Question 9 of 10: M-ary PCM Bandwidth for an Audio Channel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2014 — 07-Elec-A3 Signals and Communications. Three hours, open book, any non-communicating calculator permitted. Ten questions of equal value (20 marks each); the rubric states that five questions constitute a complete paper and only the first five presented are marked. All ten are solved here, since the set is intended as a study resource. A table of Fourier-transform pairs and properties, a table of z-transform pairs, trigonometric identities and Bessel-function graphs/tables are supplied with the paper and are used freely below.
Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (sampling, PCM, AM/FM, matched filtering); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier analysis, LTI properties); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (z-transform, regions of convergence); S. Haykin, Communication Systems, 5th ed. (digital transmission, eye diagrams).
Convention used throughout. This paper's own transform table defines the sinc function as $\operatorname{sinc}(x)=\sin x / x$ (Lathi's convention, not the normalised $\sin(\pi x)/(\pi x)$). Every bandwidth below follows from that table, in particular $2B\operatorname{sinc}(2\pi Bt)\leftrightarrow\operatorname{rect}(f/2B)$ and $B\operatorname{sinc}^2(\pi Bt)\leftrightarrow\Delta(f/2B)$.
Question 9: M-ary PCM Bandwidth for an Audio Channel (20 marks)
Given. Audio bandwidth $B = 3.2$ kHz; sampling 25% above the Nyquist rate; uniform quantizer with maximum error not exceeding 0.5% of the peak amplitude $m_p$; transmission by 4-ary pulses; roll-off factor 0.25 in part (c).
Find. The number of 4-ary pulses per sample, the minimum ISI-free transmission bandwidth, and the bandwidth with 25% roll-off.
Approach. Convert the error specification into a minimum number of quantization levels, express that count in base 4 to get the pulses per sample, form the symbol (baud) rate, and apply the Nyquist bandwidth relation with and without roll-off.
Sampling rate. The Nyquist rate is $2B = 6.4$ kHz, and sampling 25% above it gives $$f_s = 1.25\times 6400 = 8000\ \text{samples/s}$$ the familiar 8 kHz telephony rate.
Levels demanded by the error specification. A uniform quantizer spanning $\pm m_p$ with $L$ levels has step size $\Delta v = 2m_p/L$ and worst-case error $\Delta v/2 = m_p/L$. Requiring this to be at most 0.5% of $m_p$: $$\frac{m_p}{L}\le 0.005\,m_p \;\Longrightarrow\; L \ge \frac{1}{0.005} = 200\ \text{levels}$$ Note that $m_p$ cancels, so the answer does not depend on the signal amplitude — only on the relative precision demanded.
Part (a): pulses per sample in base 4. Each 4-ary pulse carries one of 4 values, so $k$ pulses distinguish $4^k$ levels. Requiring $4^k \ge 200$: $$4^3 = 64 \lt 200,\qquad 4^4 = 256 \ge 200$$ so $$k=\left\lceil\log_4 200\right\rceil=\boxed{4\ \text{4-ary pulses per sample}}$$ This realises 256 levels, comfortably exceeding the 200 required (equivalently 8 bits, since each 4-ary pulse is worth $\log_2 4 = 2$ bits).
Symbol rate on the line. Four pulses per sample at 8000 samples per second gives $$R_s = k f_s = 4\times 8000 = 32{,}000\ \text{pulses/s (baud)}$$
Part (b): minimum ISI-free bandwidth. The Nyquist criterion permits $2B_T$ independent pulses per second through a bandwidth $B_T$, regardless of how many levels each pulse carries, so $$B_T=\frac{R_s}{2}=\frac{32{,}000}{2}=\boxed{16\ \text{kHz}}$$ This floor is attained only with ideal $\operatorname{sinc}$ pulses, which are unrealisable (infinite duration, and their slow $1/t$ decay makes timing errors catastrophic).
Part (c): bandwidth with 25% roll-off. A raised-cosine pulse with roll-off factor $r$ widens the required bandwidth by the factor $(1+r)$: $$B_T=(1+r)\frac{R_s}{2}=1.25\times 16\ \text{kHz}=\boxed{20\ \text{kHz}}$$ The extra 4 kHz buys a pulse that decays as $1/t^3$ instead of $1/t$, tolerates timing jitter, and can be realised with a filter of finite complexity — a bargain in practice.
It is worth noting what the 4-ary signalling bought. Encoding 256 levels in binary would need 8 bits per sample and a bit rate of 64 kbit/s, requiring 32 kHz of bandwidth at the Nyquist floor. Packing two bits into each 4-ary pulse halves the symbol rate and therefore halves the bandwidth to 16 kHz. That saving is not free: with four amplitude levels squeezed into the same voltage range, the spacing between adjacent decision thresholds shrinks, so a higher signal-to-noise ratio is needed for the same error rate. Bandwidth and noise immunity are the two currencies of digital transmission, and M-ary signalling is the exchange rate between them.
