Fourier transform of triangular function Determine $X(\omega)$.


$g(t)$: I understand how to create a box from [-1,1] of amplitude 1/2.
$x(t) = g(t) * g(t)$
$X(\omega) = G(\omega)G(\omega)$


the solution I am seeing says that $G(\omega) = \frac{2\sin(\omega)}  {2\omega}$

I don't understand where $\sin$ came from and that the values of the 2s correlate to. I've seen proofs, but can someone provide a simple explanation of what the variables are. Thanks
A triangular function can be generated by convolving two box functions as shown below.



This is where your Step 2 comes from.

The fourier transform of a convolution $g(t) \ast g(t)$ can be calculated by multiplying the fourier transform of $g(t)$ with itself, i.e. $G(\omega)G(\omega)$.

Recall that the Fourier transform of a box function is a Sinc function ($\textrm{sinc}(x) = \frac{\textrm{sin}(x)}{x}$).



Hence, $G(w)$ is some scaled version of a sinc function, and the Fourier transform of the triangular function is $G(w)^2$.