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

Latest revision as of 11:59, 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.)

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

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.

Back to ECE438 Fall 2011 Prof. Boutin