Practice Question on "Digital Signal Processing"

Topic: Continuous-space Fourier transform of a 2D "rect" function


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.

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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".

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).

Answer 3

Write it here.

Back to ECE438 Fall 2011 Prof. Boutin

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