Question 4 of 6: PCM System Design — Sampling Rate, Bit Rate, SNR

Question 4: PCM System Design — Sampling Rate, Bit Rate, SNR (1/6 of paper)

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. Video bandwidth $W=4.5\text{ MHz}$; sampling rate $f_s=1.15\times$ Nyquist rate; uniform quantizer with maximum error $=0.05\%$ of the signal peak $V_p$; binary encoding; video signal modelled with a uniform probability density over $[-V_p,V_p]$ (part c).

Find. (a) $f_s$; (b) channel bit rate; (c) reconstructed SNR (as a ratio and in dB).

Approach. Scale the Nyquist rate by 1.15 for $f_s$; convert the maximum-quantization-error spec into a level count $L$ (max error $=$ half the step size), round up to the number of bits binary encoding needs, and multiply by $f_s$ for the bit rate; derive the SNR for a UNIFORM-pdf source from first principles (signal power $V_p^2/3$, quantization-noise power $\Delta^2/12$).

1. Part (a) — sampling rate. The minimum (Nyquist) rate is $2W$; the system oversamples by 15%: $$f_s=1.15\times2W=1.15\times2\times4.5\text{ MHz}=\boxed{10.35\text{ MHz}}$$
2. Part (b) — bit rate. For a uniform quantizer over $[-V_p,V_p]$ with $L$ levels, the step size is $\Delta=2V_p/L$ and the maximum quantization error is half a step, $\Delta/2=V_p/L$. Setting this equal to $0.05\%$ of $V_p$: $$\frac{V_p}{L}=0.0005\,V_p\ \Rightarrow\ L=\frac{1}{0.0005}=2000\text{ levels}$$ Binary encoding needs $n=\lceil\log_2L\rceil$ bits per sample; since $2^{10}=1024<2000\le2048=2^{11}$, $$n=\boxed{11\text{ bits/sample}}$$ so the channel bit rate is $$R_b=f_s\times n=10.35\text{ MHz}\times11=\boxed{113.85\text{ Mbit/s}}$$
3. Part (c) — SNR for a uniform-pdf signal. For a signal uniformly distributed over $[-V_p,V_p]$ (zero mean), the signal power is the variance of a uniform distribution of half-width $V_p$: $$P_{signal}=\frac{V_p^2}{3}$$ The quantization error is itself uniformly distributed over $[-\Delta/2,\Delta/2]$, giving noise power $\Delta^2/12$ with $\Delta=2V_p/L$: $$P_{noise}=\frac{\Delta^2}{12}=\frac{(2V_p/L)^2}{12}=\frac{V_p^2}{3L^2}$$ so the $V_p^2/3$ cancels exactly and $$SNR=\frac{P_{signal}}{P_{noise}}=\boxed{L^2=2000^2=4\times10^6}$$ $$SNR_{dB}=10\log_{10}\!\big(4\times10^6\big)=20\log_{10}(2000)=\boxed{66.02\text{ dB}}$$ (a cleaner result than the familiar sine-wave PCM formula $SNR\approx3L^2/2$, because the assumed uniform-pdf signal statistics exactly cancel the $V_p^2$ dependence).