Timeline for Why is the Fourier transform so important?
Current License: CC BY-SA 3.0
16 events
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| Apr 12, 2015 at 22:52 | comment | added | LSpice | This is a lovely answer, but a tiny correction: not every merely integrable function is the sum of its Fourier series. This definitely fails pointwise as soon as the function has discontinuities (even at the endpoints), but can even fail in the $\mathrm L^1$ sense (see en.wikipedia.org/wiki/…). | |
| Mar 9, 2015 at 5:30 | comment | added | johnwbyrd | Re the vuvuzela example: there are several adequate pure signal-domain solutions for this problem as well, so I wonder if another FFT-only example might be more illuminating here. | |
| Nov 15, 2012 at 23:47 | comment | added | Diego | would this answer be a great candidate to become a community wiki and continuously updated/worked on by the community? This seems like a pretty important question for the community and @yoda has already done a fantastic job but we can update it with the answers below, other communities, etc | |
| Mar 24, 2012 at 19:01 | comment | added | nibot | Peter K's point is really critical. Signals can be represented with respect to many different bases. Sines and cosines are special because they are the eigenfunctions of LTI systems. | |
| Sep 20, 2011 at 14:40 | history | edited | Lorem Ipsum | CC BY-SA 3.0 |
minor typos I missed the last time
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| Sep 20, 2011 at 4:42 | history | edited | Lorem Ipsum | CC BY-SA 3.0 |
changed \omega to f, to avoid confusion with missing 2\pi, as per Dilip Sawarte's comment.
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| Sep 19, 2011 at 23:59 | comment | added | Dilip Sarwate | Yoda's answer is missing factors of $\frac{1}{2\pi}$ in some key places such as the definition of the Fourier Tranform, Parseval's Theorem, etc. This really should have been a comment on his post, and he could have edited his answer to correct the errors, but I do not have enough reputation on this site to make comments, and so an answer will have to make do. | |
| Sep 8, 2011 at 13:06 | review | Suggested edits | |||
| Sep 20, 2011 at 4:30 | |||||
| Aug 20, 2011 at 3:03 | comment | added | Jonas | To elaborate on #3: Convolution is what you do when you apply a filter to an image, such as an average filter, or a Gaussian filter (though you can't Fourier-transform non-linear filters). | |
| S Aug 19, 2011 at 20:24 | history | suggested | Phonon | CC BY-SA 3.0 |
fixed a formula symbol
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| Aug 19, 2011 at 16:08 | review | Suggested edits | |||
| S Aug 19, 2011 at 20:24 | |||||
| Aug 19, 2011 at 15:22 | vote | accept | jcolebrand | ||
| Aug 19, 2011 at 15:04 | comment | added | endolith | @yoda: Yes, I miscomprehended as "broken down into a sum of simpler functions", and was just pointing out an alternative way to do that | |
| Aug 19, 2011 at 14:34 | comment | added | Lorem Ipsum | @endolith I didn't say the Fourier transform was first, just that it is powerful. Note that a Taylor series is not an expansion in terms of the constituent frequencies. For e.g., the Taylor series of $\sin(\alpha x)$ about $0$ is $\alpha x-\alpha^3x^3/3!+\alpha^5x^5/5!\ldots$, whereas the Fourier transform of $\sin(\alpha x)$ is $\left[\delta(\omega-\alpha)-\delta(\omega+\alpha)\right]/(2\jmath)$ (give or take some normalization factors). The latter is the correct frequency representation, so I'm not sure if any comparisons with Taylor series is apt here. | |
| Aug 19, 2011 at 14:10 | comment | added | endolith | "This idea that a function could be broken down into its constituent frequencies was a powerful one" I think the Taylor series was first. en.wikipedia.org/wiki/Taylor_series#History en.wikipedia.org/wiki/Fourier_analysis#History "a Fourier transform of a signal tells you what frequencies are present in your signal and in what proportions." Like a prism shows you what frequencies of light are present in a light source | |
| Aug 19, 2011 at 7:50 | history | answered | Lorem Ipsum | CC BY-SA 3.0 |