Four similar circles (radius-$R$) are packed in a simple square structure. Let's denote the area between these circles as $S$ - red area in attached image..
Now, let's think about a function $f$ that takes a value of $1$ inside $S$ and $0$ outside of $S$. No periodic boundary conditions are assumed. what analytical function describes the Fourier transform of $f$?
Here is a numerical solution (magnitude in log scale) obtained using Python.
Fraunhofer diffraction of aperture $S$ by means of Fourier Transform:
\begin{align*} F(\mathbf{k}) &= \iint_S f(x,y)e^{i\mathbf{k}\cdot \mathbf{r}} \, dA \\ &= \int_{-R}^{0} e^{ik_x x} dx \left( \int_{-R+\sqrt{-2Rx-x^2}}^{R-\sqrt{-2Rx-x^2}} e^{ik_y y} \,dy \right) \\ &\quad +\int_{0}^{R} e^{ik_x x} dx \left( \int_{-R+\sqrt{2Rx-x^2}}^{R-\sqrt{2Rx-x^2}} e^{ik_y y} dy \right) \\ &= \int_{-R}^{0} 2\sin \left[k_y \left( \sqrt{-2Rx-x^2}-R \right) \right] \frac{e^{ik_x x}}{k_y} \, dx \\ & \quad + \int_{0}^{R} 2\sin \left[k_y \left( \sqrt{2Rx-x^2}-R \right) \right] \frac{e^{ik_x x}}{k_y} \, dx \\ &= 4 \int_{0}^{R} \sin \left[k_y \left( \sqrt{2Rx-x^2}-R \right) \right] \frac{\cos k_x x}{k_y} \, dx \end{align*}
The integral doesn't have close form.