Question 5 of 5: Sampling and D/A conversion

Question 5: Sampling and D/A conversion (25 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. A CD-quality audio signal of bandwidth $W = 20$ kHz, to be sampled and encoded by pulse code modulation at $n = 16$ bits per sample; for part (d) the converter's full-scale range is $-5$ V to $+5$ V, a span of $V_{FS} = 10$ V; for part (e) a typical MP3 stream runs at 128 kbit/s.

Find. (a) the Nyquist criterion with its Fourier-transform justification, (b) the minimum sampling frequency, (c) an explanation of PCM and the resulting data rate, (d) the largest quantization error, and (e) a reason MP3 needs far less rate.

Approach. Derive the replica structure of the sampled spectrum to establish the criterion, apply it to the 20 kHz bandwidth, multiply rate by resolution for the PCM bit rate, take half a quantization step for the error bound, and finish with the perceptual-coding argument for MP3.

1. Part (a) — state the criterion. A signal $x(t)$ whose Fourier transform vanishes for $|f|\gt W$ is completely determined by samples taken at uniform intervals $T_{s}=1/f_{s}$ provided $$f_{s} \gt 2W ,$$ and $2W$ is the Nyquist rate. Uniqueness fails at or below it in general, and equality is admissible only for signals with no spectral content exactly at $W$.
2. Justify it from the Fourier transform of the sampled signal. Ideal sampling multiplies $x(t)$ by an impulse train of period $T_{s}$, whose transform is itself an impulse train of spacing $f_{s}$. Multiplication in time is convolution in frequency, so $$X_{s}(f) = f_{s}\sum_{k=-\infty}^{\infty} X(f - k f_{s}) :$$ the spectrum of the samples is a periodic repetition of the original, one copy centred on every multiple of $f_{s}$. Each copy occupies $|f|\le W$ about its centre, so neighbouring copies stay disjoint exactly when $f_{s}-W \gt W$, i.e. $f_{s}\gt 2W$. When they are disjoint, an ideal low-pass filter of cutoff $f_{s}/2$ removes every copy but the one at the origin, recovering $X(f)$ and therefore $x(t)$ exactly; in the time domain that filter is the interpolation $$x(t) = \sum_{n} x(nT_{s})\,\mathrm{sinc}\!\left(\frac{t-nT_{s}}{T_{s}}\right).$$ If $f_{s}\lt 2W$ the copies overlap, the overlapping content adds irreversibly, and the high frequencies reappear as low ones — aliasing, which no later processing can undo. This is why a practical converter is always preceded by an anti-alias low-pass filter.

3. Part (b) — minimum sampling frequency for CD-quality audio. With $W = 20$ kHz the criterion gives $$f_{s,\min} = 2W = \boxed{40\ \text{kHz}} .$$ Compact disc actually uses 44.1 kHz, and the 4.1 kHz excess is not waste: it leaves a transition band of $44.1/2 - 20 = 2.05$ kHz in which the anti-alias and reconstruction filters can roll off, since an ideal brick wall at exactly 20 kHz is unrealisable.
4. Part (c) — what pulse code modulation is. PCM converts an analogue waveform to a bit stream in three steps: the signal is sampled at $f_{s}$ (after anti-alias filtering), each sample amplitude is quantized to the nearest of $2^{n}$ discrete levels, and the level index is encoded as an $n$-bit binary word. The output is a stream of fixed-length words at the sample rate; it is the reference against which every compression scheme is measured, and it is memoryless — every sample costs the same $n$ bits whatever the signal is doing.
5. PCM data rate. The rate is simply resolution times sample rate times the number of channels: $$R_{b} = n\,f_{s} = 16 \times 40\times10^{3} = \boxed{640\ \text{kbit/s per channel}},$$ so a stereo pair needs $2\times 640 = 1.28$ Mbit/s. At the actual CD rate the figure is $16\times 44.1\times10^{3}\times 2 = 1.4112$ Mbit/s, which is the familiar uncompressed-audio number.
6. Part (d) — the quantization step. A 16-bit converter divides the 10 V span into $2^{16} = 65{,}536$ intervals, so $$\Delta = \frac{V_{FS}}{2^{n}} = \frac{10}{65{,}536} = 152.59\ \mu\text{V}.$$
7. Maximum quantization error. Rounding each sample to the nearest level means the residual can never exceed half a step, so $$|e|_{\max} = \frac{\Delta}{2} = \boxed{76.29\ \mu\text{V}},$$ and the error is bounded independently of the signal, which is what makes the bound useful. Some texts divide by $2^{n}-1$ rather than $2^{n}$, giving 152.59 µV and 76.30 µV — indistinguishable at 16-bit resolution. Expressed as a ratio, a full-scale sinusoid sees a signal-to-quantization-noise ratio of $6.02n + 1.76 = 98.1$ dB, the origin of the “about 6 dB per bit” rule.
8. Part (e) — why MP3 needs far less rate. MP3 is lossy perceptual coding, whereas PCM is a faithful sample-by-sample representation. A filter bank splits the signal into subbands and a psychoacoustic model computes, band by band and frame by frame, the masking threshold below which the ear cannot hear anything because of louder nearby content; the coder then allocates only enough bits per band to push the quantization noise just under that threshold, discards what is masked outright, exploits the redundancy between the two stereo channels through joint-stereo coding, and finally Huffman-codes the quantized coefficients. PCM, by contrast, spends its full 16 bits on every sample regardless of whether the detail is audible. A 128 kbit/s MP3 stream is therefore about a tenth of the 1.28 Mbit/s uncompressed stereo rate computed in part (c), and about an eleventh of the real 1.4112 Mbit/s CD rate, while remaining close to transparent for most listeners.