Given a stripe $X$ on the xy-plane, namely $X\subset\mathbb{R}^2$, with $X=\{(x,y)\,|\; mx-\frac{1}{2}t \le y \le mx + \frac{1}{2}t$} and its "characteristic" function $$ f(x,y) = \begin{cases} 1, & \text{if } (x,y) \in X, \\ 0, & \text{otherwise}. \end{cases} $$ What is the Fourier-transform $$ \mathcal{F}[f](u,v) = \frac{1}{2 \pi}\iint_{-\infty}^{\infty} f(x,y) e^{i(ux+vy)}dxdy$$ of this function?
I have attempted the special case $m=0$ (a stripe around the x-axis). I have constructed it from a linear combination of two Heaviside functions $\Theta(x)$: $$ f(x,y) = \frac{1}{2}[\Theta(y+\frac{t}{2}) + \Theta(-y+\frac{t}{2}) ] * 1$$ With $$\mathcal{F}[\Theta](u)=\frac{i}{u\sqrt{2\pi}} + \frac{\pi}{2}\delta(u)$$ the linearity of $\mathcal{F}$ and the shift property I get $$ \pi \delta(u) \delta(v) + \frac{\delta(u) \sin(\frac{tv}{2})}{v} $$
How to best tackle the general case $m\ne0$? Can one use something like a coordinate transformation (rotation of the plane)?
Following up on my comment, suppose we switch to coordinates $(x',y')=(x,y-mx)$. Then the stripe becomes $X'=\{(x',y')|-\frac12 t\leq y' \leq \frac12t\}$ and the Fourier transform takes the form
$$\mathcal{F}[f](u,v) = \frac{1}{2 \pi}\int_{-\infty}^{\infty} f(x',y') e^{i(ux'+v(y'+mx'))}dx'dy'=\frac{1}{2 \pi}\int_{-\infty}^{\infty} f(x',y') e^{i(u+mv)x'+ivy'}dx'dy'$$ with $f(x',y')=\frac{1}{2}[\Theta(y'+\frac{t}{2}) + \Theta(-y'+\frac{t}{2}) ] * 1$.
This is the same as above, except with $(u,v)\to (u+mv,v)$. So the result for the case $m\neq 0$ is $$ \pi \delta(u+mv)\delta(v) + \frac{\delta(u+mv)\sin(\frac{t}{2}v)}{v}, $$ which simplifies to $$ \pi \delta(u)\delta(v) + \frac{\delta(u+mv)\sin(\frac{t}{2}v)}{v}. $$