nimanm
Newbie level 1
Hi everyone,
Im trying to implement the sum-product algorithm in MATLAB for decoding LDPC codes. the problem raises up when the extrinsic information gets larger for example in high SNR region. In this case, the bit node msg, E, goes to infinity as E=log((1+A)/(1-A)) supposes with A = product(tanh(M/2)) where Ms are the check node msgs. the problem is that in MATLAB tanh of numbers greater than 18 is equal to one.
As a solution I tried to limit E to a figure like 18, I mean if log((1+A)/(1-A)) gets infinity the value of E is 18 rather than infinity. but when I used different values like 3 or 5 for this limitation (rather than 18 ) every time I got different results and surprisingly a better BER for lower limitations.
My question is that, is it a reasonable solution to limit the extrinsic values? if yes what is the optimum limitation value and if no whats the solution in the case of high extrinsic values?
thank you for considering my question. I will be grateful if you post your suggested solution to my problem.
Im trying to implement the sum-product algorithm in MATLAB for decoding LDPC codes. the problem raises up when the extrinsic information gets larger for example in high SNR region. In this case, the bit node msg, E, goes to infinity as E=log((1+A)/(1-A)) supposes with A = product(tanh(M/2)) where Ms are the check node msgs. the problem is that in MATLAB tanh of numbers greater than 18 is equal to one.
As a solution I tried to limit E to a figure like 18, I mean if log((1+A)/(1-A)) gets infinity the value of E is 18 rather than infinity. but when I used different values like 3 or 5 for this limitation (rather than 18 ) every time I got different results and surprisingly a better BER for lower limitations.
My question is that, is it a reasonable solution to limit the extrinsic values? if yes what is the optimum limitation value and if no whats the solution in the case of high extrinsic values?
thank you for considering my question. I will be grateful if you post your suggested solution to my problem.
