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  • 大小: 86KB
    文件類型: .zip
    金幣: 2
    下載: 0 次
    發布日期: 2021-05-24
  • 語言: Matlab
  • 標簽: 編譯碼??

資源簡介

關于LDPC碼編碼和BP譯碼算法的matlab仿真!

資源截圖

代碼片段和文件信息

%function?vHat?=?decodeProbDomain(rx?H?N0?iteration)
%?Probability-domain?sum?product?algorithm?LDPC?decoder
%
%??rx????????:?Received?signal?vector?(column?vector)
%??H?????????:?LDPC?matrix
%??N0????????:?Noise?variance
%??iteration?:?Number?of?iteration
%
%??vHat??????:?Decoded?vector?(0/1)?
%
%
%?Copyright?Bagawan?S.?Nugroho?2007?
%?http://bsnugroho.googlepages.com
H=[1?1?1?0?0?0?0?0;0?0?0?1?1?1?0?0;1?0?0?1?0?0?1?0;0?1?0?0?1?0?0?1];
[M?N]=size(H);rate=M/N;N0=1;
en=10^(N0/10);?
sigma=1/sqrt(2*rate*en);

c=[1?0?1?0?1?1?1?1];
rx=c+sigma*randn(1N);


%?Prior?probabilities
P1?=?ones(size(rx))./(1?+?exp(-2*rx./(N0/2)));
P0?=?1?-?P1;

%?Initialization
K0?=?zeros(M?N);
K1?=?zeros(M?N);
rji0?=?zeros(M?N);
rji1?=?zeros(M?N);
qij0?=?H.*repmat(P0?M?1);
qij1?=?H.*repmat(P1?M?1);

iteration=100;
for?n?=?1:iteration
???fprintf(‘\n‘);
???fprintf(‘\n‘);
???fprintf(‘Iteration?:?%d\n‘?n);
???
???%?-----?Horizontal?step?-----
???for?i?=?1:M
??????
??????%?Find?non-zeros?in?the?column
??????c1?=?find(H(i?:));
??????
??????for?k?=?1:length(c1)
?????????
?????????%?Get?column?products?of?drji\c1(l)
?????????drji?=?1;
?????????for?l?=?1:length(c1)
????????????if?l~=?k
???????????????drji?=?drji*(qij0(i?c1(l))?-?qij1(i?c1(l)));
????????????end
?????????end?%?for?l
?????????
?????????rji0(i?c1(k))?=?(1?+?drji)/2;
?????????rji1(i?c1(k))?=?(1?-?drji)/2;
?????????
??????end?%?for?k
??????
???end?%?for?i
???
???%?------?Vertical?step?------
???for?j?=?1:N
??????
??????%?Find?non-zeros?in?the?row
??????r1?=?find(H(:?j));
??????
??????for?k?=?1:length(r1)
????????
?????????%?Get?row?products?of?prodOfrij\ri(l)
?????????prodOfrij0?=?1;
?????????prodOfrij1?=?1;???
?????????for?l?=?1:length(r1)
????????????if?l~=?k
???????????????prodOfrij0?=?prodOfrij0*rji0(r1(l)?j);
???????????????prodOfrij1?=?prodOfrij1*rji1(r1(l)?j);
????????????end
?????????end?%?for?l
?????????
?????????%?Update?constants
?????????K0(r1(k)?j)?=?P0(j)*prodOfrij0;
?????????K1(r1(k)?j)?=?P1(j)*prodOfrij1;
?????????
?????????%?Update?qij0?and?qij1
?????????qij0(r1(k)?j)?=?K0(r1(k)?j)./(K0(r1(k)?j)?+?K1(r1(k)?j));
?????????qij1(r1(k)?j)?=?K1(r1(k)?j)./(K0(r1(k)?j)?+?K1(r1(k)?j));
???????????????
??????end?%?for?k
??????
??????%?Update?constants
??????Ki0?=?P0(j)*prod(rji0(r1?j));
??????Ki1?=?P1(j)*prod(rji1(r1?j));
??????
??????%?Get?Qj
??????Qi0?=?Ki0/(Ki0?+?Ki1);
??????Qi1?=?Ki1/(Ki0?+?Ki1);
??????if(j==1)?
??????????fprintf(‘Q1=%f???‘Qi1);
??????else
??????????fprintf(‘%f???‘Qi1);
??????end
??????%?Decode?Qj????????
??????if?Qi1?>?Qi0
?????????vHat(j)?=?1;
??????else
?????????vHat(j)?=?0;
??????end
?????????
???end?%?for?j
??fprintf(‘\nc=‘);
??
???????fprintf(‘%d???????????‘vHat);
???????
???????
???if?mod(vHat*H‘2)==0
???????break;end
???
end?%?for?n


?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????84992??2010-01-04?16:08??Ldpc?碼.doc
?????文件????????2901??2010-01-04?16:08??decodeProbDomain.m

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