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17-Phys-B4 Signals and Communications · May 2014

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. 98-Phys-B4 Communications, National Examination May 2014 — a three-hour open-book examination (any non-communicating calculator permitted). The cover page states any five of the ten questions constitute a complete paper, with only the first five as they appear in the answer book marked; every question is nonetheless answered in full below so the paper remains a complete study resource. All ten questions carry equal value (20 marks each).

Reference texts. A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Fourier transform properties, the sampling theorem, LTI eigenfunctions, z-transforms); S. Haykin and M. Moher, Communication Systems, 5th ed. (AM/FM modulation, PCM, matched filtering and eye diagrams); B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (FM Bessel spectra, M-ary baseband transmission); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (z-transform regions of convergence, partial-fraction inversion).

Question 5: PCM Design for a Television 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. Signal bandwidth $W=4.5$ MHz; sampling rate 20% above the Nyquist rate; $L=1024$ uniform quantization levels; binary coding.

Find. (a) $f_s$; (b) bits/sample $n$; (c) bit rate $R_b$ and minimum transmission bandwidth.

Approach. Chain the standard PCM design formulas: Nyquist rate $\to f_s\to n=\log_2L\to R_b=f_sn\to B_{min}=R_b/2$ (ideal Nyquist zero-ISI baseband bandwidth).

  1. Part (a) — sampling rate. Nyquist rate $=2W=2(4.5\ \text{MHz})=9$ MHz; 20% above that: $$f_s=1.2\times9\ \text{MHz}=\boxed{10.8\ \text{MHz}}.$$
  2. Part (b) — bits per sample. $L=1024=2^{10}$, so $$n=\log_2L=\boxed{10\ \text{bits/sample}}.$$
  3. Part (c) — bit rate and minimum bandwidth. $$R_b=f_s\,n=10.8\times10^6\times10=\boxed{108\ \text{Mbps}}.$$ The ideal Nyquist (zero-ISI) minimum transmission bandwidth for a binary baseband signal is half the bit rate: $$B_{min}=\frac{R_b}{2}=\boxed{54\ \text{MHz}}.$$
Final results
QuantityValue
Sampling rate $f_s$10.8 MHz
Bits per sample $n$10
Bit rate $R_b$108 Mbps
Minimum bandwidth $B_{min}$54 MHz