I am having trouble understanding this:

I have a function

$$ \delta (t_1-t_2) $$

but I want to prove that in the frequency domain, it is:

$$\delta(\omega_1+\omega_2) $$

So, we have:

$$F(t0,w_{1})=\int _{-\infty }^{\infty }\!\delta \left( {\it t_1}-{\it t_0} \right) { {\rm e}^{-iw_{{1}}t_{{1}}}}{dt_{{1}}}$$

$$ F(w_1,w_2)=\int _{-\infty }^{\infty }\!{{\rm e}^{-i \left( w_{{2}}+w_{{1}} \right) t_{{0}}}}{dt_{{0}}}$$

$$F(w_1,w_2)=2\pi \delta \left( w_{{2}}+w_{{1}} \right)$$

  • $\begingroup$ But the thing is it's not, is it? $\endgroup$ – Daniel R Dec 5 '14 at 18:22
  • $\begingroup$ Yes, I need to prove this. There are two indep. times and frequencies $\endgroup$ – Jackson Hart Dec 5 '14 at 18:24
  • $\begingroup$ @copper.hat can someone show me why it is not true? $\endgroup$ – Jackson Hart Dec 5 '14 at 18:26
  • $\begingroup$ What are $t_1,t_2$? $\endgroup$ – copper.hat Dec 5 '14 at 18:28
  • $\begingroup$ I don't know what you are trying to do. What function are you trying to take the Fourier transform of? I don't have time for chat now, it always ends up taking 10-15 mins. $\endgroup$ – copper.hat Dec 5 '14 at 18:32

This is not a formal answer, just a hint of how to proceed.

The two Fourier transform of $f:\mathbb{R}^2 \to \mathbb{R}$ is given by $\hat{f}(\omega_1,\omega_2) = \int_{\mathbb{R}^2} f(t_1,t_2) e^{-i(\omega_1 t_1 + \omega_2 t_2)} d t_1 d t_2$.

With the distribution given by $f(t_1,t_2) = \delta(t_1-t_2)$, we have $f:\mathbb{R}^2 \to \mathbb{R}$ is given by \begin{eqnarray} \hat{f}(\omega_1,\omega_2) &=& \int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(t_1,t_2) e^{-i(\omega_1 t_1 + \omega_2 t_2)} d t_1 \right) d t_2 \\ &=& \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \delta(t_1-t_2) e^{-i(\omega_1 t_1 + \omega_2 t_2)} d t_1 \right) d t_2 \\ &=& \int_{\mathbb{R}} e^{-i(\omega_1 t_2 + \omega_2 t_2)} d t_2 \\ &=& \int_{\mathbb{R}} e^{-i((\omega_1 + \omega_2) t)} d t \\ &=& 2 \pi \delta(\omega_1 + \omega_2) \end{eqnarray} The last equation follows from the fact the transform of the distribution $t \mapsto \delta(t)$ is $\omega \to 1$. Taking the inverse transform gives the desired result.

  • $\begingroup$ Was there something wrong with the edited question? $\endgroup$ – Jackson Hart Dec 8 '14 at 20:06
  • $\begingroup$ You are using the same $F$ for a function, the partially transformed function and the transformed function, so it makes it hard to guess what you want. $\endgroup$ – copper.hat Dec 8 '14 at 20:11
  • $\begingroup$ haha yeah, it is somewhat confusing. it seems like you wrote out the answer, even though you said it was just a hint? $\endgroup$ – Jackson Hart Dec 8 '14 at 20:14
  • $\begingroup$ Well, since we are dealing with distributions, there is a little more technical machinery to deal with which I have selfishly avoided by declaring 'informal' :-). $\endgroup$ – copper.hat Dec 8 '14 at 20:16
  • $\begingroup$ Well I am not fully sure of what this technical machinery is, but I was thinking the answer in the Question in satisfactory, as long as I declare the different F's like you were saying $\endgroup$ – Jackson Hart Dec 8 '14 at 20:19

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