Communicationlabmanual

Communication Lab Manual SSIT, Tumkur

COMMUNICATION LAB MANUAL
FOR
V SEMESTER B.E (E & C)
(For private circulation only)

VISHVESHWARAIAH TECHNOLOGICAL UNIVERSITY

NAME: ___________________________

DEPARTMENT OF ELECTRONICS & COMMUNICATION


SRI SIDDHARTHA INSTITUTE OF TECHNOLOGY
MARLUR, TUMKUR-572105

CONTENTS
1. II-Order Low Pass and High Pass Active Filters

2. II –Order Band Pass and Band Elimination Filters

3. Attenuators

4. Collector Amplitude Modulation & Demodulation

5. Balanced Modulator

6. Class-C Tuned Amplifier

7. Frequency Modulation and Demodulation

8. Radio Receiver Characteristics

9. Pre & De – Emphasis Networks

10. AM IC Circuit-Modulation and Demodulation

11. Pulse Amplitude Modulation

12. Pulse Width Modulation

13. Pulse Position Modulation

14. Transistor Mixer


TESTING OF EQUIPMENTS BEFORE STARTING THE
CONDUCTION
1. OP AMP
Apply sine wave of amplitude 1
volt (1 kHz) as shown in ckt
diagram, if IC is good the output
be a square wave with peaks at +
VSAT and – Vsat.

2. 555 Timer :
If IC is good for the applied 5 V D.C supply as in
ckt diagram the voltage at pin no. 5 will be 2/3
Vcc (3.3 Volts)

3. Transistor
Identify emitter, base and collector of the
transistor, with DMM in diode position, if
transistor junctions are good it should indicate a
low resistance upon forward biasing emitter base junction or collector – base
junction and should indicate either OL or 1.(depending on DMM) upon
reverse biasing EB or CB junctions.
4. Source impedance of ASG:
1. Connect the DRB with the maximum resistance to ASG as in figure.
2. Adjust the amplitude of sine wave of 5V pp at 1 KHz.
3. Start reducing the resistance of DRB this reduces the output voltage also.
Source resistance Rs is that value of DRB resistance when the amplitude of
the output signal is half of the initial value. (2.5 V pp)


CIRCUIT DIAGRAM: -
II-Order Active Low Pass Filter

II-Order Active High Pass Filter

Design:- (LPF & HPF)
Assume Pass band gain AV = 2, Cutoff frequency fC = 5KHz
Rf
1. Amplifier: AV = 1 + = 2, then Rf = R, choose Rf = R = 10KŸ
R
1
2. Filter Circuit : Cut off frequency fC = = 5KHz
2SR1C1
Choose C1 = 0.01Pf then R1 = 3.183 KŸ a 3.3 KŸ

Rf = 10KŸ, R1 = 3.3KŸ, C1 = 0.01Pf, Op-amp = PA741

1


Experiment No: DATE: __/__/____

II – Order Low Pass and High Pass Active Filters

AIM: - Design a second order Butterworth active low pass / high pass filter for a
given cut-off frequency fC = ______Hz. Conduct an experiment to draw frequency
response and verify the roll off.

PROCEDURE: -

1. Connections are made as shown in the circuit diagram.
2. Apply sine wave i/p signal of peak amplitude 5 volts.
3. Check the gain of non-inverting amplifier by keeping the frequency of the
input signal in the pass band of the filter. Note down the output voltage
VO max.
4. Keeping the input signal amplitude constant, vary the frequency until the
output voltage reduces to 0.707 Vo max, the corresponding frequency is
the cut-off frequency (fC) of the filter.

To find the Roll-off factor :-
1. For LPF :- Keeping the input signal amplitude constant, adjust the input
frequency at 10fC. Note down the output signal amplitude. The difference
in the gain of the filter at fC and 10fC gives the Roll-of factor.
2. For HPF :- Keeping the input signal amplitude constant, adjust the input
frequency at 0.1fC, note down the output signal amplitude. The difference
in the gain of the filter at fC and 0.1fC gives the Roll-of factor.

Conclusion:

2


Tabulation:

High Pass Filter Vi p-p = Volts (Constant)
I/P frequency in O/P Voltage Gain magnitude Gain magnitude in DB
Hz VO P-P (volts) (Vo/Vi) 20log(Vo/Vi)

Roll off = - (G1 - G2) db/decade =

Frequency Response for High Pass Filter

3


Tabulation:
Low Pass Filter Vi p-p = Volts (Constant)
I/P frequency in O/P Voltage Gain magnitude Gain magnitude in DB
Hz VO P-P (volts) (Vo/Vi) 20log(Vo/Vi)


Frequency Response for Low Pass Filter

Staff-in-charge:

4


CIRCUIT DIAGRAM: -
II-Order Active Band Pass Filter

II-Order Active Band Elimination Filter

Design:-

1. BPF : - R = 10KŸ, Rf = 5.86 KŸ, R1 = 1.989 KŸ, R2 = 3.3 KŸ,
C1 = 0.01Pf, C2 = 0.01Pf, Op-amp = PA741

2. BSF : - R = 10KŸ, Rf = 5.86 KŸ, Ra = 3.3 KŸ, Rb = 1.989 KŸ,
C1 = 0.01Pf, C2 = 0.01Pf, Op-amp = PA741

5



II – Order Band Pass and Band Elimination Active Filters

AIM: - Design a second order band pass and band stop active filter for a given
frequencies fC1 = ______Hz and fC2 = ______Hz. Conduct an experiment to draw
frequency response and verify the Roll off (Band Width = 3 to 5 KHz).

PROCEDURE: -

2. Apply sine wave i/p signal of peak amplitude 5 volts.
3. Check the gain of non-inverting amplifier by keeping the frequency of the
input signal in the pass band of the filter. Note down the output voltage
VO max.
4. Keeping the input signal amplitude constant, vary the frequency on either
side of pass band until the output voltage reduces to 0.707 Vo max, the
corresponding frequencies are the lower cut-off frequency (fL) and the
upper cut-off frequency (fH) of the filter.

To find the Roll-off factor :-
1. For LPF :- Keeping the input signal amplitude constant, adjust the input
frequency at 10fC, note down the output signal amplitude. The difference
in the gain of the filter at fC and 10fC gives the Roll-of factor.
2. For HPF :- Keeping the input signal amplitude constant, adjust the input
frequency at 0.1fC, note down the output signal amplitude. The difference
in the gain of the filter at fC and 0.1fC gives the Roll-of factor.

6


Design:
Specifications:
Pass band gain AV = 1.586, cut -off frequency fH = 5 KHz, fL=8 KHz, BW= 3 KHz
1. Amplifier:
Voltage gain AV = 1 + Rf / R = 1.586, choose R = 10K:,
Then Rf = 5.86 k: (use 5.6 k:+ 220 : std value)
2. Filter:
Cut - off frequency fH= 1/2S R2C2= 5 KHz
Choose C2= 0.01Pf, then R2 = 3.183 k: (Select R2 = 3.3 k:)
Cut - off frequency fL = 1/2S R1 C1 = 8 k Hz
Choose C1= 0.01Pf, then R1= 1.989 k : (Select R1 = (1.5 k: + 470:))

Tabulation:
Band Pass Filter Vi p-p = Volts (Constant)
Frequency Gain in DB
O/P Voltage VO PP (volts) Gain (Vo/Vi)
Hz 20 log (Vo/Vi)
Vomax =

fL = G1

0.1fL = 0.707 Vomax = G2
10fH=

f H= 0.707 Vomax = G2’


Frequency Response for Band Pass Filter

7


Tabulation:
Band Elimination Filter Vi p-p = Volts (Constant)
Frequency Gain in DB
O/P Voltage VO PP (volts) Gain (Vo/Vi)
Hz 20 log (Vo/Vi)
Vomax =

fL = G1

0.1fL = 0.707 Vomax = G2
10fH=

f H= 0.707 Vomax = G2’


Frequency Response for Band Elimination Filter

Conclusion:

Staff-in-charge:

8


CIRCUIT DIAGRAM: -
T-Type Attenuator S-Type Attenuator

Design:-

Specification: Vi = 5v, Vo = 2.5v, f = 1KHz
T- Type
R O (N 1) R O 2N
R1 R2
(N 1) (N 2 - 1)
RO =RS =600: (Assuming RS of ASG as 600:)
N = Attenuation factor = Vi / Vo = 2,
Therefore R1 = 200:, R2= 800:,
R1 = 200:, R2 = 800:, RL = 600:

S- Type
R O (N 2 1) R O (N 1)
R1 R2
2N (N - 1)
RO=RS=600: (Assuming Rs. of ASG as 600:)
N = attenuation factor Vi / Vo = 2,

Therefore R1 = 450:, R2 = 1.8 K:.

R1 = 450:, R2 = 1.8 K:, RL = 600:

Type Vi volts VO volts N = Vi/VO

T-Type

S-Type

9



Attenuators – T, S, Lattice and O-Pad Types

AIM: - Design the attenuation circuits using T, S, O-Pad and Lattice type
networks to attenuate a given signal of amplitude _______volts and frequency
______Hz to be reduced to 50% of the amplitude. Test the circuit and record the
results.

PROCEDURE: -

1. Find the source resistance RS of ASG.
3. Adjust the amplitude of the input signal at 5VP-P at 1KHz.
4. Measure the amplitude of the output signal.
5. Find the attenuation factor N.

Design:-

1. T-Type attenuators:-

(N - 1)
R1 RO 200
(N 1)
For N=2 and RS = RO = 600Ÿ, then
N
R2 2R O 800
(N 1)
2

2. S-Type attenuators:-

(N 2 - 1)
R1 RO 450
2N
For N=2 and RS = RO = 600 , then
(N 1)
R2 RO 1.8K
(N 1)

10


Lattice-Type Attenuator O-Pad Type Attenuator

Design:-

Specification: Vi = 5v, Vo = 2.5v, f = 1KHz
Lattice- Type
R O (N 1) R O 2N
R1 R2
(N 1) (N 2 - 1)
RO =RS =600: (Assuming RS of ASG as 600:)
N = Attenuation factor = Vi / Vo = 2,
Therefore R1 = 200:, R2= 800:,
R1 = 200:, R2 = 800:, RL = 600:

O-Pad Type
R O (N 2 1) R O (N 1)
R1 R2
2N (N - 1)
RO=RS=600: (Assuming Rs. of ASG as 600:)
N = attenuation factor Vi / Vo = 2,

Therefore R1 = 450:, R2 = 1.8 K:.

R1 = 450:, R2 = 1.8 K:, RL = 600:

Type Vi volts VO volts N = Vi/VO

Lattice-Type

O-Pad Type

11


Design:-

3. Lattice-Type attenuators:-
(N - 1)
R1 RO 200
(N 1)
N
R2 2R O 800
(N 1)
2

4. O-Pad Type attenuators:-

(N 2 - 1)
R1 RO 450
2N
(N 1)
R2 RO 1.8K
(N 1)

Conclusion:-

Staff-in-charge:-

12


CIRCUIT DIAGRAM: -
Collector AM and Demodulation using Envelop Detector

Design:-
Specifications: -
Tuned frequency = fIFT, Assume fIFT = 455 KHz, t = 2.19 Psec
RC t, i.e., RC = 100 t = 0.219 msec
Choose C = 0.01 Pf, then R = 21.97 KŸ, Select R = 22KŸ (Std. value)
1 1
Envelope detector: - ! R1 C1 !
fm fc
Let R1C1 = 100 / fc ~ 0.219 msec
Choose C1 = 1 Pf, then R1 = 219:, Select R1 = 220 : (std. value)

R1 = 220 :, C1 = 1 Pf, R = 22K:, C = 0.01Pf

Check point: -
x Ensure that AFT is not loading the ASG.
x Check the transistor (See self checking)
x Adjust the carrier frequency exactly equal to fIFT.
x Observe the clamped signal at the base of the transistor.

13



Collector AM Demodulation using Envelop Detector

AIM:- Conduct an experiment to generate an AM signal using collector
modulation for an fC = _______KHz and fm = _______Hz. Plot the variations of
modulating signal amplitude v/s modulation index.

PROCEDURE: -

1. Connections are made as shown in circuit diagram.
2. By switching off the modulating signal, find the tuned frequency of IFT by
varying the carrier signal frequency.
3. Keeping the carrier frequency the tuned frequency of IFT switch on the
modulating signal and observe the AM signal at the output of IFT.
4. Find the modulation index ‘m’, the amplitude of the carrier signal Vc and
the amplitude of the message signal Vm from the AM output by
measuring Vmax and Vmin.
Measure Vmax Vmin
(i) from the AM o/p
(ii) from the Trapezoidal w/f
5. By varying amplitude of the modulating signal note down ‘m’, ‘Vm’, ‘Vc’
from Vmax and Vmin. Make sure that Vc is remaining constant.
6. Plot graph of Vm v/s % m.

7. Connect the envelope detector ckt to the IFT o/p and observe the
demodulated signal.

Note: To obtain the trapezoidal wave from, feed the modulating signal to
Channel ‘A’ and the modulated signal to channel ‘B’ of CRO and time / Div knob
in X via A position.

14


Tabulation:-
Modulation
Tuned frequency of IFT, fIFT = ____________KHz

Vmax - Vmin Vmax - Vmin Vmax Vmin
Sl.No Vmax (V) Vmin (V) m= Vm = Vc =
Vmax Vmin 2 2

Demodulation
Sl.No Vo (V) fo (Hz)

(Vmax Vmin) (Vmax Vmin) (Vmax Vmin)
m , Vm , Vc
(Vmax Vmin) 2 2

m
L1 L2


WAVE FORMS: -

(a) Carrier wave, (b) Sinusoidal wave, (c) Amplitude modulated signal.

Conclusion:-

Staff-in-charge:-

16


CIRCUIT DIAGRAM: -
Balanced Modulator (Using Diodes)

D1, D2, D3, D4 – OA79

Waveforms-

17



Balanced Modulator (Using Diodes)

Aim:- Rig up a balanced modulator (Ring modulator) circuit. Test its operation
and record the waveforms.

Procedure: -


2. Apply the modulating signal (Sine wave) with frequency fm and the
carrier signal (square wave) with frequency fC (fC = 10 f m).

3. Observe the phase reversal of 1800 at each Zero crossing of modulating
signal in the output DSBSC signal.

Tabulation:-
Sl.No. VC Volts fC Hz Vm Volts fm Hz

Conclusion:-

Staff-in-charge:-

18


CIRCUIT DIAGRAM: -
Class-C Tuned Amplifier

2
VO PDC VDC u IC PAC
f Hz VO volts VDC volts IC mA RL ohms PAC mW
8R L mW PDC

Design:-
Specification:
Frequency f = 150 KHz, t = 6.66 usec
R1C1 t, i.e, R1C1 = 100 t
Choose C1 = 0.01Pf, the R1 = 66.6 K:.Select R1 = 68 K: (std value)
Tank ckt: f 150KHz
S
If C = 0.001Pf, then L = 1.125 mH a1mH. Then Factual = 159 KHz.

R1 = 68K:, C1 = 0.01Pf, C= 0.001Pf, L = 1mH
Check points: -
x Check the transistor (See self checking)
x Adjust i/p frequency exactly equal to tuned frequency.
x Observe the clamped signal at the base of the transistor.

19



Class-C Tuned Amplifier

Aim:- Design and test a Class-C Tuned amplifier to work at fO = ______KHz
(Center frequency). Find its maximum efficiency at optimum load.

Procedure: -
1. Connections are made as shown in circuit diagram.
2. Adjust the input frequency of the signal to get maximum output at the
load.
3. For the applied DC voltage adjust the amplitude of input sine wave signal
so that the output signal peak to peak amplitude is twice of the DC voltage
(without any distortion).
4. Vary the load resistance RL around 10 KW.
5. Note Vo, VDC, IC and RL to find PAC and PDC hence the efficiency.
(Note: While measuring Vo, short the Ammeter connection)
Ideal graph:-

Conclusion :-

Staff-in-charge:-

20


Circuit Diagram: -
Frequency Modulation Circuit: -

Frequency Demodulation Circuit: -

BT 2
Sl.No fc Hz fm Hz Vm volts fcmax Hz fcmin Hz G1 Hz G2 Hz G Hz
fm

1 f cmax - f c , 2 f c - f cmin , Max of 1 or 2

21



Frequency Modulation Demodulation

Aim:- Design and conduct a suitable experiment to generate an FM wave using
IC8038. Find the modulation index E and the bandwidth of operation BT. Display
the various waveforms.

Procedure: -

2. By switching off the modulating signal m(t), note down the carrier sine
wave of frequency of fC at pin 2 of IC 8038.

3. Apply the modulating signal m(t) with suitable amplitude to get
undistorted FM signal.

4. Note down maximum and minimum frequency of the carrier in FM signal
(i.e., fC max and fCmin)

5. Find the frequency deviation, modulation index operation band width.

6. Test the demodulator circuit by giving FM output from IC8038 as an input
for the demodulator circuit.

22


Design-1: -
1. FM modulator circuit.
Let carrier frequency fC = 3 KHz, fC = 0.3/R Ct.
Choose R = 10KŸ = Ra = Rb, then Ct = 0.01Pf.
Take RL = 10KŸ, CC = 0.01Pf.
2. Demodulator using PLL.
Let fO = fC = 3 KHz, fO = 1.2/4R1C1.
Choose C1 = 0.001Pf, then R1 = 100KŸ.
Filter design: Let fm = 1 KHz = 1/2SRC
Choose C = 0.1Pf, then R = 1.59 KŸ a 1.5 KŸ
Design - 2: -
1. FM modulator circuit.
Let carrier frequency fC = 5 KHz, fC = 0.3/R Ct.
Choose R = 10KŸ = Ra = Rb, then Ct = 0.001Pf.
Take RL = 10KŸ, CC = 0.01Pf.
2. Demodulator using PLL.
Let fO = fC = 3 KHz, fO = 1.2/4R1C1.
Choose C1 = 0.001Pf, then R1 = 100KŸ.
Filter design: Let fm = 1 KHz = 1/2SRC
Choose C = 0.1Pf, then R = 1.59 KŸ a 1.5 KŸ
Wave Form: -

23


Design:-
Specification:
0.3
Carrier frequency fC = 3 kHz, f c
RC t
Choose R= 10 KŸ, Ra = Rb, then Ct = 0.01Pf (use DCB)
Ra = Rb = 10 KŸ, RL = 10 KŸ, Ct = 0.01Pf (use DCB). R = 82 KŸ, CC = 0.01Pf.
Note: -
Usually the carrier frequency of the FM signal is in the range of 100s of
KHz, but is chosen in terms of 1s of KHz to enable proper measurement of
frequency deviating G.

Check Points: -
Ensure that a square wave and a triangular wave at pin 9 and 3 of IC 8038
respective.

Conclusion :-

Staff-in-charge:-

24


Circuit Diagram: -
Radio Receiver: -

R = 10K:, C = 0.1Pf, RL = 100:

Selectivity: -
fm = _____Hz, %m = ______
Sl.No fC Hz Vo volts

Fidility: -
fm = _____Hz, %m = ______
Sl.No fC Hz Vo volts

Sensitivity: -
fm = _____Hz, %m = ______
fC Hz Vi volts Vo volts

25



Radio Receiver Characteristics
Aim:- Plot the sensitivity/selectivity/fidelity graphs of a given AM Broadcast receiver in
MW band by conducting suitable experiment.
Procedure: -
2. Ensure the Radio Receiver is in MW band.
3. Adjust the modulation index of AM signal at 30 % fm = 400 Hz.
4. Let the receiver be tuned to 800 KHz. (can be anywhere between 540 KHz 1450
KHz).
5. Keeping the carrier frequency of the AM signal at 800 KHz, observe the
demodulated signal and note down its amplitude.
Selectivity: -
1. Repeat the step 5 by changing the carrier frequency at 805, 810, 815 and 795,
790, 785 KHz.
2. Plot a graph of carrier frequency of AM signal Vs the amplitude of the output
signal (Vo Vs fc).
Sensitivity: -
1. Repeat the steps 1 to 5.
2. Vary the amplitude of the AM signal to get a standard value of output voltage
(Volts). All the other parameters are kept constant (i.e., fc, fm, m). Note the
change in the amplitude of the output signal.
3. Repeat step 9 for different values of fc.
4. Plot a graph of amplitude of input signal v/s carrier frequency of AM signal (Vi
v/s fc).
Fidelity: -
1. Repeat the steps 1 to 5.
2. Vary the frequency of the modulating signal keeping all other parameters
constant (i.e., fc, VAM, m). Note the change in the amplitude of the output signal.

3. Plot a graph of amplitude of output signal Vs frequency of the modulating signal
(Vo Vs fm).
Conclusion:-

Staff-in-charge:-

26


Circuit Diagram: -
Pre-emphasis De-emphasis

TABULATION: - Pre-Emphasis N/W
Vo Normalized gain Normalized Gain
f Hz Vo volts Gain
Vi Gain/Go In db

De-Emphasis N/W
Vo Normalized gain Normalized Gain
f Hz Vo volts Gain
Vi Gain/Go In db

27



Pre-emphasis and De-emphasis Networks

Aim:- Design and conduct an experiment to test a pre-emphasis and de-emphasis
circuit for 75Ps between 2.1KHz to 15KHz and record the results..

Procedure: -

2. Apply a sine wave of 5Vpp amplitude, vary the frequency and note down
the gain of the circuit.

3. Plot a graph of normalized gain Vs frequency.

Design: -
1. Pre-emphasis circuit.
Given f1 = 2.1 KHz, f2 = 15KHz.
f1 = 1/2SrC, f2 = 1/2SRC
Choose C = 0.1Pf then r = 820Ÿ and R = 100Ÿ.
Also r/R = Rf/R1, then R1 = 2.2KŸ and Rf = 15KŸ.
2. De-emphasis circuit.
fC = 1/2SRdCd.
Choose Cd = 0.1Pf and fC = f1 = 2.1KHz
Then Rd = 820Ÿ.
Conclusion :-

Staff-in-charge:-

28


Circuit Diagram: - AM Modulator using MC1496

AM Demodulator using MC1496

Tabulation:-

Vmax - Vmin Vmax - Vmin Vmax Vmin
Sl.No Vmax (V) Vmin (V) m= Vm = Vc =
Vmax Vmin 2 2

29



AM – IC Circuit (Modulation Demodulation)

Aim:- Using IC1496, rig up an AM modulation and Demodulation circuit. Test its
operation and record the waveforms.

Procedure: -
a) AM Modulation
2. Give the modulating signal of 2VPP (1KHz).
3. Give the carrier signal of 1VPP (600KHz).
4. Note down the AM modulated signal at pin 6 and also at the emitter of the
buffer (emitter follower).
5. Change the amplitude levels of the modulating signal, keeping fC and fm as
constant and find the depth of modulation.
b) AM Demodulation
1. Give the AM wave to pin1 of MC1496.
2. Also give the AM wave from the buffer o/p.
3. Note the demodulated signal at pin 12 of MC1496.

Design: -
Select Vdc = +12V, IC = 3mA. RL = + Vdc/ IC = 4KŸa3.9KŸ.
Vbe = 700mV, I = 160mA, Voltage at pin 5 = 1.7V.
Vbias = (-8+1.7) = -6.3V
RS = Vbias/I = 6.3/160mA = 7KŸa6.8KŸ
Conclusion :-

Staff-in-charge:-

30


Circuit Diagram: -
Pulse amplitude modulation and demodulation

Design: -
Specifications: -
IC = 1ma, hFE = 100, VCEsat = 0.3 V, VBEsat = 0.7v (assume), fm = 100hz.
1. Biasing: - Vm(t) = IC *RC + VCEsat ----- 1
Let Vm(t) = 2.5 v w.f peak + 3v DC shift = 5.5 V peak signal
Then Rc = 5.2 kŸ, select Rc = 4.7 k Ÿ(std. Value).
Vc (t) = IB*RB + VBEsat --------2
Let Vc(t) = 2 Vpp ( 1 V peak ) , Since IB = Ic / hFE = 10uA
Then RB = 30 k Ÿ
Select RB = 22 k Ÿ (Std. Value).
2. Filter: - Cut off frequency of the filter fo fm
Choose fo = 500 Hz = 1 / 2 S RC
Choose C = 0.1 P f, then R = 3.3 k Ÿ
Rc = 4.7 K Ÿ RB = 22k Ÿ, R = 3.3k Ÿ, C = 0.1Pf
Check Points: -
1. Ensure that square wave signal at the base of the transistor should have
amplitude VJ.
2. Ensure that m (t) is having sufficient dc shift.
Tabulation: -
VC(pp) volts fC (Hz) Vm(pp) volts fm (Hz) Reconstructed output
VO volts fO (Hz)

31



Pulse Amplitude Modulation Demodulation
Aim:- Conduct an experiment to generate PAM signal and also design a circuit to
demodulate the obtained PAM signal and verify sampling theorem. Plot the
relevant waveforms.

Procedure: -
2. Apply the square wave carrier signal of 2V peak to peak amplitude
with frequency fc = 5 kHz.
3. Apply sine wave modulating signal with frequency fm = 100 Hz
with 5 Vpp amplitude and 3 V DC shift (use function generator).
4. Observe the PAM output.
5. Observe the demodulated signal at the output of the low pass filter.
6. Repeat the steps 2 to 5 for fc = 2 fm fc 2 fm.
Waveforms:

Conclusion :-

Staff-in-charge:-

32


Circuit Diagram: -
Pulse Width modulation and demodulation

Pulse Width Demodulation

33



Pulse Width Modulation Demodulation
relevant waveforms.

Procedure: -
2. Keeping the modulating signal with minimum amplitude, observe the
output of astable multivibrator with 50 % duty cycle at frequency fc.
3. Apply the modulating signal with frequency fm and the amplitude less
than the critical amplitude observe the PWM signal.
4. Verify the variation of width of the pulses with respect to clamped
modulating signal (at point A).

To find the critical amplitude: -
As the amplitude of the modulating signal is increase the width of the
pulses during the negative half of the modulating signal keeps on reducing and
that at the positive half of the modulating signal is increased the width of the
pulses during the negative half of the modulating signal keeps on reducing and
that at the positive half of the modulating signal keeps on increasing.

34


Design: -

Specifications: -
Frequency fc = 1 KHz, duty cycle: 50 %
T = 1 ms, Ton = Tb= 0.5 ms
I) Astable multivibrator: - Where RcH = charging resistance,
RDCH = Discharging Resistance,
Rf = Diode forward resistance
Ct = timing capacitor
TON = 0.69 (RCH + Rf ) Ct
Toff = 0.69 (RDCH + Rf) Ct
Ton = Toff = 0.5 ms
Choose Ct = 0.1 Pf, then (RCH + Rf) = (RDCH + Rf) = 7.246 k:
Assuming Rf of diode = 100:,
Then RCH = RDCH = 7.146 k: (use 6.8 k: + 330: std value)
II) Clamping ckt
Negative peak of the modulating signal clamped to zero
Rc 1 /fm, fm = 100Hz
RC = 100 /fm, choose C= 10 f, then R = 100K.
RCH = RDCH = (6.8K + 330 ), R = 100K , Ct = 0.1 f, C = 10 f.

Check points: -
With modulating signal zero, the voltage at pin 5 of 555 timer should be 2/3 VCC.
Ensure that modulating signal is clamped.

Tabulation: -
Unmodulated carrier PWM Output Demodulator
Dynamic Modulating
Ton Toff Max.width Min.width
fc Hz range frequency VO(V) fO(Hz)
ms ms ms ms
volts fm Hz

35


Waveforms:-

Conclusion :-

Staff-in-charge:-

36


Circuit Diagram: -
Pulse Position modulation and demodulation

Design:
m(t) = 1KHz, T = 1ms
T = RC, Let C = 0.01uf
Then R = 1Ÿ

Pulse Position Demodulator
Design: -
Specifications: -
1. Monostable Multivibrator: -
PW = 1.1 Rch Ct
Choose Ct = 0.01 Pf, then Rch = 18.18 k : (std. Value)
2. Differentiator : -
Rs * Cs 1 / fc
Choose Rs * Cs = 0.01 ms, Choose Cs = 0.001Pf, then Rs = 10k :
Rch = 18 k:. Ct = 0.01Pf, Rs = 10 k:, Cs = 0.001Pf
CHECK POINTS: -
x With modulating signal zero, the voltage at pin 5 of 555 timer should be 2 /3 Vcc.
x Ensure that wave form at pin 2 of 555 timer should have a trailing edge going below 1
/3 Vcc.

37



Pulse Position Modulation Demodulation
relevant waveforms.

Procedure: -
2. Check the working of 555 timer as a monostable multivibrator by giving an
unmodulated PWM signal. Verify the pulse width of output signal for the
designed value.
3. By applying the PWM signal note the change in the position of the pulses i.e.
PPM signal.
4. Critical amplitude of the modulating signal is that value of m(t) at which the
pulse in PPM just disappears.

Waveforms:-

Conclusion:-

Staff-in-charge:-

38

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