The following integral (perhaps fourier tranform of $\sin (t \omega)$ ) is not convergent:

$\int_{-\infty }^{\infty } e^{-i t \omega } \sin (t \omega ) \, dt$

As, $\sin (t \omega)$ is NOT an Energy Signal (but a Power Signal), then how come we get successful in finding the fourier transform of $\sin (t \omega)$ ?


You are right that such integrals are meaningless unless they are interpreted as distributions. And this is what we need to do, because - as you know - the Fourier transform of a sine function involves delta impulses. Let me try to make this a bit more intuitive:

The inverse Fourier transform of the delta function $\delta(\omega)$ (in the frequency domain) is given by

$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega)e^{i\omega t}d\omega= \frac{1}{2\pi}e^{i0\cdot t}=\frac{1}{2\pi}$$

So we have the Fourier transform relation (time domain $\Longleftrightarrow$ frequency domain)

$$1\Longleftrightarrow 2\pi\delta(\omega)$$

Using the shifting property we obtain

$$e^{i\omega_0 t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0)$$

And since

$$\sin(\omega_0t)=\frac{1}{2i}[e^{i\omega_0 t}-e^{-i\omega_0 t}]$$

we get for its Fourier transform

$$\sin(\omega_0t)\Longleftrightarrow \frac{\pi}{i}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]$$

| improve this answer | |
  • $\begingroup$ @Matt L. Do you suggest any good literature on distributions. $\endgroup$ – kaka Mar 15 '14 at 4:38
  • $\begingroup$ The best book (for non-mathematicians) I know is "The Fourier Integral and Its Applications" by A. Papoulis. It has everything you'll ever want to know about the Fourier transform, and there's a great appendix on distributions. $\endgroup$ – Matt L. Mar 15 '14 at 9:17

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.