I noticed by numerical and some explicit calculations for a few examples that for real-valued finitely supported functions $\phi \in L^2(\mathbb{R})$ we have that

$T(x):= \sum_{n \in \mathbb{Z}} |\hat{\phi}(x+2 \pi n)|^2$ is only a trigonometric polynomial $T(x) = a_0 + a_1 \cos( x) + ...+a_n \cos(n x) $ for some $n \in \mathbb{N}$ and coefficients $a_0,..,a_n.$

Now, the thing is that we apparently only have $T \in L^1([0,2\pi]),$ so it is not even clear to me that there is a Fourier series converging to $T$ at all. Despite, if there is one, then it is clear (by symmetry) that only $\cos$ terms should appear. Nevertheless, the thing that confused me most is that the Fourier expansion is apparently always finite. I guess this must be somehow related to the compact support of $\phi$. But it is not obvious to me where this actually comes from.

Does anybody know why this happens?

A very simple (I tested also harder ones) is

$$ \phi(x) = \chi_{[-1,1]}$$

then $$\sum_{n \in \mathbb{Z}} |\sin^2(x+2\pi n)(x+2 \pi n)^{-2}| = \frac{1}{2}+ \frac{1}{2} \cos(x)$$

and now notice that $$\frac{1}{\pi} \int_{-\pi}^{\pi} \sum_{n \in \mathbb{Z}} |\sin^2(x+2\pi n)(x+2 \pi n)^{-2}| \cos(kx)dx$$ is equal to $1$ if $k=0$ and $\frac{1}{2}$ if $k=1$ and zero otherwise. So also the Fourier expansion is finite and agrees with the representation (notice that we have to divide the first Fourier coefficient by $\frac{1}{2}.$)

If we consider $\phi(x) =\chi_{[-n,n]}$ then this shows that exactly the first $k<2n$ integrals are non-zero and the $2n$ th Fourier coefficient will vanish. Thus, it seems as if we could try to state: The larger the support is, the more coefficients are non-zero.