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Math Help - Inverse Z transform by inversion integral method

  1. #1
    Member moonnightingale's Avatar
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    Inverse Z transform by inversion integral method

    Hi guys i am stuck with this question.
    I have to calculate inverse Z transform by inversion integral method of

    X(Z) = 1 / (1+ z^ -2) (z raise to power -2)

    I have done it in matlab and by computation menthod and answer comes to be same as

    [1 0 -1 0 1 0 -1 - - - - - ]

    but when i do it with inversion integral my answer comes to be i^k ( i= iota)

    if i put values of k as 0 1 2 3 my answers does not match
    plz help me
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  2. #2
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    Quote Originally Posted by moonnightingale View Post
    Hi guys i am stuck with this question.
    I have to calculate inverse Z transform by inversion integral method of

    X(Z) = 1 / (1+ z^ -2) (z raise to power -2)

    I have done it in matlab and by computation menthod and answer comes to be same as

    [1 0 -1 0 1 0 -1 - - - - - ]

    but when i do it with inversion integral my answer comes to be i^k ( i= iota)

    if i put values of k as 0 1 2 3 my answers does not match
    plz help me
    I get the MATLAB answer (which is consistent with the series expansion of X(Z)). Please post all your working.
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  3. #3
    Member moonnightingale's Avatar
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    I cannot post my working here as i have done it with hand and it will take a lot time to type in Microsoft word for me

    Kindly if u can do this question by inversion integral and upload it
    the poles come to be + iota and - iota
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  4. #4
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    Quote Originally Posted by moonnightingale View Post
    I cannot post my working here as i have done it with hand and it will take a lot time to type in Microsoft word for me

    Kindly if u can do this question by inversion integral and upload it
    the poles come to be + iota and - iota
    So what you're saying is that it's too much work for you to type out here but not too much work for me.

    Wrong reply.
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