17-Phys-B4 Signals and Communications · December 2014
Question 3 of 7: PCM Sampling, Quantization and Multiplexed Bit Rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B4 Communications, National Examination
December 2014 — a three-hour closed-book examination (a standard non-programmable,
no-text-storage calculator is the only aid permitted). The cover page states any
five of the seven 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 seven
questions carry equal value.
Reference texts. A. V. Oppenheim and A. S. Willsky, Signals and
Systems, 2nd ed. (Fourier transform properties, LTI convolution, the sampling
theorem); S. Haykin and M. Moher, Communication Systems, 5th ed. (AM/DSB/FM/PM
modulation, PCM, the superheterodyne receiver); B. P. Lathi and Z. Ding, Modern
Digital and Analog Communication Systems, 4th ed. (envelope/coherent detection,
image-frequency rejection); J. G. Proakis and D. G. Manolakis, Digital Signal
Processing, 4th ed. (z-transforms, difference equations, BIBO stability).
Question 3: PCM Sampling, Quantization and Multiplexed Bit Rate (1/5 of paper)
Find. (a) minimum sampling rate $f_s$; (b) minimum bits per sample
$n$; (c) bit rate $R_b$ of one channel; (d) minimum baseband bandwidth for 10
time-multiplexed channels.
Approach. The reconstruction filter's passband must cover the full
signal bandwidth $B$ and its stopband must begin by the first spectral image at
$f_s-B$, so the sampling rate is set by requiring the transition band to fit exactly in
the gap $f_s-2B$. The bit depth follows from the uniform-quantizer noise formula
$\sigma_q=\Delta/\sqrt{12}$. The multiplexed bit rate uses ideal (zero-excess-bandwidth)
Nyquist pulse shaping, for which the minimum baseband bandwidth is $R_b/2$.
Part (a) — sampling rate. The reconstruction filter's passband
edge must sit at $B=8$ kHz (to pass the whole signal) and its stopband must begin
by $f_s-B$ (the nearest spectral image), so the transition band is $f_s-2B$. Setting this
equal to 10% of the passband bandwidth ($0.10\times8\,\text{kHz}=800$ Hz):
$$f_s-2B=0.10B \quad\Rightarrow\quad f_s=2.1B=\boxed{16{,}800\ \text{Hz} = 16.8\ \text{kHz}}$$
Part (b) — number of bits. With $L=2^n$ uniform levels spanning
the 2 V dynamic range, step size $\Delta=2/L$ and quantization-noise rms
$\sigma_q=\Delta/\sqrt{12}\le1$ mV:
$$2^n\ge\dfrac{2\ \text{V}}{\sqrt{12}\times1\ \text{mV}}=577.4 \quad\Rightarrow\quad n\ge\log_2(577.4)=9.17$$
Rounding up to the next integer, $n=9$ gives $\Delta=3.906$ mV, $\sigma_q=1.128$ mV
(fails the 1 mV spec); $n=10$ gives $\Delta=1.953$ mV, $\sigma_q=0.564$ mV (meets it), so
$$n=\boxed{10\ \text{bits/sample}}$$
Part (c) — bit rate.
$$R_b=n\,f_s=10\times16{,}800=\boxed{168{,}000\ \text{bps}=168\ \text{kbps}}$$
Part (d) — multiplexed minimum bandwidth. Time-multiplexing 10
identical PCM streams sums the bit rates, $R_{b,\text{tot}}=10\times168{,}000=1{,}680{,}000$
bps. With optimum (ideal Nyquist / zero-excess-bandwidth) pulse shaping, binary baseband
transmission needs only $R_b/2$ Hz of bandwidth:
$$B_{\min}=\dfrac{R_{b,\text{tot}}}{2}=\boxed{840{,}000\ \text{Hz}=840\ \text{kHz}}$$
