---
title: "Need for non-uniform quantization; one algorithm"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:nonuniform-quantization-algorithm"
source: https://www.bctengineer.com/q/semester_6/communication_systems/nonuniform-quantization-algorithm
site: BCT Engineering
---

# Need for non-uniform quantization; one algorithm

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2074 Chaitra (CS II), 5 marks
- Video: [Non-uniform Quantization | Basics, Importance & Calculation | Uniform Vs Non-Uniform Quantization](https://www.youtube.com/watch?v=7DoNXi4g_Bg) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/nonuniform-quantization-algorithm
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Why is non uniform quantization required, explain any one algorithms for implementing non-uniform quantization.

## Short answer

Non-uniform quantization needed because speech has wide dynamic range and peaked PDF; uniform steps give poor SQNR for weak signals. Companding algorithm: compressor → uniform quantizer → expander. µ-law: normalize x=m/mp, compress y=sgn(x)ln(1+μ|x|)/ln(1+μ) with μ=255, uniformly quantize y, transmit, expand at receiver via inverse. Small steps near zero, large steps at high amplitude yield nearly constant SQNR (~38 dB over 40 dB range) with 8 bits.

## Answer

#### Why non-uniform quantization is needed
- In uniform quantization the noise power $\Delta^2/12$ is the same for every level, so SQNR is proportional to signal power. Weak signals get poor SQNR.
- Speech has a wide dynamic range (about 40 dB) and small amplitudes are far more probable than large ones.
- To give soft talkers acceptable SQNR with uniform steps would need about 12–13 bits per sample.
- Non-uniform quantization uses **small steps for small amplitudes and large steps for large amplitudes**, making SQNR nearly constant over the range. 8 bits then give toll-quality speech (64 kbps).

#### Algorithm: companding with the µ-law
Non-uniform quantization is realised by **companding**: compressor → uniform quantizer → (channel) → expander.

```text
m(t)->[Compressor]->[Uniform quantizer+encoder]->channel
                                                    |
out<-[Expander]<-[Decoder]<-------------------------+
```

1. **Normalise** the input: $x = m/m_p$, so $|x| \le 1$.
2. **Compress** using the µ-law:
$$
y = \text{sgn}(x)\,\frac{\ln(1 + \mu|x|)}{\ln(1 + \mu)}, \qquad \mu = 255
$$
For small $x$, $y \approx \mu x/\ln(1 + \mu)$ (large gain); for large $x$ the curve is logarithmic (small gain).
3. **Uniformly quantize** $y$ into $L = 2^n$ levels and encode into $n$ bits (8 bits in telephony).
4. At the receiver **decode** and **expand** with the inverse law:
$$
|x| = \frac{(1 + \mu)^{|y|} - 1}{\mu}
$$
5. The combined effect: equal steps in $y$ correspond to small steps in $x$ near zero and large steps near $\pm 1$.

For $\mu \gg 1$, SQNR $\approx \dfrac{3L^2}{[\ln(1 + \mu)]^2}$, almost independent of signal level; with $\mu = 255$, $n = 8$ it is about 38 dB over a 40 dB input range.

In practice the curve is approximated digitally by **15 linear segments** (µ-law, T1 systems) or 13 segments for the A-law ($A = 87.6$, E1 systems).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does uniform quantization perform poorly for weak speech?
   A. The step is too large for loud sounds
   B. Noise power is fixed so SQNR is proportional to signal power
   C. It uses too many bits
   D. It causes aliasing
   Answer: B. A constant Delta^2/12 noise is relatively large against a weak signal. (Key point: Need)

2. The correct order of blocks for companding is:
   A. Expander, quantizer, compressor
   B. Compressor, uniform quantizer, channel, decoder, expander
   C. Quantizer, compressor, expander
   D. Compressor, expander, quantizer
   Answer: B. The transmitter compresses then quantizes uniformly; the receiver decodes then expands. (Key point: Companding chain)

3. The mu-law compressor is:
   A. y = sgn(x) ln(1+mu|x|)/ln(1+mu)
   B. y = mu x
   C. y = sgn(x) |x|^mu
   D. y = sgn(x) (1+mu|x|)/(1+mu)
   Answer: A. It is logarithmic for large inputs and near-linear with large gain for small ones. (Key point: mu-law formula)

4. Equal steps in the compressed variable y correspond to what steps in x?
   A. Equal everywhere
   B. Small near zero and large near +/-1
   C. Large near zero and small near +/-1
   D. Random
   Answer: B. The compressor stretches small amplitudes, so its inverse gives fine steps near zero. (Key point: Effect on steps)

## Related in Communication Systems

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