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===Answer 1=== | ===Answer 1=== | ||
| − | + | Claim that <math>CSFT \{\frac{\sin \pi x}{ \pi x}\frac{\sin \pi y }{\pi y}\} = rect(u,v)= rect(u)rect(v)</math> | |
| + | |||
| + | Proof: | ||
| + | |||
| + | <math>iCSFT\{rect(u)rect(v)\} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}rect(u)rect(v)e^{2j \pi (ux +vy) }dudv</math> | ||
| + | |||
| + | <math>=\int_{-\infty}^{\infty}rect(u)e^{2j \pi (ux) }du \int_{-\infty}^{\infty}rect(v)e^{2j \pi (vy) }dv </math> | ||
| + | |||
| + | <math>= \int_{-\frac{1}{2}}^{\frac{1}{2}}e^{2j \pi (ux) }du \int_{-\frac{1}{2}}^{\frac{1}{2}}e^{2j \pi (vy) }dv </math> | ||
| + | |||
| + | <math> = \frac{(e^{j \pi x}-e^{-j \pi x} )(e^{j \pi y}-e^{-j \pi y})}{(2j\pi x)(2j\pi y)}</math> | ||
| + | |||
| + | <math> = \frac{sin(x)sin(y)}{(\pi x)(\pi y)} = sinc(x)sinc(y)= sinc(x,y)</math> | ||
| + | |||
| + | Another way is to show by "separality", since | ||
| + | |||
| + | <math>f(x,y)=g(x)h(y),g(x) = sinc(x),h(y) = sinc(y) </math> | ||
| + | |||
| + | then <math>F(u,v)=G(u)H(v),G(u) = CTFT(f(x)),H(v) = CTFT(h(y))</math> | ||
| + | |||
| + | by CTFT pairs, <math>G(u) = rect(u),H(v) = rect(v)</math> | ||
| + | |||
| + | which shows <math>CSFT \{ sinc(x,y) \} = rect(u)rect(v) = rect(u,v)</math>, | ||
| + | |||
| + | as the same above. | ||
| + | |||
| + | --[[User:Xiao1|Xiao1]] 23:40, 12 November 2011 (UTC) | ||
| + | |||
===Answer 2=== | ===Answer 2=== | ||
Write it here | Write it here | ||
Revision as of 18:40, 12 November 2011
Contents
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.)
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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
Write it here
Answer 3
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