Difference between revisions of "Compute DSFT cosine ECE438F11" - Rhea

Latest revision as of 12:59, 26 November 2013

Question

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) ] $

Instructor's comment: Would you know how to "compute" these two Fourier transforms if asked? Recall that one cannot use the summation formula to compute the DTFT of a function whose amplitude does not decrease as t approached plus/minus infinity. -pm

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)

Instructor's comment: This is a very well intentioned answer, with proofs for almost everything that is being used. But it is a bit long? Can somebody propose a different, more straightforward approach? -pm

Answer 2

Write it here.

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