Revision as of 09:33, 14 November 2011

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) $

Instructor's comment: Not bad, except for the fact that you are dividing by zero when either x or y is zero. Technically, you should split the cases. -pm

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)

Instructor's comment: Before you use the second approach on the exam, make sure that the separability property is in the table. Otherwise, you must prove the property before using it. (But of course, proving that property is triviale.) -pm

Instructor's challenge: Can somebody answer this using duality? -pm

Answer 2

Write it here

Answer 3

Write it here.

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