---
title: "Capacity and Nyquist levels, 3 to 4 MHz, 24 dB"
subject: "Data Communication"
chapter: 1
key: "semester_5/data_communication:nyquist-shannon-bandwidth-3-4mhz"
source: https://www.bctengineer.com/q/semester_5/data_communication/nyquist-shannon-bandwidth-3-4mhz
site: BCT Engineering
---

# Capacity and Nyquist levels, 3 to 4 MHz, 24 dB

Data Communication · Chapter 1: Introduction

- Asked in: 2069 Chaitra, 2+2 marks
- Video: [Nyquist Bit Rate & Shannon Capacity | Signal-to-Noise Ratio | Computer Networks Course | Lecture 18](https://www.youtube.com/watch?v=_LxyNd-pDdk) — Mind Your Exam Channel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/nyquist-shannon-bandwidth-3-4mhz
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-1

## Question

State Nyquist's and Shannon's channel capacity formula. The spectrum of a channel is in between 3 MHz and 4 MHz. The SNR = 24 dB. Calculate the bandwidth, the maximum channel capacity using Shannon's formula. Then based on Nyquist's formula, find the number of signal levels required to reach the maximum channel capacity.

## Short answer

Bandwidth B = 4-3 = 1 MHz. Shannon capacity with SNR = 24 dB (S/N ≈ 251) is C = 1×10⁶ · log₂(252.19) ≈ 7.98 Mbps. For Nyquist, log₂L = 7.98/2 ≈ 3.99, so L ≈ 16 levels (4 bits per sample).

## Answer

**Formulas**
- Nyquist (noiseless): C = 2B log₂L
- Shannon (noisy): C = B log₂(1 + S/N)

**Given**: spectrum 3 MHz to 4 MHz, SNR = 24 dB.

1. Bandwidth B = 4 - 3 = **1 MHz**.
2. S/N = 10^(24/10) = 251.19
3. Maximum capacity C = 10⁶ × log₂(1 + 251.19) = 10⁶ × log₂(252.19) = **7.98 Mbps** (≈ 8 Mbps).
4. Number of levels: C = 2B log₂L, so log₂L = 7.98×10⁶ / (2×10⁶) = 3.99, and L = 2^3.99 = 15.9 ≈ **16 levels**.

**Answer: B = 1 MHz, C ≈ 7.98 Mbps, L ≈ 16 signal levels (4 bits per level).**

## Check yourself

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

1. B for 3 to 4 MHz is
   A. 7 MHz
   B. 4 MHz
   C. 1 MHz
   D. 3.5 MHz
   Answer: C. Upper minus lower. (Key point: Bandwidth)

2. Max capacity is about
   A. 24 Mbps
   B. 251 Mbps
   C. 1 Mbps
   D. 7.98 Mbps
   Answer: D. 10^6 x log2(252.19). (Key point: Shannon capacity)

3. Number of levels is about
   A. 16
   B. 8
   C. 32
   D. 252
   Answer: A. log2 L = 3.99. (Key point: Levels)

4. Nyquist is applied using
   A. bandwidth only
   B. C from Shannon
   C. SNR in dB
   D. N0
   Answer: B. L = 2^(C/2B). (Key point: Method)

## Related in Data Communication

- [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
- [Required SNR for 40 Mbps over 6 MHz](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acapacity-40mbps-6mhz) — 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
- [State Nyquist and Shannon capacity formulae](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Anyquist-shannon-capacity-formula) — Data Communication, chapter 1, asked 2×
- [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×

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- [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
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