Question 5 of 10: PCM Design for a Television Signal
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2014 — 07-Elec-A3 Signals and Communications. Three hours, open book, any non-communicating calculator permitted. Ten questions of equal value (20 marks each); the rubric states that five questions constitute a complete paper and only the first five presented are marked. All ten are solved here, since the set is intended as a study resource. A table of Fourier-transform pairs and properties, a table of z-transform pairs, trigonometric identities and Bessel-function graphs/tables are supplied with the paper and are used freely below.
Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (sampling, PCM, AM/FM, matched filtering); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier analysis, LTI properties); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (z-transform, regions of convergence); S. Haykin, Communication Systems, 5th ed. (digital transmission, eye diagrams).
Convention used throughout. This paper's own transform table defines the sinc function as $\operatorname{sinc}(x)=\sin x / x$ (Lathi's convention, not the normalised $\sin(\pi x)/(\pi x)$). Every bandwidth below follows from that table, in particular $2B\operatorname{sinc}(2\pi Bt)\leftrightarrow\operatorname{rect}(f/2B)$ and $B\operatorname{sinc}^2(\pi Bt)\leftrightarrow\Delta(f/2B)$.
Question 5: PCM Design for a Television Signal (20 marks)
Find. The sampling rate $f_s$, the number of bits per sample $n$, the bit rate $R_b$, and the minimum ISI-free transmission bandwidth $B_T$.
The PCM chain: sample above the Nyquist rate, quantize to 1024 levels, then binary-encode each sample into 10 bits.
Approach. Walk the PCM chain in order — sample, quantize, encode, transmit — converting each stage's specification into a rate, then apply the Nyquist signalling limit to convert the final bit rate into a bandwidth.
Sampling rate. The Nyquist rate for a 4.5 MHz baseband signal is $f_{\text{Nyq}} = 2B = 9$ MHz. Sampling 20% above it means multiplying by 1.20: $$f_s = 1.20\times 2B = 1.20\times 9\ \text{MHz} = \boxed{10.8\ \text{MHz}}$$ i.e. 10.8 million samples per second. The 20% margin is the guard band that lets a real anti-aliasing filter roll off over a finite width instead of requiring a brick wall.
Bits per sample. A binary code of $n$ bits distinguishes $2^n$ states, so representing $L = 1024$ levels needs $$n = \log_2 L = \log_2 1024 = \boxed{10\ \text{bits per sample}}$$ Since 1024 is an exact power of two, no bits are wasted — every one of the 1024 codewords maps to a distinct quantization level.
Bit rate. Each sample carries 10 bits and samples arrive at 10.8 MHz, so $$R_b = n f_s = 10 \times 10.8\times 10^{6} = \boxed{108\ \text{Mbit/s}}$$
Minimum transmission bandwidth. Nyquist's signalling criterion says a channel of bandwidth $B_T$ can carry at most $2B_T$ independent binary pulses per second without intersymbol interference, so the minimum bandwidth for $R_b$ bits per second is $$B_T = \frac{R_b}{2} = \frac{108\ \text{Mbit/s}}{2} = \boxed{54\ \text{MHz}}$$ This is a theoretical floor achieved only by ideal sinc pulses; any practical raised-cosine shaping with roll-off $r$ inflates it to $(1+r)\times 54$ MHz.
The result is worth a moment's reflection: digitizing a 4.5 MHz analogue television signal at broadcast quality costs at least 54 MHz of channel — a twelvefold expansion. This bandwidth explosion is precisely why uncompressed PCM video was never broadcast, and why every practical digital television standard places a source coder (MPEG-2, H.264) between the quantizer and the channel.
