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---- | ---- | ||
===Answer 1=== | ===Answer 1=== | ||
− | '''trigonometric identities''' | + | :'''trigonometric identities''' |
− | :By trigonometric identities(which can be proof by Eular's equations easily): | + | ::By trigonometric identities(which can be proof by Eular's equations easily): |
− | :<math>cos(\alpha+\beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)</math> | + | ::<math>cos(\alpha+\beta) = cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)</math> |
− | '''Proof of separability''' | + | :'''Proof of separability''' |
− | :<math> | + | ::<math> |
\begin{align} | \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)}\\ | DSFT(f(m) \cdot g(n)) &= \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f(m) \cdot g(n) e^{-j(mu + nv)}\\ | ||
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&= F(u) \cdot G(v) | &= F(u) \cdot G(v) | ||
\end{align}</math> | \end{align}</math> | ||
− | :where | + | ::where |
− | :<math>F(u) =\sum_{m=-\infty}^{\infty} f(m) e^{-j(mu)} = DTFT(f(m))</math> | + | ::<math>F(u) =\sum_{m=-\infty}^{\infty} f(m) e^{-j(mu)} = DTFT(f(m))</math> |
− | :<math>G(v) =\sum_{n=-\infty}^{\infty} g(n) e^{-j(nv)} = DTFT(g(n))</math> | + | ::<math>G(v) =\sum_{n=-\infty}^{\infty} g(n) e^{-j(nv)} = DTFT(g(n))</math> |
− | '''Proof of linearity''' | + | :'''Proof of linearity''' |
− | :<math> | + | ::<math> |
\begin{align} | \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)}\\ | 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)}\\ | ||
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&= F(u,v) + G(u,v) | &= F(u,v) + G(u,v) | ||
\end{align}</math> | \end{align}</math> | ||
− | :where | + | ::where |
− | :<math>F(u,v) =\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f(m,n) e^{-j(mu + nv)} = DSFT(f(m,n))</math> | + | ::<math>F(u,v) =\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} f(m,n) e^{-j(mu + nv)} = DSFT(f(m,n))</math> |
− | :<math>G(u,v) =\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} g(m,n) e^{-j(mu + nv)} = DSFT(g(m,n))</math> | + | ::<math>G(u,v) =\sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} g(m,n) e^{-j(mu + nv)} = DSFT(g(m,n))</math> |
− | '''DTFT:''' By computing DTFT or looking it up in the table, one can find | + | :'''DTFT:''' By computing DTFT or looking it up in the table, one can find |
− | :<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> |
− | with all these tools we found, one can easily show the following: | + | :with all these tools we found, one can easily show the following: |
− | :Let | + | ::Let |
− | :<math>\alpha = \frac{2\pi}{500}</math> | + | ::<math>\alpha = \frac{2\pi}{500}</math> |
− | :<math>\beta = \frac{2\pi}{200}</math> | + | ::<math>\beta = \frac{2\pi}{200}</math> |
− | :<math> | + | ::<math> |
\begin{align} | \begin{align} | ||
DSFT&(\cos \left( 2 \pi \left( \frac{m}{500}+ \frac{n}{200} \right) \right))\\ | DSFT&(\cos \left( 2 \pi \left( \frac{m}{500}+ \frac{n}{200} \right) \right))\\ | ||
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\end{align}</math> | \end{align}</math> | ||
− | :where u and v repeats in every square with 2pi length. | + | ::where u and v repeats in every square with 2pi length. |
Revision as of 20:15, 19 November 2011
Contents
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.)
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.