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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