22-Elec-A3 Signals and Communications · May 2015

Question 6 of 7: FM Signal Processing — Multiplication, Frequency Translation, Demodulation and Squaring

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Elec-A3 Signals and Communications, May 2015 — 3 hours, closed book, non-programmable calculator permitted. Seven questions, all of equal value (20 marks each); any five constitute a complete paper and only the first five in the answer book are marked. All seven are solved here, since the set is a study resource.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed.; B. P. Lathi, Linear Systems and Signals, 2nd ed.; A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed.; J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed.; S. Haykin & M. Moher, Communication Systems, 5th ed.

Check: pulse definition. The pulse is p(t) = t/a — a ramp, not a constant — and every part of Question 1 is solved against that reading. A constant-height pulse would make parts (a)–(c) trivial and would not produce the odd-harmonic structure the question is clearly aiming at.

Question 6: FM Signal Processing — Multiplication, Frequency Translation, Demodulation and Squaring (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. FM signal with $f_c = 10$ MHz, Carson bandwidth $B_T = 60$ kHz, message bandwidth $f_m = 10$ kHz.

Find. The bandwidth after a $\times 2$ multiplier; a system giving that bandwidth at a 15 MHz carrier; an FM demodulator; and whether an FM signal survives a square-law device.

Approach. Invert Carson's rule to extract the frequency deviation, which is the quantity a multiplier scales; then use the standard result that multiplication scales both carrier and deviation while mixing translates the carrier only.

Extract the deviation and modulation index. Carson's rule gives $B_T = 2(\Delta f + f_m)$, so $$\Delta f = \frac{B_T}{2} - f_m = 30 - 10 = 20\ \text{kHz}, \qquad \beta = \frac{\Delta f}{f_m} = 2$$ This is a wideband FM signal ($\beta \gt 1$); the deviation, not the bandwidth, is the physically meaningful parameter to carry forward.
Bandwidth after the $\times 2$ multiplier (part a). A frequency multiplier of order $n$ multiplies the instantaneous phase by $n$, hence both the carrier and the deviation by $n$, while the message frequency is untouched. With $n = 2$: $f_c' = 20$ MHz, $\Delta f' = 40$ kHz, $\beta' = 4$, so $$B_T' = 2(\Delta f' + f_m) = 2(40 + 10) = \boxed{\,100\ \text{kHz}\,}$$ Note the bandwidth does not simply double (that would give 120 kHz) because the $f_m$ term inside Carson's rule is not scaled.
Converting to a 15 MHz carrier at the same bandwidth (part b). Bandwidth and carrier must be set independently, which is exactly why the two operations are separate: the multiplier fixes the deviation, then a mixer slides the whole spectrum without touching it. Multiplying by 2 gives 20 MHz with the required 40 kHz deviation; mixing with a 5 MHz local oscillator produces components at 25 MHz and 15 MHz, and a band-pass filter centred at 15 MHz with a passband of at least 100 kHz selects the difference.
Figure Q6.1 — converting the FM signal to a 15 MHz carrier at 100 kHz bandwidth: doubler sets the deviation, mixer slides the carrier.
FM demodulator (part c). The standard discriminator chain converts frequency variation into amplitude variation and then detects it. A band-pass limiter first strips any amplitude variation picked up in the channel (FM carries no information in its envelope, so this is free noise rejection). A differentiator follows: applying $\frac{d}{dt}$ to $A_c\cos[\omega_ct + \phi(t)]$ yields an amplitude proportional to the instantaneous frequency $\omega_c + \dot\phi(t)$, so the signal is now AM-on-FM. An ordinary envelope detector recovers that amplitude, and a DC block removes the $\omega_c$ pedestal, leaving $\propto m(t)$. A phase-locked loop demodulator is an equally acceptable answer and performs better near threshold.
Figure Q6.2 — discriminator FM demodulator: limiter, differentiator, envelope detector, DC block.
Effect of a square-law device (part d). With $s(t) = A_c\cos[\omega_ct + \phi(t)]$, $$s^2(t) = \frac{A_c^2}{2}\Big[1 + \cos\big(2\omega_ct + 2\phi(t)\big)\Big]$$ The second term is still a pure FM signal: the carrier has moved to $2\omega_c$ and the phase deviation has doubled, so the deviation is $2\Delta f = 40$ kHz and the bandwidth 100 kHz, but the message enters $\phi(t)$ linearly and no information is lost. The signal is therefore not distorted beyond recovery — a square-law device acts on FM exactly as a $\times 2$ frequency multiplier, which is precisely how practical multipliers are built.
Recovering the message after squaring. Band-pass filter around $2f_c$ to discard the DC term, demodulate with the same discriminator chain as in part (c), and halve the result to undo the doubled deviation: $$\boxed{\,\text{squarer} \to \text{BPF at } 2f_c \to \text{limiter/discriminator} \to \text{gain } \tfrac{1}{2} \to m(t)\,}$$
Figure Q6.3 — recovering the message after a square-law device: the squared FM signal is an ordinary FM signal at 2fc with twice the deviation.
The contrast with AM is worth stating: squaring a conventional AM signal does cause irrecoverable distortion, because the message sits in the envelope and $m^2(t)$ cannot be unambiguously inverted. FM survives because its information is in the phase, and squaring is a linear operation on phase.

Quantity	Result
Frequency deviation / index	$\Delta f = 20$ kHz, $\beta = 2$
After $\times 2$ multiplier	$f_c = 20$ MHz, $\Delta f = 40$ kHz, $\beta = 4$, $B_T = 100$ kHz
To reach 15 MHz	$\times 2$ multiplier, mix with 5 MHz LO, BPF at 15 MHz (100 kHz wide)
Demodulator	Limiter → differentiator → envelope detector → DC block (or PLL)
Square-law device	Not destructive: yields FM at $2f_c$ with $2\Delta f$; recoverable

Topic. Angle modulation: Carson's rule, frequency multiplication versus frequency translation, and discriminator demodulation.

Key relations. $s(t) = A_c\cos\!\left[\omega_ct + 2\pi k_f\!\int m(\tau)d\tau\right]$; $\beta = \Delta f/f_m$; Carson $B_T = 2(\Delta f + f_m) = 2f_m(\beta+1)$; an $n$-fold multiplier maps $(f_c, \Delta f) \to (nf_c, n\Delta f)$; a mixer maps $f_c \to f_c \pm f_{LO}$ with $\Delta f$ unchanged.

Why this works. Multiplication and mixing affect an angle-modulated signal in complementary ways — one scales the phase (and therefore the deviation), the other simply shifts the spectrum — and it is this complementarity that lets a transmitter reach any wanted carrier frequency and any wanted deviation independently. The Armstrong indirect FM transmitter is built on exactly this pairing. Demodulation works because differentiation converts a frequency deviation into a proportional amplitude, reducing an FM problem to an AM one that a diode detector can finish.

Common pitfall. Doubling the bandwidth instead of the deviation when a multiplier is applied — Carson's rule is affine in $\Delta f$, not linear, so the $f_m$ term must be added back after scaling. A second is assuming a non-linearity necessarily destroys the signal; for FM the square-law device is benign, and the limiter that precedes any FM detector deliberately relies on that insensitivity to amplitude.

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, §5.2–5.4 (FM bandwidth, generation by multiplication, demodulators); Haykin & Moher, Communication Systems, Ch. 4.

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Quantity	Result
Frequency deviation / index	\(\Delta f = 20\) kHz, \(\beta = 2\)
After \(\times 2\) multiplier	\(f_c = 20\) MHz, \(\Delta f = 40\) kHz, \(\beta = 4\), \(B_T = 100\) kHz
To reach 15 MHz	\(\times 2\) multiplier, mix with 5 MHz LO, BPF at 15 MHz (100 kHz wide)
Demodulator	Limiter → differentiator → envelope detector → DC block (or PLL)
Square-law device	Not destructive: yields FM at \(2f_c\) with \(2\Delta f\); recoverable