含仿真操作录像，LDPC译码算法性能对比,对比BP译码和BF译码两种译码算法matlab仿真

function vHat = BP4(rx, H,sigma, iteration) [M N] = size(H); P0 = (1 + exp(-2*(sigma^(-2)).*rx)).^(-1); P1 = (1 + exp(2*(sigma^(-2)).*rx)).^(-1); K0 = zeros(M, N); K1 = zeros(M, N); rji0 = zeros(M, N); rji1 = zeros(M, N); a=repmat(P0, M, 1); qij0 = H.*repmat(P0, M, 1);%得到初始的概率矩阵，元素为0或P0 qij1 = H.*repmat(P1, M, 1);%得到初始的概率矩阵，元素为0或P1 % Iteration迭代 for n = 1:iteration for i = 1:M % Find non-zeros in the column c1 = find(H(i, :)); for k = 1:length(c1) 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;%由证明算出的rij0的更新方式。drji是联合概率值（乘积形式），由上面的循环得到。 rji1(i, c1(k)) = (1 - drji)/2; end % for k end % for i以上应该是将rij初始化完毕 % ------ Vertical step ------ for j = 1:N % Find non-zeros in the row r1 = find(H(:, j)); for k = 1:length(r1) 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 1 % Update constants每次更新P(a)*rij(a)，其他信息节点已知条件下，信息比特为0/1，校验式满足的概率 K0(r1(k), j) = P0(j)*prodOfrij0;%信息比特为0,校验式满足的概率 K1(r1(k), j) = P1(j)*prodOfrij1;%信息比特为1，校验式满足的概率 % Update qij0 and qij1其他校验式已知下，信息比特为0/1的概率 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第J行的k个元素，即在每个校验式里，每个信息比特为0/1的概率 % Update constants Ki0 = P0(j)*prod(rji0(r1, j));%每列的成率乘积，即每个信息比特参加的校验式的联合概率值 Ki1 = P1(j)*prod(rji1(r1, j)); Qi0 = Ki0/(Ki0 + Ki1); Qi1 = Ki1/(Ki0 + Ki1); % Decode Qj判决 if Qi1 > Qi0 vHat(j) = 1; else vHat(j) = 0; end end % for j end % for n