Question 4 of 5: Recovering the Message from an Angle-Modulated Signal

Question 4: Recovering the Message from an Angle-Modulated Signal (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. $x(t) = 10\cos(12000\pi t)$ V on $0 \lt t \le 2$ s and $x(t) = 10\cos(8000\pi t + 6000\pi)$ V on $2 \lt t \le 4$ s; carrier frequency $f_c = 5$ kHz; frequency deviation constant $k_f = 1$ kHz/V.

Find. The message $m(t)$ that produced this waveform, and a block diagram of a non-PLL FM demodulator.

Approach. Differentiate the total phase of each piece to get the instantaneous frequency, subtract the carrier to get the deviation, and divide by $k_f$ to recover $m(t)$; then confirm that the printed phase constant makes the two pieces join continuously.

1. State the FM relation. For frequency modulation the instantaneous frequency is the carrier offset in proportion to the message: $$\theta(t) = 2\pi f_c t + 2\pi k_f\!\int^{t} m(\lambda)\,d\lambda, \qquad f_i(t) = \frac{1}{2\pi}\frac{d\theta}{dt} = f_c + k_f\,m(t).$$ Inverting gives the recovery formula used in both intervals: $$\boxed{\,m(t) = \frac{f_i(t) - f_c}{k_f}\,}$$
2. Part (a) — read the instantaneous frequency on the first interval. On $0 \lt t \le 2$ the phase is $\theta = 12000\pi t$, so $$f_i = \frac{1}{2\pi}\frac{d}{dt}(12000\pi t) = \frac{12000\pi}{2\pi} = 6000\ \text{Hz},$$ a deviation of $6000 - 5000 = +1000$ Hz. Dividing by $k_f = 1000$ Hz/V gives $m = +1$ V.
3. Repeat on the second interval. On $2 \lt t \le 4$ the phase is $\theta = 8000\pi t + 6000\pi$; the constant differentiates away, so $$f_i = \frac{8000\pi}{2\pi} = 4000\ \text{Hz},$$ a deviation of $4000 - 5000 = -1000$ Hz and hence $m = -1$ V. The message is therefore a single rectangular cycle: $$\boxed{\,m(t) = \begin{cases} +1\ \text{V}, & 0 \lt t \le 2\ \text{s}\\ -1\ \text{V}, & 2 \lt t \le 4\ \text{s}\end{cases}\,}$$
4. Check that the phase is continuous at the join. A physically realisable FM signal has no phase discontinuity, and this is what the otherwise mysterious $6000\pi$ constant is for. At $t = 2$ s the first expression gives $\theta_1 = 12000\pi(2) = 24000\pi$ rad and the second gives $\theta_2 = 8000\pi(2) + 6000\pi = 22000\pi$ rad. The difference is $$\theta_1 - \theta_2 = 2000\pi = 1000\times(2\pi),$$ an exact multiple of $2\pi$, so the two cosines agree in value and slope at $t = 2$ and the waveform is continuous. Had the constant been anything else, the “FM” signal would contain a phase jump and could not have come from integrating a bounded message.

5. Quantify the modulation. The peak frequency deviation is $\Delta f = k_f|m|_{\max} = 1$ kHz. Treating the message as one cycle of a square wave of period 4 s, its fundamental is $W = 0.25$ Hz, so Carson's rule gives a transmission bandwidth $$B_T = 2(\Delta f + W) = 2(1000 + 0.25) = 2000.5\ \text{Hz} \approx 2\ \text{kHz},$$ an extreme wideband case ($\beta = \Delta f/W = 4000$) in which the bandwidth is set almost entirely by the deviation. The signal has constant envelope, so its power is $10^2/2 = 50$ W into 1 Ω regardless of the message.
6. Part (b) — a non-PLL FM demodulator. The classical answer is the frequency discriminator: a limiter followed by a differentiator, an envelope detector and a dc block. Differentiating $x(t) = A_c\cos\theta(t)$ gives $$\frac{dx}{dt} = -A_c\left[2\pi f_c + 2\pi k_f m(t)\right]\sin\theta(t),$$ whose envelope $A_c\left[2\pi f_c + 2\pi k_f m(t)\right]$ is an affine function of the message. An ordinary envelope detector therefore recovers $m(t)$ once the constant $2\pi f_c$ term is blocked. The hard limiter ahead of the chain is essential: it strips any amplitude variation picked up in the channel, so that amplitude noise cannot masquerade as message. In practice the ideal differentiator is realised as a slope circuit — a detuned resonant tank or, in the balanced form, two tanks tuned either side of $f_c$ whose detected outputs are subtracted to linearise the characteristic.