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I have some troubles finding the Laplace transform of : $~~~f(t)=~e^{a|t|} ~~~(a\le0)~~$ from it's

Fourier transform : $~~\hat{f}(s)=\int_{-\infty }^{+\infty}f(t)e^{-i2\pi st}~dt= \frac{-1}{a+2\pi is}+\frac{1}{-a+2\pi is} = \frac{-1}{a+i\omega}+\frac{1}{-a+i\omega} $ .

I do know that the Laplace transform of : $ \mathcal{L}(f(t))(p)=H(i2\pi s)=H(i\omega)$

so $H(p)$ should be equal to : $ \frac{-1}{a+p}+\frac{1}{-a+p} $

which is quite different from the result I got from wolphram :

http://www.wolframalpha.com/input/?i=laplace+transform+of+exp(a*abs(t))

Any pointers to my mistake ?

Thank you .

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  • $\begingroup$ Generally the Laplace transform is defined by $$\int_0^\infty e^{-st}f(t)\ \mathsf dt.$$ Perhaps you are thinking of the bilateral Laplace transform? $\endgroup$ – Math1000 Jun 12 '18 at 23:30
  • $\begingroup$ The problem I was treating didn't actually mention that it was the bilateral transform , it just asked for the Laplace transform without actually using the integral but instead just by using the Fourier transform and it's relation to the Laplace transform . $\endgroup$ – Hilbert Jun 12 '18 at 23:40
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    $\begingroup$ You have correctly computed the two-sided Laplace transform of $e^{a |t|}$. To get the one-sided Laplace transform, you need $\mathcal F[e^{a |t|} \Theta(t)]$ (generally, $\mathcal F[f(t) e^{-\sigma t}]$ with $\sigma > 0$, but here it is possible to take $\sigma = 0$). $\endgroup$ – Maxim Jun 16 '18 at 13:15
  • $\begingroup$ If we plug $\sigma =0$ wouldn't that just simply give us the Fourier transform that I have already evaluated above : $\hat{f}(s)= \mathcal F[f(t) ]$ ? $\endgroup$ – Hilbert Jun 16 '18 at 22:46
  • $\begingroup$ I'm saying that you need to take the Fourier transform of $e^{a |t|}$ times the unit step function, which is different from the Fourier transform of $e^{a |t|}$. Btw, use @username to make sure that person gets a notification about your comment. $\endgroup$ – Maxim Jun 17 '18 at 22:31

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