EC2259 ELECTRICAL ENGINEERING AND CONTROL SYSTEM LABORATORY LABORATORY MANUAL FOR IV SEMESTER B.E (ECE) (FOR PRIVATE CIRCULATION ONLY) ANNA UNIVERSITY -CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING, DR.NAVALAR NEDUNCHEZHIYAN COLLEGE OF ENGINEERING, THOLUDUR 606 303, CUDDALORE DIST. 1
UNIVERSITY PRACTICAL EXAMINATION ALLOTMENT OF MARKS Internal assessment Practical assessment TOTAL : 20 marks : 80 marks : 100 MARKS INTERNAL ASSESMENT (20 MARKS) Staff should maintain the assessment register and the head of the department should monitor it. SPLIT UP OF INTERNAL MARKS Observation Record note Model exam Attendance TOTAL : 3 marks : 7 marks : 5 marks : 5 marks : 20 Marks UNIVERSITY EXAMINATION The Exam will be conducted for 100 marks. Then the marks will be converted to 80 marks. SPLIT UP OF PRACTICAL EXAMINATION MARKS AIM AND PROCEDURE : 10 CIRCUIT DIAGRAM/PROGRAM : 25 CIRCUIT CONNECTION : 15 TABULATION : 15 CALCULATION/SIMULATION : 15 GRAPH AND RESULT : 10 VIVA VOCE : 10 TOTAL 100 2
LIST OF EXPERIMENTS 1. Open circuit and Load characteristics of separately excited and self excited DC generator. 2. Load test on DC Shunt motor. 3. Swinburne s test and Speed control of DC Shunt motor. 4. Load test on Single phase transformer and open circuit and short circuit test on single phase transformer. 5. Regulation of three phase alternator by EMF and MMF methods. 6. Load test on three phase Induction motor. 7. No Load and Blocked rotor tests on three phase Induction motor (Determination of equivalent circuit parameters). 8. Study of DC motor and Induction motor starters. 9. Digital Simulation of Linear systems. 10. Stability analysis of linear systems using MATLAB. 11. Study the effect P, PI, PID controllers using MATLAB. 12. Design of lead and lag compensator. 13. Transfer function of separately excited D.C. generator. 14. Transfer function of armature and field controller D.C. motor. BEYOND THE SYLLABUS 1. Load test on dc compound motor 2. Determination of transfer function parameters of field controlled dc servo motor 3. Determination of transfer function parameters of armature controlled dc servo motor 3
CONTENTS S.NO NAME OF EXPERIMENT PAGE NO 1 Open circuit and Load characteristics of separately excited and self excited DC generator. 2 Load test on DC Shunt motor. 3 Swinburne s test and Speed control of DC Shunt motor. 4 Load test on Single phase transformer and open circuit and short circuit test on single phase transformer. 5 Regulation of three phase alternator by EMF and MMF methods. 6 Load test on three phase Induction motor. 7 No Load and Blocked rotor tests on Three phase Induction motor (Determination of equivalent circuit parameters). 8 Study of DC motor and Induction motor starters. 9 Digital Simulation of Linear systems. 10 Stability analysis of Linear systems using MATLAB. 11 Study the effect P, PI, PID controllers using MATLAB. 12 Design of lead and lag compensator. 13 Transfer function of Separately excited D.C. generator. 14 Transfer function of armature and field controller D.C. motor. Signature of the Lab In Charge 4 (N.JAYACHITRA)
Exercise Number: 1(A) Title of the Experiment : OPEN CIRCUIT AND LOAD CHARACTERISTICS OF DC SEPARATELY EXCITED GENERATOR Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: a) To obtain the open circuit characteristics of a separately excited DC Generator at rated speed and to determine (i)critical field resistance at rated speed (ii)critical Speed (iii) Voltage built up by the generator at rated speed and (iv) to plot the OCC characteristic curves b) To conduct the Load test on the given separately excited DC Generator and plot its internal and external characteristics. a) FACILITIES REQUIRED: S.NO APPARATUS TYPE RANGE QUANTITY 1 Voltmeter mc (0-300) V 1 2 Voltmeter mc (0-30) V 1 3 Ammeter mc (0-20) A 1 4 Ammeter mc (0-2) A 1 5 Rheostat - 250 /1.2A 1 6 Rheostat 570 /1.2A 1 7 Tachometer Analog - 1 8 Variable Rheostat load - - 1 9 DPSTS - - 1 10 Connecting wires - - Few 5
b) PRECAUTIONS: 1. All the connections should be tight. 2. Ensure both SPST and DPST switches are in open position. 3. Check whether the load is under off position. 4. Keep the motor field rheostat at minimum resistance position during starting. 5. Keep the motor armature rheostat at maximum resistance position at starting. 6. Keep the generator field rheostat at maximum resistance position during starting. c) PROCEDURE: OPEN CIRCUIT TEST: 1. The connections are given as per the circuit diagram. 2. Verify whether field rheostat of the generator is kept at maximum position and field rheostat of motor at minimum position. 3. Switch ON the DC supply. 4. Adjust the excitation of field rheostat of the motor so as to make the motor to run at rated speed. 5. The ammeter and voltmeter readings of the generator are noted with SPSTS switch in opened position. 6. Close the SPSTS switch and excitation of the generator is varied insteps by adjusting the field rheostat of the generator, at each step the readings of field current and induced emf are noted up to its rated generator voltage. 7. The readings are tabulated and a graph of open circuit characteristics is drawn between generated voltage Vs field current. LOAD TEST: 1. Fix the armature voltage of the generator to the rated voltage by adjusting the field rheostat of the generator. 2. Close the DPSTS at the load side of the generator and increase the load in steps till the rated armature current and at each step the readings of terminal voltage, load current, and shunt field current are noted. 6
3. Finally reduce the load insteps and bring the generator and field Rheostat to its original position. d) EXPRESSION FOR THEORITICAL VALUE: 1. Critical field resistance, Rc = Slope of the tangent drawn to the linear portion of and passing through origin. Rc =. Ohms. 2. Critical Speed Nc = Rsh * Nrated / R 3. Voltage build up Egm =Voltage corresponding to the point at which the Speed and Rsh line intersect. Egm =.. volts. 4. Eg = V + IaRa in volts. Where, Eg = Generated voltage in volts. V = Terminal voltage in volts. Ia = Armature current in Amps. Ra = Armature resistance in ohms. 5. Armature current Ia = IL+ Ish in Amps. Where, IL = Load current in Amps. Ish = Shunt field current in Amps. 7
e)circuit DIAGRAM f)tabulation: OPEN CIRCUIT TEST: S. No Field current (Amps) Output voltage at no load (Volts) 8
LOAD TEST: S. No Load voltage Load current Field current Ia= IL + IF Eg=Va+IaRa Volts Amps Amps Amps Volts g) TO FIND R a Circuit Diagram Tabulation S. No Voltage (Va) Current (Ia) Ra=Va/Ia Volts Amps Ohms 9
TO FIND Ra: 1. Connections are given as per the circuit diagram. 2. Gradually vary the loading rheostat insteps and at each step note down the corresponding voltmeter and ammeter readings. 3. From the tabulated value calculate the average armature resistance. h)model GRAPH i)discussion QUESTIONS: 1. Why the armature core in d.c machines is constructed with laminated steel sheets instead of solid steel sheets? Lamination highly reduces the eddy current loss and steel sheets provide low reluctance path to magnetic field. 2. Why commutator is employed in d.c.machines? Conduct electricity between rotating armature and fixed brushes, convert alternating emf into unidirectional emf(mechanical rectifier). 3. Distinguish between shunt and series field coil construction? Shunt field coils are wound with wires of small section and have more turns. Series field coils are wound with wires of larger cross section and have less no. of turns. 10
4. How does d.c. motor differ from d.c. generator in construction? Generators are normally placed in closed room and accessed by skilled operators only. Therefore on ventilation point of view they may be constructed with large opening in the frame. Motors have to be installed right in the place of use which may have dust, dampness, inflammable gases, chemicals.etc. to protect the motors against these elements, the motor frames are made either partially closed or totally closed or flame proof. 5. How will you change the direction of rotation of d.c.motor? Either the field direction or direction of current through armature conductor is reversed. RESULT: Thus the open circuit and load characteristics of separately excited DC generator are determined. 11
Exercise Number: 01(B) Title of the Experiment : OPEN CIRCUIT AND LOAD CHARACTERISTICS SELF EXCITED DC SHUNT GENERATOR Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To obtain the open circuit characteristics of a DC Shunt Generator at rated speed and to determine (i) Critical field resistance at rated speed (ii)critical Speed (iii) Voltage built up by the generator at rated speed and (iv) to plot the OCC characteristic curves a) To conduct the Load test on the given DC Shunt Generator and plot its internal and external characteristics. a) FACILITIES REQUIRED: APPARATUS TYPE RANGE QUANTITY S.NO 1 Voltmeter mc (0-300) V 1 2 Voltmeter mc (0-30) V 1 3 Ammeter mc (0-20) A 1 4 Ammeter mc (0-2) A 1 5 Rheostat - 250 /1.2A 1 6 Rheostat - 570 /1.2A 1 7 Tachometer Analog - 1 8 Variable Rheostat load - - 1 9 DPSTS - - 1 10 Connecting wires - - Few b)exercise: 12
c) PRECAUTIONS: 1. All the connections should be tight 2. Ensure both SPST and DPST switches are in open position. 3. Check whether the load is under off position. 4. Keep the motor field rheostat at minimum resistance position during starting. 5. Keep the motor armature rheostat at maximum resistance position at starting. 6. Keep the generator field rheostat at maximum resistance position during starting. d)procedure OPEN CIRCUIT TEST 1. The connections are given as per the circuit diagram. 2. Verify whether field rheostat of the generator is kept at maximum position and field rheostat of motor at minimum position. 3. Switch ON the DC supply. 4. Adjust the excitation of field rheostat of the motor so as to make the motor to run at rated speed. 5. The ammeter and voltmeter readings of the generator are noted with SPSTS switch in opened position. 6. Close the SPSTS switch and excitation of the generator is varied insteps by adjusting the field rheostat of the generator, at each step the readings of field current and induced emf are noted upto its rated generator voltage. 7. The readings are tabulated and a graph of open circuit characteristics is drawn between generated voltage Vs field current. LOAD TEST: 13
1. Fix the armature voltage of the generator to the rated voltage by adjusting the field rheostat of the generator. 2. Close the DPSTS at the load side of the generator and increase the load in steps till the rated armature current and at each step the readings of terminal voltage, load current, and shunt field current are noted. 3. Finally reduce the load insteps and bring the generator and field rheostat to its original position. e) EXPRESSION FOR THEORITICAL VALUE: 1. Critical field resistance,rc = Slope of the tangent drawn to the linear portion of the OCC and passing through origin. Rc =. Ohms. 2. Critical Speed Nc = Rsh * Nrated / R 3. Voltage build up Egm =Voltage corresponding to the point at which the OCC at rated Speed and Rsh line intersect. Egm =.. volts. 4. Eg = V + IaRa in volts. Where, Eg = Generated voltage in volts. V = Terminal voltage in volts. Ia = Armature current in Amps. Ra = Armature resistance in ohms. 5. Armature current Ia = IL+ Ish in Amps. Where, IL = Load current in Amps. Ish = Shunt field current in Amps. f)circuit DIAGRAM 14
15
g)tabulation: OPEN CIRCUIT TEST: S. No Field current (Amps) Output voltage at no load (Volts) LOAD TEST: S. No Load voltage Load current Field current Ia= IL + IF Eg=Va+IaRa Volts Amps Amps Amps Volts 16
h)to FIND R a Circuit Diagram Tabulation S. No Voltage (Va) Current (Ia) Ra=Va/Ia Volts Amps Ohms TO FIND Ra: 1. Connections are given as per the circuit diagram. 2. Gradually vary the loading rheostat insteps and at each step note down the corresponding voltmeter and ammeter readings. 3. From the tabulated value calculate the average armature resistance. i)model GRAPH 17
j)discussion QUESTIONS: 1. What is meant by stator voltage control? The speed of the I.M can be changed by changing the stator voltage. Because the torque is proportional to square of the voltage. 2. What are the applications of stator voltage control? Fans, pump drives. 3. What are the advantages of stator voltage control method? 1. The control circuitry is simple. 2. Compact size. 3. Quick response time. 4. There is considerable savings in energy and thus it is economical method. 4. Define base speed? The synchronous speed corresponding to the rated frequency is called the base speed. RESULT : Thus the open circuit and load characteristics of self excited DC generator are determined. 18
Exercise Number: 02 Title of the Experiment: LOAD TEST ON DC SHUNT MOTOR Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT : To conduct the load test on DC shunt motor and to draw the following characteristics curves, 1. Torque Vs armature current 2. Speed Vs armature current 3. Speed Vs Torque 4. %Efficiency Vs Power output a)facilities REQUIRED: S.NO APPARATUS TYPE RANGE QUANTITY 1. Voltmeter mc (0-300) V 1 2. Ammeter mc (0-20) A 1 3. Rheostat - 250 /1.2A 1 4. Tachometer Analog - 1 5. Connecting wires - - Few 19
b)exercise: c) PRECAUTIONS: 1. Ensure that all the connections are tight. 2. Select the meters of proper range. 3. Ensures the rating of fuses. 4. The starter arm must be in OFF position at the time of starting. 5. The motor should be loaded up to rated current of the machine. d)procedure: 1. Connections are given as per the circuit diagram. 2. By closing the MCB, 230v DC supply is given to the circuit. 3. The motor is started by using three point starter. 4. Adjust the field rheostat of the motor so as to make the motor to run at rated speed. 5. Vary the load insteps and at each step note down the input voltage, input current, spring balance readings till the rated current. 6. The readings are tabulated and a graph of, Torque Vs armature current characteristics Speed Vs armature current characteristics Speed Vs Torque characteristics 20
e) EXPRESSION FOR THEORITICAL VALUE: 1. S=S1 S2 in Kg. S =Spring balance (Weight in Kg) S1, S2 = Spring balance readings in Kgs. 2. T=S*R*9.81 N mt S=Spring force in Kg. R=Radius of the brake drum in meter. T=Torque in N mt. 3. Pin =Vin * Iin in watts. Vin = Voltage input in volts. Iin =Input current in Amps. Pin =Input power in watts. 4. Pout = (2 NT) /60 in watts. Pout = output power in watts. N= Speed in rpm. T=Torque in Nm. 5. Percentage Efficiency = Pout/Pin *100 21
f)circuit DIAGRAM g)tabulation: Circumference of the Brake drum = Radius of the brake drum = S. No Input voltage Input current Spring balance Reading S1 S2 S1 S2 Speed N Torque T Input power Pin Output power Pout Efficienc y volts Amps Kg Kg Kg Rpm Nm Watts Watts % 22
h)model GRAPH: i)discussion QUESTIONS 1. What is meant by flux control method? The speed of the dc motor can be controlled by varying the field flux. This method od speed control can be used for increasing the speed of the motor is inversely proportional to the field flux 2. Under What circumstances does a dc shunt generator fails to generate? Absence of residual flux, initial flux setup by field may be opposite in direction to residual flux, shunt field circuit resistance may be higher than its critical field resistance, load circuit resistance may be less than its critical load resistance. 3. Define critical field resistance of dc shunt generator? Critical field resistance is defined as the resistance of the field circuit which will cause the shunt generator just to build up its emf at a specified field. RESULT: Thus the load tests of DC shunt motor are determined. 23
Exercise Number: 03(A) Title of the Experiment: SWINBURNES TEST Date of the Experiment: OBJECTIVE (AIM) OF THE EXPERIMENT: To conduct no load test on dc shunt motor and to predetermine efficiency while machine is running as a motor and generator. a) FACILITIES REQUIRED: S. No APPARATUS TYPE RANGE QUANTITY 1. Voltmeter mc (0-300) V 1 2. Ammeter mc (0-2) A 1 3. Ammeter mc (0-20) A 1 3. Rheostat - 250Ώ/1.2A 1 4. Tachometer - - 1 5. Connecting wires - - few b)exercise: c) PRECAUTIONS: 1. Ensure that all the connections are tight. 2. Select the meters of proper range. 3. Ensures the rating of fuses. 4. The starter arm must be in OFF position at the time of starting. 24
d)procedure: o Connections are given as per the circuit diagram. o Close the MCB switch by energizing the circuit. o Start the machine using three-point starter and allow the motor to run at its rated speed by adjusting the field rheostat. o The corresponding voltmeter and ammeter readings are noted at no-load. o Bring the field rheostat to its original minimum position and the MCB switch is switched off. o Now the armature resistance test is conducted as per the circuit diagram and the corresponding meter readings are noted. o The readings are tabulated and the efficiency of the machine when it runs as a motor and generator are determined without loading the machine e) EXPRESSION FOR THEORITICAL VALUE: FOR MOTOR: Input Power, Pi = VLIL (watts) Armature current,ia = IL IF (amps) Copper loss, Wcu = Ia 2 Ra (watts) Total losses, PL = Wc + Wcu (watts) Constant losses, Wc = ILVL - Ia 2 Ra (watts) Output Power, Po = Pin PL (watts) % Efficiency = output power (Po)/ Input power (Pi) FOR GENERATOR: Input Power, PO = VLIL (watts) Armature current,ia = IL+ IF (amps) Copper loss, Wcu = Ia 2 Ra (watts) Total losses, PL = Wc + Wcu (watts) 25
Constant losses, Wc = ILVL - Ia 2 Ra (watts) Input Power, Pi = Po + PL (watts) % Efficiency = output power (Po)/ Input power (Pi) f)circuit DIAGRAM: 26
g)tabulation TO FIND OUT CONSTANT LOSS (W co ) S. No No-load current (Io) Field Current (If) Terminal Voltage (V) No load Armature current (Iao) Constant Loss Wco=VIo- Iao 2 Amps Amps Volts Volts Watts h)circuit DIAGRAM TO FIND RA 27
TABULATION TO FIND OUT ARMATURE RESISTANCE Armature current Armature Voltage Aramture Resistance S. No (Ia) (Va) Ra = Va/Ia Amps Volts Ohms TABULATION TO FIND OUT THE EFFICIENCY RUNNING AS MOTOR Armature Resistance (Ra) = Rated current (Ir) = Constant loss (Wco) = Field current (If) = Terminal Voltage (V) = S. No Fraction of load (x) Load current IL= x*ir Armature current Ia =IL+If Armature loss Wcu=Ia 2 Ra Total loss Wt Input power Wi=VIL Output power Wo=Wi-Wt Efficiency η = Wo/Wi Amps Amps Amps Watts Watts Watts % 28
RUNNING AS GENERATOR Armature Resistance (Ra) = Rated current (Ir) = Constant loss (Wco) = Field current (If) = Terminal Voltage (V) = S. No Fraction of load (x) Load current IL= x*ir Armature current Ia =IL+If Armature loss Wcu=Ia 2 Ra Total loss Wt Output power Wo=VIL Input power Wi=Wo+Wt Efficiency η = Wo/Wi Amps Amps Amps Watts Watts Watts % 29
i)discussion QUESTIONS: 1. What are the principal advantages of rotating field type construction? Relatively small amount of power required for field system can easily supplied to rotating system using slip rings and brushes, more space is available in the stator part of the machine to provide more insulation, it is easy to provide cooling system, stationary system of conductors can easily be braced to prevent deformation. 2. What are the advantages of salient type pole construction used in sy.machines? They allow better ventilation, the pole faces are so shaped radial air gap length increases from pole center to pole tips so flux distortion in air gap is sinusoidal so emf is also sinusoidal. 3. Which type of sy.generators are used in hydroelectric plants and why? As the speed of operation is low, for hydro turbines used in hydroelectric plants, salient pole type sy.generator is used because it allows better ventilation also better than smooth cylindrical type rotor. 4. Why are alternators rated in KVA and not in KW? As load increases I2R loss also increases, as the current is directly related to apparent power delivered by generator, the alternator has only their apparent power in VA/KVA/MVA as their power rating. 5. Why the sy.impedance method of estimating voltage regulation is considered as pessimistic method? Compared to other method, the value of voltage regulation obtained by this method is always higher than the actual value so it is called as pessimistic method. RESULT: Thus the Swinburne s tests of DC shunt motor are determined. 30
Exercise Number: 03(B) Title of the Experiment: SPEED CONTROL OF DC SHUNT MOTOR Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To control the speed of the given DC shunt motor by, 1. Armature control method 2. Field control method a)facilities REQUIRED: S.NO APPARATUS TYPE RANGE QUANTITY 1. Voltmeter mc (0-300) V 1 2. Ammeter mc (0-2) A 1 3. Rheostat - 250 /1.2A 1 4. Rheostat - 50 /5A 1 5. Tachometer Analog - 1 6. Connecting wires - - few b)exercise: 31
b) PRECAUTIONS: 1. Ensure that all the connections are tight. 2. Select the meters of proper range. 3. Ensures the rating of fuses. 4. The starter arm must be in OFF position at the time of starting. d)procedure: 1. Connections are given as per the circuit diagram. 2. Switch on the 230V DC supply by closing the MCB. 3. Start the motor, using three-point starter. Armature control method (Below rated speed) 1. By keeping the field current constant, the armature rheostat is adjusted in steps and note down the corresponding armature voltage, and armature current. 2. The readings are tabulated and a graph of speed Vs armature voltage is drawn. 3. The same procedure may be repeated for some other field current. Field control method (Above the rated speed) 1. By keeping the armature voltage as constant, the field rheostat is adjusted in steps and note down the corresponding field current and speed. 2. The readings are tabulated and a graph of speed Vs field current is drawn. 3. The same procedure may be repeated for some other armature voltage. 32
e)circuit DIAGRAM: 33
f)tabulation: ARMATURE CONTROL METHOD: S. No If1 = If2 = Va (v) N (rpm) Va (v) N (rpm) FIELD CONTROL METHOD: S. No Va1 = Va2 = If (A) N (rpm) If (A) N (rpm) g)model GRAPH: 34
h)discussion QUESTIONS: 1. Why DC series motor should never be started on no-load? When the load current Ia falls to a small value, speed become dangerously high. Hence a DC series motor should never be started without some mechanical load. 2. Define slip. s= (Ns-Nr)/Ns, where, Ns is synchronous speed in rpm Nr is rotor speed in rpm. 3. Define synchronous speed. Three phase balanced power supply is fed to the three stator winding of three phase induction motor. It creates synchronously rotating magnetic field. The speed of this rotating field is called synchronous speed. It s given by Ns = 120f/p. 4. What are the conditions to be fulfilled by for a dc shunt generator to build back emf? The generator should have residual flux, the field winding should be connected in such a manner that the flux setup by field in same direction as residual flux, the field resistance should be less than critical field resistance, load circuit resistance should be above critical resistance. 5. Define armature reaction in dc machines? The interaction between the main flux and armature flux cause disturbance called as armature reaction. 35
RESULT: Thus the speed controls of DC shunt motor are determined. Exercise Number: 04(A) Title of the Experiment: OPEN CIRCUIT AND SHORT CIRCUIT TEST ON A SINGLE PHASE TRANSFORME Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT : To calculate the equivalent circuit parameter by conducting open circuit and short circuit test on single-phase transformer. i) To predetermine the efficiency and regulation of the given transformer from equivalent circuit parameter. a)facilities REQUIRED: S.NO ITEM TYPE RANGE QUANTITY 1. Voltmeter MI (0-150)V 1 2. Voltmeter M (0-300)V 1 3. Ammeter MI (0-2)A 1 4. Ammeter MI (0-5)A 1 5. Wattmeter UPF 300V/5A 1 6. Wattmeter LPF 75V/5A 1 7. Auto transformer - 2.7KVA,10A,270V 1 36
8. Connecting wires - - few b)exercise: c)precautions: 1. Ensure that all the connections are tight. 2. Secondary side of the transformer should be loaded upto rated current of the machine. 3. Calculate the multiplication factors for wattmeters. 4. Calculate the full load current of the transformer. 5. The transformer should not be loaded beyond 125% of the rated full load current. d)procedure Open circuit test 1. Connections are given as per the circuit diagram. 2. Keeping the autotransformer at its zero output position,switch ON the 3. A.C.supply. 4. 3.Adjust the auto transformer output to get the rated voltage across the LV 5. winding of the transformer. 6. Note down all the meter readings. 7. 5.Bring the autotransformer to zero output position and switch OFF the supply. Short circuit test: 1. Connections are given as per the circuit diagram. 2. Keeping the autotransformer at its zero output position, Switch ON the AC supply. 3. Gradually increase the output voltage of autotransformer till the rated full load current flow through the transformer windings. 4. Note down all the meter readings. 5. Bring the autotransformer to zero output position and switch OFF the supply. 37
e) EXPRESSION FOR THEORITICAL VALUE: Open circuit test: 1. W0 = I0 V0 cos 0 watts. Where, cos 0 = (W0 / I0 V0 ) W0 = Real power consumed V0 = Primary no load voltage in volts I0 = No load primary current 0 = Phase angle at no load 2. Magnetizing current Im = I0 sin 0 Loss compensating current Iw = I0cos 0 3. No load reactance X0 = V0/ Im 4. No load resistance R0 = Vo/Iw Short Circuit Test: 1. Z01=Vsc/Isc, R01=Wsc / Isc 2 38
X01= Z01 2 -R01 2 2. R02=k * R01 ; X02=k 2 X01; Z02=k 2 Z01 Where, Vsc =Short-circuiting voltage in volts. f)circuit DIAGRAM: 39
g)tabulation: NO LOAD TEST: S. No. Open circuit primary Voltage (V0) No load current (I0) No load power (W0) volts Amps Watts 40
SHORT CIRCUIT TEST: S. No. Vsc Isc Wsc Volts Amps Watts Efficiency: 1. Efficiency = Output power/input power = Output power/(output power + total losses) 2. Output power =(x *kva* cos ) 3. Total loss = copper loss + iron loss X = Assumed fraction of load KVA = Cos = Rating of the transformer Assumed power factor Wsc = Short circuit wattmeter reading (cu loss) W0 = Open circuit wattmeter reading (Fe loss) Regulation: % Regulation = xisc(r02 cos + X02sin )/V2 0 Where, V2 0 = Open circuit voltage at secondary + = for lagging load - = for leading load 41
h)model GRAPH: TABULATION FOR PERCENTAGE EFFICIENCY: S. No. Iron loss Cu loss Load x Total loss = W0+x 2 Wsc Output power Input power % Efficiency UPF 0.8PF UPF 0.8PF UPF 0.8PF 42
Watts Watts % Watts Watts Watts Watts Watts % % TABULATION FOR PERCENTAGE REGULATION: S. No. Power Factor 125% of load x =1.25 100% of load x = 1.00 50% of load x = 0.5 25% of load x = 0.25 cos Lagging Leading Lagging Leading Lagging Leading Lagging Leading i)discussion QUESTIONS 1. Does transformer draw any current when secondary is open? Why? 43
Yes, it (primary) will draw the current from the main supply in order to magnetize the core and to supply for iron and copper losses on no load. There will not be any current in the secondary since secondary is open. 2. List the different methods of speed control applicable to 3 phase slip ring induction motor? (i) Rotor resistance control (ii) Cascade control (iii)slip power recovery scheme 3. How speed is achieved by v/f control in 3 phase induction motor? In v/f method, motor can be varied by varying voltage and frequency but we can maintain v/f ratio is constant. 4. Define all day efficiency of a transformer? It is computed on the basis of energy consumed during a certain period, usually a day of 24 hrs. All day efficiency=output in kwh/input in kwh tor 24 hrs. RESULT: Thus the open circuit and short circuit test on single phase transformer are determined. 44
Exercise Number: 04(B) Title of the Experiment: LOAD TEST ON SINGLE PHASE TRANSFORMER Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To conduct the load test on single-phase transformer and to draw the characteristics curves. a) FACILITIES REQUIRED: S.NO ITEM RANGE TYPE QUANTITY 1. Voltmeter (0-300) V MI 1 2. Voltmeter (0-150) V MI 1 3. Ammeter (0-5) A MI 1 4. Ammeter (0-20)A MI 1 5. Wattmeter 300V/5A UPF 2 6. Transformer 1KVA - 1 7. Connecting wires - - few b)exercise: 45
c)precautions: 1. Ensure that all the connections are tight. 2. Secondary side of the transformer should be loaded upto rated current of the machine. 3. Calculate the multiplication factors for wattmeters. d)procedure: 1. Give the connections as per the circuit diagram. 2. By closing MCB, 230v single phase, 50Hz AC supply is given to the transformer. 3. Note the readings from all meters on no load. 4. Now load is applied to the transformer in step by step-up to the rated primary current and the corresponding readings are noted. 5. From the recorded values calculate the percentage efficiency and percentage regulation and draw the required graph. e) EXPRESSION FOR THEORITICAL VALUE: 1. Percentage efficiency = (Output power/input power) *100 2. Percentage Regulation =(V02 V2)/ V02 * 100 Where, V02 =Secondary voltage on no load. V2 = Secondary voltage on load. 46
f)circuit DIAGRAM: DISCUSSION QUESTIONS: 1. Mention the difference between core and shell type transformers? In core type, the windings surrounded the core considerably and in shell type the core surround the windings i.e winding is placed inside the core. 2. What is the purpose of laminating the core in a transformer? To reduce the eddy current loss in the core of the transformer. 3. Give the emf equation of a transformer and define each term? Emf induced in primary coil E1= 4.44fФmN1 volt emf induced in secondary coil E2 =4.44 fфmn2. F - Freq of AC input Фm - maximum value of flux in the core N1, N2 - Number of primary & secondary turns. 4. What are the typical uses of auto transformer? 1. To give small boost to a distribution cable to correct for the voltage drop. 2. As induction motor starter. RESULT: Thus the load test on single phase transformer is determined. 47
Exercise Number: 05 Title of the Experiment: REGULATION OF THREE PHASE ALTERNATOR BY EMF AND MMF METHOD Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To predetermine the voltage regulation of the three-phase alternator by EMF and MMF methods by conducting open circuit and short circuit tests. a) FACILITIES REQUIRED: Sl. No. Apparatus Type Range Qty. 1 Voltmeter MI (0-600) V 1 2 Ammeter MC (0-2) A 2 3 Ammeter MI (0-10) A 1 4 Rheostat Wire Wound 570 /1.2A 1 5 Rheostat Wire Wound 270 /1.2A 1 6 Tachometer Analog 1 7 SPST Switch 1 8 TPST Switch 1 9 Connecting wires 12 b)exercise: 48
c)precautions: 1. The motor field rheostat should be kept in minimum resistance position. 2. The Alternator field rheostat should be in maximum resistance position. 3. Initially all switches are kept in open position. d)procedure: OPEN CIRCUIT TEST 1. The connections are made as per the circuit diagram. 2. The D.C. motor is started and made to run at its rated speed of alternator by adjusting the field rheostat of motor. 3. The alternator field is excited by closing the SPST. 4. By varying the field rheostat the voltage across the alternator is increased upto its rated voltage. 5. The open circuit test is conducted by varying the field rheostat of the alternator and the corresponding alternator voltage and field current values are noted. 6. Open circuit characteristics are drawn by taking the field current along X axis and the phase voltage along Y axis. SHORT CIRCUIT TEST 1. The TPSTS is closed and by adjusting the field rheostat of the alternator till the voltmeter reads zero,,. 2. The short circuit current and the field current are noted. 49
3. Short circuit characteristics are drawn by taking the filed current along X axis and the short circuit current along Y axis. STATOR RESISTANCE TEST 1.The connections are made as per the circuit diagram. 2. By adjusting the load, the ammeter and voltmeter readings are tabulated. 3. From the tabulated values the stator resistance of the alternator is calculated. e) EXPRESSION FOR THEORITICAL VALUE: Let, E0=no load induced emf in volts cos = Power Factor Reff=1.6 Ra ohms Zs= E0/Isc ohms Xs= Zs 2 - Reff 2 ohms 1. E0 at unity power factor E0= (Vph+IReff) 2 + (IXs) 2 volts. 2. E0 at lagging power factor E0= (Vphcos +IReff) 2 + (Vphsin +IXs) 2 volts Vph=VL/ 3 volts 3. E0 at leading power factor 50
E0= (Vphcos +IReff) 2 + (Vphsin -IXs) 2 volts 4. % Regulation= ( E0-Vph)/ Vph *100 f)circuit DIAGRAM: 51
52
g)tabulation: OPEN CIRCUIT TEST: Sl. No. Field Current I f in Amps Open circuit voltage V L in volts Phase Voltage V PH in volts SHORT CIRCUIT TEST: Sl. No. Short circuit current ISC in Amps Field Current I f in Amps EMF METHOD: Sl. Power Eo in volts % Regulation No. Factor Lagging Leading Lagging Leading MMF METHOD: 53
Sl. No. Power Factor Field Current If in Amps Eo in volts % Regulation Lagging Leading Lagging Leading Lagging Leading h)to FIND Ra: TABULATION: 54
Sl. No. Armature Current in Amps Armature Voltage in Volts Armature resistance Ra=V/I in Ohms i)model GRAPHS: EMF Method: MMF Method: 55
j)discussion QUESTIONS: 1. Write down the emf equation for d.c.generator? E= (ФNZ/60) (P/A) V. P - No. of poles Z - Total no. of conductor Ф - Flux per pole, N-speed in rpm. 2. Why commutator is employed in d.c.machines? Conduct electricity between rotating armature and fixed brushes, convert alternating emf into unidirectional emf (mechanical rectifier) 3. Distinguish between shunt and series field coil construction? Shunt field coils are wound with wires of small section and have more no. of turns. Series field coils are wound with wires of larger cross section and have less no. of turns. 4.What is back emf in d.c. motor? As the motor armature rotates, the system of conductor come across alternate north and south pole magnetic fields causing an emf induced in the conductors. The direction of the emf induced in the conductor is in opposite to current. As this emf always opposes the flow of current in motor operation it is called as back emf. 5.Why MMF method of estimating voltage regulation is considered as optiimisticmethod? Compared to EMF method, MMF method involves more noof complex calculation steps. Further the OCC is referred twice and SCC is referred once while predetermining the voltage regulation for each load condition. Reference of OCC takes core saturation effect. As this method require more effort, final result is very close to actual value, hence this method is called as optimistic method. RESULT: Thus the three phase alternators are regulated by EMF & MMF method. 56
Exercise Number: 06 Title of the Experiment: LOAD TEST ON THREE-PHASE SQUIRREL CAGE INDUCTION MOTOR Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To determine the performance characteristics of three-phase squirrel cage induction motor by direct loading method. a)facilities REQUIRED: Sl. No. Apparatus Type Range Quantity 1 Voltmeter MI (0-600) V 1 2 Ammeter MI (0-10)A 1 3 Single element wattmeter LPF 600V,10A 2 4 3Ø Auto transformer 400V/(0-450)V 1 5 Tachometer Analog 1 6 Connecting wires As per required b)exercise: c)precautions: 57
1. At the time of starting, the motor should be in the no load condition. 2. Initially switch is in open position and the autotransformer should be kept at minimum voltage position. d)procedure: 1. The connections are made as per the circuit diagram. 2. The supply is given by closing the TPST switch. 3. The motor is started by using star delta starter. 4. The voltmeter, ammeter, wattmeter, speed and spring balance readings are noted for no load condition. 5. The load is increased in steps at each increase in step of the load the readings of voltmeter, ammeter, wattmeter, speed and spring balance are noted. e) EXPRESSION FOR THEORITICAL VALUE: 1. Torque,T=9.81*Reff*(S1-S2) N-m Where, Reff =Effective radius of the brake drum in m. S1, S2= Spring balance readings in Kg. 2. Output power,po = 2 NT/60 watts Where, N=speed of the motor in rpm 3. % Efficiency = (Output power/input power)*10 58
f)circuitdiagram: 59
g)tabulation: Sl. No. Line Voltage VL in volts Line Current IL in amps Input Power PL in watts Speed N in Rpm S1 in Kg Spring balance readings S2 in Kg S1 - S2 in Kg Torque T N-m Ouput Power watts Effi. Slip PF % % MF=--- 60
h)model GRAPH: Electrical Characteristics: DISCUSSION QUESTIONS: 1. What are the advantages of cage motor? Since the rotor has low resistance, the copper loss is low and efficiency is very high. On account of simple construction of rotor it is mechanically robust, initial cost is less, maintenance cost is less, simple starting arrangement. 2. What are the 2 types of 3phase induction motor? Squirrel cage and slip ring induction motor. 3. Write two extra features of slip ring induction motor? Rotor has 3phase winding. Extra resistance can be added in rotor circuit for improving PF with the help of three slip rings. 4. Where three point starter and four point starter are recommended? 3 Point starter- For armature speed control of DC motor(below rated speed). 4 point starter- For field speed control of DC motor(above rated speed). 5. What are the methods of starting 3 phase squirrel cage induction motor? DOL starter, primary resistance starter, star-delta starter, auto-transformer starter. RESULT: Thus the load test on three phase induction motor are determined. 61
Exercise Number: 07 Title of the Experiment: NO LOAD TEST AND BLOCKED ROTOR TEST ON THREE PHASE INDUCTION MOTOR Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To determine the equivalent circuit parameters of three phase induction motor by conducting i) No load test and ii) Blocked rotor test. a)facilities REQUIRED: Sl. No. Apparatus Type Range Quantity 1 Voltmeter MI (0-300) V 1 2 Voltmeter MI (0-600) V 1 3 Ammeter MI (0-5)A 1 4 Ammeter MI (0-10) A 1 5 Single element wattmeter UPF 600V, 5A 2 6 Double element wattmeter LPF 300V, 10A 2 7 3Ø Auto transformer 400V/(0-450) V 1 8 Connecting wires b)exercise: 62
c)precautions: 1. The autotransformer should be kept at minimum voltage position. 2. Blocking of the rotor should be done properly. 3. Making sure that only a small voltage is applied across the stator terminal with rotor blocked. NO LOAD TEST: 1. The connections are given as per the circuit diagram. 2. The supply is switched ON by closing the TPST switch. 3. The autotransformer is varied until the voltmeter reads the rated voltage of the machine and the motor starts running at no load. 4. The readings of Voltmeter, ammeter and wattmeter are noted at no load. 5. Then the TPST supply switch is opened. BLOCKED ROTOR TEST: 1. The connections are given as per the circuit diagram. 2. The rotor of the motor is blocked tightly with the help of brake drum and belt. 3. The supply is switched ON by closing the TPST switch. 4. The autotransformer is adjusted slowly until the voltmeter reads the rated current of the machine 5. The readings of voltmeter, ammeter and wattmeters are noted. 6. Now release the blocked rotor and open the TPST supply switch. 63
e) EXPRESSION FOR THEORITICAL VALUE: NO LOAD TEST: 1. No load power factor, cosф0=w0/( 3 V0I0) Where, W0=No load power in watts V0=No load voltage in volts I0=No load current in amps 2. Working component current, Iw=I0 cosф0 Amps 3. Magnetizing current,im= I0 sinф0 Amps 4. No load resistance,r0=v0/iw ohms 5. No load reactance,x0=v0/im ohms BLOCKED ROTOR TEST: 1. At short circuit power factor,cosфsc=wsc/( 3VSCISC) Where, WSC=Short circuit power in watts VSC=Short circuit voltage in volts ISC= Short circuit current in Amps 2. Motor equivalent impedance as referred to stator, Z01= VSC/( 3ISC) ohms 3. Motor equivalent resistance as referred to stator, R01= Z01 cosфsc ohms 4. Motor equivalent reacttance as referred to stator, X01= Z01 sinфsc ohms 5. R01 = R01 Reff ohms 6. Reff = 1.6 * Ra ohms Where, 64
Ra=Armature resistance offered for D.C 7. Equivalent load resistance (RL )= R01 ((1/S)-1)) ohms Where, Slip (S)=(Ns-N)/ Ns Ns=Synchronous Speed in rpm N=Speed of the rotor in rpm. 8. Effective resistance=r-01+ RL + X01 ohms 9. Rotor current referred to stator side, I2 =(V/ 3)/Effective resistance 10. Stator current, I1=I0+ I2 Amps 65
f)circuit DIAGRAM: 66
67
g)tabulation: NO LOAD TEST: Speed of the induction motor: Type of the stator connection: Multiplication factor: No load power input P0 in watts No load voltage V0 in volts No load current I0 in amps Observed reading (OR) Actual reading AR=OR*MF MF= BLOCKED ROTOR TEST: Type of the stator connection: Multiplication factor: Power input PSC in watts Line voltage VSC in volts Short circuit current ISC in amps Observed reading (OR) Actual reading AR=OR*MF MF= 68
Tabulation: Sl. No. Armature Current in Amps Armature Voltage in Volts Armature resistance Ra=V/I in Ohms 69
h) DISCUSSION QUESTIONS: 1. Why an induction motor never runs at its synchronous speed? If it runs at sy. Speed then there would be no relative speed between the two, hence no rotor emf, so no rotor current, then no rotor torque to maintain rotation. 2. What are slip rings? The slip rings are made of copper alloys and are fixed around the shaft insulating it. Through these slip rings and brushes rotor winding can be connected to external circuit. 3. What are the principal advantages of rotating field type construction? Relatively small amount of power required for field system can easily supplied to rotating system using slip rings and brushes, more space is available in the stator part of the machine to provide more insulation, it is easy to provide cooling system, stationary system of conductors can easily be braced to prevent deformation. 4. In which direction a shaded pole motor runs? The rotor starts rotation in the direction from unshaded part to the shaded part. 5. Why single phase induction motors have low PF? The current through the running winding lags behind the supply voltage by large angle so only single phase induction motors have low PF. RESULT: Thus the no load and blocked rotor test of three phase induction motor are determined. 70
Exercise Number: 09 Title of the Experiment: DIGITAL SIMULATIONS OF LINEAR SYSTEMS Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To digitally simulate the time response characteristics of a linear system without non linearities and to verify it manually. a)facilities REQUIRED Mat lab Software b)theory The time characteristics of control systems are specified in terms of time domain specifications. Systems with energy storage elements cannot respond instantaneously and will exhibit transient responses, whenever they are subjected to inputs or disturbances. The desired performance characteristics of a system of any order may be specified in terms of transient response to a unit step input signal. The transient response characteristics of a control system to a unit step input is specified in terms of the following time domain specifications Delay time td Rise time tr Peak time tp Maximum overshoot Mp Settling time ts c)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. 71
10. For the same transfer function write a matlab program to obtain the step response and verify both the results. d)program %This is a MATLAB program to find the step response num=[0 0 25]; den=[1 6 25]; sys = tf (num,den); step (sys); e)plot 72
h)discussion QUESTIONS: 1.What is meant by Delay time? It is the time to required for the response to reach 50% of final value in first attempt 2.What is meant by Rise time? It is the time to required for the response to rise from 10% to 90% of the final value 3.What is meant by over damping? It is the time to required for the response to rise from 0% to 100% of the final value 4.What is meant by peak time? It is the time to required for the response to reach peak time response 5.What is meant by setting time? It is the time to required for the response to reach an stay within a specific tolerance band RESULT: Thus the digital simulation of linear system are determined. 73
Exercise Number: 10(A) Title of the Experiment: STABILITY ANALYSIS OF LINEAR SYSTEMS (Bode Plot) Date of the Experiment : OBJECTIVE (AIM) OF THE EXPERIMENT: To obtain the bode plot for the given system whose transfer function is given as G(S)= 242(s+5) s(s+1)(s 2 +5s+121) and to find out whether the system is stable or not. a)facilities REQUIRED b)theory Mat lab Software A Linear Time-Invariant Systems is stable if the following two notions of system stability are satisfied When the system is excited by Bounded input, the output is also a Bounded output. In the absence of the input, the output tends towards zero, irrespective of the initial 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 jω-axis, then system is unstable. If one are more non-repeated roots of the characteristic equation on the jω-axis, then system is unstable. c)bode PLOT : Consider a Single-Input Single-Output system with transfer function C(s) R(s) = b0 s m + b1 s m-1 + + bm a0 s n + a1s n-1 + +an 74
Where m < n. Rule 1 A system is stable if the phase lag is less than 180 at the frequency for which the gain is unity (one). Rule 2 A system is stable if the gain is less than one (unity) at the frequency for which the phase lag is 180. The application of these rules to an actual process requires evaluation of the gain and phase shift of the system for all frequencies to see if rules 1 and 2 are satisfied. This is obtained by plotting the gain and phase versus frequency. This plot is called BODE PLOT. The gain obtained here is open loop gain. The stability criteria given above represent Limits of Stability. It is well to design a system with a margin of safety from such limits to allow for variation in components and other 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 in terms of a Gain Margin and Phase Margin from the limiting values quoted. d)algorithm 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 is 180, the system is stable. This is 5dB Gain Margin. 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. e)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 75
w=logspace(-2,4,1000); 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) f)procedure TO OBTAIN BODE PLOT 1. Rewrite the sinusoidal transfer function in the time constant form by replacing s by jω 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 phase plot. G)MANUAL CALCULATIONS i)the sinusoidal transfer function G (jω) is obtained by replacing s by jω in the given s domain transfer function G(jω)= 242(jω +5) jω (jω +1)( jω 2 +5 jω +121) On comparing the quadratic factor of G(s) with standard form of quadratic 76
factor, ζ and ωn can be evaluated. s 2 +5s+121 = s2+2ζωns + ωn 2 On comparison ωn 2 = 121 2ζωn= 5 ωn =11 rad/sec ζ = 0.227 G(jω)= 10(1+0.2jω) jω (1+jω)( 1+0.4 jω -0.0083ω 2 ) ii)corner FREQUENCIES The corner frequencies are ωc1=1rad/sec ωc2= 5 rad/sec and ωc3=11rad/sec Choose a low frequency ωl such that ωl< ωc1 and choose a high frequency ωh> ωc3. Let ωl=0.5 rad/sec and ωh=100 rad/sec Term Corner Frequency rad/sec Slope db/dec Change in slope db/dec 10-20 jω 1 1-20 -20-20= - 40 (1+jω) (1+0.2jω) 5 20-40-20 = -20 1 11-40 -40-20 = -60 ( 1+0.4 jω -0.0083ω 2 ) Iii) MAGNITUDE PLOTS Calculate A at ωl, ωc1, ωc2, ωc3, and ωh Let A= G(jω) in db At ω= ωl A= 20 log(10/0.5)=26.03db ω= ωc1, A=20log(10/1)=20db 77
ω= ωc2 A= -40log(5/1)+20=-7.96 db ω= ωc3 A = -20log(11/5) - 7.96 = -14.80 db ω= ωh A = -60log(100/11)-14.80 = - 72.3 db These values are plotted in the semilog graph sheet taking frequency along the logarithmic scale and magnitude in db along the linear scale iv) PHASE PLOT The phase angle of G(jω) as a function of ω is given by Φ = G(jω) = tan -1 0.2ω -90 tan -1 ω tan -1 0.04ω/(1 0.0083ω 2 ) Ω tan -1 0.2ω tan -1 ω tan -1 {0.04ω/ Φ = G(jω) (1 0.0083ω 2 )} 0.5 507 26.56 1.15-112 1 11.3 45 2.31-126.01 5 45 78.96 14.04-138 10 63.43 84.29 63.44-174.3 11 65.5 84.8 85.8-195.4 20 75.96 87.14 180-19.98=160-261.18 50 84.3 88.85 180-6=174-268.55 100 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 along the logarithmic scale and phase angle in degrees along the linear scale. h)output (from simulation) 242 s + 1210 ----------------------------- s^4 + 6 s^3 + 126 s^2 + 121 s 78