The Shannon limit reference article from the English Wikipedia on 24-Apr-2004 (provided by Fixed Reference: snapshots of Wikipedia from wikipedia.org)

Shannon limit

The Shannon limit or Shannon capacity of a communication channel refers to the maximum rate of error-free data that can theoretically be transferred over the link if the link is subject to random data transmission errors. The Shannon limit is a statement of Shannon's theorem for the case of a limited bandwidth communication channel.

The Shannon limit states the following: For an analog channel of signal strength S which is subject to interference by a Gaussian white noise interference signal of strength N, the best possible information transfer rate C achievable by even the most clever error correction scheme will be:

where

C = net channel capacity (in bits per second) after error correction,

W = raw channel frequency or bandwidth,

S/N = signal-to-noise ratio given as a straight ratio (not in decibels)

This shows that there is a theoretical maximum amount of information that can be transmitted over a bandwidth-limited channel in the presence of background noise. (Although Shannon's original proof applied only to analog channels subject to Gaussian noise, it is typically assumed to apply to channels with other noise characteristics as well).

The signal-to-noise ratio above is in terms of the signal power per Hz as compared to the noise power per Hz. If the bandwidth (W) is increased while the physical power available for data transmission is held constant, then the signal power per Hz must decrease. It turns out that as the bandwidth goes to infinity, the signal-to-noise ratio goes to zero and the total capacity in the model above approaches a finite limit (see spread-spectrum communication).

Examples:

1. 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.
2. 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 it may be possible to transmit using signals which are actually much weaker than the background noise level, as in spread-spectrum communications.

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