---
title: "Nyquist levels for 3 to 4 MHz channel"
subject: "Data Communication"
chapter: 1
key: "semester_5/data_communication:nyquist-noiseless-24db"
source: https://www.bctengineer.com/q/semester_5/data_communication/nyquist-noiseless-24db
site: BCT Engineering
---

# Nyquist levels for 3 to 4 MHz channel

Data Communication · Chapter 1: Introduction

- Asked in: 2078 Kartik, 5+5 marks
- Video: [DATA RATE LIMITS | Nyquist Bit Rate, Shannon Capacity](https://www.youtube.com/watch?v=A4Oz7Gnsl5E) — Prachi Surlakar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/nyquist-noiseless-24db
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-1

## Question

How is the Nyquist theorem applied for a noiseless channel? Calculate the number of discrete signal levels in the channel if a channel has a spectrum of 3 to 4 MHz with signal to noise ratio of 24 dB.

## Short answer

Bandwidth B = 1 MHz, SNR = 24 dB gives S/N ≈ 251.2, Shannon capacity C = 7.98 Mbps. Using Nyquist, C = 2B log2 L → log2 L = 3.989, so L ≈ 16 discrete signal levels (4 bits/symbol).

## Answer

#### Nyquist theorem for a noiseless channel
For a noiseless channel, the maximum bit rate depends only on the bandwidth B and the number of signal levels L:

**C = 2B log₂L** bps

At most 2B symbols per second can be sent without inter-symbol interference. By using more levels, each symbol carries log₂L bits, so the capacity rises. In a real channel, noise limits L, so Shannon's formula is used to find the practical limit.

#### Numerical
Channel spectrum 3 MHz to 4 MHz, so B = 4 - 3 = 1 MHz. SNR = 24 dB.

1. S/N = 10^(24/10) = 251.19
2. Shannon capacity: C = B log₂(1 + S/N) = 10⁶ × log₂(252.19) = 10⁶ × 7.978 = 7.978 × 10⁶ bps
3. Nyquist: C = 2B log₂L, so
   log₂L = C / 2B = 7.978×10⁶ / (2×10⁶) = 3.989
   L = 2^3.989 = 15.88 ≈ 16

**Answer: about 16 signal levels (4 bits per level); capacity ≈ 7.98 Mbps.**

## Check yourself

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

1. A channel spans 3 MHz to 4 MHz. What is B?
   A. 4 MHz
   B. 7 MHz
   C. 3.5 MHz
   D. 1 MHz
   Answer: D. B = upper - lower = 1 MHz. (Key point: Bandwidth)

2. 24 dB corresponds to S/N of about
   A. 251
   B. 24
   C. 2.4
   D. 2400
   Answer: A. 10^(24/10) = 251.19. (Key point: dB to ratio)

3. How is L found?
   A. Nyquist gives C, then Shannon gives L
   B. Shannon gives C, then Nyquist log2 L = C/2B
   C. L = S/N
   D. L = C/B
   Answer: B. Shannon capacity is used in Nyquist's formula to find L. (Key point: Method)

4. The number of levels is about
   A. 4
   B. 251
   C. 16
   D. 8
   Answer: C. log2 L = 7.978/2 = 3.99, so L is about 16. (Key point: Result)

## 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
- [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

## Related in other subjects

- [PAM, PWM, PPM; PCM of 4.8 MHz TV signal with 512 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Apam-pwm-ppm-tv-4-8mhz) — Communication Systems, chapter 4
- [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
- [E1 hierarchy; PCM of 4.2 MHz TV signal with 512 levels](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ae1-pcm-tv-4-2mhz-512) — Communication Systems, chapter 4
- [Nyquist channel capacity theorem](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Anyquist-channel-capacity) — Communication Systems, chapter 6