22-Elec-A3 Signals and Communications · December 2017

Question 4 of 5: PCM System with Uniform Quantisation

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

Question 4: PCM System with Uniform Quantisation (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.

Find. (a) $f_s$; (b) the smallest number of levels $L$; (c) the step size $\Delta$; (d) the rms quantisation noise; (e) the bit rate; (f) the rationale for non-uniform quantisation.

Approach. Apply the Nyquist criterion for the sampling rate, invert the uniform-quantiser SNR law to find the smallest integer word length that meets 40 dB, then evaluate the step size, the noise power and the bit rate in turn.

1. (a) Minimum sampling rate. The Nyquist criterion requires sampling at no less than twice the highest frequency present: $$f_s \ge 2B = 2 \times 6\ \text{kHz} \;\Longrightarrow\; \boxed{f_{s,\min} = 12\ \text{kHz} \ (12\,000 \ \text{samples/s})}$$ In practice an anti-aliasing filter with a finite roll-off forces a modest guard band and the rate is set somewhat above this; 12 kHz is the theoretical floor.
2. (b) Word length and number of levels. For a uniform quantiser loaded to full scale, the signal power is $P_s$ and the quantisation noise power is $\Delta^{2}/12$ with $\Delta = 2V_m/L$ and $L = 2^{n}$. For a source that occupies the full range (the usual design assumption, and the case for a signal specified only by its extremes), $P_s = V_m^{2}/3$ and $$\frac{S}{N} = \frac{V_m^{2}/3}{\Delta^{2}/12} = \frac{V_m^{2}/3}{(2V_m/2^{n})^{2}/12} = 2^{2n} \;\Longrightarrow\; \left(\frac{S}{N}\right)_{\text{dB}} = 6.02\,n$$ Requiring 6.02 n ≥ 40 dB gives $n \ge 6.64$, so $n = 7$ bits. Checking both candidates: $n = 6$ yields 36.1 dB (fails) and $n = 7$ yields 42.1 dB (passes). Hence $$\boxed{n = 7\ \text{bits}, \qquad L = 2^{7} = 128\ \text{levels}}$$ The same word length results from the alternative sinusoidal-loading rule $(S/N)_{\text{dB}} = 6.02n + 1.76$, which gives $n \ge 6.35 \Rightarrow n = 7$, so the answer is robust to the loading assumption.
3. (c) Quantisation step size. The 6 V span is divided into $L$ equal intervals: $$\Delta = \frac{2V_m}{L} = \frac{6\ \text{V}}{128} = \boxed{0.046875\ \text{V} = 46.875\ \text{mV}}$$
4. (d) RMS quantisation noise. The quantisation error of a fine uniform quantiser is modelled as uniformly distributed over $(-\Delta/2, +\Delta/2)$, whose variance is $\Delta^{2}/12$: $$\sigma_q = \frac{\Delta}{\sqrt{12}} = \frac{0.046875}{3.4641} = \boxed{0.01353\ \text{V} = 13.53\ \text{mV}\ \text{rms}}$$ equivalently a noise power of $1.831 \times 10^{-4}$ W in 1 Ω. Substituting back, $10\log_{10}[(9/3)/1.831\times10^{-4}] = 42.1$ dB, confirming the requirement is met with about 2 dB of margin.
5. (e) Bit rate. Each sample is encoded into $n$ bits, so $$R_b = n\,f_s = 7 \times 12\,000 = \boxed{84\ \text{kbit/s}}$$ The minimum transmission bandwidth for this stream, at the Nyquist signalling limit, is $R_b/2 = 42$ kHz — seven times the 6 kHz of the original analogue signal, which is the bandwidth expansion that digital transmission buys its noise immunity with.

6. (f) Why non-uniform quantisers are used. A uniform quantiser produces a noise power $\Delta^{2}/12$ that is independent of the signal amplitude. The signal-to-noise ratio is therefore proportional to the signal power, and it collapses for quiet passages: a talker 30 dB below full scale suffers an SNR 30 dB below the design figure. Speech makes this acute, because its amplitude probability density is strongly peaked near zero — small amplitudes are far more probable than large ones, yet a uniform quantiser spends the same resolution on both. Its dynamic range is also poor across different speakers and line levels.

   A non-uniform quantiser fixes this by making the step size proportional to the signal amplitude: fine steps near the origin where samples are dense, coarse steps near full scale where they are rare. This is realised in practice by companding — passing the signal through a logarithmic compressor (μ-law with $\mu = 255$ in North America, A-law with $A = 87.6$ in Europe), quantising uniformly, and applying the inverse expander at the receiver. The result is an SNR that stays approximately constant over a 40 dB range of input levels, which is why 8-bit companded PCM at 64 kbit/s delivers toll-quality speech where roughly 12 uniform bits would otherwise be required.