1
$\begingroup$

I am trying to evaluate the following expression $$\mathcal{F}\{\mathrm{rect}_{L_{x}}(x)\mathrm{rect}_{L_{y}}(y)\}$$ which denotes the 2-dimensional Fourier transform (reciprocal variables $k_x$, $k_y$) of a product of two rectangular functions defined by $$\mathrm{rect}_{L_{x}}(x)=\begin{cases}1, & -L_x<x<L_x \\ 0, & \text{otherwise}\end{cases}$$

This is what I tried so far ("*" denotes convolution): $$\mathcal{F}\{\mathrm{rect}_{L_{x}}(x)\mathrm{rect}_{L_{y}}(y)\} = \mathcal{F}\{\mathrm{rect}_{L_{x}}(x)\}*\mathcal{F}\{\mathrm{rect}_{L_{y}}(y)\}\\ ={L_xL_y\over\pi^2}\mathrm{sinc}(L_xk_x)*\mathrm{sinc}(L_yk_y)\\ ={L_xL_y\over\pi^2}\int_{-\infty}^{\infty}\mathrm{d}k_x'\,\mathrm{sinc}(L_xk_x)\int_{-\infty}^{\infty}\mathrm{d}k_y'\mathrm{sinc}(L_y(k_y-k_y')) $$ Now both these integrals are Dirichlet integrals and evaluate to ${\pi \over L_x}$ and ${\pi \over L_y}$ respectively which yields total result of $1$.

I suspect there's a mistake somewhere, since this is not the result I am expecting to see. Moreover when I try to do the calculation in Mathematica

FourierTransform[UnitBox[x] UnitBox[y], {x, y}, {a, b}]

I get $${\mathrm{sinc}(a/2)\mathrm{sinc}(b/2)\over 2\pi}$$ which makes much more sense for the bigger problem I am solving.

Where am I wrong?

$\endgroup$
  • 2
    $\begingroup$ The Fourier transform of $f(x)g(x)$ is $\hat f(x)*\hat g(x)$, but the Fourier transform of $f(x)g(y)$ is just $\hat f(x)\hat g(y)$. $\endgroup$ – Rahul May 18 '15 at 16:12
  • $\begingroup$ @Rahul no way, that's true! It solves my problem, thank you $\endgroup$ – Dmitry Kazakov May 18 '15 at 16:13
2
$\begingroup$

The two dimensional Fourier Transform of a function $f(x,y)$ is

$$\mathscr{F}\{f\}(k_x,k_y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{ik_xx}e^{ik_yy}dxdy$$

For $f(x,y)=\text{rect}_{L_x}(x)\text{rect}_{L_y}(y)$, the Fourier Transform is

$$\mathscr{F}\{f\}(k_x,k_y)=\int_{-L_y}^{L_y}\int_{-L_x}^{L_x}e^{ik_xx}e^{ik_yy}dxdy=2L_x\text{sinc}(k_xL_x)2L_y\text{sinc}(k_yL_y)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.