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.

Find. The minimum sampling rate for exact reconstruction, the PCM bit rate at 16 bits per sample, an illustration of aliasing, the worst-case quantization error of a 24-bit converter over a 10 V span, and a reason why compressed video needs far less rate.

Approach. Apply the sampling theorem to fix the sample rate, multiply by the word length to get the bit rate, and treat quantization as a uniform staircase whose worst-case error is half a step; parts (c) and (e) are qualitative.

1. Part (a) — apply the Nyquist criterion. A signal strictly bandlimited to $B$ hertz is completely determined by samples taken at a rate exceeding twice that bandwidth, because sampling at $f_s$ replicates the signal spectrum at every multiple of $f_s$ and the replicas remain separable only while $f_s > 2B$. With $B = 5$ MHz, $$f_s \ge 2B = 2 \times 5\ \text{MHz}$$ $$\boxed{f_{s,\min} = 10\ \text{MHz} \ (10\ \text{Msample/s})}$$ Strictly the theorem requires $f_s > 2B$, with equality permissible only for a signal having no energy exactly at $B$; in practice a real converter is run some 10–20% above the Nyquist rate so that a physically realisable anti-aliasing filter has a transition band to roll off in.
2. Part (b) — explain PCM. Pulse code modulation is the standard three-stage conversion of an analogue waveform into a binary stream. The signal is first sampled at a uniform rate satisfying the Nyquist criterion, turning a continuous-time waveform into a sequence of amplitudes; each amplitude is then quantized, that is rounded to the nearest of a finite set of $2^{n}$ levels, which is the only irreversible step in the chain; and finally each level is encoded as an $n$-bit binary word for transmission or storage. The receiver decodes the words back to levels and passes them through a reconstruction filter. Quantization is what makes the result robust: once the waveform is represented by symbols drawn from a finite set, a regenerative repeater can restore the signal exactly rather than accumulating noise as an analogue relay would.
3. Compute the PCM bit rate. Each sample is encoded independently, so the bit rate is the product of the sample rate and the word length: $$R_b = n\,f_s = 16\ \frac{\text{bits}}{\text{sample}} \times 10\times 10^{6}\ \frac{\text{samples}}{\text{s}}$$ $$\boxed{R_b = 160\ \text{Mb/s}}$$ Sampling at the minimum rate gives the minimum bit rate; running the converter 20% fast, as a practical design would, raises it to 192 Mb/s.
4. Part (c) — give an example of aliasing. Aliasing is the impersonation of a high-frequency component by a lower-frequency one when the sampling rate is too low: any component at $f_0 > f_s/2$ reappears in the reconstruction at $|f_0 - k f_s|$ folded into the band below $f_s/2$, and no subsequent processing can separate it from a genuine component at that frequency.
   Figure 5.1 — Aliasing. The samples (dots) are consistent with both the true high-frequency sinusoid (solid) and a much lower-frequency one (dashed); the reconstruction filter necessarily chooses the low-frequency alias.

   The familiar everyday example is the wagon-wheel effect in film: a wheel photographed at 24 frames per second whose spokes rotate slightly less than one spoke-spacing per frame appears to turn slowly backwards, because the frame rate undersamples the true rotation. The audio equivalent is sampling a 30 kHz ultrasonic tone with a 44.1 kHz converter and no anti-aliasing filter, which produces an audible 14.1 kHz whistle that was never present in the source. In the specific case of the video signal of part (a), sampling at 8 MHz rather than 10 MHz would fold everything between 4 and 5 MHz — the fine detail of the picture — down onto the 3 to 4 MHz region, appearing as spurious coarse patterning (moiré) across the image. The remedy in all three cases is the same: band-limit the signal with an analogue anti-aliasing filter before the sampler, because once the samples are taken the damage cannot be undone.

5. Part (d) — find the quantization step. A uniform quantizer divides the full-scale range into $2^{n}$ equal levels, so the step size is $$\Delta = \frac{V_{FS}}{2^{n}} = \frac{5 - (-5)}{2^{24}} = \frac{10\ \text{V}}{16\,777\,216} = 5.9605\times10^{-7}\ \text{V}.$$
   Figure 5.2 — The quantizer staircase against the ideal straight line. Rounding to the nearest level bounds the error at half a step in either direction.
6. Convert the step to a worst-case error. With mid-tread rounding to the nearest level, an input can never lie more than half a step from the level assigned to it, so $$|e|_{\max} = \frac{\Delta}{2} = \frac{5.9605\times10^{-7}}{2}$$ $$\boxed{|e|_{\max} = 2.980\times10^{-7}\ \text{V} = 298.0\ \text{nV}}$$ For context, the corresponding signal-to-quantization-noise ratio for a full-scale sinusoid is $6.02n + 1.76 = 146.2$ dB, far beyond what any practical analogue front end can deliver — real 24-bit converters are limited by thermal and reference noise long before quantization matters. Note that a truncating rather than rounding quantizer would double this figure to a full step, $596.0$ nV, with a bias of half a step.
7. Part (e) — explain the gap to MPEG rates. The 160 Mb/s of part (b) is the cost of representing every sample of every frame independently and exactly. MPEG does not do that: it is a lossy compression scheme that removes the redundancy and the perceptual irrelevance in the signal. The single most important mechanism is temporal redundancy removal — successive video frames are nearly identical, so motion-compensated prediction transmits only a set of motion vectors plus the small residual between the predicted and the actual frame, and full frames are sent only occasionally as reference points. Alongside this, the discrete cosine transform concentrates each block's energy into a few low-frequency coefficients, the high-frequency coefficients are quantized coarsely or discarded because the eye is far less sensitive to fine detail than to overall brightness, chrominance is subsampled for the same reason, and the surviving coefficients are entropy-coded so that common values cost fewer bits than rare ones. Any one of these is a sufficient answer to a five-mark question; together they explain a compression ratio of one to two orders of magnitude. The price is that MPEG, unlike PCM, is not an exact representation — the decoded picture is only perceptually equivalent to the original.

Final Results