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Additive white Gaussian noise
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=== Channel capacity and sphere packing === Suppose that we are sending messages through the channel with index ranging from <math>1</math> to <math>M</math>, the number of distinct possible messages. If we encode the <math>M</math> messages to <math>n</math> bits, then we define the rate <math>R</math> as: :<math> R = \frac {\log M}{n} \,\!</math> A rate is said to be achievable if there is a sequence of codes so that the maximum probability of error tends to zero as <math>n</math> approaches infinity. The capacity <math>C</math> is the highest achievable rate. Consider a codeword of length <math>n</math> sent through the AWGN channel with noise level <math>N</math>. When received, the codeword vector variance is now <math>N</math>, and its mean is the codeword sent. The vector is very likely to be contained in a sphere of radius <math display=inline>\sqrt{n(N+\varepsilon)}</math> around the codeword sent. If we decode by mapping every message received onto the codeword at the center of this sphere, then an error occurs only when the received vector is outside of this sphere, which is very unlikely. Each codeword vector has an associated sphere of received codeword vectors which are decoded to it and each such sphere must map uniquely onto a codeword. Because these spheres therefore must not intersect, we are faced with the problem of [[sphere packing]]. How many distinct codewords can we pack into our <math>n</math>-bit codeword vector? The received vectors have a maximum energy of <math>n(P+N)</math> and therefore must occupy a sphere of radius <math display=inline>\sqrt{n(P+N)}</math>. Each codeword sphere has radius <math>\sqrt{nN}</math>. The volume of an ''n''-dimensional sphere is directly proportional to <math>r^n</math>, so the maximum number of uniquely decodeable spheres that can be packed into our sphere with transmission power ''P'' is: :<math> \frac{(n(P+N))^{n/2}}{(nN)^{n/2}} = 2^{(n/2) \log\left(1+P/N \right)} \,\!</math> By this argument, the rate ''R'' can be no more than <math>\frac{1}{2} \log \left( 1+\frac P N \right)</math>.
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