How this instrument works
Channel capacity, in the Shannon-Hartley sense, is the maximum rate at which information can pass through a communication channel with an arbitrarily low error rate, given the channel's bandwidth and how much noise is mixed into the signal. Claude Shannon proved in 1948 that this ceiling depends on exactly two things: the bandwidth available in hertz, and the ratio of signal power to noise power. Widen the channel or clean up the signal and the ceiling rises; there is no third lever that raises it further.
The formula, C = B·log₂(1 + S/N), multiplies bandwidth by a logarithm of one plus the signal-to-noise ratio. That ratio must be entered as a plain linear power ratio — signal power divided by noise power — not in decibels; a 30 dB reading converts to a linear ratio of 10^(30/10) = 1000 before it goes into this formula. The log₂ term captures a genuine diminishing return: doubling the SNR does not double the capacity, because each extra bit of information needs an exponentially cleaner signal to distinguish it from noise.
Shannon capacity is a hard theoretical ceiling, not a number any real modem, Wi-Fi radio, or DSL line actually reaches. Real systems fall short because of practical constraints Shannon's proof abstracts away — imperfect coding, synchronization overhead, and non-Gaussian noise all eat into the achievable rate. Engineers use this number as a benchmark for how much headroom a real system still has, not as a design target that hardware is expected to hit exactly.
- Enter Channel bandwidth (Hz) — the width of the frequency band the channel occupies, for example 1,000,000 for a 1 MHz channel.
- Enter Signal-to-noise ratio (linear power ratio, not dB) — convert a decibel figure first using 10^(dB/10) if that's what you have on hand.
- Read Channel capacity (bits per second) directly — this is the Shannon-Hartley theoretical maximum, shown as a raw bps figure.
- Bandwidth must be zero or greater and SNR cannot be negative; the instrument will prompt you if either input falls outside that range.
Worked example — a 1 MHz channel at S/N = 1023
Enter Channel bandwidth (Hz) = 1,000,000 and Signal-to-noise ratio (linear power ratio, not dB) = 1023. Add 1 to the SNR first: 1 + 1023 = 1024, and 1024 happens to equal 2^10 exactly, so log₂(1024) = 10 with no rounding involved. Channel capacity reads 10,000,000 bits per second — multiplying the 1,000,000 Hz bandwidth by that clean exponent of 10.
10,000,000 bps is 10 Mbps, a round number that falls out of this particular pairing of bandwidth and SNR specifically because 1023 was chosen so that 1+S/N lands exactly on a power of two. Most real bandwidth and SNR combinations won't produce such a tidy log₂ result, but the underlying multiplication — bandwidth times a base-2 logarithm of one plus the ratio — works identically regardless.
Questions
Why must SNR be entered as a ratio and not in decibels?
Because the Shannon-Hartley formula is defined in terms of a linear power ratio, S/N, not a logarithmic decibel figure — decibels are themselves already a logarithmic representation of that ratio, and feeding a dB number straight into log₂(1+S/N) would double up the logarithm incorrectly. Convert first with S/N = 10^(dB/10): a 30 dB reading becomes a linear ratio of 1000 before it belongs in this formula.
Can real Wi-Fi or DSL connections actually reach their Shannon capacity?
No, not in practice. Shannon's theorem describes a theoretical ceiling assuming ideal coding and Gaussian noise; real hardware loses throughput to protocol overhead, imperfect error correction, synchronization, and noise that doesn't behave exactly like the idealized Gaussian case. Engineers use the Shannon figure as a benchmark for how much room a real system has left to improve, not as an achievable target.
Does doubling the SNR double the channel capacity?
No — capacity only grows with the logarithm of 1+S/N, so each additional bit of capacity needs an exponentially larger jump in SNR. Doubling bandwidth doubles capacity directly, since B sits outside the logarithm as a straight multiplier, but doubling SNR only adds a fixed, and comparatively small, amount to the log₂ term.
What happens if signal-to-noise ratio is zero?
Capacity drops to zero. With S/N = 0, the term 1+S/N equals 1, and log₂(1) = 0, so C = B x 0 = 0 regardless of how wide the bandwidth is — a channel with no signal power relative to noise carries no information at all, no matter how much frequency space it spans.
Is a wider channel or a cleaner signal a better way to raise capacity?
Widening the channel is generally more effective, because bandwidth B multiplies capacity directly while SNR only contributes through a logarithm. Doubling bandwidth always doubles capacity; doubling SNR typically adds far less, especially once SNR is already large, which is why engineers chase more spectrum before chasing incrementally cleaner signals.