In information theory, the Shannon–Hartley theorem is an application of the noisy channel coding theorem to the archetypal case of a continuous-time analog communications channel subject to Gaussian noise. The result establishes the maximum amount of error-free digital data (that is, information) that can be transmitted over such a communication link with a specified bandwidth in the presence of the noise interference. The law is named after Claude Shannon and Ralph Hartley. The Shannon limit or Shannon capacity of a communications channel is the theoretical maximum information transfer rate of the channel.

Introduction

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Noisy channel coding theorem

Claude Shannon's noisy channel coding theorem (1948) describes the maximum possible efficiency of error-correcting methods versus levels of noise interference and data corruption. The theory doesn't describe how to construct the error-correcting method, it only tells us how good the best possible method can be.

It establishes that given a noisy channel with information capacity C and information transmitted at a rate R, then if

there exists a coding technique which allows the probability of error at the receiver to be made arbitrarily small. This means that theoretically, it is possible to transmit information without error up to a limit, C.

The converse is also important. If

$R\; >\; C\; \backslash ,$

the probability of error at the receiver increases without bound as the rate is increased. So no useful information can be transmitted beyond the channel capacity. The theorem does not address the rare situation in which rate and capacity are equal.

Shannon–Hartley theorem

The Shannon–Hartley theorem establishes what that channel capacity is, for a finite bandwidth continuous-time channel subject to Gaussian noise.

If we had such a thing as an infinite-bandwidth, noise-free analog channel we could transmit unlimited amounts of error-free data over it per unit of time. However real life signals have both bandwidth and noise-interference limitations.

So how do bandwidth and noise affect the rate at which information can be transmitted over an analog channel?

Surprisingly, bandwidth limitations alone do not impose a cap on maximum information transfer. This is because it is still possible (at least in a thought-experiment model) for the signal to take on an infinite number of different voltage levels on each cycle, with each slightly different level being assigned a different meaning or bit sequence. If we combine both noise and bandwidth limitations, however, we do find there is a limit to the amount of information that can be transferred, even when clever multi-level encoding techniques are used. This is because the noise signal obliterates the fine differences (Hamming distance) that distinguish the various signal levels, limiting in practice the number of detection levels we can use in our scheme.

Capacity of the additive white Gaussian noise channel

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Considering all possible multi-level and multi-phase encoding techniques, the theorem gives that the theoretical maximum rate of clean (or arbitrarily low bit error rate) data C with a given average signal power that can be sent through an analog communication channel subject to additive, white, Gaussian-distribution noise interference is:

$C\; =\; BW\; \backslash times\; \backslash log\_2\; \backslash left(\; \backslash frac\{S\; +\; N\}\{N\}\; \backslash right)\; =\; BW\; \backslash times\; \backslash log\_2\; \backslash left(\; 1+\backslash frac\{S\}\{N\}\; \backslash right)$

where

C is the channel capacity in bits per second, net of error correction;

BW is the bandwidth of the channel in hertz;

S is the total signal power over the bandwidth and

N is the total noise power over the bandwidth.

S/N is the signal-to-noise ratio of the communication signal to the Gaussian noise interference expressed as a straight power ratio (not as decibels).

Normally the signal and noise are fully uncorrelated and in that case S + N is the total power of the received signal and noise together. A generalization of the above equation for the case where the additive noise is not white (or that the S/N is not constant in frequency over the bandwidth) is:

$C\; =\; \backslash int\_\{0\}^\{BW\}\; \backslash log\_2\; \backslash left(\; \backslash frac\{S(f)\; +\; N(f)\}\{N(f)\}\; \backslash right)\; df\; =\; \backslash int\_\{0\}^\{BW\}\; \backslash log\_2\; \backslash left(\; 1+\backslash frac\{S(f)\}\{N(f)\}\; \backslash right)\; df$

where

C is the channel capacity in bits per second, net of error correction;

BW is the bandwidth of the channel in Hz;

S(f) is the signal power spectrum

N(f) is the noise power spectrum

f is frequency in Hz.

For large or small and constant signal-to-noise ratios, this formula can be approximated.

If S/N >> 1, C = 0.332 · BW · SNR (in dB).

If S/N << 1, C = 1.44 · BW · S/N (in power).

The V.34 modem standard advertises a rate of 33.6 kbit/s, and V.90 claims a rate of 56 kbit/s, apparently in excess of the Shannon limit (telephone bandwidth is 3.3 kHz). In fact, neither standard actually reaches the Shannon limit. The bandwidth is not the limiting factor because it is possible and common for modems to transmit many bits per symbol. The actual limit is the signal to noise ratio which is dependent upon the underlying plant installation. V.90 uses a clever technique that assumes the local cable from the customer site to the office equipment is free of noise and that the conversion to PCM is the only disturbance. It then maps data bits onto the equivalent voltages for the PCM codecs used in the standard telephone network(s). In V.90, this only works downstream (CO to customer) and the upstream is still a V.34 variant; V.92 expands on this technique for upstream use.

Examples

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1. If the S/N is 20 dB, and the bandwidth available is 4 kHz, which is appropriate for telephone communications, then C = 4 log₂(1 + 100) = 4 log₂ (101) = 26.63 kbit/s. Note that the value of 100 is appropriate for an S/N 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 S/N required is given by 50 = 1000 log₂(1+S/N) so S/N = 2^C/W -1 = 0.035 corresponding to an S/N of -14.5 dB. This shows that it is possible to transmit using signals which are actually much weaker than the background noise level, as in spread-spectrum communications.

References

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• C. E. Shannon, The Mathematical Theory of Communication. Urbana, IL:University of Illinois Press, 1949 (reprinted 1998).
• C. E. Shannon, "Communication in the presence of noise", Proc. Institute of Radio Engineers, vol. 37, no.1, pp. 10-21, Jan. 1949.
• David J. C. MacKay. Information Theory, Inference, and Learning Algorithms Cambridge: Cambridge University Press, 2003. ISBN 0521642981
• Herbert Taub, Donald L. Schilling, Principles of Communication Systems, McGraw-Hill, 1986

See also

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• Nyquist-Shannon sampling theorem
• Turbo code

External links

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• On Shannon and Shannon's law
• The Shannon–Hartley Theorem
• The relationship between information, bandwidth and noise
• On-line textbook: Information Theory, Inference, and Learning Algorithms, by David MacKay - gives an entertaining and thorough introduction to Shannon theory, including two proofs of the noisy-channel coding theorem. This text also discusses state-of-the-art methods from coding theory, such as low-density parity-check codes, and Turbo codes.

Information theory | Mathematical theorems | Claude Shannon

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