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where p = p(X = A) Hence, the mutual information is maximized when p = 05 Note that C 1 because the digital modulator in the superchannel is a binary modulator delivering at most 1 bit per modulation symbol At nite channel noise, some of the bits transmitted will be in error However, we observe that C approaches 1 bit per channel use as s 2 0 This means that z at high SNR, nearly all of the transmitted bits could be decoded at the receiver and therefore, achieving a capacity of 1 bit per channel use In general, for M-ary modulator, the channel capacity is given by C = max[H (Y ) - H (Y X )] f ( y x) = max p ( x) log 2 y p ( x) f ( y x) x p p where p = [p1, .

, pM] and pq = p(X = q) In general, the capacity achieving distribution p has to be evaluated numerically Continuous-Input Continuous-Output Channels In this case, both the channel input X and the channel output Y are continuous values (or complex values in general) The channel is speci ed by the transition density f(y|x) In fact, we can consider a continuous-input channel as a limiting case of a discrete-input channel when we have in nitely dense constellation at the digital modulator For any input distribution p(X = x1), , p(X = xM) of an M-ary digital modulator, we could nd an equivalent input distribution f(X) in the continuous input case.

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In other words, we expect the channel capacity of continuous-input channels to be greater than or equal to the channel capacity of discrete input channels because the input distribution of the latter case is a subset of that in the former case Example 18 (Discrete Time AWGN Channel) Consider a discrete-time AWGN channel where X X is the channel input and Y Y is the channel output Both the channel input and channel output are complex numbers The channel output is given by: Yn = X n + Zn.

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where Zn is the white Gaussian channel noise with variance s 2 . Therefore, the z 1 2 1 channel transition probability is given by f(y|x) = exp y - x . The 2 2ps z 2s z channel capacity is given by C = max[H (Y ) - H (Y X )] = max H (Y ) - H (Z ) = max H (Y ) - H (Z )

f ( y) 2 sx = log 2 1 + 2 sz 2 where s x = e[|X|2]. Note that the second equality is due to the fact that H(Z) is independent of f(x) and the nal equality is due to the fact that H(Y ) is maximized if and only if Y is complex Gaussian, which is equivalent to X being complex Gaussian. f (x) f (x)

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