Chapter 3 · 7 hours
Angle Modulation
IOE past exam questions
Past questions and answers
52 questions set from this chapter, 9 of them more than once. Most asked first.
- Asked 5 times
- 2075 Bhadra (CS I) · 8 marks
- 2072 Asoj (CS I) · 6 marks
- 2071 Bhadra (CS I) · 6 marks
- 2070 Magh (CS I) · 6 marks
- 2068 Bhadra (CS I) · 8 marks
Derive the expression for single tone modulated FM signal in terms of Bessel coefficients.
Answer
In frequency modulation, the instantaneous frequency of the carrier varies linearly with the message. For a single tone, the FM wave can be expanded with Bessel functions into a carrier and an infinite set of sidebands.
Instantaneous frequency and phase
Let and . The instantaneous frequency is
where is the frequency sensitivity (Hz/V) and is the peak frequency deviation. The phase is
with modulation index . So
Complex envelope and Fourier series
The term is periodic with period , so it has a Fourier series:
Put (, limits to ):
where is the Bessel function of the first kind of order n:
Final expression
Expanded, using :
Conclusions
- The spectrum has a carrier at (amplitude ) and infinite sidebands at with amplitudes .
- Odd-order lower sidebands have reversed sign.
- The carrier amplitude depends on and becomes zero when (β = 2.405, 5.52, ...).
- Since , total power is constant: .
- Only about sidebands on each side are significant, giving Carson's rule .
- Asked 4 times
- 2081 Chaitra · 5 marks
- 2080 Chaitra (CS I) · 5 marks
- 2074 Bhadra (CS I) · 4 marks
- 2072 Asoj (CS I) · 4 marks
Write a short note on pre-emphasis and de-emphasis.
Answer
Pre-emphasis is the boosting of the high-frequency part of the message at the FM transmitter before modulation; de-emphasis is the matching attenuation of those frequencies at the receiver after demodulation.
Why they are needed
- In FM, the output noise power spectral density rises with the square of frequency (parabolic noise spectrum): .
- Audio signals have most of their energy at low frequencies; the high-frequency parts are weak.
- So the high audio frequencies have the worst SNR. Pre-emphasis raises them above the noise before transmission; de-emphasis restores the original balance and, at the same time, cuts the high-frequency noise.
Circuits
Pre-emphasis (high-pass) De-emphasis (low-pass)
+--[ R ]--+ o--[ R ]--+----o
o-+ +--+--o |
+--||-----+ | === C
C > r |
o--------------+--o o---------+----o
- Pre-emphasis: , where .
- De-emphasis: , so and the message is undistorted.
- Standard time constant s ( kHz) in USA FM broadcasting; 50 µs in Europe and many other countries.
Gain
^ pre-emphasis
| ___/
| ___/
|-----/---------- 0 dB
| \___
| \___ de-emphasis
+-----+------------> f
f1=2.1 kHz
Benefits
- Improves output SNR of FM by about 10–13 dB for audio.
- No extra bandwidth needed (for normal audio content).
- Also used in phonograph recording, tape recording and TV sound.
- Asked 3 times
- 2080 Chaitra (CS I) · 5 marks
- 2073 Magh (CS I) · 5 marks
- 2071 Magh (CS I) · 4 marks
Write a short note on stereo encoder.
Answer
A stereo encoder (stereo multiplexer) combines the left (L) and right (R) audio channels into one composite baseband signal for FM stereo broadcasting, in a way that mono receivers can still use.
Block diagram
L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
| ^ ^ ^ to FM
R --|----+ | | mod.
+-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
^ (R) ^ 38k |
[19k osc]->[x2] |
| |
+--19 kHz pilot----+
Working
- A matrix circuit (adder and subtractor) forms (mono-compatible sum) and (difference).
- Both are low-pass filtered to 15 kHz and pre-emphasized.
- is DSB-SC modulated in a balanced modulator onto a 38 kHz sub-carrier, giving sidebands from 23 to 53 kHz.
- The 38 kHz sub-carrier is made by doubling a 19 kHz crystal oscillator. A small 19 kHz pilot (about 10% deviation) is added so the receiver can regenerate the 38 kHz carrier in phase.
- The composite signal is
Spectrum of composite signal
| L+R pilot L-R DSB-SC
|______ | ____ ____
| | | | | | |
+------+------+---+----+--+----+---> f (kHz)
0 15 19 23 38 53
The total baseband bandwidth is 53 kHz, which then frequency-modulates the carrier. A mono receiver uses only 0–15 kHz (); a stereo receiver recovers as well and forms and by adding and subtracting.
- Asked 2 times
- 2080 Chaitra (CS I) · 8 marks
- 2071 Magh (CS I) · 8 marks
Explain the operation of FM superheterodyne radio receiver.
Answer
An FM superheterodyne receiver converts the received FM station (88–108 MHz) to a fixed intermediate frequency of 10.7 MHz, amplifies and limits it, and then recovers the audio with a frequency discriminator.
Block diagram
Ant
|
[RF amp]->[Mixer]->[IF amp]->[Limiter]->[Discri-]
88-108M ^ 10.7 MHz [ minator]
| | BW 200k |
| [Local osc]<------AFC---------------+
| fLO=fs+10.7M |
+--ganged--+ [De-emph]<------+
|
[AF amp]-->Spk
Working of each block
- RF amplifier: tuned to the wanted station . It gives gain at a low noise figure (important at VHF), improves sensitivity and rejects the image frequency MHz.
- Local oscillator and mixer: the oscillator is ganged with the RF tuning so that MHz. The mixer produces sum and difference frequencies; the difference, 10.7 MHz, is selected. The frequency deviation (up to ±75 kHz) is unchanged by mixing.
- IF amplifier: several fixed-tuned stages at 10.7 MHz with a bandwidth of about 200 kHz (Carson bandwidth for kHz, kHz is 180 kHz). Provides most of the gain and the selectivity.
- Amplitude limiter: clips the IF signal to a constant amplitude, removing amplitude variations caused by noise, interference and fading. The information is in the frequency, so nothing useful is lost. This is the main reason FM is less noisy than AM.
- Discriminator (FM detector): converts frequency variations into voltage variations, e.g. a Foster–Seeley discriminator, ratio detector, slope detector or PLL. Output is proportional to .
- De-emphasis network: an RC low-pass filter with 75 µs (or 50 µs) time constant. It undoes the transmitter's pre-emphasis and reduces high-frequency noise.
- AF amplifier and speaker: amplifies the audio to drive the loudspeaker. In stereo sets a stereo decoder sits between discriminator and AF stage.
- AFC (automatic frequency control): the DC part of the discriminator output shows any mistuning; it is fed back to a varactor in the local oscillator to keep the IF centred at 10.7 MHz.
Differences from AM superheterodyne
| Point | AM receiver | FM receiver |
|---|---|---|
| Signal range | 0.535–1.605 MHz | 88–108 MHz |
| IF | 455 kHz | 10.7 MHz |
| IF bandwidth | 10 kHz | 200 kHz |
| Limiter | Not used | Used |
| Detector | Envelope detector | Discriminator / PLL |
| Control loop | AGC | AFC (and AGC) |
| De-emphasis | No | Yes |
Example: for a station at 100 MHz, MHz and the image is at 121.4 MHz, outside the FM band.
- Asked 2 times
- 2079 Chaitra (CS I) · 5+5 marks
- 2067 Shrawan (CS I) · 5+5 marks
With functional block diagrams and spectral details explain the operation of stereo encoder and decoder.
Answer
Stereo FM sends two audio channels, left (L) and right (R), on one FM carrier using a composite baseband signal. It stays compatible with mono receivers because the sum is sent in the normal 0–15 kHz band.
Stereo encoder (transmitter)
L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
| ^ ^ ^ to FM
R --|----+ | | mod.
+-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
^ (R) ^ 38k |
[19k osc]->[x2] |
| |
+--19 kHz pilot----+
- Matrix: an adder and a subtractor form and .
- LPF and pre-emphasis: each signal is limited to 15 kHz and pre-emphasized.
- Balanced modulator: DSB-SC modulates a 38 kHz sub-carrier. Since the carrier is suppressed, its power goes into useful sidebands (23–53 kHz).
- Pilot: a 19 kHz crystal oscillator gives the 38 kHz sub-carrier through a frequency doubler. The 19 kHz pilot itself is also added at low level (about 10% of deviation). It lies in an empty gap (15–23 kHz), is easy to filter, and tells the receiver the exact frequency and phase of the 38 kHz sub-carrier.
- Adder: forms the composite signal that frequency-modulates the RF carrier:
Spectrum of the composite baseband
| L+R pilot L-R DSB-SC
|______ | ____ ____
| | | | | | |
+------+------+---+----+--+----+---> f (kHz)
0 15 19 23 38 53
The composite bandwidth is 53 kHz (with optional SCA services up to about 75 kHz).
Stereo decoder (receiver)
FM +->[LPF 0-15k]------------L+R--+->(+)--> 2L
demod | | ^
out --+->[BPF 23-53k]-->( X )->[LPF]-+ |
| ^ L-R | |
| | 38k +--|-->(-)--> 2R
+->[BPF 19k]-->[ x2 ] |
(pilot) L-R ---+
- The FM discriminator output is the composite signal .
- LPF (0–15 kHz) gives . A mono receiver stops here.
- Narrow BPF at 19 kHz extracts the pilot; a frequency doubler (or PLL) regenerates the 38 kHz sub-carrier, locked in phase with the transmitter.
- BPF (23–53 kHz) selects the DSB-SC signal. It is multiplied by the 38 kHz carrier and low-pass filtered (coherent detection), giving .
- De-matrixing:
- Each channel is de-emphasized and amplified to its own speaker. A stereo indicator lamp lights when the pilot is detected.
Advantages: full mono compatibility, small extra bandwidth, simple pilot-based synchronization. Drawback: the stereo SNR is somewhat lower than mono, since the channel sits at higher baseband frequencies where FM noise is larger.
- Asked 2 times
- 2076 Baisakh (CS I) · 8 marks
- 2076 Bhadra (CS I) · 2+2+2+2 marks
A sinusoidal modulating signal m(t) = 5cos(18849.55t) is applied to an FM modulator that has frequency sensitivity of 9 kHz/V. The amplitude of carrier is 25 V and the carrier frequency is 88.7 MHz. Compute a) modulation index b) Carrier frequency swing c) Bandwidth using Carson rule d) total power delivered to 10 Ω resistor.
Answer
Given: , so V and
kHz/V, V, MHz, .
Peak frequency deviation:
a) Modulation index
(wide-band FM, since )
b) Carrier frequency swing
The carrier swings from to , i.e. from 88.655 MHz to 88.745 MHz:
c) Bandwidth by Carson's rule
d) Total power in 10 Ω
The FM wave has constant amplitude, so its power does not depend on modulation:
Answer: ; carrier swing = 90 kHz; kHz; W.
- Asked 2 times
- 2076 Bhadra (CS I) · 2+4 marks
- 2070 Magh (CS I) · 2+6 marks
What is the role of amplitude limiter in limiter discriminator method? Explain FM demodulator using PLL.
Answer
Role of the amplitude limiter
In a limiter–discriminator FM detector, the amplitude limiter removes all amplitude variations of the received FM signal before it reaches the discriminator.
- Noise, interference and fading change the amplitude of the FM wave, but the information is only in its frequency.
- A discriminator's output depends on both frequency and amplitude, so amplitude changes would appear as noise and distortion.
- The limiter (a hard clipper followed by a band-pass filter at the IF) produces a constant-amplitude signal with the same zero crossings, i.e. the same instantaneous frequency.
FM + AM noise -> [Hard limiter] -> [BPF] -> constant-
/\/\/\ /\/\ square-ish at fc amplitude FM
So the discriminator sees only frequency variations, which gives FM its noise immunity.
FM demodulation using PLL
A phase-locked loop is a feedback system in which a voltage-controlled oscillator (VCO) is forced to follow the phase, and hence the frequency, of the input. The control voltage needed to do this is the demodulated message.
s(t) -->[ Phase ]-->[ Loop ]--+--> v(t) = output
[ detector ] [ filter] | (message)
^ |
| |
+------[ VCO ]<----+
Blocks
- Phase detector (multiplier): compares input and VCO phases and gives an error voltage.
- Loop filter: low-pass; removes the term and sets loop dynamics.
- VCO: its frequency is .
Analysis
Input FM wave and VCO output:
The multiplier output, after removing the term, is
When the loop is locked, is small, so (linear model). With high loop gain the loop drives , so
Differentiating both sides:
The loop filter output is therefore proportional to the message. The PLL demodulator has good threshold performance, needs no tuned circuits, and is available as a single IC (e.g. 565).
- Asked 2 times
- 2075 Bhadra (CS I) · 8 marks
- 2067 Mangsir (CS I) · 10 marks
Describe the process of demodulation of FM using PLL.
Answer
A phase-locked loop (PLL) is a negative-feedback system whose voltage-controlled oscillator (VCO) tracks the phase of the input signal. When the input is an FM wave, the voltage that steers the VCO is proportional to the message, so the PLL acts as an FM demodulator.
Block diagram
FM in +-----------+ e(t) +--------+ v(t)
s(t) --->| Phase |-------->| Loop |---+---> output
| detector | | filter | | ~ m(t)
+-----------+ | H(f) | |
^ +--------+ |
| r(t) +-------+ |
+----------| VCO |<---------+
+-------+
Blocks
- Phase detector: a multiplier comparing input and VCO output; its output depends on the phase difference.
- Loop filter: a low-pass filter that removes the double-frequency term and controls loop bandwidth and stability.
- VCO: free-running at when ; its frequency changes by Hz per volt.
Analysis
Input FM signal:
VCO output (90° offset so the error is zero at lock):
Multiplier output:
The loop filter removes the term. The error signal is
where (times the multiplier gain).
Linear model (loop in lock)
In lock is small, so :
phi1 -->(+)--> K_m --> H(f) --+--> v(t)
^ - |
| |
phi2 <--[2 pi kv / s]-+
In the frequency domain,
When the loop gain over the message band, , so :
So the loop filter output is a scaled copy of the message.
Physical explanation
- If the input frequency rises, the phase error grows; the error voltage increases and pushes the VCO frequency up until it matches.
- Thus the control voltage always follows the instantaneous input frequency, , i.e. it follows .
Lock and capture range
- Lock range: frequency range over which a locked loop stays locked.
- Capture range: range over which an unlocked loop can acquire lock (smaller than lock range).
- The peak deviation must lie within the lock range.
Advantages
- No tuned LC circuits; easy IC realization (e.g. NE565).
- Better (lower) FM threshold than a limiter–discriminator.
- Good linearity and tracking of slow carrier drift.
- Asked 2 times
- 2075 Bhadra (CS I) · 3+5 marks
- 2070 Magh (CS I) · 2+4 marks
Why pre-emphasis and de-emphasis circuits are used in commercial FM broadcasting? Explain the functional block diagram of stereo encoder.
Answer
Why pre-emphasis and de-emphasis are used
In FM, noise at the discriminator output is not flat: its power spectral density rises as (parabolic noise). Audio signals, on the other hand, have most energy at low frequencies and little at high frequencies. So the high audio frequencies have a very poor SNR.
- Pre-emphasis (at the transmitter, before the modulator): a high-pass RC network, , boosts high audio frequencies so that they stand well above the noise.
- De-emphasis (at the receiver, after the discriminator): a low-pass RC network, , brings the boosted frequencies back to their original level. It attenuates the high-frequency noise by the same amount.
- Since , the message is undistorted, but the noise is reduced. The improvement is about 10–13 dB.
- Standard time constant: 75 µs ( kHz) in USA, 50 µs in Europe and most other countries.
Pre-emph De-emph
o-+-[R]-+-o o-[R]-+-o
+-||--+ | |
C > r === C
o---------+-o o-----+-o
Stereo encoder
A stereo encoder forms one composite baseband signal from the left and right audio channels for FM stereo broadcasting.
L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
| ^ ^ ^ to FM
R --|----+ | | mod.
+-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
^ (R) ^ 38k |
[19k osc]->[x2] |
| |
+--19 kHz pilot----+
- Matrix network: adder and subtractor make (for mono compatibility) and .
- LPF (15 kHz) and pre-emphasis on both.
- Balanced modulator: DSB-SC modulates a 38 kHz sub-carrier (sidebands 23–53 kHz).
- 19 kHz pilot oscillator and doubler: gives the 38 kHz sub-carrier; the 19 kHz pilot is also added at low level so receivers can regenerate the 38 kHz carrier.
- Adder: composite signal
| L+R pilot L-R DSB-SC
|______ | ____ ____
| | | | | | |
+------+------+---+----+--+----+---> f (kHz)
0 15 19 23 38 53
This 53 kHz composite signal frequency-modulates the main carrier.
- 2080 Chaitra · 4 marks
Explain the Armstrong's method for Frequency Modulation.
Answer
Armstrong's method is an indirect method of generating wide-band FM: first a narrow-band FM (NBFM) wave is made with a crystal-controlled phase modulator, then frequency multipliers and a mixer raise the deviation and set the carrier frequency.
m(t)->[Integrator]->[Balanced]--( Σ )-> NBFM
[modulator] ^ -
^ | +
+--[-90 deg]---+ |
[Crystal osc fc1]---------------+
(NB phase modulator)
NBFM ->[ x n1 ]->[ Mixer ]->[ x n2 ]->[PA]-> WBFM
^
[Osc fc2]
Steps
- NBFM generation: the message is integrated and applied to a balanced modulator with a −90° shifted carrier. The output is added to the carrier, giving
which is NBFM. A crystal oscillator keeps very stable. 2. Frequency multiplication (×n₁): a non-linear stage multiplies both carrier and deviation: , . 3. Mixer: shifts the carrier down to a convenient value () without changing the deviation. 4. Second multiplier (×n₂): gives the final carrier and deviation (e.g. 75 kHz).
Merit: excellent carrier stability (crystal oscillator). Demerit: many multiplier stages; noise and distortion may increase.
- 2080 Chaitra · 4 marks
Why do we need pre-emphasis and de-emphasis circuits in FM? Explain.
Answer
Pre-emphasis and de-emphasis are needed in FM to improve the signal-to-noise ratio at high audio frequencies.
- Noise in FM rises with frequency. The FM discriminator output noise PSD is (triangular noise voltage, parabolic noise power). So high-frequency audio components get much more noise.
- Audio signals are weak at high frequencies. Most speech and music energy is below 1–2 kHz.
- Together, these make the SNR of the high audio frequencies very poor.
Solution
- Pre-emphasis at the transmitter: a high-pass RC network boosts high audio frequencies before modulation, .
- De-emphasis at the receiver: a low-pass RC network after demodulation, , restores the original response and attenuates the high-frequency noise at the same time.
- Since , the signal is unchanged but the output noise falls, giving about 10–13 dB SNR improvement. Standard time constant: 75 µs (USA) or 50 µs (Europe), kHz for 75 µs.
Gain pre-emph _/
| __/
|----------/--------- 0 dB
| \__
| \_ de-emph
+---------+---------> f
f1
- 2078 Chaitra · 3+7 marks
What is the relation between psdf and Autocorrelation function? Explain the Stereo FM encoder and decoder with spectral diagram.
Answer
Relation between PSD and autocorrelation
By the Wiener–Khinchin theorem, the power spectral density (PSD) of a wide-sense stationary process and its autocorrelation function form a Fourier transform pair:
Consequences:
- Total average power: .
- is real, even and non-negative because is real and even.
- Example: white noise .
Stereo FM encoder
Stereo FM sends left (L) and right (R) channels on one carrier while staying compatible with mono receivers.
L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
| ^ ^ ^ to FM
R --|----+ | | mod.
+-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
^ (R) ^ 38k |
[19k osc]->[x2] |
| |
+--19 kHz pilot----+
- Matrix forms and ; both are band-limited to 15 kHz and pre-emphasized.
- DSB-SC modulates a 38 kHz sub-carrier in a balanced modulator.
- The 38 kHz sub-carrier comes from a 19 kHz oscillator and a doubler; a small 19 kHz pilot is also added.
- Composite signal:
Spectrum
| L+R pilot L-R DSB-SC
|______ | ____ ____
| | | | | | |
+------+------+---+----+--+----+---> f (kHz)
0 15 19 23 38 53
Stereo FM decoder
FM +->[LPF 0-15k]------------L+R--+->(+)--> 2L
demod | | ^
out --+->[BPF 23-53k]-->( X )->[LPF]-+ |
| ^ L-R | |
| | 38k +--|-->(-)--> 2R
+->[BPF 19k]-->[ x2 ] |
(pilot) L-R ---+
- FM discriminator gives the composite signal.
- LPF (0–15 kHz) gives (all a mono set needs).
- A narrow BPF picks the 19 kHz pilot; a doubler makes a phase-locked 38 kHz carrier.
- BPF (23–53 kHz) selects the DSB-SC signal, which is coherently detected with the 38 kHz carrier and low-pass filtered.
- Matrix: and , then de-emphasis and audio amplifiers.
- 2080 Chaitra (CS I) · 4×1.5 marks
The equation of an angle modulated signal is v(t) = 12 sin(10⁶t + 5 sin 10⁴t). Determine the following:
a) Carrier frequency
b) Modulating frequency
c) Modulation index
d) Power dissipated in 100 Ω resistor
Answer
Compare with the standard single-tone angle-modulated wave :
V, rad/s, rad/s, .
a) Carrier frequency
b) Modulating frequency
c) Modulation index
The peak phase deviation is the coefficient of :
(Treated as FM, the peak deviation is Hz.)
d) Power in 100 Ω
The amplitude of an angle-modulated wave is constant, so
Answer: kHz, kHz, , W.
- 2079 Chaitra (CS I) · 4×2.5 marks
A message signal m(t) = 10 cos(2π 4000 t) Volt is used to frequency modulate the carrier signal c(t) = 80 cos(2π 10⁷ t) Volt. Assuming the frequency sensitivity of the frequency modulator to be 400 Hz/Volt, calculate:
a) Peak frequency deviation
b) Modulation index
c) Bandwidth of modulated signal for over 98% of FM power
d) Total modulated signal power dissipated in unit impedance
Answer
Given: V, Hz, V, Hz = 10 MHz, Hz/V.
a) Peak frequency deviation
b) Modulation index
c) Bandwidth containing over 98% of power
Carson's rule is based on keeping about 98% of the power:
Check with Bessel functions for : , , , .
| Components kept | Power fraction |
|---|---|
| 0.9728 (97.3%) | |
| 0.9992 (99.9%) |
So sidebands up to are needed for more than 98% power:
which agrees with Carson's rule.
d) Total power in unit impedance
FM has constant amplitude, so
Answer: kHz; ; kHz; W.
- 2077 Chaitra (CS I) · 2+6 marks
Differentiate between Narrow Band FM and Wide Band FM. Describe stereo FM broadcasting with its block diagram and spectral details.
Answer
NBFM versus WBFM
| Point | Narrow band FM | Wide band FM |
|---|---|---|
| Modulation index | (< 0.3) | |
| Bandwidth | (like AM) | |
| Sidebands | Carrier + one pair | Many significant pairs |
| Max deviation | Small (~5 kHz) | Large (75 kHz in broadcast) |
| Noise performance | Little better than AM | Much better SNR |
| Use | Mobile radio, police, Armstrong first stage | FM broadcast, TV sound |
Stereo FM broadcasting
Stereo FM transmits left (L) and right (R) audio on one carrier, while mono receivers can still receive the program.
Encoder (transmitter)
L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
| ^ ^ ^ to FM
R --|----+ | | mod.
+-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
^ (R) ^ 38k |
[19k osc]->[x2] |
| |
+--19 kHz pilot----+
- Matrix makes and , each limited to 15 kHz and pre-emphasized.
- DSB-SC modulates a 38 kHz sub-carrier (balanced modulator).
- 38 kHz is obtained by doubling a 19 kHz oscillator; the 19 kHz pilot is added at low level for receiver synchronization.
- The composite signal
frequency-modulates the RF carrier (peak deviation 75 kHz).
Spectrum
| L+R pilot L-R DSB-SC
|______ | ____ ____
| | | | | | |
+------+------+---+----+--+----+---> f (kHz)
0 15 19 23 38 53
Decoder (receiver)
FM +->[LPF 0-15k]------------L+R--+->(+)--> 2L
demod | | ^
out --+->[BPF 23-53k]-->( X )->[LPF]-+ |
| ^ L-R | |
| | 38k +--|-->(-)--> 2R
+->[BPF 19k]-->[ x2 ] |
(pilot) L-R ---+
- LPF gives (mono output).
- Pilot is filtered and doubled to 38 kHz; the 23–53 kHz band is coherently detected to give .
- Adding and subtracting give and .
- 2077 Chaitra (CS I) · 8 marks
For a given Armstrong FM transmitter, the NBFM output has a frequency of 200 kHz and frequency deviation of 25 Hz. This signal is then frequency multiplied by 65 and passed through mixer of oscillation frequency 10.8 MHz. The resulting signal is then fed to a frequency multiplier with n = 50. Calculate the maximum frequency deviation and the valid carrier frequency of the WBFM for commercial use. Draw the necessary diagram.
Answer
In the Armstrong method, frequency multipliers multiply both carrier frequency and deviation; the mixer shifts the carrier frequency but leaves the deviation unchanged.
Diagram
NBFM x65 Mixer x50
200 kHz -->[n1]-->13 MHz-->[X]-->2.2M-->[n2]--> WBFM
df=25 Hz df=1625 Hz ^ df=1625 110 MHz
| df=81.25k
[Osc 10.8 MHz]
Step 1: after first multiplier (n₁ = 65)
Step 2: mixer with 10.8 MHz
The mixer gives MHz, i.e. 2.2 MHz (difference) or 23.8 MHz (sum). Deviation stays 1625 Hz.
Step 3: second multiplier (n₂ = 50)
Valid carrier frequency
The sum term gives 1190 MHz (UHF), which is far outside the VHF FM broadcast range and is rejected by filtering after the mixer. The difference term is the usable one, giving
Answer: maximum frequency deviation Δf = 81.25 kHz; valid WBFM carrier f_c = 110 MHz (from the 2.2 MHz difference output).
Note: 110 MHz lies just above the 88–108 MHz broadcast band and 81.25 kHz is slightly above the 75 kHz standard, so in practice the oscillator frequency or multiplier would be adjusted slightly; with the given data, 110 MHz is the only practical (VHF) choice.
- 2076 Baisakh (CS I) · 2+6 marks
Define narrow band and wide band FM. Explain the process of generation of wide band FM wave using Armstrong method.
Answer
Narrow band and wide band FM
- Narrow band FM (NBFM): FM with a small modulation index, (practically ). Its bandwidth is about , like AM, with only the carrier and one pair of sidebands:
- Wide band FM (WBFM): FM with . It has many significant sidebands and bandwidth (Carson's rule). Example: FM broadcast with kHz, kHz, .
Generation of WBFM by Armstrong (indirect) method
Direct FM using an LC oscillator is not stable enough. Armstrong's method first makes a stable NBFM wave with a crystal oscillator, then uses frequency multiplication to obtain the required large deviation.
m(t)->[Integrator]->[Balanced]--( Σ )-> NBFM
[modulator] ^ -
^ | +
+--[-90 deg]---+ |
[Crystal osc fc1]---------------+
(NB phase modulator)
NBFM ->[ x n1 ]->[ Mixer ]->[ x n2 ]->[PA]-> WBFM
^
[Osc fc2]
Step 1: NBFM by phase modulation of the integrated message. The message is integrated (so that phase modulation of gives FM) and fed to a balanced modulator with a carrier shifted by −90°. Adding the carrier gives
with kept small (< 0.5) to limit distortion. Typical values: –200 kHz, Hz.
Step 2: frequency multiplier (×n₁). A non-linear device followed by a BPF tuned to the n-th harmonic. The output is : both carrier and deviation are multiplied by .
Step 3: mixer (frequency converter). Mixing with a crystal oscillator shifts the carrier to while the deviation stays . This lets the final carrier be set independently of the deviation.
Step 4: frequency multiplier (×n₂) to reach the final values:
Step 5: power amplifier and antenna.
Example: Hz needs for kHz.
Advantages: high carrier stability from crystal oscillators; good linearity. Disadvantages: many multiplier stages; noise is multiplied too; more complex.
- 2076 Baisakh (CS I) · 8 marks
Draw the block diagram of stereo FM encoder and decoder. Explain each block briefly.
Answer
Stereo FM carries left (L) and right (R) audio on one FM carrier as a composite baseband signal, keeping compatibility with mono receivers.
Encoder
L --+-->(+)--L+R-->[LPF]----------->( Σ )-->
| ^ ^ ^ to FM
R --|----+ | | mod.
+-->(-)--L-R-->[LPF]-->[Bal.mod]--+ |
^ (R) ^ 38k |
[19k osc]->[x2] |
| |
+--19 kHz pilot----+
- Matrix (adder/subtractor): forms and .
- LPF (15 kHz) and pre-emphasis: band-limit each audio signal and boost high frequencies.
- 19 kHz pilot oscillator: crystal-controlled reference.
- Frequency doubler (×2): gives the 38 kHz sub-carrier, phase-locked to the pilot.
- Balanced modulator: DSB-SC modulates on 38 kHz, producing 23–53 kHz.
- Adder: sums , the DSB-SC and a low-level 19 kHz pilot into the composite signal, which drives the FM modulator.
| L+R pilot L-R DSB-SC
|______ | ____ ____
| | | | | | |
+------+------+---+----+--+----+---> f (kHz)
0 15 19 23 38 53
Decoder
FM +->[LPF 0-15k]------------L+R--+->(+)--> 2L
demod | | ^
out --+->[BPF 23-53k]-->( X )->[LPF]-+ |
| ^ L-R | |
| | 38k +--|-->(-)--> 2R
+->[BPF 19k]-->[ x2 ] |
(pilot) L-R ---+
- FM discriminator: recovers the composite signal.
- LPF (0–15 kHz): extracts (mono signal).
- Narrow BPF (19 kHz): extracts the pilot.
- Doubler (×2): regenerates a 38 kHz carrier in phase with the transmitter's.
- BPF (23–53 kHz): selects the DSB-SC signal.
- Product detector and LPF: coherent detection with 38 kHz gives .
- Matrix: , .
- De-emphasis and AF amplifiers: restore flat response and drive the left and right speakers.
- 2076 Bhadra (CS I) · 2+6 marks
What are the properties of Bessel function? Show that a FM signal consists infinite number of cosine signal components centered at frequencies fc + n fm, where fc = carrier frequency, fm = message frequency and n = 0, ±1, ±2, ±3, ....
Answer
Properties of Bessel functions
- : negative-order coefficients equal positive ones, with sign change for odd .
- : total FM power is constant.
- For small : , , for (NBFM).
- For a given , becomes very small when , so only a finite number of sidebands are significant.
- is oscillatory; at β = 2.405, 5.52, 8.65, ... (carrier disappears).
FM wave as a sum of cosines at
Single-tone FM:
Write it using the complex envelope:
is periodic in , so
Put :
Hence
This proves that the FM signal is a sum of infinitely many cosine components at , , with amplitudes .
Spectrum:
| | | | |
| | | | | | |
----+---+-----+-----+-----+-----+---+---> f
fc-3fm fc-fm fc fc+fm fc+2fm
(lines spaced fm, heights A_c|J_n(beta)|)
Although infinite in theory, sidebands with are negligible, giving Carson's bandwidth .
- 2075 Baisakh (CS I) · 2+3+3 marks
What is angle modulation? Find the time domain and frequency domain expression of single tone modulated FM signal.
Answer
Angle modulation
Angle modulation is modulation in which the angle (phase) of the carrier is varied according to the message while its amplitude stays constant:
It has two forms: phase modulation (PM), , and frequency modulation (FM), .
Time-domain expression of single-tone FM
Let . Instantaneous frequency:
Instantaneous phase:
So
Frequency-domain expression
Write . The periodic term has Fourier series
Therefore
Taking the Fourier transform:
The spectrum has a carrier of amplitude and sideband pairs at with amplitudes . Total power is , and practical bandwidth is .
- 2075 Baisakh (CS I) · 3+3+3 marks
A 102.4 MHz carrier signal is frequency modulated by a 5 kHz sine wave. The resultant FM signal has frequency deviation of 75 kHz. Now, determine the followings: (a) carrier swing of FM signal, (b) the bandwidth occupied by FM signal and (c) modulation index.
Answer
Given: MHz, kHz, kHz.
(a) Carrier swing
The carrier frequency swings between MHz and MHz:
(b) Bandwidth
By Carson's rule:
(Bessel-table check: for , significant sidebands extend to about , so kHz.)
(c) Modulation index
Answer: carrier swing = 150 kHz; bandwidth = 160 kHz; .
- 2074 Bhadra (CS I) · 6+2 marks
Derive the time domain expression of single tone FM in terms of Bessel's function Jn(β). Use the result to find the expression of average power of FM signal whose significant Bessel's coefficients are taken from '−n' to '+n'.
Answer
Single-tone FM in terms of Bessel functions
For , the instantaneous frequency is , so the phase is with :
is periodic with period ; its Fourier series is
(substituting ). Therefore
i.e. a carrier at and sidebands at .
Average power with components from −n to +n
Each component is a cosine of amplitude ; the components are at different frequencies, so their powers add (1 Ω load):
using . When all components are included, , so
the same as the unmodulated carrier. The fraction of power within sidebands is . Example: for , gives , i.e. 99.9% of the power.
- 2074 Bhadra (CS I) · 9 marks
An Armstrong FM modulator is required in order to transmit an audio signal of bandwidth 50 Hz to 15 kHz. The Narrow Band (NB) phase modulator used utilizes an oscillator providing carrier frequency fc1 = 0.2 MHz. The output of the NB phase modulator is multiplied by n₁ by multiplier and then passed to mixer with a local oscillator frequency fc2 = 10.925 MHz. The desired FM wave at the transmitter output has a carrier frequency fc = 90 MHz and a frequency deviation Δf = 75 kHz, which is obtained by multiplying the mixer output frequency with n₂ by using another multiplier. Find n₁ and n₂. Assume that the NBFM signal at the output of NB phase modulator has modulation index, β = 0.5.
Answer
Given: audio 50 Hz – 15 kHz; MHz; ; MHz; final MHz, kHz.
NB PM x n1 Mixer x n2
0.2 MHz->[n1]->0.2n1 MHz->[X]->f2->[n2]-> 90 MHz
df1=25 Hz ^ df=75 kHz
[10.925 MHz]
Step 1: deviation at the NB modulator
The NB phase modulator (with integrator) must keep for every audio frequency. Since , the worst case is the lowest audio frequency, 50 Hz:
Step 2: total multiplication needed
The mixer does not change the deviation, so
Step 3: carrier frequency condition
The mixer output is the difference frequency (taking ):
Using : , so
Multipliers must be integers. Choosing and solving the frequency equation exactly:
Check
The carrier is exactly 90 MHz and the deviation is within 2.4% of 75 kHz (exactly 75 kHz is reached by reducing slightly to 0.488). Both values are practical: (six doublers) and (four doublers and a tripler).
Answer: , (ideal product 3000; continuous solution , ).
- 2073 Magh (CS I) · 4+6 marks
What is angle modulation? Explain with the help of equation and block diagram, the Armstrong method of generating FM signal.
Answer
Angle modulation
Angle modulation varies the angle of the carrier according to the message, with constant amplitude: .
- PM:
- FM: , so
For a tone , FM gives , .
Armstrong method of generating FM
Armstrong's (indirect) method generates a stable narrow-band FM wave with a crystal oscillator and then converts it to wide-band FM by frequency multiplication.
m(t)->[Integrator]->[Balanced]--( Σ )-> NBFM
[modulator] ^ -
^ | +
+--[-90 deg]---+ |
[Crystal osc fc1]---------------+
(NB phase modulator)
NBFM ->[ x n1 ]->[ Mixer ]->[ x n2 ]->[PA]-> WBFM
^
[Osc fc2]
1. NBFM generation. For small :
using , for small . This is realised by:
- integrating (so that phase modulation gives FM),
- multiplying it in a balanced modulator with the carrier shifted by −90° (),
- adding the carrier back.
Typical: kHz, , Hz.
2. Frequency multiplier ×n₁. A non-linear stage and filter; output . Carrier and deviation both multiply by .
3. Mixer. Mixing with a crystal oscillator moves the carrier to without changing deviation. This keeps the final carrier at the wanted value.
4. Frequency multiplier ×n₂ and power amplifier:
Example: Hz needs for 75 kHz deviation.
Advantages: crystal-stable carrier, good linearity. Disadvantages: many multiplier stages, multiplied noise and phase errors, complex circuit.
- 2073 Magh (CS I) · 2.5×4 marks
The angle modulated signal is given by S(t) = 20 cos(6×10⁸t + 7 sin 1250t). Determine:
i) The carrier and modulating frequency
ii) The modulation index
iii) Maximum frequency deviation
iv) Power dissipated in 10 Ω resistor
Answer
Compare with :
V, rad/s, rad/s, .
i) Carrier and modulating frequency
ii) Modulation index
iii) Maximum frequency deviation
iv) Power in 10 Ω
Answer: MHz, Hz; ; Hz; W.
- 2073 Bhadra (CS I) · 6+4 marks
Find the time domain expression for Narrowband FM signal. How NBFM can be used to generate Wideband FM signal?
Answer
Time-domain expression of NBFM
An FM wave modulated by a single tone is
Expanding with :
For narrowband FM, (in practice ). Then
so
Using :
So NBFM has a carrier and one pair of side frequencies at , like AM, but the lower side frequency has a negative sign (180° phase shift). Its bandwidth is , the same as AM.
Generating WBFM from NBFM (Armstrong / indirect method)
WBFM is obtained from NBFM using frequency multipliers and a mixer:
m(t) -> [Integrator] -> [Balanced ] -> NBFM
[modulator ] f1, df1 (b1<0.3)
crystal osc f1 ----^ (-90 deg)
NBFM -> [x n1] -> [Mixer] -> [x n2] -> WBFM
^ fc, df
[LO fLO]
- A stable crystal oscillator and a balanced modulator produce NBFM with small deviation at carrier .
- A frequency multiplier (a non-linear device + band-pass filter) multiplies the instantaneous frequency by . Both carrier and deviation are multiplied: , , so .
- Because the required deviation needs a large , the carrier would become too high. A mixer with a local oscillator shifts the carrier down () without changing the deviation.
- A second multiplier raises the deviation (and carrier) to the final values:
Example: Hz needs to give 75 kHz deviation for broadcast FM. The method gives a very stable carrier (crystal controlled) and is used in FM broadcast transmitters.
- 2072 Magh (CS I) · 6 marks
Generate a FM signal with Fc = 50 MHz and β = 1 from a NBFM signal with Fc = 1 MHz, β = 0.1 and Fm = 10 kHz.
Answer
Idea: a frequency multiplier multiplies both the carrier and the deviation (so it multiplies ); a mixer shifts only the carrier and leaves the deviation unchanged.
Given NBFM: MHz, , kHz
Required FM: MHz,
Step 1: Multiplication factor. The mixer does not change , so the multiplier must give
After a multiplier:
Step 2: Frequency translation. The carrier is now 10 MHz but 50 MHz is needed. Mix with a local oscillator and select the sum frequency with a band-pass filter:
(Choosing the difference, MHz, also works.) The deviation stays 10 kHz.
NBFM +-----+ 10 MHz +-------+ +-----+ FM out
1 MHz --->| x10 |-------->| Mixer |->| BPF |------->
df=1kHz +-----+ df=10kHz+-------+ |50MHz| 50 MHz
b=0.1 ^ +-----+ df=10kHz
LO 40 MHz b = 1
Check: , MHz.
The order can also be reversed: mix the 1 MHz NBFM with a 4 MHz LO to get 5 MHz, then multiply by 10 to get 50 MHz with kHz.
Answer: frequency multiplier followed by a mixer with MHz (sum taken) gives MHz, kHz, .
- 2072 Magh (CS I) · 2+4 marks
Define lock range, capture range in PLL. Explain the demodulation of FM using PLL.
Answer
Lock range and capture range
- Lock range (tracking range): the range of input frequencies over which a PLL that is already locked stays in lock as the input frequency is slowly changed. It is set mainly by the VCO range and loop gain.
- Capture range (acquisition range): the range of input frequencies over which an unlocked PLL can acquire lock. It is limited by the loop filter bandwidth.
Capture range is always less than or equal to lock range: , both centred on the VCO free-running frequency .
FM demodulation using PLL
FM in +-------+ e(t) +------+ v(t) = output
------->| Phase |------->| Loop |----+---------->
s(t) | det. | | LPF | |
+-------+ +------+ |
^ |
| +-------+ |
+--------| VCO |<------+
r(t) +-------+
Working:
- Input FM: , where .
- The VCO free-running frequency is set to . Its output is with .
- The phase detector (multiplier) output, after the loop filter removes the term, is
- This error voltage drives the VCO so that its phase follows the input phase. When the loop is locked, is small and .
- Hence
So the VCO control voltage is directly proportional to the message. The PLL demodulator needs no tuned circuits, is linear, and gives good threshold performance, so it is widely used in IC FM receivers.
- 2072 Magh (CS I) · 2+4 marks
Explain Carson's rule for determining the bandwidth of an FM wave. Under what condition the bandwidth of FM signal is same as that of AM? Explain.
Answer
Carson's rule
In theory an FM wave has infinitely many sidebands at , with amplitudes . In practice, sidebands beyond about are very small. Carson's rule says that about 98% of the total power of an FM wave lies within
where is the peak frequency deviation and the highest modulating frequency.
- For large (WBFM): (set by the deviation).
- For small (NBFM): .
Example: commercial FM with kHz and kHz gives kHz (200 kHz channels are allotted).
When FM bandwidth equals AM bandwidth
When (narrowband FM, ), , so
which is the same as the bandwidth of AM ().
Reason: for small , , , and for . So the NBFM wave is
It has only the carrier and one pair of sidebands, exactly like AM, so it occupies . The only difference is that the lower sideband is phase-reversed, so the resultant changes the phase (not the amplitude) of the carrier.
- 2072 Magh (CS I) · 3+2 marks
List the major differences between AM and FM superheterodyne receiver. Why are superheterodyne receivers provided with automatic gain control (AGC) mechanism?
Answer
Differences between AM and FM superheterodyne receivers
| Point | AM receiver | FM receiver |
|---|---|---|
| RF band | MW/SW, 0.54–1.65 MHz (MW) | VHF, 88–108 MHz |
| Intermediate frequency | 455 kHz | 10.7 MHz |
| IF bandwidth | about 10 kHz | about 200 kHz |
| Limiter | Not used | Used before detector to remove amplitude noise |
| Detector | Envelope (diode) detector | Discriminator, ratio detector or PLL |
| De-emphasis | Not used | De-emphasis network (75 μs) after detector |
| Gain control | AGC | AFC common; limiter does most amplitude control |
| Noise immunity | Poor | Good (capture effect, limiter) |
Why AGC is used
Automatic gain control (AGC) automatically adjusts the gain of the RF and IF amplifiers according to the strength of the received signal, using a DC voltage taken from the detector output.
It is needed because:
- Signals from near and far stations differ widely in strength (microvolts to hundreds of millivolts). Without AGC, the volume changes when tuning between stations.
- Fading makes the strength of one station vary with time; AGC keeps the output level nearly constant.
- Strong signals would overload the RF/IF stages and the detector and cause distortion; AGC reduces gain for strong signals.
- Weak signals get full gain, which improves sensitivity.
- 2072 Asoj (CS I) · 6 marks
A carrier of frequency 10⁶ Hz and amplitude 3 volts is frequency modulated by a sinusoidal modulating signal frequency 500 Hz and peak amplitude 1 volt. The frequency deviation is 1 kHz. The level of the modulating waveform is changed to 5 V peak and the modulating frequency is changed to 2 kHz. Write the expression for the new modulated waveform.
Answer
Given: Hz, V, original V, Hz, kHz.
Step 1: Frequency sensitivity of the modulator. In FM, the deviation is proportional to the modulating amplitude:
Step 2: New deviation. New V, new kHz. The deviation depends only on amplitude, not on :
Step 3: New modulation index.
(The original index was .)
Step 4: New FM waveform. Taking the modulating signal as a cosine, :
(If a sine carrier is used, write with the same argument.)
Answer: V, with kHz and .
- 2072 Asoj (CS I) · 4 marks
Write a short note on FM radio receiver.
Answer
An FM broadcast receiver is a superheterodyne receiver working in the 88–108 MHz band with an IF of 10.7 MHz. It differs from an AM receiver mainly in the limiter, the FM detector and the de-emphasis network.
Ant
| +----+ +-----+ +-------+ +---------+ +-------+
+->| RF |->|Mixer|->| IF |->| Limiter |->| FM |
|amp | | | |10.7MHz| | | | det. |
+----+ +-----+ +-------+ +---------+ +-------+
^ |
+---------+ +-------+ +-------+ v
| Local |<---| AFC | | Audio |<-[De-emph]
| osc. | +-------+ | amp |-> Speaker
+---------+ +-------+
Blocks:
- RF amplifier: selects and amplifies the weak VHF signal; improves image rejection and noise figure.
- Mixer and local oscillator: convert every station to the fixed IF of 10.7 MHz ( MHz).
- IF amplifier: provides most of the gain and selectivity, with a bandwidth of about 200 kHz.
- Limiter: clips the IF signal to a constant amplitude, removing amplitude noise and interference.
- FM detector: discriminator, ratio detector or PLL converts frequency variations into audio voltage.
- De-emphasis (75 μs RC network): attenuates high audio frequencies that were boosted by pre-emphasis at the transmitter, which reduces high-frequency noise.
- AFC: a DC voltage from the detector corrects drift of the local oscillator.
- Audio amplifier and speaker: reproduce the sound. In stereo receivers a stereo decoder is placed after the detector.
- 2071 Magh (CS I) · 2+2+2+2 marks
The equation of an angle modulation voltage is E = 10 sin(10⁸t + 3 sin 10⁴t). Calculate the carrier and modulating frequency, modulating index and power dissipated in 100 Ω resistor.
Answer
Given: V, .
Comparing with the standard form :
Carrier frequency
Modulating frequency
Modulation index
Power in 100 Ω
The amplitude of an angle-modulated wave is constant, so its power equals the unmodulated carrier power:
Answer: MHz, kHz, , W.
- 2071 Magh (CS I) · 6+2 marks
Explain demodulation of FM using limiter-discriminator method. Why pre-emphasis is needed in FM during transmission?
Answer
FM demodulation by limiter–discriminator
A limiter–discriminator converts frequency variations of the FM wave into amplitude variations and then detects the envelope.
FM in +---------+ +-----+ +---------+ +--------+ m(t)
----->| Limiter |>| BPF |>|Differen-|>|Envelope|---->
|(hard | | | |tiator | |detector|
| clipper)| +-----+ |(slope) | +--------+
+---------+ +---------+
-
Limiter and BPF: the received FM wave has unwanted amplitude variations due to noise and fading. The hard limiter clips it to a square-like wave of constant amplitude; the BPF centred at keeps only the fundamental. Output: , .
-
Discriminator (differentiator): its output is proportional to frequency:
This is a wave that is both FM and AM; its envelope is , which follows the message.
- Envelope detector: recovers . A DC-blocking capacitor removes the constant , leaving an output proportional to .
In practice the differentiator is a slope detector (a tuned circuit operated on the slope of its response), a balanced slope detector (two tuned circuits, one above and one below , giving better linearity), or a Foster–Seeley discriminator. The limiter is essential: without it, amplitude noise would pass straight to the envelope detector.
Why pre-emphasis is needed
- In FM, the output noise power spectral density after detection increases with the square of frequency (triangular noise spectrum), so high audio frequencies suffer more noise.
- High-frequency components of speech and music have small amplitudes, so their SNR is poorest.
- Pre-emphasis (an RC high-pass network, time constant 75 μs, corner 2.1 kHz) boosts high audio frequencies before modulation. At the receiver a matching de-emphasis network attenuates them back, and also attenuates the high-frequency noise. The result is an SNR improvement of about 10–13 dB with no overall distortion of the message.
- 2071 Magh (old course) · 3+5 marks
Derive the general expression for frequency modulation. Explain with the block diagram how can you generate FM using phase modulator.
Answer
General expression for FM
Let the carrier be . A general angle-modulated wave is
The instantaneous frequency is .
In FM, the instantaneous frequency varies linearly with the message :
where is the frequency sensitivity (Hz/V). Integrating,
Hence the general FM expression:
Single-tone case: for ,
FM generation using a phase modulator
A PM wave is : its phase follows the input . Comparing with the FM expression, if the input to a phase modulator is the integral of the message, the output is FM.
m(t) +------------+ x(t)=int m +-----------+ FM wave
----->| Integrator |------------>| Phase |---------->
+------------+ | modulator |
+-----------+
^
carrier Ac cos wc t
With :
This is FM with . This is the basis of the indirect (Armstrong) method: the phase modulator is built from a balanced modulator and a 90° phase shift of a crystal oscillator, giving NBFM with high carrier stability. Frequency multipliers and a mixer then convert it to wideband FM. (Conversely, a differentiator followed by a frequency modulator gives PM.)
- 2071 Bhadra (CS I) · 7 marks
Describe the limiter-discriminator method for demodulation of FM wave.
Answer
The limiter–discriminator is a direct method of FM demodulation. It first removes amplitude variations (limiter), then converts frequency variations into amplitude variations (discriminator), and finally detects the envelope.
FM in +-------+ +-----+ +-----------+ +--------+ m(t)
----->| Hard |>| BPF |>|Discrimina-|>|Envelope|---->
|limiter| | fc | |tor (d/dt) | |detector|
+-------+ +-----+ +-----------+ +--------+
1. Limiter and band-pass filter
The received FM signal is , where varies due to noise and fading. The hard limiter clips it into a rectangular wave of fixed amplitude whose zero crossings carry the frequency information. The BPF centred at removes the harmonics. The output is a constant-amplitude FM wave:
2. Discriminator
An ideal discriminator has a transfer function over the FM band, i.e. it acts as a differentiator:
The output is a hybrid AM–FM wave whose envelope varies linearly with (provided , so the envelope never becomes negative).
Practical discriminators:
- Slope detector: a tuned circuit detuned so that sits on the linear slope of its response; output amplitude rises with frequency. Simple but not very linear.
- Balanced slope detector: two tuned circuits at and , each with a diode detector; outputs subtracted. Gives a wider linear range and cancels the DC term.
- Foster–Seeley discriminator / ratio detector: phase-shift discriminators with good linearity.
3. Envelope detector
A diode detector recovers . A DC-blocking capacitor removes the term, giving an output proportional to .
Why the limiter is essential: the discriminator output amplitude depends on . Without the limiter, any amplitude noise on the received signal would appear directly in the output. The limiter makes the detector respond only to frequency changes, which is the main reason for FM's noise immunity.
- 2071 Bhadra (CS I) · 2×4 marks
A modulating signal m(t) = 5 cos 18849.55t is applied to an FM modulator that has a frequency sensitivity of 9 kHz/V. Compute (i) peak frequency deviation, (ii) modulation index, (iii) frequency swing and (iv) Carson's bandwidth.
Answer
Given: , kHz/V.
So V and
(i) Peak frequency deviation
(ii) Modulation index
(iii) Frequency swing
The carrier swings from to :
(iv) Carson's bandwidth
Answer: kHz, , swing = 90 kHz, kHz.
- 2070 Magh (CS I) · 4+4 marks
In an FM system a baseband signal band limited to 10 kHz modulates 100 MHz carrier wave so that the frequency deviation is 75 kHz. Find:
i) Carrier frequency swing in the FM signal and modulation index
ii) The practical bandwidth of the FM signal
Answer
Given: kHz (highest baseband frequency), MHz, kHz.
i) Carrier frequency swing and modulation index
The instantaneous frequency varies between
Modulation index (deviation ratio, since is the highest message frequency):
ii) Practical bandwidth
By Carson's rule (contains about 98% of the FM power):
Check with Bessel functions: with , Carson's rule keeps sidebands up to ; the power in these is , i.e. 98.1% of the total.
If the stricter 1% rule is used (keep sidebands with ), and , so and
Answer: swing = 150 kHz, , practical (Carson) bandwidth = 170 kHz (220 kHz by the 1% rule).
- 2070 Bhadra (CS I) · 2+5 marks
How is the spectrum of Narrow band FM similar to and different from the spectrum of conventional AM? Explain how NBFM is generated by using Armstrong's method.
Answer
NBFM spectrum compared with AM
For single-tone modulation:
Similarities: both have a carrier and one pair of sidebands at ; both have bandwidth ; sideband amplitudes are proportional to the modulation index.
Differences: in NBFM the lower sideband is inverted (180° phase shift). In AM the sideband phasors add to the carrier in phase, changing its amplitude; in NBFM they add at 90° to the carrier, changing mainly its phase. NBFM also has small amplitude variation and some higher-order terms that AM does not have.
Generation of NBFM by Armstrong's method
From the NBFM expression with general message:
The second term is a DSB-SC wave of the integrated message on a carrier shifted by 90°. This leads to the block diagram:
m(t) +------------+ +----------+ -A.phi.sin wct
---->| Integrator |---->| Balanced |----------+
+------------+ | modulator| |
+----------+ v
^ +---+ NBFM
+----------+ | + |------>
| -90 deg | +---+
| shift | ^
+----------+ |
^ |
+-------------+ | Ac cos wct |
| Crystal osc |---------+----------------+
| fc |
+-------------+
Working:
- The message is integrated (so that the phase modulator produces FM rather than PM).
- A crystal oscillator gives a very stable carrier .
- The carrier is shifted by 90° and fed to a balanced modulator with the integrated message, producing (DSB-SC).
- The carrier is added to this output, giving NBFM.
The deviation is kept small (, about 25 Hz in practice) to limit distortion. Wideband FM is then obtained by frequency multipliers and a mixer. The main merit is crystal-controlled carrier stability.
- 2070 Bhadra (CS I) · 6 marks
Explain with necessary mathematical relations, the demodulation of FM wave using non-synchronous method.
Answer
Non-synchronous (non-coherent) FM demodulation recovers the message without generating a local carrier in phase with the received carrier. The standard method is differentiation followed by envelope detection (the discriminator method).
FM +-------+ +-----+ +------------+ +--------+ m(t)
--->|Limiter|>| BPF |>|Differentia-|>|Envelope|---->
+-------+ +-----+ |tor j2pi af | |detector|
+------------+ +--------+
Received FM (after limiter and BPF):
Step 1: Differentiation. A circuit whose gain rises linearly with frequency, , acts as a differentiator:
is still FM, but its amplitude now varies as , i.e. it is an AM–FM wave.
Step 2: Envelope detection. Since , the envelope is always positive:
Step 3: DC removal. A blocking capacitor removes :
Practical circuits:
- Slope detector: a tuned circuit with resonance slightly above ; the carrier sits on its sloping skirt, so amplitude varies with frequency. Simple but non-linear.
- Balanced slope detector: two tuned circuits at with diode detectors connected back to back; the output is the difference, giving an S-shaped, more linear response and no DC.
- Foster–Seeley discriminator and ratio detector: use the phase shift of a double-tuned transformer.
- Zero-crossing detector: counts zero crossings; the average of a pulse train triggered at each crossing is proportional to the instantaneous frequency.
Role of the limiter: the envelope detector also responds to amplitude noise. The limiter makes the input amplitude constant so that only frequency variations reach the output.
- 2069 Bhadra (CS I) · 6+4 marks
Find the time domain expression for single tone FM modulated wave, in terms of Bessel coefficients. Derive the expression for estimating practical bandwidth of a FM signal.
Answer
Single-tone FM in terms of Bessel coefficients
For :
The term is periodic with period , so it can be written as a complex Fourier series:
with coefficients (put )
which is the Bessel function of the first kind of order . Substituting:
So the FM wave has a carrier component and infinitely many side frequencies at with amplitudes , using .
Properties: , so total power , independent of . The carrier amplitude falls as increases and is zero at
Practical bandwidth of FM
Although the spectrum is infinite, becomes very small when . For a fixed , decreases rapidly once exceeds about . Keeping the significant sidebands up to :
Carson's rule: taking (which keeps about 98% of the power):
Two limiting cases support this:
- (WBFM): (the carrier sweeps over ).
- (NBFM): only and are significant, so .
1% rule (universal curve): keep all sidebands with ; this gives a slightly larger bandwidth, e.g. for , and versus Carson's .
For a general message with highest frequency , use the deviation ratio : .
- 2069 Bhadra (CS I) · 2×5 marks
A sinusoidal modulating signal m(t) = 5cos 18849.55t is applied to an FM modulator that has a frequency sensitivity of 9 kHz/V. The amplitude of the carrier is 25 V and the frequency is 88.7 MHz. Compute:
i) Peak frequency deviation
ii) Modulation index
iii) Frequency swing
iv) Carson's bandwidth and
v) Total power delivered in 10 Ω resistor.
Answer
Given: , kHz/V, V, MHz, .
i) Peak frequency deviation
ii) Modulation index
iii) Frequency swing
The carrier moves between MHz and MHz.
iv) Carson's bandwidth
v) Power in 10 Ω
FM has constant amplitude, so total power equals the carrier power:
Answer: kHz, , swing = 90 kHz, kHz, W.
- 2068 Bhadra (CS I) · 8 marks
For the following Armstrong FM transmitter, compute maximum frequency deviation and the carrier frequency fc if f1 = 200 kHz, fLO = 10.8 MHz, Δf1 = 24 Hz, n1 = 65 and n2 = 50.
[Figure: m(t) → NBFM (output f1, Δf1) → frequency multiplier ×n1 → mixer with local oscillator fLO → frequency multiplier ×n2 → S_WBFM(t) with fc, Δfc]
Answer
Given: kHz, Hz, , MHz, .
Rules: a multiplier multiplies both carrier and deviation; a mixer changes only the carrier (deviation unchanged). The mixer output is taken as the difference frequency (normal Armstrong design, to bring the carrier down before the second multiplier).
Step 1: After the first multiplier ()
Step 2: After the mixer
Step 3: After the second multiplier ()
Check: kHz.
| Stage | Carrier | Deviation |
|---|---|---|
| NBFM output | 200 kHz | 24 Hz |
| After ×65 | 13 MHz | 1.56 kHz |
| After mixer | 2.2 MHz | 1.56 kHz |
| After ×50 | 110 MHz | 78 kHz |
(If the sum frequency 23.8 MHz were selected, would be 1190 MHz with the same deviation, which is not a practical FM carrier.)
Answer: maximum frequency deviation kHz, carrier frequency MHz.
- 2067 Mangsir (CS I) · 2×5 marks
A modulated signal has the expression:
Z(t) = 50 cos{98.6 × 10⁶ × 2πt + 15 sin(2 × 10³ × 2πt)} volts
Determine: a) type of modulation, b) frequency deviation, c) frequency sensitivity of the modulator d) modulation index and e) power delivered to a 50 Ohms impedance transmitting antenna.
Answer
Given: V, .
Comparing with : V, MHz, kHz, .
a) Type of modulation
The amplitude is constant and the phase varies sinusoidally, so it is angle modulation. If the message is , the phase term is its integral, so it is frequency modulation (FM); the carrier at 98.6 MHz is also in the FM broadcast band. (If the message were , the same wave would be PM; the expression alone cannot separate the two.)
b) Frequency deviation
c) Frequency sensitivity
. The message amplitude is not given, so taking V:
(For any other , kHz/V.)
d) Modulation index
e) Power delivered to 50 Ω antenna
Answer: FM; kHz; kHz/V (for V); ; W.
- 2067 Shrawan (CS I) · 2.5×4 marks
A harmonic signal m(t) = 20 cos(2π 2000 t) Volt is used to frequency modulate the carrier signal c(t) = 50 cos(2π 10⁷ t) Volt. Assuming the frequency sensitivity of the frequency modulator to be 200 Hz/Volt, calculate:
a) peak frequency deviation
b) modulation index
c) bandwidth of modulated signal for over 98% of FM power
d) total modulated signal power
Answer
Given: V, V, Hz/V.
So V, kHz, V, MHz.
a) Peak frequency deviation
b) Modulation index
c) Bandwidth containing over 98% of power
Bessel coefficients for :
| 0 | 1 | 2 | 3 | 4 | |
|---|---|---|---|---|---|
| 0.2239 | 0.5767 | 0.3528 | 0.1289 | 0.0340 |
Fraction of power in carrier + sideband pairs :
- up to : (96.4%, not enough)
- up to : (99.8%)
So 3 pairs of sidebands are needed:
This agrees with Carson's rule: kHz.
d) Total modulated signal power
Taking a 1 Ω load (normalised power):
(FM power equals unmodulated carrier power, independent of .)
Answer: kHz, , kHz, W (normalised to 1 Ω).
- 2065 Kartik (CS I) · 6+2 marks
How can you generate FM wave using Armstrong modulator (Indirect method)? Explain with the help of block diagram. Why pre-emphasis and de-emphasis networks are used in FM?
Answer
Armstrong (indirect) FM modulator
In the indirect method, a narrowband FM wave is first generated using a crystal-controlled phase modulator, and then frequency multipliers and a mixer convert it to wideband FM.
m(t) +----------+ +----------+ NBFM
---->|Integrator|-->| Balanced |--+
+----------+ | modulator| |
+----------+ v
^ +---+ f1, df1
+--------+ +------+ | + |---------+
|Crystal |->|-90deg| +---+ |
| osc f1 | +------+ ^ v
+--------+--------------+ +-------+
| x n1 |
WBFM +-------+ +-------+ +-------+
<-------| x n2 |<--| Mixer |<---------------+
fc, df +-------+ +-------+
^
[LO fLO]
Working:
- NBFM generation: for small ,
The message is integrated, multiplied with the 90°-shifted crystal carrier in a balanced modulator (giving the DSB-SC term), and the carrier is added. Deviation is kept very small (e.g. Hz, ) to avoid distortion. 2. First multiplier (): a non-linear device and BPF multiply the instantaneous frequency, so carrier becomes and deviation . 3. Mixer: shifts the carrier to without changing the deviation. This keeps the final carrier at the required value. 4. Second multiplier (): gives the final deviation and carrier:
Example: kHz, Hz, , MHz, gives MHz MHz and kHz.
Merits: excellent carrier stability (crystal oscillator, no AFC needed). Demerits: many multiplier stages, more complex, multiplied phase noise.
Why pre-emphasis and de-emphasis are used
- After FM detection, the output noise power spectral density is proportional to , so noise is heaviest at high audio frequencies.
- High-frequency audio components have low amplitude, so their SNR is poor.
- Pre-emphasis (high-pass RC, 75 μs) at the transmitter boosts high audio frequencies before modulation; de-emphasis (low-pass RC, 75 μs) at the receiver restores the original balance and at the same time cuts the high-frequency noise. The overall SNR improves by about 10–13 dB without distorting the message.
- 2065 Kartik (CS I) · 2+2+2+2 marks
A modulating signal m(t) = 3cos(2000t) modulates the carrier signal c(t) = 9cos(70000t) to produce the modulated signal s(t) = 9cos(70000t + 16 sin2000t). Calculate: the total modulated signal power, modulation index, peak frequency deviation and the bandwidth of modulated signal.
Answer
Given: , , .
So V, rad/s, .
Total modulated signal power
FM has constant amplitude, so (normalised to 1 Ω):
Modulation index
From the expression directly:
Peak frequency deviation
(In rad/s: rad/s. The frequency sensitivity is Hz/V.)
Bandwidth (Carson's rule)
(Equivalently rad/s.)
Answer: W, , kHz, kHz.
- 2065 Kartik (CS I) · 6+2 marks
Explain stereo FM transmitter and receiver with the help of block diagrams. If mono FM receiver is used to receive the signal from stereo FM transmitter, what will be the output of FM receiver? Explain.
Answer
Stereo FM sends two audio channels, left (L) and right (R), on one FM carrier in a way that is compatible with mono receivers. It uses frequency-division multiplexing of the sum and difference signals.
Stereo FM transmitter
L,R->+--------+ L+R ---------------->+-----+
| Matrix | | |
| (+, -) | L-R ->[Balanced ]--->| Sum |->[FM ]-> out
+--------+ [modulator] | | [mod]
^ 38 kHz | |
[19 kHz osc]->[x2]--+ | |
| | |
+---- pilot 19 kHz ----->+-----+
- A matrix forms and (both 30 Hz – 15 kHz).
- DSB-SC modulates a 38 kHz subcarrier, occupying 23–53 kHz.
- A 19 kHz pilot (half the subcarrier frequency, about 10% deviation) is added so the receiver can regenerate 38 kHz in phase.
- The composite baseband signal is
and it frequency-modulates the carrier (total deviation 75 kHz).
Baseband spectrum: L+R: 0-15 kHz | pilot 19 kHz | L-R DSB-SC: 23-53 kHz.
Stereo FM receiver
+--------+ +--------+ L+R +--------+--> L
FM ->| FM |--+->| LPF |----->| Matrix |
|receiver| | | 15 kHz | | (+,-) |--> R
+--------+ | +--------+ +--------+
| +--------+ +-----+ ^ L-R
+->| BPF |->|Prod.|--+ (via LPF)
| |23-53kHz| | det.|
| +--------+ +-----+
| +--------+ +----+ ^ 38 kHz
+->|BPF 19k |->| x2 |---+
+--------+ +----+
- The FM detector gives the composite baseband signal.
- A 15 kHz LPF gives .
- A 23–53 kHz BPF and a synchronous (product) detector with the regenerated 38 kHz carrier (pilot ×2) give .
- The matrix forms and , fed to the two speakers.
Output of a mono receiver
A mono FM receiver's audio stage passes only up to about 15 kHz. The signal at 23–53 kHz and the 19 kHz pilot are outside its audio band (and are removed by its de-emphasis and audio filtering). So the output is only
i.e. a normal monophonic signal containing both channels. This is why the system is called compatible: stereo broadcasts can be received on old mono sets without loss of programme content.
- 2065 Kartik (CS I) · 4 marks
Write a short note on phase modulation.
Answer
Phase modulation (PM) is angle modulation in which the instantaneous phase of the carrier varies linearly with the message signal, while the amplitude stays constant.
where is the phase sensitivity (rad/V).
Instantaneous frequency:
so in PM the frequency deviation depends on the rate of change of the message.
Single tone: for ,
The PM index is independent of , while the frequency deviation increases with .
Relation with FM:
- PM of = FM of (differentiator + FM modulator gives PM).
- FM of = PM of (integrator + phase modulator gives FM, used in Armstrong's method).
Features: constant envelope and power ; spectrum given by Bessel functions as in FM; bandwidth by Carson's rule . PM needs coherent (phase-reference) detection. Digital forms of PM (BPSK, QPSK) are widely used in modern data communication.
- 2064 Shrawan (CS I) · 6+2 marks
Explain how frequency modulated signals can be generated by direct method using a varactor diode. What are the disadvantages of such a method?
Answer
In the direct method, the frequency of an LC oscillator is varied directly by the message, using a voltage-variable reactance. A varactor diode is a reverse-biased p-n junction whose depletion capacitance changes with reverse voltage.
Varactor diode FM generator
+Vcc (bias)
|
RFC +-----------+
| | LC tank |
m(t) --||--+-+---||----+---------| L || C0 |--> FM out
Cc Cb | | Hartley / |
--- Cv | Colpitts |
/\ varactor| osc. |
| +-----------+
GND
- A DC reverse bias sets the operating point; the message is added through a coupling capacitor and RF choke.
- The varactor capacitance is in parallel with the tank capacitance .
Analysis. The junction capacitance is (n = 1/2 for abrupt junction). For a small message, it varies nearly linearly:
The oscillator frequency is
For , using :
with and . The instantaneous frequency varies linearly with the message, i.e. FM.
Disadvantages
- Poor carrier stability: an LC oscillator (not crystal) drifts with temperature and supply; an AFC loop is needed.
- Non-linearity: varies non-linearly with voltage, so large deviations cause distortion; only small is usable.
- Varactor capacitance is temperature-sensitive.
- The crystal oscillator cannot be used directly, since its frequency cannot be pulled much.
- 2064 Shrawan (CS I) · 4×2 marks
A low frequency signal m(t) = 2cos(5000t) modulates the carrier signal c(t) = 10cos(100,000t) to produce the modulated signal u(t) = 10cos(100,000t + 10sin5000t). Calculate: the total modulated signal power; modulation index; peak frequency deviation; the bandwidth of modulated signal.
Answer
Given: , , .
So V, rad/s, V, rad/s, .
Total modulated signal power
FM has constant amplitude; normalised to 1 Ω:
Modulation index
From the expression: .
Peak frequency deviation
( rad/s; rad/s/V, i.e. 3978.9 Hz/V.)
Bandwidth (Carson's rule)
(In rad/s: rad/s.)
Answer: W, , kHz, kHz.
- 2064 Shrawan (CS I) · 6+2 marks
Show that PLL can be used to demodulate FM signal with a neat diagram. What are the basic uses of PLL?
Answer
A phase-locked loop (PLL) is a negative-feedback system in which a voltage-controlled oscillator (VCO) is forced to track the phase (and so the frequency) of the input signal. When it tracks an FM wave, the VCO control voltage is the message.
FM in +--------+ e(t) +-------+ v(t) output
s(t) --->| Phase |----->| Loop |---+----> m(t)
|detector| |filter | |
+--------+ | H(f) | |
^ +-------+ |
| r(t) +-------+ |
+----------| VCO |<---+
+-------+
Proof that PLL demodulates FM
- Input FM: , with .
- VCO output (free-running at ): , with .
- Phase detector (multiplier) output:
The loop filter rejects the term, leaving
- Linearised model: when locked, is small, so . The loop adjusts to keep near zero. Then
- Differentiating:
So the loop-filter output is proportional to the message: the PLL is an FM demodulator. For good tracking the loop gain must be high and the loop bandwidth must cover the message bandwidth.
Basic uses of PLL
- FM and FSK demodulation
- Carrier recovery for coherent (synchronous) detection, e.g. stereo pilot, DSB-SC, PSK
- Frequency synthesisers (with a divider in the feedback path) in radios and phones
- Clock and bit-timing recovery in digital receivers
- Frequency multiplication/division and tracking filters
Questions from Old Question Collection (BEI EX 656) (BEI Communication Systems (EX 656) exam papers, 2078 to 2081 Chaitra), Communication System I (EX 652) (BEX Communication System I (EX 652) papers 2064 to 2080, plus two old BCT Communication Systems papers (2068, 2071)) and Communication System II (EX 702) (BEX Communication System II (EX 702) exam papers, 2069 to 2081). Answers are written for this site; check them against your class notes.
Chapter titles and hours from the IOE syllabus ↗