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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 m(t)=Amcos⁡ωmtm(t) = A_m\cos\omega_mt and c(t)=Accos⁡ωctc(t) = A_c\cos\omega_ct. The instantaneous frequency is

fi(t)=fc+kfm(t)=fc+kfAmcos⁡ωmt=fc+Δfcos⁡ωmtf_i(t) = f_c + k_fm(t) = f_c + k_fA_m\cos\omega_mt = f_c + \Delta f\cos\omega_mt

where kfk_f is the frequency sensitivity (Hz/V) and Δf=kfAm\Delta f = k_fA_m is the peak frequency deviation. The phase is

θi(t)=2π∫0tfi(τ) dτ=ωct+Δffmsin⁡ωmt=ωct+βsin⁡ωmt\theta_i(t) = 2\pi\int_0^t f_i(\tau)\,d\tau = \omega_ct + \frac{\Delta f}{f_m}\sin\omega_mt = \omega_ct + \beta\sin\omega_mt

with modulation index β=Δf/fm\beta = \Delta f/f_m. So

s(t)=Accos⁡(ωct+βsin⁡ωmt)s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt)

Complex envelope and Fourier series

s(t)=Re[Ac ejβsin⁡ωmt ejωct]s(t) = \text{Re}\left[A_c\,e^{j\beta\sin\omega_mt}\,e^{j\omega_ct}\right]

The term s~(t)=Acejβsin⁡ωmt\tilde s(t) = A_ce^{j\beta\sin\omega_mt} is periodic with period Tm=1/fmT_m = 1/f_m, so it has a Fourier series:

s~(t)=∑n=−∞∞cnejnωmt\tilde s(t) = \sum_{n=-\infty}^{\infty} c_n e^{jn\omega_mt} cn=1Tm∫−Tm/2Tm/2Acejβsin⁡ωmte−jnωmtdtc_n = \frac{1}{T_m}\int_{-T_m/2}^{T_m/2} A_c e^{j\beta\sin\omega_mt}e^{-jn\omega_mt}dt

Put x=ωmtx = \omega_mt (dt=dx/ωmdt = dx/\omega_m, limits −π-\pi to π\pi):

cn=Ac2π∫−ππej(βsin⁡x−nx)dx=AcJn(β)c_n = \frac{A_c}{2\pi}\int_{-\pi}^{\pi} e^{j(\beta\sin x - nx)}dx = A_cJ_n(\beta)

where Jn(β)J_n(\beta) is the Bessel function of the first kind of order n:

Jn(β)=12π∫−ππej(βsin⁡x−nx)dxJ_n(\beta) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j(\beta\sin x - nx)}dx

Final expression

s~(t)=Ac∑n=−∞∞Jn(β)ejnωmt\tilde s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)e^{jn\omega_mt} s(t)=Re[Ac∑nJn(β)ej(ωc+nωm)t]=Ac∑n=−∞∞Jn(β)cos⁡[(ωc+nωm)t]s(t) = \text{Re}\left[A_c\sum_n J_n(\beta)e^{j(\omega_c + n\omega_m)t}\right] = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[(\omega_c + n\omega_m)t]

Expanded, using J−n(β)=(−1)nJn(β)J_{-n}(\beta) = (-1)^nJ_n(\beta):

s(t)=AcJ0(β)cos⁡ωct+AcJ1(β)[cos⁡(ωc+ωm)t−cos⁡(ωc−ωm)t]+AcJ2(β)[cos⁡(ωc+2ωm)t+cos⁡(ωc−2ωm)t]+…\begin{aligned} s(t) &= A_cJ_0(\beta)\cos\omega_ct \\ &\quad + A_cJ_1(\beta)[\cos(\omega_c+\omega_m)t - \cos(\omega_c-\omega_m)t] \\ &\quad + A_cJ_2(\beta)[\cos(\omega_c+2\omega_m)t + \cos(\omega_c-2\omega_m)t] + \dots \end{aligned}

Conclusions

  1. The spectrum has a carrier at fcf_c (amplitude AcJ0(β)A_cJ_0(\beta)) and infinite sidebands at fc±nfmf_c \pm nf_m with amplitudes Ac∣Jn(β)∣A_c|J_n(\beta)|.
  2. Odd-order lower sidebands have reversed sign.
  3. The carrier amplitude depends on β\beta and becomes zero when J0(β)=0J_0(\beta) = 0 (β = 2.405, 5.52, ...).
  4. Since ∑Jn2(β)=1\sum J_n^2(\beta) = 1, total power is constant: P=Ac2/2P = A_c^2/2.
  5. Only about β+1\beta + 1 sidebands on each side are significant, giving Carson's rule BT≈2(β+1)fm=2(Δf+fm)B_T \approx 2(\beta+1)f_m = 2(\Delta f + f_m).
  • 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): SN(f)∝f2S_N(f) \propto f^2.
  • 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: Hpe(f)=1+jff1H_{pe}(f) = 1 + j\dfrac{f}{f_1}, where f1=12πRCf_1 = \dfrac{1}{2\pi RC}.
  • De-emphasis: Hde(f)=11+jf/f1H_{de}(f) = \dfrac{1}{1 + jf/f_1}, so HpeHde=1H_{pe}H_{de} = 1 and the message is undistorted.
  • Standard time constant RC=75 μRC = 75\ \mus (f1≈2.1f_1 \approx 2.1 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

  1. A matrix circuit (adder and subtractor) forms L+RL+R (mono-compatible sum) and L−RL-R (difference).
  2. Both are low-pass filtered to 15 kHz and pre-emphasized.
  3. L−RL-R is DSB-SC modulated in a balanced modulator onto a 38 kHz sub-carrier, giving sidebands from 23 to 53 kHz.
  4. 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.
  5. The composite signal is
m(t)=(L+R)+(L−R)cos⁡(2π 38000 t)+Kcos⁡(2π 19000 t)m(t) = (L+R) + (L-R)\cos(2\pi\,38000\,t) + K\cos(2\pi\,19000\,t)

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 (L+RL+R); a stereo receiver recovers L−RL-R as well and forms LL and RR 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

  1. RF amplifier: tuned to the wanted station fsf_s. It gives gain at a low noise figure (important at VHF), improves sensitivity and rejects the image frequency fs+2(10.7)f_s + 2(10.7) MHz.
  2. Local oscillator and mixer: the oscillator is ganged with the RF tuning so that fLO=fs+10.7f_{LO} = f_s + 10.7 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.
  3. IF amplifier: several fixed-tuned stages at 10.7 MHz with a bandwidth of about 200 kHz (Carson bandwidth for Δf=75\Delta f = 75 kHz, W=15W = 15 kHz is 180 kHz). Provides most of the gain and the selectivity.
  4. 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.
  5. 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 kfm(t)k_fm(t).
  6. 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.
  7. AF amplifier and speaker: amplifies the audio to drive the loudspeaker. In stereo sets a stereo decoder sits between discriminator and AF stage.
  8. 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

PointAM receiverFM receiver
Signal range0.535–1.605 MHz88–108 MHz
IF455 kHz10.7 MHz
IF bandwidth10 kHz200 kHz
LimiterNot usedUsed
DetectorEnvelope detectorDiscriminator / PLL
Control loopAGCAFC (and AGC)
De-emphasisNoYes

Example: for a station at 100 MHz, fLO=110.7f_{LO} = 110.7 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 L+RL+R 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----+
  1. Matrix: an adder and a subtractor form L+RL+R and L−RL-R.
  2. LPF and pre-emphasis: each signal is limited to 15 kHz and pre-emphasized.
  3. Balanced modulator: L−RL-R DSB-SC modulates a 38 kHz sub-carrier. Since the carrier is suppressed, its power goes into useful sidebands (23–53 kHz).
  4. 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.
  5. Adder: forms the composite signal that frequency-modulates the RF carrier:
m(t)=[L(t)+R(t)]+[L(t)−R(t)]cos⁡(2π 38k t)+Apcos⁡(2π 19k t)m(t) = [L(t)+R(t)] + [L(t)-R(t)]\cos(2\pi\,38\text{k}\,t) + A_p\cos(2\pi\,19\text{k}\,t)

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 ---+
  1. The FM discriminator output is the composite signal m(t)m(t).
  2. LPF (0–15 kHz) gives L+RL+R. A mono receiver stops here.
  3. 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.
  4. BPF (23–53 kHz) selects the L−RL-R DSB-SC signal. It is multiplied by the 38 kHz carrier and low-pass filtered (coherent detection), giving L−RL-R.
  5. De-matrixing:
(L+R)+(L−R)=2L,(L+R)−(L−R)=2R(L+R) + (L-R) = 2L, \qquad (L+R) - (L-R) = 2R
  1. 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 L−RL-R 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: m(t)=5cos⁡(18849.55t)m(t) = 5\cos(18849.55t), so Am=5A_m = 5 V and

fm=18849.552π=3000 Hz=3 kHzf_m = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}

kf=9k_f = 9 kHz/V, Ac=25A_c = 25 V, fc=88.7f_c = 88.7 MHz, R=10 ΩR = 10\ \Omega.

Peak frequency deviation:

Δf=kfAm=9×5=45 kHz\Delta f = k_fA_m = 9 \times 5 = 45\ \text{kHz}

a) Modulation index

β=Δffm=453=15\beta = \frac{\Delta f}{f_m} = \frac{45}{3} = 15

(wide-band FM, since β≫1\beta \gg 1)

b) Carrier frequency swing

The carrier swings from fc−Δff_c - \Delta f to fc+Δff_c + \Delta f, i.e. from 88.655 MHz to 88.745 MHz:

Carrier swing=2Δf=2×45=90 kHz\text{Carrier swing} = 2\Delta f = 2 \times 45 = 90\ \text{kHz}

c) Bandwidth by Carson's rule

BT=2(Δf+fm)=2(45+3)=96 kHzB_T = 2(\Delta f + f_m) = 2(45 + 3) = 96\ \text{kHz}

d) Total power in 10 Ω

The FM wave has constant amplitude, so its power does not depend on modulation:

P=Ac22R=2522×10=62520=31.25 WP = \frac{A_c^2}{2R} = \frac{25^2}{2 \times 10} = \frac{625}{20} = 31.25\ \text{W}

Answer: β=15\beta = 15; carrier swing = 90 kHz; BT=96B_T = 96 kHz; P=31.25P = 31.25 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 2fc2f_c term and sets loop dynamics.
  • VCO: its frequency is fc+kvv(t)f_c + k_vv(t).

Analysis

Input FM wave and VCO output:

s(t)=Acsin⁡[ωct+ϕ1(t)],ϕ1(t)=2πkf∫0tm(τ)dτs(t) = A_c\sin[\omega_ct + \phi_1(t)], \quad \phi_1(t) = 2\pi k_f\int_0^t m(\tau)d\tau r(t)=Avcos⁡[ωct+ϕ2(t)],ϕ2(t)=2πkv∫0tv(τ)dτr(t) = A_v\cos[\omega_ct + \phi_2(t)], \quad \phi_2(t) = 2\pi k_v\int_0^t v(\tau)d\tau

The multiplier output, after removing the 2ωc2\omega_c term, is

e(t)=Kmsin⁡[ϕ1(t)−ϕ2(t)]=Kmsin⁡ϕe(t)e(t) = K_m\sin[\phi_1(t) - \phi_2(t)] = K_m\sin\phi_e(t)

When the loop is locked, ϕe\phi_e is small, so sin⁡ϕe≈ϕe\sin\phi_e \approx \phi_e (linear model). With high loop gain the loop drives ϕe→0\phi_e \to 0, so

ϕ2(t)≈ϕ1(t)  ⇒  2πkv∫v(t)dt≈2πkf∫m(t)dt\phi_2(t) \approx \phi_1(t) \;\Rightarrow\; 2\pi k_v\int v(t)dt \approx 2\pi k_f\int m(t)dt

Differentiating both sides:

v(t)≈kfkvm(t)v(t) \approx \frac{k_f}{k_v}m(t)

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

  1. Phase detector: a multiplier comparing input and VCO output; its output depends on the phase difference.
  2. Loop filter: a low-pass filter that removes the double-frequency term and controls loop bandwidth and stability.
  3. VCO: free-running at fcf_c when v=0v = 0; its frequency changes by kvk_v Hz per volt.

Analysis

Input FM signal:

s(t)=Acsin⁡[ωct+ϕ1(t)],ϕ1(t)=2πkf∫0tm(τ) dτs(t) = A_c\sin[\omega_ct + \phi_1(t)], \qquad \phi_1(t) = 2\pi k_f\int_0^t m(\tau)\,d\tau

VCO output (90° offset so the error is zero at lock):

r(t)=Avcos⁡[ωct+ϕ2(t)],ϕ2(t)=2πkv∫0tv(τ) dτr(t) = A_v\cos[\omega_ct + \phi_2(t)], \qquad \phi_2(t) = 2\pi k_v\int_0^t v(\tau)\,d\tau

Multiplier output:

s(t)r(t)=AcAv2{sin⁡[2ωct+ϕ1+ϕ2]+sin⁡[ϕ1−ϕ2]}s(t)r(t) = \frac{A_cA_v}{2}\left\{\sin[2\omega_ct + \phi_1 + \phi_2] + \sin[\phi_1 - \phi_2]\right\}

The loop filter removes the 2ωc2\omega_c term. The error signal is

e(t)=Kmsin⁡ϕe(t),ϕe=ϕ1−ϕ2e(t) = K_m\sin\phi_e(t), \qquad \phi_e = \phi_1 - \phi_2

where Km=AcAv/2K_m = A_cA_v/2 (times the multiplier gain).

Linear model (loop in lock)

In lock ϕe\phi_e is small, so sin⁡ϕe≈ϕe\sin\phi_e \approx \phi_e:

 phi1 -->(+)--> K_m --> H(f) --+--> v(t)
          ^ -                  |
          |                    |
         phi2 <--[2 pi kv / s]-+
v(t)=Km [ϕe(t)∗h(t)],ϕ2(t)=2πkv∫v dtv(t) = K_m\,[\phi_e(t) * h(t)], \qquad \phi_2(t) = 2\pi k_v\int v\,dt

In the frequency domain,

Φe(f)=Φ1(f)1+L(f),L(f)=KmkvH(f)jf\Phi_e(f) = \frac{\Phi_1(f)}{1 + L(f)}, \qquad L(f) = \frac{K_mk_vH(f)}{jf}

When the loop gain ∣L(f)∣≫1|L(f)| \gg 1 over the message band, Φe≈0\Phi_e \approx 0, so ϕ2(t)≈ϕ1(t)\phi_2(t) \approx \phi_1(t):

2πkv∫v(t) dt=2πkf∫m(t) dt  ⇒  v(t)=kfkv m(t)2\pi k_v\int v(t)\,dt = 2\pi k_f\int m(t)\,dt \;\Rightarrow\; v(t) = \frac{k_f}{k_v}\,m(t)

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, fc+kfm(t)f_c + k_fm(t), i.e. it follows m(t)m(t).

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 Δf\Delta f 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 f2f^2 (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, Hpe(f)=1+jf/f1H_{pe}(f) = 1 + jf/f_1, 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, Hde(f)=1/(1+jf/f1)H_{de}(f) = 1/(1 + jf/f_1), brings the boosted frequencies back to their original level. It attenuates the high-frequency noise by the same amount.
  • Since HpeHde=1H_{pe}H_{de} = 1, the message is undistorted, but the noise is reduced. The improvement is about 10–13 dB.
  • Standard time constant: 75 µs (f1=2.12f_1 = 2.12 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----+
  1. Matrix network: adder and subtractor make L+RL+R (for mono compatibility) and L−RL-R.
  2. LPF (15 kHz) and pre-emphasis on both.
  3. Balanced modulator: L−RL-R DSB-SC modulates a 38 kHz sub-carrier (sidebands 23–53 kHz).
  4. 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.
  5. Adder: composite signal
m(t)=(L+R)+(L−R)cos⁡(2π 38k t)+Apcos⁡(2π 19k t)m(t) = (L+R) + (L-R)\cos(2\pi\,38\text{k}\,t) + A_p\cos(2\pi\,19\text{k}\,t)
 |  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

  1. 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
sNB(t)≈Accos⁡ωc1t−βAcsin⁡ωmt sin⁡ωc1t,β<0.3s_{NB}(t) \approx A_c\cos\omega_{c1}t - \beta A_c\sin\omega_mt\,\sin\omega_{c1}t, \quad \beta < 0.3

which is NBFM. A crystal oscillator keeps fc1f_{c1} very stable. 2. Frequency multiplication (×n₁): a non-linear stage multiplies both carrier and deviation: n1fc1n_1f_{c1}, n1Δf1n_1\Delta f_1. 3. Mixer: shifts the carrier down to a convenient value (n1fc1−fc2n_1f_{c1} - f_{c2}) without changing the deviation. 4. Second multiplier (×n₂): gives the final carrier fc=n2(n1fc1−fc2)f_c = n_2(n_1f_{c1} - f_{c2}) and deviation Δf=n1n2Δf1\Delta f = n_1n_2\Delta f_1 (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.

  1. Noise in FM rises with frequency. The FM discriminator output noise PSD is SN(f)∝f2S_{N}(f) \propto f^2 (triangular noise voltage, parabolic noise power). So high-frequency audio components get much more noise.
  2. Audio signals are weak at high frequencies. Most speech and music energy is below 1–2 kHz.
  3. 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, Hpe(f)=1+jf/f1H_{pe}(f) = 1 + jf/f_1.
  • De-emphasis at the receiver: a low-pass RC network after demodulation, Hde(f)=1/(1+jf/f1)H_{de}(f) = 1/(1 + jf/f_1), restores the original response and attenuates the high-frequency noise at the same time.
  • Since HpeHde=1H_{pe}H_{de} = 1, 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), f1=1/(2πRC)≈2.1f_1 = 1/(2\pi RC) \approx 2.1 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) SX(f)S_X(f) of a wide-sense stationary process and its autocorrelation function RX(τ)R_X(\tau) form a Fourier transform pair:

SX(f)=∫−∞∞RX(τ) e−j2πfτdτ,RX(τ)=∫−∞∞SX(f) ej2πfτdfS_X(f) = \int_{-\infty}^{\infty}R_X(\tau)\,e^{-j2\pi f\tau}d\tau, \qquad R_X(\tau) = \int_{-\infty}^{\infty}S_X(f)\,e^{j2\pi f\tau}df

Consequences:

  • Total average power: P=RX(0)=∫SX(f) dfP = R_X(0) = \int S_X(f)\,df.
  • SX(f)S_X(f) is real, even and non-negative because RX(τ)R_X(\tau) is real and even.
  • Example: white noise R(τ)=N02δ(τ)⇔S(f)=N02R(\tau) = \frac{N_0}{2}\delta(\tau) \Leftrightarrow S(f) = \frac{N_0}{2}.

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----+
  1. Matrix forms L+RL+R and L−RL-R; both are band-limited to 15 kHz and pre-emphasized.
  2. L−RL-R DSB-SC modulates a 38 kHz sub-carrier in a balanced modulator.
  3. The 38 kHz sub-carrier comes from a 19 kHz oscillator and a doubler; a small 19 kHz pilot is also added.
  4. Composite signal:
m(t)=(L+R)+(L−R)cos⁡(2π 38k t)+Apcos⁡(2π 19k t)m(t) = (L+R) + (L-R)\cos(2\pi\,38\text{k}\,t) + A_p\cos(2\pi\,19\text{k}\,t)

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 ---+
  1. FM discriminator gives the composite signal.
  2. LPF (0–15 kHz) gives L+RL+R (all a mono set needs).
  3. A narrow BPF picks the 19 kHz pilot; a doubler makes a phase-locked 38 kHz carrier.
  4. BPF (23–53 kHz) selects the L−RL-R DSB-SC signal, which is coherently detected with the 38 kHz carrier and low-pass filtered.
  5. Matrix: (L+R)+(L−R)=2L(L+R)+(L-R) = 2L and (L+R)−(L−R)=2R(L+R)-(L-R) = 2R, 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(t)=Acsin⁡(ωct+βsin⁡ωmt)v(t) = A_c\sin(\omega_ct + \beta\sin\omega_mt):

Ac=12A_c = 12 V, ωc=106\omega_c = 10^6 rad/s, ωm=104\omega_m = 10^4 rad/s, β=5\beta = 5.

a) Carrier frequency

fc=ωc2π=1062π=159.15 kHzf_c = \frac{\omega_c}{2\pi} = \frac{10^6}{2\pi} = 159.15\ \text{kHz}

b) Modulating frequency

fm=ωm2π=1042π=1591.55 Hz≈1.59 kHzf_m = \frac{\omega_m}{2\pi} = \frac{10^4}{2\pi} = 1591.55\ \text{Hz} \approx 1.59\ \text{kHz}

c) Modulation index

The peak phase deviation is the coefficient of sin⁡104t\sin 10^4t:

β=5 rad\beta = 5\ \text{rad}

(Treated as FM, the peak deviation is Δf=βfm=5×1591.55=7957.7\Delta f = \beta f_m = 5 \times 1591.55 = 7957.7 Hz.)

d) Power in 100 Ω

The amplitude of an angle-modulated wave is constant, so

P=Ac22R=1222×100=144200=0.72 WP = \frac{A_c^2}{2R} = \frac{12^2}{2 \times 100} = \frac{144}{200} = 0.72\ \text{W}

Answer: fc=159.15f_c = 159.15 kHz, fm=1.59f_m = 1.59 kHz, β=5\beta = 5, P=0.72P = 0.72 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: Am=10A_m = 10 V, fm=4000f_m = 4000 Hz, Ac=80A_c = 80 V, fc=107f_c = 10^7 Hz = 10 MHz, kf=400k_f = 400 Hz/V.

a) Peak frequency deviation

Δf=kfAm=400×10=4000 Hz=4 kHz\Delta f = k_fA_m = 400 \times 10 = 4000\ \text{Hz} = 4\ \text{kHz}

b) Modulation index

β=Δffm=40004000=1\beta = \frac{\Delta f}{f_m} = \frac{4000}{4000} = 1

c) Bandwidth containing over 98% of power

Carson's rule is based on keeping about 98% of the power:

BT=2(Δf+fm)=2(4+4)=16 kHzB_T = 2(\Delta f + f_m) = 2(4 + 4) = 16\ \text{kHz}

Check with Bessel functions for β=1\beta = 1: J0=0.7652J_0 = 0.7652, J1=0.4401J_1 = 0.4401, J2=0.1149J_2 = 0.1149, J3=0.0196J_3 = 0.0196.

Components keptPower fraction J02+2∑Jn2J_0^2 + 2\sum J_n^2
n=0,±1n = 0, \pm10.9728 (97.3%)
n=0,±1,±2n = 0, \pm1, \pm20.9992 (99.9%)

So sidebands up to n=2n = 2 are needed for more than 98% power:

B=2nfm=2×2×4=16 kHzB = 2nf_m = 2 \times 2 \times 4 = 16\ \text{kHz}

which agrees with Carson's rule.

d) Total power in unit impedance

FM has constant amplitude, so

P=Ac22R=8022×1=3200 WP = \frac{A_c^2}{2R} = \frac{80^2}{2 \times 1} = 3200\ \text{W}

Answer: Δf=4\Delta f = 4 kHz; β=1\beta = 1; B=16B = 16 kHz; P=3200P = 3200 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

PointNarrow band FMWide band FM
Modulation indexβ≪1\beta \ll 1 (< 0.3)β≫1\beta \gg 1
Bandwidth≈2fm\approx 2f_m (like AM)≈2(Δf+fm)\approx 2(\Delta f + f_m)
SidebandsCarrier + one pairMany significant pairs
Max deviationSmall (~5 kHz)Large (75 kHz in broadcast)
Noise performanceLittle better than AMMuch better SNR
UseMobile radio, police, Armstrong first stageFM 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----+
  1. Matrix makes L+RL+R and L−RL-R, each limited to 15 kHz and pre-emphasized.
  2. L−RL-R DSB-SC modulates a 38 kHz sub-carrier (balanced modulator).
  3. 38 kHz is obtained by doubling a 19 kHz oscillator; the 19 kHz pilot is added at low level for receiver synchronization.
  4. The composite signal
m(t)=(L+R)+(L−R)cos⁡(2π 38k t)+Apcos⁡(2π 19k t)m(t) = (L+R) + (L-R)\cos(2\pi\,38\text{k}\,t) + A_p\cos(2\pi\,19\text{k}\,t)

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 L+RL+R (mono output).
  • Pilot is filtered and doubled to 38 kHz; the 23–53 kHz band is coherently detected to give L−RL-R.
  • Adding and subtracting give 2L2L and 2R2R.
  • 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)

f1=65×200 kHz=13 MHzΔf1=65×25=1625 Hz\begin{aligned} f_1 &= 65 \times 200\ \text{kHz} = 13\ \text{MHz} \\ \Delta f_1 &= 65 \times 25 = 1625\ \text{Hz} \end{aligned}

Step 2: mixer with 10.8 MHz

The mixer gives 13±10.813 \pm 10.8 MHz, i.e. 2.2 MHz (difference) or 23.8 MHz (sum). Deviation stays 1625 Hz.

Step 3: second multiplier (n₂ = 50)

Δf=50×1625=81250 Hz=81.25 kHzfc=50×2.2=110 MHz(difference)fc=50×23.8=1190 MHz(sum)\begin{aligned} \Delta f &= 50 \times 1625 = 81250\ \text{Hz} = 81.25\ \text{kHz} \\ f_c &= 50 \times 2.2 = 110\ \text{MHz} \quad (\text{difference}) \\ f_c &= 50 \times 23.8 = 1190\ \text{MHz} \quad (\text{sum}) \end{aligned}

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, β=Δf/fm≪1\beta = \Delta f/f_m \ll 1 (practically β<0.3\beta < 0.3). Its bandwidth is about 2fm2f_m, like AM, with only the carrier and one pair of sidebands:
s(t)≈Accos⁡ωct−βAcsin⁡ωmt sin⁡ωcts(t) \approx A_c\cos\omega_ct - \beta A_c\sin\omega_mt\,\sin\omega_ct
  • Wide band FM (WBFM): FM with β≫1\beta \gg 1. It has many significant sidebands and bandwidth BT≈2(Δf+fm)B_T \approx 2(\Delta f + f_m) (Carson's rule). Example: FM broadcast with Δf=75\Delta f = 75 kHz, W=15W = 15 kHz, β=5\beta = 5.

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 ∫m dt\int m\,dt gives FM) and fed to a balanced modulator with a carrier shifted by −90°. Adding the carrier gives

s1(t)=Accos⁡ωc1t−Acβ1sin⁡ωmt sin⁡ωc1t≈Accos⁡(ωc1t+β1sin⁡ωmt)s_1(t) = A_c\cos\omega_{c1}t - A_c\beta_1\sin\omega_mt\,\sin\omega_{c1}t \approx A_c\cos(\omega_{c1}t + \beta_1\sin\omega_mt)

with β1\beta_1 kept small (< 0.5) to limit distortion. Typical values: fc1=100f_{c1} = 100–200 kHz, Δf1≈25\Delta f_1 \approx 25 Hz.

Step 2: frequency multiplier (×n₁). A non-linear device followed by a BPF tuned to the n-th harmonic. The output is cos⁡(n1ωc1t+n1β1sin⁡ωmt)\cos(n_1\omega_{c1}t + n_1\beta_1\sin\omega_mt): both carrier and deviation are multiplied by n1n_1.

Step 3: mixer (frequency converter). Mixing with a crystal oscillator fc2f_{c2} shifts the carrier to n1fc1−fc2n_1f_{c1} - f_{c2} while the deviation stays n1Δf1n_1\Delta f_1. This lets the final carrier be set independently of the deviation.

Step 4: frequency multiplier (×n₂) to reach the final values:

fc=n2(n1fc1−fc2),Δf=n1n2Δf1f_c = n_2(n_1f_{c1} - f_{c2}), \qquad \Delta f = n_1n_2\Delta f_1

Step 5: power amplifier and antenna.

Example: Δf1=25\Delta f_1 = 25 Hz needs n1n2=3000n_1n_2 = 3000 for Δf=75\Delta f = 75 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 L+RL+R and L−RL-R.
  • 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 L−RL-R on 38 kHz, producing 23–53 kHz.
  • Adder: sums L+RL+R, the DSB-SC L−RL-R and a low-level 19 kHz pilot into the composite signal, which drives the FM modulator.
m(t)=(L+R)+(L−R)cos⁡(2π 38k t)+Apcos⁡(2π 19k t)m(t) = (L+R) + (L-R)\cos(2\pi\,38\text{k}\,t) + A_p\cos(2\pi\,19\text{k}\,t)
 |  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 L+RL+R (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 L−RL-R DSB-SC signal.
  • Product detector and LPF: coherent detection with 38 kHz gives L−RL-R.
  • Matrix: (L+R)+(L−R)=2L(L+R)+(L-R) = 2L, (L+R)−(L−R)=2R(L+R)-(L-R) = 2R.
  • 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 Jn(β)J_n(\beta)

  1. J−n(β)=(−1)nJn(β)J_{-n}(\beta) = (-1)^nJ_n(\beta): negative-order coefficients equal positive ones, with sign change for odd nn.
  2. ∑n=−∞∞Jn2(β)=1\sum_{n=-\infty}^{\infty}J_n^2(\beta) = 1: total FM power is constant.
  3. For small β\beta: J0(β)≈1J_0(\beta) \approx 1, J1(β)≈β/2J_1(\beta) \approx \beta/2, Jn(β)≈0J_n(\beta) \approx 0 for n≥2n \ge 2 (NBFM).
  4. For a given β\beta, Jn(β)J_n(\beta) becomes very small when n>β+1n > \beta + 1, so only a finite number of sidebands are significant.
  5. Jn(β)J_n(\beta) is oscillatory; J0(β)=0J_0(\beta) = 0 at β = 2.405, 5.52, 8.65, ... (carrier disappears).

FM wave as a sum of cosines at fc+nfmf_c + nf_m

Single-tone FM:

s(t)=Accos⁡(2πfct+βsin⁡2πfmt),β=Δffms(t) = A_c\cos(2\pi f_ct + \beta\sin2\pi f_mt), \qquad \beta = \frac{\Delta f}{f_m}

Write it using the complex envelope:

s(t)=Re[Acejβsin⁡ωmtejωct]s(t) = \text{Re}\left[A_ce^{j\beta\sin\omega_mt}e^{j\omega_ct}\right]

s~(t)=Acejβsin⁡ωmt\tilde s(t) = A_ce^{j\beta\sin\omega_mt} is periodic in Tm=1/fmT_m = 1/f_m, so

s~(t)=∑n=−∞∞cnejnωmt,cn=fm∫−1/2fm1/2fmAcejβsin⁡ωmte−jnωmtdt\tilde s(t) = \sum_{n=-\infty}^{\infty}c_ne^{jn\omega_mt}, \qquad c_n = f_m\int_{-1/2f_m}^{1/2f_m}A_ce^{j\beta\sin\omega_mt}e^{-jn\omega_mt}dt

Put x=ωmtx = \omega_mt:

cn=Ac2π∫−ππej(βsin⁡x−nx)dx=AcJn(β)c_n = \frac{A_c}{2\pi}\int_{-\pi}^{\pi}e^{j(\beta\sin x - nx)}dx = A_cJ_n(\beta)

Hence

s(t)=Re[∑n=−∞∞AcJn(β)ej2π(fc+nfm)t]=Ac∑n=−∞∞Jn(β)cos⁡[2π(fc+nfm)t]s(t) = \text{Re}\left[\sum_{n=-\infty}^{\infty}A_cJ_n(\beta)e^{j2\pi(f_c + nf_m)t}\right] = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c + nf_m)t]

This proves that the FM signal is a sum of infinitely many cosine components at fc+nfmf_c + nf_m, n=0,±1,±2,…n = 0, \pm1, \pm2, \dots, with amplitudes AcJn(β)A_cJ_n(\beta).

Spectrum:

S(f)=Ac2∑nJn(β)[δ(f−fc−nfm)+δ(f+fc+nfm)]S(f) = \frac{A_c}{2}\sum_{n}J_n(\beta)[\delta(f - f_c - nf_m) + \delta(f + f_c + nf_m)]
         |     |     |     |     |
     |   |     |     |     |     |   |
 ----+---+-----+-----+-----+-----+---+---> 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 ∣n∣>β+1|n| > \beta + 1 are negligible, giving Carson's bandwidth 2(β+1)fm2(\beta + 1)f_m.

  • 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:

s(t)=Accos⁡θi(t)s(t) = A_c\cos\theta_i(t)

It has two forms: phase modulation (PM), θi=ωct+kpm(t)\theta_i = \omega_ct + k_pm(t), and frequency modulation (FM), θi=ωct+2πkf∫m(τ)dτ\theta_i = \omega_ct + 2\pi k_f\int m(\tau)d\tau.

Time-domain expression of single-tone FM

Let m(t)=Amcos⁡2πfmtm(t) = A_m\cos2\pi f_mt. Instantaneous frequency:

fi(t)=fc+kfAmcos⁡2πfmt=fc+Δfcos⁡2πfmtf_i(t) = f_c + k_fA_m\cos2\pi f_mt = f_c + \Delta f\cos2\pi f_mt

Instantaneous phase:

θi(t)=2π∫0tfi(τ)dτ=2πfct+Δffmsin⁡2πfmt\theta_i(t) = 2\pi\int_0^tf_i(\tau)d\tau = 2\pi f_ct + \frac{\Delta f}{f_m}\sin2\pi f_mt

So

s(t)=Accos⁡(2πfct+βsin⁡2πfmt),β=Δffm=kfAmfms(t) = A_c\cos(2\pi f_ct + \beta\sin2\pi f_mt), \qquad \beta = \frac{\Delta f}{f_m} = \frac{k_fA_m}{f_m}

Frequency-domain expression

Write s(t)=Re[Acejβsin⁡ωmtejωct]s(t) = \text{Re}[A_ce^{j\beta\sin\omega_mt}e^{j\omega_ct}]. The periodic term has Fourier series

ejβsin⁡ωmt=∑n=−∞∞Jn(β)ejnωmt,Jn(β)=12π∫−ππej(βsin⁡x−nx)dxe^{j\beta\sin\omega_mt} = \sum_{n=-\infty}^{\infty}J_n(\beta)e^{jn\omega_mt}, \quad J_n(\beta) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j(\beta\sin x - nx)}dx

Therefore

s(t)=Ac∑n=−∞∞Jn(β)cos⁡[2π(fc+nfm)t]s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c + nf_m)t]

Taking the Fourier transform:

S(f)=Ac2∑n=−∞∞Jn(β)[δ(f−fc−nfm)+δ(f+fc+nfm)]S(f) = \frac{A_c}{2}\sum_{n=-\infty}^{\infty}J_n(\beta)\left[\delta(f - f_c - nf_m) + \delta(f + f_c + nf_m)\right]

The spectrum has a carrier of amplitude AcJ0(β)A_cJ_0(\beta) and sideband pairs at fc±nfmf_c \pm nf_m with amplitudes Ac∣Jn(β)∣A_c|J_n(\beta)|. Total power is Ac2/2A_c^2/2, and practical bandwidth is 2(β+1)fm2(\beta + 1)f_m.

  • 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: fc=102.4f_c = 102.4 MHz, fm=5f_m = 5 kHz, Δf=75\Delta f = 75 kHz.

(a) Carrier swing

The carrier frequency swings between fc−Δf=102.325f_c - \Delta f = 102.325 MHz and fc+Δf=102.475f_c + \Delta f = 102.475 MHz:

Carrier swing=2Δf=2×75=150 kHz\text{Carrier swing} = 2\Delta f = 2 \times 75 = 150\ \text{kHz}

(b) Bandwidth

By Carson's rule:

BT=2(Δf+fm)=2(75+5)=160 kHzB_T = 2(\Delta f + f_m) = 2(75 + 5) = 160\ \text{kHz}

(Bessel-table check: for β=15\beta = 15, significant sidebands extend to about n=β+1=16n = \beta + 1 = 16, so B=2×16×5=160B = 2 \times 16 \times 5 = 160 kHz.)

(c) Modulation index

β=Δffm=755=15\beta = \frac{\Delta f}{f_m} = \frac{75}{5} = 15

Answer: carrier swing = 150 kHz; bandwidth = 160 kHz; β=15\beta = 15.

  • 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 m(t)=Amcos⁡ωmtm(t) = A_m\cos\omega_mt, the instantaneous frequency is fi=fc+kfAmcos⁡ωmtf_i = f_c + k_fA_m\cos\omega_mt, so the phase is ωct+βsin⁡ωmt\omega_ct + \beta\sin\omega_mt with β=kfAm/fm=Δf/fm\beta = k_fA_m/f_m = \Delta f/f_m:

s(t)=Accos⁡(ωct+βsin⁡ωmt)=Re[Acejβsin⁡ωmtejωct]s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt) = \text{Re}\left[A_ce^{j\beta\sin\omega_mt}e^{j\omega_ct}\right]

ejβsin⁡ωmte^{j\beta\sin\omega_mt} is periodic with period 1/fm1/f_m; its Fourier series is

ejβsin⁡ωmt=∑n=−∞∞cnejnωmt,cn=12π∫−ππej(βsin⁡x−nx)dx=Jn(β)e^{j\beta\sin\omega_mt} = \sum_{n=-\infty}^{\infty}c_ne^{jn\omega_mt}, \qquad c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j(\beta\sin x - nx)}dx = J_n(\beta)

(substituting x=ωmtx = \omega_mt). Therefore

s(t)=Ac∑n=−∞∞Jn(β)cos⁡[(ωc+nωm)t]s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[(\omega_c + n\omega_m)t]

i.e. a carrier AcJ0(β)A_cJ_0(\beta) at fcf_c and sidebands AcJn(β)A_cJ_n(\beta) at fc+nfmf_c + nf_m.

Average power with components from −n to +n

Each component is a cosine of amplitude AcJk(β)A_cJ_k(\beta); the components are at different frequencies, so their powers add (1 Ω load):

Pn=∑k=−nnAc2Jk2(β)2=Ac22[J02(β)+2∑k=1nJk2(β)]P_n = \sum_{k=-n}^{n}\frac{A_c^2J_k^2(\beta)}{2} = \frac{A_c^2}{2}\left[J_0^2(\beta) + 2\sum_{k=1}^{n}J_k^2(\beta)\right]

using J−k2=Jk2J_{-k}^2 = J_k^2. When all components are included, ∑k=−∞∞Jk2(β)=1\sum_{k=-\infty}^{\infty}J_k^2(\beta) = 1, so

P=Ac22P = \frac{A_c^2}{2}

the same as the unmodulated carrier. The fraction of power within ±n\pm n sidebands is J02+2∑1nJk2J_0^2 + 2\sum_{1}^{n}J_k^2. Example: for β=1\beta = 1, n=2n = 2 gives 0.76522+2(0.44012+0.11492)=0.99920.7652^2 + 2(0.4401^2 + 0.1149^2) = 0.9992, 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; fc1=0.2f_{c1} = 0.2 MHz; β1=0.5\beta_1 = 0.5; fc2=10.925f_{c2} = 10.925 MHz; final fc=90f_c = 90 MHz, Δf=75\Delta f = 75 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 β1≤0.5\beta_1 \le 0.5 for every audio frequency. Since β1=Δf1/fm\beta_1 = \Delta f_1/f_m, the worst case is the lowest audio frequency, 50 Hz:

Δf1=β1fm,min=0.5×50=25 Hz\Delta f_1 = \beta_1f_{m,min} = 0.5 \times 50 = 25\ \text{Hz}

Step 2: total multiplication needed

The mixer does not change the deviation, so

n1n2=ΔfΔf1=75×10325=3000n_1n_2 = \frac{\Delta f}{\Delta f_1} = \frac{75 \times 10^3}{25} = 3000

Step 3: carrier frequency condition

The mixer output is the difference frequency (taking n1fc1>fc2n_1f_{c1} > f_{c2}):

fc=n2(n1fc1−fc2)  ⇒  90=n2(0.2n1−10.925)f_c = n_2(n_1f_{c1} - f_{c2}) \;\Rightarrow\; 90 = n_2(0.2n_1 - 10.925)

Using n1n2=3000n_1n_2 = 3000: 0.2n1n2=6000.2n_1n_2 = 600, so

90=600−10.925n2n2=51010.925=46.68,n1=300046.68=64.27\begin{aligned} 90 &= 600 - 10.925n_2 \\ n_2 &= \frac{510}{10.925} = 46.68, \qquad n_1 = \frac{3000}{46.68} = 64.27 \end{aligned}

Multipliers must be integers. Choosing n1=64n_1 = 64 and solving the frequency equation exactly:

0.2×64=12.8 MHzf2=12.8−10.925=1.875 MHzn2=901.875=48\begin{aligned} 0.2 \times 64 &= 12.8\ \text{MHz} \\ f_2 &= 12.8 - 10.925 = 1.875\ \text{MHz} \\ n_2 &= \frac{90}{1.875} = 48 \end{aligned}

Check

fc=48×1.875=90 MHzΔf=n1n2Δf1=64×48×25=76800 Hz=76.8 kHz\begin{aligned} f_c &= 48 \times 1.875 = 90\ \text{MHz} \\ \Delta f &= n_1n_2\Delta f_1 = 64 \times 48 \times 25 = 76800\ \text{Hz} = 76.8\ \text{kHz} \end{aligned}

The carrier is exactly 90 MHz and the deviation is within 2.4% of 75 kHz (exactly 75 kHz is reached by reducing β1\beta_1 slightly to 0.488). Both values are practical: 64=2664 = 2^6 (six doublers) and 48=24×348 = 2^4 \times 3 (four doublers and a tripler).

Answer: n1=64n_1 = 64, n2=48n_2 = 48 (ideal product 3000; continuous solution n1≈64.3n_1 \approx 64.3, n2≈46.7n_2 \approx 46.7).

  • 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: s(t)=Accos⁡θi(t)s(t) = A_c\cos\theta_i(t).

  • PM: θi(t)=ωct+kpm(t)\theta_i(t) = \omega_ct + k_pm(t)
  • FM: θi(t)=ωct+2πkf∫0tm(τ)dτ\theta_i(t) = \omega_ct + 2\pi k_f\int_0^tm(\tau)d\tau, so fi=fc+kfm(t)f_i = f_c + k_fm(t)

For a tone m=Amcos⁡ωmtm = A_m\cos\omega_mt, FM gives s(t)=Accos⁡(ωct+βsin⁡ωmt)s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt), β=kfAm/fm\beta = k_fA_m/f_m.

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 β\beta:

s(t)=Accos⁡(ωct+βsin⁡ωmt)=Accos⁡ωctcos⁡(βsin⁡ωmt)−Acsin⁡ωctsin⁡(βsin⁡ωmt)≈Accos⁡ωct−βAcsin⁡ωmt sin⁡ωct\begin{aligned} s(t) &= A_c\cos(\omega_ct + \beta\sin\omega_mt) \\ &= A_c\cos\omega_ct\cos(\beta\sin\omega_mt) - A_c\sin\omega_ct\sin(\beta\sin\omega_mt) \\ &\approx A_c\cos\omega_ct - \beta A_c\sin\omega_mt\,\sin\omega_ct \end{aligned}

using cos⁡x≈1\cos x \approx 1, sin⁡x≈x\sin x \approx x for small xx. This is realised by:

  • integrating m(t)m(t) (so that phase modulation gives FM),
  • multiplying it in a balanced modulator with the carrier shifted by −90° (sin⁡ωct\sin\omega_ct),
  • adding the carrier Accos⁡ωctA_c\cos\omega_ct back.

Typical: fc1=200f_{c1} = 200 kHz, β<0.5\beta < 0.5, Δf1≈25\Delta f_1 \approx 25 Hz.

2. Frequency multiplier ×n₁. A non-linear stage and filter; output cos⁡(n1ωc1t+n1βsin⁡ωmt)\cos(n_1\omega_{c1}t + n_1\beta\sin\omega_mt). Carrier and deviation both multiply by n1n_1.

3. Mixer. Mixing with a crystal oscillator fc2f_{c2} moves the carrier to n1fc1−fc2n_1f_{c1} - f_{c2} without changing deviation. This keeps the final carrier at the wanted value.

4. Frequency multiplier ×n₂ and power amplifier:

fc=n2(n1fc1−fc2),Δf=n1n2Δf1f_c = n_2(n_1f_{c1} - f_{c2}), \qquad \Delta f = n_1n_2\Delta f_1

Example: Δf1=25\Delta f_1 = 25 Hz needs n1n2=3000n_1n_2 = 3000 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 s(t)=Accos⁡(ωct+βsin⁡ωmt)s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt):

Ac=20A_c = 20 V, ωc=6×108\omega_c = 6 \times 10^8 rad/s, ωm=1250\omega_m = 1250 rad/s, β=7\beta = 7.

i) Carrier and modulating frequency

fc=6×1082π=95.49×106 Hz=95.49 MHzfm=12502π=198.94 Hz\begin{aligned} f_c &= \frac{6 \times 10^8}{2\pi} = 95.49 \times 10^6\ \text{Hz} = 95.49\ \text{MHz} \\ f_m &= \frac{1250}{2\pi} = 198.94\ \text{Hz} \end{aligned}

ii) Modulation index

β=7\beta = 7

iii) Maximum frequency deviation

Δf=βfm=7×198.94=1392.6 Hz≈1.39 kHz\Delta f = \beta f_m = 7 \times 198.94 = 1392.6\ \text{Hz} \approx 1.39\ \text{kHz}

iv) Power in 10 Ω

P=Ac22R=2022×10=20 WP = \frac{A_c^2}{2R} = \frac{20^2}{2 \times 10} = 20\ \text{W}

Answer: fc=95.49f_c = 95.49 MHz, fm=198.94f_m = 198.94 Hz; β=7\beta = 7; Δf=1392.6\Delta f = 1392.6 Hz; P=20P = 20 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 m(t)=Amcos⁡2πfmtm(t) = A_m \cos 2\pi f_m t is

s(t)=Accos⁡[2πfct+βsin⁡2πfmt],β=kfAmfm=Δffms(t) = A_c \cos\left[2\pi f_c t + \beta \sin 2\pi f_m t\right], \qquad \beta = \frac{k_f A_m}{f_m} = \frac{\Delta f}{f_m}

Expanding with cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B) = \cos A\cos B - \sin A \sin B:

s(t)=Accos⁡2πfct cos⁡(βsin⁡2πfmt)−Acsin⁡2πfct sin⁡(βsin⁡2πfmt)s(t) = A_c \cos 2\pi f_c t \,\cos(\beta \sin 2\pi f_m t) - A_c \sin 2\pi f_c t\, \sin(\beta \sin 2\pi f_m t)

For narrowband FM, β≪1\beta \ll 1 (in practice β<0.3\beta < 0.3). Then

cos⁡(βsin⁡2πfmt)≈1,sin⁡(βsin⁡2πfmt)≈βsin⁡2πfmt\cos(\beta \sin 2\pi f_m t) \approx 1, \qquad \sin(\beta \sin 2\pi f_m t) \approx \beta \sin 2\pi f_m t

so

s(t)≈Accos⁡2πfct−βAcsin⁡2πfct sin⁡2πfmts(t) \approx A_c \cos 2\pi f_c t - \beta A_c \sin 2\pi f_c t \,\sin 2\pi f_m t

Using sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\sin A \sin B = \tfrac12[\cos(A-B) - \cos(A+B)]:

sNBFM(t)≈Accos⁡2πfct+βAc2cos⁡2π(fc+fm)t−βAc2cos⁡2π(fc−fm)ts_{NBFM}(t) \approx A_c \cos 2\pi f_c t + \frac{\beta A_c}{2}\cos 2\pi (f_c+f_m)t - \frac{\beta A_c}{2}\cos 2\pi (f_c-f_m)t

So NBFM has a carrier and one pair of side frequencies at fc±fmf_c \pm f_m, like AM, but the lower side frequency has a negative sign (180° phase shift). Its bandwidth is 2fm2f_m, 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]
  1. A stable crystal oscillator and a balanced modulator produce NBFM with small deviation Δf1\Delta f_1 at carrier f1f_1.
  2. A frequency multiplier (a non-linear device + band-pass filter) multiplies the instantaneous frequency by nn. Both carrier and deviation are multiplied: fc→nf1f_c \to n f_1, Δf→nΔf1\Delta f \to n \Delta f_1, so β→nβ1\beta \to n\beta_1.
  3. Because the required deviation needs a large nn, the carrier would become too high. A mixer with a local oscillator shifts the carrier down (fLO−n1f1f_{LO} - n_1 f_1) without changing the deviation.
  4. A second multiplier n2n_2 raises the deviation (and carrier) to the final values:
Δf=n1n2Δf1,fc=n2 ∣fLO−n1f1∣\Delta f = n_1 n_2 \Delta f_1, \qquad f_c = n_2\,|f_{LO} - n_1 f_1|

Example: Δf1=25\Delta f_1 = 25 Hz needs n1n2=3000n_1 n_2 = 3000 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 β\beta); a mixer shifts only the carrier and leaves the deviation unchanged.

Given NBFM: fc1=1f_{c1} = 1 MHz, β1=0.1\beta_1 = 0.1, fm=10f_m = 10 kHz

Δf1=β1fm=0.1×10 kHz=1 kHz\Delta f_1 = \beta_1 f_m = 0.1 \times 10\ \text{kHz} = 1\ \text{kHz}

Required FM: fc=50f_c = 50 MHz, β=1\beta = 1

Δf=βfm=1×10 kHz=10 kHz\Delta f = \beta f_m = 1 \times 10\ \text{kHz} = 10\ \text{kHz}

Step 1: Multiplication factor. The mixer does not change β\beta, so the multiplier must give

n=ββ1=10.1=10n = \frac{\beta}{\beta_1} = \frac{1}{0.1} = 10

After a ×10\times 10 multiplier:

fc2=10×1 MHz=10 MHz,Δf2=10×1 kHz=10 kHzf_{c2} = 10 \times 1\ \text{MHz} = 10\ \text{MHz}, \qquad \Delta f_2 = 10 \times 1\ \text{kHz} = 10\ \text{kHz}

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:

fLO+10 MHz=50 MHz  ⇒  fLO=40 MHzf_{LO} + 10\ \text{MHz} = 50\ \text{MHz} \;\Rightarrow\; f_{LO} = 40\ \text{MHz}

(Choosing the difference, fLO=60f_{LO} = 60 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: β=Δf/fm=10 kHz/10 kHz=1\beta = \Delta f / f_m = 10\ \text{kHz}/10\ \text{kHz} = 1, fc=50f_c = 50 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 Δf=10\Delta f = 10 kHz.

Answer: frequency multiplier n=10n = 10 followed by a mixer with fLO=40f_{LO} = 40 MHz (sum taken) gives fc=50f_c = 50 MHz, Δf=10\Delta f = 10 kHz, β=1\beta = 1.

  • 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: 2fcapture≤2flock2f_{capture} \le 2f_{lock}, both centred on the VCO free-running frequency f0f_0.

FM demodulation using PLL

 FM in   +-------+  e(t)  +------+  v(t) = output
 ------->| Phase |------->| Loop |----+---------->
 s(t)    | det.  |        | LPF  |    |
         +-------+        +------+    |
             ^                        |
             |        +-------+       |
             +--------|  VCO  |<------+
              r(t)    +-------+

Working:

  1. Input FM: s(t)=Acsin⁡[2πfct+ϕ1(t)]s(t) = A_c \sin[2\pi f_c t + \phi_1(t)], where ϕ1(t)=2πkf∫0tm(τ)dτ\phi_1(t) = 2\pi k_f \int_0^t m(\tau)d\tau.
  2. The VCO free-running frequency is set to fcf_c. Its output is r(t)=Avcos⁡[2πfct+ϕ2(t)]r(t) = A_v \cos[2\pi f_c t + \phi_2(t)] with ϕ2(t)=2πkv∫0tv(τ)dτ\phi_2(t) = 2\pi k_v \int_0^t v(\tau)d\tau.
  3. The phase detector (multiplier) output, after the loop filter removes the 2fc2f_c term, is
e(t)≈12AcAvsin⁡[ϕ1(t)−ϕ2(t)]=12AcAvsin⁡ϕe(t)e(t) \approx \tfrac12 A_c A_v \sin[\phi_1(t) - \phi_2(t)] = \tfrac12 A_c A_v \sin\phi_e(t)
  1. This error voltage drives the VCO so that its phase follows the input phase. When the loop is locked, ϕe\phi_e is small and ϕ2(t)≈ϕ1(t)\phi_2(t) \approx \phi_1(t).
  2. Hence
2πkv∫0tv(τ) dτ≈2πkf∫0tm(τ) dτ  ⇒  v(t)≈kfkv m(t)2\pi k_v \int_0^t v(\tau)\,d\tau \approx 2\pi k_f \int_0^t m(\tau)\,d\tau \;\Rightarrow\; v(t) \approx \frac{k_f}{k_v}\, m(t)

So the VCO control voltage v(t)v(t) 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 fc±nfmf_c \pm n f_m, with amplitudes AcJn(β)A_c J_n(\beta). In practice, sidebands beyond about n=β+1n = \beta + 1 are very small. Carson's rule says that about 98% of the total power of an FM wave lies within

BT≈2(Δf+fm)=2fm(β+1)B_T \approx 2(\Delta f + f_m) = 2 f_m(\beta + 1)

where Δf\Delta f is the peak frequency deviation and fmf_m the highest modulating frequency.

  • For large β\beta (WBFM): BT≈2ΔfB_T \approx 2\Delta f (set by the deviation).
  • For small β\beta (NBFM): BT≈2fmB_T \approx 2 f_m.

Example: commercial FM with Δf=75\Delta f = 75 kHz and fm=15f_m = 15 kHz gives BT=2(75+15)=180B_T = 2(75 + 15) = 180 kHz (200 kHz channels are allotted).

When FM bandwidth equals AM bandwidth

When β≪1\beta \ll 1 (narrowband FM, β<0.3\beta < 0.3), Δf≪fm\Delta f \ll f_m, so

BT=2(Δf+fm)≈2fmB_T = 2(\Delta f + f_m) \approx 2 f_m

which is the same as the bandwidth of AM (2fm2f_m).

Reason: for small β\beta, J0(β)≈1J_0(\beta) \approx 1, J1(β)≈β/2J_1(\beta) \approx \beta/2, and Jn(β)≈0J_n(\beta) \approx 0 for n≥2n \ge 2. So the NBFM wave is

s(t)≈Accos⁡2πfct+βAc2[cos⁡2π(fc+fm)t−cos⁡2π(fc−fm)t]s(t) \approx A_c\cos 2\pi f_c t + \frac{\beta A_c}{2}\left[\cos 2\pi(f_c + f_m)t - \cos 2\pi(f_c - f_m)t\right]

It has only the carrier and one pair of sidebands, exactly like AM, so it occupies 2fm2f_m. 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

PointAM receiverFM receiver
RF bandMW/SW, 0.54–1.65 MHz (MW)VHF, 88–108 MHz
Intermediate frequency455 kHz10.7 MHz
IF bandwidthabout 10 kHzabout 200 kHz
LimiterNot usedUsed before detector to remove amplitude noise
DetectorEnvelope (diode) detectorDiscriminator, ratio detector or PLL
De-emphasisNot usedDe-emphasis network (75 μs) after detector
Gain controlAGCAFC common; limiter does most amplitude control
Noise immunityPoorGood (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:

  1. Signals from near and far stations differ widely in strength (microvolts to hundreds of millivolts). Without AGC, the volume changes when tuning between stations.
  2. Fading makes the strength of one station vary with time; AGC keeps the output level nearly constant.
  3. Strong signals would overload the RF/IF stages and the detector and cause distortion; AGC reduces gain for strong signals.
  4. 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: fc=106f_c = 10^6 Hz, Ac=3A_c = 3 V, original Am=1A_m = 1 V, fm=500f_m = 500 Hz, Δf=1\Delta f = 1 kHz.

Step 1: Frequency sensitivity of the modulator. In FM, the deviation is proportional to the modulating amplitude:

kf=ΔfAm=1 kHz1 V=1 kHz/Vk_f = \frac{\Delta f}{A_m} = \frac{1\ \text{kHz}}{1\ \text{V}} = 1\ \text{kHz/V}

Step 2: New deviation. New Am=5A_m = 5 V, new fm=2f_m = 2 kHz. The deviation depends only on amplitude, not on fmf_m:

Δf′=kfAm′=1 kHz/V×5 V=5 kHz\Delta f' = k_f A_m' = 1\ \text{kHz/V} \times 5\ \text{V} = 5\ \text{kHz}

Step 3: New modulation index.

β′=Δf′fm′=5 kHz2 kHz=2.5\beta' = \frac{\Delta f'}{f_m'} = \frac{5\ \text{kHz}}{2\ \text{kHz}} = 2.5

(The original index was β=1000/500=2\beta = 1000/500 = 2.)

Step 4: New FM waveform. Taking the modulating signal as a cosine, m(t)=5cos⁡(2π×2000 t)m(t) = 5\cos(2\pi \times 2000\,t):

s(t)=Accos⁡[2πfct+β′sin⁡2πfm′t]=3cos⁡[2π×106 t+2.5sin⁡(2π×2000 t)] V=3cos⁡[6.283×106 t+2.5sin⁡(12566 t)] V\begin{aligned} s(t) &= A_c \cos\left[2\pi f_c t + \beta' \sin 2\pi f_m' t\right] \\ &= 3\cos\left[2\pi \times 10^6\, t + 2.5 \sin(2\pi \times 2000\, t)\right]\ \text{V} \\ &= 3\cos\left[6.283\times10^6\, t + 2.5\sin(12566\, t)\right]\ \text{V} \end{aligned}

(If a sine carrier is used, write 3sin⁡[…]3\sin[\ldots] with the same argument.)

Answer: s(t)=3cos⁡[2π×106t+2.5sin⁡(4000πt)]s(t) = 3\cos[2\pi\times10^6 t + 2.5\sin(4000\pi t)] V, with Δf=5\Delta f = 5 kHz and β=2.5\beta = 2.5.

  • 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 (fLO=fs+10.7f_{LO} = f_s + 10.7 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: E=10sin⁡(108t+3sin⁡104t)E = 10\sin(10^8 t + 3\sin 10^4 t) V, R=100 ΩR = 100\ \Omega.

Comparing with the standard form E=Acsin⁡(ωct+βsin⁡ωmt)E = A_c \sin(\omega_c t + \beta \sin \omega_m t):

Ac=10 V,ωc=108 rad/s,ωm=104 rad/s,β=3A_c = 10\ \text{V}, \quad \omega_c = 10^8\ \text{rad/s}, \quad \omega_m = 10^4\ \text{rad/s}, \quad \beta = 3

Carrier frequency

fc=ωc2π=1082π=15.915×106 Hz≈15.92 MHzf_c = \frac{\omega_c}{2\pi} = \frac{10^8}{2\pi} = 15.915 \times 10^6\ \text{Hz} \approx 15.92\ \text{MHz}

Modulating frequency

fm=ωm2π=1042π=1591.5 Hz≈1.59 kHzf_m = \frac{\omega_m}{2\pi} = \frac{10^4}{2\pi} = 1591.5\ \text{Hz} \approx 1.59\ \text{kHz}

Modulation index

β=3(peak deviation Δf=βfm=3×1591.5=4774.6 Hz)\beta = 3 \quad (\text{peak deviation } \Delta f = \beta f_m = 3 \times 1591.5 = 4774.6\ \text{Hz})

Power in 100 Ω

The amplitude of an angle-modulated wave is constant, so its power equals the unmodulated carrier power:

P=Vrms2R=(Ac/2)2R=1022×100=0.5 WP = \frac{V_{rms}^2}{R} = \frac{(A_c/\sqrt2)^2}{R} = \frac{10^2}{2 \times 100} = 0.5\ \text{W}

Answer: fc=15.92f_c = 15.92 MHz, fm=1.59f_m = 1.59 kHz, β=3\beta = 3, P=0.5P = 0.5 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)  | +--------+
      +---------+         +---------+
  1. 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 fcf_c keeps only the fundamental. Output: s(t)=Accos⁡[2πfct+ϕ(t)]s(t) = A_c\cos[2\pi f_c t + \phi(t)], ϕ(t)=2πkf∫0tm(τ)dτ\phi(t) = 2\pi k_f\int_0^t m(\tau)d\tau.

  2. Discriminator (differentiator): its output is proportional to frequency:

ds(t)dt=−Ac[2πfc+2πkf m(t)]sin⁡[2πfct+ϕ(t)]\frac{ds(t)}{dt} = -A_c\left[2\pi f_c + 2\pi k_f\, m(t)\right]\sin\left[2\pi f_c t + \phi(t)\right]

This is a wave that is both FM and AM; its envelope is 2πAc[fc+kfm(t)]2\pi A_c [f_c + k_f m(t)], which follows the message.

  1. Envelope detector: recovers 2πAc[fc+kfm(t)]2\pi A_c[f_c + k_f m(t)]. A DC-blocking capacitor removes the constant 2πAcfc2\pi A_c f_c, leaving an output proportional to m(t)m(t).

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 fcf_c, 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 c(t)=Accos⁡2πfctc(t) = A_c\cos 2\pi f_c t. A general angle-modulated wave is

s(t)=Accos⁡θi(t)s(t) = A_c\cos\theta_i(t)

The instantaneous frequency is fi(t)=12πdθidtf_i(t) = \frac{1}{2\pi}\frac{d\theta_i}{dt}.

In FM, the instantaneous frequency varies linearly with the message m(t)m(t):

fi(t)=fc+kf m(t)f_i(t) = f_c + k_f\, m(t)

where kfk_f is the frequency sensitivity (Hz/V). Integrating,

θi(t)=2π∫0tfi(τ) dτ=2πfct+2πkf∫0tm(τ) dτ\theta_i(t) = 2\pi\int_0^t f_i(\tau)\,d\tau = 2\pi f_c t + 2\pi k_f \int_0^t m(\tau)\,d\tau

Hence the general FM expression:

sFM(t)=Accos⁡[2πfct+2πkf∫0tm(τ) dτ]s_{FM}(t) = A_c\cos\left[2\pi f_c t + 2\pi k_f\int_0^t m(\tau)\,d\tau\right]

Single-tone case: for m(t)=Amcos⁡2πfmtm(t) = A_m\cos 2\pi f_m t,

θi(t)=2πfct+kfAmfmsin⁡2πfmt\theta_i(t) = 2\pi f_c t + \frac{k_f A_m}{f_m}\sin 2\pi f_m t s(t)=Accos⁡[2πfct+βsin⁡2πfmt],β=kfAmfm=Δffms(t) = A_c\cos\left[2\pi f_c t + \beta\sin 2\pi f_m t\right], \qquad \beta = \frac{k_f A_m}{f_m} = \frac{\Delta f}{f_m}

FM generation using a phase modulator

A PM wave is Accos⁡[2πfct+kp x(t)]A_c\cos[2\pi f_c t + k_p\, x(t)]: its phase follows the input x(t)x(t). 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 x(t)=∫0tm(τ)dτx(t) = \int_0^t m(\tau)d\tau:

s(t)=Accos⁡[2πfct+kp∫0tm(τ) dτ]s(t) = A_c\cos\left[2\pi f_c t + k_p\int_0^t m(\tau)\,d\tau\right]

This is FM with kp=2πkfk_p = 2\pi k_f. 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 A(t)cos⁡[2πfct+ϕ(t)]A(t)\cos[2\pi f_c t + \phi(t)], where A(t)A(t) 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 fcf_c removes the harmonics. The output is a constant-amplitude FM wave:

s(t)=Accos⁡[2πfct+2πkf∫0tm(τ) dτ]s(t) = A_c\cos\left[2\pi f_c t + 2\pi k_f\int_0^t m(\tau)\,d\tau\right]

2. Discriminator

An ideal discriminator has a transfer function H(f)=j2πafH(f) = j2\pi a f over the FM band, i.e. it acts as a differentiator:

ds(t)dt=−Ac[2πfc+2πkf m(t)]sin⁡[2πfct+2πkf∫0tm(τ) dτ]\frac{ds(t)}{dt} = -A_c\left[2\pi f_c + 2\pi k_f\, m(t)\right]\sin\left[2\pi f_c t + 2\pi k_f\int_0^t m(\tau)\,d\tau\right]

The output is a hybrid AM–FM wave whose envelope 2πAc[fc+kfm(t)]2\pi A_c[f_c + k_f m(t)] varies linearly with m(t)m(t) (provided fc>kf∣m(t)∣maxf_c > k_f |m(t)|_{max}, so the envelope never becomes negative).

Practical discriminators:

  • Slope detector: a tuned circuit detuned so that fcf_c 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 fc+δf_c + \delta and fc−δf_c - \delta, 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 2πAc[fc+kfm(t)]2\pi A_c[f_c + k_f m(t)]. A DC-blocking capacitor removes the 2πAcfc2\pi A_c f_c term, giving an output proportional to m(t)m(t).

Why the limiter is essential: the discriminator output amplitude depends on AcA_c. 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: m(t)=5cos⁡(18849.55 t)m(t) = 5\cos(18849.55\,t), kf=9k_f = 9 kHz/V.

So Am=5A_m = 5 V and

fm=ωm2π=18849.552π=3000 Hz=3 kHzf_m = \frac{\omega_m}{2\pi} = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}

(i) Peak frequency deviation

Δf=kfAm=9 kHz/V×5 V=45 kHz\Delta f = k_f A_m = 9\ \text{kHz/V} \times 5\ \text{V} = 45\ \text{kHz}

(ii) Modulation index

β=Δffm=45 kHz3 kHz=15\beta = \frac{\Delta f}{f_m} = \frac{45\ \text{kHz}}{3\ \text{kHz}} = 15

(iii) Frequency swing

The carrier swings from fc−Δff_c - \Delta f to fc+Δff_c + \Delta f:

Swing=2Δf=2×45=90 kHz (±45 kHz)\text{Swing} = 2\Delta f = 2 \times 45 = 90\ \text{kHz}\ (\pm 45\ \text{kHz})

(iv) Carson's bandwidth

BT=2(Δf+fm)=2(45+3)=96 kHzB_T = 2(\Delta f + f_m) = 2(45 + 3) = 96\ \text{kHz}

Answer: Δf=45\Delta f = 45 kHz, β=15\beta = 15, swing = 90 kHz, BT=96B_T = 96 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: fm=10f_m = 10 kHz (highest baseband frequency), fc=100f_c = 100 MHz, Δf=75\Delta f = 75 kHz.

i) Carrier frequency swing and modulation index

The instantaneous frequency varies between

fmax=100 MHz+75 kHz=100.075 MHz,fmin=99.925 MHzf_{max} = 100\ \text{MHz} + 75\ \text{kHz} = 100.075\ \text{MHz}, \qquad f_{min} = 99.925\ \text{MHz} Carrier swing=2Δf=2×75=150 kHz\text{Carrier swing} = 2\Delta f = 2 \times 75 = 150\ \text{kHz}

Modulation index (deviation ratio, since fmf_m is the highest message frequency):

β=Δffm=7510=7.5\beta = \frac{\Delta f}{f_m} = \frac{75}{10} = 7.5

ii) Practical bandwidth

By Carson's rule (contains about 98% of the FM power):

BT=2(Δf+fm)=2(75+10)=170 kHzB_T = 2(\Delta f + f_m) = 2(75 + 10) = 170\ \text{kHz}

Check with Bessel functions: with β=7.5\beta = 7.5, Carson's rule keeps sidebands up to n=β+1≈8n = \beta + 1 \approx 8; the power in these is J02+2∑n=18Jn2(7.5)=0.981J_0^2 + 2\sum_{n=1}^{8}J_n^2(7.5) = 0.981, i.e. 98.1% of the total.

If the stricter 1% rule is used (keep sidebands with ∣Jn(β)∣>0.01|J_n(\beta)| > 0.01), ∣J11(7.5)∣=0.015|J_{11}(7.5)| = 0.015 and ∣J12(7.5)∣=0.005|J_{12}(7.5)| = 0.005, so nmax=11n_{max} = 11 and

B=2nmaxfm=2×11×10=220 kHzB = 2 n_{max} f_m = 2 \times 11 \times 10 = 220\ \text{kHz}

Answer: swing = 150 kHz, β=7.5\beta = 7.5, 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:

sAM(t)=Accos⁡ωct+μAc2cos⁡(ωc+ωm)t+μAc2cos⁡(ωc−ωm)tsNBFM(t)≈Accos⁡ωct+βAc2cos⁡(ωc+ωm)t−βAc2cos⁡(ωc−ωm)t\begin{aligned} s_{AM}(t) &= A_c\cos\omega_c t + \frac{\mu A_c}{2}\cos(\omega_c+\omega_m)t + \frac{\mu A_c}{2}\cos(\omega_c-\omega_m)t \\ s_{NBFM}(t) &\approx A_c\cos\omega_c t + \frac{\beta A_c}{2}\cos(\omega_c+\omega_m)t - \frac{\beta A_c}{2}\cos(\omega_c-\omega_m)t \end{aligned}

Similarities: both have a carrier and one pair of sidebands at fc±fmf_c \pm f_m; both have bandwidth 2fm2f_m; 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:

s(t)≈Accos⁡2πfct−Ac ϕ(t)sin⁡2πfct,ϕ(t)=2πkf∫0tm(τ) dτs(t) \approx A_c\cos 2\pi f_c t - A_c\,\phi(t)\sin 2\pi f_c t, \qquad \phi(t) = 2\pi k_f\int_0^t m(\tau)\,d\tau

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:

  1. The message is integrated (so that the phase modulator produces FM rather than PM).
  2. A crystal oscillator gives a very stable carrier Accos⁡ωctA_c\cos\omega_c t.
  3. The carrier is shifted by 90° and fed to a balanced modulator with the integrated message, producing −Acϕ(t)sin⁡ωct-A_c\phi(t)\sin\omega_c t (DSB-SC).
  4. The carrier is added to this output, giving NBFM.

The deviation is kept small (β<0.3\beta < 0.3, 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):

s(t)=Accos⁡[2πfct+2πkf∫0tm(τ) dτ]s(t) = A_c\cos\left[2\pi f_c t + 2\pi k_f\int_0^t m(\tau)\,d\tau\right]

Step 1: Differentiation. A circuit whose gain rises linearly with frequency, H(f)=j2πafH(f) = j2\pi a f, acts as a differentiator:

v1(t)=ads(t)dt=−2πaAc[fc+kfm(t)]sin⁡[2πfct+2πkf∫0tm(τ) dτ]v_1(t) = a\frac{ds(t)}{dt} = -2\pi a A_c\left[f_c + k_f m(t)\right]\sin\left[2\pi f_c t + 2\pi k_f\int_0^t m(\tau)\,d\tau\right]

v1(t)v_1(t) is still FM, but its amplitude now varies as fc+kfm(t)f_c + k_f m(t), i.e. it is an AM–FM wave.

Step 2: Envelope detection. Since fc≫kf∣m(t)∣f_c \gg k_f|m(t)|, the envelope is always positive:

v2(t)=2πaAc[fc+kfm(t)]v_2(t) = 2\pi a A_c\left[f_c + k_f m(t)\right]

Step 3: DC removal. A blocking capacitor removes 2πaAcfc2\pi a A_c f_c:

y(t)=2πaAckf m(t)  ∝  m(t)y(t) = 2\pi a A_c k_f\, m(t) \;\propto\; m(t)

Practical circuits:

  • Slope detector: a tuned circuit with resonance slightly above fcf_c; the carrier sits on its sloping skirt, so amplitude varies with frequency. Simple but non-linear.
  • Balanced slope detector: two tuned circuits at fc±δff_c \pm \delta f 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 m(t)=Amcos⁡2πfmtm(t) = A_m\cos 2\pi f_m t:

s(t)=Accos⁡[2πfct+βsin⁡2πfmt]=Re{Ac ej2πfct ejβsin⁡2πfmt}s(t) = A_c\cos\left[2\pi f_c t + \beta\sin 2\pi f_m t\right] = \text{Re}\left\{A_c\, e^{j2\pi f_c t}\, e^{j\beta\sin 2\pi f_m t}\right\}

The term ejβsin⁡2πfmte^{j\beta\sin 2\pi f_m t} is periodic with period 1/fm1/f_m, so it can be written as a complex Fourier series:

ejβsin⁡2πfmt=∑n=−∞∞cn ej2πnfmte^{j\beta\sin 2\pi f_m t} = \sum_{n=-\infty}^{\infty} c_n\, e^{j2\pi n f_m t}

with coefficients (put x=2πfmtx = 2\pi f_m t)

cn=12π∫−ππej(βsin⁡x−nx) dx=Jn(β)c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{j(\beta\sin x - nx)}\,dx = J_n(\beta)

which is the Bessel function of the first kind of order nn. Substituting:

s(t)=Ac∑n=−∞∞Jn(β)cos⁡[2π(fc+nfm)t]s(t) = A_c\sum_{n=-\infty}^{\infty} J_n(\beta)\cos\left[2\pi(f_c + n f_m)t\right]

So the FM wave has a carrier component AcJ0(β)A_cJ_0(\beta) and infinitely many side frequencies at fc±nfmf_c \pm nf_m with amplitudes AcJn(β)A_cJ_n(\beta), using J−n(β)=(−1)nJn(β)J_{-n}(\beta) = (-1)^n J_n(\beta).

Properties: ∑nJn2(β)=1\sum_n J_n^2(\beta) = 1, so total power =Ac2/2= A_c^2/2, independent of β\beta. The carrier amplitude J0(β)J_0(\beta) falls as β\beta increases and is zero at β=2.405,5.52,…\beta = 2.405, 5.52, \ldots

Practical bandwidth of FM

Although the spectrum is infinite, Jn(β)J_n(\beta) becomes very small when n>βn > \beta. For a fixed β\beta, Jn(β)J_n(\beta) decreases rapidly once nn exceeds about β+1\beta + 1. Keeping the significant sidebands up to nmaxn_{max}:

B=2 nmax fmB = 2\, n_{max}\, f_m

Carson's rule: taking nmax=β+1n_{max} = \beta + 1 (which keeps about 98% of the power):

BT=2(β+1)fm=2(Δf+fm)B_T = 2(\beta + 1)f_m = 2(\Delta f + f_m)

Two limiting cases support this:

  • β≫1\beta \gg 1 (WBFM): BT≈2βfm=2ΔfB_T \approx 2\beta f_m = 2\Delta f (the carrier sweeps over ±Δf\pm\Delta f).
  • β≪1\beta \ll 1 (NBFM): only J0J_0 and J1J_1 are significant, so BT≈2fmB_T \approx 2f_m.

1% rule (universal curve): keep all sidebands with ∣Jn(β)∣>0.01|J_n(\beta)| > 0.01; this gives a slightly larger bandwidth, e.g. for β=5\beta = 5, nmax=8n_{max} = 8 and B=16fmB = 16 f_m versus Carson's 12fm12 f_m.

For a general message with highest frequency WW, use the deviation ratio D=Δf/WD = \Delta f/W: BT=2(Δf+W)=2W(D+1)B_T = 2(\Delta f + W) = 2W(D+1).

  • 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: m(t)=5cos⁡(18849.55 t)m(t) = 5\cos(18849.55\,t), kf=9k_f = 9 kHz/V, Ac=25A_c = 25 V, fc=88.7f_c = 88.7 MHz, R=10 ΩR = 10\ \Omega.

Am=5 V,fm=18849.552π=3000 Hz=3 kHzA_m = 5\ \text{V}, \qquad f_m = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}

i) Peak frequency deviation

Δf=kfAm=9×5=45 kHz\Delta f = k_f A_m = 9 \times 5 = 45\ \text{kHz}

ii) Modulation index

β=Δffm=453=15\beta = \frac{\Delta f}{f_m} = \frac{45}{3} = 15

iii) Frequency swing

Swing=2Δf=90 kHz\text{Swing} = 2\Delta f = 90\ \text{kHz}

The carrier moves between 88.7−0.045=88.65588.7 - 0.045 = 88.655 MHz and 88.74588.745 MHz.

iv) Carson's bandwidth

BT=2(Δf+fm)=2(45+3)=96 kHzB_T = 2(\Delta f + f_m) = 2(45 + 3) = 96\ \text{kHz}

v) Power in 10 Ω

FM has constant amplitude, so total power equals the carrier power:

P=Ac22R=2522×10=62520=31.25 WP = \frac{A_c^2}{2R} = \frac{25^2}{2 \times 10} = \frac{625}{20} = 31.25\ \text{W}

Answer: Δf=45\Delta f = 45 kHz, β=15\beta = 15, swing = 90 kHz, BT=96B_T = 96 kHz, P=31.25P = 31.25 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: f1=200f_1 = 200 kHz, Δf1=24\Delta f_1 = 24 Hz, n1=65n_1 = 65, fLO=10.8f_{LO} = 10.8 MHz, n2=50n_2 = 50.

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 (×65\times 65)

f2=n1f1=65×200 kHz=13 MHzΔf2=n1Δf1=65×24=1560 Hz\begin{aligned} f_2 &= n_1 f_1 = 65 \times 200\ \text{kHz} = 13\ \text{MHz} \\ \Delta f_2 &= n_1 \Delta f_1 = 65 \times 24 = 1560\ \text{Hz} \end{aligned}

Step 2: After the mixer

f3=f2−fLO=13−10.8=2.2 MHz,Δf3=1560 Hzf_3 = f_2 - f_{LO} = 13 - 10.8 = 2.2\ \text{MHz}, \qquad \Delta f_3 = 1560\ \text{Hz}

Step 3: After the second multiplier (×50\times 50)

fc=n2f3=50×2.2 MHz=110 MHzΔfc=n2Δf3=50×1560=78 000 Hz=78 kHz\begin{aligned} f_c &= n_2 f_3 = 50 \times 2.2\ \text{MHz} = 110\ \text{MHz} \\ \Delta f_c &= n_2 \Delta f_3 = 50 \times 1560 = 78\,000\ \text{Hz} = 78\ \text{kHz} \end{aligned}

Check: Δfc=n1n2Δf1=3250×24=78\Delta f_c = n_1 n_2 \Delta f_1 = 3250 \times 24 = 78 kHz.

StageCarrierDeviation
NBFM output200 kHz24 Hz
After ×6513 MHz1.56 kHz
After mixer2.2 MHz1.56 kHz
After ×50110 MHz78 kHz

(If the sum frequency 23.8 MHz were selected, fcf_c would be 1190 MHz with the same deviation, which is not a practical FM carrier.)

Answer: maximum frequency deviation Δfc=78\Delta f_c = 78 kHz, carrier frequency fc=110f_c = 110 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: Z(t)=50cos⁡{2π(98.6×106)t+15sin⁡(2π×2×103 t)}Z(t) = 50\cos\{2\pi(98.6\times10^6)t + 15\sin(2\pi\times 2\times10^3\, t)\} V, R=50 ΩR = 50\ \Omega.

Comparing with Accos⁡[2πfct+βsin⁡2πfmt]A_c\cos[2\pi f_c t + \beta\sin 2\pi f_m t]: Ac=50A_c = 50 V, fc=98.6f_c = 98.6 MHz, fm=2f_m = 2 kHz, β=15\beta = 15.

a) Type of modulation

The amplitude is constant and the phase varies sinusoidally, so it is angle modulation. If the message is m(t)=Amcos⁡2πfmtm(t) = A_m\cos 2\pi f_m t, the phase term sin⁡\sin 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 sin⁡\sin, the same wave would be PM; the expression alone cannot separate the two.)

b) Frequency deviation

fi(t)=12πdθdt=fc+βfmcos⁡2πfmtf_i(t) = \frac{1}{2\pi}\frac{d\theta}{dt} = f_c + \beta f_m\cos 2\pi f_m t Δf=βfm=15×2 kHz=30 kHz\Delta f = \beta f_m = 15 \times 2\ \text{kHz} = 30\ \text{kHz}

c) Frequency sensitivity

kf=Δf/Amk_f = \Delta f / A_m. The message amplitude is not given, so taking Am=1A_m = 1 V:

kf=30 kHz1 V=30 kHz/Vk_f = \frac{30\ \text{kHz}}{1\ \text{V}} = 30\ \text{kHz/V}

(For any other AmA_m, kf=30/Amk_f = 30/A_m kHz/V.)

d) Modulation index

β=Δffm=302=15\beta = \frac{\Delta f}{f_m} = \frac{30}{2} = 15

e) Power delivered to 50 Ω antenna

P=Ac22R=5022×50=25 WP = \frac{A_c^2}{2R} = \frac{50^2}{2 \times 50} = 25\ \text{W}

Answer: FM; Δf=30\Delta f = 30 kHz; kf=30k_f = 30 kHz/V (for Am=1A_m = 1 V); β=15\beta = 15; P=25P = 25 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: m(t)=20cos⁡(2π 2000 t)m(t) = 20\cos(2\pi\, 2000\, t) V, c(t)=50cos⁡(2π107t)c(t) = 50\cos(2\pi 10^7 t) V, kf=200k_f = 200 Hz/V.

So Am=20A_m = 20 V, fm=2f_m = 2 kHz, Ac=50A_c = 50 V, fc=10f_c = 10 MHz.

a) Peak frequency deviation

Δf=kfAm=200×20=4000 Hz=4 kHz\Delta f = k_f A_m = 200 \times 20 = 4000\ \text{Hz} = 4\ \text{kHz}

b) Modulation index

β=Δffm=40002000=2\beta = \frac{\Delta f}{f_m} = \frac{4000}{2000} = 2

c) Bandwidth containing over 98% of power

Bessel coefficients for β=2\beta = 2:

nn01234
Jn(2)J_n(2)0.22390.57670.35280.12890.0340

Fraction of power in carrier + nn sideband pairs =J02+2∑k=1nJk2= J_0^2 + 2\sum_{k=1}^{n}J_k^2:

  • up to n=2n = 2: 0.0501+2(0.3326+0.1245)=0.9640.0501 + 2(0.3326 + 0.1245) = 0.964 (96.4%, not enough)
  • up to n=3n = 3: 0.964+2(0.0166)=0.9980.964 + 2(0.0166) = 0.998 (99.8%)

So 3 pairs of sidebands are needed:

B=2nfm=2×3×2 kHz=12 kHzB = 2 n f_m = 2 \times 3 \times 2\ \text{kHz} = 12\ \text{kHz}

This agrees with Carson's rule: BT=2(Δf+fm)=2(4+2)=12B_T = 2(\Delta f + f_m) = 2(4 + 2) = 12 kHz.

d) Total modulated signal power

Taking a 1 Ω load (normalised power):

P=Ac22=5022=1250 WP = \frac{A_c^2}{2} = \frac{50^2}{2} = 1250\ \text{W}

(FM power equals unmodulated carrier power, independent of β\beta.)

Answer: Δf=4\Delta f = 4 kHz, β=2\beta = 2, B=12B = 12 kHz, P=1250P = 1250 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:

  1. NBFM generation: for small β\beta,
s(t)≈Accos⁡ω1t−Ac ϕ(t)sin⁡ω1t,ϕ(t)=2πkf∫0tm(τ)dτs(t) \approx A_c\cos\omega_1 t - A_c\,\phi(t)\sin\omega_1 t, \quad \phi(t) = 2\pi k_f\int_0^t m(\tau)d\tau

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. Δf1≈25\Delta f_1 \approx 25 Hz, β<0.3\beta < 0.3) to avoid distortion. 2. First multiplier (×n1\times n_1): a non-linear device and BPF multiply the instantaneous frequency, so carrier becomes n1f1n_1 f_1 and deviation n1Δf1n_1\Delta f_1. 3. Mixer: shifts the carrier to ∣fLO−n1f1∣|f_{LO} - n_1 f_1| without changing the deviation. This keeps the final carrier at the required value. 4. Second multiplier (×n2\times n_2): gives the final deviation and carrier:

Δf=n1n2Δf1,fc=n2∣fLO−n1f1∣\Delta f = n_1 n_2\Delta f_1, \qquad f_c = n_2|f_{LO} - n_1 f_1|

Example: f1=200f_1 = 200 kHz, Δf1=25\Delta f_1 = 25 Hz, n1=64n_1 = 64, fLO=10.8f_{LO} = 10.8 MHz, n2=48n_2 = 48 gives fc=48×2f_c = 48 \times 2 MHz =96= 96 MHz and Δf=76.8\Delta f = 76.8 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 f2f^2, 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: m(t)=3cos⁡(2000t)m(t) = 3\cos(2000t), c(t)=9cos⁡(70000t)c(t) = 9\cos(70000t), s(t)=9cos⁡(70000t+16sin⁡2000t)s(t) = 9\cos(70000t + 16\sin 2000t).

So Ac=9A_c = 9 V, ωm=2000\omega_m = 2000 rad/s, β=16\beta = 16.

fm=20002π=318.31 Hz,fc=700002π=11.14 kHzf_m = \frac{2000}{2\pi} = 318.31\ \text{Hz}, \qquad f_c = \frac{70000}{2\pi} = 11.14\ \text{kHz}

Total modulated signal power

FM has constant amplitude, so (normalised to 1 Ω):

P=Ac22=922=40.5 WP = \frac{A_c^2}{2} = \frac{9^2}{2} = 40.5\ \text{W}

Modulation index

From the expression directly:

β=16\beta = 16

Peak frequency deviation

Δf=βfm=16×318.31=5092.96 Hz≈5.093 kHz\Delta f = \beta f_m = 16 \times 318.31 = 5092.96\ \text{Hz} \approx 5.093\ \text{kHz}

(In rad/s: Δω=16×2000=32000\Delta\omega = 16 \times 2000 = 32000 rad/s. The frequency sensitivity is kf=Δf/Am=5092.96/3=1697.65k_f = \Delta f/A_m = 5092.96/3 = 1697.65 Hz/V.)

Bandwidth (Carson's rule)

BT=2(Δf+fm)=2(5092.96+318.31)=10822.5 Hz≈10.82 kHzB_T = 2(\Delta f + f_m) = 2(5092.96 + 318.31) = 10822.5\ \text{Hz} \approx 10.82\ \text{kHz}

(Equivalently 2(β+1)ωm=2×17×2000=680002(\beta+1)\omega_m = 2 \times 17 \times 2000 = 68000 rad/s.)

Answer: P=40.5P = 40.5 W, β=16\beta = 16, Δf=5.093\Delta f = 5.093 kHz, BT≈10.82B_T \approx 10.82 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 ----->+-----+
  1. A matrix forms L+RL+R and L−RL-R (both 30 Hz – 15 kHz).
  2. L−RL-R DSB-SC modulates a 38 kHz subcarrier, occupying 23–53 kHz.
  3. A 19 kHz pilot (half the subcarrier frequency, about 10% deviation) is added so the receiver can regenerate 38 kHz in phase.
  4. The composite baseband signal is
m(t)=[L+R]+[L−R]cos⁡(2π 38k t)+Kcos⁡(2π 19k t)m(t) = [L+R] + [L-R]\cos(2\pi\, 38\text{k}\, t) + K\cos(2\pi\, 19\text{k}\, t)

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 |---+
                    +--------+  +----+
  1. The FM detector gives the composite baseband signal.
  2. A 15 kHz LPF gives L+RL+R.
  3. A 23–53 kHz BPF and a synchronous (product) detector with the regenerated 38 kHz carrier (pilot ×2) give L−RL-R.
  4. The matrix forms (L+R)+(L−R)=2L(L+R)+(L-R) = 2L and (L+R)−(L−R)=2R(L+R)-(L-R) = 2R, 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 L−RL-R 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

L+RL + R

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.

sPM(t)=Accos⁡[2πfct+kp m(t)]s_{PM}(t) = A_c\cos\left[2\pi f_c t + k_p\, m(t)\right]

where kpk_p is the phase sensitivity (rad/V).

Instantaneous frequency:

fi(t)=fc+kp2πdm(t)dtf_i(t) = f_c + \frac{k_p}{2\pi}\frac{dm(t)}{dt}

so in PM the frequency deviation depends on the rate of change of the message.

Single tone: for m(t)=Amcos⁡2πfmtm(t) = A_m\cos 2\pi f_m t,

s(t)=Accos⁡[2πfct+βpcos⁡2πfmt],βp=kpAms(t) = A_c\cos\left[2\pi f_c t + \beta_p\cos 2\pi f_m t\right], \qquad \beta_p = k_p A_m

The PM index βp\beta_p is independent of fmf_m, while the frequency deviation Δf=kpAmfm\Delta f = k_p A_m f_m increases with fmf_m.

Relation with FM:

  • PM of m(t)m(t) = FM of dmdt\frac{dm}{dt} (differentiator + FM modulator gives PM).
  • FM of m(t)m(t) = PM of ∫m dt\int m\,dt (integrator + phase modulator gives FM, used in Armstrong's method).

Features: constant envelope and power Ac2/2A_c^2/2; spectrum given by Bessel functions as in FM; bandwidth by Carson's rule 2(Δf+fm)2(\Delta f + f_m). 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 V0V_0 sets the operating point; the message m(t)m(t) is added through a coupling capacitor and RF choke.
  • The varactor capacitance is in parallel with the tank capacitance C0C_0.

Analysis. The junction capacitance is Cv=K(V0+VR)nC_v = \frac{K}{(V_0 + V_R)^{n}} (n = 1/2 for abrupt junction). For a small message, it varies nearly linearly:

C(t)=C0−ΔCcos⁡2πfmtC(t) = C_0 - \Delta C\cos 2\pi f_m t

The oscillator frequency is

fi(t)=12πLC(t)=12πLC0[1−ΔCC0cos⁡2πfmt]−1/2f_i(t) = \frac{1}{2\pi\sqrt{L C(t)}} = \frac{1}{2\pi\sqrt{LC_0}}\left[1 - \frac{\Delta C}{C_0}\cos 2\pi f_m t\right]^{-1/2}

For ΔC≪C0\Delta C \ll C_0, using (1−x)−1/2≈1+x/2(1-x)^{-1/2} \approx 1 + x/2:

fi(t)≈fc[1+ΔC2C0cos⁡2πfmt]=fc+Δfcos⁡2πfmtf_i(t) \approx f_c\left[1 + \frac{\Delta C}{2C_0}\cos 2\pi f_m t\right] = f_c + \Delta f\cos 2\pi f_m t

with fc=12πLC0f_c = \frac{1}{2\pi\sqrt{LC_0}} and Δf=ΔC2C0fc\Delta f = \frac{\Delta C}{2C_0}f_c. The instantaneous frequency varies linearly with the message, i.e. FM.

Disadvantages

  1. Poor carrier stability: an LC oscillator (not crystal) drifts with temperature and supply; an AFC loop is needed.
  2. Non-linearity: CvC_v varies non-linearly with voltage, so large deviations cause distortion; only small ΔC/C0\Delta C/C_0 is usable.
  3. Varactor capacitance is temperature-sensitive.
  4. 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: m(t)=2cos⁡(5000t)m(t) = 2\cos(5000t), c(t)=10cos⁡(100000t)c(t) = 10\cos(100000t), u(t)=10cos⁡(100000t+10sin⁡5000t)u(t) = 10\cos(100000t + 10\sin 5000t).

So Am=2A_m = 2 V, ωm=5000\omega_m = 5000 rad/s, Ac=10A_c = 10 V, ωc=105\omega_c = 10^5 rad/s, β=10\beta = 10.

fm=50002π=795.77 Hz,fc=1052π=15.915 kHzf_m = \frac{5000}{2\pi} = 795.77\ \text{Hz}, \qquad f_c = \frac{10^5}{2\pi} = 15.915\ \text{kHz}

Total modulated signal power

FM has constant amplitude; normalised to 1 Ω:

P=Ac22=1022=50 WP = \frac{A_c^2}{2} = \frac{10^2}{2} = 50\ \text{W}

Modulation index

From the expression: β=10\beta = 10.

Peak frequency deviation

Δf=βfm=10×795.77=7957.7 Hz≈7.96 kHz\Delta f = \beta f_m = 10 \times 795.77 = 7957.7\ \text{Hz} \approx 7.96\ \text{kHz}

(Δω=10×5000=50000\Delta\omega = 10 \times 5000 = 50000 rad/s; kf=Δω/Am=25000k_f = \Delta\omega/A_m = 25000 rad/s/V, i.e. 3978.9 Hz/V.)

Bandwidth (Carson's rule)

BT=2(Δf+fm)=2(β+1)fm=2×11×795.77=17507 Hz≈17.51 kHzB_T = 2(\Delta f + f_m) = 2(\beta + 1)f_m = 2 \times 11 \times 795.77 = 17507\ \text{Hz} \approx 17.51\ \text{kHz}

(In rad/s: 2×11×5000=1100002 \times 11 \times 5000 = 110000 rad/s.)

Answer: P=50P = 50 W, β=10\beta = 10, Δf=7.96\Delta f = 7.96 kHz, BT≈17.51B_T \approx 17.51 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

  1. Input FM: s(t)=Acsin⁡[2πfct+ϕ1(t)]s(t) = A_c\sin[2\pi f_c t + \phi_1(t)], with ϕ1(t)=2πkf∫0tm(τ)dτ\phi_1(t) = 2\pi k_f\int_0^t m(\tau)d\tau.
  2. VCO output (free-running at fcf_c): r(t)=Avcos⁡[2πfct+ϕ2(t)]r(t) = A_v\cos[2\pi f_c t + \phi_2(t)], with ϕ2(t)=2πkv∫0tv(τ)dτ\phi_2(t) = 2\pi k_v\int_0^t v(\tau)d\tau.
  3. Phase detector (multiplier) output:
s(t) r(t)=12AcAvsin⁡[ϕ1−ϕ2]+12AcAvsin⁡[4πfct+ϕ1+ϕ2]s(t)\,r(t) = \tfrac12 A_cA_v\sin[\phi_1 - \phi_2] + \tfrac12 A_cA_v\sin[4\pi f_c t + \phi_1 + \phi_2]

The loop filter rejects the 2fc2f_c term, leaving

e(t)=12AcAvsin⁡ϕe(t),ϕe=ϕ1−ϕ2e(t) = \tfrac12 A_cA_v\sin\phi_e(t), \qquad \phi_e = \phi_1 - \phi_2
  1. Linearised model: when locked, ϕe\phi_e is small, so sin⁡ϕe≈ϕe\sin\phi_e \approx \phi_e. The loop adjusts v(t)v(t) to keep ϕe\phi_e near zero. Then
ϕ2(t)≈ϕ1(t)  ⇒  2πkv∫0tv(τ) dτ≈2πkf∫0tm(τ) dτ\phi_2(t) \approx \phi_1(t) \;\Rightarrow\; 2\pi k_v\int_0^t v(\tau)\,d\tau \approx 2\pi k_f\int_0^t m(\tau)\,d\tau
  1. Differentiating:
v(t)≈kfkv m(t)v(t) \approx \frac{k_f}{k_v}\,m(t)

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.

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