t - VTU e-Learning Centre
t - VTU e-Learning Centre
t - VTU e-Learning Centre
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C 2 . log<br />
2(<br />
1<br />
)<br />
<br />
2(<br />
1<br />
)<br />
<br />
Simplifying you get,<br />
1 1 <br />
. log<br />
2(<br />
1<br />
)<br />
<br />
2(<br />
1<br />
)<br />
<br />
<br />
1<br />
log 2<br />
2(<br />
1<br />
)<br />
<br />
<br />
<br />
C = log [2 (1+)] – log <br />
or<br />
2(<br />
1<br />
)<br />
<br />
C = log2 <br />
<br />
Remember, = 2 and<br />
= - (p log p + q log q)<br />
What are the extreme values of p?<br />
Case – I: Say p = 1<br />
What does this case correspond to?<br />
What is C for this case/<br />
Note: = 0, = 2 0 = 1<br />
2( 1<br />
1)<br />
<br />
C log2 <br />
log<br />
1 <br />
<br />
2<br />
[ 4]<br />
If rs = 1/sec, C = 2 bits/sec.<br />
Can this case be thought of in practice?<br />
1<br />
Case – II: p =<br />
2<br />
bits/sec. -----(13)<br />
<br />
2 bits / sym.<br />
1 1 1 <br />
Now, = – log log 1<br />
2 2 2 <br />
= 2 = 2<br />
2( 2 1)<br />
<br />
log<br />
<br />
log 2 [ 3]<br />
1.<br />
585 bits / sec.<br />
2
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