4
$\begingroup$

Suppose I know the difference between an unknown function $f$ and its delayed version, can I use LaplaceTransform to find $f$?

I didn't succeed in this toy example: let $f(t)=e^{-t}$ but we pretend we don't know this. So we have

$$ f(t)-f(t-a)=e^{-t}-e^{-(t-a)}\overset{\mathcal{L}}{\Longrightarrow} (1-e^{-as})\mathcal{F}(s)=\frac{1-e^a}{1 + s} $$

However, InverseLaplaceTransform can't go back to $f(t)=e^{-t}$:

In[1]:= Assuming[a>0,
            InverseLaplaceTransform[
                LaplaceTransform[Exp[-t]-Exp[-(t-a)], t, s]/(1-Exp[-a s]), s, t]]
Out[1]= InverseLaplaceTransform[(1/(1+s)-E^a/(1+s))/(1-E^(-s a)), s, t]

Do I understand Laplace transform correctly or do I need more tweaks to find the original $f(t)$?

Update: I think I made a simple mistake in Laplace transform: $e^{-as}\mathcal{F}(s)=\mathcal{L}[f(t-a)u(t-a)]\ne \mathcal{L}[f(t-a)]$, where $u(\cdot)$ is the unit step function. In fact, we have

enter image description here

and thus inverse Laplace transform recovers the original $f(t)$ from $f(t)-f(t-a)u(t-a)$, but not from $f(t)-f(t-a)$ (which can be done by Fourier transform, as @AccidentalFourierTransform showed).

$\endgroup$
2
$\begingroup$

FWIW, if you use the Fourier transform instead of the Laplace one, the method works perfectly well. For example, if we let $f(t)=\exp(-t^2)$, we get

InverseFourierTransform[-((E^(-(s^2/4)) (-1 + E^(I a s)))/(Sqrt[2] (1 - Exp[I a s]))), s, t]
(* E^(-t^2) *)

I am not entirely sure why InverseLaplaceTransform is unable to transform back, but in principle the method should work as well.

$\endgroup$
  • $\begingroup$ I actually made a mistake about Laplace transform. And in order to recover $f(t)$ from $f(t)-f(t-a)$ we do need Fourier transform. $\endgroup$ – ziyuang Dec 15 '17 at 21:25
2
$\begingroup$

It looks like you have something extra in the denominator.

LaplaceTransform[Exp[-t] - Exp[-(t - a)], t, s] // Simplify
(* (1 - E^a)/(s + 1) *)

InverseLaplaceTransform[%, s, t]
(* E^-t - E^(a - t) *)
$\endgroup$
  • $\begingroup$ The denominator is for solving $f$. $\endgroup$ – ziyuang Dec 12 '17 at 22:17
  • $\begingroup$ Sorry for misreading your question. But your LT identity doesn't work for any function I've tried. MMa seems to be correctly finding the LT of these functions. Where does your formula come from? $\endgroup$ – Bill Watts Dec 13 '17 at 0:57
  • $\begingroup$ You are right. I made a mistake. See my update. Now everything makes sense. $\endgroup$ – ziyuang Dec 15 '17 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.