Re: encoding ldpc
On Nov 24, 3:44*pm, "ChinookPass" <james.a.laudo...@saic.com> wrote:
> Hi, I'm a new poster to comp.dsp. *Recently I've been digging in to LDPCs
> out of general curiosity and after reading several papers and trying out
> some techniques, I have a question about encoding LDPCs. *There are two
> methods that seem to have the most general application, but I have not had
> success with either method. *
>
> Both methods involve taking the original parity check matrix and
> manipulating it so that the square parity portion of the matrix is upper
> triangular (or nearly so) or decomposing the square parity portion of the
> matrix into complete upper and lower triangular matrices.
>
> The second decomposition is preferred to the first because the decoder
> becomes well defined performing back substitution to calculate the parity
> bits. *The first manipulation (also in Richardson and Urbanke's book Modern
> Coding Theory) results in some submatrices and the encoder is more
> complicated.
>
> My difficulty is in manipulating the check matrix so that the parity
> portion is non-singular. *I've tried several different regular code
> generation techniques and LU decomposition techniques which involve
> permuting the H matrix and none of the algorithms converge to a solution so
> that the parity portion is non-singular. *Is this a problem that involves
> extensive trial and error to find an H-matrix that will yield a
> non-singular parity portion or are there algorithms available that yield a
> higher probability of success? *
>
> Of course my troubles so far do not even get into the question of whether
> or not the codes I have been generating are good codes. *I'm just trying to
> come up with the encoder!
>
> Thanks.
> Jim
There are some classes of LDPC codes that have very simple encoders.
One type that I believe is actually used in some contemporary wireless
standards (like 802.11n) are systematic repeat-accumulate codes. The
encoder for these is very simple, basically just an interleaver and a
one-bit accumulator.
Jason
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