如何解码 LDPC答案

【问题标题】：How to decode LDPC如何解码 LDPC
【发布时间】：2019-10-17 13:37:20
【问题描述】：

我生成一个奇偶校验矩阵，并使用 comm.LDPCEncoder 对码字进行编码。但是，当我使用 comm.LDPCDecoder 解码 LLR 时，解码后的字与源不同。你能帮我找出哪里错了吗？

K = 4;N = 8;
H1 =  [0 0 0 1 1 0 1 1;
    0 0 0 0 1 1 0 0
    1 1 0 0 0 0 0 1
    0 0 1 0 0 1 0 1];
H = sparse(H1);
hEnc = comm.LDPCEncoder(H);
hDec = comm.LDPCDecoder(H);
msg_source = logical(randi([0 1], K, 1));
msg_coded = step(hEnc, msg_source);  
LLR = double(msg_coded);
msg_decoded = step(hDec,LLR);
errors = (sum(xor(msg_source,msg_decoded)));

非常感谢。

【问题讨论】：

标签： matlab

【解决方案1】：

看起来LLR = double(msg_coded); 是问题的根源。

我不是通信专家，从未使用过 LDPC 编码器和解码器。
我尝试了 MATLAB 代码示例，发现在调制和解调之后，1s（一）得到负值，0s（零）得到正值。

而不是LLR = double(msg_coded);
您可以使用以下转换：LLR = 1 - double(msg_coded)*2;.
0 --> 1
1 --> -1

这里是修改后的代码：

K = 4;N = 8;
H1 =  [0 0 0 1 1 0 1 1;
    0 0 0 0 1 1 0 0
    1 1 0 0 0 0 0 1
    0 0 1 0 0 1 0 1];
H = sparse(H1);
hEnc = comm.LDPCEncoder(H);
hDec = comm.LDPCDecoder(H);
msg_source = logical(randi([0 1], K, 1));
msg_coded = step(hEnc, msg_source);  
%LLR = double(msg_coded);

%Convert from logical to double where 0 goes to -1 and 1 goes to 1.
LLR = 1 - double(msg_coded)*2;
msg_decoded = step(hDec,LLR);
errors = (sum(xor(msg_source,msg_decoded)));

这是我用于查找问题的调制和解调代码（基于 MATLAB 示例）：

K = 4;N = 8;
H1 =  [0 0 0 1 1 0 1 1;
    0 0 0 0 1 1 0 0
    1 1 0 0 0 0 0 1
    0 0 1 0 0 1 0 1];
H = sparse(H1);
hEnc = comm.LDPCEncoder(H);
hDec = comm.LDPCDecoder(H);

hMod = comm.PSKModulator(4, 'BitInput',true);
hChan = comm.AWGNChannel(...
        'NoiseMethod','Signal to noise ratio (SNR)','SNR',10);
hDemod = comm.PSKDemodulator(4, 'BitOutput',true,...
        'DecisionMethod','Approximate log-likelihood ratio', ...
        'Variance', 1/10^(hChan.SNR/10));

msg_source = logical(randi([0 1], K, 1));
msg_coded = step(hEnc, msg_source);  

modSignal      = step(hMod, msg_coded);
receivedSignal = step(hChan, modSignal);
demodSignal    = step(hDemod, receivedSignal);
msg_decoded   = step(hDec, demodSignal);

errors = (sum(xor(msg_source,msg_decoded)));

【讨论】：

【解决方案2】：

为什么要将逻辑编码数据转换为双精度？编码后尝试解码：

K = 4;N = 8;
H1 =  [0 0 0 1 1 0 1 1;
    0 0 0 0 1 1 0 0
    1 1 0 0 0 0 0 1
    0 0 1 0 0 1 0 1];
H = sparse(H1);
hEnc = comm.LDPCEncoder(H);
hDec = comm.LDPCDecoder(H);
msg_source = logical(randi([0 1], K, 1));
msg_coded = step(hEnc, msg_source);  
msg_decoded = step(hDec,LLR);
errors = (sum(xor(msg_source,msg_decoded)));

【讨论】：