t - VTU e-Learning Centre
t - VTU e-Learning Centre
t - VTU e-Learning Centre
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Q . p<br />
P(<br />
x 1 / y 1)<br />
p<br />
Q<br />
Similarly calculate other p(x/y)’s<br />
They are<br />
(ii) P(x=1 / y=2) = q<br />
(iii) P(x=2 / y=1) = q<br />
(iv) P(x=2 / y=2) = p<br />
Equation (5) becomes:<br />
H(x/y) = [Q . p log p + Q . q log q + Q . q log q + Q . p log p]<br />
= [Q . p log p + Q (1 - p) log q + Q (1 - p) . log q + Q . p log p]<br />
= + [2Q . p log p + 2Q . q log q]<br />
= + 2Q [p log p + q log q]<br />
H(x/y) = 2Q . ----- (6)<br />
Step – 4: By definition we have,<br />
Dt = H(x) – H(x/y)<br />
Substituting for H(x) & H(x/y) we get,<br />
Dt = - 2P log2 P – 2Q . log2 Q – 2Q . ----- (7)<br />
Recall, 2P + 2Q = 1, from which,<br />
Q = ½ –P<br />
1 1 1 <br />
D t - 2P log 2 P - 2<br />
P<br />
log<br />
P<br />
2<br />
P<br />
2 2 2 <br />
1 <br />
i.e., Dt - 2P log2<br />
P (1-<br />
2P) log P<br />
2P<br />
----- (8)<br />
2 <br />
Step – 5: By definition of channel capacity we have,<br />
C <br />
Max<br />
P(<br />
x)<br />
Q(<br />
X)<br />
D <br />
t
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