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Thread: Help for Z transform of t^2

  1. #1
    Member moonnightingale's Avatar
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    Help for Z transform of t^2

    I am unable to calculate Z transform of t^2 ( t Square)
    Can any body show me step by step method to calculate its Z transform
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by moonnightingale View Post
    I am unable to calculate Z transform of t^2 ( t Square)
    Can any body show me step by step method to calculate its Z transform
    The (one sided) z-transform is defined as:

    $\displaystyle
    \displaystyle
    \mathcal{Z}(t^2)=\sum_{t=0}^{\infty} \frac{t^2}{z^n}
    $

    Now you proceed by finding the sum of the series on the right (or just looking $\displaystyle n^2$ up in a table of z-transforms).

    CB
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  3. #3
    MHF Contributor chisigma's Avatar
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    The [one side] z-transform is usually applicable to a discrete sequence $\displaystyle a_{n}$ and is defined as...

    $\displaystyle \displaystyle \mathcal{Z} \{a_{n}\} = \sum_{n=0}^{\infty} a_{n} z^{-n}$ (1)

    According to Z-Transform -- from Wolfram MathWorld, if You have a continous function $\displaystyle f(t)$ , its z-transform is the z-tranform of the sequence $\displaystyle a_{n} = f(nT)$ , where T is the so called 'sampling time'. In your case is $\displaystyle f(t)= t^{2}$ so that is...

    $\displaystyle \displaystyle \mathcal{Z} \{f(nT)\} = T^{2} \sum_{n=0}^{\infty} n^{2} z^{-n}= T^{2} \frac{z (z+1)}{(z-1)^{3}}$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    Member moonnightingale's Avatar
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    yes i know answer but how did u perform last step
    Kindly explain how u wrote (z-1)^3 in denominator
    I am well conversant with Z transform but this step involves some math series
    Kindly explain ur last step
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  5. #5
    Grand Panjandrum
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    You differentiate:

    $\displaystyle
    \displaystyle
    f(z)=\frac{z}{z-1}=1+z^{-1}+z^{-2}+...+z^{-n}+ ... \ \ ;\ z \in (-1,1)
    $

    twice and manipulate the results to get the series $\displaystyle 1+z^{-1}+2 z^{-2}+...+n z^{-n}+ ... \ \ $ on the right hand side.

    CB
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  6. #6
    MHF Contributor chisigma's Avatar
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    In detail You start from the expansion...

    $\displaystyle \displaystyle \frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}$ , $\displaystyle |z|<1$ (1)

    ... and then You derivate (1) and obtain ...

    $\displaystyle \displaystyle \frac{d}{dz} \frac{1}{1-z} = - \frac{1}{(1-z)^{2}} = \frac{1}{z} \sum_{n=0}^{\infty} n \cdot z^{n}$ (2)

    ... so that is...

    $\displaystyle \displaystyle \sum_{n=0}^{\infty} n\cdot z^{n} = -\frac{z}{(1-z)^{2}}$ (3)

    Now You derivate (3) and obtain...

    $\displaystyle \displaystyle \frac{d}{dz} \{- \frac{z}{(1-z)^{2}}\} = - \frac{1+z}{(1-z)^{3}} = \frac{1}{z} \sum_{n=0}^{\infty} n^{2} z^{n}$ (4)

    ... so that is ...

    $\displaystyle \displaystyle \sum_{n=0}^{\infty} n^{2} z^{n} = \frac{z\cdot (z+1)}{(z-1)^{3}}$ (5)

    ... and, replacing $\displaystyle z$ with $\displaystyle z^{-1}$, ...

    $\displaystyle \displaystyle \sum_{n=0}^{\infty} n^{2} z^{-n} = \frac{z^{-1}\cdot (1+ z^{-1})}{(z^{-1}-1)^{3}}$ (6)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
    Member moonnightingale's Avatar
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    Thanks a lot for detailed reply
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