keep all the switches in the OFF position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch ON the power and the SPST switches S1 and S2. 5. Apply 50% of the rated field voltage. 6. Apply the 50% of the rated armature voltage. 7. Apply load and note the field current and spring balance readings. 8. Vary the field voltage and repeat the previous step. 9. Calculate torque and plot the graph between torque and field current. 10. Determine KmKf from graph. ii) To determine Rf 1. Initially keep all the switches in the OFF position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch ON the power and the SPST switch S2. 5. Note the field currents for various field voltages. 6. Calculate Rf. iii) To determine Lf 1. Initially keep all the switches in the OFF position. 2. Keep all the voltage adjustment knobs in the minimum position. 3. Give connections. 4. Switch ON the power . 5. Apply AC voltage to field windings 6. Note the currents for various input AC voltages. 7. Calculate Lf.
16 Tabulation to determine KmKf S.No Field Spring Balance Torque Current Readings (kg) T = 9.81(S1-S2)r If (A) S1 S2 (Nm) Where r is the radius of the brake drum. r = _____________m Tabulation to determine Rf S.No. Field Field Rf Voltage Current (Ω) Ef (V) If (A)Calculation by least square methodRf = [V1I1 +V2I2 +V3I3+V4I4 ] / (I12+I22+I32+I42) Tabulation to determine Zf S.No. Field Field Zf Voltage Current (Ω) Ef (V) If (A)Average Zf = ________Ω
17 MODEL GRAPH To find KmKfTorque(Nm) If (A) MODEL CALCULATIONS 1. Rf = ……..Ohms 2. Zf = …….. Ohms 3. Lf = √(Zf2 –Rf2 ) / 2πf = ……. H 4. f = 50 Hz 5. J = 0.074 kg/m2, B = 0.001Nm/rad/s 6. From Graph, KmKf = ∆T / ∆If = ………… Nm / ARESULT:INFERENCE:
18EXPT. NO:DATE :DETERMINATION OF TRANSFER FUNCTION PARAMETERS OF AC SERVO MOTORAIM: To derive the transfer function of the given AC servomotor andexperimentally determine the transfer function parameters.APPRATUS REQUIRED:FORMULA: 1. Motor transfer function θ(s) Km = Eo (s) s (1+s.τm) 2. Motor gain constant Km = K1 K2 + B 3. Motor time constant τm= J K2 + B Where K1 = slope of torque - control phase voltage characteristics K2= slope of torque -speed characteristics J = Moment of inertia of load and the rotor B= viscous frictional coefficient of load and the rotorTHEORY When the objective of a system is to control the position of an object, then thesystem is called a servomechanism. The motors that are used in automatic controlsystems are called servomotors. Servomotors are used to convert an electrical signal (control voltage) into anangular displacement of the shaft. In general, servomotors have the followingfeatures. 1. Linear relationship between speed and electrical control signal 2. Steady state stability 3. Wide range of speed control
19 4. Linearity of mechanical characteristics throughout the entire speed range 5. Low mechanical and electrical inertia 6. Fast responseDerivation of Transfer Function:Let Tm = Torque developed by the servomotor θ = angular displacement of the rotor ω = dθ / dt = angular speed TL = torque required by the load J = Moment of inertia of the load and the rotor B = Viscous frictional coefficient of the load and the rotor K1 = slope of the control phase voltage and torque characteristics. K2 = slope of the speed and torque characteristics.The transfer function of the AC servomotor can be obtained by torque equation. Themotor developed torque is given by Tm = K1 e c – K2 dθ…………………………………(1) dtThe rotating part of the motor and the load can be modeled by TL = J d2θ + B.dθ …………………………. …………(2) dt2 dtAt equilibrium, the motor torque is equal to load torque. Hence, K1 e c – K2 dθ /dt = J d 2θ + B dθ …………....(3) dt dtTaking Laplace Transform K1 Ec (s) – K2 s θ(s) = J s2 θ(s) + B s θ(s)………………(4) θ (s) K1 Km T.F = = = Ec (s) s (K2+s J+B) s (1 +s τm) K1Where motor gain constant Km = B + K2 Jand motor time constant τm = B + K2
20PROCEDURE:I. DETERMINATION OF TORQUE SPEED CHARACTERISTICS 1. Give the connections. 2. Connect voltmeter or a digital Multimeter across the control winding. 3. Apply rated voltage to the reference phase winding and control phase winding. 4. Note the no load speed. 5. Apply load in steps. For each load, note the speed. 6. Repeat steps 4,5 for various control voltage levels and tabulate the readings.II. DETERMINATION OF TORQUE – CONTROL VOLTAGE CHARACTERISTICS 1. Make connections. 2. Connect voltmeter or a digital Multimeter across the control phase winding 3. Apply rated Voltage to Reference phase winding. 4. Apply a certain voltage to the control phase winding and make the motor run at low speed. Note the voltage and the no load speed. 5. Apply load to motor. Motor speed will decrease. Increase the control voltage until the motor runs at same speed as on no-load. Note the control voltage and load. 6. Repeat steps 5 for various loads 7. Repeat 4-6 for various speeds and tabulate.Torque Speed Characteristics Radius of brake drum =________ Vc = Vc = Vc = Load N Torque Load N Torque Load N Torque g rpm N-m g rpm N-m g rpm N-mModel Graph Torque N-m Speed (rpm)
21Torque –Control Voltage Characteristics N1 = N2 = N3 =Load Vc Torque Load Vc Torque Load Vc Torque g V N-m g V N-m g V N-mModel Graph Torque N-m Control voltage (Volts)From Graph , K1 = K2 =Given , B= J=From Calculations, Km = τm =
23EXPT. NO:DATE : ANALOG SIMULATION OF TYPE-0 AND TYPE-1 SYSTEM AIM: To simulate the time response characteristics of I order and II order, type 0and type-1 systems.APPARATUS REQUIRED:THEORY: Order of the system: The order of the system is given by the order of the differential equationgoverning the system. The input-output relationship of a system can be expressed bytransfer function. Transfer function of a system is obtained by taking Laplacetransform of the differential equation governing the system and rearranging them asratio of output and input polynomials in ‘s’. The order is given by the maximumpower of ‘s’ in denominator polynomial Q(s) T(s) = P(s) / Q(s) P(s) --- Numerator polynomial Q(s) --- Denominator polynomial Q(s) =ao sn + a1sn-1 + a2 sn-2 + ………….+ an-1 s + a nIf n=0, then system is Zero-Order system.If n=1, then system is First-Order system.If n=0, then system is Second-Order system. Type of the systemType of the system is given by the number of poles of the loop transfer function at theorigin. G(s)H(s) = K P(s) / Q(s) (s+z1) (s+z2) (s+z3) …….. = sN (s+p1) (s+p2) (s+p3) ……..If N=0, the system is a Type Zero system.If N=1, the system is a Type One system.If N=0, the system is a Type Two system.
24First –Order Type ‘0’ system The generalized transfer function for first order Type –0 system is T(s) = C(s) / R(s) = 1/(1+sτ) --------------------------------------------------------(1) C(s) ---- Output of the system R(s) ----- Reference input to the system.If input is a Step input R(s) = 1/s ----------------------------------------------------- (2)From eqn (1) 1 C(s) = R(s) ---------------------------------------(3) (1+sτ)substituting for R(s), 1 1 C(s) = ------------------------------------(4) s (1+sτ)To find C(t) , Take Inverse Laplace Transform of eqn (4), C(t) = 1 – e-t/τ ------------------(5)PROCEDURE: 1. Give the connections as per the block diagram in the process control simulator using the front panel diagram . 2. Set the Input (set point) value using the set value knob. 3. Observe the Output (process value or PV) using CRO and plot it in the graph. 4. Tabulate the reading and calculate the % error. 5. Repeat the procedure in closed loop condition.
25 TABULATION FOR FIRST ORDER SYSTEM: (a)Type Zero systemLoop type Set Point Process Settling Error % Error SP variable Time SP-PV SP-PV x 100% (V) PV (s) (V) SP (V)Open LoopClosed Loop (b)Type One System Loop type Set Process Settling Error % Error Point variable Time SP-PV SP-PV x 100% SP PV (s) (V) SP (V) (V) Open loop Closed loop TABULATION FOR SECOND ORDER SYSTEM (a)Type Zero systemLoop type Set Point Process variable Settling Error % Error SP PV Time SP-PV SP-PV x 100% (V) (V) (s) (V) SPOpen loopClosed loop (b)Type One systemLoop type Set Point Process Settling Error % Error SP variable Time SP-PV SP-PV x 100% (V) PV (s) (V) SP (V)Open loopClosed loop
27EXPT. NO:DATE : DIGITAL SIMULATION OF FIRST ORDER SYSTEMS(i) Digital Simulation of first order Linear and Non Linear SISO Systems AIM: To digitally simulate the time response characteristics of Linear and Non Linear SISO systems using state variable formulation. APPARATU REQUIRED: A PC with MATLAB package. THEORY: SISO linear systems can be easily defined with transfer function analysis. The transfer function approach can be linked easily with the state variable approach. The state model of a linear-time invariant system is given by the following equations: X(t) = A X(t) + B U(t) State equation Y(t) = C X(t) + D U(t) Output equation Where A = n x n system matrix, B = n x m input matrix, C= p x n output matrix and D = p x m transmission matrix,PROGRAM/ SIMULINK MODEL:
30 (ii) Digital Simulation of Multi-Input Multi-Output Linear SystemsAIM: To digitally simulate the time response characteristics of MIMO Linear systemusing state-variable formulation.APPARATUS REQUIRED: • PC • MATLAB Package.THEORY: State Variable approach is a more general mathematical representation of asystem, which, along with the output, yields information about the state of the systemvariables at some predetermined points along the flow of signals. It is a direct time-domain approach, which provides a basis for modern control theory and systemoptimization.u1(t) y1(t)u2(t) Controlled system y2(t) U Controlled Y . State variables (n) . system . .um(t) yp(t) . ....... X x1(t) x2(t) xn(t). X(t) = A X(t) + B U(t) State equation Y(t) = C X(t) + D U(t) Output equationThe state vector X determines a point (called state point) in an n - dimensional space,called state space. The state and output equations constitute the state model of thesystem.
33 Expt. No.: Date: DIGITAL SIMULATION OF SECOND ORDER SYSTEMS AIM: To digitally simulate the time response characteristics of second order linear and non-linear system with saturation and dead zone. APPARATU REQUIRED: A PC with MATLAB package.PROGRAM / SIMULINK MODEL:
35Expt. No.:Date: STABILITY ANALYSIS OF LINEAR SYSTEMS AIM: To analyze the stability of linear system using Bode plot/ Root Locus / Nyquist Plot. APPARATUS REQUIRED: A PC with MATLAB package. THEORY: A Linear Time-Invariant Systems is stable if the following two conditions ofsystem stability are satisfied When the system is excited by a bounded input, the output is also bounded. In the absence of the input, the output tends towards zero, irrespective of the initial conditions. PROCEDURE: 1. Write a Program to obtain the Bode plot / Root locus / Nyquist plot for the given system. 2. Determine the stability of given system using the plots obtained. PROGRAM:
37EXPT. NO:DATE : CLOSED LOOP DC POSITION CONTROL SYSTEMAIM: To study the operation of closed loop position control system (DCServomotor) with a PI controller.APPARATUS REQUIRED:THEORY: A pair of potentiometers is used to convert the input and output positions intoproportional electrical signals. The desired position is set on the input potentiometerand the actual position is fed to feedback potentiometer. The difference between thetwo angular positions generates an error signal, which is amplified and fed toarmature circuit of the DC motor. If an error exists , the motor develops a torque torotate the output in such a way as to reduce the error to zero. The rotation of the motorstops when the error signal is zero, i.e., when the desired position is reached. 1/T1 KP Chopper Motor Gear + - Position Sensor Fig. 1 Block Diagram
38 Fig.(2.).Front PanelPROCEDURE: 1. Switch on the system. Keep the pulse release switch in OFF position. 2. Vary the set point with the pulse release switch in the ON position and note the output position. 3. Note SP voltage , PV voltage, P voltage and PI output voltage. 4. Calculate KP using the formula KP = P/(SP-PV).
39 TABULATION S.NO. POSITION (degrees) Error (set – output) in degrees set outputRESULT:INFERENCE:
40EXPT. NO:DATE : CLOSED LOOP AC POSITION CONTROL SYSTEMAIM: To study the closed loop operation of AC position control system (ACServomotor) with PI controller.APPARATUS REQUIRED:THEORY: CONSTRUCTIONAL DETAILS The AC servomotor is a two-phase induction motor with some special designfeatures. The stator consists of two pole pairs (A-B and C-D) mounted on the innerperiphery of the stator, such that their axes are at an angle of 90o in space. Each polepair carries a winding, one winding is called the reference winding and other windingis called the control winding. The exciting currents in the two windings should have aphase displacement of 90 o. The supply used to drive the motor is single-phase andhence a phase advancing capacitor is connected to one of the phases to produce aphase difference of 90 o. The stator constructional features of AC servomotor areshown in fig.1. The rotor construction is usually of squirrel cage or drag-cup type. Thesquirrel cage rotor is made of laminations. The rotor bars are placed on the slots andshort-circuited at both ends by end rings. The diameter of the rotor is kept small inorder to reduce inertia and to obtain good accelerating characteristics. Drag cupconstruction is employed for very low inertia applications. In this type ofconstruction, the rotor will be in the form of hollow cylinder made of aluminium. Thealuminium cylinder itself acts as short-circuited rotor conductors. WORKING PRINCIPLES The stator windings are excited by voltages of equal rms magnitude and 90 ophase difference. This results in exciting currents i1 and i2 displaced in phase by 90 oand having identical rms values. These currents give rise to a rotating magnetic fieldof constant magnitude. The direction of rotation depends on the phase relationship ofthe two currents (or voltages). The exciting current shown in fig.2 produces aclockwise rotating magnetic field. When i1 is shifted by 180 o, an anticlockwiserotating magnetic field is produced. This rotating magnetic field sweeps over the rotorconductors. The rotor conductor experience a change in flux and so voltages are
41induced in rotor conductors. This results in circulating currents in the short-circuitedrotor conductors resulting in rotor flux. Due to the interaction of stator & rotor flux, a mechanical force (or Torque) isdeveloped in the rotor and the rotor starts moving in the same direction as that ofrotating magnetic field. Fig 1 Stator Construction of AC Servomotor Fig 2.Waveforms of Stator & Rotor Excitation Current Fig.3 Basic Block Diagram of AC Position Control System
42PROCEDURE: 1. Switch ON the system. Keep the pulse release switch in the OFF position. 2. Vary the set point with the pulse release switch in the ON and note the output position. 3. Note the SP voltage, PV voltage, P voltage and PI output voltage. 4. Calculate KP using the formula KP = P/(SP-PV).TABULATION S.NO. SET OUTPUT ERROR=(set position-output POSITION POSITION position) (degrees) (degrees) (degrees)RESULT:INFERENCE:
43 Ex. No: Date: STEPPER MOTOR Aim: To study the Stepper motor Theory: Stepper motors are highly accurate pulse-driven motors that change their angular position in steps, in response to input pulses from digitally controlled systems. A stepper or stepping motor converts electronic pulses into proportionate mechanical movement. Each revolution of the stepper motors shaft is made up of a series of discrete individual steps. A step is defined as the angular rotation produced by the output shaft each time the motor receives a step pulse. These types of motors are very popular in digital control circuits, such as robotics, because they are ideally suited for receiving digital pulses for step control. Each step causes the shaft to rotate a certain number of degrees. A step angle represents the rotation of the output shaft caused by each step, measured in degrees. Figure.1. illustrates a simple application for a stepper motor. Each time the controller receives an input signal, the paper is driven a certain incremental distance. Fig.1In addition to the paper drive mechanism in a printer, stepper motors are also popularin machine tools, process control systems, tape and disk drive systems, andprogrammable controllers.The Common Features of stepper motors are • Brushless – Stepper motors are brushless. The commentator and brushes of conventional motors are some of the most failure-prone components, and they create electrical arcs that are undesirable or dangerous in some environments.
44 • Load Independent – Stepper motors will turn at a set speed regardless of load as long as the load does not exceed the torque rating for the motor. • Open Loop Positioning – Stepper motors move in quantified increments or steps. As long as the motor runs within its torque specification, the position of the shaft is known at all times without the need for a feedback mechanism. • Holding Torque – Stepper motors are able to hold the shaft stationary. • Excellent response to start-up, stopping and reverse.Types of Stepper Motor 1. Permanent-magnet stepper motor The permanent-magnet stepper motor operates on the reaction between a permanent-magnet rotor and an electromagnetic field. Figure shows a basic two-pole PM stepper motor. The rotor shown in Figure (a) has a permanent magnet mounted at each end. The stator is illustrated in Figure (b). Both the stator and rotor are shown as having teeth Fig.2 The teeth on the rotor surface and the stator pole faces are offset so that there will be only a limited number of rotor teeth aligning themselves with an energized stator pole. The number of teeth on the rotor and stator determine the step angle that will occur each time the polarity of the winding is reversed. The greater the number of teeth, the smaller the step angle.
45 Fig.3The holding torque is defined as the amount of torque required to move the rotorone full step with the stator energized.An important characteristic of the PM stepper motor is that it can maintain theholding torque indefinitely when the rotor is stopped.Figure (a) shows a permanent magnet stepper motor with four stator windings.By pulsing the stator coils in a desired sequence, it is possible to control the speedand direction of the motor. Figure (b) shows the timing diagram for the pulses required to rotate the PMstepper motor.2.Variable-reluctance (VR) stepper motor The variable-reluctance (VR) stepper motor differs from the PM stepper inthat it has no permanent-magnet rotor and no residual torque to hold the rotor atone position when turned off. When the stator coils are energized, the rotor teeth will align with theenergized stator poles. This type of motor operates on the principle of minimizingthe reluctance along the path of the applied magnetic field. By alternating thewindings that are energized in the stator, the stator field changes, and the rotor ismoved to a new position. The stator of a variable-reluctance stepper motor has a magnetic coreconstructed with a stack of steel laminations. The rotor is made of unmagnetizedsoft steel with teeth and slots. The relationship among step angle, rotor teeth, and stator teeth is expressed usingthe following equation:
46 N s − Nr ψ= × 360° ------(1) Ns × NrIn this circuit, the rotor is shown with fewer teeth than the stator. This ensures thatonly one set of stator and rotor teeth will align at any given instant. The stator coils are energized in groups referred to as phases.According to above Eq., the rotor will turn 30° each time a pulse is applied.Figure (a) shows the position of the rotor when phase A is energized. As long asphase A is energized, the rotor will be held stationary. Fig.4• When phase A is switched off and phase B is energized, the rotor will turn 30° until two poles of the rotor are aligned under the north and south poles established by phase B.
47 • By repeating this pattern, the motor will rotate in a clockwise direction. The direction of the motor is changed by reversing the pattern of turning ON and OFF each phase. • The disadvantage of this design for a stepper motor is that the steps are generally quite large (above 15°). • Multistack stepper motors can produce smaller step sizes because the motor is divided along its axial length into magnetically isolated sections, or stacks.Result:
48EXPT. NO.:DATE:DETERMINATION OF TRANSFER FUNCTION OF SEPRATELY EXITED DC GENERATORAIM: To determine the transfer function of separately exited generator.APPARATUS REQUIRED:Ammeter MC (0-1A), (0-10A)Ammeter MI (0-5A),(0-50mA)Voltmeter MC (0-300V)Voltmeter MI (0-300V)Rheostat 1000 / 1ARheostat 50 / 5AAuto Transformer 1¢ 230V/270VTHEORY:The transfer function of a separately excited generator can be represented in blockdiagram format as shown belowThe transfer function isIL(s)/Vf(s) = Kg /(Rf+sLf)(Rl+sLa)Where Vf(s)- Excitation VoltageRf, Lf - Field resistance & InductanceIf(s) - Field CurrentKg – Induced emf constant in V/AmpLl – Total load InductanceRl – Total load resistanceKg can be obtained by conducting open circuit testRf, Lf, Ra, Lf can be found out by voltmeter- Ammeter method
49DC GENERATORCircuit DiagramDETERMINATION OF Ra :Circuit DiagramTabulationS.No Va (V) Ifa (A) Ra =Va/Ia( )12345678910 Mean value of Ra =
50DETERMINATION OF Rf :Circuit DiagramTabulationS.No Vf (V) If (A) Rf ( )=Vf/If12345678910 Mean value of Rf =
51Determination of LaCircuit DiagramDetermination of LfCircuit DiagramOpen circuit characteristics
52PROCEDURE:Determination of Kg: 1. The connections are made as shown in fig. 2. DPST switch is closed 3. The motor is started with help of starter 4. The motor is brought to the rated speed by adjusting the motor field rheostat. The drives the generated at rated speed. 5. Note down the field current If and the open circuit voltage Eo. 6. By adjusting the Rf, the field current is increased in convenient steps up to the rated field current. 7. In each step the readings of Eo and If are noted. Throughout the experiment the speed is maintained at constant 8. A plot of Eo Vs If is drawn by taking If on X axis and Eo on Y-axis. 9. A tangent to the linear portion of the curve is dran through the origin. The slope of this line ,Eo Vs If gives Kg.V-A method to obtain Ra, Rf, La & Lf 1. Give the connection as shown in fig to measure Ra & Rf and note down the V & I 2. To measure La &Lf give the connection as shown in fig. 3. Apply an AC voltage & measure the field reactance Zf & armature reactance Za. 4. Calculate Lf= Sqrt(Zf2 – Rf2) /2πf 5. Calculate La = Sqrt(Za2 – Ra2) /2πf Where f= supply frequency (50Hz)RESULT:
53EXPT. NO.:DATE: STUDY OF SYNCHROSAIM: To study the characteristics of Synchros.APPARATUS REQUIRED:THEORY: A Synchro is an electro-magnetic transducer used to convert an angularposition of a shaft into an electrical signal. It is commercially known as a Selsyn or anAutosyn. The basic element of a synchro is a synchro transmitter whose constructionis very similar to that of a 3 phase Alternator. The stator is of concentric coil type, inwhich three identical coils are placed with their axis 120˚ apart, and is star connected.The rotor is of dumb bell shaped construction and is wound with a concentric coil. ACvoltage is applied to the rotor winding through slip rings. Fig.1 Constructional Features of Synchro TransmitterThe constructional features and schematic diagram of a synchro transmitter andreceiver is shown in fig.1. Let an AC voltage Vc(t) = Vr Sin ωt be applied to the rotorof the synchro transmitter. The applied voltage causes a flow of a magnetizing currentin the rotor coil, which produces a sinusoidal time varying flux directed along its axisand distributed nearly sinusoidally in the air gap along the stator periphery. Becauseof transformer action, voltages are induced in each of the stator coils. As the air gapflux is sinusoidally distributed, the flux linking any stator coil is proportional to thecosine of the angle between the rotor and stator coil axis, and so is the voltageinduced in the stator coil. Thus, we see that synchro transmitter acts like a single-
54phase transformer in which the rotor coil is the primary and stator coil is thesecondary. Fig .2 Schematic Diagram of Synchro Transmitter Let Vs1, Vs2 & Vs3 be the voltage induced in the stator coils S1,S2 and S3 with respectto the neutral. Then, for the rotor position of the synchro transmitter shown in the fig.2 where the rotor axis makes an angle θ with the axis of the stator coil S2 Vs1 = K Vr Sinωt Cos ( θ + 120 ) ---------------------------- (1) Vs2 = K Vr Sinωt Cos ( θ ) ----------------------------------(2) Vs3 =K Vr Sinωt Cos ( θ + 240 ) ----------------------------(3)The three terminal voltages of stator are Vs1s2 = Vs1 - Vs2 =√3 KVr Sin( θ + 240 ) Sin ωt -----------------------(4) Vs2s3 = Vs2 - Vs3 = √3 KVr Sin( θ + 120 ) Sin ωt ----------------------(5) Vs3s1 = Vs3 - Vs1 =√3 KVr Sin(θ) Sin ωt -----------------------(6)When θ = 0, from equations (1), (2) and (3), it is seen that the maximum voltage isinduced in the stator coil S2 , while it follows from the equation (6) from that theterminal voltage Vs3s1 is zero . This position of the rotor is defined as the “electricalzero” of the transmitter and is used as reference for specifying the angular position ofthe rotor. The input to the synchro transmitter is the angular position of its rotor shaftand the output is a set of 3 single-phase voltages given by equations (4) to (6). Themagnitude of these voltages is function of the shaft position.The outputs of the synchro transmitter are applied to the stator windings of a “synchrocontrol transformer”. The rotor of the control transformer is cylindrical in shape sothat the air gap is practically uniform. The system acts as an error detector.Circulating currents of the same phase but of different magnitude flow through thetwo sets of stator coils. This results in the establishment of an identical flux pattern inthe gap at the control transformer as the voltage drop in resistances and leakagereactances of the two sets of stator coils are usually small. The voltage induced in the
55control transformer rotor is proportional to the cosine of the angle between the tworotors (φ) and is given by E(t) = K1 Vr Cos φ Sin ωtWhen φ =900, the voltage induced in the control transformer is zero. This position isknown as electrical zero position of the control transformer. Fig. 3 Synchro Error DetectorPROCEDURE: Tabulation 1: 1. Give connections as given in the circuit diagram. 2. Vary the input position and note the output position. 3. Plot the variation in output position with respect to the input position. Tabulation 2: 1. Give excitation to the rotor winding. 2. Measure the output voltage across S1-S2, S2-S3 and S3-S1 of stator windings for different rotor positions. 3. Plot the voltage Vs. angle characteristics.
57EXPT. NO:DATE : DESIGN OF COMPENSATOR NETWORKSAIM:To design a compensator network for the process given in the Process ControlSimulator.APPARATUS REQUIRED:THEORY: Practical feedback control systems are often required to satisfy designspecification in the transient as well as steady state regions. This is not possible byselecting good quality components alone (due to basic limitations and characteristicsof these components). Cascade compensation is most commonly used for this purposeand design of compensation networks figures prominently in any course in automaticcontrol systems. In general, there are two situations in which compensation is required. In thefirst case the system is absolutely unstable and the compensation is required tostabilize it as well as to achieve a specified performance. In the second case thesystem is stable but the compensation is required to obtain the desired performance.The systems which are of type 2 or higher are usually unstable. For these systems,lead compensator is required, because the lead compensator increases the margin ofstability. For type 1 and type 0 systems stable operation is always possible. If the gainis sufficiently reduced, in such cases, any of three components viz. Lag, Lead, Lag –Lead must be used to obtain the desired performance. The simulation of this behaviorof the Lead – Lag Compensator can be done with the module (VLLN – OI). An electronic Lead - lag network using Operational amplifiers is givenfigure 1. C2 C1 R2 R4 - - R1 + R3 + Fig.1 LEAD -LAG NETWORK USING OPERATIONAL -AMPLIFIER
58 The transfer function for this circuit can be obtained as follows : Let Z1 = R1 C1 The second op-amp acts as a sign inverter with a variable gain to compensate for the magnitude. The transfer function of the entire system is given by G(jω) = ( R4 R2/ R3 R1 ) (1+R1C1s) / (1+R2 C2 s) G(jω) = ( R4 R2/ R3 R1 ) ( √1+T12ω2) / ( √1+T22ω2 ) where T1 = R1 C1 ; T2 = R2 C2 φ = angle G(jω) = - tan-1(T1ω) – tan-1(T2ω). Thus steady state output is For an input =X sinωt, Yss(t) =X (R4 R2/ R3 R1) (( √1+T12ω2) / ( √1+T22ω2 ))sin(wt – tan-1 T1ω –tan-1 T2ω ) From this expression, we find that if T1 > T2, then tan-1 T1ω –tan-1 T2ω > 0. Thus if T1 > T2 , then the network is a LEAD NETWORK. If T1 < T2 , then the network is a LAG NETWORK. DETERMINATION OF VALUES FOR ANGLE COMPENSATION: Frequency of sine wave = 20 Hz Angle to be compensated = 70º φ= tan-1 (2π f * T1) –tan-1 (2π f * T2) T1 = 10, then substituting in above equation 70 - tan-1 (2 * π * 20 * 10) –tan-1 (2 * π * 20 *T2) solving for T2 T2 = 0.003 .Hence, the values of T1 and T2 are chosen from which values of R1 ,C1 , R2 and C2can be determined .For example, T1 =R1 C1 = 10 ; If C1= 1µF, then R1 = 10 M . T2 = 0.003 = R2 C2 then C2 =1 µF, and hence R2 = 3 M .These values produce a phase lead of 70º which is the desired compensation angle.Nominal Value for R1 =1 M C1 = 0.1 µF R2 =20 K C2 = 0.01 µF
59PROCEDURE: 1. Switch ON the power to the instrument. 2. Connect the individual blocks using patch chords. 3. Give a sinusoidal input as the set value . 4. Measure the amplitude and frequency of the input signal. 5. Measure the amplitude and phase shift of the output signal with respect to the input sine wave using CRO. 6. Draw the magnitude versus frequency plot and phase versus frequency plot. 7. Using the technique explained previously, calculate the values of R1, C1, R2 and C2 to compensate for the phase shift of the output signal. 8. Connect the components at the points provided. 9. Now include the compensation block in the forward path before the process using patch chords. 10. Now measure the phase shift φ of the output signal with the input and verify for compensation. 11. Draw the magnitude versus frequency plot and phase versus frequency plot for the designed compensator. Table – 1 (for the process without compensation): S.No. Input Output Gain (dB) Phase shift φ Freq (Hz) Voltage(V) 20 log(Vo/Vin) º (º)
60 Table – I1 (for the process with compensation): Vin = V S.No. Input Output Gain (dB) P Phase shift φ Freq (Hz) Voltage(V) 20 log(Vo/Vin) º (º)A – Amplitude of input sine wave (V)F –Frequency of the input sine wave (Hz)Φ – Phase shift (Degrees)RESULT :INFERENCE:
61EXPT. NO.:DATE: STUDY OF P, PI, PID CONTROLLERSAIM: To study the P, PI, PID controller using MATLAB software .APPARATUS REQUIRED:THEORY: The transient response of a practical control system often exhibits dampedoscillation before reaching steady state value. In specifying the transient responsecharacteristics of control systems to unit step input, it is common to specify thefollowing i) Delay Time(Td) ii) Rise time(Tr) iii) Peal time( Tp) iv) Max. overshoot (Mp) v) Settling time( Ts)Proportional control: The output of the controller is proportional to input U(t) = Kp e(t) E(t) = error signal U(t) controller out[put Kp = proportional constant • It amplifies the error signal and increases loop gain. Hence steady state tracking accuracy , disturbance signal rejection and relative stability are improved. • Its drawbacks are low sensitivity to parameter variation and it produces constant steady state error.Proportional + Integral Control: The output of the PI controller is given by t U(t) = Kp [ e(t) + (1/Ti )∫ e(t) dt ] 0 where Kp is the proportionality constant and Ti is called the integral time. • This controller is also called RESET controller. • It introduces a zero in the system and increases the order by 1. • The type number of open loop system is increased by 1
62 • It eliminates steady state error. Damping ratio remains same. • Increase in order decreases the stability of system.Proportional + Integral Control + Differential Control: The output of a PID controller is given by t U(t) = Kp [ e(t) + (1/Ti )∫ e(t) dt + Td de(t)/dt ] 0 The PID controller introduces a zero in the system and increases the damping.This reduces peak overshoot and reduces rise time. Due to increase in damping,ultimately peak overshoot reduces. The stability of the system improves. In PID controller, all effects are combined. Proportional control stabilizes gainbut produces steady state error. Integral control eliminates error. Derivative controllerreduces rate of change of error. TUNING OF PID CONTROLLERS Proportional-integral-differential (PID) controllers are commonly employed inprocess control industries. Hence we shall present various techniques of tuning PIDcontrollers to achieve certain performance index for systems dynamic response. Thetechnique to be adopted for determining the proportional, integral and derivativeconstants of the controller depends upon the dynamic response of the plant. In presenting the various tuning techniques we shall assume the basic controlconfiguration, wherein the controller input is the error between the desired output andthe actual output. This error is manipulated by the controller (PID) to produce acommand signal for the plant according to the relationship. U(s)=Kp (1+(1/sτi)+sτd) Where Kp= proportional gain constant τI= integral time constant. τd= Derivative time constant.PROCEDURE: 1. Give the step input to the system selected and obtain the response using CRO. 2. For the obtained response (S-shaped curve), draw a tangent at the inflection point and find its intersection with the time axis and the line corresponding to the steady-state value of the output. 3. Find the dead time L where the tangent cutting X- axis, and the time constant T which is specified in model graph. 4. From the value of L and T, find the value of Kp, τI and τd settings by using the following formulas: Kp = 1.2(T/L) , τI = 2L and τd = 0.5L. 5. Connect the unknown system in closed loop with the help of a PID controller and substitute all those values obtained in the previous step.
63 6. Simulate the system with a step input and view the response using CRO. 7. Comment on the response obtained using controller.(I)General Block Diagram(II)Block Diagram for P Controller(III)OP –Amp P Controller Using Inverting Amplifier
64(IV) Block Diagram For PI Controller(V) PI Controller Using Op-Amp(VI) Block Diagram Of PID Controller
65Block Diagram Of Closed Loop Control Using PID Controller M(s) R(s) E(s) C(s) PID(KP,Ti,Td ) Transfer Function C(s)RESULT:INFERENCE: