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Z-Transform of finite series.

  1. Apr 9, 2015 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data
    Let ## x_j = \begin{Bmatrix}
    {1, 0 \leq j \leq N-1} \\
    {0, else} \\
    \end{Bmatrix}
    ##

    Show that ##\hat{x}(\phi) = \frac{e^{-i\frac{N-1}{2}\phi}sin(\frac{N}{2}\phi)}{sin(\frac{1}{2}\phi)}##


    2. Relevant equations

    ##\hat{x}(\phi) := \sum_{j = -\infty}^{\infty} x_j e^{-ij\phi}##

    3. The attempt at a solution

    So I get ##\hat{x}(z) = \sum_{j = 0}^{N-1} z^{-j} = \sum_{0}^{N-1} (\frac{1}{z})^{j}##

    I believe this is a geometric series, with sum ##\hat{x}(z) = \frac{1-z^{-N}}{1-z^{-1}} = \frac{1-e^{-iN\phi}}{1-e^{-i\phi}}##

    Of course this makes the assumption that ##|\frac{1}{z}| < 1 ## which I am not entirely sure about. Any tips appreciated.
     
  2. jcsd
  3. Apr 9, 2015 #2

    Orodruin

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    Yes, it is a geometric series. Since it is finite you do not need the condition on z for convergence.

    What remains is to recast it on the form given in the assignment. I suggest working backwards if you have problems.
     
  4. Apr 9, 2015 #3

    ElijahRockers

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    Sorry, I should have added that's where I am stuck. I can play around with ##e^{ix} = cos(x) + isin(x)## but it doesn't seem to be getting any closer to the final answer.
     
  5. Apr 9, 2015 #4

    Orodruin

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    This is why I suggest working backwards. How can you express sin(x) in terms of exponents of ix?
     
  6. Apr 9, 2015 #5

    ElijahRockers

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    Got it! Thank you!
     
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