Dirac delta distribution is defined as
$f(t_{0})=\int_{-\infty }^{\infty } \! f(t)\delta(t-t_{0}) \, dt $ where $f(t)$ is smooth function. Then my question is:
:Calculate Fourier transform $\hat \delta(\omega)$ from $\delta (t-t_{0})$
Solution:
$$\hat \delta(\omega)=\frac{1}{\sqrt{2 \pi} }\int_{-\infty }^{\infty } \! \delta (t-t_{0}) e^{-j \omega t}\, dt $$
$$\hat \delta(\omega)=\frac {1}{\sqrt{2 \pi}}e^{-j \omega t_{0}}$$
Can someone explain me how they got this solution and write what are the steps between? On internet I always find some general formulas and I don't know how to use them.
Best Answer
Let $f$ be a smooth integrable function as you commented. Then we can define the Fourier Transform as
$$\mathcal{F}[f](y) := \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-iyx}\mathrm{d}x$$
We can also derive the inverse of Fourier Transform as:
$$\mathcal{F}^c[f](y):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{+iyx}\mathrm{d}x$$
Now lets see what happen wen we apply both to a function
$$f(y) = \mathcal{F}^c[\mathcal{F}[f]](y) = \frac{1}{2\pi}\int_\mathbb{R}\left(\int_{\mathbb{R}} f(\omega)e^{-i\omega x}\mathrm{d}\omega \right)e^{+iyx}\mathrm{d}x = \int_{\mathbb{R}}f(\omega)\left(\int_{\mathbb{R}}\frac{1}{2\pi}e^{-i(\omega-y)x}\mathrm{d}x\right)\mathrm{d}\omega$$
We can see in this last equality that the function in brakets acts as a Dirac Delta. So we can relate this as de Dirac as a notation (because the Dirac Delta just have formal sense in distribution theory)
$$\delta(\omega-y) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-i(\omega-y)x}\mathrm{d}x$$
This is an intuitive constructive way of seen that the Dirac is related to Fourier Transforms. Now for your question we can use the definition on how an Dirac Delta acts
$\hat{\delta}(\omega) = \frac{1}{(2\pi)^{1/2}}\int \delta(t-t_0)e^{-i\omega t}\mathrm{d}t = \frac{1}{(2\pi)^{1/2}}e^{-\omega t_0}$
Where we just used the above
$$f(y) = \int f (\omega)\delta(\omega-y)\mathrm{d}\omega$$
For $f(t):=e^{-i\omega t}/\sqrt{2\pi}$