LABORATORY MANUALExperiments in Network Analysis

ByMASHAALLAH BHAI

Peon of Holy Prophet [P.B.U.H]Dedicated to: Prof. Dr. Sohail Aftab Qureshi

Sr.# List of experiments Page #1. To verify Thevenin’s theorem in a simple dc circuit.2. To verify maximum power transfer theorem.3. To examine frequency response of a parallel RC circuit.4. To examine the step and pulse response of a series RC circuit.5. To examine the sinusoidal frequency response of a series RL circuit.6. To examine the pulse response of a series RL circuit. 7. To examine the series resonance and to measure the resonant

frequency of a RLC circuit. 8. To prove the superposition theorem in circuits. 9. The magnitude and phase response of a low pass filter.10. The magnitude and phase response of a high pass filter.11. Frequency characteristics of a band pass filter.12. Frequency characteristics of a band stop filter.Experiment#1:To verify thevenin’s theorem in a simple d.c. circuit. Equipment:DC power supply (0 - 10 V)DVM or VOMResistors (1/4 watt)1k 3k 6.2k 5 percentProcedure:

1. For the circuit shown in the figure, use thevenin’s theorem to calculate the values of Vth and Rth and record them in the observations table.

2. Measure the open circuit voltage VAB and record this as Vth under the measured column of the observations table.

3. Replace the source with a short circuit and measure the resistance between the terminals A and B. Record this as Rth under the measured column.

4. Calculate the voltage across and current through a 3k load that is to be placed across the terminals A and B. Perform the calculations for both the actual circuit and its thevenin equivalent. The results should be identical. Record the results in the column headed “calculated”.

5. Connect a 3k load to the terminals A and B of the circuit in the figure. Measure the resulting load current and voltage.

6. Construct the circuit of figure B with the calculated values of Vth and Rth. Connect a 3k load to the terminals A and B. Measure the resulting load current and voltage, and record them in the table. They should agree closely with those in the adjacent columns.

Circuit diagram:

A

R1= 3k R3= 1k

Vd.c.=8.94V R2=6.2k 3k

B

Vth

Vth = [R2/R1 + R2]Vd.c.

= [6.2k/6.2k + 3k]8.94 = 6.025V

Rth

Rth = R3 + [R1R2/R1 + R2] = 3.022k Thevenin parameters Calculated MeasuredVth 6.025V Vth

Rth 3.022k Rth

Loaded circuit parameters

s

5.431kVd.c.=8.94V

IL = 8.94/5.431k = 1.646mAVL = 3.0015V

Rth = 3.022k

Vth = 6.025V RL = 3k

IL = Vth/Rth + RL = 6.025/3.022k + 3k = 1mA

Loaded circuit parametersCalculated Measured

Actual circuit Thevenin equivalentIL 1.646mA IL IL

VL 3.0015V VL VL

Experiment#2:To verify maximum power transfer theorem.Equipment:DC power supply (0 - 10 V)DVM or VOMResistors (1/4 watt)1k 1.5k 2k 2.7k 3k

(two)3.3k 4.3k 5.6k 7.5k 10k 5%

Procedure:1. Given VS = 10V and RS = 3k, calculate the load voltage VL, load current IL

and load power PL for each value of RL given in the table. 2. Setup the source with VS = 10V and RS = 3k. For each value of RL, measure

and record the load voltage VL and the load current IL. From this calculate the power PL. Complete the table for measured data.

3. Plot a graph of load power PL versus load resistance RL from measured data. Determine where is the load power a maximum and its value.

Calculated dataRL (k) 1 1.5 2 2.7 3 3.3 4.3 5.6 7.5 10VL (V) 2.5 3.3 4.2 4.7 5 5.1 5.8 6.5 7.3 7.6IL (mA) 2.5 2.2 1.9 1.7 1.6 1.6 1.3 1.16 0.88 0.7PL (mW) 6.25 7.3 8.1 8.2 8.3 8.3 8.01 7.55 6.4 5.8

Measured dataRL (k) 1 1.5 2 2.7 3 3.3 4.3 5.6 7.5 10VL (V)IL (mA)PL (mW)Circuit diagram:

IL

VS = 10VRL = 3k

Experiment#3:To examine the sinusoidal frequency response of a series RL circuit.Equipment:Signal generatorOscilloscope and time 1X probesDVMInductor 0.5-32H (as resistance < 500 at 1kHz)

Resistor 36k5%

Circuit diagram:

R

VR

L VL

Procedure:1. Setup the apparatus as shown in the figure. 2. Set f = 10 kHz and note I, VR and Vcoil. By increasing and decreasing f above

and below 10 kHz, find out the variational effects on all the above three quantities.

3. Now vary the frequency from 2 kHz to 40 kHz and record VR and Vcoil for all the values of f in the table.

4. Complete the table by calculating I, Zt and Zcoil.Freq. EFFECTSf I VR Vcoil f I VR Vcoil

R = 36 kFREQ (kHz)

VR (volts) Vcoil (volts) IR = VR/R Zt = Vin/I Zcoil = Vcoil/I

2 510203040From the figure,I = Vin/R + jLI= Vin/(R)2 + (L)2

VR = IRVcoil = I(jL)Vcoil= L(Vin/(R)2 + (L)2)

Frequency VS VR

Vin

Frequency VS Vcoil

Experiment#4:

To examine frequency response of a parallel RC circuit. Equipment: Signal generatorOscilloscope and two 1X probesHigh impedance voltmeterCapacitors, 0.001F 10%Resistors (1/4W): 16 k 10%, three 100 precisionProcedure:

1. Construct the circuit shown.2. With the aid of oscilloscope, adjust the source to provide a convenient

terminal voltage at 10 kHz. 3. Use the oscilloscope to monitor the voltages V1 and V2. By observing the

oscilloscope traces, note any variations in V1 and V2 as you vary the frequency around 10 kHz. A decade below and above 10 kHz is a good range. Recall that IR is proportional to V1 and Ic is proportional to V2.

Observations & calculations:Freq. V1 V2 IR IC Z XC I V251020Circuit diagram:

R

IR

V1

IV2

IC C

Freq. VARIATIONAL EFFECTSf IR _ IC I f IR _ IC I

A.C.

Experiment#5: To examine the pulse response of a series RL ciorcuit.Equipment: Signal generatorOscilloscope and time 1X probesDVMInductor: 73.8 mH (ac resistance < 500 at 1 kHz)Resistor: 30 k 5%Procedure:

1. Calculate and record quantities Zcoil, Z, I, VR and Vcoil for a terminal voltage of 5V and frequency of 1 kHz.

2. Use the DVM to measure voltages Vcoil and VR.3. Calculate and sketch the following. Circuit diagram:

R

VR

L Vcoil

I

Theory:Q: A voltage pulse of magnitude 6 volts and duration 3 seconds extending from t = 3 seconds to t = 6 seconds is applied to a series R-L circuit consisting of R = 6 and L = 2H. Obtain the current i(t). Also calculate the voltage across L and R. Solution: R

L

i(t)

v(t)

v(t)

1 2 3 4 5 6 7 tApplication of kirchhoff’s voltage law to the circuit yields, Ldi/dt + Ri = 6[u(t - 3) – u(t - 6)]On Laplace transformation, we getL[s.I(s) – i(0+)] + R.I(s) = 6[e-3s/s – e-6s/s]But i(0+) = 0.Hence L[s.I(s)] + R.I(s) = 6[e-3s/s – e-6s/s]I(s)[Ls + R] = (6/s)[e-3s – e-6s]I(s) = (6/s)[e-3s – e-6s]/[Ls + R]L = 2HR = 6I(s) = (6/s)[e-3s – e-6s]/[2s + 6]I(s) = 6[e-3s – e-6s]/s[2s + 6]I(s) = 6[e-3s – e-6s]/s[2(s + 3)]I(s) = 3[e-3s – e-6s]/s(s + 3)Consider: Using partial fraction:3/s(s + 3) = A/s + B/(s + 3)3 = A(s + 3) + BsPut s = 03 = 3AA = 1Put s = -33 = A(s + 3) + Bs3 = -3BB = -1I(s) = 3[e-3s – e-6s]/s(s + 3)I(s) = [e-3s – e-6s][A/s + B/(s + 3)]I(s) = [e-3s – e-6s][1/s - 1/(s + 3)]I(s) = e-3s/s – e-3s/(s + 3) - e-6s/s + e-6s/(s + 3) On inverse Laplace transformationi(t) = u(t - 3) – u(t - 6) – e-3(t - 3)u(t - 3) + e-3(t - 6)u(t - 6)i(t) = u(t - 3)[1 - e-3(t - 3)] - u(t - 6)[1 - e-3(t - 6)]Voltage across the inductor is given by vL = Ldi/dt = Ld/dt{u(t - 3)[1 - e-3(t - 3)] - u(t - 6)[1 - e-3(t - 6)]}d/dt{1 - e-3(t - 3)} = 0 - e-3(t - 3)d/dt{ -3(t - 3)} = -e-3(t - 3)d/dt{ -3t + 9)}= -e-3(t - 3){-3} = 3 e-3(t - 3)

d/dt{1 - e-3(t - 6)} = 0 - e-3(t - 6)d/dt{ -3(t - 6)} = -e-3(t - 6)d/dt{ -3t + 18)}= -e-3(t - 6){-3} = 3 e-3(t - 6)

vL = Ldi/dt = Ld/dt{u(t - 3)[1 - e-3(t - 3)] - u(t - 6)[1 - e-3(t - 6)]}vL = Ldi/dt = 2{u(t - 3) 3 e-3(t - 3) - u(t - 6) 3 e-3(t - 6)}vL = Ldi/dt = 6{u(t - 3)e-3(t - 3) - u(t - 6)e-3(t - 6)}t=0 vL = 0V1 0V2 0V3 6V4 0.299V5 0.015V6 -6V

Voltage across resistorvR = i(t)R = 6u(t - 3)[1 - e-3(t - 3)] - 6u(t - 6)[1 - e-3(t - 6)]t=0 01 02 03 04 5.75 5.9856 0

Experiment#6:The magnitude of a low pass filter.

Equipment:Signal generatorOscilloscope, DVMResistors: 15 k, 1.5 kCapacitors: 0.047 F.Circuit diagram:

R

Vin Vout

C

Low-Pass FiltersFigure shows a simple RC circuit used as a low-pass filter. The bandwidth of the passband is BW = f2 – f1

BW = fc – 0 = fc

Frequency fc is the cutoff frequency; it is the frequency above, which the output voltage drops below 70.7 percent of the input voltage. Vout = 0.707Vin

According to voltage-divider equation:

Vout = Vin{XC/R2 + XC2}

If the rms input voltage to the filter is 10 V, determine the output voltage at200 Hz 600 Hz 1200 Hz 5000 Hz 10,000 HzAssume R = 1 kC = 0.047 Ff Hz Vout

200 9.983V600 9.8471200 9.426V5000 5.60710000 3.207

Experiment#7:

The magnitude of a high pass filter. Equipment:Signal generatorOscilloscope, DVMResistors: 15 k, 1.5 kCapacitors: 0.047 F.Circuit diagram:

C

Vin R Vout

A.C.

Observations:Frequency Hz

Vin

VVout

100 2.4 0.9200 2.4 1.4500 2.4 21000 2.4 2.22000 2.4 2.25000 2.4 2.210000 2.4 2.220000 2.4 2.2

Experiment#8:

Frequency characteristics of a bandstop filter.Equipment:Signal generatorOscilloscope and 1X probeHigh impedance millivoltmeterCapacitors: 0.01F, 0.01FInductor: 100 – 200 H radio frequency coilResistors: (1/4 watt): 100, 1k, 2k 5% Observations:

Component values Resonant frequency Q ΔfL C R Calculated Measured

Circuit diagram:a

R

Lcoil Vout

Vin Rcoil

C

bFor input frequencies within the stop-band region, Vout < 0.707 Vin. The resonance frequency and bandwidth of the bandstop filter may be determined from fr = 1/2LCQs = fr/BW = rLcoil/R1 + Rcoil

Assuming the current drawn by the load is insignificant (unimportant),Vout = Vin(Zab/Zcircuit)Zab = Rcoil + j(XL - XC)Zcircuit = R1 + Rcoil + j(XL - XC)Zab = R2

coil + (XL - XC)2

Zcircuit = (R1 + Rcoil)2 + (XL - XC)2

Vout = Vin(R2coil + (XL - XC)2/(R1 + Rcoil)2 + (XL - XC)2)

Δf = f2 – f1

Q = fr/Δf(BW)Experiment#9:To study the frequency characteristics of a band pass filter. Equipment:

Audio signal generatorOscilloscopeHigh impedance milli ohmmeter, capacitorsInductors, resistorsCircuit diagram:

R0

Vout

LVin

C

A bandpass filter that uses a series RLC circuit whose resonance frequency and bandwidth provide the desired passband is shown in figure. Resonant frequency:

fr = 1/2LCBW = fr/Qs

Qs = rL/R0

Vout = VinR0/R02 + (XL - XC)2

Observations:Component values Resonant frequency Q ΔfL C R Calculated Measured

Experiment#10:To be able to predict and verify voltages in circuit containing d.c. and a.c. Equipment:Function generator with d.c. OffsetMultimeter DVMD.C. power supply 0 – 10 V (if d.c. offset not available on function generator)Capacitor 0.001 FResistor 20 k

A.C. = 1V(rms)D.C. = 2VCircuit diagram:

C

VC

Vin R VR

A.C. & D.C.

VR = VR(D.C.) + VR(A.C.) = 0 + 1 = 1VVc = Vc(D.C.) + Vc(A.C.) = 2 + 0 = 2VObservations:

VC VR

dc ac dc acCalculatedData

2 0 0 1

Measured DataExperiment#11:To examine step and pulse response of series RC circuit. Equipment:Square wave generatorOscilloscope and time 1X probesCapacitors: 0.001 F 10%Resistors: two 10 k 5%Circuit diagram:

R

C

Observations:R C Τ = RC Τ = 5RC fmax

12

Charging curve:

Number of time constants Calculated voltage Measured voltage12345

Discharging curve:

Number of time constants Calculated voltage Measured voltage12345

Voltage across C:

Experiment#12:To show that the resonant frequency of a series RLC circuit is given by 1/2LC and plot the frequency response of an RLC circuit. Equipment:Audio signal generatorOscilloscope Capacitors: 0.001 F Resistance: 100 5%Inductor.Circuit diagram:

R

VR

L Vin

C

Theory:

Z = R2 + (XL - XC)2

I = V/ZFor resonance frequencyXL = XC

2frL = 1/2frCfr = 1/2LCObservations:Frequency vR

500 Hz1 kHz2 kHzfr = 3 kHz5 kHz10 kHz20 kHz50 kHz

Frequency XC XL Z I vR

500 Hz1 kHz2 kHzfr = 3 kHz5 kHz10 kHz20 kHz50 kHzFrequency versus vR