Thread: Parity Check Matrix

  1. #1
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    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

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    Registered User VirtualAce's Avatar
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