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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?

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    $\begingroup$ 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 $\hat f(x)\hat g(y)$. $\endgroup$
    – user856
    Commented May 18, 2015 at 16:12
  • $\begingroup$ @Rahul no way, that's true! It solves my problem, thank you $\endgroup$ Commented May 18, 2015 at 16:13

1 Answer 1

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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

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

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