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===Answer 1=== | ===Answer 1=== | ||
| − | + | <math> x[n] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-2j \pi (ux +vy) }dxdy | |
| + | </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=== | ||
Write it here | Write it here | ||
Revision as of 16:15, 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
Write it here
Answer 3
Write it here.
