---
title: "PCM SQNR increases 6 dB per bit"
subject: "Communication Systems"
chapter: 4
key: "semester_6/communication_systems:pcm-sqnr-6db-per-bit"
source: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-sqnr-6db-per-bit
site: BCT Engineering
---

# PCM SQNR increases 6 dB per bit

Communication Systems · Chapter 4: Pulse Modulation

- Asked in: 2075 Asoj (CS II), 6 marks
- Video: [SNR in PCM | Quantization noise in PCM | Sgnal to noise ratio in PCM](https://www.youtube.com/watch?v=NHlhFWeFebM) — Techno Tutorials ( e-Learning)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/pcm-sqnr-6db-per-bit
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-4

## Question

With necessary derivations show that in case of PCM, SQNR increases approximately by 6dB for each extra bit used.

## Short answer

For PCM with n bits, L=2^n levels, step Δ=2mp/L. Quantization noise power Nq=Δ²/12. For full-load sinusoid, signal power S=mp²/2. SQNR = S/Nq = 1.5L² = 1.5×2^(2n). In dB: SQNRdB = 1.76 + 6.02n dB. Increasing n by 1 adds 6.02 dB ≈ 6 dB because L doubles, Δ halves, Nq drops by factor 4 (6 dB). This holds for any signal with constant peak-to-power ratio.

## Answer

In PCM with $n$ bits per sample, the quantizer has $L = 2^n$ levels. We find SQNR in terms of $n$.

1. **Step size** for a message in the range $\pm m_p$:
$$
\Delta = \frac{2m_p}{L} = \frac{2m_p}{2^n}
$$
2. **Quantization noise**: the error $q$ is uniformly distributed in $(-\Delta/2, \Delta/2)$:
$$
N_q = \int_{-\Delta/2}^{\Delta/2} q^2\frac{1}{\Delta}\,dq = \frac{\Delta^2}{12} = \frac{m_p^2}{3L^2}
$$
3. **Signal power**: take a full-load sinusoid $m(t) = m_p\cos\omega_mt$, so $S = m_p^2/2$.
4. **SQNR**:
$$
\begin{aligned}
\text{SQNR} &= \frac{S}{N_q} = \frac{m_p^2/2}{m_p^2/(3L^2)} = \frac{3}{2}L^2 = \frac{3}{2}\,2^{2n} \\
\text{SQNR}_{dB} &= 10\log_{10}\frac{3}{2} + 10\log_{10}2^{2n} \\
&= 1.76 + 20n\log_{10}2 \\
&= 1.76 + 6.02n\ \text{dB}
\end{aligned}
$$
5. **Effect of one extra bit**: going from $n$ to $n + 1$ bits,
$$
\text{SQNR}_{dB}(n + 1) - \text{SQNR}_{dB}(n) = 6.02(n + 1) - 6.02n = 6.02 \approx 6\ \text{dB}
$$
In ratio terms, $L$ doubles, $\Delta$ halves, and $N_q = \Delta^2/12$ falls by a factor of 4, i.e. $10\log_{10}4 = 6.02$ dB.

For a general signal, $\text{SQNR} = 3L^2\,\overline{m^2}/m_p^2$, i.e. $\text{SQNR}_{dB} = \alpha + 6.02n$, where $\alpha$ depends only on the signal's power-to-peak ratio. The 6 dB/bit rule holds for any signal.

**Price paid:** each extra bit increases bit rate $R_b = nf_s$ and transmission bandwidth $B_T \approx nW$ in proportion, so PCM trades bandwidth for SNR exponentially: SQNR $\propto 2^{2B_T/W}$.

Example: 7 bits give $1.76 + 42.1 = 43.9$ dB; 8 bits give $49.9$ dB, about 6 dB more.

## Check yourself

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

1. The quantization noise power N_q = Delta^2/12 with Delta = 2mp/L equals:
   A. mp^2/(3L^2)
   B. mp^2/(12L)
   C. 4mp^2/L
   D. mp^2/L^2
   Answer: A. Delta^2/12 = (4mp^2/L^2)/12 = mp^2/(3L^2). (Key point: Noise power)

2. For a full-load sinusoid, SQNR in dB is:
   A. 6.02 + 1.76n
   B. 1.76 + 6.02n
   C. 3.01n
   D. 1.5 + 6n^2
   Answer: B. SQNR = 1.5 x 2^(2n), so dB = 1.76 + 20n log10(2) = 1.76 + 6.02n. (Key point: SQNR in dB)

3. Adding one bit increases SQNR by about 6 dB because:
   A. L doubles, Delta halves, and N_q falls by a factor of 4
   B. N_q falls by a factor of 2
   C. The signal power doubles
   D. Bandwidth halves
   Answer: A. Noise power is proportional to Delta^2, so halving Delta gives a factor 4, i.e. 6.02 dB. (Key point: Why 6 dB)

4. The 6 dB per bit rule applies to:
   A. Only sinusoidal signals
   B. Any signal, since only the constant alpha depends on power-to-peak ratio
   C. Only speech
   D. Only 8-bit PCM
   Answer: B. SQNR_dB = alpha + 6.02n, where alpha depends on the signal; the slope is the same. (Key point: Generality)

5. What is the cost of using more bits per sample?
   A. Lower bit rate
   B. Proportionally higher bit rate and bandwidth
   C. Less SQNR
   D. None
   Answer: B. Rb = n fs and BT ~ nW, so bandwidth grows with n. (Key point: Price of extra bits)

## Related in Communication Systems

- [E1 hierarchy; PCM SQNR at 32 kbps](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-pcm-4khz-1-1v-32kbps) — Communication Systems, chapter 5
- [SQNR of uniformly quantized PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Asqnr-uniform-pcm-bits) — Communication Systems, chapter 4
- [E1 digital hierarchy; SQNR in uniformly quantized PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-hierarchy-sqnr-uniform-pcm) — Communication Systems, chapter 5, asked 2×
- [PCM, DPCM and DM comparison; SQNR of PCM](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-dpcm-dm-compare-sqnr) — Communication Systems, chapter 4
- [PCM vs DM SQNR for 6cos(5000πt) with 128 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apcm-dm-6cos5000pit-128) — 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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