Question 2 of 5: Uniform PCM of a Speech Signal — Sampling, Step Size, Levels, SNR and Bit Rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Engineers Canada national examination
16-Elec-A3 — Signals and Communications, May 2018. Closed book (one approved Casio or
Sharp calculator permitted), 3 hours, 4 pages. Five questions, all compulsory, all of equal value
(20 marks each). All five are solved in full below, every sub-part answered.
Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication
Systems, 5th ed. (AM modulation and detection, PCM quantisation, frequency conversion);
B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier series of pulse trains, signal power);
A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed. (discrete-time LTI systems,
convolution, DTFT); J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed.
(difference equations, frequency response); S. Haykin, Communication Systems, 5th ed.
(envelope and coherent detection, mixers and image products).
Check: a note on the Question 3 figure. The plotted message and the printed equation agree: the plotted waveform has period 1 in normalised time with
peaks of $+1.5$ at $t = 0, \pm 1, \pm 2$ and troughs of $-1.5$ midway between them, which is exactly
$m(t) = \cos(2\pi f_m t) + \tfrac{1}{2}\cos(6\pi f_m t)$ evaluated with $f_m t$ as the abscissa. The
solution below therefore uses the printed equation, and the peak value $|m|_{\max} = 1.5$ read from it.
The paper's "$f_m = 4$ Kz" is read as 4 kHz.
Question 2: Uniform PCM of a Speech Signal — Sampling, Step Size, Levels, SNR and Bit Rate (20 marks)
Find. The Nyquist sampling rate; the largest step size that meets the error budget; the
smallest number of levels; the signal-to-quantisation-noise ratio under the sinusoid / triangular-error model;
and the resulting bit rate.
Figure Q2.1 — the PCM transmitter chain for the speech signal, with the values established in parts (a) to (e).
Approach. Apply the sampling theorem for (a); translate the error specification into a
step size through the rounding relation $|e|_{\max} = \Delta/2$ for (b); divide the span by the step for (c);
form the ratio of the sinusoid's mean square to the triangular error's mean square for (d); and finally
multiply the word length by the sampling rate for (e).
Minimum sampling rate. The signal is low-pass with highest frequency $W = 8$ kHz, so the
Nyquist criterion requires
$$f_s \ge 2W = 2(8\ \text{kHz}) \quad\Rightarrow\quad \boxed{\,f_s^{\min} = 16\ \text{kHz}\,}$$
In a real telephone-grade codec one would leave a guard band and sample somewhat faster, but "minimum" means
exactly the Nyquist rate here. Note this is a wideband speech channel — ordinary toll telephony bands
speech to 3.4 kHz and samples at 8 kHz.
Quantisation step size. A uniform quantiser that rounds to the nearest level has an
error confined to half a step: $|e| \le \Delta/2$. The specification $|e| < 2$ mV therefore fixes
$$\Delta \le 2\,|e|_{\max} = 2(2\ \text{mV}) \quad\Rightarrow\quad \boxed{\,\Delta = 4\ \text{mV}\,}$$
This is the largest admissible step, and taking the largest admissible step is what makes the number
of levels in part (c) the smallest possible — the two parts are one question asked twice.
Smallest number of quantisation levels. The levels must cover the whole $-2$ V to
$+2$ V range in steps of $\Delta$:
$$L = \frac{V_{pp}}{\Delta} = \frac{4\ \text{V}}{4\times10^{-3}\ \text{V}} \quad\Rightarrow\quad
\boxed{\,L = 1000\ \text{levels}\,}$$
Because $1000$ comes out as an exact integer, no rounding-up is needed; had it not, one would round
up, since rounding down would widen the step past the error budget.
Signal power under the sinusoid model. The largest sinusoid that fits the range has
peak $A_m = V_{pp}/2 = 2$ V, so
$$P_s = \frac{A_m^{2}}{2} = \frac{(2)^2}{2} = 2\ \text{W (into 1 }\Omega).$$
This "fully loaded" assumption is the standard one; real speech has a much lower crest-factor-corrected
power, which is why practical PCM uses companding.
Noise power under the triangular model. Modelling the error as a triangular wave of peak
$\Delta/2$, its mean square is one third of the square of its peak:
$$P_q = \frac{(\Delta/2)^{2}}{3} = \frac{\Delta^{2}}{12}
= \frac{(4\times10^{-3})^{2}}{12} = 1.333\times10^{-6}\ \text{W}.$$
This is the same $\Delta^2/12$ that a uniform probability density over one step would give — the
triangular-wave picture is just the deterministic route to it, and it is the reason the "$\Delta^2/12$"
formula is quoted without proof in most texts.
Form the signal-to-noise ratio. Dividing,
$$\mathrm{SNR} = \frac{P_s}{P_q} = \frac{2}{1.333\times10^{-6}} = 1.5\times10^{6},$$
$$\boxed{\,\mathrm{SNR} = 1.5\times10^{6} = 61.76\ \text{dB}\,}$$
A useful cross-check: substituting $A_m = L\Delta/2$ into $P_s/P_q$ gives $\mathrm{SNR} = 1.5L^2$, and
$1.5(1000)^2 = 1.5\times10^6$ exactly. In decibels this is the familiar
$6.02n + 1.76$ dB with $n = \log_2 1000 = 9.97$ bits.
Bit rate. Each sample must be coded into a binary word long enough to address all
$L = 1000$ levels:
$$n = \lceil \log_2 L\rceil = \lceil 9.966\rceil = 10\ \text{bits/sample},$$
and at the Nyquist rate of part (a),
$$R_b = n f_s = 10 \times 16\ \text{kHz} \quad\Rightarrow\quad \boxed{\,R_b = 160\ \text{kb/s}\,}$$
Ten bits actually address 1024 levels, so the realised step is slightly finer than required (3.91 mV) and
the realised SNR slightly better (62.0 dB) — the small bonus one always gets from rounding a level
count up to a power of two.
Figure Q2.2 — the triangular-wave model of the quantisation error used in part (d); its mean-square value is (step/2)²/3 = step²/12.
Check: rounding convention. The numbers above assume a rounding (mid-tread)
quantiser, for which the peak error is $\Delta/2$. If the quantiser truncated instead, the peak error would
be a full $\Delta$, forcing $\Delta = 2$ mV, $L = 2000$ and $n = 11$ bits ($R_b = 176$ kb/s). Rounding is the
standard assumption for uniform PCM and is what the "$\pm$" phrasing of the error budget implies.
