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[[Category:problem solving]] | [[Category:problem solving]] | ||
[[Category:discrete-space Fourier transform]] | [[Category:discrete-space Fourier transform]] | ||
− | = [[ | + | <center><font size= 4> |
+ | '''[[Digital_signal_processing_practice_problems_list|Practice Question on "Digital Signal Processing"]]''' | ||
+ | </font size> | ||
+ | |||
+ | Topic: Discrete-space Fourier transform computation | ||
+ | |||
+ | </center> | ||
+ | ---- | ||
+ | ==Question== | ||
Compute the discrete-space Fourier transform of the following signal: | Compute the discrete-space Fourier transform of the following signal: | ||
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::<math>DTFT(cos(w_0n))=\pi[ \frac{}{}\delta(w-w_0)+\delta(w+w_0) ]</math> | ::<math>DTFT(cos(w_0n))=\pi[ \frac{}{}\delta(w-w_0)+\delta(w+w_0) ]</math> | ||
::<math>DTFT(sin(w_0n))=\frac{\pi}{j}[ \delta(w-w_0)-\delta(w+w_0) ]</math> | ::<math>DTFT(sin(w_0n))=\frac{\pi}{j}[ \delta(w-w_0)-\delta(w+w_0) ]</math> | ||
− | + | :::<span style="color:green"> 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 </span> | |
:with all these tools we found, one can easily show the following: | :with all these tools we found, one can easily show the following: | ||
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--[[User:Xiao1|Xiao1]] 23:03, 19 November 2011 (UTC) | --[[User:Xiao1|Xiao1]] 23:03, 19 November 2011 (UTC) | ||
+ | :<span style="color:green"> 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 </span> | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
---- | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] |
Latest revision as of 12:59, 26 November 2013
Practice Question on "Digital Signal Processing"
Topic: Discrete-space Fourier transform computation
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.)
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.