22-Elec-A3 Signals and Communications · May 2017

Question 4 of 5: Non-Uniform PCM Quantizer Design for a Voice Signal

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

Question 4: Non-Uniform PCM Quantizer Design for a Voice 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.

Find. The two step sizes and the complete set of decision thresholds and output levels; the number of bits per sample; and the resulting bit rate.

Approach. For a mid-riser uniform quantizer the worst-case error is half a step, so each zone's error specification fixes its step size directly; count the levels each zone needs, add them, and round the total up to the next power of two.

1. Step size in the fine (inner) zone. With rounding to the nearest level the maximum error is $\Delta/2$, so $$\frac{\Delta_1}{2}\lt 0.001\,m_p \ \Longrightarrow\ \boxed{\Delta_1 = 0.002\,m_p}$$ The inner zone spans $-m_p/2$ to $+m_p/2$, a width of $m_p$, so it needs $$L_1=\frac{m_p}{0.002\,m_p}=500\ \text{levels}$$ (250 on each side of the origin).
2. Step size in the coarse (outer) zones. Similarly $\Delta_2/2\lt 0.004\,m_p$ gives a nominal $\Delta_2=0.008\,m_p$. Each outer zone spans $m_p/2$ to $m_p$, a width of $0.5\,m_p$, which would require $0.5/0.008=62.5$ steps — not an integer, so we round up to keep the error inside the specification: $$L_2 = \lceil 62.5\rceil = 63\ \text{levels per side},\qquad \boxed{\Delta_2 = \frac{0.5\,m_p}{63}=0.0079365\,m_p}$$ giving a worst-case error of $0.003968\,m_p \lt 0.004\,m_p$. Rounding down to 62 steps would give $\Delta_2=0.008065\,m_p$ and an error of $0.004032\,m_p$, which violates the specification — the direction of rounding is the graded point here.
3. (a) Quantizer characteristic and threshold list. The characteristic is the odd, staircase-shaped mapping sketched below. Written out explicitly, with all values in units of $m_p$:

   Zone	Decision thresholds	Output (reconstruction) levels	Count
   Inner, 0 to +1/2	0, 0.002, 0.004, …, 0.500 (i.e. 0.002k, k = 0…250)	0.001, 0.003, …, 0.499 (0.002k + 0.001)	250
   Inner, −1/2 to 0	mirror image: −0.002k	−(0.002k + 0.001)	250
   Outer, +1/2 to +1	0.500, 0.507937, 0.515873, …, 1.000 (0.5 + 0.0079365j, j = 0…63)	0.503968, 0.511905, …, 0.996032	63
   Outer, −1 to −1/2	mirror image: −(0.5 + 0.0079365j)	−(0.503968 + 0.0079365j)	63
   Total number of quantization levels			626

   A sample is assigned the output level at the midpoint of whichever interval it falls in; samples at exactly $\pm m_p/2$ belong to the outer zone by the wording of the specification. The characteristic is therefore a mid-tread staircase that is fine near the origin and four times coarser near the peaks — the same idea that A-law and µ-law companding implement with a smooth compressor curve.

4. (b) Bits per sample. A binary code must address all 626 levels, and the number of codewords available with $n$ bits is $2^{n}$: $$2^{n}\ge 626\ \Longrightarrow\ n\ge \log_2 626 = 9.29\ \Longrightarrow\ \boxed{n = 10\ \text{bits/sample}}$$ Ten bits provide 1024 codewords, so 398 remain spare (they can be used for signalling, or the zones can be made slightly finer than required). Nine bits would give only 512 — not enough.
5. (c) Bit rate. The Nyquist rate for a 10 kHz bandwidth is 20 kHz; a 20 % oversampling allowance to ease the anti-aliasing filter gives $$f_s = 1.2\times 2W = 1.2\times 20\ \text{kHz} = \boxed{24\ \text{kHz}}$$ and therefore $$R_b = n f_s = 10 \times 24\times10^{3} = \boxed{240\ \text{kbit/s}}$$ The oversampling costs 40 kbit/s but relaxes the anti-aliasing filter's transition band from zero width to 4 kHz, which is the whole point of the allowance.