1
$\begingroup$

In our Signals and Systems class, we defined the Fourier transform and its inverse pair as such:

$X(jw) = \int_{-\infty}^{+\infty}x(t)\times e^{-j\omega t}dt $

$x(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}X(jw)\times e^{j\omega t}d\omega$

I sort of wrapped my head around as to how the $2\pi$ factor is arbitrarily divided among the pairs.

Since we aren't dealing the mathematics of distributions, when it comes to pairs such as:

$ \sin(\omega_0 t) \longleftrightarrow \frac{j}{\pi} \left[\ \delta(\omega+\omega_0) - \delta(\omega - \omega_0)\right]$

I don't know how one gets rigorously from the time to the frequency domain, but working backwards from the frequency domain, I understand why the $\frac{j}{\pi}$ factor is there.

However, in the section of the materials that deals with the properties of the Fourier transform, we are told that this holds:

$\int_{-\infty}^tx(\tau)d\tau \longleftrightarrow \frac{1}{j\omega}X(j\omega) + \pi X(0)\delta(\omega)$

What I don't get is why the factor next to the seccond term of the Fourier transform is $\pi$. Supposedly, this term relates to the dc component of the original function, and it would make sense to me that it's equal to simply $X(0)\delta(f)$ (such as in this link), or since the definition of the transform pairs is slightly different, to be $2\pi X(0)\delta(\omega)$, but neither of these happens.

What am I missing? The materials we have state that one can get this term by taking the integral of the Fourier series and then take the limit as the period tends to infinity, but I fiddled with that and I still couldn't get this factor.

$\endgroup$
1
$\begingroup$

I'm not sure if that answer your question and I can't add comment (not enough reputation) but this might help :

The integral $\int_{-\infty}^t x(\tau)d\tau$ can be seen as a convolution of function $x(t)$ with the step function $u(t)$.

Therefore $\int_{-\infty}^t x(\tau)d\tau=x(t)\ast u(t)$.

The Fourier transform of a convolution being the product of the respective transform, taking the Fourier transform will give $X(j\omega)(\frac{1}{j\omega}+\pi\delta(\omega))=\frac{X(j\omega)}{j\omega}+\pi X(0)\delta(\omega)$.

So the $\pi$ factor comes from the transform of the step function.

$\endgroup$
  • $\begingroup$ I had a hunch at first that these two are related, but in the materials we have, the transform of the Step function actually comes immediately after the integral property, and they use the integral property to find it. I suppose if I knew how to do the Fourier transform of the step signal in another way that I could get this result. $\endgroup$ – John Doe Feb 6 '19 at 13:23
  • $\begingroup$ @JohnDoe The way you defined $\mathcal{F},\mathcal{F}^{-1}$ together with the Fourier inversion theorem (for tempered distributions) then $\mathcal{F}^{-1}[2\pi\delta(\omega)] = 1 \implies \mathcal{F}[1] = 2\pi \delta(\omega)$. Then $u(t) = sign(t) + \frac12$ so $\mathcal{F}[u(t)] =\mathcal{F}[sign(t)] + \pi \delta(\omega)$. Also $i \omega \mathcal{F}[sign(t)] = \mathcal{F}[sign'(t)] = \mathcal{F}[\delta(t)] = 1$ and $i\omega pv(\frac{1}{i \omega}) = 1$ implies $\mathcal{F}[sign(t)]=pv(\frac{1}{i \omega})+c\delta(\omega)$ ($pv$ for principal value) and $c = 0$ since $sign(t)$ is real and odd. $\endgroup$ – reuns Feb 6 '19 at 15:18
  • $\begingroup$ The Fourier transforms of distributions are made rigorous simply by pairing them by muliplication and convolution with Gaussians, that is $\mathcal{F}[x(t)]$ is the unique distribution such that $\mathcal{F}[x (t) e^{-\pi t^2/n^2}] = \mathcal{F}[x(t)] \ast n e^{-\pi \omega^2 n^2}$. The Fourier inversion theorem for gaussians automatically transfers to distributions. $\endgroup$ – reuns Feb 6 '19 at 15:22
  • $\begingroup$ @reuns Thanks, that allows me to get the given relation, but it still seems to be in a roundabout way. It seems to me that there should be a way to get it without resorting to other properties of the Fourier transform (convolution->multiplication). $\endgroup$ – John Doe Feb 6 '19 at 18:18
  • $\begingroup$ @reuns I guess the question is why do we take half of the DC value $\left.\int_{-\infty}^{+\infty}x(t)dt\right.$ as opposed to the whole value. $\endgroup$ – John Doe Feb 6 '19 at 19:39

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.