t - VTU e-Learning Centre
t - VTU e-Learning Centre
t - VTU e-Learning Centre
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d<br />
(i) Find D t <br />
dP<br />
d d <br />
1 <br />
i.e., D t <br />
2P<br />
log 2 P (<br />
1<br />
2P)<br />
log<br />
P<br />
2P<br />
dP dP<br />
<br />
<br />
2 <br />
<br />
1 <br />
<br />
<br />
2<br />
P<br />
log2<br />
e<br />
2 . P log<br />
<br />
2 e<br />
<br />
2 <br />
1 <br />
<br />
( 2)<br />
log <br />
2 log P<br />
2<br />
<br />
2 P<br />
P<br />
1 <br />
2 <br />
<br />
P<br />
<br />
2 <br />
<br />
<br />
Simplifying we get,<br />
d<br />
dP<br />
D log P log Q <br />
t<br />
Setting this to zero, you get<br />
log2 P = log2 Q + <br />
OR<br />
P = Q . 2 = Q . ,<br />
Where, = 2 <br />
How to get the optimum values of P & Q?<br />
2<br />
Substitute, P = Q in 2P + 2Q = 1<br />
i.e., 2Q + 2Q = 1<br />
OR<br />
1<br />
Q =<br />
2(<br />
1 )<br />
1<br />
Hence, P = Q . = . =<br />
2(<br />
1 )<br />
Step – 6: Channel capacity is,<br />
2<br />
<br />
2( 1 )<br />
2P<br />
log P 2Q<br />
log Q 2 <br />
C Max<br />
Q<br />
P(<br />
x)<br />
Q(<br />
X)<br />
Substituting for P & Q we get,<br />
-----(9)<br />
-----(10)<br />
-----(11)<br />
-----(12)
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