22-Elec-A3 Signals and Communications · December 2018

Question 2 of 5: Uniform PCM — Signal-to-Quantisation-Noise Ratio and Bit Rate

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

Notes on this paper

Paper format. National Examinations, December 2018 — 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 state any assumption made where a question admits more than one reading.

Reference texts.

B. P. Lathi and R. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 2–4 (Fourier series and transforms), Ch. 4 (amplitude modulation, DSB/SSB), 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 (discrete-time systems).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 3 (Fourier series), Ch. 7 (sampling), Ch. 10 (z-transform).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 3 (z-transform), Ch. 5 (frequency-domain analysis of LTI systems), Ch. 9 (filter structures).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 5 (pulse-code modulation and quantisation noise).

Note on the Question 4 figure. The block diagram shows both summing junctions drawn as plain adders — there are no minus signs at any input — so the printed arm gains are the multipliers exactly as shown: −3/4 on the first delay output into the input adder, +1/2 on the second delay output into the same adder, and +1/2 on the first delay output feeding forward to the output adder. The consequence, developed in Question 4, is that one pole falls outside the unit circle. That is a genuine property of the printed network, and the answer below treats it as such.

Question 2: Uniform PCM — Signal-to-Quantisation-Noise Ratio 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	$B$	8 kHz
Signal range	—	−2 V to +2 V
Peak-to-peak span	$V_{pp}$	4 V
Signal peak amplitude (sinusoid model)	$A$	2 V
Quantisation noise model	—	sinusoid of amplitude $A_n$
Required signal-to-noise ratio	$\text{SNR}$	60 dB

Find. (a) SNR as a function of $A_n$; (b) the $A_n$ that meets 60 dB; (c) the number of quantisation levels $L$; (d) the word length $n$ in bits per sample; and (e) the resulting bit rate $R_b$.

Approach. Both signal and noise are modelled as sinusoids, so both powers are simply (amplitude)$^2/2$ and the ratio reduces to a ratio of squared amplitudes. Convert the decibel specification to a linear ratio, solve for $A_n$, then link $A_n$ to the step size through the peak quantisation error, and finish with Nyquist sampling.

(a) Express both powers. Modelling the message as a full-scale sinusoid $m(t)=A\cos\omega t$ with $A=2$ V, its mean-square value is $$S = \frac{A^{2}}{2} = \frac{2^{2}}{2} = 2\ \text{W}.$$ The quantisation noise is likewise modelled as a sinusoid of amplitude $A_n$, so $$N = \frac{A_n^{2}}{2}.$$ The ratio is therefore $$\boxed{\ \text{SNR} = \frac{S}{N} = \frac{A^{2}/2}{A_n^{2}/2} = \frac{A^{2}}{A_n^{2}} = \frac{4}{A_n^{2}}\ }$$ The factors of one-half cancel, which is precisely why the question tells you to model both as sinusoids — it removes any dependence on the waveform's crest factor.
(b) Impose the 60 dB requirement. Converting from decibels, $$\text{SNR}_{\text{lin}} = 10^{60/10} = 10^{6}.$$ Substituting into the result of Step 1 and solving, $$\frac{4}{A_n^{2}} = 10^{6} \;\Longrightarrow\; A_n^{2} = 4\times10^{-6} \;\Longrightarrow\; \boxed{A_n = 2\times10^{-3}\ \text{V} = 2\ \text{mV}}$$
(c) Relate the noise amplitude to the step size. With rounding to the nearest level, the quantisation error of a uniform quantiser of step $\Delta$ is bounded by $\pm\Delta/2$. Modelling that error as a sinusoid means its peak value is $A_n$, so $$A_n = \frac{\Delta}{2} \;\Longrightarrow\; \Delta = 2A_n = 4\times10^{-3}\ \text{V} = 4\ \text{mV}.$$ The quantiser must cover the full 4 V span in steps of $\Delta$, so the number of levels is $$L = \frac{V_{pp}}{\Delta} = \frac{4}{4\times10^{-3}} \;\Longrightarrow\; \boxed{L = 1000\ \text{levels}}$$
(d) Convert levels to bits. An $n$-bit word addresses $2^{n}$ levels, so $n$ must satisfy $2^{n} \ge L$: $$n = \lceil \log_2 1000 \rceil = \lceil 9.966 \rceil = \boxed{10\ \text{bits per sample}}$$ Ten bits provide 1024 levels, comfortably covering the 1000 required (and delivering a slightly better SNR than the 60 dB specified). Rounding down to 9 bits would give only 512 levels and miss the specification by some 5.9 dB — the ceiling is mandatory.
(e) Sample and compute the bit rate. By the sampling theorem the minimum (Nyquist) sampling rate for a message of bandwidth $B=8$ kHz is $$f_s = 2B = 16\ \text{kHz},$$ and each sample carries $n$ bits, so $$R_b = n f_s = 10 \times 16\,000 \;\Longrightarrow\; \boxed{R_b = 160\ \text{kbit/s}}$$

Check: modelling assumptions stated explicitly. Two conventions were adopted where the question leaves a choice. (i) The quantiser is assumed to round to the nearest level, giving a peak error of $\Delta/2$; a truncating quantiser has a peak error of $\Delta$, which would halve the step and double $L$ to 2000, requiring 11 bits and a 176 kbit/s bit rate. (ii) Sampling is assumed to be at exactly the Nyquist rate; any practical guard band raises $f_s$ and hence $R_b$ proportionally. Both assumptions follow the question's own framing ("the required SNR", "the bit rate of the quantized signal") and are the standard textbook readings.

Quantity	Symbol	Result
(a) SNR in terms of $A_n$	$\text{SNR}$	$A^{2}/A_n^{2} = 4/A_n^{2}$
(b) Noise amplitude for 60 dB	$A_n$	2 mV
(c) Quantiser step size	$\Delta$	4 mV
(c) Number of levels	$L$	1000
(d) Word length	$n$	10 bits/sample (1024 levels)
(e) Sampling rate	$f_s$	16 kHz
(e) Bit rate	$R_b$	160 kbit/s

Topic. Uniform pulse-code modulation: quantisation noise, word length and bit rate.

Key relations. $\text{SNR}=S/N$; for sinusoids $S=A^{2}/2$ and $N=A_n^{2}/2$; peak rounding error $A_n=\Delta/2$; levels $L=V_{pp}/\Delta$; word length $n=\lceil\log_2 L\rceil$; Nyquist rate $f_s = 2B$; bit rate $R_b = n f_s$.

Why this works. The whole chain is a sequence of one-way conversions: a decibel specification fixes a linear power ratio, which fixes the tolerable error amplitude, which fixes the step size, which fixes the number of levels, which fixes the word length, which combined with the sampling rate fixes the bit rate. Each link is forced, so there is no design freedom once the SNR and bandwidth are stated — this is why PCM system sizing is so mechanical in practice. The familiar rule of thumb $\text{SNR}_{\text{dB}} \approx 6.02n + 1.76$ for a full-scale sinusoid is exactly this calculation done symbolically; here the question routes you through $A_n$ instead, which is equivalent but keeps the algebra visible.

Common pitfall. The most frequent error is dividing the peak amplitude rather than the full peak-to-peak span by $\Delta$ when counting levels, which halves $L$ and loses a bit. The second is rounding the word length down: $\log_2 1000=9.97$ looks close enough to 10 that candidates sometimes write 9, but any fractional part must be rounded up or the SNR specification is violated.

Source. Haykin, Communication Systems, 5th ed., §5.3–5.5 (quantisation noise and PCM); Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., §6.2.

← Question 1 Next: Question 3 →