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# Scribe Notes 6-5

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on April 15, 2009 at 7:10:01 pm

*Under Construction*

Recall that the capacity of a AWGN channel with noise and power constraint :

Suppose now we have two channels with  and total power constraint , how should we use our power?

Suppose, for example, , we can get a capacity of  if we only use channel 1 and  if we only use channel 2. Since , we can acheive a larger capacity if we use channel 1 alone. But what if we divide the power half-half? The new capacity would be , which is larger than either of the cases when we use only one channel.

Intuitively, when power is already large, we have to double the power to add an extra bit, so it is easier to use two channels with , .

It is obivious to see that the maximum capacity can only be acheived when we use all the power we get, i.e. . Use Lagrange Multiplier Method to find the maximum capacity with the constraints above, there exists  such that the equation below equals 0 when we take partial differentiation with respect to  and :

We get:

This leads to the water-filling strategy when we divide our power to different channels.

The image below shows the idea of water-filling strategy and how it works when the total power constraint shrink, with  as variables and  the total power constraints.

The same idea can also be applied to when there are multiple parallel channels.