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Theorem: Let $$f(z) = \sum_{n=0}^{+\infty}a_nz^n$$ where $z\in\mathbb{C}$. If $f(z_0)$ exists for some $z_0\in\mathbb{C}$ then it converges for all $z\in\mathbb{C}$ such that $|z|\lt|z_0|$. Proof: It follows from the hypothesis that $\exists M\ge0 : |a_nz_0^n|\le M$ for all $n\in\mathbb{N}$. We have $|a_nz^n|=|a_nz_0^n||\frac{z}{z_0}|^n\le M|\frac{z}{z_0}|^n$. By the comparison theorem and the behavior of geometric series, we conclude that $f(z)$ converges for $|z|<|z_0|$.

So if we choose $z_0$ such that $f(z_0) = 0$, then $f(z)$ converges for $|z|\lt|z_0|$(Because $f(z_0) = 0$ means that $f(z_0)$ exists). In other words, ROC of $\mathcal{Z}$-Transform depends on the location of poles as well as zeros of $f(z)$ but according to the literature, ROC depends only on the location of poles. So what's my mistake here? Does the location of zeros affect the ROC?

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  • $\begingroup$ Why do you think you get to choose $z_0$? $\endgroup$
    – Jazzmaniac
    Oct 6 '21 at 11:28
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    $\begingroup$ @Jazzmaniac: I think the OP means to say that $z_0$ simply is a zero of $X(z)$, so $X(z_0)=0$ is satisfied, and, hence - at least according to the OP - $X(z)$ should converge for any $z$ satisfying $|z|<|z_0|$ (because it apparently converges for $z=z_0$). $\endgroup$
    – Matt L.
    Oct 6 '21 at 11:44
  • $\begingroup$ I'm confused. As far as I understand poles and zeros are defined for systems but the Z-transform is applied to signals. What's the pole of a signal ? $\endgroup$
    – Hilmar
    Oct 6 '21 at 12:29
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    $\begingroup$ @Hilmar: $x[n]$ is just a sequence; its interpretation is irrelevant here, isn't it? We usually compute Z-transforms of signals and of impulse responses, so both interpretations are fine I guess. $\endgroup$
    – Matt L.
    Oct 6 '21 at 13:40
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    $\begingroup$ @Jazzmaniac: I'm not sure I understand it, but that's my interpretation of the question. I think the OP's argument goes as follows: if $X(z)$ converges for some $z_0$ then it must converge for all $z$ satisfying $|z|<|z_0|$. If $z_0$ is a zero of $X(z)$ then $X(z_0)=0$ holds, and, consequently, $X(z)$ converges for $z_0$, hence it converges for all $z$ satisfying $|z|<|z_0|$. $\endgroup$
    – Matt L.
    Oct 6 '21 at 16:09
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Just because $f(z)$ converges for $|z|\le|z_0|$ for your choice of $\{ z_0 : f(z_0) = 0\}$ doesn't imply that $f(z)$ can't converge for $|z| \gt |z_0|$ also... up until the pole of next greatest magnitude.

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  • $\begingroup$ Thanks. So this theorem doesn't restrict ROC in any way? I mean it seems to me that this theorem shows ROC and location of zeros are related and should be considered. $\endgroup$
    – S.H.W
    Oct 6 '21 at 19:12
  • $\begingroup$ @S.H.W Right: it tells you part of where the ROC is, it just doesn't tell you everything about where the ROC is. $\endgroup$
    – Peter K.
    Oct 6 '21 at 20:47

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