analog communication manual tce
TRANSCRIPT
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No.132, AECS Layout, I.T.P.L. Road, Kundalahalli, Bangalore- 560 037
A LAB MANUAL ON
ANALOG COMMUNICATION + LIC
Subject Code: 06ECL58
(As per VTU Syllabus)
PREPARED BY
STUDENTS - Dept. of TCE
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 2
CONTENTS
EXPT.
NO.NAME OF THE EXPERIMENT
1 Active low pass & high pass filters second order
2 Active band pass & band reject filters second order
3Schmitt trigger design and test a Schmitt trigger circuit for thegiven values of UTP and LTP
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.
7 Class-C single tuned amplifier
8Amplitude modulation using Transistor/FET (Generation andDetection)
9 Pulse Amplitude modulation and Detection
10 PWM and PPM
11 Frequency modulation using 8038/2206
12 Precision Rectifiers- both Full Wave and Half Wave
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CYCLE WISE EXPERIMENTS
SEM: V EXAM MARKS: 50
BRANCH: TCE IA MARKS: 25
SUBJECT: ANALOG COMMUNICATION & LIC LAB
SUB CODE: 06ECL58
CYCLE - 1
1) Active low pass & high pass filterssecond order
2) Active band pass & band reject filterssecond order
3) Schmitt trigger design and test a Schmitt trigger circuit for the given values of UTP and
LTP
CYCLE - 24) 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
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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, opampA 741
DESIGN
For a 2nd order Filter, F H = 1 / 2 RC Hz
Let FH = 2 KHz and R = 33 K
2 10 3 = 1 / 2 33 10 3 C
The pass band gain of the filter, AF = (1+Rf / R1)
For a second order filter, AF = 1.586, Let R1 = 10K
RF = 5.86 k
C = 2200 pF
0
0
0
0
uA741
2
3
7
4
-
+
V+
V-
C2200Pf
R1
10K
R
33k
V1
R
33k
Rf
10k
C2200Pf
Vo
Low pass circuit Diagram
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DEPARTMENT OF TCE, CMRIT 5
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 fon X-axis and gain in dB on Y axis.
RESULT:
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 6
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, opampA 741
DESIGN:
For a 2nd order Filter, FL= 1 / 2 RC 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+Rf / R1)
For a second order filter, AF = 1.586, Let R1 = 10K
RF = 5.86 k
0
0
0
0
R
33k
V1
R1
10K
Rf
10k
uA741
2
3
7
4
-
+
V+
V-
C
2200Pf
R
33k
C
2200Pf
Vo
High pass circuit Diagram
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT
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 fon X-axis and gain in dB on Y axis.
RESULT:
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
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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, opampA 741
DESIGN:
For a 2nd order Filter, F= 1 / 2 RC Hz
(i) 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+Rf / 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
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
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CIRCUIT DIAGRAM:-
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 fon X-axis and gain in dB on Y axis.
RESULT:
0
0
0
00
0
uA741
2
3
7
4
-
+
V+
V-
R1
10k
R R
C'C'
5.8k
uA741
2
3
7
4
-
+
V+
V-
R1
10k
V1
0V
5.8k
C
R R
C
Vo
6
6
Vo
Rf Rf
BAND PASS FILTER
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 10
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.
COMPONENTSREQUIRED:
Resistors- 33K, 10K,5.86 K
Capacitors 2200pF , opampA 741
DESIGN:
For a 2nd order Filter, F= 1 / 2 RC Hz
(ii) For High pass section
Let FL = 10 KHz and C = 0.01 F, F L = 1 / 2 RC 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+Rf / R1)
For a second order filter, AF = 1.586, Let R1 = 10K
RF = 5.86 k
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
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CIRCUITDIAGRAM:-
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
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 12
RESULT:
EXPERIMENT NO. 3
DESIGN AND TEST A SCHMITT TRIGGER CIRCUIT FOR THE GIVEN VALUES
OF UTP AND LTP
AIM:
Design a square wave generator for a given UTP and LTP.
COMPONENTS REQUIRED:
Op-Amp - A7411
Resistors1k - 1, 2.2k - 1
THEORY:
Schmitt Trigger is also known as Regenerative Comparator. This is a square wave
generator which generate a square based on the positive feedback applied. As shown in the
fig. below, the feedback voltage is Va. The input voltage is applied to the inverting terminal
and the feedback voltage is applied to the non-inverting terminal. In this circuit the op-amp
acts as a comparator. It compares the potentials at two input terminals. Here the output shifts
between
+ Vsat and Vsat. When the input voltage is greater than Va, the output shifts to Vsat and
when the input voltage is less than Va, the output shifts to + Vsat. Such a comparator circuit
exhibits a curve known asHysterisis curve which is a plot of Vin vs V0. The input voltage at
which the output changes from + Vsat toVsat is called Upper Threshold Point (UTP) and the
input voltage at which the output shifts from Vsat to + Vsat is calledLower Threshold Point
(LTP). The feedback voltage Va depends on the output voltage as well as the reference
voltage.
A Zero Cross Detector is also a comparator where op-amp compares the input voltage
with the ground level. The output is a square wave and inverted form of the input.
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
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CIRCUIT DIAGRAM:
Schmitt Trigger
Zero Cross Detector
DESIGN:
Given UTP = + 4V and LTP = - 2V
Let I1 be the current through R1 and I2 be the current through R2.
W.K.T the current into the input terminal of an op-amp is zero.
I1 + I2 = 0
I1 = ( V0Va ) / R1
I2 = ( VrefVa ) / R2
( V0Va ) / R1 + ( VrefVa ) / R2 = 0
Va = ( V0 R2 + VrefR1 ) / ( R1 + R2 )
When V0 = + Vsat, Va = UTP
When V0 = - Vsat, Va = LTP
[ ( Vsat R2 ) / ( R1 + R2) ] + [ ( VrefR1 ) / ( R1 + R2 ) ] = UTP ------- (1)
[ ( - Vsat R2 ) / ( R1 + R2 ) ] + [ ( VrefR1 / (R1 + R2 ) ] = LTP -------(2)
+
-
U1
UA741
3
26
7 1
4 5R1
2.2k
R21k
+ Vref
+ Vcc
- Vcc
Vo
Vin
+
-
U2
UA741
3
26
7 1
4 5
Vo
+ Vcc
- Vcc
Vin
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 14
(1)(2)
( 2 Vsat R2 ) / ( R1 + R2 ) = UTPLTP = 6V
Simplifying this equation we get,
7 R2 = 3 R1
Assume R2 = 1k
R1 = 2.2k
(1)+ (2)
( 2 VrefR1 ) / ( R1 + R2 ) = UTP + LTP = 2V
Simplifying the above equation, we get
Vref= 1.4V
PROCEDURE:
1. Rig up the connections as shown in the circuit diagram.
2. Give a sinusoidal input of 10V peak to peak and 500 Hz from a signal generator.
3. Check the output at pin no. 6 (square wave).
4. Coincide the point where the output shifts from + Vsat toVsat with any point on
the input wave.
5.
Measure the input voltage at this point. This voltage is UTP.6. Coincide the point where the output shifts fromVsat to + Vsat with any point on
the input wave.
7. Measure the input voltage at this point. This voltage is LTP.
8. Another method of measuring UTP and LTP is using the Hysterisis Curve.
9. To plot the hysterisis curve give channel 1 of CRO to the output and channel 2 of
CRO to the input.
10.Press the XY knob. Adjust the grounds of both the knobs.
11.Measure UTP and LTP as shown in the fig. and check whether it matches with the
designed value.
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 15
WAVEFORMS:
Vin
5
4
0 t
- 2
- 5
Schmitt Trigger
V0
100 t
- 10
Zero Cross Detector
V0
10
0 t
- 10
HYSTERISIS CURVE:
V0
+ Vsat
Vin
LTP UTP
- Vsat
NOTE: The same circuit can be designed for different values of UTP and LTP.
For UTP = 4V and LTP = 2V, R1= 10k , R 2 = 1k and Vref = 3.3V. Check whether the
circuit works properly for these values.
RESULT:
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 16
EXPERIMENT N0 4
CLASSC - TUNED AMPLIFIER
AIM: To design and test a class ctuned amplifier to work at f0 = 734 kHz and to find its
max efficiency at optimum load
COMPONENTS REQUIRED
SLNO COMPONENTS RANGE QUANTITY
1. Dc Regulated Power Supply +5V 1
2. Ammeter 0 -10MA 1
3 Inductor 100MH 1
4.
Capacitors 470Pf 1
1000mf 1
0.01mf 1.
5 Resistors 15k 1
22 1
6 Transistor BF194 1
7 CR0 Probe Springs - 1
Springs - 10
THEORY: Class C Tuned Amplifier Amplify Large signal at radio frequency with
better frequency response. Efficiency is more than 78% and it increases with decrease in
conduction angle. It is used in radio transmitters and receivers with class c operation the
collector current flows for less than half a cycle. A parallel resonant circuit can filter the
pulses of collector current and produce a pure sine wave of output voltage. The max
efficiency of a tuned class c amplifier is 100% the Ac voltage drives the base and an
amplified and inverted signal is then capacitive coupled to the load resistance. Because of the
parallel resonant circuit, the output voltage is max at resonant frequency f0 = 1/2xLC
On either side of the voltage gain drops off shown class C is always intended to amplify a
narrow ban of frequency.
DESIGN:
F O = 1/2 LC
Let L = 100 F and C = 470 pF
F O = 1 / 2 3.142 100 10-6 470 10 -12
F O = 734 KHz
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T = 1/F O
T = 1/ 734 10 3
T = 1.36
S
RB C B 10 T OWhere T O = 1/F O
RB C B = 10 1/ F O
C B = 10 / F O RB
Let RB = 15 k
C B = 10 / 734 103 15 10 3
C B = 908 pF
Use Standard Value
CONDUCTION ANGLE = T e / T 3600
Where Te = Time period across emitter
T = time period across collector
DUTY CYCLE D = / T
Or D = / 3600
TABULAR COLUMN: ( Vin = 5 volts)
Sl.No RL( ) V
OUT
(V)
I dc
(mA)
P ac = V O2/8RL P dc =VCC I dc = Pac/ Pdc %
PROCEDURE
1. Make the connections as shown in circuit diagram set input signal frequency to the tuned
circuit resonant frequency
2. Vary input voltage to get an undistorted approx sine wave by keeping load resistance to
a fixed value by varying load resistance note down the output voltage and calculate current
Iac
3. Tabulate the reading in tabular column\
4. Plot the graph of rl along xaxis and n across yaxis
From the graph, determine optimum load to calculate conduction angle the output is
taken across emitter
C B = 1000 Pf
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 18
RESULTA classC tuned amplifier Is designed to work at a reasonable frequency fo 734kHz.
The max optimum load is 400 and conduction angle 0 = 77.
EXPERIMENT N0 5
R-2R DAC USING OP-AMP
AIM:
Demonstrate Digital to Analog conversion for digital (BCD) inputs using R-2R
network.
COMPONENTS REQUIRED:
Op-amp - A741
Resistors10k - 4
22k - 6
Dual power supply, Multimeter, bread board, connecting wires.
THEORY:
Nowadays digital systems are used in many applications because of their increasingly
efficient, reliable and economical operation. Since digital systems such as microcomputers
use a binary system of ones and zeros, the data to be put into the microcomputer have to be
converted from analog form to digital form. The circuit that performs this conversion and
reverse conversion are called A/D and D/A converters respectively.
D/A converter in its simplest form uses an op-amp and resistors either in the binary
weighted form or R-2R form.
The fig. below shows D/A converter with resistors connected in R-2R form. It is so
called as the resistors used here are R and 2R. The binary inputs are simulated by switches b 0
to b3 and the output is proportional to the binary inputs. Binary inputs are either in high (+5V)
or low (0V) state.
The analysis can be carried out with the help of Thevenins theorem. The output
voltage corresponding to all possible combinations of binary inputs can be calculated as
below.
V0 = - RF [ (b3/2R) + (b2/4R) + (b1/8R) + (b0/16R) ]
Where each inputs b3, b2, b1 and b0 may be high (+5V) or low (0V).
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The great advantage of D/A converter of R-2R type is that it requires only two sets of
precision resistance values. In weighted resistor type more resistors are required and the
circuit is complex. As the number of binary inputs is increased beyond 4 even D/A converter
circuits get complex and their accuracy degenerates. Therefore in critical applications IC D/A
converter is used.
Some of the parameters must be known with reference to converters. They re
resolution, linearity error, settling time etc.
Resolution = 0.5V / 28 = 5 / 256 = 0.0195
WAVEFORMS:
DESIGN:
The equation for output voltage is given by
V0 = - RF [ (b3/2R) + (b2/4R) + (b1/8R) + (b0/16R) ]
V0 = - RF . Vref[ (b3/2R) + (b2/4R) + (b1/8R) + (b0/16R) ]
Case (i) If b0 b1 b2 b3 = 1 0 0 0 for 0.5 volts change in output for LSB change
- 0.5 = - 20 x 103.Vref[ (1 / (16 x 1.103)) + 0 + 0 + 0)
Vref= 4V
Case (ii) If Vref= 5V and b0b1b2b3 = 0 1 0 0, then
V0 = - 20.103 . 5 [ (0 + (1/ (8.1.103)) + 0 + 0) ]
V0 = - 1.25V
0
R
0b0
+
-
U1
UA741
3
26
4 1
7 5
Vref
RR
2R-vcc
Vo
LSBb1
R 2R
2R 2R
R- 2R LADDER NETWORK
b3
2R
MSB
0
RF = 2k
21
b2
+vcc
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TABULAR COLUMN:
Inputs Output voltage
b3 b2 b1 b0 Theoretical Practical
0 0 0 0
.
.
.
.
.
1 1 1 1
PROCEDURE:
1. Test the op-amp and other components before rigging up the circuit.
2. Rig up the circuit as shown in the fig.
3. Apply different combination of binary inputs using switches.
4. Observe the output at pin no. 6 of op-amp using multimeter or CRO.
5. Tabulate the readings as shown.
6. Calculate the resolution of the converter.
RESULT:
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 21
Exp 6 - ASTABLE MULTIVIBRATOR
AIM:
To design and verify the operation of astable multivibrator using 555 Timer for given
frequency and duty cycle.
APPARATUS REQUIRED:
Timer - 555
Resistors10k - 1
4.5k - 2
7.25k- 1
Capacitors0.01 F-1
0.1 F-1
Signal Generator, DC power supply, CRO and connecting wires
THEORY:
A 555 timer is a monolithic timing circuit that can produce accurate and highly stable
time delays or oscillations, some of the applications of 555 are square wave generator, astable
and monostable multivibrator.
Astable multivibrator is a free running oscillator has two quasi stable state in one state o/p
voltage remains low for a time interval of Toff and then switches over to other state in which
the o/p remains high for an interval of Ton the time interval Ton and Toff are determined by
the external resistors a capacitor and it does not require an external trigger, when the power is
switched on the timing capacitor begins to charge towards 2/3 Vcc through RA & RB, when
the capacitor voltage has reached this value, the upper comparator of the timer triggers the
flip flop in it and the capacitor begins to discharge through RB when the capacitor voltage
reaches 1/3 Vcc the lower comparator is triggered and another cycle begins, the charging and
discharging cycle repeats between 2/3 Vcc and 1/3Vcc for the charging and discharging
periods t1and t2 respectively. Since the capacitor charges through RA and RB and discharges
through RB only the charge and discharge are not equal as a consequence the output is not a
symmetrical square wave and the multivibrator is called an asymmetric astable multivibrator
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CIRCUIT DIAGRAM:
ASSYMETRIC MULTIVIBRATOR
SYMMETRIC MULTIVIBRATOIR
0
C1
555D
1
2
3
4
5
6
7
8
GN
DTRIGGER
OUTPUT
RE
SE
T
CO
NT
RO
LTHRESHOLD
DISCHARGE
VC
C
Vcc+5V
D1
C2
Ra
O/PRbD2
0
555
1
2
3
4
5
6
7
8
GND
TRIGGER
OUTPUT
RESET
CONTROL
THRESHOLD
DISCHARGE
VCC
RB
C
0.01uf
RAVcc
CRO
+5V
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DESIGN:
ASSYMETRIC: Given f = 1khz
Duty cycle = 60%
T = 0.693(Ra + 2Rb) C
F = 1.45/ [(Ra + 2Rb) C]
Duty cycle = (Ra + Rb)/Ra + 2Rb)
1K = 1.45/ [(Ra + 2Rb) 0.1*10-6]
Ra + 2Rb = 14.5K
Ra + Rb = 8.7K
Assume Ra = 4.7K
Rb = 9.57K 10K
SYMETRIC Given f = 1khz
Duty cycle = 50%
Charging time = Discharging time
T = TON =TOFF
T = 0.693(Ra + Rb) C
F = 1.45/ [(Ra + Rb) C]
Duty cycle = (Ra + Rb)/Ra + Rb)
1K = 1.45/ [(Ra + Rb) 0.1*10-6]
Ra + 2Rb = 14.5K
Since the duty cycle is 50%, Ra = Rb
2Ra = 14.5K
Ra = 7.25K 6.8K
Rb = 7.25K 6.8K
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PROCEDURE:
1. Connections are made as shown in the circuit diagram
2. Switch on the DC power supply unit
3. Observe the wave form on CRO at pin 3 and measure the o/p pulse amplitude
4. Observe the wave form on CRO at pin 6 and measure Vcmax and Vc min
5. Verify that Vcmax=2/3Vcc and Vc min=1/3 Vcc
6. Calculate the duty cycle D, o/p frequency and verify with specified value
TABULAR COLUMN
Ra Rb CF(theo) =
1.45/(Ra + Rb)Ton Toff T F DY
WAVEFORMS:
Result:
Lower
Vc at pin 62/3VCC
1/3VCC0V
Vout 5V
at pin 3
0V
Ton
Toff
t
t
Upper threshold
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Expt 7- MONOSTABLE MULTIVIBRATOR
AIM:
To design and verify the operation of monostable multivibrator using 555 Timer for given
Pulse width.
APPARATUS REQUIRED:
Timer - 555
Resistors10k - 1
Capacitors 0.01 F-1
0.1 F-1
Signal Generator, DC power supply, CRO and connecting wires
THEORY:
Monostable multivibrator has a stable state and a quasi stable state, the output
of it is normally low and it corresponds to reset of the flip flop in the timer, on the application
of external negative trigger pulse at pin 2 the circuit is triggered and the flip flop in the timer
is set which in turn releases the short across C and pushes the output high, At the same time
the voltage across C rises exponentially with the time constant RAC and remains in this state
for a period RAC even if it is triggered again during this interval, When the voltage across the
capacitor reaches 2/3 Vcc, the threshold comparator resets the flip flop in the timer which
discharges C and the output is driven low the circuit will remain in this state until the
application of the next trigger pulse.
CIRCUIT DIAGRAM:
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DESIGN:
Given
Tp = 1ms
F = 1KHz
T = 1.1R C
Let C = 0.1uFR = (1*10-3 ) / (1.1*0.1*10
-6 )
R = 9.09 K 10K
PROCEDURE:
1. Rig up the circuit as shown in the figure after checking all the components.
2. Apply suitable inputs to the astable multivibrator (DC & Trigger inputs)
3. Observe the waveform across the timing capacitor in one channel and the output in the
other channel..
4. Verify the designed values and the repeat the above procedure for different set of values.
TABULAR COLUMN
R C Tp = 1.1RC Tp(prac)
0
0
C555
1
2
3
4
5
6
7
8
GND
TRIGGER
OUTPUT
RESET
CON
TROL
THRESHOLD
DISCHARGE
VCC
0.01uf
RARt
Vcc
CRO
+5V
BY127
CtInput
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DEPARTMENT OF TCE, CMRIT 2
WAVEFORMS:
RESULT:
Result:
VccT t
t
t
t
Input trigger pulses
Trigger pulses at pin 2
Upper
Capacitor voltage Vc at pin 5
TpT
Output pulse at pin 3
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 28
EXPERIMENT N0 7
COLLECTOR MODULATION
Aim: - To generate AM signal, information signal given the collector. Also, demodulate it.
Measure the modulation index using two different methods.
Components Required:- IFT, AFT, SL 100/BF 194 transistor, resistors, capacitors, diode
0A79, connecting board, connecting wires and CRO.
Circuit Diagram:-
0.01microF
MESSAGE SIGNAL
FM = 2kHzVAMPL = 5v(p-p)
470k
2
1
AFT (GREEN)
BF 194
VCC
OPEN
IFT (RED)
COLLECTOR AMPLITUDE MODULATION
120
2
1
+6v
o/p AMwave
-6v
0.01microF
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
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Theory: -The modulator is a linear power amplifier that takes the low-level modulating
signal and amplifies it to a high power level. The modulating output signal is coupled through
a modulating transformer to the Class C amplifier. The secondary winding of the modulation
transformer is connected in series with collector supply voltage Vcc of the Class C amplifier.
This means that modulating signal is applied in series with the collector power supply supply
voltage of the Class C amplifier applying collector modulation.
In the absence of the modulating input signal, there will be zero modulation voltage
across the secondary of the transformer. Therefore, the collector supply voltage will be
applied directly to the Class C amplifier generating current pulses of equal amplitude and
output of the tuned circuit will be a steady sine wave.
When the modulating signal occurs, the a.c. voltage across the secondary of the
modulating transformer will be added to and subtracted from the collector supply voltage.
This varying supply voltage is then applied to the Class C amplifier resulting in variation inthe amplitude of the carrier sine wave in accordance with the modulating signal. The tuned
circuit then converts the current pulses into an amplitude-modulated wave.
Design:-
Let fm= kHz
m=
RC>>tc or RC (1/mm)
Or RC/3= (1/mm)
m=2fm
Assuming value of C=0.01FSubstituting value of C and fm=1 kHz,
we get R=9.5k 10k
m= (Vmax-Vmin)/ (Vmax+Vmin)
Vm= (Vmax-Vmin)/2
Waveform:-
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DEPARTMENT OF TCE, CMRIT 30
Procedure:-
1. Design the collector modulator circuit assuming fm=1 kHz and m=0.5 take C=0.01F.
2. Before wiring, check all components using multimeter.
3. Make connections as shown in figure.
4. Set the carrier frequency to 2v and 455 kHz.
5. Set the modulating signal to 5v and 1 kHz.
6. Keep carrier amplitude constant and vary the modulating voltage in steps and measureVmax 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:-
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 31
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.
Components required -0A79 diodes, resistors, capacitors, function generator, connecting
board and CRO.
Circuit Diagram:-
ENVELOPE DETECTOR
Vm
m
6k
2
1
0.6v
0A79
0.1microF
AM SIGNAL
FM = 2kHz
VAMPL = 2vFC = 455kHzMOD = 0.5
output
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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 theinput signal becomes greater than the voltage across capacitor, the diode conducts again and
the process is repeated.
Waveform:-
Design:-
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 33
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 bym= (Vmax-Vmin)/ (Vmax+Vmin)
6. Plot the graph Vo versus modulation index m.
Tabular column:-
Modulation index m Output in volts Vo
Graph:-
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DEPARTMENT OF TCE, CMRIT 34
Result:
m
Vo
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 35
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 COMPONENTS RANGE QUANTITY
1.
2.
3.
4.
5.
6.
7.
IC 8038
Signal generator
Resistors
CRO probes
Voltage supply
Capacitors
Mother board
(0-100)MHz
10 k ohms
4.7k ohms,
22k ohms,
82k ohms.
12 V
0.01micro F
1.00 micro F
1
1
4
1
1
1
2
1
1
1
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 36
DESIGN:
Let R=Ra=Rb Let f=33 kHz
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:
f= 3*(2*Ra-Rb)/10*Rac*Ra
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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.
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DEPARTMENT OF TCE, CMRIT 38
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) triggersand 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.
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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.
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 40
EXPERIMENT NO 11
PULSE AMPLITUDE MODULATION
AIM: To conduct an experiment to generate PAM signal and design a circuit to
demodulate the PAM signal
COMPONENTS REQUIRED:
SLNO COMPONENTS RANGE QUANTITY
1 Transistor SL 100 1
2 Resistor 22 K 3
4.7K 1
10 K 1
680 1
3 Capacitor Function Generator 0.1 f 1
4 Diode 0A79 1
5 Signal Generator - 2
6 CRO 30MHZ 1
THEORY:
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.
Q1
PULSE AMPLITUDE MODULATION
Vcc +5V
Vo
SL 100BE
C
4.7K
22K
m(t)
C(t)
10K
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DEPARTMENT OF TCE, CMRIT 41
DESIGNFc>> 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.
D1
R
680
C
0.1uF
OA 79
PAM I/P DEMOD O/P
DEMODULATION
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 42
EXPERIMENT NO 12
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 COMPONENTS RANGE QUANTITY
1 Op- Amp ( A741) 12 V 3
2 Resistors 10 K 3
15K 1
3 Capacitor Function Generator 0.01 f 2
4 DC Regulated Power supply 12 V 1
5 Signal Generator - 2
6 CRO Probes - 3
7 CRO 30MHZ 1
8 Springs 15 15
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.
0 0
uA741
2
3
7
4
-
+
V+
V-
10k 10K
R1
10k
uA741
2
3
7
4
-
+
V+
V-
6
m(t)
c(t)
1kHz
>1kHz
6 PWMO/P
PWM MODULATION
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 43
DESIGNRC >>T
Time Period Tp=0.1msR1C1=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.
0 0
uA741
2
3
7
4
-
+
V+
V-
R1
10kC2
1n
R2 = 15k
C2
6 PWMO/P
PWMI/P
m(t)0.01uF
0.01uF
DEMODULATION
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 44
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 COMPONENTS RANGE QUANTITY
1 Opamplifier Ma741 12 555 - Timer - 1
3 Resistors 10 K 1
18 K 1
5 Capacitor 0.01mf 2
6 Dc Regulated Power Supply + 5v 1
7 Function Generator - 2
8 CRO 30mhz 1
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
DESIGNPulse Width = 200s
0
0
C555
1
2
3
4
5
6
7
8
GND
TRIGGER
OUTPUT
RESET
CONTROL
THRESHOLD
DISCHARGE
VCC
0.01uf
RARt
Vcc
CRO
+5V
BY127
CtInput
0 0
uA741
2
3
7
4
-
+
V+
V-
10k 10K
R1
10k
uA741
2
3
7
4
-
+
V+
V-
6
m(t)
c(t)
1kHz
>2KHz
6 PWMO/P
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
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Tp=1.1 RC
Let C=0.01F
Therefore R=18K
PROCEDURE1. Make the connections as shown in the circuit diagram.
2. Set the carrier amplitude to around 4v (p-p) and frequency = 1khz.
3. se the signal amplitude to around 2v (p-p) and frequency around (< 1khz)
4. 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.
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 46
EXPT. 11 - PRECISION RECTIFIER
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 non-
inverting 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:
GND
R = 10kR1 = 22k
+
-
UA741
3
26
4
7
+
-
UA7413
26
4
7
R = 10k
R2 = 3.3K
R = 10k
D1
D2
VinVinVout
+Vcc
-Vcc -Vcc
+Vcc
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DESIGN:
Given : Vo = +0.5V in the +ve cycle
= +0.1V in the -ve cycle
During the +ve half cycle the simplified circuit will be as shown below.
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.
GND
+
-
UA741
3
26
4
7
R1 R
R = 10k
R
+
-
UA7413
26
4
7
R2
VinVinVout
+Vcc
-Vcc -Vcc
+Vcc
I2
I3
V
V
A
B
v
I1
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ANALOG COMMUNICATION LAB MANUAL, V SEM TCE
DEPARTMENT OF TCE, CMRIT 48
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) )
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) )
0.1R2 + 3k = R2
0.9 R2 = 3k
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 t
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DEPARTMENT OF TCE, CMRIT 49
- 0.25
V0
0.5
0.1
0 t
RESULT:
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Viva questions for analog communication lab
1. Define the word communication.2. What are the basic components of electronic communication.
3. What is Transmitter.
4. What is receiver5. 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?
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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?48.What is carsons rule?49.State advantages of PWM?
50.State various Pulse modulation methods?