Question 3 of 7: PCM design: sampling rate, word length, bit rate and multiplexed bandwidth (20 marks)

Question 3 — PCM design: sampling rate, word length, bit rate and multiplexed bandwidth (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) the sampling rate $f_s$; (b) the smallest word length $b$; (c) the PCM bit rate $R_b$; (d) the minimum baseband bandwidth for ten multiplexed streams.

Approach. Treat the reconstruction filter as the real constraint: its passband must reach $W$, its transition region is fixed at $0.1W$, and the first spectral image must begin no lower than the filter’s stopband edge. Then size the quantiser from the uniform-quantisation noise power, multiply to get the bit rate, and close with the Nyquist bandwidth for binary signalling.

1. Fix the reconstruction filter’s stopband edge. The filter must pass the whole signal, so its passband width is $W = 8\ \text{kHz}$; the transition region is 10 % of that: $$\Delta f = 0.10 \times 8\ \text{kHz} = 0.8\ \text{kHz} \qquad\Longrightarrow\qquad f_{\text{stop}} = W + \Delta f = 8.8\ \text{kHz}.$$ Everything at or above 8.8 kHz is fully rejected; everything below 8 kHz passes with constant gain, as the question states.
2. Place the first image above the stopband edge. Sampling replicates the baseband spectrum about every multiple of $f_s$; the lowest-frequency component of the first image sits at $f_s - W$. Zero distortion (other than quantisation) requires that image to land in the filter’s stopband: $$f_s - W \ge f_{\text{stop}} = W + \Delta f \qquad\Longrightarrow\qquad f_s \ge 2W + \Delta f = 16 + 0.8\ \text{kHz}$$ $$\boxed{\;f_s = 16.8\ \text{kHz}\;}$$ This exceeds the bare Nyquist rate of 16 kHz by exactly the 800 Hz of guard band that a real (non-brick-wall) filter demands.
3. Size the quantiser from the noise specification. For uniform quantisation with step $\Delta$ the noise is uniformly distributed on $[-\Delta/2, \Delta/2]$, giving mean-square value $\Delta^{2}/12$. Requiring $\Delta/\sqrt{12} \lt 1\ \text{mV}$, $$\Delta \lt \sqrt{12}\times 10^{-3} = 3.464\ \text{mV}.$$ With $L = 2^{b}$ levels spanning the 2 V range, $\Delta = V_{pp}/2^{b}$, so $$2^{b} \gt \frac{2}{3.464\times10^{-3}} = 577.4 \qquad\Longrightarrow\qquad b \ge \log_{2}(577.4) = 9.17 .$$ Word lengths are integers, hence $$\boxed{\;b = 10\ \text{bits per sample}\ (L = 1024\ \text{levels})\;}$$ Verifying: $\Delta = 2/1024 = 1.953\ \text{mV}$ and $\sigma_q = \Delta/\sqrt{12} = 0.564\ \text{mV}$, comfortably under the limit; 9 bits would give $1.13\ \text{mV}$ and fail. The resulting signal-to-quantisation-noise ratio is $1.76 + 6.02b = 61.9\ \text{dB}$.
4. Compute the PCM bit rate. Every sample carries $b$ bits, so $$R_b = f_s\, b = 16.8\ \text{kHz} \times 10\ \text{bits} \qquad\Longrightarrow\qquad \boxed{\;R_b = 168\ \text{kbit/s}\;}$$
5. Multiplex ten streams and apply the Nyquist bandwidth limit. Time-division multiplexing simply adds the bit rates (framing overhead is neglected, as the question intends): $$R_{\text{total}} = 10 \times 168\ \text{kbit/s} = 1.68\ \text{Mbit/s}.$$ For binary signalling with optimum (ideal Nyquist, zero roll-off) pulse shaping, the minimum baseband bandwidth is one half the symbol rate, and for a binary scheme the symbol rate equals the bit rate: $$B_{\min} = \frac{R_{\text{total}}}{2} \qquad\Longrightarrow\qquad \boxed{\;B_{\min} = 840\ \text{kHz}\;}$$ A practical raised-cosine filter with roll-off $\alpha$ would need $B = (1+\alpha)R/2$, i.e. about 1.05 MHz at $\alpha = 0.25$.