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Math Help - z-transform, some guestion

  1. #1
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    z-transform, some guestion

    I sorta know how to transform this sum, but only by heart.
    \sum_{k=0}^n a_k 3^{n-k} =[z-transform]= A(z) \frac{z}{z-3}

    I'd really like to understand why it transforms like it does. Can anyone explain it to me..?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: z-transform, some guestion

    Quote Originally Posted by liquidFuzz View Post
    I sorta know how to transform this sum, but only by heart.
    \sum_{k=0}^n a_k 3^{n-k} =[z-transform]= A(z) \frac{z}{z-3}

    I'd really like to understand why it transforms like it does. Can anyone explain it to me..?
    If You have two sequences a_{n} and b_{n}, then their 'correlation product' is by definition the sequence that has general term...

    c_{n}= \sum_{k=0}^{n} a_{k}\ b_{n-k} (1)

    If You denote with A(z) the Z-transform of a_{n}, B(z) the Z-transform of b_{n} and C(z) the Z-transform of c_{n}, then You have the well known relation...

    C(z)=A(z)\ B(z) (2)

    If You have c_{n}=\sum_{k=0}^{n} a_{k}\ 3^{n-k}, that means that is b_{n}= 3^{n} and is...

    B(z)= \sum_{n=0}^{\infty} b_{n}\ z^{-n}= \sum_{n=0}^{\infty} 3^{n}\ z^{-n}=   \frac{1}{1-3\ z^{-1}} (3)

    ... so that (2) becomes...

    C(z)=\frac{A(z)}{1-3\ z^{-1}} (4)



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    Re: z-transform, some guestion

    Thanks!
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  4. #4
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    Re: z-transform, some guestion

    I just realised, it's convolution, right?

    I've seen differential equations described with series. Can the transform of for instance a_{n+2} in a series can be written like this?

    If we have \displystyle A(z) = \sum_{n=0}^\infty a_n z^{-n}.

    \displystyle \sum_{n=0}^\infty a_{n+2} z^{-n} = \sum_{k=2}^\infty a_{k} z^{-k+2} = z^2 \sum_{n=0}^\infty \left( a_{n} z^{-n}-a_0 z^0 - a_1 z^{-1} \right) = z^2 \left( A(z)-a_0 z^0 - a_1 z^{-1} \right)

    Is this the way of setting up a solution using z-transform to for instance: \displystyle a_{n+2} + a_n = n
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