Fourier transform of a pyramid

Question

Has anyone calculated the Fourier coefficients for a pyramid function?
Let us define the pyramid function as, $z = f(x,y)$.
We are looking at 5 planes making up the pyramid. The 4 base points and apex are located at $ z(-X,-Y) = 0, \quad z(X,-Y) = 0, \quad z(X,Y) = 0, \quad z(-X,Y) = 0, \quad z(0,0) = 1 $ respectively. Then the equation for the planes will be,
$$ \begin{align} base \quad &: \quad z = 0 \\ front(-X \leq x \leq X, -Y \leq y \leq 0) \quad &: \quad z = 1 + \frac{y}{Y} \\ back(-X \leq x \leq X, 0 \leq y \leq Y) \quad &: \quad z = 1 - \frac{y}{Y} \\ left(-X \leq x \leq 0, -Y \leq y \leq Y) \quad &: \quad z = 1 + \frac{x}{X} \\ front(0 \leq x \leq X, -Y \leq y \leq Y) \quad &: \quad z = 1 - \frac{x}{X} \end{align} $$

Answer 1

Let $f(x,y)$ be the function supported on $[-1,1]\times[-1,1]$ and there given by $\min\left(1-|x|,1-|y|\right)$. We have:

$$\widehat{f}(u,v) = \frac{2 v \cos v \sin u-2 u\cos u\sin v}{\pi u^3 v-\pi u v^3}$$ and you just need to adjust the above transform by scaling and translating.