Continuous-space Fourier transform of the 2D "sinc" function (Practice Problem)

Compute the Continuous-space Fourier transform (CSFT) of

$ f(x,y)= \frac{\sin \pi x}{ \pi x}\frac{\sin \pi y }{\pi y}. $

(Justify all your steps.)

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Answer 1

Claim that $ CSFT \{\frac{\sin \pi x}{ \pi x}\frac{\sin \pi y }{\pi y}\} = rect(u,v)= rect(u)rect(v) $

Proof:

$ iCSFT\{rect(u)rect(v)\} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}rect(u)rect(v)e^{2j \pi (ux +vy) }dudv $

$ =\int_{-\infty}^{\infty}rect(u)e^{2j \pi (ux) }du \int_{-\infty}^{\infty}rect(v)e^{2j \pi (vy) }dv $

$ = \int_{-\frac{1}{2}}^{\frac{1}{2}}e^{2j \pi (ux) }du \int_{-\frac{1}{2}}^{\frac{1}{2}}e^{2j \pi (vy) }dv $

$ = \frac{(e^{j \pi x}-e^{-j \pi x} )(e^{j \pi y}-e^{-j \pi y})}{(2j\pi x)(2j\pi y)} $

$ = \frac{sin(x)sin(y)}{(\pi x)(\pi y)} = sinc(x)sinc(y)= sinc(x,y) $

Another way is to show by "separality", since

$ f(x,y)=g(x)h(y),g(x) = sinc(x),h(y) = sinc(y) $

then $ F(u,v)=G(u)H(v),G(u) = CTFT(f(x)),H(v) = CTFT(h(y)) $

by CTFT pairs, $ G(u) = rect(u),H(v) = rect(v) $

which shows $ CSFT \{ sinc(x,y) \} = rect(u)rect(v) = rect(u,v) $,

as the same above.

--Xiao1 23:40, 12 November 2011 (UTC)

Answer 2

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Answer 3

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