Wolfson
Newbie level 2
Hello,
The Shannon limit is Eb/N0=(2^
-1)/n where n=2*r
So the Shannon limit for r=1/2 should be Eb/N0=1 and in dB 0dB.
In the literature it is often said to be 0.97869 (0,19dB)!
For example here:
eBook-turbo and Trellis Coding
and here:
http://wicl.kaist.ac.kr/pdf/sychung%20phd%20thesis.pdf
This link will only work if you are member of IEEE
**broken link removed**
I also found a hint in one document that the shannon limit is in deed 1 (0dB), but only for Codes of rate 1/2 defined over GF(q) and 0.97869 (0,19dB) for binary Codes
Could anybody explain that to me?
Thanks in advance,
Wolfson
The Shannon limit is Eb/N0=(2^
-1)/n where n=2*rSo the Shannon limit for r=1/2 should be Eb/N0=1 and in dB 0dB.
In the literature it is often said to be 0.97869 (0,19dB)!
For example here:
eBook-turbo and Trellis Coding
and here:
http://wicl.kaist.ac.kr/pdf/sychung%20phd%20thesis.pdf
This link will only work if you are member of IEEE
**broken link removed**
I also found a hint in one document that the shannon limit is in deed 1 (0dB), but only for Codes of rate 1/2 defined over GF(q) and 0.97869 (0,19dB) for binary Codes
Could anybody explain that to me?
Thanks in advance
,Wolfson
