analog communication manual

A LAB MANUAL ON

ANALOG COMMUNICATION + LICSubject Code: 06ECL58(As per VTU Syllabus)

PREPARED BY

Staff members - Dept. of E&C

No.132, AECS Layout, I.T.P.L. Road, Kundalahalli, Bangalore- 560 037

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE

CONTENTSEXPT. NO. 1 2 3 4 5 6 NAMEOF THE

EXPERIMENT

Active low pass & high pass filters second order Active band pass & band reject filters second order Schmitt trigger design and test a Schmitt trigger circuit for the given values of UTP and LTP Frequency synthesis using PLL Design and test R-2R DAC using OP-AMP. Design and test the following circuits using IC 555 a) Astable multivibrator for given frequency and duty cycle b) Monostable multivibrator for given pulse width W.

7 8 9 10 11 12

Class-C single tuned amplifier Amplitude modulation using Transistor/FET (Generation and Detection) Pulse Amplitude modulation and Detection PWM and PPM Frequency modulation using 8038/2206 Precision Rectifiers- both Full Wave and Half Wave

DEPARTMENT OF E&C, CMRIT

2

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE CYCLE WISE EXPERIMENTS SEM: V EXAM MARKS: 50 BRANCH: ECE IA MARKS: 25 SUBJECT: ANALOG COMMUNICATION & LIC LAB SUB CODE: 06ECL58 CYCLE - 1 1) Active low pass & high pass filters second order 2) Active band pass & band reject filters second order 3) Schmitt trigger design and test a Schmitt trigger circuit for the given values of UTP and LTP CYCLE - 2 4) Frequency synthesis using PLL 5) Design and test R-2R DAC using OP-AMP. 6) Design and test the following circuits using IC 555 (a) Astable multivibrator for given frequency and duty cycle (b) Monostable multivibrator for given pulse width W. CYCLE - 3 7) Class-C single tuned amplifier 8) Amplitude modulation using Transistor/FET (Generation and Detection) 9) Pulse Amplitude modulation and Detection CYCLE - 4 10) PWM and PPM 11) Frequency modulation using 8038/2206 12) Precision Rectifiers- both Full Wave and Half Wave


3

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE EXPERIMENT N0. 1(A) SECOND ORDER ACTIVE LOW PASS FILTER AIM: To obtain the frequency response of an active low pass filter for the desired cut off frequency. COMPONENTS REQUIRED: Resistors- 33K, 10K, 5.86 K Capacitors 2200pF, opamp A 741 DESIGN For a 2nd order Filter, F H = 1 / 2RC Hz

Let FH = 2 KHz and R = 33 K 2 10 3 = 1 / 2 33 10 3 C C = 2200 pF

The pass band gain of the filter, AF = (1+R f / R1) For a second order filter, AF = 1.586, Let R1 = 10K RF = 5.86 k

L o w

p a s sR 1 1 0 K

c i r c u i t

D i a g r a mR 1 f 0 k

0R 3 3 k R 3 3 k

u A 3 +

V+ 7 4 V1 V o 4

2

-

7

0

V

1 C 2 2 0 0 2 P f C 2 0 0 P f

0 0


4

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE PROCEDURE: 1. Before wiring the circuit, check all the components. 2. Design the filter for a gain of 1.586 and make the connections as shown in the circuit diagram. 3. Set the signal generator amplitude to 10V peak to peak and observe the input voltage and output voltage on the CRO 4. By varying the frequency of input from Hz range to KHz range, note the frequency and the corresponding output voltage across pin 6 of the op amp with respect to the gnd. 5. The output voltage (VO) remains constant at lower frequency range. 6. Tabulate the readings in the tabular column. 7. Plot the graph with f on X-axis and gain in dB on Y axis. RESULT:


5

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE EXPERIMENT N0. 1(B) SECOND ORDER ACTIVE HIGH PASS FILTER AIM: To obtain the frequency response of an active high pass filter for the desired cut off frequency. COMPONENTS REQUIRED: Resistors- 33K, 10K, 5.86 K Capacitors 2200pF, opamp A 741 DESIGN: For a 2nd order Filter, FL= 1 / 2RC Hz

Let FL = 2 KHz and R = 33 K 2 10 3 = 1 / 2 33 10 3 C C = 2200 pF The pass band gain of the filter, AF = (1+R f / R1) For a second order filter, AF = 1.586, Let R1 = 10K

RF = 5.86 k

H i g h

p a s sR 1 1 0 K

c i r c u i t

D i a g r a mR 1 f 0 k

0C C

u A 3 + f R

V+ 7 4 V1

2

-

7

V o

2 V 1

2 0

0 P

f2 2

0 0

P

4

0

R 3 3 k

3

3 k

0

0


6

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE PROCEDURE: 1. Before wiring the circuit, check all the components. 2. Design the filter for a gain of 1.586 and make the connections as shown in the circuit diagram. 3. Set the signal generator amplitude to 10V peak to peak and observe the input voltage In addition, output voltage on the CRO. 4. By varying the frequency of input from HZ range to KHA range, note the frequency And the corresponding output voltage across pin 6 of the op amp with respect to the gnd. 5-.The output voltage (VO) remains constant at lower frequency range. 6. Tabulate the readings in the tabular column. 7. Plot the graph with f on X-axis and gain in dB on Y axis. RESULT:


7

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE EXPERIMENT N0. 2(A) SECOND ORDER ACTIVE BAND PASS FILTER AIM: To obtain the frequency response of an active band pass filter for the desired cut off frequency and to verify the roll off. COMPONENTS REQUIRED: Resistors- 33K, 10K, 5.86 K Capacitors 2200pF, opamp A 741 DESIGN: For a 2nd order Filter, (i) F= 1 / 2RC Hz

For High pass section

Let FL = 2 KHz and R = 33 K 2 10 3 = 1 / 2 33 10 3 C C = 2200 pF (ii) For low pass section Let FH = 10 KHz And R = 33 k 10 10 3 = 1 / 2 33 10 3 C C = 470 pF The pass band gain of the filter, AF = (1+R f / R1) For a second order filter, AF = 1.586, Let R1 = 10K

RF = 5.86 k The Center frequency FC = FH FL Hence FC = 4.5 KHz


8

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE CIRCUIT DIAGRAM:B AR 1 0 k 7 2

N1

D

P A

S

SR f

F IL T E RR 1 1 0 k 2 u 1 6 R V o C ' A 7 R f

5 . 8 k

5 . 8 k

V+ 7 4 V16 V o 4

03 C V 0 V 1 R C R

u A +

V+

7 4 V-

0R

3

+

C

'

0

4

0 0 0

PROCEDURE: 1. Before wiring the circuit, check all the components. 2. Design the two filters for the desired cut off frequencies and make the connections as shown in the circuit diagram. 3. Set the signal generator amplitude to 10V peak to peak and observe the input voltage And output voltage on the CRO. 4. By varying the frequency of input from Hz range to KHz range, note the frequency And the corresponding output voltage across pin 6 of the op amp with respect to the gnd. 5-.The output voltage (VO) remains constant at lower frequency range. 6. Tabulate the readings in the tabular column. 7. Plot the graph with f on X-axis and gain in dB on Y axis.

RESULT:


9

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE EXPERIMENT N0 2(B) SECOND ORDER ACTIVE BAND REJECT FILTER AIM: To obtain the frequency response of an active band reject filter for the desired cut off frequency and to verify the roll off. COMPONENTS REQUIRED: Resistors- 33K, 10K, 5.86 K Capacitors 2200pF , opamp A 741 DESIGN: For a 2nd order Filter, (ii) F= 1 / 2RC Hz

For High pass section

Let FL = 10 KHz and C = 0.01 F, F L = 1 / 2RC Hz 10 10 3 = 1 / 2 R 0.01 10 -6 R = 1.59 k (ii) For low pass section Let FH = 2 KHz And R = 33 k 2 10 3 = 1 / 2 33 10 3 C C = 2200 pF The pass band gain of the filter, AF = (1+R f / R1) For a second order filter, AF = 1.586, Let R1 = 10K

RF = 5.86 k


10

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE CIRCUIT DIAGRAM:-

B A N DR 1 1 0 k 2

R E J E C TR f 5 7 . 8 K

F I L T E R

0

S U M M E R7 4 16 R 2 = 1 0 k R 4 = 1 0 k

u A

H I G H P A S S S E C TC I O N C0 . 0 1 u R F 0 . 0

V-

3 1 u R

+ F

V+

4

V+ u A 7 4 V4 16

2

-

3

0R 1 1 0 k 2 R 7 f = 5 . 8 K

R 5 = 3 .3 K

+

0

0

L O WR '

P A S SR '

S E- C T I O NV+ u A 7 4 16 V3 +

R

3

=

1 0

k

C '

C '

0

PROCEDURE: 1. Before wiring the circuit, check all the components. 2. Design the two filters for the desired cut off frequencies and make the connections as shown in the circuit diagram. 3. To simplify the design, set R2=R3=R and C2=C3=C then choose a value of C tc or RC (1/mm) Or RC/3= (1/mm) m=2fm Assuming value of C=0.01F Substituting value of C and fm=1 kHz, we get R=9.5k 10k m= (Vmax-Vmin)/ (Vmax+Vmin) Vm= (Vmax-Vmin)/2 Waveform:-

Procedure:1. 2. 3. 4. 5. Design the collector modulator circuit assuming fm=1 kHz and m=0.5 take C=0.01F. Before wiring, check all components using multimeter. Make connections as shown in figure. Set the carrier frequency to 2v and 455 kHz. Set the modulating signal to 5v and 1 kHz.


29

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE 6. Keep carrier amplitude constant and vary the modulating voltage in steps and measure Vmax and Vmin, and calculate modulation index. 7. Tabulate the reading taken. 8. Feed AM output to Y-plates and modulation signal yo X-plates of CRO. Obtain trapezoidal pattern. 9. Plot the graph of modulating signal versus modulation index. Observations:Vmax in volts Vmin in volts (mod index) Vm in volts

Graph:m

Vm

Result:-

EXPERIMENT N0 8 ENVELOPE DETECTOR Aim: - conduct an experiment to demonstrate envelope detector for an input AM signal. Plot variation of output signal amplitude versus depth of modulation. DEPARTMENT OF E&C, CMRIT 30

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE Components required -0A79 diodes, resistors, capacitors, function generator, connecting board and CRO. Circuit Diagram:-

0A79 0.6vAM SIGNAL 2

VAMPL = 2vFC = 455kHz MOD = 0.5 FM = 2kHz

0.1microF

1

6k

output

ENVELOPE DETECTOR Theory: An envelope detector is a simple and highly effective device that is well suited for the demodulation of a narrow band AM wave, for which the percentage modulation is less than 100%. In an envelope detector, the output of the detector follows the envelope of the modulated signal, hence the name to it. Figure above shows the circuit of an envelope detector. It consists of a diode and a resistor-capacitor filter. This circuit is also known as diode detector. In the positive, half cycle of the AM signal diode conducts and current flows through R whereas in the negative half cycle, diode is reverse biased and no current flows through R. As a result, only positive half of the AM wave appears across RC. During the positive half cycle, the diode is forward biased and the capacitor C charges up rapidly to the peak value of the input signal. When the input signal falls below this value, the diode becomes reverse biased and the capacitor C slowly discharges through the load resistor RL. The discharging process continues until the next positive half cycle when the input signal becomes greater than the voltage across capacitor, the diode conducts again and the process is repeated.

Waveform:-


31


Design:Let fm=1 kHz m= fc=455 kHz C=0.01f Let Rc>>fc Or RC=3/ (mm) Substituting value of C and m in above equation we get, Therefore, R=10 k Procedure:1. Before wiring the circuit, check all the components using the multimeter. 2. Make the connections as shown in the figure. 3. From the function generator apply the AM wave to the input. 4. Vary the modulation index knob, note down the Vmax and Vmin simultaneously, and note down the output voltage the output VO in steps. 5. Modulation index is given by m= (Vmax-Vmin)/ (Vmax+Vmin) 6. Plot the graph Vo versus modulation index m.

Tabular column:Modulation index m Output in volts Vo

Graph:DEPARTMENT OF E&C, CMRIT 32


Vo

m

Result:


33


EXPERIMENT N0 9 FREQUENCY MODULATION USING IC 8038

AIM:To design and conduct an experiment to generate FM wave IC8038 with f= 33 kHz. COMPONENTS REQUIRED:

SL. NO 1. 2. 3.

COMPONENTS IC 8038 Signal generator Resistors

RANGE (0-100)MHz 10 k ohms 4.7k ohms, 22k ohms, 82k ohms. 12 V 0.1 micro F 1.00 micro F

QUANTITY 1 1 4 1 1 1 2 1 1 1

4. 5. 6. 7.

CRO probes Voltage supply Capacitors Mother board


34


DESIGN: Let R=Ra=Rb Let f=33 kHz

f= 3*(2*Ra-Rb)/10*Rac*Ra Substituting for R&C in above equation, we get f=0.3/RC Let R=10k ohms Therefore C =0.001*10-6F Calculation Frequency deviation =Fmax-Fmin Modulation index= Frequency deviation / fm GRAPH:


35


THEORY: Frequency modulation: FM is that form of angle modulation in which the instantaneous frequency is varied linearly with the message signal. The IC 8038 waveform generator is a monolithic integrated circuit capable of producing high accuracy sine square , triangular, saw tooth and pulse waveforms with a minimum number of external components.


36


Block diagram of ICL 8038 Basic principle of IC 8038 The operation of IC 8038 is based on charging and discharging of a grounded capacitor C, whose charging and discharging rates are controlled by programmable current generators Ia and Ib. When switch is at position A, the capacitor charges at a rate determined by current source Ia . Once the capacitor voltage reaches Vut, the upper comparator (CMP 1) triggers and reset the flip-flop out put. This causes a switch position to change from position A to B. Now, capacitor charge discharging at the rate determined by the current sink Ib . Once the capacitor reaches lower threshold voltage, the lower comparator (CMP 2) triggers and set the flip-flop output. This causes the switch position to change from position B to A. And this cycle repeats. As a result, we get square wave at the output of Flip flop and triangular wave across capacitor. The triangular wave is then passed through the on chip wave shaper to generate sign wave. To allow automatic frequency controls, currents Ia and Ib are made programmable through an external control voltage Bi. For equal magnitudes of Ia and Ib, output waveforms are symmetrical conversely, when two currents are unequal, output waveforms are asymmetrical. By making, one of the currents much larger than other we can get saw tooth waveform across capacitor and rectangular waveform at the output of flip-flop.


37


Working The frequency of the waveform generator is direct function of the dc voltage at terminal 8. By altering this voltage, frequency modulation is performed. For small deviations, the modulating signal can be applied to pins, merely providing dc-dc coupling with a capacitor. An external resistor between pins 7and 8 is not necessary but it can be used to increase input impedance from about 8k. The sine wave has relatively high output impedance. The circuit may use a simple op_amp follower to provide a buffering gain and amplitude adjustments. The IC 8038 is fabricated with advanced monolithic technology, using Schottky-barrier diodes and thin film resistors, and the output is stable over a wide range of temperatures and supply variations. PROCEDURE: 1. Rig up the circuit as shown in the figure. 2. Apply +12,-12V from the supply. 3. Observe the sinusoidal waveform at pin 2.It should be same as design carrier frequency. 4. Switch on signal generator and apply the signal amplitude of 0.5V and frequency of 1 kHz. 5. Observe the output between pin 2 and ground. 6. Sketch the waveforms. Show the graph of message carrier and modulation signal. RESULT: The frequency modulation is seen and the transmission bandwidth was found to be kHz.

EXPERIMENT NO 11 DEPARTMENT OF E&C, CMRIT 38

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE PULSE AMPLITUDE MODULATION AIM: To conduct an experiment to generate PAM signal and design a circuit to demodulate the PAM signal COMPONENTS REQUIRED: SLNO 1 2 COMPONENTS Transistor Resistor RANGE SL 100 22 K 4.7K 10 K 680 0.1 f 0A79 30MHZ QUANTITY 1 3 1 1 1 1 1 2 1

3 4 5 6 THEORY:

Capacitor Function Generator Diode Signal Generator CRO

In PAM the amplitude of the pulses are varied in accordance with the modulating signal. (Denoting the modulating signal as m (t). PAM is achieved simply by multiplying the carrier with the m (t) signal. The balanced modulators are frequency used as multipliers for this purpose. The Output is a series of pulses, the amplitude of which vary in proportion to the modulating signal. The form of pulse Amplitude modulation shown in the circuit diagram is referred to as natural PAM because the tops of the pulses follow the shape of the modulating signal. As shown in fig, the samples are taken at regular interval of time. If enough samples are taken, a reasonable approximation of the signal being sampled can be constructed at the receiving end. This is known as PAM.P U L S E A M P L I T U D E M O D U L A T I O N

V c c + 5 VQ 1

C E S L 1 0 0

1 0 K

B 2 2 K

C ( t ) m ( t )

4 . 7 K V o


39

ANALOG COMMUNICATIONDLABAMANUAL, V SEM ECE D E M O U L T I O ND 1

O A P A M

7 9 R C I / P 6 8 0 0 . 1 u F

D E M O D

O / P

DESIGN Fc>> 1/RC i.e., R>1/FcC Let Fc =15 kHz and C=0.1F Therefore R~680 PROCEDURE: 1. Make the Connections as shown in circuit diagram. 2. Set the carrier amplitude to 2 Vpp and in the frequency of 5 kHz to 15 kHz. 3. Set the i/p Signal amplitude to around 1V (p-p) and frequency to 2 kHz. 4. Connect the CRO at the emitter of the transistor and observe the Pam waveform. 5. Now connect the O/p(i.e. PAM) signal to the demodulation circuit and observe the signal if it matched plot the waveform

RESULT: The circuit to generate PAM signal and to demodulate the PAM signal were designed and the waveform were observed.

EXPERIMENT NO 12 DEPARTMENT OF E&C, CMRIT 40

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE PULSE WIDTH MODULATION AIM : To Conduct an Experiment to generate a PWM Signal for the given analog signal of frequency less than 1 kHz and to design a demodulation circuit. COMPONENTS REQUIRED SLNO 1 2 3 4 5 6 7 8 THEORY: Pulse width Modulation (PWM) is also known as Pulse duration Modulation (PDM). Three variations of PWM are possible. In One variation, the leading edge of the pulse is held constant and change in the pulse width with signal is measured with respect to the leading edge. In other Variable, the tail edge is held in constant and w.r.t to it, the pulse width is measured in the third variation, the centre of the pulse is held constant and pulse width changes on either side of the centre of the pulse. The PWM has the disadvantage when compared to PDM that its pulses are of varying width and therefore of varying power content, this means the transmitter must be powerful enough to handle the max width pulses.P m ( t ) z 1 0 kR 1 V+ 2 u 3 + A 7 1 0 K

COMPONENTS Op- Amp ( A 741) Resistors Capacitor Function Generator DC Regulated Power supply Signal Generator CRO Probes CRO Springs

RANGE 12 V 10 K 15K 0.01 f 12 V 30MHZ 15

QUANTITY 3 3 1 2 1 2 3 1 15

W

M

M

O

D

U

L

A

T

I O

N

1 k H c ( t )

V+ 7 4 16 V-

> 1 k H 1z 0 k

7 4 V-

16

2

u A

7

3 4

P W M O / P

+

0

0

4


41

ANALOG COMMUNICATION LAB MANUAL, V SEM ECED E M O D U L A T I O NR 1 2 u A 2 7 V+

P W I/ P

1

0 . 0 1 u 3F n + 1

M0 k C

7 4 16 V-

P W M O / P

R

2

=

1

5 k

C

2

m

( t )

0 . 0 1 u F4

0

0

DESIGN RC >>T Time Period Tp=0.1ms R1C1=Tp Let R1=10K C1=0.01F Fc=1/2R2C2 Fc=1KhZ Let R2=15K C2=0.01F

PROCEDURE: 1. Make the connections as shown in the circuit diagram, 2. Set the carrier amplitude to 2vpp and frequency 1 KHz (Say 1 5khz) 3. Set the signal amplitude to 2 Vpp and frequency < 1khz (Say 560 kHz) 4. Observe the o/p signal at pin 6of 2nd op-amp and observe the variation in pulse width by varying the modulating signal amplitude. 5. Draw PWM Waveform 6. Now connect the output to the demodulate circuit and observe the signal it matches with m(t) RESULT: The circuit to generate a PWM signal is designed and the output waveforms are observed. In addition, a circuit to demodulate the PWM signal is designed and the output is observed.


42

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE EXPERIMENT NO 13 PULSE POSITION MODULATION AIM : To conduct an experiment to generate PPM signal of pulse width(between 100 ms and 200ms) for a given modulating signal. COMPONENTS REQUIRED: SLNO 1 2 3 5 6 7 8 THEORY : In this type of modulation , the amplifier and width of the pulses is kept constant while the position of each pulse with reference to the position of a reference pulse is changed according to the instantaneous sampled value of the modulating signal. Pulse position modulation is observed from pulse width modulation. Any pulse has a leading edge and trailing edge in this system the leading edge is held in fixed position while the trailing edge varies towards or away from the leading edge in accordance to the instantaneous value of sampled signal COMPONENTS Op amplifier 555 - Timer Resistors Capacitor Dc Regulated Power Supply Function Generator CRO RANGE Ma 741 10 K 18 K 0.01mf + 5v 30mhz QUANTITY 1 1 1 1 2 1 2 1

m

( t ) z 1 0 kR 1 2 u 3 + A V+ 7 1 0 KVcc +5V

1 k H c ( t ) > 2 K

V+

H 1 z0 k

7 4 16 V-

2

7

8

u A

BY127

4

RtVCC 2 RESET T R IG G E R T H R E S H O LD 6

RA

7 4 16 V-

3 4

P W M O / P

+

Input

Ct 555D IS C H A R G E 7

C0CONTROL

4

CRO

3

OU T P U T GND 1

0

0

0.01u f0

DESIGN Pulse Width = 200s Tp=1.1 RC Let C=0.01F Therefore R=18K PROCEDURE DEPARTMENT OF E&C, CMRIT 43

5

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE 1. 2. 3. 4. Make the connections as shown in the circuit diagram. Set the carrier amplitude to around 4v (p-p) and frequency = 1khz. se the signal amplitude to around 2v (p-p) and frequency around (< 1khz) Observe the output signal at pin no : 3 of the 555 timer and also observe the variation in pulse position by varying the modulating signal amplitude 5. Draw the PPM waveform RESULT The circuit to generate a PPM signal of pulse width 200 ms is designed and the output waveform of PPM was observed.

EXPT. 11 - PRECISION RECTIFIERDEPARTMENT OF E&C, CMRIT 44


AIM: Design and test the working of Full Wave Precision Rectifier using op-amp. COMPONENTS REQUIRED: OP-Amp - A741 -1 Resistors - 10k - 3 22k - 1 3.3k - 1 Diodes - BY127 - 2 THEORY: Precision Rectifier name itself suggests that it rectifies even lower input voltages i.e. voltages less than 0.7v (diode drop). A rectifier is a device, which converts AC voltage to DC voltage. Precision rectifier converts AC to pulsating DC. Normal rectifiers using transformers cannot rectify voltages below 0.7v, so we go for precision rectifiers. In this circuit the diodes are placed in such a way that one diode is forward biased in the positive half cycle and the other in the negative half cycle. Consider the circuit diagram shown below. Here in the positive half cycle D1 is forward biased and D2 is reverse biased. The simplified circuit will act as two inverted amplifiers connected in series. Hence the total gain will be the product of individual gains. During the negative half cycle, D1 is reverse biased and D2 is forward biased. Hence the simplified circuit is an inverting amplifier connected in series with a noninverting amplifiers. Hence the output will be inverted and a DC output (unidirectional) is obtained .The precision rectifier we are using is a full wave rectifier. CIRCUIT DIAGRAM:R 1 = 2 2 k D + V c c 6 U A 7 4 1 D 3 . 3 K 2 2 3 7 V i n R = 1 0 2 k3 1 + V c c 6 U A 7 4 V o u t 1 7 R = 1 0 k R = 1 0 k

G

N

D

DESIGN:

R

2

=

Given : Vo = +0.5V in the +ve cycle = +0.1V in the -ve cycle DEPARTMENT OF E&C, CMRIT 45

4

- V c c

4

+

+

- V c c

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE During the +ve half cycle the simplified circuit will be as shown below. v

V = (-R1 / R) Vin V0 = (-R / R)V = (-R / R) (-R1 / R) Vin V0 = (R1/R) Vin As V0 = 0.5V, Vin = 0.25V R1 / R = 0.5 / 0.25 = 2

Assume R = 10k , then R1 = 20k NOTE: A DRB can be used in the place of R1 and that resistance can be adjusted to 20K or 22K resistance can be used. During the negative half cycle, the simplified circuit will be shown below.V

I1V i n R = A 1 0

R I 2

1

R

R

+ V c c 6 U A 7 4 1 2 3

+ V c c 6 U A 7 4 V o u t 1

7

I 3

G

N

D

R

2 V

Applying KCL at point A I1 = I2 + I3 From virtual ground concept VA = 0 ( VB = 0) I1 = Vin / R , I2 = -V / ( R1+R ), I3 = -V / R2 Vin / 10k = - V ( (1 / 30k) + (1 / R2) ) DEPARTMENT OF E&C, CMRIT 46

4

- V c c

4

+

+

k B

3

- V c c

-

2

7

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE As Vin = - 0.25 - 0.25 / 10k = -V ( (1 / 30k) + (1 / R2) ) --------(1) As the second Op-Amp works as a non inverting amplifier V0 = (1 + (R / R1) + R) V = (1 + (10k / 30k) ) V ---------(2) From (1) V = - 0.25 / 10k = - V ( (R2 + 30k) / (30k x R2 ) ) V = 0.75 R2 / ( R2 + 30k ) Substituting this in the equation (2) we get V0 = (1 + 1 / 3) (0.75 R2 / (R2 + 30k) ) 0.1 R2 + 3k = R2 R2 = 3.3k PROCEDURE: 1. Rig up the circuit as shown in the circuit diagram. 2. Give an input of 0.5V peak to peak (sine wave). 3. Check and verify the designed values. 4. Design the same circuit for a different set of values. WAVEFORMS: Vin 0.25 0 - 0.25 t 0.9 R2 = 3k

V0 0.5 0.1 DEPARTMENT OF E&C, CMRIT 47

ANALOG COMMUNICATION LAB MANUAL, V SEM ECE 0 t

RESULT:

Viva questions for analog communication lab 1. Define the word communication. 2. What are the basic components of electronic communication. 3. What is Transmitter.DEPARTMENT OF E&C, CMRIT 48


4. What is receiver 5. What is communication channel? 6. State two types of communication? 7. What is baseband signal? 8. What is baseband transmission? 9. What is the need for modulation? 10.Define the carrier signal? 11.What is the classification of modulation? 12.What is frequency deviation? 13.Define noise? 14.Define the basic sources of noise? 15.What is shot noise? 16.Define signal to noise ratio? 17.What is noise factor? 18.State the equation for noise factor for cascade connection? 19.Define amplitude modulation? 20.Define modulation index? 21.State the bandwidth required for amplitude modulation? 22. What is frequency domain display? 23. What is time domain display? 24. What is maximum power of sideband of AM? 25. What is the maximum total power of AM wave? 26. Define a high level modulation? 27. Define a low level modulation? 28.Why amplitude modulation is used for broadcasting? 29.What is the position of the operating point of class-C? 30.What is the advantage of SSB over DSB-SC? 31.What is the function of Transistor mixer? 32.What is the principle of Envelope detector? 33.Where SSB transmission is used? 34.State sampling theorem? 35.What is Nyquist criteria? 36.What is Roll-off factor? 37.Define the order of the filter? 38.What are the classification of filters? 39.Differentiate between butter-worth and cheybeshev filter? 40.Define selectivity? 41.What is quadrature null effect? 42.Define FM? 43.What is percentage modulation? 44.Define pre-emphasis and De-emphasis? 45.What are the advantages of using pre and de-emphasis? 46.List of some advantages of FM over AM? 47.Define wideband FM?DEPARTMENT OF E&C, CMRIT 49


48.What is carsons rule? 49.State advantages of PWM? 50.State various Pulse modulation methods?


50

analog communication manual

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