---
title: "Shannon-Hartley channel capacity theorem"
subject: "Data Communication"
chapter: 1
key: "semester_5/data_communication:shannon-hartley-theorem"
source: https://www.bctengineer.com/q/semester_5/data_communication/shannon-hartley-theorem
site: BCT Engineering
---

# Shannon-Hartley channel capacity theorem

Data Communication · Chapter 1: Introduction

- Asked in: 2075 Asoj, 4 marks; 2072 Chaitra, 4 marks; 2068 Chaitra, 4 marks (asked 3 times)
- Video: [Channel Capacity by Shannon-Hartley | Basics, Proof & Maximum Bandwidth Condition](https://www.youtube.com/watch?v=3ekWsXeZ8TM) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/shannon-hartley-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-1

## Question

State and explain Shannon-Hartley channel capacity theorem with example.

## Short answer

Shannon-Hartley theorem: C = B log2(1 + S/N) bps, where B is bandwidth and S/N is signal-to-noise ratio. Example: For a 3000 Hz channel with 30 dB SNR, S/N = 1000, so capacity C ≈ 3000×9.97 ≈ 29.9 kbps.

## Answer

The Shannon-Hartley theorem gives the maximum error-free data rate (channel capacity) of a communication channel that has bandwidth B and Gaussian noise:

**C = B log₂(1 + S/N)** bits per second

where C = channel capacity (bps), B = bandwidth (Hz), S/N = signal-to-noise power ratio (not in dB). If SNR is given in dB, convert first: S/N = 10^(SNR_dB/10).

#### Explanation
- Capacity grows linearly with bandwidth but only logarithmically with signal power.
- It is a theoretical upper limit. Any rate below C can be sent with an arbitrarily small error using suitable coding; above C, errors are unavoidable.
- It depends on noise, so it applies to real (noisy) channels, unlike Nyquist's formula.
- Increasing the number of signal levels cannot beat this limit.

#### Example
A telephone line has B = 3000 Hz and SNR = 30 dB.

S/N = 10^(30/10) = 1000

C = 3000 × log₂(1 + 1000) = 3000 × 9.967 ≈ 29,901 bps

**Answer: about 29.9 kbps.** No modem can exceed this on such a line, whatever the modulation. (Real lines with a slightly higher SNR, about 35 dB, allow roughly 33.6 kbps, which is why analog modems stopped near that rate.)

## Check yourself

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

1. What is the Shannon-Hartley capacity?
   A. C = B log2(1 + S/N)
   B. C = 2B log2 L
   C. C = B log2(S/N)
   D. C = B(1 + S/N)
   Answer: A. Capacity depends on bandwidth and the signal-to-noise power ratio. (Key point: Formula)

2. SNR is 30 dB. What S/N goes into the formula?
   A. 30
   B. 1000
   C. 3
   D. 300
   Answer: B. S/N = 10^(30/10) = 1000; the formula needs the ratio, not dB. (Key point: dB conversion)

3. What does C represent?
   A. The rate every modem achieves
   B. The minimum rate of a channel
   C. The maximum error-free rate; above C errors are unavoidable
   D. The rate with no noise
   Answer: C. Rates below C can be sent with arbitrarily small error using suitable coding. (Key point: Meaning of capacity)

4. For B = 3000 Hz and SNR 30 dB, the capacity is about
   A. 3 kbps
   B. 90 kbps
   C. 9.97 kbps
   D. 29.9 kbps
   Answer: D. C = 3000 x log2(1001) = 3000 x 9.967, about 29,901 bps. (Key point: Example result)

## Related in Data Communication

- [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×
- [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
- [Entropy, information and channel capacity](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Ainformation-entropy-capacity) — Data Communication, chapter 8
- [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
- [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
- [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

## Related in other subjects

- [Shannon-Hartley theorem; capacity limit 1.44S/N₀](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-hartley-1-44) — Communication Systems, chapter 6, asked 2×
- [Shannon-Hartley theorem; energy/power of e^(−2t)u(t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-hartley-energy-e2t) — Communication Systems, chapter 1
- [Implications and limitations of Shannon-Hartley theorem](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-hartley-implications-limits) — Communication Systems, chapter 6
- [Shannon channel capacity theorem and its limitations](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Ashannon-capacity-limitations) — Communication Systems, chapter 6, asked 2×