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Shannon-Hartley channel capacity theorem

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.

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

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.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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