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Communication Lab Manual SSIT, Tumkur

COMMUNICATION LAB MANUALFOR

V SEMESTER B.E (E & C)(For private circulation only)

VISHVESHWARAIAH TECHNOLOGICAL UNIVERSITY

NAM E: ___________________________

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 1volt (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)




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 f C = ______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 (f C) of the filter.

To find the Roll-off factor :-

1. For LPF :- Keeping the input signal amplitude constant, adjust the input

frequency at 10f C. Note down the output signal amplitude. The difference

in the gain of the filter at f C and 10f C gives the Roll-of factor.

2. For HPF :- Keeping the input signal amplitude constant, adjust the input

frequency at 0.1f C, note down the output signal amplitude. The difference

in the gain of the filter at f C and 0.1f C gives the Roll-of factor.

Conclusion:

2




Tabulation:

High Pass Filter Vi p-p = Volts (Constant)

I/P frequency in

Hz

O/P Voltage

VO P-P (volts)

Gain magnitude

(Vo/Vi)

Gain magnitude in DB

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 inHz

O/P VoltageVO P-P (volts)

Gain magnitude(Vo/Vi)

Gain magnitude in DB20log(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 = 10K , Rf = 5.86 K , R1 = 1.989 K , R2 = 3.3 K ,

C1 = 0.01 f, C2 = 0.01 f, Op-amp = A741

2. BSF : - R = 10K , Rf = 5.86 K , Ra = 3.3 K , Rb = 1.989 K ,

C1 = 0.01 f, C2 = 0.01 f, Op-amp = A741

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 f C1 = ______Hz and f C2 = ______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 (f L) and the

upper cut-off frequency (f H) of the filter.

To find the Roll-off factor :-

1. For LPF :- Keeping the input signal amplitude constant, adjust the input

frequency at 10f C, note down the output signal amplitude. The difference

in the gain of the filter at f C and 10f C gives the Roll-of factor.

2. For HPF :- Keeping the input signal amplitude constant, adjust the input

frequency at 0.1f C, note down the output signal amplitude. The difference

in the gain of the filter at f C and 0.1f C gives the Roll-of factor.

6




Design:

Specifications:

Pass band gain A V = 1.586, cut -off frequency f H = 5 KHz, f L=8 KHz, BW= 3 KHz

1. Amplifier: Voltage gain A V = 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 f H= 1/2 R2C2= 5 KHz

Choose C2= 0.01f, then R2 = 3.183 k (Select R2 = 3.3 k)

Cut - off frequency f L = 1/2 R1 C1 = 8 k Hz

Choose C1= 0.01f, then R1= 1.989 k (Select R1 = (1.5 k + 470))

Tabulation:

Band Pass Filter Vi p-p = Volts (Constant)

FrequencyHz

O/P Voltage VO PP (volts) Gain (Vo/Vi)Gain in DB

20 log (Vo/Vi) Vomax =

f L = G1

0.1f L = 0.707 Vomax = G210f H=

f H= 0.707 Vomax = G2’


Frequency Response for Band Pass Filter

7




Tabulation:

Band Elimination Filter Vi p-p = Volts (Constant)

FrequencyHz

O/P Voltage VO PP (volts) Gain (Vo/Vi)Gain in DB

20 log (Vo/Vi)

Vomax =

f L = G1

0.1f L = 0.707 Vomax = G210f H=

f H= 0.707 Vomax = G2’


Frequency Response for Band Elimination Filter

Conclusion:

Staff-in-charge:

8




CIRCUIT DIAGRAM: -

T-Type Attenuator -Type Attenuator

Design:-

Specification: Vi = 5v, Vo = 2.5v, f = 1KHz

T- Type

1)N

2NR

1)N

1)NR

2O

2O

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

- Type

1)N1)NR

2N1)NR O

2

2O

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

-Type

-Type

9





Attenuators – T, , Lattice and O-Pad Types

AIM: - Design the attenuation circuits using T, , 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.2. Connections are made as shown in the circuit diagram.

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:-

For N=2 and RS = RO = 600, then

8001)N

N2R

2001)N

1)NR

2

O

2. -Type attenuators:-


1.8K1)N

1)NR

4502N

1)N

R

O

2

O

10




Lattice-Type Attenuator O-Pad Type Attenuator

Design:-

Specification: Vi = 5v, Vo = 2.5v, f = 1KHz

Lattice- Type

1)N

2NR 1)N

1)NR 2O2O1

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

1)N

1)NR

2N

1)NR O

2

2O

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

attice-Type

-Pad Type

11




Design:-

3. Lattice-Type attenuators:-

For N=2 and RS = RO = 600, then800

1)N

N2R

200

1)N

1)NR

2

O

4. O-Pad Type attenuators:-


1.8K1)N

1)NR

4502N

1)NR

O

2

O

Conclusion:-

Staff-in-charge:-

12




CIRCUIT DIAGRAM: -

Collector AM and Demodulation using Envelop Detector

Design:-

Specifications: -

Tuned frequency = f IFT, Assume f IFT = 455 KHz, t = 2.19 sec

RC >> t, i.e., RC = 100 t = 0.219 msecChoose C = 0.01 f, then R = 21.97 K, Select R = 22K (Std. value)

Envelope detector: -fc

1 CR

fm

111

Let R1C1 = 100 / fc ~ 0.219 msec

Choose C1 = 1 f, then R1 = 219, Select R1 = 220 (std. value)

R1 = 220 , C1 = 1 f, R = 22K , C = 0.01f

Check point: -

Ensure that AFT is not loading the ASG. Check the transistor (See self checking) Adjust the carrier frequency exactly equal to f IFT. 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 f C = _______KHz and f m = _______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 byvarying 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, f IFT = ____________KHz

Sl.No Vmax (V) Vmin (V) m =minmax

minmax

VV

V-V

Vm =

2

V-V minmax Vc =2

VV minmax

Demodulation

Sl.No Vo (V) fo (Hz)

2

Vmin)(Vmax Vc,

2

Vmin)(Vmax Vm,

Vmin)(Vmax

Vmin)(Vmaxm

21

21

LL

LLm

15




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 f C (f C = 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

f Hz VO volts VDC volts IC mA RL ohms mW8R

VP

L

2O

AC mW

I VP CDCDC DC

AC

P

P

Design:-

Specification:

Frequency f = 150 KHz, t = 6.66 usec

R1C1 >> t, i.e, R1C1 = 100 t

Choose C1 = 0.01f, the R1 = 66.6 K .Select R1 = 68 K (std value)

Tank ckt:

f 150KHz

If C = 0.001f, then L = 1.125 mH 1mH. Then Factual = 159 KHz.

R1 = 68K, C1 = 0.01 f, C= 0.001 f, L = 1mH

Check points: - Check the transistor (See self checking) Adjust i/p frequency exactly equal to tuned frequency. 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 f O = ______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 theload.

3. For the applied DC voltage adjust the amplitude of input sine wave signalso 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 P AC 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: -

Sl.No fc Hz fm Hz Vm volts fcmax Hz fcmin Hz 1 Hz 2 Hz Hzmf

T 2B

21cminc2ccmax1 orof Max ,f -f ,f -f

21





Frequency Modulation & Demodulation

Aim:- Design and conduct a suitable experiment to generate an FM wave using

IC8038. Find the modulation index 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 f C 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., f C max and f Cmin)

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 f C = 3 KHz, f C = 0.3/R Ct.

Choose R = 10K = Ra = Rb, then Ct = 0.01f.

Take RL = 10K, CC = 0.01f.2. Demodulator using PLL.

Let f O = f C = 3 KHz, f O = 1.2/4R1C1.

Choose C1 = 0.001f, then R1 = 100K.

Filter design: Let f m = 1 KHz = 1/2RC

Choose C = 0.1f, then R = 1.59 K 1.5 K

Design - 2: -

1. FM modulator circuit.

Let carrier frequency f C = 5 KHz, f C = 0.3/R Ct.

Choose R = 10K = Ra = Rb, then Ct = 0.001f.

Take RL = 10K, CC = 0.01f.

2. Demodulator using PLL.

Let f O = f C = 3 KHz, f O = 1.2/4R1C1.

Choose C1 = 0.001f, then R1 = 100K.

Filter design: Let f m = 1 KHz = 1/2RC

Choose C = 0.1f, then R = 1.59 K 1.5 K

Wave Form: -

23




Design:-

Specification:

Carrier frequency f C = 3 kHz,t

cRC

0.3 f

Choose R= 10 K, Ra = Rb, then Ct = 0.01f (use DCB)

Ra = Rb = 10 K, RL = 10 K, Ct = 0.01f (use DCB). R = 82 K, CC = 0.01f.

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 .

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 = 0.1f, RL = 100

Selectivity: -

fm = _____Hz, %m = ______

Sl.No f C Hz Vo volts

Fidility: -

fm = _____Hz, %m = ______

Sl.No f C Hz Vo volts

Sensitivity: -

fm = _____Hz, %m = ______

f C 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 1450KHz).

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 outputsignal (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 thechange 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 (Viv/s fc).

Fidelity: -

1. Repeat the steps 1 to 5.

2. Vary the frequency of the modulating signal keeping all other parametersconstant (i.e., fc, V AM, 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

f Hz Vo volts Gain Vi

Vo Normalized gainGain/Go

Normalized GainIn db

De-Emphasis N/W

f Hz Vo volts Gain Vi

Vo Normalized gainGain/Go

Normalized GainIn db

27





Pre-emphasis and De-emphasis Networks

Aim:- Design and conduct an experiment to test a pre-emphasis and de-emphasis

circuit for 75s 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 f 1 = 2.1 KHz, f 2 = 15KHz.

f 1 = 1/2rC, f 2 = 1/2RC

Choose C = 0.1f then r = 820 and R = 100.

Also r/R = Rf /R1, then R1 = 2.2K and Rf = 15K.

2. De-emphasis circuit.

f C = 1/2RdCd.

Choose Cd = 0.1f and f C = f 1 = 2.1KHz

Then Rd = 820.

Conclusion :-

Staff-in-charge:-

28




Circuit Diagram: - AM Modulator using MC1496

AM Demodulator using MC1496

Sl.No Vmax (V) Vmin (V) m =minmax

minmax

VV

V-V

Vm =

2

V-V minmax Vc =

2

VV minmax

Tabulation:-

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 thebuffer (emitter follower).

5. Change the amplitude levels of the modulating signal, keeping f C and f m asconstant 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 = 4K3.9K.

Vbe = 700mV, I = 160mA, Voltage at pin 5 = 1.7V.

Vbias = (-8+1.7) = -6.3V

RS = Vbias/I = 6.3/160mA = 7K6.8K

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 k, select Rc = 4.7 k (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

Select RB = 22 k (Std. Value).

2. Filter: - Cut off frequency of the filter fo >> fmChoose fo = 500 Hz = 1 / 2 RC

Choose C = 0.1 f, then R = 3.3 k

Rc = 4.7 K RB = 22k , R = 3.3k , C = 0.1f

Check Points: -

1. Ensure that square wave signal at the base of the transistor should haveamplitude > V.

2. Ensure that m (t) is having sufficient dc shift.

Tabulation: -

Reconstructed output VC(pp) volts f C (Hz) Vm(pp) volts f m (Hz)

VO volts f O (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 therelevant waveforms.

Procedure: -


2. Apply the square wave carrier signal of 2V peak to peak amplitudewith frequency f c = 5 kHz.

3. Apply sine wave modulating signal with frequency fm = 100 Hzwith 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 f c = 2 f m & fc < 2 f m.

Waveforms:

Conclusion :-

Staff-in-charge:-

32




Circuit Diagram: -

Pulse Width modulation and demodulation

Pulse Width Demodulation

33





Pulse Width Modulation & Demodulation

Aim:- Conduct an experiment to generate PAM signal and also design a circuit todemodulate the obtained PAM signal and verify sampling theorem. Plot therelevant waveforms.

Procedure: -


2. Keeping the modulating signal with minimum amplitude, observe theoutput of astable multivibrator with 50 % duty cycle at frequency f c.

3. Apply the modulating signal with frequency f m and the amplitude less

than the critical amplitude observe the PWM signal.

4. Verify the variation of width of the pulses with respect to clampedmodulating signal (at point A).

To find the critical amplitude: -

As the amplitude of the modulating signal is increase the width of thepulses during the negative half of the modulating signal keeps on reducing andthat 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 andthat at the positive half of the modulating signal keeps on increasing.

34




Design: -

Specifications: -

Frequency f c = 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 f, 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

Ton

ms

Toff

ms fc Hz

Max.width

ms

Min.width

ms

Dynamic

rangevolts

Modulating

frequencyfm Hz VO(V) f O(Hz)

35




Waveforms:-

Conclusion :-

Staff-in-charge:-

36




Circuit Diagram: -

Pulse Position modulation and demodulation

Pulse Position Demodulator

Design:

m(t) = 1KHz, T = 1ms

T = RC, Let C = 0.01uf

Then R = 1

Design: -

Specifications: -

1. Monostable Multivibrator: -

PW = 1.1 Rch Ct

Choose Ct = 0.01 f, then Rch = 18.18 k (std. Value)

2. Differentiator : -

Rs * Cs <<1 / f c

Choose Rs * Cs = 0.01 ms, Choose Cs = 0.001f, then Rs = 10k

Rch = 18 k. Ct = 0.01f, Rs = 10 k, Cs = 0.001f

CHECK POINTS: -

With modulating signal zero, the voltage at pin 5 of 555 timer should be 2 /3 Vcc.

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

Aim:- Conduct an experiment to generate PAM signal and also design a circuit to

demodulate the obtained PAM signal and verify sampling theorem. Plot therelevant waveforms.

Procedure: -


2. Check the working of 555 timer as a monostable multivibrator by giving anunmodulated PWM signal. Verify the pulse width of output signal for thedesigned 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 thepulse in PPM just disappears.

Waveforms:-

Conclusion:-

Staff-in-charge:-

38




Circuit Diagram: -

Transistor Mixer

Design: -

Specifications: -

VCC = 6V, VCE = 5V, assume IC = 1ma, hFE = 100, VBEsat = 0.6v.

VCC = VCE + IE * RE

Since IE ~ IC, then RE = 1 K

Vb = IB * RB + VBEsat + IE * RE ------ (1)

Where Vb = VCC * R2 /R1 + R2, RB = R1 + R2, RB = R1 * R2 /R1 + R2

From eq (1) it can be found that R1 ~ 2.5 R2, Choose R2 = 18 K , then R1 45 K R1 = 47 K , R2 = 18 K , RE = 1K , CC = 0.1f, CE = 10f

Check points: -

During the mixer operation under on circumstance the frequency of local oscillator orfrequency of the carrier should be kept at tuned frequency of IFT. (i.e., f LO = f IFT = f S)

Tabulation: -

f IFT = __________, Vmax = ___________, Vmin = _________, %m = __________

Operation fS Hz fLO Hz fO Hz fLO + fS Hz

Up Conversion

Down Conversion

39





Transistor Mixer

Aim:- Conduct an experiment to generate PAM signal and also design a circuit todemodulate the obtained PAM signal and verify sampling theorem. Plot therelevant waveforms.

Procedure: -


2. Keeping the amplitude of the local oscillator in minimum position, find thetuned frequency of IFT (f IFT) by varying the carrier frequency (fs) of the

input AM signal.

Down conversion: -

3. Adjust the carrier frequency of the AM signal more than the tunedfrequency of IFT, now adjusting the local oscillator frequency (f LO = fs +

f IFT)

(Note: local oscillator amplitude 0). Observe the output AM signal withcarrier frequency at f IFT.

4. Repeat the step 3 for different carrier frequencies.

Up conversion: -

5. Adjust the carrier frequency of the AM signal less than the frequency ofIFT, now adjusting the local oscillator frequency (f LO = f S + f IFT) (Note :

local oscillator amplitude 0). Observe the output AM signal with carrierfrequency at f IFT.

6. Repeat the step 5 for different carrier frequencies.

Formulas: Zo = Vo rms/Io, gc = Vo rms/ (Vi rms.Zo)

Conclusion:-

Staff-in-charge:-

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