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How can I prove that the ROC (region of convergence) of the Laplace Transform of any finite-duration signal is the entire complex plane?

Lets think of a "friendly" function $f(t)$ such is zero outside the interval $t \in (a,\ b)$ with $0\leq a<b<\infty$ (so, the function is compact-supported in the time variable $t$).

I want to understand why its Laplace Transform's region of convergence (ROC) is the whole complex plane, and how it is proved (or disproved).

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  • $\begingroup$ if you had written the Laplace transform formula for your function you would have seen immediately the answer $\endgroup$
    – reuns
    Feb 18, 2016 at 1:48
  • $\begingroup$ I don't have a specific function, the question is to generally prove the ROC for ANY finite duration signal $\endgroup$
    – dms94
    Feb 18, 2016 at 1:59
  • $\begingroup$ I believe it is due the Paley–Wiener theorem, (and maybe the accepted answer is wrong - but I am not really sure) $\endgroup$
    – Joako
    Nov 15, 2023 at 2:50

1 Answer 1

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Suppose $f(t)$ is the function that represents the signal "in the time domain." The Laplace transform (by definition) is $$\int_0^\infty f(t)e^{-st}\,dt$$ If $f\in L^1[0,\infty)$, this exists for $\mathrm{Re}\,s\ge 0$. Now suppose that $f$, in addition to being $L^1$, is zero outside the interval $[0,a]$, which corresponds to the signal having finite duration, namely from $t=0$ up to $t=a$. This implies that in the above integral, $|e^{-st}|=e^{-t\mathrm{Re}\,s}$ $\le e^{|s|a}$ for $t$ with $f(t)\ne 0$, which in turn gives $$\int_0^\infty |f(t)e^{-st}|\,dt\le e^{|s|a}\int_0^\infty |f(t)|\,dt$$ This shows that the Laplace transform is finite for all $s$.

Note this answer uses some "mathematical lingo," which may not be the usual language of signals and systems. Just in case, $f\in L^1$ just means that $\int |f|<\infty$. If the function $f$ has finite duration and is bounded, then it's easy to see that it's $L^1$.

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  • $\begingroup$ I have a doubt about your answer: if a function $f(t)$ is of finite duration, then in the time variable $t$ it is compact-supported, and the Fourier Transform of a compacted-supported function is given by an Analytic function due the Paley–Wiener theorem, so its spectra cannot be of also of compact-support since due the Identity Theorem a power series cannot match a constant value in a non-zero measure interval if it is not just a constant value for all $t$. (...) $\endgroup$
    – Joako
    Nov 15, 2023 at 2:43
  • $\begingroup$ (...) Given the similarities among the Fourier Transform and the Laplace Transform, I wonder why in this case it is possible to be of compact support in the $s$ variable? Since $s=\sigma+iw$ I am expecting it to be unbounded at least in the complex variable at $\sigma = 0$... maybe I have misunderstood everything but it do surprise me the Laplace Transform could be finite when the signal in the time domain is also finite. $\endgroup$
    – Joako
    Nov 15, 2023 at 2:46

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