# Parity Check Matrix

• 08-18-2007
Ken JS
Parity Check Matrix
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
• 08-19-2007
VirtualAce
What?