Question 9 of 10: 4-ary PCM Design for an Audio Signal
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B4 Communications, National Examination
May 2014 — a three-hour open-book examination (any non-communicating calculator
permitted). The cover page states any five of the ten questions constitute a
complete paper, with only the first five as they appear in the answer book marked; every
question is nonetheless answered in full below so the paper remains a complete study resource.
All ten questions carry equal value (20 marks each).
Reference texts. A. V. Oppenheim and A. S. Willsky, Signals and
Systems, 2nd ed. (Fourier transform properties, the sampling theorem, LTI eigenfunctions,
z-transforms); S. Haykin and M. Moher, Communication Systems, 5th ed. (AM/FM
modulation, PCM, matched filtering and eye diagrams); B. P. Lathi and Z. Ding, Modern
Digital and Analog Communication Systems, 4th ed. (FM Bessel spectra, M-ary baseband
transmission); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed.
(z-transform regions of convergence, partial-fraction inversion).
Question 9: 4-ary PCM Design for an Audio Signal (20 marks)
Given. $W=3.2$ kHz; $f_s=1.25\times2W$; max quantization error
$\le0.5\%$ of peak amplitude $V_p$; encoding uses 4-ary pulses (2 bits/pulse).
Find. (a) 4-ary pulses per sample $M$; (b) minimum zero-ISI bandwidth;
(c) bandwidth with 25% raised-cosine roll-off.
Approach. Convert the error spec into a MINIMUM number of quantizer levels
$L$, then find the smallest integer number of 4-ary pulses $M$ with $4^M\ge L$; bandwidth then
follows from the SYMBOL (4-ary pulse) rate $R_s=f_sM$, not the bit rate.
Part (a) — pulses per sample. $f_s=1.25\times2(3.2\ \text{kHz})=1.25
\times6.4=\boxed{8\ \text{kHz}}$. For a uniform quantizer with step $\Delta=2V_p/L$, the
maximum quantization error is $\Delta/2=V_p/L$. Requiring $V_p/L\le0.005V_p$ gives
$$L\ge\frac1{0.005}=200\ \text{levels}.$$
The smallest number of 4-ary pulses $M$ with $4^M\ge200$: $4^3=64<200\le256=4^4$, so
$$M=\boxed{4\ \text{4-ary pulses per sample}}.$$
Part (b) — minimum zero-ISI bandwidth. The 4-ary symbol (pulse)
rate is
$$R_s=f_s\,M=8000\times4=\boxed{32{,}000\ \text{symbols/s (baud)}}.$$
The ideal Nyquist criterion for zero ISI needs bandwidth equal to HALF the symbol rate,
regardless of the pulse alphabet size:
$$B_{min}=\frac{R_s}{2}=\boxed{16\ \text{kHz}}.$$
Part (c) — 25% roll-off bandwidth. A raised-cosine Nyquist pulse
with roll-off factor $\alpha$ needs bandwidth $(R_s/2)(1+\alpha)$:
$$B=\frac{R_s}2(1+0.25)=16{,}000\times1.25=\boxed{20\ \text{kHz}}.$$
