Additive White Gaussian Noise - Channel Capacity

Channel Capacity

The AWGN channel is represented by a series of outputs at discrete time event index . is the sum of the input and noise, where is independent and identically distributed and drawn from a zero-mean normal distribution with variance (the noise). The are further assumed to not be correlated with the .


Z_i \sim N(0, n)
\,\!

Y_i = X_i + Z_i\sim N(X_i, n).
\,\!

The capacity of the channel is infinite unless the noise n is nonzero, and the are sufficiently constrained. The most common constraint on the input is the so-called "power" constraint, requiring that for a codeword transmitted through the channel, we have:


\frac{1}{n}\sum_{i=1}^k x_i^2 \leq P,

where represents the maximum channel power. Therefore, the channel capacity for the power-constrained channel is given by:


C = \max_{f(x) \text{ s.t. }E \left( X^2 \right) \leq P} I(X;Y)
\,\!

Where is the distribution of . Expand, writing it in terms of the differential entropy:


\begin{align}
I(X;Y) = h(Y) - h(Y|X)
&= h(Y)-h(X+Z|X)
&= h(Y)-h(Z|X)
\end{align}
\,\!

But and are independent, therefore:


I(X;Y) = h(Y) - h(Z)
\,\!

Evaluating the differential entropy of a Gaussian gives:


h(Z) = \frac{1}{2} \log(2 \pi e n)
\,\!

Because and are independent and their sum gives :


E(Y^2) = E(X+Z)^2 = E(X^2) + 2E(X)E(Z)+E(Z^2) = P + n
\,\!

From this bound, we infer from a property of the differential entropy that


h(Y) \leq \frac{1}{2} \log(2 \pi e(P+n))
\,\!

Therefore the channel capacity is given by the highest achievable bound on the mutual information:


I(X;Y) \leq \frac{1}{2}\log(2 \pi e (P+n)) - \frac {1}{2}\log(2 \pi e n)
\,\!

Where is maximized when:


X \sim N(0, P)
\,\!

Thus the channel capacity for the AWGN channel is given by:


C = \frac {1}{2} \log\left(1+\frac{P}{n}\right)
\,\!

Read more about this topic:  Additive White Gaussian Noise

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