---
title: "Non-uniform quantization; PCM numerical (4 kHz, ±2.4 V, 64 kHz)"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:nonuniform-companding-pcm-4khz-64khz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/nonuniform-companding-pcm-4khz-64khz
site: BCT Engineering
---

# Non-uniform quantization; PCM numerical (4 kHz, ±2.4 V, 64 kHz)

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2072 Kartik (CS II), 3+3+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/nonuniform-companding-pcm-4khz-64khz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Explain basic process of Non-uniform quantization including companding technique of its realization. An audio signal of frequency 4 kHz and maximum dynamic range of ±2.4 V is digitized by PCM system with its bit rate of 64 kHz. Calculate numbers of bits per sample, quantization noise power and SQNR_dB. Estimate the minimum bandwidth required for TDM of 10 such audio signals (assume no extra framing and synchronization bits).

## Short answer

Non-uniform quantization uses small steps for low amplitudes and large steps for high amplitudes, realized by companding (compressor + uniform quantizer + expander). For 4 kHz audio at 64 kbps, fs = 8 kHz, bits per sample = 8, quantization noise power Nq = 2.93×10⁻⁵ V², SQNR = 49.93 dB. TDM of 10 signals requires minimum bandwidth 320 kHz.

## Answer

#### Non-uniform quantization and companding
In non-uniform quantization the step size is **small for low amplitudes and large for high amplitudes**. Since speech has mostly small amplitudes and a wide dynamic range, this keeps SQNR nearly constant for loud and soft talkers.

It is realised by **companding**:
```text
m(t)->[Compressor]->[Uniform Q + encoder]->channel
out <-[Expander]<---[Decoder]<-------------+
```
1. **Compressor** applies a logarithmic gain: weak signals amplified more than strong ones. µ-law: $y = \text{sgn}(x)\dfrac{\ln(1 + \mu|x|)}{\ln(1 + \mu)}$, $\mu = 255$; or A-law with $A = 87.6$.
2. **Uniform quantizer** then acts on the compressed signal; equal steps in $y$ are small steps in $x$ near zero.
3. **Expander** at the receiver applies the inverse law, restoring the original amplitude relation.
In practice the curve is a 13-segment (A-law) or 15-segment (µ-law) piecewise linear approximation.

#### Numerical
**Given:** $W = 4$ kHz, range $\pm 2.4$ V ($m_p = 2.4$ V), $R_b = 64$ kbps. Assume sampling at the Nyquist rate and a full-scale sinusoidal signal.

**Bits per sample**
$$
\begin{aligned}
f_s &= 2W = 8\ \text{kHz} \\
n &= \frac{R_b}{f_s} = \frac{64\,000}{8000} = 8\ \text{bits}, \quad L = 2^8 = 256
\end{aligned}
$$

**Quantization noise power**
$$
\begin{aligned}
\Delta &= \frac{2m_p}{L} = \frac{4.8}{256} = 0.01875\ \text{V} \\
N_q &= \frac{\Delta^2}{12} = \frac{(0.01875)^2}{12} = 2.93 \times 10^{-5}\ \text{V}^2
\end{aligned}
$$

**SQNR**
$$
\begin{aligned}
S &= \frac{m_p^2}{2} = \frac{2.4^2}{2} = 2.88\ \text{V}^2 \\
\text{SQNR} &= \frac{2.88}{2.93 \times 10^{-5}} = 98\,304 \\
\text{SQNR}_{dB} &= 10\log_{10}98\,304 = 49.93\ \text{dB}
\end{aligned}
$$
(Check: $1.76 + 6.02 \times 8 = 49.92$ dB.)

**Minimum bandwidth for TDM of 10 signals**
$$
\begin{aligned}
R_{b,total} &= 10 \times 64 = 640\ \text{kbps} \\
B_{T,min} &= \frac{R_{b,total}}{2} = 320\ \text{kHz}
\end{aligned}
$$

**Answer: $n = 8$ bits; $N_q = 2.93 \times 10^{-5}$ V² (normalised to 1 Ω); SQNR = 49.93 dB; $B_T = 320$ kHz.**

## Check yourself

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

1. Why is non-uniform quantization used for speech?
   A. Speech has mostly small amplitudes and wide dynamic range, so it keeps SQNR nearly constant
   B. It uses fewer samples
   C. It removes aliasing
   D. It reduces the sampling rate
   Answer: A. Finer steps at low amplitudes give nearly constant SQNR for loud and soft talkers. (Key point: Why non-uniform)

2. The compressor in companding:
   A. Amplifies strong signals more than weak ones
   B. Applies a logarithmic law amplifying weak signals more than strong ones
   C. Is placed at the receiver
   D. Applies a linear gain
   Answer: B. The mu-law/A-law compressor is logarithmic, so equal steps after it are small steps at low input levels. (Key point: Compressor action)

3. With W = 4 kHz and Rb = 64 kbps, bits per sample are:
   A. 4
   B. 16
   C. 6
   D. 8
   Answer: D. fs = 8 kHz, n = 64000/8000 = 8. (Key point: Bits per sample)

4. For +/-2.4 V and 8 bits, the step size and noise power are:
   A. 0.01875 V and 2.93x10^-5 V^2
   B. 0.0094 V and 7.3x10^-6 V^2
   C. 0.01875 V and 0.0188 V^2
   D. 0.15 V and 1.9x10^-3 V^2
   Answer: A. Delta = 4.8/256 = 0.01875 V and Nq = Delta^2/12 = 2.93x10^-5. (Key point: Quantization noise)

## 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
- [Uniform vs non-uniform quantization; companding laws](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Auniform-nonuniform-companding-laws) — Communication Systems, chapter 4
- [PCM bit rate for 4.5 MHz TV signal with 1024 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-tv-4-5mhz-1024) — Communication Systems, chapter 4
- [Companding; delta modulator numerical (5 kHz, 200 kHz)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Acompanding-dm-5khz-200khz) — Communication Systems, chapter 4, asked 2×
- [PCM 7-bit at 50 Mbps: maximum message bandwidth](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-7bit-50mbps-bandwidth) — Communication Systems, chapter 4
- [Non-uniform quantization using uniform quantizer; μ-law](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-quantization-mu-law) — Communication Systems, chapter 4