Analog Communications Lab PDF
Analog Communications Lab PDF
Analog Communications Lab PDF
INDEX
S.No
1.
2.
3.
Page
No
Date of
Performance
Date Of
Submission
Assessment
of marks
(Max 10M)
Sign. Of
Faculty
4.
Balanced Modulator
5.
6.
Mixer Characteristics
7.
8.
9.
Synchronous Detector
10
AGC Characteristics
11
Frequency Synthesizer
12.
Squelch Circuit
Average marks
Signature of Lab-In-charge
HOD
INSTRUCTIONS TO STUDENTS
Dept. of ECE
8. Student should develop the habit of pocket discussion / group discussion related to the experiments/
exercises so that exchanges of knowledge / skills could take place.
9. Student should develop habit to submit the practical, exercise continuously and progressively
on the
Dept. of ECE
Exp No: 1
Date:
AMPLITUDE MODULATION & DEMODULATION
modulation & over modulation) and also to calculate the modulation index.
APPARATUS REQUIRED:
S. No.
1
Component
Transistor
Resistors
Capacitors
4
5
6
7
8
9
Diode
Functions Generator
Regulated Power Supply
Bread Board
Cathode Ray Oscilloscope
Connecting Wires
Specification
BC107
100K
4.7K
270
33
4.7F
0.001F
0A79
1Mhz
(0-30)V
0-20MHz
Single Strand
Quantity
2
2
2
1
1
2
1
1
2
1
1
1
As Required
THEORY:
Modulator section illustrates the circuit of modulating amplifier employing a transistor as an active device
in CE mode. R1 & R2 establish a quiescent forward bias for the transistor. The modulating signal fed at the
emitter section causes the bias to increase or decrease in accordance with the modulating signal. C3 is bypass
capacitor for carrier. Thus the carrier signal applied at the base gets amplified more when the amplitude of
the modulating signal is at its maximum and less when the amplitude of the modulating signal is small. C 2
couples the modulated signal to output of the modulator. Demodulation involves two operations.
i)
ii)
The diode in the circuit diagram of demodulator does half wave rectification. The rectified signal is applied
to a low pass filter to extract the modulating signal.
CIRCUIT DIAGRAM:
VCC = 12V
Dept. of ECE
R3
C2
Output
C1
RF Input
L1 100 H
BC 107
R2
E
R4
C3
AF Input
Demodulator:
D1
AM Input
AF Output
OA 79
1K
1nF
EXPECTED WAVEFORMS:-
PROCEDURE:
1. Made the connections according to Circuit Diagram.
2. Measure and note down the frequency & amplitude (p-p) of the fixed carrier signal.
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RESULT:
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QUESTIONS
Tm
Turbomachinery Institute Science & Technology
Dept. of ECE
1.
AM is Defined as ____________
2.
4.
5.
6.
7.
8.
9.
AM Demodulator is ___________
10.
11.
12.
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Date:
DIODE DETECTOR CHARACTERISTICS
Component
Transistor
Resistors
Capacitors
4
5
6
7
8
9
10
Diode
Functions Generator
Regulated Power Supply
Bread Board
Cathode Ray Oscilloscope
Potentiometer
Connecting Wires
THEORY:
Specification
BC107
100K
4.7K
270
33
4.7F
0.001F
Variable
0A79
1Mhz
(0-30)V
0-20MHz
100K
Single Strand
Quantity
2
2
2
1
1
2
1
2
1
2
1
1
1
1
As Required
(ii)
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EXPECTED WAVEFORMS:-
AM Modulated signal
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approximately 300mv p-p of 1MHz frequency and AF voltage is 10V p-p 2KHz frequency.
3. Now connect the modulator output to the simple diode detector input.
4. Observe the AF signal at the output to the simple diode detector at approximately 50%
modulation using CRO.
5. Compare it with the original AF and observe that the detected signal is same as the AF signal
applied. Thus no information is lost in the process of modulation. (Note: Only wave shape and
frequency will be same, amplitude will be attenuated and phase may change)
6. To observe AM wave at different frequencies, connect AF signal from external signal generator
to the input of modulator and observe demodulated wave at different frequencies.
7. Repeat the experiment using practical diode detector circuit.
PRECAUTIONS:
RESULT:
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QUESTIONS
Tm
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ECE
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Date:
FREQUENCY MODULATION AND DEMODULATION
AIM: To study the functioning of frequency modulation & demodulation and to calculate the modulation
index.
APPARATUS REQUIRED
S. No.
1
Component
ICs
Resistors
Capacitors
4
5
6
7
8
Functions Generator
Regulated Power Supply
Bread Board
Cathode Ray Oscilloscope
Connecting Wires
Specification
XR 2206, LM 565
100K, 4.7K
10K
220
47K
0.01F,0.001F
0. 1F, 470pF, 0.01pF
1F,10F
1Mhz
(0-12)V
0-20MHz
Single Strand
Quantity
1 Each
2 Each
1
1
1
2
1
1
2
1
1
1
As Required
THEORY:
1. AF Generator:
This is an op-amp placed wein bridge oscillator. A FET input quad Op-Amp (ICTL084) is used here
to generate low frequency signals of 500 Hz and 5KHz to use as modulating signal. In this
experiment, a switch is provided to change the frequency. Required amplification is provided to
avoid loading effect.
2. Regulated power supply:
This consists of bridge rectifier, capacitor filters and three terminal regulators to provide required dc
voltages in the circuit i.e. +15 V, -15 V, +5V .
3. Modulator:
This has been developed using XR-2206 integrated circuit. The IC XR-2206 is a monolithic
Function generator; the output waveforms can be both amplitude and frequency modulated by an
external voltage. Frequency of operation can be selected externally over a range of 0.01 MHz. The
circuit is ideally suited for communications, instrumentations and function generator applications
requiring
CIRCUIT DIAGRAM:
Frequency Modulation Circuit Diagram:
+12v
Turbomachinery Institute Science & Technology
ECE
4.7K
19
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100K
47K
AF I/F
10 F/63v
FM O/P
2
10K
4.7
K
0.1 F
XR 2206
14
220
5
0.01 F
6
13
1
10
1 F/63v
10
10KPF
0.1 F
AF Input
8
680
1KPF
LM 565
7
AF Output
4
3
5
1
680
470PF
5V
sinusoidal tone, AM, FM or FSK generation. In this experiment, IC XC-2206 is connected to
generate sine wave, which is used as a carrier signal. The amplitude of carrier signal is 5vPP of 100
KHz frequencies.
4. Demodulator:
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change of frequency f.
This error voltage pulls the VCO to the new point. This error voltage will be the demodulated
version of the frequency modulated input signal.
PROCEDURE:
1. Connect the circuit as per the given circuit diagram.
2. Switch on the power supply.
3. Measure the frequency of the carrier signal at the FM output terminal with input terminals open and plot
the same on graph.
4. Apply the modulating signal of 500HZ with 1Vp-p.
5. Trace the modulated wave on the C.R.O & plot the same on graph.
6. Find the modulation index by measuring minimum and maximum frequency deviations from the carrier
frequency using the CRO.
Mr =
7. Repeat the steps 5& 6 by changing the amplitude and /or frequency of the modulating Signal.
8. For demodulation apply the modulated signal as an input to demodulator circuit and compare the
demodulated signal with the input modulating signal & also draw the same on the graph.
NOTE: Note down all the input and output wave forms of the signals applied and obtained respectively.
PRECAUTIONS:
RESULT:
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QUESTIONS
1. Frequency modulation is defined as ____
2. The advantages of indirect method of FM generation are __
3. Modulation index and frequency deviation of FM is ____
4. The advantages of FM are ______
5. Narrow band FM is _____
6. Compare narrow band FM and wide band FM?
7. Differentiate FM and AM _____
8. FM wave can be converted into PM wave _____
9, State the principle of reactance tube modulator _____
10. The bandwidth of FM system is ___
11. The function of FM discriminator is_________
12. Ratio detector differ from foster-seely discriminator _____
13. Linear detector is ________
14. The drawbacks of slope detector are ___________
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E
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Date:
BALANCED MODULATOR
Component
Modulator IC
Resistors
Capacitors
5
6
7
8
9
Functions Generator
Regulated Power Supply
Bread Board
Cathode Ray Oscilloscope
Connecting Wires
Specification
MC1496
10K
4.7K
270
33
0.1F
0.001F
1Mhz
(0-30)V
0-20MHz
Single Strand
Quantity
2
3
2
1
1
4
1
2
1
1
1
As Required
THEORY:
1. RF Generator:
Colpitts oscillator using FET is used here to generate RF signal of approximately 100 KHz Frequency to use
as carrier signal in this experiment. Adjustments for Amplitude and Frequency are provided in panel for
ease of operation.
2. AF Generator:
Low Frequency signal of approximately 5KHz is generated using OP-AMP based Wein- Bridge oscillator.
IC TL 084 is used as an active component; TL 084 is FET input general purpose quad OP-AMP integrated
circuit. One of the OP-AMP has been used as amplifier to improve signal level. Facility is provided to
change output voltage.
3. Regulated Power Supply:
This consists of bridge rectifier, capacitor filters and three terminal regulators to provide required DC
voltage in the circuit i.e. +12v, -8v @ 150 MA each.
CIRCUIT DIAGRAM:
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4. Modulator:
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EXPECTED WAVEFORMS:
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RESULT:
QUESTIONS
1. The two ways of generating DSB_SC are ________
2. The applications of balanced modulator are ________
3. The advantages of suppressing the carrier ________
4. The advantages of balanced modulator __________
5. The advantages of Ring modulator __________
6. The expression for the output voltage of a balanced modulator is _________
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Date:
PRE-EMPHASIS & DE-EMPHASIS
Component
Resistors
Capacitors
3
4
5
6
7
Functions Generator
Regulated Power Supply
Bread Board
Cathode Ray Oscilloscope
Connecting Wires
Specification
1K
0.1F
0.001F
1Mhz
(0-30)V
0-20MHz
Single Strand
Quantity
2
2
1
2
1
1
1
As Required
THEORY:
Frequency modulation is much immune to noise than amplitude modulation and significantly more
immune than phase modulation. A single noise frequency will affect the output of the receiver only if it falls
with in its pass band.
The noise has a greater effect on the higher modulating frequencies than on lower ones. Thus, if the
higher frequencies were artificially boosted at the transmitter and correspondingly cut at the receiver,
improvement in noise immunity could be expected. This booting of the higher frequencies, in accordance
with a pre-arranged curve, is termed pre-emphasis, and the compensation at the receiver is called deemphasis. If the two modulating signals have the same initial amplitude, and one of them is pre-emphasized
to (say) twice this amplitude, whereas the other is unaffected (being at a much lower frequency) then the
receiver will naturally have to de-emphasize the first signal by a factor of 2, to ensure that both signals have
the same amplitude in the output of the receiver. Before demodulation, i.e. while susceptible to noise
interference the emphasized signal had twice the deviation it would have had without pre-emphasis, and was
thus more immune to noise. Alternatively, it is seen that when this signal is de-emphasized any noise
sideband voltages are de-emphasized with it, and therefore have a correspondingly lower amplitude than
they would have had without emphasis again their effect on the output is reduced. The amount of preemphasis in U.S FM broadcasting, and in the sound transmissions accompanying television, has been
standardized at 75 microseconds, whereas a number of other
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The usage of microseconds for defining emphasis is standard. 75 microseconds de-emphasis corresponds to
a frequency response curve that is 3 db down at the frequency whose time constant is RC is 75
microseconds. This frequency is given by f=1/2RC and it is therefore 2120 Hz; with 50-microseconds deemphasis it would have been 3180 Hz. Figure I shows pre emphasis and de-emphasis curves for a 7
microseconds emphasis, as used in the united states. If emphasis is applied to amplitude modulation, some
improvement will also result, but it is not as great as in FM because the highest modulating frequencies in
AM are no more affected by noise than any others.
Apart from that, it would be difficult to introduce pre-emphasis and de-emphasis in existing AM
services since extensive modifications would be needed, particularly in view of the huge numbers is
receivers in use.
PROCEDURE:
I-PRE-EMPHASIS
1. Connect the circuit as per the circuit diagram
2. Apply a sine wave to the input terminals of 2 VP-P (Vi)
3. By varying the input frequency with fixed amplitude, note down the output amplitude (Vo) with
respect to the input frequency.
4. Calculate the gain using the formula
Gain = 20 log (VO/ VI) db
Where VO = output voltage in volts.
VI = Input voltage in volts.
And plot the frequency response.
II-DE-EMPHASIS
1. Connect the circuit as per circuit diagram.
Repeat steps 2, 3 & 4 of Pre-Emphasis to de-emphasis also.
PRECAUTIONS:
RESULT:
Turbomachinery Institute Science & Technology
ECE
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Dept. of
ECE
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QUESTIONS
1. The need for pre-emphasis __________
2. The operation of pre-emphasis circuit __________
3. Pre-emphasis operation is similar to high pass filter explain how_______
4. De-emphasis operation is similar to low pass filter justify _______
5. De-emphasis is _______
6. Draw the frequency response of a pre-emphasis circuit_______
7. Draw the frequency response of a de-emphasis circuit ______
8. Give the formula for the cutoff frequency of the pre-emphasis circuit ______
9. The significance of the 3db down frequency is ______
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Date:
CHARACTERISTICS OF MIXER
Component
Resistors
Capacitors
3
4
5
6
7
8
Transistor
Functions Generator
Regulated Power Supply
Bread Board
Cathode Ray Oscilloscope
Connecting Wires
Specification
4.7K, 22K,3.3K
27K, 6.2K
10K
0.1F, 50pF
0.1F
100pF
BC 547
1Mhz
(0-30)V
0-20MHz
Single Strand
Quantity
1
3
3
2
2
4
1
2
1
1
1
As Required
THEORY:
The fundamental principle of modulation involves the mixing or multiplying of a low frequency
signal with a higher frequency signal such as an AM or FM carrier. This enables the information contained
in the low frequency signal to be transmitted through space as high frequency electromagnetic waves.
Commercial radios use as intermediate frequency (455 khz for commercial AM). These if carrier contain all
the information available to the receiver but in order to obtain this information mixing must take place to
obtain the lower frequency signals riding onto IF or RF frequency. This principle is also used in mixing
low frequency signals up to IF or RF frequencies. The purpose of this experiment is to observe the effect of
mixing two frequencies using a nonlinear single transistor mixer and to demonstrate the use of a band stop
and a 2-pole low pass filter in a practical application.
BLOCK DIAGRAM DESCRIPTION:
1. RF Generator 1:
Colpitts oscillator using FET is used here to generate RF signal of approximately 500 KHz to
use as local oscillator in this experiment. Adjustments for amplitude and Frequency are
provided on panel for ease of operation.
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50pF
3.3K
22K
4.7K
27K
6.2K
6.2K
27K
Output
C
0.1 F
B
BC 547
100pF
100pF
100pF
VX
E
0.1 F
VY
10K
10K
10K
EXPECTED WAVEFORMS:
2. RF Generator 2:
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TABULAR COLOUMN:
S.NO
Inputs
Vx sin (2fxt)
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Filter Output
Vy sin (2fyt)
49
Vo sin (2fot)
Dept. of
RESULT:
QUESTIONS:
1. The need for a frequency mixer is _________
2. Heterodyning is
3. ___________filter is used at the o/p of transistor circuit in a frequency mixer?
4. The frequency components that appear at the collector of the transistor in the mixer circuit are __
5. The transistor operated in the nonlinear region in a frequency mixer because ________________
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