t - VTU e-Learning Centre
t - VTU e-Learning Centre
t - VTU e-Learning Centre
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P(y=0) = P(x=0) . P00 + P(x=1) P10<br />
= p + (1 – p) (1 – )<br />
Similarly, P(Y=1) = p (1 – ) + (1 – p) <br />
H(Y) p ( 1<br />
p)<br />
( 1<br />
)<br />
log p ( 1<br />
p)<br />
( 1<br />
)<br />
<br />
p ( 1<br />
)<br />
( 1<br />
)<br />
log p ( 1<br />
)<br />
( 1<br />
)<br />
<br />
H(Y) = - Q1(p) . log Q1(p) – Q2(p) log Q2(p)<br />
Substituting the values, H(y) = 0.82<br />
H(y/x) = P ( x<br />
Simplifying, you get,<br />
i<br />
j<br />
i,<br />
y j)<br />
log P ( y j / x i)<br />
H(y/x) = – [p . . log + (1 – p) (1 – ) log (1 – ) + p(1 – ) log (1 – )]<br />
+ (1 – p) log ]<br />
= – p [ . log – (1 – ) log (1 – ) – log – (1 – ) log (1 – )]<br />
H(y/x) = PK + C<br />
Compute ‘K’<br />
+ terms not dependent on p<br />
Rateat<br />
whichinformnissupplied<br />
Rateat<br />
whichinformn<br />
<br />
<br />
<br />
<br />
by<br />
thechannel<br />
to theReceiver<br />
is<br />
receivedby<br />
theReceiver<br />
i.e., H(x) – H(x/y) = H(y) – H(y/x)<br />
Home Work:<br />
Calculate Dt = H(y) – H(y/x)<br />
d<br />
Find Dt<br />
dp<br />
d<br />
Set Dt to zero and obtain the condition on probability distbn.<br />
dp<br />
Finally compute ‘C’, the capacity of the channel.<br />
Answer is: 344.25 b/second.
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