Revision as of 19:13, 19 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.)

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) ] $

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.

Answer 2

Write it here.

Back to ECE438 Fall 2011 Prof. Boutin