Link between Z-transform and Taylor series expansion

In summary, the conversation discusses the relationship between the Z-transform and Taylor series expansion in order to compute the inverse Z-transform. The textbook suggests using the Taylor series expansion of ##z^{-n_0} F(z^{-1})## around ##z=0## to find the ##f[n]## values. This was a new concept for the person, who was only familiar with other methods of finding the inverse Z-transform. They express interest in expanding their understanding of this reasoning.
  • #1

fatpotato

Homework Statement
Inverting a Z-transform using Taylor series expansion
Relevant Equations
See embedded image in my post
Hello,

I am reading a course on signal processing involving the Z-transform, and I just read something that leaves me confused.

Let ##F(z)## be the given Z-transform of a numerical function ##f[n]## (discrete amplitudes, discrete variable), which has a positive semi-finite support and finite ROC. My textbook says that we should be able to compute the inverse Z-transform of ##F(z)##, so the ##f[n]## values, using the fact that "##f[n]## values are given by the Taylor series expansion of ##z^{-n_0} F(z^{-1})## around ##z=0##", using the following equation:

1654787498356.png

This sounds fascinating, but I don't understand! How are Z-transform and Taylor series expansion linked, and why? I only ever learned about inverting the Z-transform using tables, long division or using the inverse transform definition involving contour integral in the complex plane (which I never had the chance to use).

Any information is welcome!
 
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  • #2
Look at the definition: [tex]
F(z) = \sum_{n=0}^\infty f[n]z^{-n}[/tex] Setting [itex]t = z^{-1}[/itex] this is [tex]
F(t^{-1}) = \sum_{n=0}^\infty f[n]t^{n}.[/tex] This is a power series in [itex]t[/itex], so if [itex]G(t) = F(t^{-1})[/itex] has a convergent Taylor series at 0 this is [tex]
\sum_{n=0}^\infty f[n]t^{n} = \sum_{n=0}^\infty \frac{1}{n!} G^{(n)}(0)t^n[/tex] and hence [tex]f[n] = \frac{1}{n!} G^{(n)}(0) = \frac{1}{n!}\left.\frac{d^n}{dt^n}F(t^{-1})\right|_{t=0}.[/tex]
 
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  • #3
Beautiful, thank you for the derivation!

In a classical engineering program setting, I tend to believe that applying Taylor series would be limited to Calculus and I would have never thought of using it here.

Would you happen to have a suggestion on where I can expand my horizons to come up with such reasonings myself?
 

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