Timeline for How can I find the following Fourier Transform without directly using FT pairs?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2023 at 22:28 | comment | added | robert bristow-johnson | Oh, you're right. I just looked at it wrong. Multiplying by 1/2 is not the same as adding 1/2. | |
| Jul 6, 2023 at 22:26 | comment | added | robert bristow-johnson | Matt, it samples at $t=-\frac12$ and at $t=+\frac12$. I don't see it sampling at $t=0$. | |
| Jul 6, 2023 at 20:59 | comment | added | Matt L. | @robertbristow-johnson: Sampling of the discontinuity happens at $n=0$. That's also where $\frac12\delta(t)$ comes from. | |
| Jul 6, 2023 at 20:31 | comment | added | robert bristow-johnson | In your Eq. (1), I don't see where that $\frac12 \delta(t)$ comes from. | |
| Jul 6, 2023 at 20:27 | comment | added | robert bristow-johnson | I didn't go through the math, but it seems to me that this should be straight forward since the dirac impulse train was offset by 1/2 meaning that there is no sampling of the edge of $e^{-t} u(t)$. I don't see any jump discontinuity getting sampled. | |
| Jul 6, 2023 at 18:48 | comment | added | Ahsan Yousaf | + 1 Very nice. I actually had no idea about this correction. Thanks for bringing it to our attention! | |
| Jul 6, 2023 at 18:40 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Jul 6, 2023 at 16:12 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Jul 6, 2023 at 16:07 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Jul 6, 2023 at 15:57 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Jul 6, 2023 at 15:49 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Jul 6, 2023 at 15:37 | history | answered | Matt L. | CC BY-SA 4.0 |