Editing Shannon–Hartley theorem (section)

==Examples==
# At a SNR of 0&nbsp;dB (Signal power = Noise power) the Capacity in bits/s is equal to the bandwidth in hertz. 
# If the SNR is 20&nbsp;dB, and the bandwidth available is 4&nbsp;kHz, which is appropriate for telephone communications, then C = 4000 log<sub>2</sub>(1 + 100) = 4000 log<sub>2</sub> (101) = 26.63&nbsp;kbit/s. Note that the value of S/N = 100 is equivalent to the SNR of 20&nbsp;dB.
# If the requirement is to transmit at 50 kbit/s, and a bandwidth of 10&nbsp;kHz is used, then the minimum S/N required is given by 50000 = 10000 log<sub>2</sub>(1+S/N) so C/B = 5 then S/N = 2<sup>5</sup> − 1 = 31, corresponding to an SNR of 14.91&nbsp;dB (10 x log<sub>10</sub>(31)).
# What is the channel capacity for a signal having a 1&nbsp;MHz bandwidth, received with a SNR of −30&nbsp;dB ? That means a signal deeply buried in noise. −30&nbsp;dB means a S/N = 10<sup>−3</sup>. It leads to a maximal rate of information of 10<sup>6</sup> log<sub>2</sub> (1 + 10<sup>−3</sup>) = 1443 bit/s. These values are typical of the received ranging signals of the GPS, where the navigation message is sent at 50 bit/s (below the channel capacity for the given S/N), and whose bandwidth is spread to around 1&nbsp;MHz by a pseudo-noise multiplication before transmission.
# As stated above, channel capacity is proportional to the bandwidth of the channel and to the logarithm of SNR. This means channel capacity can be increased linearly either by increasing the channel's bandwidth given a fixed SNR requirement or, with fixed bandwidth, by using [[higher-order modulation]]s that need a very high SNR to operate. As the modulation rate increases, the [[spectral efficiency]] improves, but at the cost of the SNR requirement. Thus, there is an exponential rise in the SNR requirement if one adopts a [[Quadrature amplitude modulation|16QAM or 64QAM]]; however, the spectral efficiency improves.