Analog comm lab manual

Analog Communications Lab
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Contents
Contents.........................................................................................................................................2
1. Amplitude Modulation and Demodulation.............................................................................4
2.Frequency Modulation and Demodulation............................................................................13
3.Balanced Modulator................................................................................................................21
4.Pre-emphasis and De-emphasis..............................................................................................27
5.Characteristics of mixer..........................................................................................................34
6.Phase detection and Measurement using PLL......................................................................39
7.Synchronous Detector..............................................................................................................45
8.SSB System...............................................................................................................................49
MATLAB PROGRAMS.............................................................................................................52
1.Amplitude Modulation............................................................................................................53
2.Demodulation of AM wave using Hilbert transform............................................................57
3.Demodulation using Diode detector.......................................................................................59
4.DSBSC Modulation..................................................................................................................61
5.DSBSC Demodulation.............................................................................................................63
6.Generation DSBSC using Balanced Modulator....................................................................65
7.SSBSC Modulation..................................................................................................................67
8.Demodulation of SSBSC..........................................................................................................69
9.Frequency Spectrum of Amplitude Modulated Wave..........................................................71
10.Frequency Spectrum of Amplitude Modulated Wave........................................................73
Output Waveform.......................................................................................................................73
11.Frequency Spectrum of SSBSC............................................................................................75
12.Performing Pre-emphasis and De-emphasis........................................................................77
13.Frequency Modulation..........................................................................................................79
14.Demodulation of FM wave....................................................................................................81
15.Frequency Spectrum of FM..................................................................................................83
16.Construction of SQUELCH circuit......................................................................................85

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1. Amplitude Modulation and Demodulation

Aim: To study amplitude modulation and demodulation and to calculate modulation index by
changing the modulating signal’s amplitude.

Equipment required:
Analog Communication Kit, CRO, Connecting wires.

Components required:
1. Transistors: BC107 (2 nos.)
2. Diodes: A79 (1 no.)
3. Resistors: 8 K-1 No, 10K-(2nos), 100 K-(4nos)
4. Capacitors:0.1µF-(2nos),-0.0015µ-(1no), 1KpF-(2nos),22µF-(1no),0.01 µF-(1no)
5. Inductors -2.5mH-(1no)

Theory:
In Amplitude modulation of a carrier signal is varied by the modulating voltage whose
frequency is invariably, lower than that of the carrier frequency. In practice, the carrier
frequency may be high frequency (HF), while the modulating frequency is audio frequency.
Formally AM is defined as a system of modulation in which the amplitude of the carrier signal
is made proportional to the instantaneous amplitude of the modulating voltage.
Let the carrier voltage and the modulating voltage, Vc and Vm respectively be represented by
Vc = VcSinωct
Vm= VmSinωmt
Note that phase angle has been ignored in both expressions since it is unchanged by the
amplitude modulation process. Its conclusion here would merely complicate the preceding
without affecting the result. However, it will certainly not be possible to ignore phase angle
when we deal with frequency and phase modulation.
From the definition of AM, it follows that the (max amplitude V c of the unmodulated
carrier will have to be made proportional to the instantaneous modulating voltage V mSinωmt
when the carrier is amplitude modulated.
Frequency spectrum of the AM wave:
We shall show mathematically that the frequencies present in the AM wave are the
carrier frequency and the first pair of side band frequencies, where a sideband frequency is
defined as
fSB = fc ±n fm and in first pair n=1
When a carrier signal is amplitude modulated, the proportionality constant is made to unity, and
the instantaneous modulating voltage variations are superimposed onto carrier amplitude. Thus,
when there is temporarily no modulation, the amplitude of the carrier is equal to its unmodulated
value. When modulation is present, the amplitude of the carrier is varied by its instantaneous
value. This situation is illustrated in fig 1, which shows how the max amplitude of the
amplitude- modulated voltage is made to vary in accordance with modulating voltage
changes.Fig1 also shows that something unusual (distortion, as it happens) will occur if Vm is

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greater than Vc.This and the fact that the ratio Vm/Vc often occurs, leads to the following
definition of modulation index.
m=Vm/Vc → Eq. 1

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The modulation index is a number lying between 0 and 1 it is very often expressed as a
percentage and called the percentage modulation.
It is possible to write an equation for the amplitude of the amplitude-modulated voltages, thus
we have
A = Vc+Vm = Vc+VmSinωmt+Vc+ VcSinωct
A=Vo(1+m Sinωmt)
The instantaneous value of the resulting amplitude modulated wave is
V =A Sinωct = Vc(1+m Sinωmt) Sinωct →Eq. 2
Eq 2 may be expanded by means of the trignometrical relation
Sin x Sin y = ½ {Cos(x-y) – Cos(x+y)} to give
V = VcSinωct + mVc/2Cos (ωct-ωmt) - mVc/2Cos (ωct-ωmt) → Eq. 3
It has been shown that the eq of an amplitude modulated wave contains three terms. The
first term is identical to eq1 and represents the unmodulated carrier. It is thus apparent that the
process of amplitude modulation has the effect of loading to the unmodulated wave , rather than
changing it .The two additional terms produced are the two side bands out end . The frequency
of the lower side band (LSB) is fc-fm and the frequency of the upper side band (USB) is f c+fm .
The very important conclusion to be made at this stage is that the band width required for
amplitude modulation is twice the frequency of the modulating signal. In modulation by several
sine waves simultaneously, as in the AM broadcasting service, the band width required is twice
the highest modulated frequency.

Representation of AM
Amplitude modulation may be represented in any of three waves depending on the point of
view . Fig 2 shows the frequency spectrum and so illustrates equation 3 . AM is shown simply as
consisting three discrete frequencies , of these the central frequency , i.e. the carrier has the
highest amplitude , and other two are disposed symmetrically about it , having amplitude which
are equal to each other , but which can never exceed half the carrier amplitude.
The appearance of the amplitude modulated wave is of great interest, and it is shown in fig2 for
one cycle of the modulating sine wave. It is derived from fig1 , which showed an amplitude , or
what may now be called the top envelope , is given by the relation A = (V c + VmSinωmt )
.Similarly , the maximum negative amplitude , or bottom envelope , is given by - A = - (V c +
VmSinωmt ). The modulated wave extends between these two limiting envelopes and has a
repretion rate equal to the unmodulated carrier frequency.
It will be recalled that Vm= mVc and it is now possible to use this relation to calculate the index
(or depth ) of modulation from the wave form of fig 2 as follows.
Vm = (Vmax – Vmin)/2
Vc = Vmax – Vm = Vmax – ( (Vmax – Vmin)/2 ) = (Vmax – Vmin) / 2
M = Vm / Vc = ( (Vmax – Vmin) / 2) / ( (Vmax – Vmin) / 2)
M=(Vmax – Vmin) / (Vmax + Vmin)

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Circuit Diagram :
AM Modulator

AM Demodulator

Procedure:
1) Connect the circuit as shown in the circuit diagram.
2) Apply the 100 KHz carrier signal and amplitude of 6V(p-p) to the input of AM modulator at
100 KΩ pot and 1 KHz of modulating signal to the AM modulator at 100 KΩ pot as shown in
the circuit diagram.
3) Apply the power supply of 12V as shown in the circuit diagram.
4) Observe the amplitude modulated wave synchronization with the modulating signal on a
dual trace CRO following figure shown the connections.

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5) Adjust the 10 KΩ linear pot for carrier suppression and 100KΩ linear pot for proper
modulation.i.e. 100%.
6) Now by varying the amplitude of the modulating signal , the depth of modulation varies.
Calculate the maxima and minima points of modulated wave on the CRO and calculate the
depth of modulation using formula.

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Waveforms:

Result:
Amplitude modulated signal is generated and original signal is demodulated from AM
signal Depth of modulation is calculated for various amplitude levels of modulating signals.

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Attach one Graph Sheet here

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2.Frequency Modulation and Demodulation
Aim:
To study the process of frequency modulation and demodulation and to calculate the
depth of modulation by varying the modulating voltage.

Equipment Required:
Analog communication kit , CRO, connecting wires.

Components Required:

1. IC’s:8083,LM-565,TL08IC(OR)TL084
2. Resistors:4.7kΩ-(2nos),10kΩ-(5nos),15kΩ-(1nos),30kΩ-(1nos),300kΩ-(2nos)
3. Capacitors:0.1µF-(2nos),0.001µF-(3nos),0.0022µF-(1no)

Theory:
The general equation of an unmodulated wave, or carrier, may be written as X=A sin (ωt+ф)
Where X = instantaneous value of voltage or current.
A = (maximum) amplitude
ω = angular velocity (rad/s)
ф = phase angle, rad
Note that we represent an angle in radians.
If any one of these three parameters is varied in accordance with another signal, normally of a
lower frequency, then the second signal is called the modulation and the first is said to
modulated by the second. In the frequency modulation, frequency of the carrier is made to vary.
For simplicity, it is again assumed that the modulation, signal is sinusoidal. This signal has two
important parameters which must be represented by the modulation process without distortion:
namely, its amplitude and frequency. It is assumed that the phase relations of a complex
modulation signal will be preserved. By the definition of frequency modulation, the amount by
which the carrier frequency is varied from its unmodulated value, called the deviation, is made
proportional to the instantaneous value of the modulating voltage. The rate at which this
frequency variations or oscillations takes place is naturally equal to the modulating frequency.
The situation is illustrated in fig 1, which shows the modulating voltage and the resulting
frequency-modulate wave. Fig -1 also shows the frequency variation with time which is seen to
be identical to the variation with time of the modulating voltage. As an example of FM, all
signals having the same amplitude will deviate the frequency by the same amount, says 45 kHz,
no matters what their frequencies. Similarly, all signals of the same frequency, says 2 kHz, will
deviate the carrier at the same rate of 2000 times per second, no matter what their individual
amplitudes. The amplitudes of the frequency modulated wave remains constants at all times, this
is , in fact the greatest signal advantage of FM.

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Fig 1:

Mathematical Representation of FM:
The instantaneous frequency of the frequency modulated wave is given by
f = fc (1+k Vmcosωmt)
Where fc = unmodulated (or average carrier frequency)
K = proportional constant
Vmcosωmt = instantaneous modulating voltage (cosine being preferred for simplicity in
calculations)
The maximum deviation for this particular signal will occur when the cosine term has its
maximum value, that is +(or)-1. under these conditions, the instantaneous frequency will be
f =fc(1+KVm) so that the maximum deviation
δ=KVmfc
The instantaneous amplitude of the FM signal will be given by a formula of the form
V = A sin {F ωc ωm} = A sin ф equation1
Where F (ωc ωm), is some function, as yet undetermined, of the carrier and modulating
frequencies. This function represents an angle and will be called for convenience. The problem
now is to determine the instantaneous value for this angle.

Fig -2

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As fig-2 shows, ф is the angle traced out by the vector A in time t.If A were rotating with
a constant angular velocity, say p.This angle ф would be given by P t in radians.In this
instance,however the angular velocity is anything but constant. It is, in fact, governed by the
formula for ω=ωc(1+kVmcos ωmt).in order to find f, ω must be integrated with respect to time.
Thus
θ = ∫ωdt= ωc (1+KVmcosωmt)dt
θ = ωc∫(1+KVmcosωmt)dt
θ = ωct+ KVmsinωmt/ωm
θ = ωct+ KVmsinωmt/fm
θ = ωct+ δ /fmsinωmt
The deviation utilized, in turn the fact that ωc is constanst,the formula.
∫cosnx dx = (sinnx/n), which had shown that K Vm f c = δ.Equation 1 may now be
subdivided to give the instantaneous value of the FM voltage, thus
V = A Sin(ωct+δ/fm Sinωct)
Mf= max.frequency deviation / modulating frequency= δ/fm
V = A Sin (ωct+mf Sinωct)
It is important to note as the modulated frequency decrease and the modulating voltage
amplitude i.e. δ remains constant, the modulation index increase. This will be basis for
distinguishing frequency modulation from phase modulation. Note also that m f,which is the ratio
of two frequencies is measured in radians.

Circuit description:
In this kit, frequency modulation is generated by using IC
8038.The frequency of the waveform generator is a direct function of the DC voltage at terminal
8 (measured from V+).By altering this voltage, frequency modulation is performed. For small
deviations (e.g.+ (or) – 10%) the modulating signal can be applied directly to pin 8,merely
providing DC decoupling with a capacitor as in Fig 3.
An external resistor between pins 7 and 8 is not necessary, but it can be used to increase input
impedance from about 8KΩ (pin 7 and 8 connected together) to about (R=8KΩ).

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For larger FM deviations or for frequency sweeping, the modulating signal is applied between
the positive supply voltage at pin 8.
During the demodulation,FM output is given to a phase lock loop(565).We have seen that,
during lock, the average dc level of the phase comparator output is directly proportional to the
frequency of the input signal.
As the frequency shifts, it is the output which causes the VCO to shift and key tracking. In other
words,the phase comparator output is an exact replica of the original modulating audio signal
.Fig-4 shows connections of 565 as FM Demodulator. The component values shown are for a
carrier
frequency of 82KHZ approx. The demodulated output is followed by a three stage filter to
remove Rf component. A small capacitor of 0.01µF as connected between pins 7 and 8 to
eliminate possible oscillations in the current source.

Circuit Diagram:

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Procedure:
Step 1: connect the circuit as shown in the circuit diagram.
Step 2: check the circuit properly and apply the power supply to the circuit
Step 3: observe the carrier signal from the FM modulator at pin 2 of the
IC 8083, which is 82KHZ.
Step 4: apply the modulating signal frequency of 4KHZ, 6Volts(p-p) from the function generator
to the FM input at pin 8 as shown in the figure below.
Fig:

Step 5: Trigger CRO with respect to CH1.adjust amplitude of the modulating signal until we get
undistorted FM output. It is difficult to trigger FM on analog CRO.That is why you adjust
modulating signal amplitude until small distortions notified in Fm output.

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Step 6: Calculate maximum frequency and minimum frequency from the FM output and
calculate modulating index using table shown in below

Demodulation fig:
FM Demodulation

STEP7:Connect the FM output to the demodulator input as shown in figure.

Step8 :In this condition decrease the amplitude if the modulating signal generator until we get
undistorted demodulated output(1 Vp-p).

Observations:
S.NO AMP fmax fmin(Hz) ∆f(Hz)
(p-pV) fc+∆f fc+∆f β=

Result:
The frequency modulated signal is generated and original signal is demodulated and modulation
index is calculated for Fm signal

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Attach one normal graph sheet here

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3.Balanced Modulator
Aim:
1. To construct and properly adjust a balanced modulator and study its operation.
2. To observe that the output is a double sideband with a suppressed carrier signal.
3. To adjust it for optimum carrier suppression.
4. To verify that the output audio level directly affects the double side-band output
amplitude.
5. To observe that the output is maximum with zero audio input.
6. To measure carrier only output and the peak sideband output and to calculate the carrier
suppression.
Equipment Required:
Analog communication kit, CRO connecting wires.
Component Required:
1. IC’S:1496- (1no)
2. Resistors:100Ω-(1no),820Ω-(1no),1kΩ-(1no)1.2kΩ-(1no),2.7kΩ-(1no),10kΩ-(3nos),47kΩ-
(1nos),5OkΩ-POT
3. Capacitors: 1µF-(1no),22µF-(1no),100µF-(2nos)
Theory:
One circuit that lends itself extremely well to balanced modulator application is the differential
amplifier. A simplified diagram of a differential amplifier is shown in figure 1.

FIG 1: A Differential amplifier used as a balanced modulator

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Q3 acts as the current source for Q1 and Q2 .If the RF input is applied to the bases of Q1 and Q2
in phase, current through both transistors will be identical and the voltage difference across the
output will be zero. This is the common mode rejection of the differential amplifier and it has
balanced modulator out the carrier.
The audio input is applied to the base of Q3.This upsets the circuits balance. As a result, the
audio mixing and RF signals are mixed across Q1 and Q2 .This is non linear mixing and,
therefore, side bands appear at the output. However, the carrier or RF input does not. Since it is
a common mode signal, it is rejected.
A differential amplifier must be constructed of transistors whose characteristics are very closely
matched. Forming the transistors on a single chip of silicon as done with ICs ensure this
necessary match.
Therefore, the differential amplifier is ideally suited to integrated circuit construction.

Fig 2

IC Balanced Modulators:
Fig-2 shows IC that has been specifically designed for use as balanced modulators. Fig-2 is the
1496 balanced modulator, which is manufactured by Motorola, National and Signetics. This
device uses a differential amplifier configuration similar to what was previously described. Its
carrier suppression is rated at a minimum of -5db with a typical value of -65db at 500 kHz.

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Step-By–Step Procedure:
A balanced modulator using a 1496 integrated circuit .You will verify that it does suppress the
carrier and also adjust it for optimum carrier suppression.
1. Balanced modulation circuit is designed using 1496 integrated circuit, and connect balanced
modulator circuit as per the circuit diagram.
2. Apply the power supply =12 to the pin No.8 of IC 1496.
3. Give the Audio frequency of 200 Hz sine wave from Function Generator 1 and 100KHZ Sine
wave form from the Function Generator 2 which are provided on the Analog Communication
Laboratory Kit as a inputs. Adjust R1 (1k Ω linear pot) and R2 (50k Ω linear pot) for mid –
range which are provided on the Analog Communication Kit. Connect your oscilloscope to the
output and set the vertical input control to 1v/cm and the sweep to 1ms.cm.
4. Check the circuit diagram properly and switch on the Analog Communication Kit. Adjust
Oscilloscope’s trigger control for a stable display. You may use square as carrier wave and
observe the output waveform.
5. Vary R1 (1K) both clockwise, counter clockwise, what effect does it have on the output
=_____________
6.Disconnect the modulating wave input R1 (1KΩ) .The output should now be close to zero. Set
your Oscilloscope vertical input to 0.2 v/cm. Now adjust R2 for minimum output. If possible,
increase the Oscilloscope’s vertical input sensitivity to measure the output voltage.
EOUT CARRIER ONLY =___________________
7. Set the vertical input control to 1v/cm. Connect the sine input to R1(1K) and adjust R1 for
maximum output without producing clipping. Measure the peak sideband output voltage.
EPK SIDEBANDS =____________________
8.Calculate the carrier suppression in DB.

DB =20LOG

DB =______________________
9.Turn off your experiment and disconnected your circuit.
OBSERVATIONS:
1. AF signal frequency= 200 Hz.
2. RF signal frequency = 100 KHZ.
3. Varying R1 ↑, DSBC amplitude (p-p) ↑ proportionally.
4. After disconnecting Sine input to R1.
5. Eout carrier only = 20 mV (p-p).
6. Epk sidebands = 2.4V (p-p).
7. Carrier suppression in db = 20 log EPK Sideband / Eout carrier only =41.5

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Waveforms:

Input and output waveforms of balanced modulator
Result:
1. The Double side band suppressed carrier signal is obtained (Balanced output i.e.,100%
modulation is obtained).
2. Carrier Suppression in DB is calculated
i.e.,41.5db

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Attach normal graph sheet here

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4.Pre-emphasis and De-emphasis
Aim: To study the frequency response of the Pre-emphasis and De-emphasis circuits and draw
the graphs.

Equipment Required:
Analog communication kit, C.R.O, connecting wires

1. Resistors: 47Ω, 100Ω, 1kΩ, 50k Ω-pot
2. Capacitors: 0.1µf, 0.01µf, 0.001µf, 0.047µf

Theory:
Frequency modulation is much more to noise than amplitude modulation and is
significantly more immune than phase modulation. A single noise frequency will effect the out
put of a receiver only if it falls with in its pass band .The carrier and noise voltages will mix and
if the difference is audible, it will naturally interfere with the reception of wanted signals.
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 boosting of higher
modulating frequencies, in accordance with a prearranged curve, is termed pre-emphasis, and
the compensation at the receiver is called de-emphasis.
If two modulating signals have the same initial amplitude, and one of them is pre-
emphasized to (say) twice this amplitude, where as the other is unaffected (being at a much
lower frequency), then the receiver will naturally have to do 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. Side band voltages are de-emphasized with it, and therefore have correspondingly lower
amplitude than they would have had without emphasis again their effect on the output is
reduced.
The amount of pre emphasis U.S FM broadcasting, and in the sound transmission
accompanying television, has been standardized at 75 micro seconds, whereas a number of the
other services, notably CCIR and Australian TV sound transmission, use 50 micro seconds. The
usage of microseconds for defining emphasis is standard. A 75 microsecond’s de-emphasis
Corresponds to a frequency response curve that is 3 db down at the frequency whose
time constant RC is 75 microseconds. This frequency is given by f=1/2ΠRC and is therefore
2120HZ: with 50-micro seconds de-emphasis it would have been 3180HZ.
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 of receivers in use.

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Circuit Diagrams:

Procedure:
1. Construct the circuit as shown in the circuit diagram.
2. Observe the
3. I/P waveform on the CRO in channel 1.
4. Adjust the amplitude of the sine wave using the amplitude knob to a particular voltage, say
4V or 6V or 10V etc.
5. Measure the O/P voltage in the CRO and note down in the observation table.
6. Calculate the attenuation and Log f values as shown in the observation table.
7. Draw the graph frequency (X-axis) and attenuation in db (Y-axis) to show the emphasis
curves on a semi log graph.
8. Various values of R and C are available so that the time constant in suitably selected
depending upon the application.
Pre-Emphasis
i/p voltage=4volts

Frequency Output Log f Attenuation (db)
(Hz) (volts) 20 log eo/ei

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De-Emphasis
I/p Voltage=10volts
Frequency in Hz Output in volts Logf Attenuation in db
20 log eo/ei

Figure 1Frequency response of Pre-emphasis &De-emphasis

Result: The frequency response of Pre-emphasis and De-emphasis circuits obtained.

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Attach one semilog graph Sheet here

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5.Characteristics of mixer
Aim:
To measure the following parameters of mixer of the signal band (Medium wave: 550 kHz to
1.5MHz) Radio Receiver.
1 Conversion gain
2 Image suppression (or) rejection
3 Sensitivity of the mixer
Equipment Required:
1 AM-FM signal generator (AP-LAB)
2 Audio signal generator
3 Oscilloscope
4 Radio Receiver Dynamic demonstrator
5 Wires for subsystems interconnection
6 Spectrum analyzers.
Theory:
A Frequency changer(mixer or converter) is a nonlinear resistance having two sets of
input frequency (FR and LO) and one set of output (IF)a mixer is used to shift the input received
RF signal (Frequency band of 550kHZ to 1.5KHz) to low frequency Intermediate Frequency
(IF) of 455kHz.The block diagram of the mixer is as follows.
Fig:

Image frequency and its rejection:
The frequency of the local oscillator f0 in majority of receivers, is kept higher than the
signal frequency fs and is given by f0 = fs + fi (where fi is the intermediate frequency) of fs = f0-fi.
When fs and f0 are mixed together in the mixer, they produce the difference frequency f i and it is
the only frequency that is amplified by the IF stage.
If still higher frequency fsi=f0+fi manages to reach the mixer, then this will also produce f i
when mixer with f0. This spurious If signal will also be amplified by the IF stage, and will
therefore produce interference. This results in two stations being received simultaneously, which
is undesirable fsi is called the image frequency and is defined by
fsi=fs+2fi
The rejection of this image frequency by a single tuned circuit is measured by ‘α’ the rejection
ratio. it is the ratio of gain at the signal frequency to the gain at the image frequency and given
by α =√1+q²ρ²
Where ρ=(fsi/fs)-(fs/fsi)

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If the receiver has an RF stage, then there will be two tuned circuits tuned to fs,the rejection will
be calculated by the same formula and the total rejection will be the product of the two.
image rejection depends on the front end selectivity of the receiver and must be achieved before
the IF stage, because after the unwanted frequency enters the IF amplifier, it is impossible to
remove it from the unwanted signal. For short wave range ,an RF stage is essential for frequency
rejection.
RadioReceiverDynamicDemonstrator

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Procedure:
(i)Conversion Gain:
(a) set the RF signal level to about 100mv at a frequency of 100KHZ from an AM-FM
signal generator and noted as v1 volts.
(b) Apply the RF signal at the antenna input of the antenna stage.
(c) As transistor T1 acts both as LO stage with sufficient output power and mixer, the IF
output is selected by IF-T in IF amplifier stage.
(d) Tune the gang condenser to tune antenna stage and also tune the coiling mixer-Local
oscillator stage for maximum output at IF-T1 of IF stages.
(e) Measure the IF level(V2 volts) at the point of IF-T1 without connection to the base of
T2.If the level of IF is low at this point, measure the IF level at the output coil of IF-T3(without
connecting detector diode OA79) and subtract the IF amplifier gain to get the IF detected
level(V2 volts).
(f) The conversion gain/loss is calculated as 20logV2/V1db (if V2 is less than v1then it is
conversion loss).
(g) The level of V1 and V2 can be measured either with oscilloscope or spectrum analyzer.

(ii) Image Suppression/Rejection:
(a) As L.O frequency in radio receiver is above the RF signal by IF frequency (455
KHz), the image frequency is above L.O frequency by IF frequency.
Image frequency = L.O frequency + IF frequency.
(b) Apply the RF signal of 1000KHz,from AM-FM signal generator at a level of about
50 or 100mV at the antenna terminal of the antenna stag and note the level as V1.
(c) Measure the IF output level at IF-T1 of IF amplifier as V2.
(d) Calculate RF(or desired input) gain as 20logV2/V1db.
(e) Select the image signal (RF+2IF =1000+2*455=1910KHZ)from AM-FM signal
generator and repeat steps (b) and (c) above and measure v2(Im)
(f) Calculate Receiver gain/loss at image frequency, as
20log V2 (Im)/V1(Im) is a level of input signal of 1910KHz to the receiver at image
frequency. This value should be less than RF gain for better performance of the receiver

(iii) Sensitivity of The Receiver:
(a) Apply 1000 KHz RF signal at level of 100mV at the input of the receiver(either at antenna
input or base input of T1)
(b) Measure the (S/N)ratio of the audio output across the loud speaker
(c) Decrease the RF input level of AM-FM signal generator till (S/N)ratio of step(b) above,
decrease by 3db or till the unacceptable level of audio tone is heard in the speaker. The RF level
at this point is the sensitivity of the mixture and measured as db µV or dbmV.
Observations:
A. Conversion gain:
RF (100 KHz) input = 150Mv
(a)If output (at IF – T3) = 9.2V = A
(b)If input level (at IF – T1) at 455 KHz required to get the same IF output of 9.2 V = 150 mV
= B.
If B ≥A then conversion gain = 20 log (A/B)
If B <A then conversion gain = 20 log (A/B)

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As same levels are obtained in case (a) and (b) above, the conversion gain = 0 db.
B.Image suppression (or) rejection:
RF input = 50mV (at 1000KHz)
IF output voltage = 9V.
Image = 1000 + 910 = 1910KHz
IF output (at input of 910 KHz) = 0.2V.
g1 = RF to IF gain = 20 log (9V/50mV) = 45.1db
g2 = Image to IF gain = 20 log (0.2V/50mV) = 12db
Image rejection = (g1 – g2)db = 33.1db
C.Sensitivity of the mixer:
RF input level at 1000 KHz = 50mV.Measure AF output level impedance switch position of AF
power meter at 6Ω position.AF output level = 4vp-p (or) 0.35watts.
RF (100KHz) input level required for half power level of AF p-p measured above (2Vp-p or
0.175mW)= 2.5mV.
Sensitivity = 20log (2.5mV/1µV) = 67.95dbµV.
Result:
Characteristics of mixer are studied. Conversion gain, Image suppression and Sensitivity of
mixer are obtained.

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6.Phase detection and Measurement using PLL
Aim: To determine the phase difference in capture range using PLL.

Equipment Required:
Analog communication kit, CRO, connecting wires.

I.C’S:4046-(1no).
Resistors:560Ω-(1no),4k7Ω-(1no),100kΩ-(1no),27kΩ-(1no)
Capacitors: 0.1µ F-(1no), 0.01µF-(1no), 0.001µF-(1no), 4.7µF-1No.

Theory:
As same as the theory written in voltage Controlled Oscillator using PLL. (I.E., Experiment no:
2)
Circuit Diagram :

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Procedure:
Step 1:Connect the circuit as per the circuit diagram as shown in figure give the input at pin 14,
i.e. sine wave of 1KHZ and peak to peak of 6 volts . And observe the input sine wave in channel
1 of CRO.
Step 2:Now connect the pin 3 and pin 4 to channel 2 of CRO which is the output of the IC 4046.
And measure the output frequency on CRO (channel 2) which is equal to the input frequency
(i.e. 1KHZ).
Step 3:Now connect the pin -9 to ground of the 4046 with piece of wire, and record the resultant
output frequency of the phase locked loop.
F1 =____________Hz
This output frequency is the lower range of VCO which is determined by the 0.1µF capacitor
connected between pins 6 and 7 and 100KW resistor connected between pin 12 and ground.
Step 4:Now with same wire connect the pin 9 to +5 volts supply, record the output which is
higher than the one you measured in step 3.
Fh=_____________Hz
This frequency is the upper range of VCO which is determined by the 0.1µF capacitor
connected between pin 6 and 7 and the 560Ω resistor connected pins 11 and ground.
Step 5:Now remove the connection between pin 9 and the +5 volts supply and measure the
output which is the same as the input.

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Step 6:Now slowly increase the input frequency; you should observe that, the output
frequencies also increase. In fact, the output frequencies follow the changes of the input
frequency and should be exactly equal. Check the input frequency to confirm this.
Step 7:Continue to slowly increase the input frequency of PLL and stop when the output
frequency of the PLL does not continue to increase. Measure the input frequency and record
result Fin(H)=___________Hz. This is the upper range of VCO which is same as measured in
step4 .The phase locked loop then follows input frequency changes for frequencies below this
upper range.
Step 8:Now decrease the input frequency, At some point the output frequency will remain
constant. Measure the input frequency and record the result.
Fin (L) =____________Hz
This frequency is lower range of VCO and which is same as measured in step
3.Consequqntly the PLL circuit follow changes in the input frequency for any frequency
between the lower and upper range of the VCO. There for the loop is locked.
Fin(H)- Fin(L). Which is lock range, =_____________Hz
Step 9:Measure the frequency just below the lowest range of the lock range, which is termed as
capture range.
Step 10:In capture range observe and measure the phase difference between the i/p and o/p
waveforms at the same frequencies.
Δф =t1-t2/T *360°=_______
OBSERVATIONS:

RESULT:
Lock range=_________________
Capture range=_______________
Phase detection=______________

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7.Synchronous Detector
Aim:To obtain the demodulating (or) message signal using DSB or SSB synchronous or
coherent detector.

Equipment Required:
Analog communication kit , CRO , Connecting wires.

1. IC’s:1496–2No.
2.Resistors:100Ω-(2nos),10Ω-(1no),820Ω-(2nos)47KΩ-(1nos),1KΩ-(2nos),1K-
Pot(2nos),1.2KΩ(3nos),50KΩPot-(2nos),2.7KΩ-(2nos),7.3296KΩ-(1nos)
3. Capacitors: 0.047µF, 0.1mF, 22mF, 100mF.

Block Diagram:

Procedure:
1. A connection was made as per circuit diagram.
2. A balanced modulator circuit was connected for both Transmitter and Receiver.
3. Carrier signal was same and was applied to two balanced modulator circuits.
4. Apply the power +12V to pin 8 of both the circuits.
5. The output of the Receiver balanced modulator circuit was connected to low pass filter.
6. Low pass filter circuit was designed according to the requirement.
7. And the modulating signal was obtained at the LPF.

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Observations:
Transmitter Receiver
fc=15 KHz fc=15 KHz
fm=462Hz T=2.3msec.

After Filter:
T=2.3msec.
fm = 434.78 Hz.
Designing LPF
fm = 462HZ = 1/2∏RC
Let c = 0.047µF.
R= 1/( 2∏ ×0.047×10 -6× 462)
R = 7.3296K Ω

Result:
Thus the modulating signal obtained from the DSB detector.

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8.SSB System
Aim:
To generate a SSB signal using Balanced Modulator.

Equipment Required:
Analog communication kit, CRO, Connecting wires.
1. IC’s 1496–1No.
2. Resistors: 100Ω-1, 10Ω-1820Ω-1, 47KΩ-1, 1KΩ-1, 1KΩ-Pot-1, 2.7KΩ-1.
3. Capacitors: 0.047µF-1, 0.1mF-1, 22mF-1,100mF–1.

Block Diagram:

Procedure:
1. Obtain a DSB-SC signal using balanced modulator
2. Design a Band pass filter (BPF) for a given specifications to eliminate one of the side band
from DSB signal.
3. Connect the output of balanced modulator to the input of BPF.
4. Obtain the output from BPF to get SSB signal which is either LSB or USB.
5. Note down its frequency.
Wide Band BPF Circuit Diagram :

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Observations:
Design of BPF
We consider wide band BPF, Which is a cascade of HPF & LPF
fc=6.3 KHz fm = 200Hz
Let us consider USB/SSB = fc + fm = 6.5 Hz.

fL=6.3 KHz
fO=6.5KHz
fH=6.7 KHz

HPF LPF
fL= 1/2∏RC=6.3 kHz Fh= 1/2∏RC=6.7 KHz
Choose C=.047uf Let C=.01uf
RL=537 ohms Rh=2.37K ohms

Result:
Thus the SSB signal was generated using balanced modulator

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Attach semilog graph sheet here

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MATLAB PROGRAMS

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1.Amplitude Modulation
t=0:0.0001:0.02
msg=10*cos(2*pi*100*t)
carr=20*cos(2*pi*1000*t)
amw=(20+msg).*cos(2*pi*1000*t)
subplot(2,1,1)
plot(t,msg)
title('Message signal')
subplot(2,1,2)
plot(t,carr)
title('Carrier signal')
figure;
subplot(3,1,1)
plot(t,amw)
title('Under Modulation')
subplot(3,1,2)
plot(t,amw)
title('over modulation')
subplot(3,1,3)
plot(t,amw)
title('100 % modulation')

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OUTPUT WAVEFORMS

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2.Demodulation of AM wave using Hilbert transform
t=0:0.0001:0.02
fc=1000
Ec=7
Carr=Ec*sin(2*pi*fc*t)
fm=100
Em=7
Mod=Em*sin(2*pi*fm*t)
Am=(Ec+Mod).*(sin(2*pi*fc*t))
disp('Performing Amplitude Demodulation using Hilbert transform');
Am_hil=hilbert(Am)
Am_abs=abs(Am_hil)
Am_Demod=Am_abs-mean(Am_abs)
disp('plotting the results');
figure;subplot(4,1,1);plot(t,Mod);
title('Message Waveform');
%xlabel('Time(sec)');ylabel('Amplitude');
subplot(4,1,2);plot(t,Carr);title('carrier waveform');
%xlabel('Time(Sec)');ylabel('Amplitude');
subplot(4,1,3);plot(t,Am);title('amplitude modulated wave form');
subplot(4,1,4);
plot(t,Am_Demod);
title('Amplitude demodulated waveform');

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