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

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

Jump to PostClick ME

I dunno, what were you expecting as a response?

Jump to PostIn 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

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
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Veteran Poster
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Ken JS
0
Newbie Poster

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
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SITI HARYANTI
0
Newbie Poster

guite useful... thnx ya 4 ur infOrmatiOn. may GOD always bless u... :)

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