Timeline for Multiples of frequencies in Fourier transforms [closed]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2016 at 4:27 | history | closed |
DanielSank user36790 ACuriousMind♦ Gert knzhou |
Needs details or clarity | |
| Jun 18, 2016 at 18:20 | history | protected | Qmechanic♦ | ||
| Jun 18, 2016 at 18:19 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 6 characters in body; edited tags
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| Jun 18, 2016 at 18:14 | vote | accept | user1936752 | ||
| Jun 18, 2016 at 13:40 | answer | added | garyp | timeline score: 2 | |
| Jun 18, 2016 at 11:07 | answer | added | Selene Routley | timeline score: 2 | |
| Jun 18, 2016 at 7:32 | review | Close votes | |||
| Jun 19, 2016 at 4:27 | |||||
| Jun 18, 2016 at 7:12 | comment | added | DanielSank | Consider a signal $x(t) = \exp(i \Omega t)$. Now Fourier transform: $\tilde{x}(\omega) \equiv \int dt x(t) \exp(-i \omega t) = \int dt \, \exp(i (\Omega - \omega) t ) = 2 \pi \delta(\Omega - \omega)$. So that's it, you get a delta function at $\Omega$, not at $2 \Omega$. In other words, $\exp(i \Omega t)$ has frequency $\Omega$; it does not have frequency $2 \Omega$. | |
| Jun 18, 2016 at 6:34 | history | edited | user1936752 | CC BY-SA 3.0 |
added 385 characters in body
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| Jun 17, 2016 at 19:58 | answer | added | hsinghal | timeline score: 1 | |
| Jun 17, 2016 at 12:21 | comment | added | Selene Routley | I'm not sure I understand; if I read this right, why does $e^{i\,\omega\,t}$ not show a spectral component the higher harmonics? Is that your question? Actually, in the discrete Fourier transform it sometimes can owing to aliasing. So if your question's about this kind of thing, then please tell us the sampling interval, and number of points in the FFT. This is the minimum information needed to know what an FFT will do. | |
| Jun 17, 2016 at 8:39 | history | asked | user1936752 | CC BY-SA 3.0 |