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With integral transforms both the transform and its inverse are integrals.

In the case of the Z-Transform the transform is a sum.

My question
Why do you need an integral (instead of another sum) to invert the transform? Are there cases where a sum would suffice?

vonjd
  • 8,810
  • Note that $x[n]$ is defined on a discrete domain, wheras its z-Transform is not. In terms of the link you provided, the z-Transform is denoted by $X(z)$, which is a complex function, i.e. the independent variable $z$ of the function is a complex number, and in particular, there are more than countably many values which you could insert here. Thus, an action which involves 'summing over the values in the domain' will not be a sum, but an integral. I know that this comment is rather vague and does not answer why, but only points out a significant difference. – Roland May 16 '14 at 14:38

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