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It is easy to define an (ideal) LTI system that would have an infinite number of poles - for instance, if the transfer function is $$ H(z)=\frac{1}{\cos(z)-1} $$ However, this would only define a countably infinite set of poles.

I am curious: are there systems with uncountably infinite sets of poles? I would guess that the very definition of a pole would prohibit a function from having a contiguous region of poles, but what about a hypothetical transfer function that would have its poles defined on something like Cantor set?

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    $\begingroup$ Can you define an analytic function that is zero for a contiguous segment of $x$ having non-zero measure and is non-zero outside of that contiguous segment? $\endgroup$
    – robert bristow-johnson
    Commented Nov 28, 2020 at 19:25
  • $\begingroup$ Wow that's an interesting idea with the Cantor set. I guess one could as well ask the question about zeros instead of poles. $\endgroup$
    – Olli Niemitalo
    Commented Nov 28, 2020 at 19:45
  • $\begingroup$ It seems to me that it is possible to have a non-negative and $C_\infty$ function that vanishes only on the Cantor set. $\endgroup$
    – Laurent Duval
    Commented Nov 28, 2020 at 23:45
  • $\begingroup$ isn't the Cantor set a countably infinite set of points? $\endgroup$
    – robert bristow-johnson
    Commented Nov 29, 2020 at 1:19
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    $\begingroup$ @robert No, being uncountably infinite is one of the most important properties of the Cantor set. $\endgroup$
    – Mr_Tusk
    Commented Nov 29, 2020 at 1:22

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It turns out that there's no way a function's Laplace transform can have uncountably many poles. The reasoning should be based on the fact that a Laplace transform of a function must be holomorphic (which can be derived using Morera's theorem), and since its poles must be, by definition, isolated points, there must be a finite number of them for any compact subset of the transform's ROC (that's why my original idea with a function having its poles defined on the Cantor set would fail). And because of that, the total set of poles is a union of countably many finite sets - so it must be countable as well. For the z-transform, the reasoning should be the same. The Math Stack Exchange had several related questions.

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This really comes down to how you define LTI.

Far as I can tell, a time-discrete system being LTI means its in- and output are related by a countably large system of ordinary differential equations.

Since each of these transforms into a polynomial through the Laplace-, Fourier- or z-transform, the transfer function, in which ever domain it exists, is always a quotient of polynomials. Coefficients of polynomials are countable.

Since polynomials have, including their multiplicity, exactly the number of roots as their degree, and poles are just the roots of the denominator polynomial, we'll always have countably many roots – and thus, poles.

Not quite sure you'd want to relax the constraint that poles be countable, anyway: If you do that, transfer functions cease to be meromorphic, and that means the residue theorem doesn't apply, and as far as I can intuit, that leads to unsolvable integrals that you'd need to define the $z$-transform to begin with...

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  • $\begingroup$ As a German, you know well that your oriental neighbors are about 40 millions, hence countable. $\endgroup$
    – Laurent Duval
    Commented Nov 29, 2020 at 13:41
  • $\begingroup$ The class of LTI systems is larger than the ones that are described by systems of differential (or difference) equations. As long as input and output are related by convolution, the system is LTI. E.g., in continuous-time, an ideal delay $h(t)=\delta(t-T)$ is LTI (but not described by a differential equation). In discrete time, the same is true if the delay is not an integer. Other examples of LTI systems not described by (a finite number of) differential (difference) equations are ideal filters. $\endgroup$
    – Matt L.
    Commented Nov 29, 2020 at 17:44
  • $\begingroup$ @MattL. ah, I need to unwrap your comment: If the delay is not an integer, we just sum up an infinite (but countable) set of interpolation points, right, to get the coefficients that go with each integer $z$ power. that doesn't change the fact that we've got a polynomial that can only have a countable set of roots. # $\endgroup$
    – Marcus Müller
    Commented Nov 29, 2020 at 17:51
  • $\begingroup$ I feel a bit uncomfortable with transfer functions that are a ratio of two polynomials with infinite order ... Following that logic, an ideal low pass filter should have such a transfer function. However, the corresponding power series doesn't converge, which is the reason why an ideal lowpass filter actually has no transfer function. I'm not sure though what bearing that has on the question. $\endgroup$
    – Matt L.
    Commented Nov 30, 2020 at 12:48

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