22-Elec-A3 Signals and Communications · December 2015

Question 4 of 6: PCM Transmission of an Audio Signal

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

Notes on this paper

Paper format: 07-Elec-A3 Signals and Communications, December 2015 — 3 hours, closed book, a standard non-programmable calculator permitted. Six questions of equal value (20 marks each); any five constitute a complete paper, and only the first five presented are marked. All six are solved here, since the set is intended as a study resource.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (AM/FM, PCM, frequency conversion); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Fourier series, LTI systems); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (z-transform, difference equations, stability); S. Haykin, Communication Systems, 5th ed. (modulation and detection).

Question 4: PCM Transmission of an Audio Signal (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
Signal bandwidth	$W$	8 kHz
Dynamic range (peak to peak)	$V_{pp}$	2 V
Filter transition region	—	10% of passband = 0.8 kHz
Maximum quantization noise	$\sigma_{q}$	1 mV rms
Multiplexed channels	—	10

Find. (a) the sampling rate; (b) the minimum number of bits per sample; (c) the resulting bit rate; (d) the minimum channel bandwidth for a 10-channel binary multiplex with optimum filtering.

Approach. A practical reconstruction filter is not a brick wall, so the sampling rate must exceed the Nyquist rate by enough to fit the filter's transition region between the signal band and its first image. The bit count follows from the uniform-quantizer noise formula, and the multiplex bandwidth from the Nyquist signalling rate.

Size the reconstruction filter. The filter must pass the whole signal band, so its passband is $W = 8$ kHz wide. The transition region is 10% of that: $$\text{transition} = 0.10 \times 8\ \text{kHz} = 0.8\ \text{kHz}$$ so the filter reaches full attenuation at $8 + 0.8 = 8.8$ kHz.
Place the first image. Sampling at $f_{s}$ creates an image whose lower edge sits at $f_{s} - W$. For the filter to reject that image completely, the image must begin no lower than the filter's stopband edge: $$f_{s} - W \ge W + 0.8\ \text{kHz}$$
Solve for the sampling rate. $$f_{s} \ge 2W + 0.8\ \text{kHz} = 16 + 0.8 = 16.8\ \text{kHz} \quad\Longrightarrow\quad \boxed{\,f_{s} = 16.8\ \text{kHz}\,}$$ The 800 Hz excess over the 16 kHz Nyquist rate is precisely the guard band the non-ideal filter needs; sampling at exactly 16 kHz would leave no room for the roll-off and the image would fold into the passband.

Figure 4.1 — Spectrum of the sampled signal with the reconstruction filter mask overlaid. The first image starts at $f_{s} - W = 8.8$ kHz, exactly where the filter reaches its stopband.

Relate bits to quantization noise. With $N$ bits the $V_{pp} = 2$ V range is divided into $2^{N}$ steps of size $\Delta = V_{pp}/2^{N}$, and a uniform quantizer produces noise of rms value $$\sigma_{q} = \frac{\Delta}{\sqrt{12}} = \frac{V_{pp}}{2^{N}\sqrt{12}}$$
Impose the 1 mV limit. Requiring $\sigma_{q} \lt 10^{-3}$ V: $$2^{N} \gt \frac{2}{10^{-3}\sqrt{12}} = 577.4 \quad\Longrightarrow\quad N \ge 10$$ Checking both candidates: $N = 9$ gives $\Delta = 3.906$ mV and $\sigma_{q} = 1.128$ mV, which fails; $N = 10$ gives $\Delta = 1.953$ mV and $\sigma_{q} = 0.564$ mV, which passes. $$\boxed{\,N = 10\ \text{bits per sample}\,}$$
Compute the bit rate. Each sample carries 10 bits and samples arrive at 16.8 kHz: $$\boxed{\,R_{b} = f_{s}N = 16\,800 \times 10 = 168\ \text{kbps}\,}$$
Aggregate the multiplex. Ten such streams time-division multiplexed give $$R_{\text{mux}} = 10 \times 168\ \text{kbps} = 1.68\ \text{Mbps}$$
Apply the Nyquist signalling limit. With optimum (ideal Nyquist) filtering, a baseband channel of bandwidth $B$ supports $2B$ binary symbols per second without intersymbol interference, so $$\boxed{\,B_{\min} = \frac{R_{\text{mux}}}{2} = \frac{1.68\ \text{Mbps}}{2} = 840\ \text{kHz}\,}$$

It is worth noting the cost of digitisation: an 8 kHz analogue signal has been converted into a stream requiring 840 kHz of baseband channel once ten of them share the link — a bandwidth expansion of more than a hundredfold per channel, bought in exchange for regenerative repeating, error control and immunity to accumulated noise.

Quantity	Result
Filter transition width	0.8 kHz (stopband edge at 8.8 kHz)
Sampling rate	$f_{s} = 16.8$ kHz
Bits per sample	$N = 10$ ($\Delta = 1.953$ mV, $\sigma_{q} = 0.564$ mV rms)
Bit rate, one channel	$R_{b} = 168$ kbps
Aggregate rate, 10 channels	1.68 Mbps
Minimum channel bandwidth	$B_{\min} = 840$ kHz

Topic: Pulse-code modulation: practical sampling with a non-ideal reconstruction filter, uniform quantization noise, and Nyquist-limited baseband signalling.

Key relations: $f_{s} \ge 2W + (\text{transition width})$; $\Delta = V_{pp}/2^{N}$; $\sigma_{q} = \Delta/\sqrt{12}$; $\mathrm{SQNR} \approx 6.02N + 1.76$ dB for a full-scale sinusoid; $R_{b} = Nf_{s}$; Nyquist signalling $R_{\text{sym}} = 2B$ for binary transmission.

Why this works: Sampling replicates the spectrum every $f_{s}$, so reconstruction is a matter of isolating the baseband copy. An ideal brick-wall filter would allow $f_{s} = 2W$ exactly, but no realisable filter has zero transition width; the gap between $W$ and $f_{s} - W$ is exactly the room the roll-off needs, which is why real telephone systems sample 3.4 kHz speech at 8 kHz rather than 6.8 kHz. The $\sqrt{12}$ in the noise formula is the rms of a uniform distribution over one step: the quantization error is equally likely anywhere in $\pm\Delta/2$ when the signal is busy relative to the step size. Finally, the Nyquist criterion caps the symbol rate at twice the bandwidth, so bits and hertz are related by a factor of two for binary signalling.

Common pitfall: Setting $f_{s} = 2W$ and ignoring the transition region is the most common error; the question deliberately specifies the filter's roll-off so this cannot be skipped. Also frequent: computing the "10% transition" as 10% of $f_{s}$ or of the whole sampled band rather than of the filter passband. In part (b), rounding $N$ down (or solving with $\Delta$ rather than $\Delta/\sqrt{12}$) yields 9 bits, which genuinely misses the noise specification — always verify the chosen $N$ against the requirement rather than trusting the inequality algebra alone.

Source: Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., Ch. 6 (sampling and PCM) and Ch. 7 (baseband digital transmission, Nyquist criterion); Haykin, Communication Systems, 5th ed., Ch. 7.

← Question 3 Next: Question 5 →