# frequency domain of a delta function

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:

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

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

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

• But the thing is it's not, is it? – Daniel R Dec 5 '14 at 18:22
• Yes, I need to prove this. There are two indep. times and frequencies – Jackson Hart Dec 5 '14 at 18:24
• @copper.hat can someone show me why it is not true? – Jackson Hart Dec 5 '14 at 18:26
• What are $t_1,t_2$? – copper.hat Dec 5 '14 at 18:28
• 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. – copper.hat Dec 5 '14 at 18:32

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.
• 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. – copper.hat Dec 8 '14 at 20:11