Question 2 of 5: Non-Uniform (Two-Range) PCM Quantizer Design

Question 2: Non-Uniform (Two-Range) PCM Quantizer Design (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 $W = 10$ kHz; peak value $m_p$ (the answer is expressed as a fraction of $m_p$, which is never given numerically); maximum quantization error $0.001\,m_p$ for $|m_s| \le m_p/2$ and $0.004\,m_p$ otherwise; 20% oversampling above the Nyquist rate.

Find. The step size, threshold list and level count in each zone; the number of bits per sample; and the resulting bit rate.

Approach. A rounding (nearest-level) quantizer of step $\Delta$ has a worst-case error of $\Delta/2$, so each error specification fixes a step size directly. Divide each zone's width by its step to count levels, round the count up where it is not an integer, add the zones, and take $n = \lceil \log_2 L \rceil$ bits.

1. Convert each error specification into a step size. For uniform quantization with rounding to the nearest level, a sample can lie at most half a step from its representation level, so $e_{\max} = \Delta/2$ and $\Delta = 2e_{\max}$: $$\Delta_{\text{in}} = 2(0.001\,m_p) = 0.002\,m_p, \qquad \Delta_{\text{out}} = 2(0.004\,m_p) = 0.008\,m_p.$$ The inner (fine) step is four times smaller than the outer (coarse) step, which is exactly the companding idea: spend resolution where the amplitude density is highest.
2. Count the levels in the inner zone. The fine zone runs from $-m_p/2$ to $+m_p/2$, a total width of $m_p$: $$L_{\text{in}} = \frac{m_p}{\Delta_{\text{in}}} = \frac{m_p}{0.002\,m_p} = 500\ \text{levels}.$$ This divides exactly, so the inner step stays at its specified value $0.002\,m_p$ and the inner worst-case error is exactly $0.001\,m_p$.
3. Count the levels in each outer zone, and round in the safe direction. Each outer zone (from $m_p/2$ to $m_p$, and its mirror image) has width $m_p/2$: $$\frac{0.5\,m_p}{0.008\,m_p} = 62.5\ \text{levels},$$ which is not an integer. The count must be rounded up, not to the nearest integer: taking 62 levels would give $\Delta = 0.5/62 = 0.008065\,m_p$ and an error of $0.004032\,m_p$, which violates the 0.4% specification. Taking 63 levels gives $$\Delta_{\text{out}} = \frac{0.5\,m_p}{63} = 0.0079365\,m_p, \qquad e_{\max} = 0.0039683\,m_p \lt 0.004\,m_p. \checkmark$$
4. State the quantizer characteristic (part a). With the origin at the centre of the fine zone, the decision thresholds and output levels are:

   Input range	Step	Thresholds	Output levels	Count
   $-m_p \le m_s \lt -m_p/2$	$0.0079365\,m_p$	$-m_p + k\Delta_{\text{out}}$, $k = 1\ldots62$	midpoints $-m_p + (k-\tfrac12)\Delta_{\text{out}}$	63
   $-m_p/2 \le m_s \le +m_p/2$	$0.002\,m_p$	$-m_p/2 + k\Delta_{\text{in}}$, $k = 1\ldots499$	midpoints $-m_p/2 + (k-\tfrac12)\Delta_{\text{in}}$	500
   $+m_p/2 \lt m_s \le +m_p$	$0.0079365\,m_p$	$m_p/2 + k\Delta_{\text{out}}$, $k = 1\ldots62$	midpoints $m_p/2 + (k-\tfrac12)\Delta_{\text{out}}$	63

   Each output level is the midpoint of its interval, which is what makes the worst-case error equal to half the step.

5. Total the levels and size the codeword (part b). $$L = L_{\text{in}} + 2L_{\text{out}} = 500 + 2(63) = 626\ \text{levels}.$$ A unique binary code needs $n$ bits with $2^n \ge L$: $$n = \lceil \log_2 626 \rceil = \lceil 9.29 \rceil = 10, \qquad 2^9 = 512 \lt 626 \le 1024 = 2^{10},$$ $$\boxed{\,n = 10\ \text{bits per sample}\,}$$ A natural assignment is to let the most significant bits identify the zone and the remainder index the level within it; 1024 − 626 = 398 codewords are simply left unused.
6. Set the sampling rate and bit rate (part c). The Nyquist rate for a 10 kHz signal is $2W = 20$ kHz, and 20% oversampling raises it to $$f_s = 1.20 \times 2W = 1.20 \times 20\ \text{kHz} = 24\ \text{kHz}.$$ Each sample carries 10 bits, so $$R_b = n f_s = 10 \times 24{,}000,$$ $$\boxed{\,R_b = 240\ \text{kbit/s}\,}$$