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Math Help - Find Equation of Sequence with Z-transform

  1. #1
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    Find Equation of Sequence with Z-transform

    I have the following sequence: a_{n+1}=a_n i - p where i and p are constants. I need to find an equation for the sequence that is independent of the previous elements except a_0.

    The Z-transform gives: Z^{-1}A - Z^{-1}a_0=iA - \frac{p}{1-Z^{-1}} or A = \frac{p}{i}\frac{1}{(1-Z^{-1})(1-i^{-1}Z^{-1})} - \frac{a_0}{i}\frac{Z^{-1}}{1-i^{-1}Z^{-1}}.

    Doing partial fractions yields: A = \frac{p}{i-1}\frac{1}{1 - Z^{-1}} - \frac{p}{i(i-1)}\frac{1}{1 - i^{-1}Z^{-1}} - \frac{a_0}{i}\frac{Z^{-1}}{1-i^{-1}Z^{-1}}

    Now when I take the reverse Z-transform with \frac{1}{1-i^{-1}Z^{-1}} \rightarrow i^{-n} gives me the wrong answer. If I substitute \frac{1}{1-i^{-1}Z^{-1}} \rightarrow i^{n} instead, I get the correct answer. Any ideas what I've done wrongly?

    P.S.
    I thought I had made a mistake and it should be i instead of i^{-1}, but I double checked everything and cannot find any mistakes. That would suggest I've probably made a mistake in the Z-transform.
    Last edited by fobos3; December 17th 2013 at 07:17 AM.
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  2. #2
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    Re: Find Equation of Sequence with Z-transform

    It's been a long time since I played with Z transforms but the formula you started with appears incorrect.

    I have that the advance theorem shows

    a_{n+1}\to zA[z]-za_0

    not

    a_{n+1}\to z^{-1}A[z]-z^{-1}a_0 as you have listed.
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  3. #3
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    Re: Find Equation of Sequence with Z-transform

    were you able to solve this? I worked it through and your approach works if you correct that small error.

    one small note. Don't use i as a constant unless it's an iterator. It's just too easily mistaken for sqrt{-1}
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  4. #4
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    Re: Find Equation of Sequence with Z-transform

    Yes, it worked out. i\in \mathbb{R} is just a constant. In engineering we generally use j^2 = -1 so that is why I used i.
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  5. #5
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    Re: Find Equation of Sequence with Z-transform

    yeah, don't use j either :P
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