22-Elec-A3 Signals and Communications · May 2017

Question 5 of 5: Angle-Modulated Signal — Message Recovery, Deviation and Bandwidth

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada national examination 16-Elec-A3 — Signals and Communications, May 2017. Closed book (one non-programmable calculator permitted), 3 hours, 4 pages. Five questions, all compulsory, all of equal value (20 marks each). All five are solved in full below.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed. (AM/DSB power and efficiency, PCM, FM/Carson); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier series, filtering, average power); A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed. (LTI convolution, BIBO stability); J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed. (FIR structures, transfer functions); C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (first-order RC response).

Question 5: Angle-Modulated Signal — Message Recovery, Deviation and Bandwidth (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.

Quantity	Value
Carrier amplitude	A_c = 5 V
Phase deviation	φ(t) = a cos(2πf₁t) + b cos(2πf₂t)
Phase coefficients	a = 1 rad, b = 2 rad
Tone frequencies	f₁ = 2000 Hz, f₂ = 2f₁ = 4000 Hz
Frequency deviation constant	k_f = 5 kHz/V

Find. The message m(t) that produced this phase; the average power of s(t); the peak frequency deviation; the transmission bandwidth; and a demodulator block diagram.

Approach. In FM the phase is the integral of the message, so differentiating $\phi(t)$ recovers m(t). The power of a constant-envelope signal is independent of the modulation. The peak deviation must be maximised properly — because $f_2=2f_1$ the two sinusoidal terms are harmonically related and their peaks do not coincide.

(a) Recover the message. For frequency modulation $\phi(t)=2\pi k_f\!\int^t m(\lambda)\,d\lambda$, so $$m(t)=\frac{1}{2\pi k_f}\frac{d\phi}{dt} =\frac{1}{2\pi k_f}\left[-2\pi a f_1\sin(2\pi f_1 t)-2\pi b f_2\sin(2\pi f_2 t)\right]$$ $$m(t)=-\frac{a f_1}{k_f}\sin(2\pi f_1 t)-\frac{b f_2}{k_f}\sin(2\pi f_2 t) =-\frac{(1)(2000)}{5000}\sin(2\pi f_1t)-\frac{(2)(4000)}{5000}\sin(2\pi f_2t)$$ $$\boxed{m(t)=-0.4\sin\!\left(2\pi(2000)t\right)-1.6\sin\!\left(2\pi(4000)t\right)\ \text{V}}$$ The message is itself a two-tone signal, dominated by the 4 kHz component because that term contributes both a larger phase amplitude and a higher frequency.

Figure Q5.1 - Recovered message m(t) over two cycles of the 2 kHz tone.

(b) Average power. An angle-modulated signal has a constant envelope, so the modulation moves power around in frequency but never changes the total: $$P=\frac{A_c^{2}}{2}=\frac{5^{2}}{2}=\boxed{12.5\ \text{W}}$$ This is one of the defining advantages of FM — transmitters can run at constant, and therefore efficient, output power.
(c) Peak frequency deviation. The instantaneous frequency deviation is $$\Delta f(t)=\frac{1}{2\pi}\frac{d\phi}{dt}=-\left[a f_1\sin(2\pi f_1t)+b f_2\sin(2\pi f_2t)\right] =-\left[2000\sin\theta+8000\sin 2\theta\right]$$ with $\theta=2\pi f_1 t$. It is tempting to add the two peaks and quote 10 000 Hz, but the sine terms peak at different instants. Substituting $\sin2\theta=2\sin\theta\cos\theta$ collapses the expression to a single variable $c=\cos\theta$: $$g(\theta)=\sin\theta\left(2000+16000\cos\theta\right),\qquad \frac{dg}{d\theta}=0\ \Longrightarrow\ 16c^{2}+c-8=0$$ $$c=\frac{-1+\sqrt{1+512}}{32}=0.67655\quad(\text{the other root, } c=-0.73905,\ \text{gives only } 6618\ \text{Hz})$$ With $\sin\theta=\sqrt{1-c^{2}}=0.73640$, $$\Delta f = 0.73640\left(2000+16000\times0.67655\right)=0.73640\times12\,824.8$$ $$\boxed{\Delta f = 9444\ \text{Hz} \approx 9.44\ \text{kHz}}$$ which is 5.6 % below the naive sum of the peaks. A numerical sweep of $|g(\theta)|$ over a full cycle confirms the maximum to five figures.
(d) Transmission bandwidth. Carson's rule uses the peak deviation and the highest message frequency, here $W=f_2=4$ kHz: $$B_T = 2\left(\Delta f + W\right)=2\left(9444+4000\right)$$ $$\boxed{B_T = 26.89\ \text{kHz}}$$ The effective deviation ratio is $\beta=\Delta f/W = 9444/4000 = 2.36$, comfortably in the wideband regime, so Carson's rule (rather than the narrowband approximation $B_T\approx 2W$) is the right tool. Using the naive 10 kHz deviation would have over-stated the bandwidth by about 1.1 kHz.
(e) Demodulator. A conventional frequency discriminator recovers the message in four stages: a band-pass filter centred on $f_c$ with bandwidth $B_T$ rejects out-of-band noise; a hard limiter strips any amplitude variation picked up in the channel (essential, since all the information is in the zero crossings); a differentiator converts frequency variation into amplitude variation, producing $A_c[\omega_c+2\pi k_f m(t)]\sin(\cdot)$; and an envelope detector followed by a dc block and a low-pass filter of bandwidth $W$ delivers $m(t)$ scaled by $A_ck_f$. A phase-locked loop is the modern alternative: its loop filter output is directly proportional to $m(t)$ and it performs better at low signal-to-noise ratio.

Figure Q5.2 - Frequency-discriminator demodulator.

Result	Value
Message signal (a)	m(t) = −0.4 sin(2π·2000t) − 1.6 sin(2π·4000t) V
Average power (b)	P = A_c²/2 = 12.5 W
Peak frequency deviation (c)	Δf = 9444 Hz (naive sum would give 10 000 Hz)
Deviation ratio	β = Δf/W = 2.36
Carson bandwidth (d)	B_T = 2(9444 + 4000) = 26.89 kHz
Demodulator (e)	BPF → limiter → differentiator → envelope detector → dc block + LPF (or PLL)

Topic. Frequency modulation: message recovery from the phase, constant-envelope power, peak deviation of a multi-tone message and Carson's rule.

Key relations. $\phi(t)=2\pi k_f\int m$, so $m=\frac{1}{2\pi k_f}\dot\phi$; $P=A_c^{2}/2$; $\Delta f=\max\left|\frac{1}{2\pi}\dot\phi\right|$; deviation ratio $\beta=\Delta f/W$; Carson's rule $B_T=2(\Delta f+W)$.

Why this works. Angle modulation stores information in the instantaneous phase, so the message is obtained by differentiation — which is precisely what a discriminator does electrically. Since the amplitude is fixed, the power is fixed too, and all the "modulation energy" appears as sideband structure. Carson's rule is an empirical statement that essentially all the power lies within the frequency excursion plus one message bandwidth on either side; it interpolates correctly between the narrowband limit $2W$ and the wideband limit $2\Delta f$.

Common pitfall. Adding the individual peak deviations when the tones are harmonically related. Here $2000+8000=10\,000$ Hz overstates the true 9444 Hz because $\sin\theta$ and $\sin2\theta$ cannot peak simultaneously; the maximisation must be done properly. A second trap is using $f_1$ as the message bandwidth in Carson's rule instead of the highest tone $f_2$.

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, 5th ed., §5.1–5.3 (angle modulation, bandwidth of FM, demodulators); Haykin, Communication Systems, 5th ed., Ch. 4.

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