22-Elec-A3 Signals and Communications · December 2019

Question 3 of 5: PCM System Design — Quantisation Levels, Noise Power and Bit Rate

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

Notes on this paper

Paper format. National Examinations, December 2019 — 16-Elec-A3 Signals and Communications. Three hours, closed book (an approved Casio or Sharp calculator is permitted). Answer all 5 questions; all 5 questions are of equal value (20 marks each, 100 marks total). Candidates are urged to submit a clear statement of any assumption made where a question admits more than one reading.

Reference texts.

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 2–3 (Fourier series and transforms), Ch. 4 (amplitude modulation, DSB and SSB), Ch. 5 (angle modulation, Carson's rule), Ch. 6 (sampling and PCM).
B. P. Lathi, Linear Systems and Signals, 2nd ed. — Ch. 6–7 (Fourier analysis of periodic and aperiodic signals), Ch. 5 and 9 (discrete-time systems and the z-transform).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 3 (Fourier series), Ch. 4 (Fourier transform), Ch. 8 (modulation), Ch. 10 (z-transform).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 2 (LTI systems and convolution), Ch. 5 (frequency-domain analysis), Ch. 9 (FIR filter structures).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 4 (angle modulation), Ch. 5 (pulse-code modulation and quantisation noise).

Note on scope. This sitting carries no printed figures: every system in Questions 4 and 5 is specified analytically, so all block diagrams and plots below are constructed from the given equations. Where the examiner asks for a "plot", a labelled sketch with the correct line positions, weights and signs is what earns the marks — each figure here is drawn to that standard.

Question 3: PCM System Design — Quantisation Levels, Noise Power and Bit Rate (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
Message bandwidth	$W$	8 kHz
Signal range	—	−3 V to +3 V
Peak-to-peak span	$V_{pp}$	6 V
Peak value	$V_p$	3 V
Maximum quantisation error	$e_{\max}$	0.5 % of $V_p$
Level constraint	$L$	a power of 2

Find. The smallest admissible $L$; the quantisation noise power; the reconstructed SNR under a sinusoidal signal model; the SNR change when $L$ is doubled; the word length; and the resulting bit rate.

Approach. Convert the 0.5 % accuracy specification into a bound on the step size, round up to the next power of two, then apply the standard uniform-quantiser noise model $N_q = \Delta^2/12$ and the Nyquist sampling rate.

Translate the accuracy specification into a step-size bound. A uniform quantiser with step $\Delta$ produces an error uniformly distributed on $[-\Delta/2,\ +\Delta/2]$, so the worst-case error is $\Delta/2$. Requiring this to stay within 0.5 % of the 3 V peak, $$\frac{\Delta}{2} \le 0.005\,V_p = 0.005(3) = 0.015\ \text{V} \quad\Longrightarrow\quad \Delta \le 0.030\ \text{V}$$
Convert the step-size bound into a level count. The $L$ levels must span the full 6 V range, so $\Delta = V_{pp}/L$ and $$L \ge \frac{V_{pp}}{\Delta} = \frac{6}{0.030} = 200$$ The smallest power of two that satisfies this is $2^8 = 256$ (since $2^7 = 128 \lt 200$): $$\boxed{L = 256\ \text{levels}}$$
Recompute the realised step size and verify the specification. $$\Delta = \frac{V_{pp}}{L} = \frac{6}{256} = 0.023438\ \text{V} \qquad \frac{\Delta}{2} = 0.011719\ \text{V} \le 0.015\ \text{V}\ \checkmark$$ The realised worst-case error is 0.39 % of peak, comfortably inside the 0.5 % budget.
Compute the quantisation noise power. For a uniformly distributed error the mean-square value is $$N_q = \frac{1}{\Delta}\int_{-\Delta/2}^{\Delta/2} e^2\,de = \frac{\Delta^2}{12} = \frac{(0.023438)^2}{12}$$ $$\boxed{N_q = 4.578\times10^{-5}\ \text{W}\ \ (\text{i.e. V}^2)}$$
Compute the signal power under the sinusoidal model. A full-scale sinusoid has amplitude equal to the peak, $A = V_p = 3$ V, so $$S = \frac{A^2}{2} = \frac{3^2}{2} = 4.5\ \text{W}$$
Form the signal-to-quantisation-noise ratio. $$\text{SNR} = \frac{S}{N_q} = \frac{4.5}{4.578\times10^{-5}} = 98\,304$$ $$\boxed{\text{SNR} = 10\log_{10}(98\,304) = 49.93\ \text{dB}}$$ This agrees exactly with the textbook rule for a full-scale sinusoid, $\text{SNR}_{\text{dB}} = 6.02n + 1.76 = 6.02(8) + 1.76 = 49.92$ dB, which is a useful independent check on both $L$ and $\Delta$.
Determine the effect of doubling the number of levels. Doubling $L$ halves $\Delta$ and therefore quarters $N_q$, while $S$ is unchanged: $$\Delta \text{SNR} = 10\log_{10}(4) = 6.02\ \text{dB}$$ $$\boxed{\text{SNR increases by } 6.02\ \text{dB (from 49.93 dB to 55.95 dB)}}$$ Equivalently: every additional bit buys about 6 dB, the single most quoted result in PCM design.
Determine the word length. $$n = \log_2 L = \log_2 256$$ $$\boxed{n = 8\ \text{bits per sample}}$$
Determine the sampling rate and bit rate. Nyquist requires $f_s \ge 2W = 2(8\ \text{kHz}) = 16$ kHz; taking the minimum, $$R_b = n f_s = 8 \times 16\,000$$ $$\boxed{R_b = 128\ \text{kbit/s}}$$ In practice a modest guard band would push $f_s$ slightly above 16 kHz to allow a realisable anti-aliasing filter; the ideal-filter figure of 128 kbit/s is the answer expected here.

Quantity	Result
(a) Smallest number of levels	$L = 256$
Realised step size	$\Delta = 23.44$ mV
(b) Quantisation noise power	$N_q = 4.578\times10^{-5}$ V$^2$
(c) Signal power (sinusoid)	$S = 4.50$ W
(c) SNR	$98\,304 = 49.93$ dB
(d) SNR change on doubling $L$	$+6.02$ dB (to 55.95 dB)
(e) Bits per sample	$n = 8$
(f) Sampling rate / bit rate	$f_s = 16$ kHz, $R_b = 128$ kbit/s

Topic. Uniform quantisation, quantisation-noise power, PCM signal-to-noise ratio and bit-rate budgeting.

Key relations. $\Delta = V_{pp}/L$; worst-case error $\Delta/2$; $N_q = \Delta^2/12$; full-scale sinusoid $S = V_p^2/2$; $\text{SNR}_{\text{dB}} = 6.02n + 1.76$ for a sinusoid; $f_s \ge 2W$ (Nyquist); $R_b = n f_s$.

Why this works. The whole question turns on one translation: an accuracy specification stated as a percentage of peak is really a statement about the step size, because the largest error a uniform quantiser can make is half a step. Once $\Delta$ is bounded, the level count follows from the requirement that $L$ steps cover the full peak-to-peak span, and the power-of-two constraint forces a round-up rather than a round-down. The $N_q = \Delta^2/12$ result is the variance of a uniform distribution over one step and assumes only that the signal traverses many levels between samples, which speech comfortably does. The $6.02n + 1.76$ formula is not an independent fact but the same calculation written in decibels for a full-scale sinusoid: the 1.76 dB is precisely the $10\log_{10}(3/2)$ crest-factor bonus a sinusoid enjoys over a uniformly distributed source, so it must not be applied when the question specifies a different loading.

Common pitfall. The classic error is dividing the peak rather than the peak-to-peak span by $L$, which halves $\Delta$ and delivers $L = 128$ — one bit short and a failing design. The second is rounding $L$ down to the nearest power of two because 200 is "closer" to 128 than to 256; the constraint is an inequality and only rounding up satisfies it. A third is forgetting that doubling the level count improves SNR by 6.02 dB, not 3 dB, because the noise power falls by a factor of four when the step size halves.

Source. Haykin, Communication Systems, 5th ed., Ch. 5 (§5.3–5.5, quantisation noise and PCM SNR); Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., Ch. 6.

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