22-Elec-A3 Signals and Communications · December 2016

Question 4 of 6: PCM Transmission of a Video Signal

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada national examination 07-Elec-A3 — Signals and Communications, December 2016. Three hours, closed book; a standard non-programmable calculator is the only permitted aid. Six questions of equal value (20 marks each); any five constitute a complete paper. All six are solved below, because this set is a study resource rather than an examination script.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (Ch. 2–4 Fourier analysis, Ch. 4 amplitude modulation, Ch. 5 angle modulation, Ch. 6 sampling and pulse-code modulation); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Ch. 3 Fourier series, Ch. 10 the z-transform); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (Ch. 3 z-transform, Ch. 7 realisation structures); S. Haykin and M. Moher, Communication Systems, 5th ed. (Ch. 3 amplitude modulation, Ch. 4 angle modulation, Ch. 7 digital transmission).

Notation used throughout. The normalised sinc function is $\operatorname{sinc}(x)=\dfrac{\sin \pi x}{\pi x}$, so that $\operatorname{sinc}(0)=1$ and the zeros sit at the non-zero integers. Average power of a periodic signal is $P=\frac{1}{T_0}\int_{T_0}|x(t)|^{2}\,dt$, and the exponential Fourier series is $x(t)=\sum_{n=-\infty}^{\infty} c_n e^{j n \omega_0 t}$ with $P=\sum_n |c_n|^{2}$ (Parseval).

Question 4: PCM Transmission of a Video 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. Message bandwidth $W = 4.5\ \text{MHz}$; sampling rate 15% above the Nyquist minimum; uniform quantiser whose maximum error is $0.05\%$ of the signal peak $V_p$; binary (one bit per binary digit) encoding; source amplitude uniformly distributed over $[-V_p, +V_p]$.

Find. The sampling rate, the channel bit rate, and the reconstruction signal-to-quantisation-noise ratio.

Approach. Apply the sampling theorem with the stated guard factor, convert the fractional error specification into a word length by way of the step size, then evaluate the signal-to-noise ratio from the actual source statistics rather than the sinusoidal formula.

Sampling rate. The Nyquist rate for a low-pass signal of bandwidth $W$ is $f_{\text{Nyq}} = 2W = 9\ \text{MHz}$. The practical rate carries a 15% guard band to accommodate the finite roll-off of a real anti-aliasing filter: $$f_s = 1.15 \times 2W = 1.15 \times 9\ \text{MHz}$$ $$\boxed{f_s = 10.35\ \text{MHz (Msamples/s)}}$$
Convert the error specification into a step size. For a uniform quantiser of $L = 2^{n}$ levels spanning the peak-to-peak range $2V_p$, the step is $\Delta = 2V_p/2^{n}$ and the worst-case error is half a step: $$\varepsilon_{\max} = \frac{\Delta}{2} = \frac{V_p}{2^{n}} \le 0.0005\,V_p \quad\Longrightarrow\quad 2^{n} \ge 2000$$
Choose the word length. Since $2^{10} = 1024 \lt 2000 \le 2048 = 2^{11}$, the smallest integer satisfying the specification is $$\boxed{n = 11\ \text{bits/sample}}$$ Ten bits would give a maximum error of $0.098\%$, which violates the requirement; eleven bits give $1/2048 = 0.0488\%$, comfortably inside it. The word length must always be rounded up.
Channel bit rate. Each sample is encoded into $n$ binary digits, so $$R_b = n f_s = 11 \times 10.35\ \text{Mbit/s}$$ $$\boxed{R_b = 113.85\ \text{Mbit/s}}$$ The corresponding minimum (Nyquist) channel bandwidth for binary signalling is $R_b/2 = 56.9\ \text{MHz}$ — more than twelve times the original video bandwidth, which is the familiar bandwidth-for-robustness trade that PCM makes.
Signal power for a uniform source. If the amplitude is uniformly distributed on $[-V_p, V_p]$, its mean square is $$P_s = \int_{-V_p}^{V_p} \frac{v^{2}}{2V_p}\,dv = \frac{V_p^{2}}{3}.$$
Quantisation noise power and the SNR. Quantisation error is uniform on $[-\Delta/2, \Delta/2]$, so $P_q = \Delta^{2}/12$. With $\Delta = 2V_p/2^{n}$, $$\text{SNR} = \frac{V_p^{2}/3}{\Delta^{2}/12} = \frac{V_p^{2}/3}{\frac{1}{12}\left(\frac{2V_p}{2^{n}}\right)^{2}} = 2^{2n}$$ $$\boxed{\text{SNR} = 2^{22} = 4.194 \times 10^{6} \equiv 66.23\ \text{dB}}$$ Equivalently $\text{SNR}_{\text{dB}} = 6.02n = 6.02 \times 11$. The uniform-density assumption is what removes the familiar $+1.76\ \text{dB}$ term, which belongs specifically to a full-scale sinusoid.

Figure Q4.1 — the PCM transmitter chain with the values derived in this answer.

Check: the quantiser is assumed to span exactly the signal's peak-to-peak range $2V_p$ (no headroom), which is the standard reading of “quantization error is a maximum of 0.05% of the peak value”. If instead the quantiser were designed with headroom, $n$ would rise by the corresponding number of bits and $R_b$ with it.

Quantity	Result
Nyquist rate	$2W = 9\ \text{MHz}$
(a) Sampling rate	$f_s = 10.35\ \text{MHz}$
Word length	$n = 11$ bits/sample ($L = 2048$ levels)
(b) Bit rate	$R_b = 113.85\ \text{Mbit/s}$
(c) SNR (uniform pdf)	$2^{22} = 4.19\times 10^{6}$, i.e. $66.23\ \text{dB}$

Topic. Pulse-code modulation: the sampling theorem with a practical guard factor, uniform quantisation, word-length selection, and signal-to-quantisation-noise ratio.

Key relations. $f_s \ge 2W$; step size $\Delta = 2V_p/2^{n}$ with worst-case error $\Delta/2$; quantisation noise $P_q = \Delta^{2}/12$; bit rate $R_b = nf_s$; and $\text{SNR}_{\text{dB}} = 6.02n + C$, where $C = 1.76\ \text{dB}$ for a full-scale sinusoid but $C = 0$ for a uniformly distributed source.

Why this works. The three parts form a single chain: the sampling theorem fixes how often, the error specification fixes how finely, and their product fixes the channel rate. The quantisation-noise model treats the error as a random variable uniformly spread across one step, which is accurate whenever the step is small compared with the signal's excursions — certainly true at 2048 levels. The $6.02\ \text{dB}$-per-bit rule then falls out because each extra bit halves $\Delta$ and therefore quarters the noise power.

Common pitfall. Three traps dominate. Rounding $n$ down (to 10 here) silently violates the accuracy specification. Applying the memorised $\text{SNR} = 6.02n + 1.76\ \text{dB}$ when the question has deliberately specified a uniform density inflates the answer by $1.76\ \text{dB}$; the question states the distribution precisely so that the constant vanishes. Finally, the 15% oversampling applies to the Nyquist rate $2W$, not to the bandwidth $W$.

Source. Lathi & Ding, Modern Digital and Analog Communication Systems, 4th ed., §6.2–6.3 (PCM, quantisation noise); Haykin & Moher, Communication Systems, 5th ed., §7.3–7.5.

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