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資源簡介

多進制LDPC擴展最小和譯碼(EMS),已經成功 希望對后面研究的小伙伴有幫助

資源截圖

代碼片段和文件信息

function?[xhat?success?kk]?=?decoder_EMS(fHqx_encode)??%還沒改好?稍后再改
???tic?
???clc
???clear
???load?H_djinzhi_8_bian.mat;
???load?PP.mat;
???load?x_encode;
???tt=234;
???q=8;
???H=H_djinzhi_8_bian;
???[mn]?=?size(H);?if?m>n?H=H‘;?[mn]?=?size(H);?end
???
???f=zeros(qn);
???f(2:q:)=PP;

???
%????[iijjsH]?=?find(H);?
%????W?=?sparse(iijjones(size(ii))mn);?%indicator?function
%????nvars?=?full(sum(W2));?????????%?每行的非零個數?528
%????min_vars?=?min(nvars);??????????%?最小行重
%????max_vars?=?max(nvars);??????????%?最大行重
%????
%????nfuns?=?full(sum(W1));????????%?每列的非零個數?1056
%????min_funs?=?min(nfuns);??????????%?最小列重
%????max_funs?=?max(nfuns);??????????%?最大列重


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%?先確定Nm、補償因子R、offset?R=B[Nm]-offset?off

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件???????6167??2014-07-29?12:13??decoder_EMS.m

-----------?---------??----------?-----??----

?????????????????6167????????????????????1


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