Timeline for Convolution in frequency domain
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2015 at 14:24 | comment | added | Matt L. | @DilipSarwate: Note that it's an odd length DFT, that's why there is no bin at Nyquist. The two 0's are the (conjugate complex) elements at indices $-2$ and $2$, just below Nyquist. | |
| Nov 20, 2015 at 14:20 | comment | added | Dilip Sarwate |
Why does A equal ` [0,-j,0,0,j]` ? I am puzzled by the two 0's in a row.
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| Nov 20, 2015 at 13:17 | history | edited | Matt L. | CC BY-SA 3.0 |
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| Nov 9, 2015 at 19:36 | comment | added | Dole | L Thanks, I forgot that positive imaginary part actually reflects inverted sine wave, that threw me off a bit there. Indeed I had the spectrum from negative to positive and you have it from 0 up. | |
| Nov 9, 2015 at 18:35 | comment | added | Matt L. | @Dole: They are symmetric: the first element is the zeroth bin (DC), the second element is bin number 1 (the frequency of the sinusoid); and the last element is bin number -1 (the negative frequency of the sinusoid). Try to see it by periodic continuation of the vector: (...,0,j,[0,-j,0,0,j],0,-j,...) | |
| Nov 9, 2015 at 17:25 | comment | added | Dole | Where does that frequency domain representation of the sine wave come from? Don't the negative and positive parts always need to be symmetric around the DC offset bin? | |
| Nov 9, 2015 at 16:43 | vote | accept | Dole | ||
| Nov 9, 2015 at 16:43 | vote | accept | Dole | ||
| Nov 9, 2015 at 16:43 | |||||
| Nov 8, 2015 at 20:23 | history | edited | Matt L. | CC BY-SA 3.0 |
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| Nov 8, 2015 at 19:05 | history | edited | Matt L. | CC BY-SA 3.0 |
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| Nov 8, 2015 at 19:00 | history | answered | Matt L. | CC BY-SA 3.0 |