# Math Help - z-transform, some guestion

1. ## 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..?

2. ## Re: z-transform, some guestion

Originally Posted by liquidFuzz
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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Thanks!

4. ## 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: $\displystyle a_{n+2} + a_n = n$