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Information Theory: Fall 2017

Solutions to Homework #
Jeremy Pons.
November 6, 2017
Solution to problem *.

1

Exercise 3.2 : Z channel

We have to find the capacity of a Zchannel.

The Capacity C is define as :
C = max(I(X; Y )) = max(H(Y ) − H(Y |X)
We find that :
P (X = 0) = 1 − p
P (X = 1) = p
P (Y = 0) = (1 − p) +
P (Y = 1) = p2

p
2

=1−

p
2

So now we can find that :
H(Y |X) = P (X = 0)H(Y |X = 0) + P (X = 1)H(Y |X = 1) = p
H(Y ) = (1 − p2 )Log( 1−1 p ) + p2 Log( p2 ) = H( p2 )
2

So we can find the mutual information I(X;Y) :

1

Information Theory: Fall 2017
I(X; Y ) = H( p2 ) − p)
To find the maximum we need to derivate the mutual information and look when it is
equal to zero. It is equal to zero when p=1.
If p=1, I(X; Y ) = log(2)
Then we have Capacity = log(2)

2

Exercise 3.13 Water-filling

We have : Y j = gjXj + Zj
We now that the power constraint P is equal to the sum of the powers. P = P 1 + P 2 If
we use only one channel, we will need to use a second channel if the power contraints P
is superior or equal to λ(water-level).
Or we know that P j = [λ −
If P &gt; P j +

1
gj 2

1
gj 2

]

; We need to use the both channels.

We know that Eb(R) =

P
R

For every sequence of (2nR,n) codes with average power P we know that
R &lt; 21 Log(1 + g 2 P )
Then we can get the minimum energy-per-bit : Eb(R) =

1
(22 R
g2 R

− 1)

En(R) is monotonically increasing convex in R and when R goes to 0 we have :
LimEb(R) = 2lb2
g2

2