# Laplace transform region of Convergence of a finite duration signal [closed]

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 $$0\leq a (so, the function is compact-supported in the time variable $$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).

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

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 $L^1$.
• 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 $t$. (...) Nov 15, 2023 at 2:43
• (...) 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. Nov 15, 2023 at 2:46