22-Elec-A3 Signals and Communications · December 2015

Question 5 of 6: Frequency Modulation with a Two-Tone Phase Deviation

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

Notes on this paper

Paper format: 07-Elec-A3 Signals and Communications, December 2015 — 3 hours, closed book, a standard non-programmable calculator permitted. Six questions of equal value (20 marks each); any five constitute a complete paper, and only the first five presented are marked. All six are solved here, since the set is intended as a study resource.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (AM/FM, PCM, frequency conversion); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Fourier series, LTI systems); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (z-transform, difference equations, stability); S. Haykin, Communication Systems, 5th ed. (modulation and detection).

Question 5: Frequency Modulation with a Two-Tone Phase Deviation (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	Symbol	Value
Carrier amplitude	$A_{c}$	5 V
Phase deviation	$\phi(t)$	$a\cos(2\pi f_{1}t) + b\cos(2\pi f_{2}t)$
Phase coefficients	$a$, $b$	1, 2 rad
Tone frequencies	$f_{1}$, $f_{2} = 2f_{1}$	2000 Hz, 4000 Hz
Frequency deviation constant	$k_{f}$	5 kHz/V

Find. (a) the message $m(t)$; (b) the average power; (c) the peak frequency deviation; (d) the transmission bandwidth; (e) a demodulator block diagram.

Approach. In FM the phase is the integral of the message, so differentiating the given $\phi(t)$ recovers $m(t)$ directly. Average power depends only on the carrier amplitude. The peak deviation requires maximising a two-tone sum — which, crucially, is not the sum of the individual peaks — and Carson's rule then gives the bandwidth.

Relate phase to message. For frequency modulation with deviation constant $k_{f}$ (in Hz per volt), $$\phi(t) = 2\pi k_{f}\int_{-\infty}^{t} m(\lambda)\,d\lambda \quad\Longrightarrow\quad m(t) = \frac{1}{2\pi k_{f}}\frac{d\phi}{dt}$$
Differentiate the given phase. $$\frac{d\phi}{dt} = -2\pi a f_{1}\sin(2\pi f_{1}t) - 2\pi b f_{2}\sin(2\pi f_{2}t)$$
Divide by $2\pi k_{f}$ and substitute values. With $af_{1} = 1 \times 2000 = 2000$ and $bf_{2} = 2 \times 4000 = 8000$, and $k_{f} = 5000$ Hz/V: $$\boxed{\,m(t) = -\left[0.4\sin(2\pi \times 2000\,t) + 1.6\sin(2\pi \times 4000\,t)\right]\ \text{V}\,}$$ The negative sign is a consequence of the cosine phase given; a sign inversion of the message is immaterial to the modulation, but the amplitudes 0.4 V and 1.6 V are not.
Compute the average power. An angle-modulated carrier has constant envelope, so modulation does not change its power at all: $$\boxed{\,P_{\text{avg}} = \frac{A_{c}^{2}}{2} = \frac{25}{2} = 12.5\ \text{W}\,}$$ This constant-envelope property is the reason FM tolerates non-linear (and hence efficient) power amplifiers, unlike AM.
Set up the peak frequency deviation. The instantaneous frequency deviation is $$\Delta f(t) = \frac{1}{2\pi}\frac{d\phi}{dt} = -\left[af_{1}\sin\theta + bf_{2}\sin 2\theta\right], \qquad \theta = 2\pi f_{1}t$$ using $f_{2} = 2f_{1}$ so that both terms share the single variable $\theta$. The peak deviation is the maximum of the magnitude of this expression.
Maximise the two-tone sum. Setting the derivative to zero, $af_{1}\cos\theta + 2bf_{2}\cos 2\theta = 0$. Substituting $\cos 2\theta = 2\cos^{2}\theta - 1$ and $c = \cos\theta$ gives $2000c + 16000(2c^{2} - 1) = 0$, i.e. $$16c^{2} + c - 8 = 0 \quad\Longrightarrow\quad c = \frac{-1 + \sqrt{513}}{32} = 0.67655$$ Then $\sin\theta = 0.73640$, $\sin 2\theta = 0.99637$, and $$\Delta f_{\text{peak}} = 2000(0.73640) + 8000(0.99637) = 1472.8 + 7970.9 \quad\Longrightarrow\quad \boxed{\,\Delta f = 9444\ \text{Hz} \approx 9.44\ \text{kHz}\,}$$

The naive answer — adding the two individual peak deviations, $af_{1} + bf_{2} = 2000 + 8000 = 10\,000$ Hz — overstates the result by about 6%, because the two sinusoids do not attain their maxima at the same instant. The bound is only reached when the tones are harmonically unrelated over a long enough observation; here $f_{2} = 2f_{1}$ locks their relative phase permanently.

Figure 5.1 — Instantaneous frequency deviation over one period of the lower tone. The true peak of 9444 Hz falls short of the naive 10 000 Hz sum because the two components never peak together.

Apply Carson's rule. With the highest message frequency $W = f_{2} = 4$ kHz, $$B_{T} = 2\left(\Delta f + W\right) = 2(9444 + 4000) = 26\,888\ \text{Hz} \quad\Longrightarrow\quad \boxed{\,B_{T} \approx 26.9\ \text{kHz}\,}$$ The effective modulation index is $\beta = \Delta f/W = 2.36$, comfortably in the wideband regime where Carson's rule is the appropriate estimate.

For part (e), the standard non-coherent FM receiver converts frequency variation into amplitude variation and then detects the envelope. A limiter is essential because any amplitude disturbance picked up on the channel would otherwise pass straight through the differentiator into the output.

Figure 5.2 — Frequency discriminator. The band-pass filter selects the channel, the limiter removes amplitude variation, the differentiator converts frequency deviation into amplitude, and the envelope detector recovers the message. A phase-locked loop is the common alternative and performs better at low signal-to-noise ratio.

Quantity	Result
Message signal	$m(t) = -[0.4\sin(2\pi\,2000t) + 1.6\sin(2\pi\,4000t)]$ V
Average power	12.5 W
Peak frequency deviation	$\Delta f = 9444$ Hz (naive sum would give 10 000 Hz)
Effective modulation index	$\beta = \Delta f/f_{2} = 2.36$
Carson bandwidth	$B_{T} = 26.9$ kHz
Demodulator	BPF → limiter → differentiator → envelope detector → DC block

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

Key relations: $\phi(t) = 2\pi k_{f}\int m\,dt$ and $m(t) = \dfrac{1}{2\pi k_{f}}\dfrac{d\phi}{dt}$; instantaneous deviation $\Delta f(t) = \dfrac{1}{2\pi}\dfrac{d\phi}{dt}$; $P = A_{c}^{2}/2$ regardless of modulation; $\beta = \Delta f/W$; Carson $B_{T} = 2(\Delta f + W)$.

Why this works: The defining property of FM is that instantaneous frequency, not amplitude, follows the message; the phase is therefore its integral, and one differentiation undoes the modulation exactly. Because the envelope is constant, the average power is fixed by $A_{c}$ alone — no power flows from carrier to sidebands as the index changes; instead the distribution among the Bessel-weighted sidebands changes while the total stays put. Carson's rule then captures the empirical fact that essentially all of that power lies within $\Delta f + W$ of the carrier on each side.

Common pitfall: Adding the individual peak deviations is the dominant error, and it is wrong whenever the tones are harmonically related, as they are here. A second trap is confusing the phase coefficients $a = 1$ and $b = 2$ (in radians) with modulation indices; the deviation of each component is $af_{1}$ and $bf_{2}$, so the higher tone with the larger coefficient dominates by a factor of four. Third, in Carson's rule $W$ is the highest frequency present in the message (4 kHz here), not the lowest tone and not the sum of the tones.

Source: Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., Ch. 5 (angle modulation, Carson's rule, demodulation); Haykin, Communication Systems, 5th ed., Ch. 4.

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