22-Elec-A3 Signals and Communications · December 2019

Question 5 of 5: FM Modulator Driven by a Rectangular Message — Deviation, Signal Form, Carson Bandwidth and Demodulation

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

Question 5: FM Modulator Driven by a Rectangular Message — Deviation, Signal Form, Carson Bandwidth and Demodulation (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.

Find. The peak frequency deviation; an explicit plottable expression for $s(t)$; Carson-rule bandwidths for three pulse durations spanning six decades; and the demodulator output for a general angle-modulated input.

Approach. Resolve the step combination into a simple two-level rectangular message, scale by $k_f$ to obtain the instantaneous frequency, integrate to get the phase, and apply Carson's rule with the message bandwidth estimated as the reciprocal of the pulse width.

1. Resolve the message into levels. Adding the three steps interval by interval: $$m(t) = \begin{cases} 0, & t \lt 0\\ +2\ \text{V}, & 0 \lt t \lt T\\ 2 - 4 = -2\ \text{V}, & T \lt t \lt 2T\\ -2 + 2 = 0, & t \gt 2T \end{cases}$$ so the message is a single positive rectangular pulse of amplitude 2 V followed immediately by a negative pulse of amplitude −2 V, each of duration $T$. The message has zero net area, $\int m\,dt = 2T - 2T = 0$, which will matter in part (b).
2. (a) Compute the peak frequency deviation. The instantaneous frequency is $f_i(t) = f_c + k_f m(t)$, so the deviation is largest where $|m(t)|$ is largest: $$\Delta f = k_f\,|m(t)|_{\max} = (5\ \text{kHz/V})(2\ \text{V})$$ $$\boxed{\Delta f = 10\ \text{kHz}}$$ The instantaneous frequency therefore swings between 110 kHz and 90 kHz.
3. (b) Fix the amplitude from the power constraint. An FM signal has constant envelope, so its average power is $A_c^2/2$ regardless of the modulation. Setting this to 1 W, $$\frac{A_c^2}{2} = 1 \quad\Longrightarrow\quad A_c = \sqrt{2} = 1.414\ \text{V}$$
4. Integrate the message to obtain the phase. The FM phase is $\phi(t) = 2\pi k_f \int_{-\infty}^{t} m(\tau)\,d\tau$. Integrating the two-level message piecewise gives the triangular phase function $$\int_{-\infty}^{t} m(\tau)\,d\tau = \begin{cases} 0, & t \lt 0\\ 2t, & 0 \le t \le T\\ 4T - 2t, & T \le t \le 2T\\ 0, & t \gt 2T \end{cases}$$ which rises linearly to $2T$ at $t = T$ and returns to zero at $t = 2T$ — the phase is continuous throughout and the modulator ends exactly where it started.
5. Write the plottable FM expression. Substituting $A_c = \sqrt2$, $f_c = 100$ kHz and $k_f = 5$ kHz/V, $$\boxed{\;s(t) = \sqrt{2}\,\cos\!\left(2\pi(100\,000)t + 2\pi(5000)\!\int_{-\infty}^{t}\! m(\tau)\,d\tau\right)\ \text{V}\;}$$ which, evaluated interval by interval, becomes the explicitly plottable piecewise form $$s(t) = \begin{cases} \sqrt{2}\cos\!\left(2\pi(100\,000)t\right), & t \lt 0\\ \sqrt{2}\cos\!\left(2\pi(110\,000)t\right), & 0 \le t \le T\\ \sqrt{2}\cos\!\left(2\pi(90\,000)t + 2\pi(20\,000)T\right), & T \le t \le 2T\\ \sqrt{2}\cos\!\left(2\pi(100\,000)t\right), & t \gt 2T \end{cases}$$ The constant $2\pi(20\,000)T$ in the third line is exactly what keeps the phase continuous at $t = T$ — an FM modulator never produces a phase jump, only a slope change.

6. (c) Estimate the message bandwidth for each pulse duration. A rectangular pulse of width $T$ has a $\mathrm{sinc}$ spectrum whose first null — the usual engineering bandwidth estimate — sits at $W = 1/T$. Carson's rule then gives $$B_T = 2\left(\Delta f + W\right) = 2\left(10\ \text{kHz} + \frac{1}{T}\right)$$ Working the three cases:

   Case	$T$	$W = 1/T$	$\beta = \Delta f/W$	$B_T = 2(\Delta f + W)$	Regime
   (i)	1 s	1 Hz	$10^{4}$	20.002 kHz $\approx$ 20 kHz	Wideband: $B_T \to 2\Delta f$
   (ii)	1 ms	1 kHz	10	22 kHz	Wideband, $W$ now visible
   (iii)	1 µs	1 MHz	0.01	2.02 MHz $\approx$ 2 MHz	Narrowband: $B_T \to 2W$

   $$\boxed{B_T \approx 20\ \text{kHz},\quad 22\ \text{kHz},\quad 2.02\ \text{MHz}}$$ The three cases deliberately span the whole range of the modulation index. For a very slow message the deviation dominates and the bandwidth is set entirely by $2\Delta f$; for a very fast message the deviation is irrelevant and the signal behaves like narrowband FM, occupying $2W$ just as an AM signal would.
7. (d) Write the demodulator output. For $x(t) = A\cos(\omega_c t + \phi(t))$ with $\omega_c = 100\,000$ rad/s (note this is a radian frequency, so $f_c = 100\,000/2\pi = 15.92$ kHz), the instantaneous frequency deviation from the carrier is $$f_i(t) - f_c = \frac{1}{2\pi}\frac{d\phi(t)}{dt}$$ An ideal FM demodulator (a limiter followed by a discriminator) produces an output proportional to that deviation. Scaling by the same constant the modulator used, $k_f$, returns the message itself: $$\boxed{\;y(t) = \frac{1}{2\pi k_f}\,\frac{d\phi(t)}{dt} = \frac{1}{2\pi (5000)}\,\frac{d\phi(t)}{dt}\ \text{volts}\;}$$ Since $\phi(t)$ is baseband with bandwidth below 2 kHz and the carrier sits at 15.92 kHz, the spectra do not overlap and the differentiation-plus-envelope-detection operation recovers $d\phi/dt$ cleanly; a post-detection low-pass filter of cut-off 2 kHz removes the residual carrier-frequency products. The demodulated signal is therefore, up to the constant $1/(2\pi k_f)$, simply the derivative of the phase — which is the defining statement of what an FM discriminator does.