How do I find the Parity Check Matrix, when I’m given the generator matrix? e.g.
| 1 0 0 1 1 |
G = | 0 1 0 1 2 |
| 0 0 1 1 3 |
Can anybody teach me how to find the Parity Check Matrix or any website that will teach me?
Lots of help,
Ken JS
Ken JS 0 Newbie Poster
Ken JS 0 Newbie Poster
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
darkagn 315 Veteran Poster Featured Poster
In binary, -1=1, so -P^t=P^t. What you have done looks good to me, but maybe someone else can check for you also? It's a long time since I did this sort of thing...
Ken JS 0 Newbie Poster
In binary, -1=1, so -P^t=P^t. What you have done looks good to me, but maybe someone else can check for you also? It's a long time since I did this sort of thing...
Oic... Thanks darkagn...
For this example, Generator Martix is not fully in binary:
| 1 0 0 1 1 |
| 0 1 0 1 2 | = G
| 0 0 1 1 3 |
will the P become this:
| 1 1 |
| 1 2 | = P
| 1 3 |
| 1 2 3 |
| 1 1 1 | = P^t
or this (mod2)?
| 1 1 |
| 1 0 | = P
| 1 1 |
| 1 0 1 |
| 1 1 1 | = P^t
lots of help
Ken JS
darkagn 315 Veteran Poster Featured Poster
Sorry there Ken JS. I'm struggling to remember this stuff but I thought that the parity check matrix was generally binary. Because your G is not in binary, I don't think you can apply (mod 2) to it, so I would say that your first P^t is correct. I must stress that I'm not 100% sure about this though, is there someone else out there that knows this for sure?
SITI HARYANTI 0 Newbie Poster
guite useful... thnx ya 4 ur infOrmatiOn. may GOD always bless u... :)
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