A couple of confusions have been occurred. The Signal I'm considering is f(t) = sin(t)*u(t)

  1. Fourier Transform of it can be derived. $-i \pi (\delta (\omega -1)-\delta (\omega +1))$

  2. According to my mathematica code, the ROC of LaplaceTransformation didn't have the j$\omega$ axis in it's region of convergence.( Re{s}>0 )

So it's not stable. (Makes sense. sin(t) is not absolutely summable) https://www.wolframcloud.com/objects/ramithuh/Published/misc_sin_laplace.nb

  1. So, can a Fourier Transform exist even if the j$\omega$ axis is not in the Region of Convergence in it's Laplace Transform?

Things got worse when I considered f(t) = sin(t). It's Laplace Transform Integral didn't converge. So considering sin(t)*u(t) and sin(t)*u(-t) separately, I got two different ROC which doesn't have an overlap. Re{s} > 0 and Re{s} < 0. So it means that the Laplace transform of Sin(t) doesn't exist right? Initially what i thought is Laplace Transform can cover all the signals which FourierTransform covers. Turns out it's not the case?

Please point out in which step of my reasoning is wrong...

Thanks a bunch! :)

Update: Thank you for pointing out my mistake. :D $ \mathcal{F(sin(t)*u(t))} = -\frac{1}{2} i \pi \delta (\omega -1)+\frac{1}{2} i \pi \delta (\omega +1)-\frac{1}{2 (\omega -1)}+\frac{1}{2 (\omega +1)}$


1 Answer 1


You're right that the Laplace transform is not more general than the Fourier transform. They are just different. There are several (theoretically) important functions for which the Laplace transform doesn't exist, but the Fourier transform does. A few examples are

  1. $x(t)=e^{j\omega_0t}$
  2. $x(t)=\sin(\omega_0t+\phi)$
  3. $x(t)= \textrm{sinc}(\omega_0t)$
  4. $x(t)=\frac{1}{\pi t}$
  5. $x(t)=\textrm{sign}(t)$

The first two involve Dirac impulses in their Fourier transform, whereas the third and the fourth have discontinuous Fourier transforms. I hope it is obvious why the complex exponential and the sine are important. The $\textrm{sinc}$ function is needed to represent the impulse response of ideal frequency-selective filters, such as low pass and high pass filters, and $1/(\pi t)$ is the impulse response of an ideal Hilbert transformer. Note that those functions are non-causal (i.e., they are not zero for $t<0$). The class of non-causal functions for which the (two-sided) Laplace transform exists is quite restricted because damping through multiplication with $e^{st}$ only works either for $t>0$ (if $\textrm{Re}\{s\}<0$) or for $t<0$ (if $\textrm{Re}\{s\}>0$), but not for both.

In your first example ($\sin(t)u(t)$) you got the Fourier transform wrong, even though it does contain Dirac impulses. The Fourier transform has Dirac impulses whenever the Laplace transform (if it exists) has poles on the imaginary axis.

  • $\begingroup$ Thank you for your detailed answer! Just to clarify a point, so for the above listed functions, Bilateral Laplace Transform doesn't exist right? $\endgroup$ Commented Dec 3, 2018 at 14:35
  • $\begingroup$ I was actually confused initially with this. quora.com/… $\endgroup$ Commented Dec 3, 2018 at 14:36
  • $\begingroup$ @RamithHettiarachchi: You're right, that was the point: for those functions the (bilateral) Laplace transform doesn't exist, but the Fourier transform does (if we allow discontinuous transforms and Dirac delta impulses). $\endgroup$
    – Matt L.
    Commented Dec 3, 2018 at 15:50
  • 1
    $\begingroup$ @RamithHettiarachchi: I can understand that you got confused with the answer in the quora link ... :) $\endgroup$
    – Matt L.
    Commented Dec 3, 2018 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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