22-Elec-A3 Signals and Communications · May 2015

Question 4 of 7: PCM of a Speech Signal — Sampling, Quantization and Bit Rate

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

Question 4: PCM of a Speech Signal — Sampling, Quantization 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. Message bandwidth \(W = 6\) kHz; required reconstructed signal-to-quantization-noise ratio \(\mathrm{SNR}_q = 40\) dB; uniform quantizer; binary encoding.

Find. Minimum sampling rate, smallest admissible number of quantization levels, resulting bit rate, and the rationale for companding.

Approach. Apply the Nyquist criterion for the rate, invert the uniform-quantizer SNR law for the word length, round up to a whole number of bits, and multiply.

1. Minimum sampling rate. The signal is band-limited to \(W = 6\) kHz, so by the sampling theorem $$f_s \ge 2W = \boxed{\,12\ \text{kHz}\,}$$ In a practical codec a guard band is left for the anti-aliasing filter (which is why telephony samples 3.4 kHz speech at 8 kHz rather than 6.8 kHz), but the theoretical minimum asked for is 12 kHz.
2. Word length from the SNR requirement. For a uniform quantizer with \(n\) bits driven at full scale, $$\mathrm{SNR}_q\ [\text{dB}] = 1.76 + 6.02\,n$$ Setting this to 40 dB gives $$n \ge \frac{40 - 1.76}{6.02} = 6.35$$ Since \(n\) must be an integer, \(n = 7\) bits, and the smallest number of levels is $$\boxed{\,L = 2^n = 128\ \text{levels}\,}$$ which actually delivers \(1.76 + 6.02(7) = 43.9\) dB — comfortably above the requirement, whereas 6 bits would give only 37.9 dB and fail. The alternative textbook form \(\mathrm{SNR}_q = L^2\) for a signal whose amplitude is uniformly distributed gives \(L \ge 10^{40/20} = 100\), and rounding up to the next power of two returns the same 128 levels.
3. Bit rate. Each sample is encoded into \(n\) bits, so $$R_b = n f_s = 7 \times 12{,}000 = \boxed{\,84\ \text{kb/s}\,}$$ The corresponding minimum transmission bandwidth for baseband binary signalling is \(R_b/2 = 42\) kHz — seven times the original message bandwidth, the characteristic bandwidth expansion of PCM bought in exchange for noise immunity and regeneration.
   Figure Q4.1 — the PCM encoder chain for this specification: 12 kHz sampling, 128-level uniform quantizer, 7-bit words, 84 kb/s.
4. Why non-uniform quantizers (part d). A uniform quantizer has a fixed step \(\Delta\), so its noise power \(\Delta^2/12\) is the same whether the sample is loud or quiet. The signal-to-noise ratio therefore falls decibel for decibel as the signal level falls, and speech spends most of its time near the low end of its range while the quantizer must be scaled for the rare loud peaks. Quiet passages — which is where intelligibility lives — would be reproduced with an unacceptably low SNR, and a talker further from the microphone would sound noisier than one close to it.

A non-uniform quantizer instead uses fine steps near zero and coarse steps near full scale, which is implemented in practice by companding: the sample is passed through a logarithmic compressor (the A-law used in Canada's international interfaces and Europe, or the μ-law used in North American telephony), quantized uniformly, and expanded again at the receiver. Because the step size then grows in proportion to the amplitude, the ratio of signal to quantization noise becomes almost constant over a 40 dB dynamic range of input levels. Equivalently, companding buys about 24 dB of extra SNR for weak signals at no increase in bit rate, or lets an 8-bit companded codec match the perceptual quality of a 12-bit uniform one — which is exactly why the 64 kb/s telephone channel is companded rather than uniform.