- C = W log
_{2}(1 + S/N)

- C = W log

**Examples**:

- If the SNR is 20 dB, and the bandwidth available is 4 kHz, which is appropriate for telephone communications, then C = 4 log
_{2}(1 + 100) = 4 log_{2}(101) = 26.63 kbit/s. Note that the value of 100 is appropriate for an SNR of 20 dB. - If it is required to transmit at 50 kbit/s, and a bandwidth of 1 MHz is used, then the minimum SNR required is given by 50 = 1000 log
_{2}(1+S/N) so S/N = 2^{C/W}-1 = 0.035 corresponding to an SNR of -14.5 dB. This shows that in some sense it may be possible to transmit using signals below the noise level, using wide bandwidth communication, as in spread spectrum communications.

See also Shannon's theorem