- #1
electronic engineer
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- 3
I need to find the inverse z-transform for this function:
[tex] \frac {Z^2-Zr Cos W0} {Z^2-r^2 Sin^2 W0} [/tex]
[tex] \frac {Z^2-Zr Cos W0} {Z^2-r^2 Sin^2 W0} [/tex]
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You are right. It should be:electronic engineer said:thanks but I see that you didn't pay attention to the first order term (zr cos w0), I wonder whether the partial funtions expanded should be changed or it might be the same?!
You are right.electronic engineer said:and i think your previous fractions is wrong , check this website:
http://mathworld.wolfram.com/PartialFractionDecomposition.html
if you want more details about my solution or more discussing about , do post here please!
thanks.
The inverse Z-transform of \frac{Z^2-Zr\cos\omega_0}{Z^2-r^2\sin^2\omega_0}
is e^{r\sin\omega_0n}\cos(\omega_0n)
, where n
is the sample index.
The inverse Z-transform can be calculated using the inverse power series method, partial fraction decomposition, or residue theorem.
The inverse Z-transform is used to convert a discrete-time signal from the Z-domain to the time domain. It is essential in analyzing and designing digital filters and control systems.
Yes, the inverse Z-transform can be calculated for any rational Z-transform that has a finite region of convergence.
No, the inverse Z-transform is not unique. Different methods of calculation can result in different expressions that are still equivalent in the Z-domain.