1
$\begingroup$

I have been asked to create a parity check matrix for a code made up of the codewords C=(0000,1110,1011,0101). I created a generator set {0101,1011} this set creates 1110 when the codewords in the set are added together and 0000 if the they are summed with themselves. I put the set into a generator matrix -

\begin{vmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \end{vmatrix}

I then transpose to create the Parity check matrix.

\begin{vmatrix} 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{vmatrix}

Would this Parity check matrix be correct?

The reason why i am checking is the next part of the question, gives a codeword 1101 which has been received over a noisy channel. Using the nearest neighbour decoding I can see the nearest codeword is 0101. I then have to decode the word 1101 using error syndrome.

So I take the code word and parity check

\begin{vmatrix} 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{vmatrix}

\begin{vmatrix} 1\\ 1 \\ 0 \\ 1 \end{vmatrix}

Then i sum out the two matrices

  • $1+1+0+1+1+0+0+1$
  • $1+1+1+1+0+0+1+1$

I get the error syndrome

\begin{vmatrix} 1\\0 \end{vmatrix}

Looking at the parity check matrix this shows the error is in the third column but then that would mean the codeword received is meant to be $1111$ instead of $1101$, which isnt a codeword in the original code. Where have i gone wrong.

$\endgroup$
1
  • $\begingroup$ Thanks folks, looks like where i went wrong was adding instead of multiplying. Thanks $\endgroup$
    – user445714
    Commented Mar 16, 2014 at 8:06

2 Answers 2

2
$\begingroup$

Normally, you want to work with codes in standard form - that is, you want to write the generator matrix in the form $[I | P]$. Then, a parity check matrix corresponding to this code is $[-P^T | I]$ (i.e. it is a generator of the dual code - the code consisting of all vectors orthogonal to the original code).

$\endgroup$
1
$\begingroup$

I don't get the Then I sum out the two matrices part. You're supposed to compute $\begin{bmatrix} 1 & 1 & 0 & 1\end{bmatrix} \times \text{Parity matrix}^T$ to find the syndrome.

Then you get the syndrome $\begin{bmatrix} 1 & 1\end{bmatrix}^T$ which yields the correction $0101$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .