Timeline for Why are the basis functions for DFT so?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2016 at 18:40 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 29, 2016 at 18:08 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 30, 2016 at 17:36 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 8, 2016 at 9:20 | comment | added | Gilles | You may also want to look at this question and this one. They got some nice info on this as well. | |
| Apr 1, 2016 at 4:26 | comment | added | robert bristow-johnson | Jazz is right, the basis functions are $e^{j2\pi kn/N}$ and they could have just as well been $e^{-j2\pi kn/N}$. by convention, we call the imaginary unit "$j$" (except everyone else in the world, other than EEs call it "$i$"), but we could just as well as declared $-j$ to be the imaginary unit. there is no difference, in any of their properties, between "$j$" and "$-j$". both have equal claim to squaring to be $-1$. if we went to all of our textbooks and replaced every instance of $j$ with $-j$ (and vise versa), every theorem would continue to be just as valid. | |
| Mar 31, 2016 at 6:52 | answer | added | Behind The Sciences | timeline score: 1 | |
| Mar 30, 2016 at 15:18 | comment | added | Jazzmaniac | In fact, your basis functions are $\exp(+2\pi i kn/N)$, the minus sign stems from the sesquilinear product on complex vector spaces: It is antilinear in the first argument and linear in the second. So the basis you expand into is conjugated. | |
| Mar 30, 2016 at 8:50 | comment | added | Derek Elkins left SE | Whether we use the conjugate or not is essentially a convention. As far as other functions, we do use all kinds of other functions. The discrete cosine transform, the Z transform, wavelet transforms, number theoretic transforms, etc. The Fourier transform is just the basis that most resolves frequency (at the cost of temporal locality). | |
| Mar 30, 2016 at 8:39 | answer | added | spectre | timeline score: 1 | |
| Mar 30, 2016 at 8:11 | comment | added | MaximGi | Because it's the definition of the Fourier's transform : represent any function as a sum of complex exponentials. see en.wikipedia.org/wiki/Fourier_analysis | |
| Mar 30, 2016 at 6:48 | history | asked | GrowinMan | CC BY-SA 3.0 |