The basics are that GxP^t == 0 mod 2 (for even parity) or GxP^t == 1 mod 2
(for odd parity). The notation '^t' denotes the transposed of a matrix. If G is
given you can compute P^t and hence P.
kind regards,
Jos
ps. Google knows about efficient methods to compute P given the generator
I have this example with the answer, but I’m sure the way I use to find the Parity Check Matrix is correct. So please help me along, if I’m wrong.
Example, the generator matrix for a [7,4] linear block code is given as
| 1 0 0 0 1 0 1 |
| 0 1 0 0 1 1 1 |
| 0 0 1 0 0 1 0 | = G
| 0 0 0 1 0 1 0 |
----------------------------------------------------------------------------------------
Am I correct that G is the first 4 bit and P is the last 3 bit because of the [7,
4]???
Is it the same for [5, 3], G is the first 3 bit???
| 1 0 0 0 1 0 1 |
| 0 1 0 0 1 1 1 |
| 0 0 1 0 0 1 0 | = G
| 0 0 0 1 0 1 0 |
|___G__|__P_|
If is correct, I have the P and can compute P^t.
| 1 0
1 |
| 1 1
1 |
| 0 1
0 | = P
| 0 1
0 |
So
|
1 1 0 0 |
| 0 1 1 1 | = P^t
| 1 1 0 0 |
The Parity Check Matrix formula I’m given is H = [
-P^t | I]
So I take the P^t and add on with I behind. Which the result is:
| 1 1 0 0 1 0 0 |
| 0 1 1 1 0 1 0 | = H
| 1 1 0 0 0 0 1 |
|__P^t__|__I__|
But where did the
‘–‘P^t go to??
As my given answer is this without the
‘–‘.
I’m not sure how did the I come from but it looks like a method by putting a bit at a time.
Lot of help
Ken JS
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