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It's well known that you can estimate the Fourier Transform $X(f)$ of a signal $x(t)$ via its Laplace Transform $X(s)$, just by setting $s = j2\pi f$ to the latter, as long as the region of convergence includes the imaginary axis.

However, I do not have a clear view of how (and when) we can obtain the Laplace Transform via the Fourier Transform of a signal (which is the opposite of what I've stated before).

For example, $x(t) = e^{-at}u(t)$ has a Fourier Transform $X(f) = \frac{1}{a+j2\pi f}$ as long as $a > 0$. Its Laplace Transform is $X(s) = \frac{1}{a+s}$, for any value of $a$, as long as $\mathrm{Re}\{s\} > -a$. We can see that if we set $j2\pi f = s$ to the Fourier Transform, we can directly obtain the Laplace Transform. The same holds for any rational function of $j2\pi f$.

Is there a theorem or something that can be clearly stated about it? It looks to me that it has something to do with the convergence of the Fourier integral (if it does converge, then the Laplace Transform converges as well).

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    $\begingroup$ i guess it would be the same as long as the Laplace Transform is the two-sided integral, like the Fourier integral and "the region of convergence includes the imaginary axis". then it's just a substitution of symbols: $ s \leftrightarrow j 2 \pi f $ because the integrals are otherwise identical. $\endgroup$ Mar 23, 2019 at 3:02
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    $\begingroup$ So it sounds like X(s) is the analytic continuation of X(f), given the proper restrictions, is that right? $\endgroup$
    – GKH
    Mar 25, 2019 at 16:06

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You need to distinguish three cases:

  1. There are Dirac impulses in the expression for the Fourier transform. In this case you can't just replace $j\omega$ by $s$ to obtain the Laplace transform. The Laplace transform might not exist, or its form is different from the expression for the Fourier transform. A simple example for which the Laplace transform doesn't exist is $x(t)=e^{j\omega_0t}$ with Fourier transform $X(j\omega)=2\pi\delta(\omega-\omega_0)$. An example for which the Laplace transform exists but for which it cannot be obtained by setting $j\omega=s$ is $x(t)=u(t)$ with Fourier transform $X(j\omega)=\pi\delta(\omega)+\frac{1}{j\omega}$ and Laplace transform $X_L(s)=\frac{1}{s}$.

  2. There are no Dirac impulses in the expression for the Fourier transform, but replacing $j\omega$ by $s$ results in poles on the imaginary axis. In this case, replacing $j\omega$ by $s$ results in a valid expression for the Laplace transform, but it's the Laplace transform of a different time domain function. Example: $X(j\omega)=\frac{1}{j\omega}$ corresponds to $x(t)=\frac12\textrm{sgn}(t)$, whereas $X(s)=\frac{1}{s}$ corresponds to $x(t)=u(t)$.

  3. There are no Dirac impulses in the expression for the Fourier transform, and replacing $j\omega$ by $s$ does not result in any poles on the imaginary axis. In that case, the Laplace transform can be found by replacing $j\omega$ by $s$. However, the expression for the Laplace transform generally corresponds to several different time-domain functions, depending on the chosen region of convergence (ROC). Only the one with the ROC including the imaginary axis corresponds to the time-domain function described by the given Fourier transform.

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  • $\begingroup$ Could you please give an example of case #3? Given an explicit expression $X(s)$ for a Laplace transform, isn't the region of convergence just $\mathrm{Re}(s) > \alpha$, where $\alpha \in \bar{\mathbb{R}}$ is the least (extended) real number such that $X(s)$ is analytic on the domain $\mathrm{Re}(s) > \alpha$? $\endgroup$
    – tparker
    Jan 26 at 5:15
  • $\begingroup$ @tparker: If you consider the general case, i.e., the bilateral Laplace transform, which can also be applied to functions that don't vanish for $t<0$, then the ROC is generally a vertical strip. For a given expression $X(s)$, there are several possible ROCs, and as many inverse transforms. $\endgroup$
    – Matt L.
    Jan 26 at 12:09
  • $\begingroup$ I gave an example in my answer at dsp.stackexchange.com/a/91873/71032. I may have gotten some signs wrong. $\endgroup$
    – tparker
    Jan 27 at 20:47
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So it sounds like X(s) is the analytic continuation of X(f), given the proper restrictions, is that right?

that's what i would say. be careful about expressing the function and its argument.
use a different "$X(\cdot)$" if you need to. like: $$ X_\mathscr{L}(j 2 \pi f) = X_\mathscr{F}(f) $$
where $$ X_\mathscr{L}(s) \triangleq \mathscr{L}\Big\{ x(t) \Big\} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-st} \, \mathrm{d}t $$ and $$ X_\mathscr{F}(f) \triangleq \mathscr{F}\Big\{ x(t) \Big\} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2\pi f t} \, \mathrm{d}t $$

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