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Given that $|a|<1$, then which of those inverse-Z-transform equations are we to use?

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I am leaning towards the first because (as I understand it), $z$ is merely a complex number that is evaluated on the unit circle, and thus $|z|$ must always equal 1? I guess my question is, when is $|z|=1$? Always?

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Which one you choose is quite arbitrary: it depends on which region of convergence (ROC) you choose.

As Example 2 and Example 3 of that wikipedia page shows, if you want the signal to be causal, then choose the first one. If you want the signal to be anti-causal, choose the second.

Another way to think about it is what happens when $|z| = 1$. On the unit circle is where the Fourier transform of the signal is defined. If you want that to be finite, then you need to choose the inverse that includes the unit circle.

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  • $\begingroup$ Thanks Peter. Hmm... well I arrived at $\frac{1}{1-az^{-1}}$ because that is the z-transform, $H(z)$ of a filter that can generate $x[n] = ax[n-1] + w[n]$, given that is is fed by $w[n]$, where $w[n]$ is an standard gaussian r.v. So when I came to take the inverse, I did not know which one to pick. I suppose for this case, the causal one then... $\endgroup$ Commented Sep 5, 2013 at 1:29
  • $\begingroup$ I would say you usually want the causal one. In your case, as $|a| < 1$, the unit circle is only included in the ROC by the causal one. $\endgroup$
    – Peter K.
    Commented Sep 5, 2013 at 12:15

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