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Hi, I am trying to make a parity check matrix from non-systematic to systematic. Hence I am attaching my code below. Somewhat it is correct, but there are some problems. It would be really great if someone could help me in this. Thanks

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sb
sb 2016 年 6 月 9 日
コメント済み: SP22 2017 年 11 月 20 日
Information theory and coding. I am working on LDPC coding and decoding. Please check the code below
  2 件のコメント
Walter Roberson
Walter Roberson 2016 年 6 月 9 日
Do we get some hints about what the problems are ?
sb
sb 2016 年 6 月 10 日
編集済み: Walter Roberson 2017 年 5 月 25 日
For e.g.
This is my H matrix
H =
1 0 1 1 0
0 0 1 0 1
1 0 0 1 0
1 0 1 1 1
I want to have an identity matrix inside the matrix. the dimension on H matrix here is (mxn) which is (4x5).
I should have matrix as this in the result:
Hsys =
0 1 0 0 0
0 0 1 0 0
1 0 0 1 0
0 0 0 0 1
I should have an identity matrix of dimension 'm'

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回答 (1 件)

freebil
freebil 2016 年 6 月 10 日
You have to do gauss jordan elimination to convert a parity check matrix to upper triangular form. For example,
H=[1 1 0 1 1 0 0 1 0 0;
0 1 1 0 1 1 1 0 0 0;
0 0 0 1 0 0 0 1 1 1;
1 1 0 0 0 1 1 0 1 0;
0 0 1 0 0 1 0 1 0 1];
There is rref() in matlab and you have to do it in GF(2). So,
HH = mod(rref(H),2)
gives
HH = 1 0 0 0 0 0 1 1 1 0
0 1 0 0 0 1 0 1 0 0
0 0 1 0 0 1 0 1 0 1
0 0 0 1 0 0 0 1 1 1
0 0 0 0 1 1 1 0 0 1
  2 件のコメント
Aitor López Hernández
Aitor López Hernández 2017 年 5 月 25 日
Hello there,
Wouldn't the systematic form of a parity check matrix be of the form H = [A In-k]?
SP22
SP22 2017 年 11 月 20 日
This might help you to get parity check matrix in form H=[A In-k].
temp=HH(:,1:5) %Access the identity matrix through column
temp2=HH(:,6:10) %Acces the parity through column
Hsyst=horzcat(temp2,temp) %Actual systematic matrix in the form H((n-k)*n)=[P' :I(n-k)]

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