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

# Need for non-uniform quantization; SQNR of DM

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2076 Chaitra (CS II), 2+5 marks; 2073 Shrawan (CS II), 2+6 marks (asked 2 times)
- 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-dm-sqnr
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Explain the need for non uniform quantization. Determine the SQNR for delta modulation with no slope overload condition.

## Short answer

Non-uniform quantization needed because uniform steps give poor SQNR for weak signals; companding provides constant SQNR over wide dynamic range with fewer bits. For delta modulation without slope overload, SQNR = 3fs³/(8π²fm²W) or 30log fs - 20log fm - 10log W - 14.2 dB.

## Answer

#### Need for non-uniform quantization

In uniform quantization the step size $\Delta$ is fixed, so the quantization noise power $\Delta^2/12$ is the same for all signal levels. Hence:

- Weak signals (low amplitude) get a **poor SQNR**, while loud signals get a high SQNR.
- Speech has a wide dynamic range (about 40 dB) and small amplitudes occur much more often than large ones.
- To keep weak signals acceptable with uniform steps, many more bits would be needed.

**Non-uniform quantization** uses small steps for small amplitudes and large steps for large amplitudes. This gives nearly **constant SQNR** over a wide range of input levels with fewer bits (8 bits instead of about 12 for telephony). It is implemented by **companding** ($\mu$-law, A-law).

#### SQNR of delta modulation (no slope overload)

Let the message be a single tone $m(t) = A_m\cos 2\pi f_m t$, step size $\Delta$, sampling rate $f_s = 1/T_s$, and the receiver LPF bandwidth $W$.

**1. Condition for no slope overload.** The staircase can rise by at most $\Delta$ per $T_s$:

$$
\left|\frac{dm}{dt}\right|_{max} = 2\pi f_m A_m \le \frac{\Delta}{T_s} \;\Rightarrow\; A_{m,max} = \frac{\Delta f_s}{2\pi f_m}
$$

**2. Signal power** (with maximum amplitude):

$$
S = \frac{A_m^2}{2} = \frac{\Delta^2 f_s^2}{8\pi^2 f_m^2}
$$

**3. Granular noise power.** With no slope overload, the error $q(t)$ lies within $\pm\Delta$ and is uniformly distributed:

$$
E[q^2] = \frac{1}{2\Delta}\int_{-\Delta}^{\Delta} q^2\,dq = \frac{\Delta^2}{3}
$$

This power is spread uniformly over $0$ to $f_s$. The output LPF passes only $W$, so

$$
N_q = \frac{\Delta^2}{3}\cdot\frac{W}{f_s}
$$

**4. SQNR:**

$$
\text{SQNR} = \frac{S}{N_q} = \frac{\Delta^2 f_s^2/(8\pi^2 f_m^2)}{\Delta^2 W/(3f_s)} = \frac{3 f_s^3}{8\pi^2 f_m^2 W}
$$

In dB:

$$
\text{SQNR}_{dB} = 30\log_{10} f_s - 20\log_{10} f_m - 10\log_{10} W - 14.2
$$

So SQNR improves by 9 dB each time $f_s$ is doubled. Example: $f_s = 64$ kHz, $f_m = 1$ kHz, $W = 4$ kHz gives SQNR $= 2490 \approx 33.96$ dB.

## Check yourself

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

1. Why is uniform quantization poor for speech?
   A. Noise grows with signal amplitude so loud signals suffer
   B. It needs fewer bits than non-uniform
   C. Noise power Delta^2/12 is the same for all levels, so weak signals get poor SQNR
   D. It cannot represent negative values
   Answer: C. Fixed step gives fixed noise, so weak (frequent) speech amplitudes have low SQNR. (Key point: Need for non-uniform)

2. What is the condition for no slope overload in DM for a tone?
   A. Am <= Delta / fs only
   B. fm >= fs
   C. 2 pi fm Am <= Delta fs
   D. 2 pi fm Am >= Delta fs
   Answer: C. The staircase can rise at most Delta per Ts, i.e. slope Delta*fs, which must exceed the maximum tone slope. (Key point: No slope overload)

3. In the DM SQNR derivation, what is the total granular noise power and how much reaches the output?
   A. Delta^2/12 times fs/W
   B. Delta^2/12; all of it reaches the output
   C. Delta^2/3 spread over 0 to fs; the LPF passes the fraction W/fs
   D. Delta^2/3; all of it reaches the output
   Answer: C. Error is uniform in +-Delta giving Delta^2/3, and only W/fs of it falls in the LPF band. (Key point: Granular noise power)

4. What happens to DM SQNR when fs is doubled?
   A. It is unchanged
   B. It rises by 3 dB
   C. It rises by 9 dB since SQNR is proportional to fs^3
   D. It rises by 6 dB
   Answer: C. SQNR = 3 fs^3/(8 pi^2 fm^2 W); doubling fs multiplies by 8, about 9 dB. (Key point: DM SQNR result)

## Related in Communication Systems

- [Need for non-uniform quantization; one algorithm](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-quantization-algorithm) — 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
- [SQNR for uniform quantization in terms of levels and bits](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asqnr-uniform-levels-bits) — Communication Systems, chapter 4, asked 2×
- [SQNR of delta modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-sqnr-derive) — Communication Systems, chapter 4, asked 2×
- [Aperture effect, DPCM and DM slope overload numerical](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aaperture-effect-dpcm-dm-slope) — Communication Systems, chapter 4
- [Delta modulation encoder/decoder; PCM vs DM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-encoder-decoder-pcm-vs-dm) — Communication Systems, chapter 4

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