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::where u and v repeats in every square with 2pi length.
 
::where u and v repeats in every square with 2pi length.
  
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--[[User:Xiao1|Xiao1]] 23:03, 19 November 2011 (UTC)
  
 
===Answer 2===
 
===Answer 2===

Revision as of 11:38, 20 November 2011

Practice Problem on Discrete-space Fourier transform computation

Compute the discrete-space Fourier transform of the following signal:

$ f[m,n]= \cos \left( 2 \pi \left( \frac{m}{500}+ \frac{n}{200} \right) \right) $

(Write enough intermediate steps to fully justify your answer.)


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Answer 1

trigonometric identities
By trigonometric identities(which can be proof by Eular's equations easily):
$ cos(\alpha+\beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta) $
Proof of separability
$ \begin{align} DSFT(f(m) \cdot g(n)) &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f(m) \cdot g(n) e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} f(m) e^{-j(mu)} \sum_{n=-\infty}^{\infty} g(n) e^{-j(nv)}\\ &= F(u) \cdot G(v) \end{align} $
where
$ F(u) =\sum_{m=-\infty}^{\infty} f(m) e^{-j(mu)} = DTFT(f(m)) $
$ G(v) =\sum_{n=-\infty}^{\infty} g(n) e^{-j(nv)} = DTFT(g(n)) $
Proof of linearity
$ \begin{align} DSFT(f(m,n) + g(m,n)) &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} [f(m,n) + g(m,n)] e^{-j(mu + nv)}\\ &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f(m,n) e^{-j(mu + nv)} + \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} g(m,n) e^{-j(mu + nv)}\\ &= F(u,v) + G(u,v) \end{align} $
where
$ F(u,v) =\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f(m,n) e^{-j(mu + nv)} = DSFT(f(m,n)) $
$ G(u,v) =\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} g(m,n) e^{-j(mu + nv)} = DSFT(g(m,n)) $
DTFT: By computing DTFT or looking it up in the table, one can find
$ DTFT(cos(w_0n))=\pi[ \frac{}{}\delta(w-w_0)+\delta(w+w_0) ] $
$ DTFT(sin(w_0n))=\frac{\pi}{j}[ \delta(w-w_0)-\delta(w+w_0) ] $
with all these tools we found, one can easily show the following:
Let
$ \alpha = \frac{2\pi}{500} $
$ \beta = \frac{2\pi}{200} $
$ \begin{align} DSFT&(\cos \left( 2 \pi \left( \frac{m}{500}+ \frac{n}{200} \right) \right))\\ &= DSFT[\cos \left( \alpha m + \beta n \right)] \\ &= DSFT[\cos(\alpha m)\cos(\beta n) - \sin(\alpha m)\sin(\beta n)]\\ &= DSFT[\cos(\alpha m)\cos(\beta n)] - DSFT[\sin(\alpha m)\sin(\beta n)]\\ &= DSFT[\cos(\alpha m)] \cdot DSFT[\cos(\beta n)] - DSFT[\sin(\alpha m)] \cdot DSFT[\sin(\beta n)]\\ &= \pi[ \delta(u-\alpha)+\delta(u+\alpha) ]\cdot\pi[ \delta(v-\beta)+\delta(v+\beta) ] + \frac{\pi}{j}[ \frac{}{}\delta(u-\alpha)-\delta(u+\alpha) ]\cdot\frac{\pi}{j}[ \frac{}{}\delta(v-\beta)-\delta(v+\beta) ]\\ &= \pi^2\{[ \delta(u-\alpha)+\delta(u+\alpha) ]\cdot[ \delta(v-\beta)+\delta(v+\beta) ] - [\delta(u-\alpha)-\delta(u+\alpha) ]\cdot[ \delta(v-\beta)-\delta(v+\beta) ]\}\\ &= 2\pi^2\{\delta(u-\alpha)\delta(v+\beta) + \delta(u+\alpha)\cdot\delta(v-\beta)\}\\ &= 2\pi^2\{\delta(u-\alpha,v+\beta) + \delta(u+\alpha,v-\beta)\}\\ \end{align} $
where u and v repeats in every square with 2pi length.

--Xiao1 23:03, 19 November 2011 (UTC)

Answer 2

Write it here.


Back to ECE438 Fall 2011 Prof. Boutin

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang