22-Elec-A3 Signals and Communications · May 2016

Question 3 of 6: PCM — Bit Rate, Quantizer Resolution and Analog Bandwidth

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

Notes on this paper

Paper format: Professional Engineers Ontario / EGBC national examination 07-Elec-A3 — Signals and Communications, May 2016. Closed book (a standard non-programmable calculator is the only aid permitted), 3 hours. Six questions of equal value (20 marks each); any five constitute a complete paper. All six are solved below, because the set is intended as a study resource.

Reference texts.

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed., Oxford University Press — Ch. 3 (signal transmission), Ch. 4 (amplitude modulation), Ch. 5 (angle modulation), Ch. 6 (sampling and PCM).
S. Haykin and M. Moher, Communication Systems, 5th ed., Wiley — Ch. 3 (amplitude modulation), Ch. 4 (angle modulation), Ch. 5 (pulse modulation and PCM).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed., Prentice Hall — Ch. 4 (continuous-time Fourier transform), Ch. 10 (the z-transform).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed., Pearson — Ch. 3 (z-transform), Ch. 7 (structures for discrete-time systems).

Question 3: PCM — Bit Rate, Quantizer Resolution and Analog Bandwidth (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.

Quantity	Symbol	Value
Channel bit rate	$R_b$	90 kb/s
Required reconstruction SNR	$\mathrm{SNR}_{\min}$	> 50 dB
Quantization	—	uniform, $L=2^{n}$ levels over $[-V_m,V_m]$
Source amplitude distribution	$p(v)$	uniform on $[-V_m,V_m]$

Find. The largest analog bandwidth $B$ that can be carried by this channel while still meeting the 50 dB reconstruction requirement.

Approach. Fix the number of bits per sample from the SNR requirement, divide the bit rate by that to get the sampling rate, and halve the sampling rate to obtain the maximum signal bandwidth permitted by the sampling theorem.

Quantization SNR for a uniformly distributed source. With $L=2^{n}$ levels spanning $2V_m$, the step size is $\Delta=2V_m/2^{n}$ and the quantization-error power is $\Delta^{2}/12$. A source uniform on $[-V_m,V_m]$ has mean-square value $V_m^{2}/3$, so $$\mathrm{SNR}_q=\frac{V_m^{2}/3}{\Delta^{2}/12}=\frac{V_m^{2}/3}{4V_m^{2}/(12\cdot 2^{2n})}=2^{2n},$$ that is, in decibels, $$\mathrm{SNR}_q\big|_{\text{dB}}=20n\log_{10}2=6.02\,n\ \text{dB}.$$ Note that the familiar “$6.02n+1.76$” expression applies to a sinusoidal input; for the uniform distribution assumed here the additive constant is exactly zero, which is why the question specifies the distribution.
Choose the number of bits per sample. Requiring $6.02n\gt 50$ dB gives $n\gt 8.31$, and $n$ must be an integer, so $$\boxed{\;n=9\ \text{bits/sample}\;}$$ Checking both neighbours confirms the choice: $n=8$ yields only $6.02\times 8=48.2$ dB, which fails, while $n=9$ yields $6.02\times 9=54.2$ dB, comfortably above the 50 dB floor.
Convert the bit rate into a sampling rate. A PCM stream carries $n$ bits for every sample, so $R_b=n f_s$ and $$f_s=\frac{R_b}{n}=\frac{90\times 10^{3}\ \text{b/s}}{9\ \text{b/sample}} =\boxed{\;10\ \text{kHz}\;}$$
Apply the sampling theorem. Alias-free reconstruction requires $f_s\ge 2B$, so the analog bandwidth is bounded by $$B\le\frac{f_s}{2}=\frac{10\ \text{kHz}}{2} \;\Rightarrow\;\boxed{\;B_{\max}=5\ \text{kHz}\;}$$ This is the binding answer. Spending a tenth bit would raise the SNR to 60.2 dB but drop the sampling rate to 9 kHz and the bandwidth to 4.5 kHz — the classic PCM trade of resolution against bandwidth at fixed channel capacity.
Sanity-check the design. At $n=9$, $f_s=10$ kHz and $B=5$ kHz, the delivered bit rate is $9\times 10\ \text{kHz}=90$ kb/s, exactly the channel capacity, and the delivered SNR is 54.2 dB, exceeding the specification by 4.2 dB. In practice a guard band would be added (a real anti-aliasing filter is not ideal), so a designer would quote something like 4.5 kHz of usable audio bandwidth with 0.5 kHz of transition band.

Quantity	Result
Required bits per sample	$n=9$ (54.2 dB; 8 bits gives only 48.2 dB)
Sampling rate	$f_s=R_b/n=10$ kHz
Maximum analog bandwidth	$B_{\max}=f_s/2=5$ kHz
Delivered SNR	54.2 dB (specification 50 dB)

Topic. Pulse-code modulation: quantization noise, the 6 dB-per-bit rule and the bandwidth–resolution trade at fixed bit rate.

Key relations. Step size $\Delta=2V_m/2^{n}$; quantization noise power $\Delta^{2}/12$; uniform-source SNR $=2^{2n}\Rightarrow 6.02n$ dB (sinusoid: $6.02n+1.76$ dB); PCM bit rate $R_b=nf_s$; Nyquist $f_s\ge 2B$.

Why this works. Uniform quantization error is, to a good approximation, a random variable spread evenly over one step, so its power depends only on the step size — not on the signal. Adding a bit halves the step and therefore quarters the noise power, which is the 6.02 dB per bit. Because the channel delivers a fixed number of bits per second, every extra bit of resolution must be paid for with fewer samples per second, and the sampling theorem converts samples per second directly into usable analog bandwidth. The design therefore reduces to a single integer choice: the smallest $n$ that satisfies the SNR floor is also the one that leaves the most bandwidth.

Common pitfall. Using the sinusoidal formula $6.02n+1.76$ when the question explicitly states a uniformly distributed source: that shortcut gives $n=9$ here as well, but on a paper with a tighter specification it produces an $n$ one bit too small. The second pitfall is rounding $n$ down — $n$ must be rounded up, since 8.31 bits is not realisable and 8 bits misses the specification.

Source. Haykin & Moher, Communication Systems, 5th ed., §5.5–5.7 (quantization noise and PCM); Lathi & Ding, 4th ed., §6.2–6.3.

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