PCM — Sampling Rate, Quantization Bits, Bit Rate, TDM Bandwidth: 17-Phys-B4 Signals and Communications, December 2015, Q4

Question 4: PCM — Sampling Rate, Quantization Bits, Bit Rate, TDM Bandwidth (1/6 of paper)

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 $W=8$ kHz; dynamic range $2$ V p-p; LPF transition width $=10\%$ of its own passband width; required quantization noise $<1$ mV rms; 10 PCM streams to be TDM-multiplexed.

Find. (a) minimum $f_s$; (b) minimum bits/sample $n$; (c) bit rate; (d) minimum baseband channel bandwidth for the 10-stream mux.

Approach. Add a guard band equal to the reconstruction filter's own transition width on top of the Nyquist rate; size the quantizer from the uniform-quantizer noise formula $\sigma_q=\Delta/\sqrt{12}$; multiply bits by sampling rate for the bit rate; use the Nyquist minimum-bandwidth rule $B_{\min}=R_b/2$ for optimum (e.g. raised-cosine at zero rolloff / sinc) baseband pulses.

Part (a) — sampling rate. The reconstruction filter's passband must cover the full signal band $W$, so its transition width is $0.10\times8=0.8$ kHz, and its stopband edge sits at $8+0.8=8.8$ kHz. To avoid the first spectral image (centred at $f_s-W$) overlapping that stopband edge, $$f_s-W\ge W+0.8\ \text{kHz}\ \Rightarrow\ f_s\ge2W+0.8=\boxed{16.8\text{ kHz}}$$
Part (b) — quantization bits. With an $n$-bit quantizer over the $2$ V p-p range, step size $\Delta=2/2^n$ and rms quantization noise $\sigma_q=\Delta/\sqrt{12}$. Requiring $\sigma_q<1$ mV: $$n=9:\ \Delta=3.906\text{ mV},\ \sigma_q=1.128\text{ mV (fails)}\qquad n=10:\ \Delta=1.953\text{ mV},\ \sigma_q=0.564\text{ mV (passes)}$$ $$\boxed{n=10\text{ bits/sample}}$$
Part (c) — bit rate. $$R_b=n\,f_s=10\times16.8\text{ kHz}=\boxed{168\text{ kbps}}$$
Part (d) — TDM bandwidth. Ten multiplexed streams give a combined bit rate $R_{b,\text{tot}}=10\times168=1680$ kbps; with optimum (Nyquist) filtering of the baseband pulses the minimum channel bandwidth is half the bit rate: $$B_{\min}=\dfrac{R_{b,\text{tot}}}{2}=\dfrac{1680}{2}=\boxed{840\text{ kHz}}$$

Question 4 — final results
Quantity	Result
Sampling rate $f_s$	$16.8$ kHz
Bits per sample $n$	$10$
Bit rate $R_b$	$168$ kbps
10-stream TDM bandwidth	$840$ kHz

Concept & Theory

Topic. PCM system design: practical (non-ideal-filter) sampling rate, uniform quantizer sizing, bit rate, and Nyquist minimum bandwidth for TDM.

Key relations. Nyquist criterion with guard band $f_s\ge2W+\Delta f_{\text{transition}}$; uniform quantization noise $\sigma_q=\Delta/\sqrt{12}$, $\Delta=\text{range}/2^n$; $R_b=nf_s$; Nyquist minimum bandwidth $B_{\min}=R_b/2$ (achieved only with an ideal/raised-cosine pulse shape at zero excess bandwidth).

Why this works. A real reconstruction filter cannot cut off infinitely sharply at $W$, so its finite transition band must be given room before the first alias image begins — sampling at exactly $2W$ (as an ideal filter would allow) would let the image and the signal band overlap. Bits are added one at a time until the quantization step is fine enough to meet the noise spec, since noise falls by a factor of 2 for each extra bit ($6.02$ dB/bit).

Common pitfall. Using the ideal Nyquist rate $f_s=2W=16$ kHz here (ignoring the filter's transition band, which the question explicitly supplies to prevent that shortcut); rounding the bit count down instead of up when the noise requirement is not quite met.

Source. S. Haykin and M. Moher, Communication Systems, 5th ed., §6.2 (sampling and reconstruction with practical filters), §6.6 (PCM, quantization noise); B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed., §8.4 (TDM and Nyquist bandwidth).

Question 4 of 6: PCM — Sampling Rate, Quantization Bits, Bit Rate, TDM Bandwidth

Question 4: PCM — Sampling Rate, Quantization Bits, Bit Rate, TDM Bandwidth (1/6 of paper)