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I am reading a book on Laplace transform, and in the section on the convergence of Laplace transform for various signals the following theorem is stated, without any proof :

If a signal's Fourier transform be zero in some frequencies, or have discontinuities, it will not have Laplace transform (like sinc function, or periodic signals)

I can not see the proof. For signals with a discontinuous Fourier transform I know that $\int_{-\infty}^{\infty} |x(t)|dt$ may not converge leading to discontinuities in the Fourier spectra (again, like sinc), but I don't think it will lead to not having a Laplace transform at all.

What is the proof for the above statement? (if correct)

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    $\begingroup$ The theorem should read "zero in some frequency range", because isolated zeros are no problem. $\endgroup$
    – Matt L.
    Jan 17 '15 at 22:00
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First of all, it is important to distinguish between the unilateral and the bilateral Laplace transform. For causal signals we can use the unilateral Laplace transform. If for such a signal the Fourier transform exists, then also its Laplace transform exists. This is simply the case because here the Fourier transform is equivalent to the Laplace transform for $s=j\omega$, and if the Fourier integral converges, the Laplace transform must exist with a region of convergence $\Re\{s\}\ge a$, for some real-valued $a\le 0$.

If a two-sided signal is considered, we have to use the bilateral Laplace transform. Here the problem is that for $\Re\{s\}>0$ the signal is damped for $t>0$ but amplified for $t<0$ (and vice versa). Consequently, for two-sided signals with a constant envelop (e.g. sinusoids), the Laplace transform does not exist, even if the Fourier transform exists, because the Laplace integral cannot converge for signals with an exponentially increasing envelope. The same is true for all periodic signals (because they are sums of sinusoids). But even for two-sided signals which decay for increasing $|t|$ the bilateral Laplace transform might not exist if they decay too slowly, which is the case for the sinc function (decaying only as $1/t$).

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  • $\begingroup$ Thanks, but how about the "signals with their Fourier spectra being zero in some frequency ranges" ? $\endgroup$
    – user215721
    Jan 18 '15 at 12:21
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    $\begingroup$ @user215721: The corresponding time-domain signals are two-sided and decay as $1/t^n$, so the Laplace integral does not converge, because the factor $e^{-st}$ grows faster for $\Re\{s\}\neq 0$. $\endgroup$
    – Matt L.
    Jan 18 '15 at 15:34

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