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[[Category:problem solving]] | [[Category:problem solving]] | ||
[[Category:continuous-space Fourier transform]] | [[Category:continuous-space Fourier transform]] | ||
| + | [[Category:2D rect]] | ||
| − | = Continuous-space Fourier transform of | + | <center><font size= 4> |
| + | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
| + | </font size> | ||
| + | |||
| + | Topic: Continuous-space Fourier transform of a 2D "rect" function | ||
| + | |||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
Compute the Continuous-space Fourier transform (CSFT) of | Compute the Continuous-space Fourier transform (CSFT) of | ||
| Line 16: | Line 25: | ||
Justify your answer. | Justify your answer. | ||
---- | ---- | ||
| − | |||
==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! | ||
---- | ---- | ||
===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> | ||
| + | :<span style="color:green">Instructor's comment: </span> | ||
| + | |||
| + | ::<span style="color:green"> a) You could perhaps simplify your answer a bit. (It's actually a sinc!). </span> | ||
| + | |||
| + | ::<span style="color:green"> b) The parenthesis around the u and v in the denominator of the answer are a bit confusing, but you would not lose any point for that of course. </span> | ||
| + | |||
| + | ::<span style="color:green"> c) You should add a few extra steps before writing the answer. If you do the steps in your head, there is a high likelihood of making a mistake. If you make a mistake in the answer and your wrote no intermediate steps, then you would get very little partial credit. </span> | ||
| + | |||
| + | ::<span style="color:green"> d) Actually, these is a slight problem with your answer at u=0 or v=0. So technically, you should split your solution into three cases: "u=0", "v=0", and "neither u nor v equal to zero". </span> | ||
| + | |||
| + | :::<span style="color:green"> pm </span> | ||
| + | ---- | ||
===Answer 2=== | ===Answer 2=== | ||
| − | + | ||
| + | <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> | ||
| + | |||
| + | <math> = \frac{sin(u)sin(v)}{(\pi u)(\pi v)} = sinc(u)sinc(v)= sinc(u,v)</math> | ||
| + | --[[User:Xiao1|Xiao1]] 23:26, 12 November 2011 (UTC) | ||
| + | ---- | ||
| + | |||
| + | :<span style="color:green">Instructor's comment: </span> | ||
| + | |||
| + | ::<span style="color:green"> a) This is a good amount of intermediate steps: not too much, not too little, altough adding an extra step to do the actual integration would not hurt. </span> | ||
| + | |||
| + | ::<span style="color:green"> b) Technically, your reasoning is not valid at u=0 or v=0 (because you would be dividing by zero). </span> | ||
| + | |||
| + | :::<span style="color:green"> pm </span> | ||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
Write it here. | Write it here. | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Latest revision as of 11:57, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Continuous-space Fourier transform of a 2D "rect" function
Question
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)}] $
- Instructor's comment:
- a) You could perhaps simplify your answer a bit. (It's actually a sinc!).
- b) The parenthesis around the u and v in the denominator of the answer are a bit confusing, but you would not lose any point for that of course.
- c) You should add a few extra steps before writing the answer. If you do the steps in your head, there is a high likelihood of making a mistake. If you make a mistake in the answer and your wrote no intermediate steps, then you would get very little partial credit.
- d) Actually, these is a slight problem with your answer at u=0 or v=0. So technically, you should split your solution into three cases: "u=0", "v=0", and "neither u nor v equal to zero".
- pm
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) $ --Xiao1 23:26, 12 November 2011 (UTC)
- Instructor's comment:
- a) This is a good amount of intermediate steps: not too much, not too little, altough adding an extra step to do the actual integration would not hurt.
- b) Technically, your reasoning is not valid at u=0 or v=0 (because you would be dividing by zero).
- pm
Answer 3
Write it here.
