Timeline for In a periodogram, why is the modulus squared of the DFT taken rather than the DFT itself? [duplicate]
Current License: CC BY-SA 4.0
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| when toggle format | what | by | license | comment | |
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| May 24, 2018 at 3:59 | history | closed |
Marcus Müller MBaz user28715 A_A lennon310 |
Duplicate of Why so many methods of computing PSD? | |
| May 23, 2018 at 21:09 | answer | added | hotpaw2 | timeline score: 0 | |
| May 23, 2018 at 17:46 | review | Close votes | |||
| May 24, 2018 at 3:59 | |||||
| May 23, 2018 at 17:28 | comment | added | Marcus Müller | Because that's how it's defined. Also, how would you visualize a complex-valued 2D object? "How accurate" and "when is it very accurate" would require us to first define accuracy, and for that define signal classes, possibly introduce you to the theory of stochastic signals, what an autocorrelation function is and so on... too broad, I'm afraid! | |
| May 23, 2018 at 17:18 | history | edited | user29 | CC BY-SA 4.0 |
added 46 characters in body
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| May 23, 2018 at 17:10 | review | First posts | |||
| May 24, 2018 at 4:03 | |||||
| May 23, 2018 at 17:08 | history | asked | user29 | CC BY-SA 4.0 |