22-Elec-A3 Signals and Communications · December 2016

Question 6 of 6: VCO-Based FM Modulator, Demodulator and Carson 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 07-Elec-A3 — Signals and Communications, December 2016. Three hours, closed book; a standard non-programmable calculator is the only permitted aid. Six questions of equal value (20 marks each); any five constitute a complete paper. All six are solved below, because this set is a study resource rather than an examination script.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (Ch. 2–4 Fourier analysis, Ch. 4 amplitude modulation, Ch. 5 angle modulation, Ch. 6 sampling and pulse-code modulation); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Ch. 3 Fourier series, Ch. 10 the z-transform); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (Ch. 3 z-transform, Ch. 7 realisation structures); S. Haykin and M. Moher, Communication Systems, 5th ed. (Ch. 3 amplitude modulation, Ch. 4 angle modulation, Ch. 7 digital transmission).

Notation used throughout. The normalised sinc function is $\operatorname{sinc}(x)=\dfrac{\sin \pi x}{\pi x}$, so that $\operatorname{sinc}(0)=1$ and the zeros sit at the non-zero integers. Average power of a periodic signal is $P=\frac{1}{T_0}\int_{T_0}|x(t)|^{2}\,dt$, and the exponential Fourier series is $x(t)=\sum_{n=-\infty}^{\infty} c_n e^{j n \omega_0 t}$ with $P=\sum_n |c_n|^{2}$ (Parseval).

Question 6: VCO-Based FM Modulator, Demodulator and Carson 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. A voltage-controlled oscillator whose characteristic is plotted as control voltage $V$ against output frequency $f$, a straight line of slope $10^{-4}\ \text{V/Hz}$ passing through $V = 0$ at $f = f_c$. Required modulator output power is 10 W. For part (c) the message has bandwidth $W = 10\ \text{kHz}$ and peak value $|m|_{\max} = 2\ \text{V}$.

Find. The FM output expression at the prescribed power, a demodulator block diagram, and an approximate transmission bandwidth.

[Figure not reproduced: Figure Q6.1 — the VCO characteristic redrawn from the exam figure: control voltage against output frequency, zero volts at f₊. See the official exam paper.]

Approach. Invert the plotted slope to obtain the frequency sensitivity in hertz per volt, write the standard FM expression with that sensitivity, fix the amplitude from the power requirement, and apply Carson's rule for the bandwidth.

Read the frequency sensitivity. The figure plots volts on the vertical axis against hertz on the horizontal axis, so its slope has units of V/Hz and the modulator sensitivity is its reciprocal: $$k_f = \frac{1}{10^{-4}\ \text{V/Hz}} = \boxed{k_f = 10^{4}\ \text{Hz/V}}$$ That is, one volt of control input shifts the instantaneous frequency by $10\ \text{kHz}$.
Instantaneous frequency and phase. Driving the VCO with $m(t)$ gives $$f_i(t) = f_c + k_f\,m(t),\qquad \theta(t) = 2\pi\!\int_{0}^{t} f_i(\lambda)\,d\lambda = 2\pi f_c t + 2\pi k_f\!\int_{0}^{t} m(\lambda)\,d\lambda .$$
Fix the amplitude from the power (part a). An FM signal has constant envelope $A_c$, so its average power is $A_c^{2}/2$ regardless of the message. Setting this to 10, $$A_c = \sqrt{2 \times 10} = \sqrt{20} = 4.4721\ \text{V}$$ $$\boxed{\,s_{\text{FM}}(t) = 4.4721\cos\!\left[2\pi f_c t + 2\pi\times 10^{4}\!\int_{0}^{t} m(\lambda)\,d\lambda\right]}$$ with the phase constant $2\pi k_f = 6.2832\times 10^{4}\ \text{rad/(V}\cdot\text{s)}$.
Demodulator (part b). The classical answer is the limiter–discriminator chain drawn below: a band-pass filter and hard limiter first strip any amplitude disturbance (legitimate because the wanted information is entirely in the zero crossings); a differentiator then converts frequency deviation into amplitude, producing $$\frac{d}{dt}s_{\text{FM}}(t) = -A_c\left[2\pi f_c + 2\pi k_f m(t)\right]\sin\!\left[\theta(t)\right],$$ whose envelope is linear in $m(t)$; an envelope detector recovers that envelope; and a DC block plus a gain of $1/(2\pi k_f)$ removes the $2\pi f_c$ pedestal and restores the scale. A phase-locked loop whose loop-filter output is taken as the demodulated signal is an equally acceptable answer and performs better at low carrier-to-noise ratio.

Figure Q6.2 — limiter-discriminator FM demodulator (part b).

Peak frequency deviation. With $|m|_{\max} = 2\ \text{V}$, $$\Delta f = k_f\,|m|_{\max} = 10^{4} \times 2 = 20\ \text{kHz},\qquad \beta = \frac{\Delta f}{W} = \frac{20}{10} = 2 .$$ A deviation ratio of 2 places this firmly in the wide-band FM regime, so the narrow-band approximation $B \approx 2W$ would be badly wrong here.
Carson's rule (part c). $$B_T \approx 2\left(\Delta f + W\right) = 2\left(20 + 10\right)\ \text{kHz}$$ $$\boxed{B_T \approx 60\ \text{kHz}}$$ Equivalently $B_T = 2W(\beta + 1) = 2(10)(3)\ \text{kHz}$. Carson's rule captures roughly 98% of the transmitted power; the true FM spectrum has infinitely many sidebands, which is why the question asks only for an approximate value.

Quantity	Result
Frequency sensitivity	$k_f = 10^{4}\ \text{Hz/V}$
Carrier amplitude for 10 W	$A_c = \sqrt{20} = 4.4721\ \text{V}$
(a) FM output	$s_{\text{FM}}(t) = 4.4721\cos\!\left[2\pi f_c t + 2\pi\!\cdot\!10^{4}\int_0^t m\,d\lambda\right]$
(b) Demodulator	Limiter → differentiator → envelope detector → DC block (or a PLL)
Peak deviation / deviation ratio	$\Delta f = 20\ \text{kHz}$; $\beta = 2$
(c) Carson bandwidth	$B_T \approx 60\ \text{kHz}$

Topic. Frequency modulation with a voltage-controlled oscillator: frequency sensitivity, constant-envelope power, discriminator demodulation and Carson's bandwidth rule.

Key relations. $f_i = f_c + k_f m(t)$; $s_{\text{FM}} = A_c\cos\left[2\pi f_c t + 2\pi k_f\int m\right]$; $P = A_c^{2}/2$ independent of $m(t)$; $\Delta f = k_f|m|_{\max}$; $\beta = \Delta f / W$; and $B_T \approx 2(\Delta f + W) = 2W(\beta+1)$.

Why this works. FM places the information in the argument of the cosine, so the envelope carries none of it. That single fact does most of the work here: it makes the power question trivial (amplitude alone sets $P$), it justifies the limiter in the demodulator, and it is why FM buys noise immunity by spending bandwidth. Carson's rule is the practical summary of that trade — the significant Bessel sidebands extend to roughly $\beta + 1$ on either side of the carrier.

Common pitfall. The dominant error on this question is using the plotted slope directly as $k_f$. The figure's vertical axis is volts and its horizontal axis is hertz, so $10^{-4}\ \text{V/Hz}$ must be inverted to give $10^{4}\ \text{Hz/V}$; taking $k_f = 10^{-4}$ would yield a deviation of $2\times 10^{-4}\ \text{Hz}$ and a nonsensical bandwidth of $20\ \text{kHz}$ driven entirely by $W$. A second trap is trying to compute FM power from the message statistics as in the AM questions — the constant envelope makes that unnecessary and wrong. A third is applying $B \approx 2W$ (the narrow-band result) when $\beta = 2$.

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., §5.1–5.4 (angle modulation, generation with a VCO, Carson's rule, discriminator demodulation); Haykin & Moher, Communication Systems, 5th ed., §4.3–4.7.

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