---
title: "SQNR for uniform quantization in terms of levels and bits"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:sqnr-uniform-levels-bits"
source: https://www.bctengineer.com/q/semester_6/communication_systems/sqnr-uniform-levels-bits
site: BCT Engineering
---

# SQNR for uniform quantization in terms of levels and bits

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2074 Asoj (CS II), 7 marks; 2070 Asar (CS II), 8 marks (asked 2 times)
- Video: [SNR of Quantization | Waveforms of Quantization | Derivation of SNR in Quantization | Quantization](https://www.youtube.com/watch?v=lGZhdKUGFoU) — 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/sqnr-uniform-levels-bits
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

Derive expression for evaluating signal-to-quantization noise ratio (SQNR) for uniform quantization in terms of number of levels and number of bits per source symbol.

## Short answer

For uniform quantization with L levels over [-mp, mp], step size Δ=2mp/L, noise power Δ²/12. SQNR = 3L²P/mp² = (3P/mp²)4ⁿ where n=log₂L bits. For full-scale sinusoid, SQNR = 1.5L² = 1.5×2²ⁿ, SQNRdB = 1.76 + 6.02n dB.

## Answer

**SQNR** is the ratio of signal power to quantization noise power at the output of a quantizer.

#### Step 1: Step size

Let the message $m(t)$ lie in $[-m_p, m_p]$ and be quantized uniformly into $L$ levels. Then the step size is

$$
\Delta = \frac{2m_p}{L}
$$

#### Step 2: Quantization noise power

The quantization error $q = m - m_q$ lies in $[-\Delta/2, \Delta/2]$. For a fine quantizer it is uniformly distributed, with pdf $f(q) = 1/\Delta$:

$$
\begin{aligned}
N_q = E[q^2] &= \int_{-\Delta/2}^{\Delta/2} q^2\cdot\frac{1}{\Delta}\,dq = \frac{1}{\Delta}\left[\frac{q^3}{3}\right]_{-\Delta/2}^{\Delta/2} \\
&= \frac{\Delta^2}{12} = \frac{(2m_p/L)^2}{12} = \frac{m_p^2}{3L^2}
\end{aligned}
$$

#### Step 3: SQNR in terms of levels

With message power $P = \overline{m^2(t)}$:

$$
\text{SQNR} = \frac{P}{N_q} = \frac{3L^2\, P}{m_p^2}
$$

#### Step 4: SQNR in terms of bits

With binary coding of $n$ bits per sample, $L = 2^n$:

$$
\text{SQNR} = \frac{3P}{m_p^2}\,2^{2n} = \frac{3P}{m_p^2}\,4^n
$$

$$
\text{SQNR}_{dB} = 10\log_{10}\!\left(\frac{3P}{m_p^2}\right) + 6.02\,n\ \text{dB}
$$

#### Special case: full-scale sinusoid

For $m(t) = m_p\cos\omega_m t$, $P = m_p^2/2$:

$$
\text{SQNR} = \frac{3L^2}{2} = 1.5L^2 = 1.5\times 2^{2n}
$$

$$
\text{SQNR}_{dB} = 1.76 + 6.02\,n\ \text{dB}
$$

Example: $n = 8$ bits gives SQNR $= 1.76 + 48.16 = 49.92$ dB.

**Conclusions:**

- SQNR increases as the **square of the number of levels**.
- SQNR increases **exponentially with bits**: each extra bit adds about **6 dB**.
- But each extra bit increases the bit rate $n f_s$ and the bandwidth $n f_s/2$, so there is a trade-off between SQNR and bandwidth.

## Check yourself

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

1. For a message in [-mp, mp] with L levels, what is the step size?
   A. 2mp/L
   B. mp/(2L)
   C. mp/L
   D. L/(2mp)
   Answer: A. The total range 2mp is divided into L equal steps. (Key point: Step size)

2. What is the quantization noise power for uniform error in (-Delta/2, Delta/2)?
   A. Delta^2/12 = mp^2/(3L^2)
   B. Delta^2/3
   C. Delta^2/6
   D. Delta/12
   Answer: A. Integrating q^2/Delta over the range gives Delta^2/12. (Key point: Noise power)

3. What is SQNR in dB for n-bit coding?
   A. 10 log(P/mp^2) + 3.01 n
   B. 6.02 n only, for any signal
   C. 10 log(3P/mp^2) + 6.02 n
   D. 20 log(3P/mp^2) + 3 n
   Answer: C. SQNR = (3P/mp^2) 2^(2n), so each bit adds 20 log 2 = 6.02 dB. (Key point: SQNR in bits)

4. What is SQNR in dB for a full-scale sinusoid?
   A. 1.76 + 3.01 n
   B. 6.02 n
   C. 1.76 + 6.02 n
   D. 4.77 + 6.02 n
   Answer: C. P = mp^2/2 gives SQNR = 1.5 L^2, i.e. 1.76 + 6.02n dB. (Key point: Full-scale sine)

## Related in Communication Systems

- [Quantization; SQNR for linear quantization](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aquantization-sqnr-linear) — Communication Systems, chapter 4, asked 2×
- [Need for non-uniform quantization; SQNR of DM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anonuniform-quantization-dm-sqnr) — Communication Systems, chapter 4, asked 2×
- [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
- [Bits per sample and information rate (4.5 kHz, 8 levels)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Abits-info-rate-4-5khz-8-levels) — Communication Systems, chapter 6
- [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
- [SQNR of delta modulation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adm-sqnr-derive) — Communication Systems, chapter 4, asked 2×

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