$\sin (t \omega)$ is not an Energy Signal, then how come its Fourier transform do exist? - Signal Processing Stack Exchange most recent 30 from dsp.stackexchange.com 2019-12-14T03:01:55Z https://dsp.stackexchange.com/feeds/question/14990 https://creativecommons.org/licenses/by-sa/4.0/rdf https://dsp.stackexchange.com/q/14990 1 $\sin (t \omega)$ is not an Energy Signal, then how come its Fourier transform do exist? kaka https://dsp.stackexchange.com/users/5049 2014-03-14T07:09:07Z 2014-03-14T09:04:46Z <p>The following integral (perhaps fourier tranform of $\sin (t \omega)$ ) is not convergent:</p> <p>$\int_{-\infty }^{\infty } e^{-i t \omega } \sin (t \omega ) \, dt$</p> <p>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 $\sin (t \omega)$ ?</p> https://dsp.stackexchange.com/questions/14990/-/14991#14991 3 Answer by Matt L. for $\sin (t \omega)$ is not an Energy Signal, then how come its Fourier transform do exist? Matt L. https://dsp.stackexchange.com/users/4298 2014-03-14T09:04:46Z 2014-03-14T09:04:46Z <p>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:</p> <p>The inverse Fourier transform of the delta function $\delta(\omega)$ (in the frequency domain) is given by</p> <p>$$\frac{1}{2\pi}\int_{-\infty}^{\infty}\delta(\omega)e^{i\omega t}d\omega= \frac{1}{2\pi}e^{i0\cdot t}=\frac{1}{2\pi}$$</p> <p>So we have the Fourier transform relation (time domain $\Longleftrightarrow$ frequency domain)</p> <p>$$1\Longleftrightarrow 2\pi\delta(\omega)$$</p> <p>Using the shifting property we obtain</p> <p>$$e^{i\omega_0 t}\Longleftrightarrow 2\pi\delta(\omega-\omega_0)$$</p> <p>And since</p> <p>$$\sin(\omega_0t)=\frac{1}{2i}[e^{i\omega_0 t}-e^{-i\omega_0 t}]$$</p> <p>we get for its Fourier transform</p> <p>$$\sin(\omega_0t)\Longleftrightarrow \frac{\pi}{i}[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)]$$</p>