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Exercise 1: Frequency and Phase Modulation
EXERCISE OBJECTIVE
When you have completed this exercise, you will be able to describe frequency modulation and an FM circuit. You will also be able to
describe phase modulation and a PM circuit. You will use an oscilloscope and a multimeter to make measurements.
DISCUSSION
As mentioned before, the carrier possesses the following parameters:
·
an amplitude,
·
a total phase (angle),
· an instantaneous frequency,
Any one of these parameters may be modulated by a message. Any one of these parameters is chosen, then:
·
its rate of variation should be directly proportional to the message signal frequency,
·
its amount of variation (shift) should be directly proportional to the message signal amplitude,
In angle modulation, the total phase is a function of message, that is:
There are two options:
Ø FREQUENCY MODULATION (FM) in which:
or
where:
-
or
is the (angular) frequency deviation constant.
Therefore, the total phase of a FM signal is:
, the next results are obtained:
In case of sine wave modulator signal,
·
where
is the maximum frequency deviation of the FM signal
and:
·
where:
is the modulation index.
Ø PHASE MODULATION (PM), in which:
and, consequently:
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In case of sine wave modulator signal,
, the next result are obtained:
Following are the three frequency modulation (FM) concepts you need to remember.
1. The carrier signal frequency deviates only with the message signal amplitude.
2. The message signal’s frequency does not deviate the carrier signal’s frequency but does affect the rate of deviation.
3. There are no desired amplitude variations of the FM carrier; only frequency deviations contain the message signal intelligence.
The maximum carrier frequency deviation (plus or minus) occurs when the message signal’s amplitude is at what value?
q 01
When the message signal is at its zero reference, the carrier frequency deviation is zero because the carrier is at its center frequency.
If the message signal’s amplitude is constant but the frequency increases (for example, from 2 kHz to 4 kHz), the amount of the carrier
signal’s frequency deviation does not change.
However, the same frequency deviation will occur 4000 times per second (4 kHz) instead of 2000 times per second (2 kHz).
Because FM amplitude variations do not contain any message signal intelligence, the FM carrier’s amplitude can be limited within
desired values.
Consequently, noise amplitude spikes are reduced by limiter circuit. Efficient class C amplifiers, which may affect amplitude but not
frequency, can be used in FM equipment.
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The percentage of modulation (%Mod) describes the extent of the carrier frequency deviations relative to a standardized value which
is: 75 kHz as specified by the Federal Communications Commission (FCC) for a 100% modulation.
The carrier signal’s center frequency is its unmodulated frequency: the frequency with no message signal (or zero amplitude).
When the carrier deviation is less than the
kHz (3/4 of 75 kHz) is 75% modulation.
, the percentage of modulation is also reduced. For example, a deviation of 56.25
What percentage of modulation (% Mod.) is a deviation of … kHz?
q 02
Like AM, the FM carrier signal sidebands contain the message signal intelligence. If the frequency deviation is held to a minimum, only
two FM sidebands are produced.
However, as the message signal amplitude increases, additional significant sidebands above and below the carrier center frequency are
produced.
The bandwidth required for an FM signal depends upon two factors: the message signal amplitude and the frequency.
A sideband pair are two sidebands that are spaced equally above and below the carrier’s center frequency. The energy contained in each
sideband pair decreases as the sideband pairs are removed from the center frequency.
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A point is reached at which a sideband pair contains so little energy that they can be disregarded.
The point is determined by the modulation index.
The FM modulation index (MI) is the ratio of the amount of carrier frequency deviation (fcd) to the message signal frequency (fm).
MI = β = fcd/fm
For example, if a 5 kHz message signal (fm) caused a carrier frequency deviation of ±10 kHz (fcd), the MI would be 2.
The table shows the significant sideband pairs for FM signals with modulation index up to 14.
The FM signal with a modulation index of 2 would have 4 significant pairs.
An FM signal with 4 significant pairs and modulated by 5 kHz message signal would have the sidebands spaced 5 kHz apart for 20 kHz
on each side of the FM center frequency.
If the message signal frequency is …. and the percentage of modulation is …, what is the FM
bandwidth?
q 03
In PM, carrier signal modulation occurs when the carrier signal changes its phase (and frequency) with the changes in the message
signal amplitude and frequency.
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The amount of phase shift is proportional to the message signal amplitude.
When the carrier’s phase changes (shifts), frequency deviations also occur.
The amount of frequency deviation is directly proportional to the rate and the total amount of phase shift.
The rate of phase shift is directly proportional to the message signal’s frequency.
Therefore, the amount of frequency deviation in PM is directly proportional to the message signal’s amplitude and frequency.
In FM, do changes in the message signal’s amplitude and frequency cause frequency deviations in the carrier signal?
q 04
In FM, only message signal amplitude changes cause carrier frequency deviations.
Advantage and Disadvantages of FM and PM
In addition to good signal-to-noise ratio in frequency and phase modulation, more efficient class C amplifiers can be used because
amplitude distortion does not affect the signal quality.
The PM and broadcast FM disadvantages are wide bandwidths and the necessity of line-of-sight signal propagation for FM frequencies.
PROCEDURE
Frequency Modulation (FM)
In this PROCEDURE section, you will frequency modulate a carrier signal, measure its parameters, and observe its characteristics.
From Analog Communications Directory, bring into the Editor Window, the next m files:
a_scope.m, a_multimeter.m, signal_gen.m, vco_lo.m, quadrature_detector.m.
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An FM signal is generated with the VCO-LO circuit block.
In this PROCEDURE section, you will place the vco_lo_jumper variable in the 452 kHz position. (Do not make this setting yet.)
The potentiometer knob on the VCO-LO circuit block adjusts the output amplitude.
To adjust the VCO-LO output frequency, you will adjust the NEGATIVE SUPPLY.
A simplified schematic of the VCO-LO circuit is shown.
The oscillator consists of two transistors that are connected in a cross-coupled oscillator configuration.
The oscillator’s frequency is determined by the tuning of the LC network.
You can tune the LC network by changing the value of the NEGATIVE SUPPLY voltage at the anode of varactor diode CR2.
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The value of the negative voltage affects the CR2 capacitance, which, in turn, affects the tuning of the LC network.
As the NEGATIVE SUPPLY voltage becomes more negative, VCO-LO’s output frequency increases.
At 0 Vdc, the output frequency is about 310 kHz.
At 10 Vdc, the output frequency is about 510 kHz.
A buffer is between the oscillator and the VCO-LO potentiometer, which you adjust to set the VCO-LO output amplitude.
At test point T, you can measure the dc voltage at CR2’s anode.
The sine wave message signal input is at (M), and it causes the voltage at CR2 to vary.
You can observe the FM signal at (FM) OUT.
1.
On the VCO-LO circuit block, select the 452 kHz position.
Adjust the NEGATIVE SUPPLY for - 4.0 Vdc and the VCO-LO output amplitude to 600 mVpkpk.
2.
Connect the oscilloscope channel 2 probe to (FM) OUT on VCO-LO.
Set channel 2 for 200 mV/DIV and sweep to 0.5 µs/DIV.
3.
Connect the MULTIMETER to T on the VCO-LO circuit block.
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4.
5.
Connect the oscilloscope channel 1 probe to T.
Set channel 1 for 2.0 V/DIV and the vertical mode to dc.
6.
Adjust the channel 1 and 2 oscilloscope traces so they appear as show.
The channel 1 dc trace, which indicates - 4.0 Vdc at T, should be on the second division line from the top of the oscilloscope display.
What is the value for the ch1vertical_position variable from the scope.m file?
q 05
7.
Accurately measure the period (T) between the peaks of the unmodulated FM carrier signal on channel 2.
8.
From the period (T), calculate the center frequency (fc) of the unmodulated FM carrier signal. You can also use MATH MENU
from scope. The result is around (kHz):
q 06
9.
You will vary the voltage to simulate the voltage change caused by the message signal, which will be a sine wave.
While observing your oscilloscope screen, adjust the NEGATIVE SUPPLY knob CW and then CCW so that the dc voltage on channel 1
varies by about ±1 Vdc.
On channel 2, does the FM carrier frequency change as the dc voltage on channel 1 varies?
q 07
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10.
You will determine the frequency deviation of the FM carrier when the message signal amplitude changes by 1 Vdc.
Adjust the NEGATIVE SUPPLY knob CW to change the voltage at T on the VCO-LO circuit block to -5.0 Vdc.
11.
Accurately measure the period (T) between the peaks of the modulated FM carrier signal on channel 2.
12.
From the period (T), calculate the frequency (f) of the modulated FM carrier signal. The result is around (kHz):
q 08
13.
The center frequency (fc) of the unmodulated FM carrier signal is around 455 kHz, for NEGATIVE SUPPLY equal to -4 Vdc, and the
frequency (f) of the modulated FM carrier signal is around 470 kHz, for NEGATIVE SUPPLY equal to -5 Vdc.
Calculate FM frequency deviation when the message signal’s amplitude changes by 1 Vdc.
q 09
The FM frequency deviation when the message signal’s amplitude changes by 1 Vdc. equals 15 kHz.
The frequency deviation constant is equal to:
q 10
14.
To return the carrier frequency to the 455 kHz center frequency, adjust the NEGATIVE SUPPLY knob CCW to change the voltage at T
on the VCO-LO circuit block from - 5.0 Vdc to - 4.0 Vdc.
15.
You will now observe the effect of a 2 Vpk-pk, 5 kHz message signal on the FM carrier frequency.
Connect the SIGNAL GENERATOR to the (M) input on the VCO-LO circuit block.
16.
Adjust the SIGNAL GENERATOR for a 2.0 Vpk-pk, 5 kHz sine wave. This adjustment is equivalent to varying the voltage at T ±1 V.
17.
Set the sweep to 0.05 ms/DIV. Observe channel 2, which shown an FM signal.
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When the message signal amplitude was -1 V, the voltage at T decreased to - 5 V and the frequency increased to 470 kHz.
When the message signal amplitude is 1 V, the frequency will decrease to:
q 11
18.
Calculate the modulation index (MI).
q 12
19.
With an MI of 3, select the number of significant sideband pairs (SSP) from the table.
(If the MI is not a whole number, use the next highest MI to select the number of SSPs)
20.
As shown, the sideband pairs are spaced evenly on each side of the center frequency and are spaced 5 kHz apart.
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Calculate the bandwidth (BW) of the FM signal.
q 13
CONCLUSION
• In FM, only changes in the message signal amplitude cause carrier frequency deviations.
• In PM, changes in the message signal amplitude and frequency cause carrier frequency deviations.
• In FM and PM, any variations in the carrier amplitude do not contain message intelligence; the amplitude can be limited to reduce
noise.
• The modulation index is the ratio of frequency deviation and message signal frequency.
• The number of significant sideband pairs of an FM signal increase with the modulation index (MI).
• The FM signal bandwidth equals the message signal frequency times the number of significant sideband pairs times two (2).
REVIEW QUESTIONS
1. In FM, the carrier frequency deviates with changes in what parameter(s) of the message signal?
q 14
2. In PM, the carrier frequency deviates with changes in what parameter(s) of the message signal?
q 15
3. Why can a limiter circuit be used to cut off the peaks and valleys of an FM or PM signal?
q 16
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4. A 4 kHz message signal causes an FM carrier signal to have a frequency deviation of ±20 kHz. What is the MI?
q 17
5. An FM carrier signal with an MI of 4 contains a 3 kHz message signal. Use the table to determine the bandwidth of the FM carrier.
q 18
Exercise 2: Demodulation (Quadrature Detector)
EXERCISE OBJECTIVE
When you have completed this exercise, you will be able to explain demodulation of an FM signal and describe the operation of a
quadrature detector. You will use an oscilloscope to make signal measurements.
DISCUSSION
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FM demodulators are referred to as discriminators or frequency detectors.
A quadrature detector is one of several circuits that can demodulate FM signals.
Other FM demodulator circuits include the Foster-Seeley discriminator, the ratio detector, the pulsecounting detector, and the phaselocked loop detector.
All of these circuits convert FM frequency variations into the amplitude and frequency of the message signal.
The frequency deviations caused by 1 kHz and 2 kHz message signals with equal amplitudes are shown.
The X-axis is the FM frequency deviation, and the Y-axis is time. The intersection of the X and Y axes is the center frequency of the
FM carrier, or 0 frequency deviation.
The time span on the Y-axis is 1 ms
The left plot shows the frequency deviations caused by a 1 kHz sine wave message signal.
The right plot shows the frequency deviations caused by a 2 kHz sine wave message signal.
Because the 1 kHz and 2 kHz message signals have equal amplitudes, the magnitudes of the frequency deviations are equal.
A 1 kHz message signal causes two maximum frequency deviations (one positive and one negative) per millisecond.
A 2 kHz message signal causes how many maximum frequency deviations per millisecond?
q 19
An FM discriminator converts the magnitude and rate of the frequency deviations into the amplitude and frequency of the message
signal.
The FM discriminator on the circuit board is a quadrature detector.
The QUADRATURE DETECTOR circuit block includes a PHASE SHIFTER/LIMITER, a PHASE DETECTOR, and a FILTER (see
the image below).
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At the quadrature detector’s input, the FM signal takes two paths.
One FM signal is input to a phase shifter/limiter.
The phase shifter/limiter converts frequency deviations into phase deviations at about a 90o phase shift.
Signals are in quadrature when they have a phase difference of 90o.
The original FM signal and the phase shifter/limiter are input to phase detector which is a balanced modulator.
The phase detector outputs a signal with a frequency that is twice the FM frequency and a dc voltage that varies with the phase
difference between the two inputs.
Because the phase difference between the phase detector’s input signals varies with the FM signal’s frequency deviations, the baseband
output voltage changes with the amplitude and frequency of what signal?
q 20
A low-pass filter at the phase detector’s output removes the high frequency signal and passes the varying dc output voltage as the
recovered message signal.
PROCEDURE
The following PROCEDURE on the quadrature detector is divided into two sections:
• Phase Shifter and Limiter
• Phase Detector and Filter
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Each section starts with an explanation of the signals that you will observe and the parameters that you will measure and calculate.
Phase Shifter and Limiter
In this PROCEDURE section, you will observe how a phase shifter changes the phase of an FM carrier signal and how a limiter reduces
the amplitude of the phase-shifted signal.
The phase shifter/limiter circuit is composed of an amplifier, an LC network and a limiter circuit.
The transfer function of the phase shifter is:
Calculate the frequency at which the relative phase difference between FM signal (Uin) and the phase shifter output signal (Uout) is 90o.
q 21
At the center frequency, fr, the relative phase difference between signals is 90o.
However, frequencies greater or less than fr are shifted less or more than 90o, respectively, from the original FM signal
Less than 90o degrees if the frequency is higher than the tuned circuit and greater than 90o if the frequency is lower than the tuned
circuit.
In reference to the original FM signal, frequency deviations on each side of the FM center frequency will be…
q 22
The phase-shifted FM signal is input to the limiter.
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The limiter has two diodes connected from the output to ground with their polarities reversed: anodes connect to cathodes.
The reversed-polarity diodes limit the output amplitude and minimize any AM modulation that the phase shifter may cause.
1.
On the VCO-LO circuit block, select the 452 kHz position.
Adjust the unmodulated FM carrier signal at (FM) output for 300 mVpk-pk.
Disconnect SIGNAL GENERATOR from the (M) input on the VCO-LO circuit block.
2.
Connect (FM) OUT on the VCO-LO circuit block to FM on the QUADRATURE DETECTOR circuit block.
3.
Connect the oscilloscope channel 1 probe to (FM) out on VCO-LO circuit block.
Set channel 1 vertical mode/coupling to ac and scale to 100 mV/DIV.
4.
5.
6.
Connect the oscilloscope channel 2 probe to the output of the PHASE SHIFTER/LIMITER on the QUADRATURE DETECTOR circuit
block. Set channel for 500 mv/DV.
Set sweep for 0.5 µs/DIV.
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Adjust the FM frequency by turning the NEGATIVE SUPPLY knob on the left side of the base unit until the waveform on channel 2
has a maximum amplitude (You can also use multimeter, as shown in the figure above).
What is the neg_supply value?
q 23
7.
If not, set the neg_supply variable to
Vdc.
When the output of the PHASE SHIFTER/LIMITER is maximum, the FM center frequency is equal to what frequency?
q 24
8.
The FM center frequency is equal to the LC network resonant frequency (fr), that is 450 kHz.
Find the sweep value so that one cycle (360o) of the unmodulated FM signal cover 8 horizontal divisions, as shown in the image below.
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q 25
If not, set the sweep to
µs/DIV. (1 / fr / horizontal_division_number).
What is the phase difference between the carrier signal on channel 1 and the PHASE SHIFTER/LIMITER output signal on channel 2.
q 26
9.
In the PHASE SHIFTER circuit what component causes a phase shift of 90o between the input and output signals?
(The signals are shown in the image below.)
Recall that the transfer function of the phase shifter is:
q 27
At the central frequency (fr), the transfer function is:
10.
Adjust the NEGATIVE SUPPLY knob CW (for example, by -2 V) and then CCW (for example, by +2 V) to vary the FM frequency.
Why did the phase difference between the input and output signals increase and decrease, while the amplitude always decreased?
q 28
Magnitude response and phase response of the PHASE SHIFTER circuit are shown in the image below.
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11.
If a message signal were applied to the FM carrier signal, would the PHASE SHIFTER/LIMITER output change phase with the
frequency deviations of the modulated FM signal?
q 29
12.
Why does the PHASE SHIFTER/LIMITER on channel 2 have flattened peaks and valleys?
q 30
13.
While observing the PHASE SHIFTER/LIMITER output signal (channel 2), reduce the amplitude of the FM input signal (channel 1) to
about 100 mVpk-pk and then return to 300 mVpk-pk
14.
When the input signal was reduced to 100 mVpk-pk, did the PHASE SHIFTER/LIMITER output signal (channel 2) become less or more
flattened?
q 31
Phase Detector and Filter
In this PROCEDURE section, you will observe how the phase detector and filter recover the message signal.
The original FM signal and the 90o phase-shifted FM from the limiter are combined in the phase detector.
The phase detector is a balanced modulator.
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A balanced modulator combines FM input frequencies to produce sum and difference frequencies. Because the FM inputs have equal
frequencies, the sum frequency is two times the FM frequency.
However, when the input frequencies to a balanced modulator are equal, the difference frequency component becomes a dc voltage that
varies with phase deviations from 90o.
Because the message signal caused the FM frequency deviations, which are converted to phase differences, the phase detector’s
difference dc voltage component varies directly with the message signal.
Therefore, the output of the phase detector contains the sum frequency and the message signal.
The RC network at the output of the phase detector is a low pass filter that removes the sum frequency and pass the message signal.
1.
Connect the circuit and channel 1 and 2 oscilloscope probes as shown.
Set AC couple for both channels.
Adjust the channel 1 and channel 2 oscilloscope signals as shown in the graph below.
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2.
3
What is the PHASE DETECTOR’s output signal on channel 2?
q 32
4.
What is the PHASE DETECTOR output difference component?
q 33
5.
What varies the PHASE DETECTOR’s dc output voltage (difference component)?
q 34
6.
Connect the channel 2 probe to the output of the PHASE SHIFTER/LIMITER.
If necessary, adjust the FM frequency with the NEGATIVE SUPPLY voltage to -3.8 Vdc, so that the phase difference between the input
signals on channel 1 and 2 is 90o.(sweep should be 0.278μs/DIV)
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7.
Connect the multimeter to the PHASE DETECTOR’s output.
With a 90o phase difference between the input signals, measure the dc voltage of th PHASE DETECTOR’s output.
q 35
8.
Set the phase difference between the signals on channel1 and 2 to 135o (refer to the graph below) by adjusting the
NEGATIVE SUPPLY voltage from VCO-LO.
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The value of the neg_supply variable is around:
q 36
If not, set the neg_supply variable to
Vdc.
With a 135o phase difference between the input signals, what is the dc voltage at the PHASE DETECTOR’s output.
q 37
9
Set the phase difference between the signals on channel1 and 2 to 45o (refer to the graph below) by adjusting the
NEGATIVE SUPPLY voltage from VCO-LO.
The value of the neg_supply variable is around:
q 38
If not, set the neg_supply variable to
Vdc.
With a 45o phase difference between the input signals, what is the dc voltage at the PHASE DETECTOR’s output.
q 39
10
When the phase difference between the signals on channel 1 and 2 increased or decreased, did the dc output voltage change?
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q 40
11.
Set the phase difference between the signals on channel 1 and 2 back to 90o (refer to the graph below) by adjusting the neg_supply
value to – 3.8 Vdc.
12.
Connect the SIGNAL GENERATOR to (M) on the VCO-LO circuit block.
13.
14.
Adjust the SIGNAL GENERATOR for a 5 V pk-pk, 45 kHz sine wave.
15.
Connect the channel 1 probe to the (FM) output on VCO-LO.
Connect the channel 2 probe to PHASE SHIFTER/LIMITER output.
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Set channel 1 for 100 mV/DIV, and set channel 2 for 500mV/DIV. Set sweep to 2.5 μs/DIV.
Set the coupling to AC on channel 2.
Set vertical position to 0 for both oscilloscope channels.
16.
Compare the signals at the two inputs to the PHASE DETECTOR.
Is the phase of the FM signal on channel 2 varying with respect to the signal on CH 1?
q 41
17.
Connect the channel 2 probe to the PHASE DETECTOR’s output to observe the sum frequency signal.
On channel 2, is the dc level of the sum frequency signal, changing?
q 42
18.
Adjust the SIGNAL GENERATOR for a 300 mVpk-pk, 3 kHz sine wave.
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Set the oscilloscope time sweep to 0.1 ms/DIV.
Connect the channel 1 probe to the output of the SIGNAL GENERATOR to observe the message signal.
19.
Observe the dc variations of the PHASE DETECTOR output signal on channel 2.
20.
Compare the message signal on channel 1 with the dc variations of the PHASE DETECTOR output on channel 2.
21.
Do the dc variations of the PHASE DETECTOR output have the same frequency as the message signal?
q 43
22.
Connect the channel 2 probe to the output of the FILTER. Set channel 2 to 100 mV/DIV.
23.
Does the recovered message signal on channel 2 vary with the message signal amplitude and frequency on channel 1?
q 44
24.
What frequency does the FILTER remove?
q 45
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CONCLUSION
• A quadrature detector is an FM demodulator or discriminator that converts the FM frequency variations into the message signal.
• A quadrature detector consists of a phase shifter, a limiter, a phase detector and a filter.
• Signals are in quadrature when they have a phase difference of 90o.
• The phase shifter shifts the FM center frequency by 90o; frequencies on each side of the center frequency are shifted more or less than
90o.
• The phase detector is a balanced modulator that outputs a dc voltage that varies with the phase difference of the input signals.
• A filter at the phase detector’s output removes the higher output frequencies and pass the message signal.
Remove all connections from your circuit board.
REVIEW QUESTIONS
1. What function does a discriminator or frequency detector perform?
q 46
2. What does the term quadrature mean?
q 47
3. What function does the quadrature detector first perform?
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q 48
4. What is the quadrature detector’s phase detector?
q 49
5. What are the output signals of the quadrature detector’s phase detector?
q 50
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