My friend and I had an argument upon the Laplace transform of $\sin \omega t$. He's saying that its Laplace transform does not exist and only the Laplace transform of $u(t) \sin \omega t$ exist. But when I checked in my mathematics book ("Advanced Engineering Mathematics" by Greenberg), the transform of $\sin \omega t$ was given. Who is right, then?


Your friend sort of have a point. If you're talking about one-sided laplace transform it is in effect the function $\sin(\omega t)u(t)$ that is transformed. The laplace transform is the integral:

$$\int_0^\infty sin(\omega t) e^{-st}dt = \int_{-\infty}^\infty \sin(\omega t)u(t)e^{-st} = {\omega\over s^2+\omega^2}$$

If you're on the other hand talking about double sided transform you've got to use distribution interpretation of the transform. The definition:

$$\mathcal L sin(\omega t) = \mathcal F sin(\omega t)e^{-\sigma t}$$

Which is only valid for $\sigma=0$ and the result is

$$\mathcal L sin(\omega t) = \mathcal F sin(\omega t) = {\delta_{\omega}-\delta_{-\omega}\over 2i}$$

  • $\begingroup$ So, who is right? Is the proof given in book wrong? $\endgroup$
    – Vedanshu
    Oct 15 '15 at 5:06
  • $\begingroup$ The proof in the book is probably correct given the definition that's in the book. I suspect it gives something a result like $\omega/(s^2+\omega^2)$ and in that case it's the unilateral transform it uses. For the bilateral transform you'll either have to restrict the function by multiplying with $u(t)$ or use distribution variant of it. Note that the unilateral transform will when inversely transformed give back a function only defined on one side. $\endgroup$
    – skyking
    Oct 15 '15 at 6:03
  • $\begingroup$ What's the distribution interpretation of the transform ? $\endgroup$
    – Vedanshu
    Nov 8 '15 at 15:35
  • $\begingroup$ @AnshKumar It's based on the Fourier transform for distribution (a definition that will allow Fourier transforming $e^{i\omega t}$, resulting in $\delta_\omega$). The Laplace transform is defined by Fourier transforming after multiplying with $e^{-\sigma t}$. $\endgroup$
    – skyking
    Nov 9 '15 at 6:46
  • $\begingroup$ Could you have a look at my another question $\endgroup$
    – Vedanshu
    Dec 8 '15 at 12:15

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.