Difference between revisions of "Compute CSFT of 2D sinc function ECE438F11" - Rhea

Latest revision as of 11:58, 26 November 2013

Question

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.)

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!

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) $

Instructor's comment: Not bad, except for the fact that you are dividing by zero when either x or y is zero. Technically, you should split the cases. -pm

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)

Instructor's comment: Before you use the second approach on the exam, make sure that the separability property is in the table. Otherwise, you must prove the property before using it. (But of course, proving that property is triviale.) -pm

Instructor's challenge: Can somebody answer this using duality? -pm

Answer 2

Using duality we start with a 2-D rect and take the Fourier Transform.

$ \begin{align} F(u,v) &=& \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} {\color{red}rect(x,y)} 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(\pi u)sin(\pi v)}{(\pi u)(\pi v)} \text{ (Eq. a)} \\ &=& sinc(u)sinc(v) \\ &=& sinc(u,v) \\ \\ \end{align} $

In Eq. a this is valid for all values of u and v because of the property that $ \lim_{x\to0}\frac{sin(x)}{x} = 1 $.

Now we can use the duality property that states $ F(x,y) \to f(-u,-v) $ Also using the fact that $ sin(-x)=-sin(x) $ and since there is two sine functions multiplied together we get that

$ F(x,y)=sinc(x,y)=sinc(-x,-y)=F(-x,-y)\to f(u,v)=rect(u,v) $

So we get that the Fourier transform of sinc(x,y) is rect(u,v)

Instructor's comment: The "rect" function in red inside the integral on the first line was added by me. -pm

Yeah that is kind of important. Thanks

Answer 3

Write it here.

Back to ECE438 Fall 2011 Prof. Boutin