Balanced Modulatopr

Communication Lab Manual

SSIT, Tumkur

COMMUNICATION LAB MANUALFOR

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

VISHVESHWARAIAH TECHNOLOGICAL UNIVERSITY

NAME: ___________________________

DEPARTMENT OF ELECTRONICS & COMMUNICATION


SSIT, Tumkur

SRI SIDDHARTHA INSTITUTE OF TECHNOLOGYMARLUR, TUMKUR-572105

CONTENTS1. II-Order Low Pass and High Pass Active Filters 2. II Order Band Pass and Band Elimination Filte rs 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


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TESTING OF EQUIPMENTS BEFORE STARTING THE CONDUCTION1. OP AMPApply 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)


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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 1. Amplifier: AV = 1 + R

Rf = 2, then Rf = R, choose Rf = R = 10K

2. Filter Circuit : Cut off frequency fC = Choose C1 = 0.01Pf then R1 = 3.183 K a 3.3 K

1 2S R C = 5KHz1 1

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

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Experiment No:

DATE: __/__/____

II Order Low Pass and High Pass Active FiltersAIM: - Design a second order Butterworth active low pass / high pass filter for agiven 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 voltageVO max.

4. Keeping the input signal amplitude constant, vary the frequency until the output voltage reduces to 0.707 Vo max, the corresponding frequency isthe cut-off frequency (fC) of the filter.

To find the Roll-off factor :1. For LPF :- Keeping the input signal amplitude constant, adjust the inputfrequency 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 inputfrequency 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:

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

I/P frequency in Hz

O/P VoltageVO 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

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Tabulation:Low Pass Filter Vi p-p = Volts (Constant)

I/P frequency in Hz

O/P VoltageVO P-P (volts)

Gain magnitude (Vo/Vi)

Gain magnitude in DB 20log(Vo/Vi)

Roll off = - (G1 - G2) db/decade = Frequency Response for Low Pass Filter

Staff-in-charge:

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

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Experiment No:

DATE: __/__/____

II Order Band Pass and Band Elimination Active FiltersAIM: - Design a second order band pass and band stop active filter for a givenfrequencies fC1 = ______Hz and fC2 = ______Hz. Conduct an experiment to draw

frequency response and verify the Roll off (Band Width = 3 to 5 KHz).

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 voltageVO 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, thecorresponding 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 inputfrequency 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 inputfrequency 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.

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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)

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

20 log (Vo/Vi)

Vomax =fL =

G1

0.1fL = 0.707 Vomax = 10fH= fH= 0.707 Vomax =

G2

G2

Roll off = - (G1 - G2) db/decade = Frequency Response for Band Pass Filter

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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 =fL =

G1

0.1fL = 0.707 Vomax = 10fH= fH= 0.707 Vomax =

G2

G2

Roll off = - (G1 - G2) db/decade = Frequency Response for Band Elimination Filter

Conclusion:

Staff-in-charge:

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CIRCUIT DIAGRAM: T-Type Attenuator S-Type Attenuator

Design:Specification: Vi = 5v, Vo = 2.5v, f = 1KHz T- Type R1

R (N _ 1) O (N _ 1)

R2

R 2N O (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- TypeR1

R (N 2 _ 1) O 2N

R2

R (N _ 1) O (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

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Experiment No:

DATE: __/__/____

Attenuators T, S, Lattice and O-Pad TypesAIM: - 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.

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:RFor N=2 and RS = RO = 600 , then1

R

R2

(N - 1) 200 O (N _ 1) N 2R 800 O (N 2 _ 1)

2. S-Type attenuators:RFor N=2 and RS = RO = 600 , then1

R

R2

(N 2 -1) O 2N (N _ 1) R O (N _ 1)

450 1.8K

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Lattice-Type Attenuator

O-Pad Type Attenuator

Design:Specification: Vi = 5v, Vo = 2.5v, f = 1KHz Lattice- Type R1

R (N _ 1) O (N _ 1)

R2

R 2N O (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 TypeR1

R (N 2 _ 1) O 2N

R2

R (N _ 1) O (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

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Design:3. Lattice-Type attenuators:RFor N=2 and RS = RO = 600 , then1

R

R2

(N - 1) 200 O (N _ 1) N 2R 800 O (N 2 _ 1)

4. O-Pad Type attenuators:RFor N=2 and RS = RO = 600 , then1

R

R2

(N 2 -1) O 2N (N _ 1) R O (N _ 1)

450 1.8K

Conclusion:-

Staff-in-charge:-

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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 K , Select R = 22K (Std. value) Envelope detector: fc

1 ! R1 !C 1 fm

1

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.

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Experiment No:

DATE: __/__/____

Collector AM & Demodulation using Envelop DetectorAIM:- Conduct an experiment to generate an AM signal using collectormodulation 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 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. m, Vm, Vc

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

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Tabulation:ModulationTuned frequency of IFT, fIFT = ____________KHz Sl.No Vmax (V) Vmin (V) m =

V - V max V mi n V _m ax min

- V Vm = V mi n 2 max Vc = 2

V mi n

max _

V

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

m

(Vmax _ Vmin) , (Vmax _ Vmin)

Vm

(Vmax _ Vmin) , 2

Vc

(Vmax _ Vmin) 2

m_

L1 _ L _ _L 2 L __1 2

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WAVE FORMS: -

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

Conclusion:-

Staff-in-charge:-

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CIRCUIT DIAGRAM: Balanced Modulator (Using Diodes)

D1, D2, D3, D4 OA79

Waveforms-

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Experiment No:

DATE: __/__/____

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

Procedure: 1. Connections are made as shown in the circuit diagram. 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:-

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CIRCUIT DIAGRAM: Class-C Tuned Amplifier

f Hz VO volts VDC volts IC mA RL ohms mW

PAC

V2 O 8RL

PD Cu C

V I DC mW

P PACDC

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


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Experiment No:

DATE: __/__/____

Class-C Tuned AmplifierAim:- 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:-

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Circuit Diagram: Frequency Modulation Circuit: -

Frequency Demodulation Circuit: -

Sl.No fc Hz fm Hz Vm volts fc ma x Hz fc min Hz G1 Hz G2 Hz G Hz

f

T2m

B

1

fcmax - fc ,

2

fc - fcmin ,21

Max of

1

or

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Experiment No:

DATE: __/__/____

Frequency Modulation & DemodulationAim:- Design and conduct a suitable experiment to generate an FM wave usingIC8038. Find the modulation index E and the bandwidth of operation BT. Display

the various waveforms. Procedure: 1. Connections are made as shown in the circuit diagram. 2. By switching off the modulating signal m(t), note down the carrier sinewave 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.

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Design-1: 1. FM modulator circuit.Let carrier frequency fC = 3 KHz, fC = 0.3/R Ct. Choose R = 10K = Ra = Rb, then Ct = 0.01Pf. Take RL = 10K , CC = 0.01Pf.

2. Demodulator using PLL.Let fO = fC = 3 KHz, fO = 1.2/4R1C1. Choose C1 = 0.001Pf, then R1 = 100K . Filter design: Let fm = 1 KHz = 1/2SRC

Choose C = 0.1Pf, then R = 1.59 K a 1.5 K Design - 2: 1. FM modulator circuit.Let carrier frequency fC = 5 KHz, fC = 0.3/R Ct. Choose R = 10K = Ra = Rb, then Ct = 0.001Pf. Take RL = 10K , CC = 0.01Pf.

2. Demodulator using PLL.Let fO = fC = 3 KHz, fO = 1.2/4R1C1. Choose C1 = 0.001Pf, then R1 = 100K . Filter design: Let fm = 1 KHz = 1/2SRC

Choose C = 0.1Pf, then R = 1.59 K a 1.5 K Wave Form: -

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Design:Specification:Carrier frequency fC = 3 kHz,

f c RC

0.3t

Choose R= 10 K , Ra = Rb, then Ct = 0.01Pf (use DCB)Ra = Rb = 10 K , RL = 10 K , Ct = 0.01Pf (use DCB). R = 82 K , 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:-

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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

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Experiment No:

DATE: __/__/____

Radio Receiver CharacteristicsAim:- Plot the sensitivity/selectivity/fidelity graphs of a given AM Broadcast receiver in MW band by conducting suitable experiment. Procedure: 1. Connections are made as shown in the circuit diagram. 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:-

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Circuit Diagram: Pre-emphasis De-emphasis

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

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

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Experiment No:

DATE: __/__/____

Pre-emphasis and De-emphasis NetworksAim:- 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: 1. Connections are made as shown in the circuit diagram. 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 = 820 and R = 100 .Also r/R = Rf/R1, then R1 = 2.2K and Rf = 15K .

2. De-emphasis circuit.fC = 1/2SRdCd. Choose Cd = 0.1Pf and fC = f1 = 2.1KHz Then Rd = 820 .

Conclusion :-

Staff-in-charge:-

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Circuit Diagram: - AM Modulator using MC1496

AM Demodulator using MC1496

Tabulation:Sl.No Vm a x (V) Vm in (V) m =

V - V max V m i n _ Vm = 2 Vmax mi n

Vmin

- Vc m ax V = 2

V m in

max _

V

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Experiment No:

DATE: __/__/____

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 1. Connections are made as shown in the circuit diagram.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 = 4K a3.9K . Vbe = 700mV, I = 160mA, Voltage at pin 5 = 1.7V. Vbias = (-8+1.7) = -6.3V RS = Vbias/I = 6.3/160mA = 7K a6.8K

Conclusion :-

Staff-in-charge:-

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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 ----- 1Let 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. F ilter: - 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 kRc = 4.7 K RB = 22k , R = 3.3k , 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 outputVO volts fO (Hz)

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Experiment No:

DATE: __/__/____

Pulse Amplitude Modulation & DemodulationAim:- 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: 1. Connections are made as shown in the circuit diagram. 2. Apply the square wave carrier signal of 2V peak to peak amplitudewith 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:-

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Circuit Diagram: Pulse Width modulation and demodulation

Pulse Width Demodulation

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Experiment No:

DATE: __/__/____

Pulse Width Modulation & DemodulationAim:- 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: 1. Connections are made as shown in the circuit diagram. 2. Keeping the modulating signal with minimum amplitude, observe theoutput 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.

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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 capacitorTON = 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 Ton ms Toff ms fc Hz Max.width ms Min.width ms Dynamic range volts Modulating frequency fm Hz DemodulatorVO(V) fO(Hz)

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Waveforms:-

Conclusion :-

Staff-in-charge:-

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Circuit Diagram: Pulse Position modulation and demodulation

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

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

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Balanced Modulatopr