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Math Help - Z-transform: what is Z{n * u[n - N]} ?

  1. #1
    Member courteous's Avatar
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    Z-transform: what is Z{n * u[n - N]} ?

    These two Z-transform properties are relevant
    x[n-n_0] \Longleftrightarrow z^{-n_0}X(z)
    nx[n] \Longleftrightarrow -z \frac{dX(z)}{dz}

    yet I don't know how (in which order) one should apply them on the following:
    Z(n u[n - N])

    Even using the z-transform definition, I cannot get a closed form solution:
    Z(n u[n - N]) = \sum_{n=-\infty}^{\infty} n u[n-N] z^{-n} = \sum_{n=N}^{\infty} n z^{-n} = \sum_{n=0}^{\infty} n z^{-n} - \sum_{n=0}^{N-1} n z^{-n} = \frac{z^{-1}}{(1-z^{-1})^2 } - \sum_{n=0}^{N-1} n z^{-n} = ?

    Is there a closed form of \sum_{n=0}^{N-1} n z^{-n} = z^{-1} + 2z^{-2} + ... + (N-1)z^{-(N-1)} ? What is it?


    PS. Is there a more suitable sub-forum I should be posting these DSP questions? Or is this it?
    Last edited by courteous; October 10th 2012 at 03:24 AM.
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  2. #2
    Member courteous's Avatar
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    Re: Z-transform: what is Z{n * u[n - N]} ?

    How silly of me, should've just take a derivative of \sum_{n=0}^{\infty} z^n which gives (considering the linearity of a differentiation op.)
    \left(\sum_{n=0}^{\infty} z^n\right)' = \frac{1}{(1-z)^2} = \sum_{n=0}^{\infty} n z^{n-1}. Therefore, \sum_{n=0}^{\infty} n z^n = \frac{z}{(1-z)^2}. And then some more paper-gymnastics for the partial sum of the same thing.

    I really need to take this "math side of physics" more seriously ... yes, I'm writing mostly to myself, but who cares anyway. Thanks for the help.
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    Forum Admin topsquark's Avatar
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    Re: Z-transform: what is Z{n * u[n - N]} ?

    Quote Originally Posted by courteous View Post
    I really need to take this "math side of physics" more seriously ... yes, I'm writing mostly to myself, but who cares anyway. Thanks for the help.
    Yeah, I do that all the time!

    -Dan
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