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Justify your answer. | Justify your answer. | ||
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==Share your answers below== | ==Share your answers below== | ||
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! | 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! | ||
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<math> = \frac{1}{2j\pi(u)}\frac{1}{2j\pi(v)}[-e^{-j \pi (u)} + e^{j \pi (u)}][-e^{-j \pi (v)} + e^{j \pi (v)}]</math> | <math> = \frac{1}{2j\pi(u)}\frac{1}{2j\pi(v)}[-e^{-j \pi (u)} + e^{j \pi (u)}][-e^{-j \pi (v)} + e^{j \pi (v)}]</math> | ||
| + | ---- | ||
===Answer 2=== | ===Answer 2=== | ||
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| + | <math> F(u,v) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2j \pi (ux +vy) }dxdy = \int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-2j \pi (ux +vy) }dxdy</math> | ||
| + | <math> = \frac{(e^{j \pi u}-e^{-j \pi u} )(e^{j \pi v}-e^{-j \pi v})}{(2j\pi u)(2j\pi v)}</math> | ||
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| + | <math> = \frac{sin(u)sin(v)}{(\pi u)(\pi v)} = sinc(u)sinc(v)= sinc(u,v)</math> | ||
| + | ---- | ||
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===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
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[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 18:25, 12 November 2011
Contents
Continuous-space Fourier transform of the 2D "rect" function (Practice Problem)
Compute the Continuous-space Fourier transform (CSFT) of
$ f(x,y)= \left\{ \begin{array}{ll} 1, & \text{ if } |x|<\frac{1}{2} \text{ and } |y|<\frac{1}{2}\\ 0, & \text{ else}. \end{array} \right. $
Justify your answer.
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
$ x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2j \pi (ux +vy) }dxdy $
$ = \frac{1}{2j\pi(u)}\frac{1}{2j\pi(v)}[-e^{-j \pi (u)} + e^{j \pi (u)}][-e^{-j \pi (v)} + e^{j \pi (v)}] $
Answer 2
$ F(u,v) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2j \pi (ux +vy) }dxdy = \int_{-\frac{1}{2}}^{\frac{1}{2}}\int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-2j \pi (ux +vy) }dxdy $ $ = \frac{(e^{j \pi u}-e^{-j \pi u} )(e^{j \pi v}-e^{-j \pi v})}{(2j\pi u)(2j\pi v)} $
$ = \frac{sin(u)sin(v)}{(\pi u)(\pi v)} = sinc(u)sinc(v)= sinc(u,v) $
Answer 3
Write it here.
