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

• Oct 10th 2012, 03:08 AM
courteous
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?
• Oct 17th 2012, 08:16 AM
courteous
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. :)
• Oct 17th 2012, 08:24 AM
topsquark
Re: Z-transform: what is Z{n * u[n - N]} ?
Quote:

Originally Posted by courteous
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! (Nod)

-Dan