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Shannon Channel Capacity

Maximum error-free data rate of a channel given bandwidth and signal-to-noise ratio.

Formula

C=Blog⁡2(1+SNR)C = B \log_2(1 + SNR)

Variables

CCapacity (bps)
BBandwidth (Hz)
SNRSignal-to-noise ratio

Example

B=1 MHz, SNR=1000: C = 10⁶×9.97 ≈ 9.97 Mbps

How it works

Shannon channel capacity gives the theoretical upper limit on how fast information can be sent over a noisy communication link if the only impairments are bandwidth limitation and additive white Gaussian noise. It comes from information theory: as bandwidth increases or the signal-to-noise ratio improves, the number of reliably distinguishable signal states per second rises, so the maximum error-free rate increases logarithmically rather than linearly. In practice, this is an ideal benchmark, not a guarantee, because real systems also suffer from coding losses, interference, fading, and hardware imperfections.

When to use it

Use this formula when estimating the absolute data-rate limit of a communications channel, such as a radio link, fiber segment, or modem link, from its bandwidth and SNR. It is especially useful for comparing whether a target throughput is theoretically feasible before accounting for protocol overhead and implementation losses.

Worked example, step by step

Given: B = 20 MHz, SNR = 15 Step 1: Substitute into Shannon's formula: C = B log2(1 + SNR) Step 2: C = 20 × 10^6 × log2(1 + 15) Step 3: C = 20 × 10^6 × log2(16) Step 4: log2(16) = 4 Step 5: C = 20 × 10^6 × 4 = 80 × 10^6 bps Final answer: C = 80 Mbps

Common mistakes to avoid

  • Using SNR in dB directly instead of converting to a linear ratio first
  • Forgetting that bandwidth B must be in hertz, not kHz or MHz, unless converted consistently
  • Treating Shannon capacity as a guaranteed real-world throughput instead of a theoretical maximum

Did You Know?

Shannon's 1948 limit tells engineers the absolute ceiling no clever coding can ever beat.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026