---
title: "Required SNR for 40 Mbps over 6 MHz"
subject: "Data Communication"
chapter: 1
key: "semester_5/data_communication:capacity-40mbps-6mhz"
source: https://www.bctengineer.com/q/semester_5/data_communication/capacity-40mbps-6mhz
site: BCT Engineering
---

# Required SNR for 40 Mbps over 6 MHz

Data Communication · Chapter 1: Introduction

- Asked in: 2071 Chaitra, 3+2 marks
- Video: [L17: Nyquist Limit Theorem, Shannon's Channel Capacity, Maximum Bit Rate with Example | DCN Lectures](https://www.youtube.com/watch?v=0O6jxk5i_Lc) — Easy Engineering Classes
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/capacity-40mbps-6mhz
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-1

## Question

Given a channel with an intended capacity of 40 Mbps. The bandwidth of the channel is 6 MHz. What signal-to-noise ratio is required in order to achieve this capacity? Also find number of bits/sample if channel becomes noiseless.

## Short answer

Shannon gives S/N = 2^(40/6) - 1 ≈ 100.6, or 20 dB. For a noiseless channel, Nyquist yields log₂L = C/(2B) = 40/12 ≈ 3.33 bits/sample, L ≈ 10 levels.

## Answer

#### Required SNR (Shannon)
C = B log₂(1 + S/N)

40 × 10⁶ = 6 × 10⁶ × log₂(1 + S/N)

log₂(1 + S/N) = 40/6 = 6.667

1 + S/N = 2^6.667 = 101.59

S/N = 100.59

In dB: SNR = 10 log₁₀(100.59) = 20.03 dB

**Answer: S/N ≈ 100.6, that is about 20 dB.**

#### Bits per sample if the channel is noiseless (Nyquist)
C = 2B log₂L, so log₂L = C/2B = 40×10⁶ / (12×10⁶) = 3.33

L = 2^3.33 = 10.08 levels

**Answer: about 3.33 bits per sample.** Since L must be a whole number in practice, 3 bits (L = 8) would give only 36 Mbps and 4 bits (L = 16) would give 48 Mbps; L ≈ 10 is the exact figure for 40 Mbps.

## Check yourself

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

1. For C = 40 Mbps, B = 6 MHz, 1 + S/N equals
   A. 40/6
   B. 2^6
   C. 40 x 6
   D. 2^(40/6) = 101.6
   Answer: D. log2(1+S/N) = 6.667. (Key point: Required SNR)

2. The required SNR is about
   A. 20 dB
   B. 100 dB
   C. 6.7 dB
   D. 40 dB
   Answer: A. 10 log10(100.6) = 20.03 dB. (Key point: SNR in dB)

3. In the noiseless case, bits per sample is
   A. 6.67
   B. 3.33
   C. 20
   D. 1.67
   Answer: B. log2 L = C/2B = 40/12 = 3.33. (Key point: Noiseless bits)

4. 3 bits (L = 8) gives
   A. 40 Mbps
   B. 48 Mbps
   C. 36 Mbps
   D. 24 Mbps
   Answer: C. 2 x 6 x 3 = 36 Mbps. (Key point: Whole levels)

## Related in Data Communication

- [Capacity and Nyquist levels, 3 to 4 MHz, 24 dB](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Anyquist-shannon-bandwidth-3-4mhz) — Data Communication, chapter 1
- [Nyquist levels for 3 to 4 MHz channel](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Anyquist-noiseless-24db) — Data Communication, chapter 1
- [Capacity for 3.1 kHz, SNR 0 dB](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acapacity-3-1khz-0db) — Data Communication, chapter 1
- [Channel capacity for 6 kHz, 6 dB](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acapacity-6khz-6db) — Data Communication, chapter 1
- [Shannon-Hartley channel capacity theorem](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Ashannon-hartley-theorem) — Data Communication, chapter 1, asked 3×
- [Entropy, information and channel capacity](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Ainformation-entropy-capacity) — Data Communication, chapter 8

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