When one analyzes an LDPC code, it is usual to plot the BER versus the $E_b / N_0$. My questions are:

  • Why are some LDPC codes better than others in terms of noise immunity? After all, they are just defined by matrices $\mathbf H$.
  • What properties or parameters of this parity-check matrix have influence on the BER for a given SNR?

1 Answer 1


As you mention, the LDPC code is completely determined by its generator matrix H. Hence, the properties of H define the (theoretic) performance of the code.

Since the LDPC code is a linear code, the ultimate measure for the (theoretic) code performance is its minimum distance between two codewords, that is:

$$ d = \min_{c_1,c_2\in\mathcal{C}}\|c_1-c_2\|_0 $$

where $\mathcal{C}$ is the set of existing codewords (i.e. bit patterns that fulfill the parity check) and $\|\ldots\|_0$ denotes the hamming distance. This is equal to the expression (since it's a linear code)

$$ d = \min_{c\in\mathcal{C}}\|c\|_0 $$

where $\|\|_0$ is the Hamming weight.

I wrote the word "theoretically" in parenthesis, since the finally obtained performance does also depend on the decoding algorithm. Due to the low-density of the parity check nodes, the LDPC is suitable for iterative decoding, which can yield big gains.

Another improtant measure is the length of the shortest cycle of the LDPC Tanner Graph. The shorter the cycle, the worse is the performance of the iterative decoding.

  • $\begingroup$ So could one affirm that the bigger the codewords (i.e. the more number of bits ($n$) used to encode a message), the better the noise-immunity? Since if more bits are used, it's more probable that $d$ will increase (although this would not be true in general, maybe there is a certain correlation). Also, isn't it less likely to have short cycles if the parity-check matrix is bigger? $\endgroup$
    – Tendero
    Dec 29, 2016 at 15:37
  • $\begingroup$ Yes, longer codewords in general would yield better performance (i.e. going closer to the Shannon capacity). There is no real "probability that d will increase", because you design H such that d is big. But, with higher dimensions, you have more degrees of freedom to accomplish a high d and long cycles. $\endgroup$ Dec 29, 2016 at 15:39

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