---
title: "Uniform vs non-uniform quantization; companding laws"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:uniform-nonuniform-companding-laws"
source: https://www.bctengineer.com/q/semester_6/communication_systems/uniform-nonuniform-companding-laws
site: BCT Engineering
---

# Uniform vs non-uniform quantization; companding laws

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2075 Asoj (CS II), 2+2+2 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/uniform-nonuniform-companding-laws
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Differentiate between uniform quantization and non-uniform quantization. Why is non-uniform quantization done for speech signal? Explain about companding laws.

## Short answer

Uniform quantization has equal step size, constant noise power, but SQNR falls for weak signals. Non-uniform uses small steps near zero, large steps at high amplitude, giving nearly constant SQNR. Speech has wide dynamic range and peaked PDF near zero, so non-uniform quantization (companding) maintains quality with fewer bits. µ-law (μ=255) and A-law (A=87.6) compress logarithmically for large amplitudes, expand at receiver. Both are implemented as piecewise linear segments.

## Answer

#### Uniform vs non-uniform quantization

| Point | Uniform | Non-uniform |
|---|---|---|
| Step size | Equal everywhere | Small near zero, large at high amplitude |
| Quantization noise | Same $\Delta^2/12$ for all levels | Small for weak, larger for strong signals |
| SQNR | Falls for weak signals | Nearly constant over range |
| Implementation | Simple ADC | Compressor + uniform ADC + expander |
| Suited to | Signals with uniform pdf | Speech, wide dynamic range |

#### Why non-uniform quantization for speech
- Speech has a large dynamic range (about 40 dB between loud and soft talkers).
- Its amplitude pdf is peaked near zero: small amplitudes occur most often.
- With uniform steps, SQNR $\propto$ signal power, so quiet talkers and soft sounds get poor SQNR.
- Non-uniform steps give small steps where samples are most likely, so SQNR becomes almost independent of level, and 8 bits give quality that would need about 12 bits uniformly.

#### Companding laws
Companding = compressing at the transmitter + expanding at the receiver. With $x = m/m_p$ normalised input ($|x| \le 1$):

**µ-law** (North America, Japan; $\mu = 255$):
$$
y = \text{sgn}(x)\,\frac{\ln(1 + \mu|x|)}{\ln(1 + \mu)}
$$

**A-law** (Europe, ITU-T E1 systems; $A = 87.6$):
$$
y = \begin{cases} \text{sgn}(x)\dfrac{A|x|}{1 + \ln A}, & 0 \le |x| \le \dfrac{1}{A} \\[2mm] \text{sgn}(x)\dfrac{1 + \ln(A|x|)}{1 + \ln A}, & \dfrac{1}{A} \le |x| \le 1 \end{cases}
$$

- $\mu = 0$ or $A = 1$ gives uniform quantization.
- Both laws are nearly linear for small inputs and logarithmic for large inputs.
- In practice they are implemented as 13-segment (A-law) or 15-segment (µ-law) piecewise linear approximations.

## Check yourself

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

1. In uniform quantization the noise power is the same for every signal, so:
   A. SQNR is constant
   B. SQNR falls for weak signals
   C. SQNR is infinite for weak signals
   D. SQNR does not depend on L
   Answer: B. Noise is fixed at Delta^2/12, so a smaller signal power gives a lower SQNR. (Key point: Uniform quantization weakness)

2. Why is non-uniform quantization used for speech?
   A. Speech is always loud
   B. Speech has a wide dynamic range and small amplitudes are most probable
   C. Speech is band-limited to 100 Hz
   D. Uniform quantizers cannot be built
   Answer: B. Small steps near zero serve the most likely samples and make SQNR nearly level-independent. (Key point: Why for speech)

3. The A-law is linear for small inputs and logarithmic for large ones; the break is at |x| equal to:
   A. A
   B. 1/A
   C. 1/(1+ln A)
   D. mu
   Answer: B. For |x| <= 1/A the curve is A|x|/(1+ln A), and above 1/A it is logarithmic. (Key point: A-law region)

4. Which pairing of law and region is correct?
   A. mu-law for Europe, A-law for North America
   B. mu-law (mu = 255) for North America/Japan, A-law (A = 87.6) for Europe/E1
   C. Both only for Japan
   D. A-law uses mu = 255
   Answer: B. mu-law is used in North America/Japan and A-law in ITU-T E1 systems. (Key point: Standards)

## Related in Communication Systems

- [Uniform vs non-uniform quantization; 16-level PCM numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Auniform-nonuniform-16-level-pcm) — Communication Systems, chapter 4
- [Non-uniform quantization; PCM numerical (4 kHz, ±2.4 V, 64 kHz)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-companding-pcm-4khz-64khz) — Communication Systems, chapter 4
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