Timeline for "Instantaneous impulse response" in a linear time-varying system
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2022 at 21:00 | history | tweeted | twitter.com/StackSignals/status/1579215137457999872 | ||
| Oct 9, 2022 at 14:55 | vote | accept | XYZT | ||
| S Oct 7, 2022 at 12:31 | history | edited | lennon310 | CC BY-SA 4.0 |
small typo in title
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| S Oct 7, 2022 at 12:31 | history | suggested | Gab | CC BY-SA 4.0 |
small typo in title
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| Oct 7, 2022 at 9:22 | answer | added | Matt L. | timeline score: 5 | |
| Oct 7, 2022 at 7:32 | review | Suggested edits | |||
| S Oct 7, 2022 at 12:31 | |||||
| Oct 6, 2022 at 22:43 | comment | added | Ash | Your analysis agrees with this blog post. | |
| Oct 6, 2022 at 22:24 | comment | added | XYZT | Hmm, but it does not look like a convolution - which I thought would be the case if it was correct. | |
| Oct 6, 2022 at 22:11 | comment | added | Ash | Your last equation is correct. The response for an impulse at a delay of $\tau'$ is only valid for the impulse response defined at that time delay (i.e. $h(\tau, t-\tau')$). A response $h(\tau, t)$ would only be realized by an impulse occuring at $\delta(t)$, not $\delta(t-\tau')$. Variable overload, adding prime for input delay | |
| Oct 6, 2022 at 22:08 | history | edited | XYZT | CC BY-SA 4.0 |
added 76 characters in body
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| Oct 6, 2022 at 22:01 | history | edited | XYZT | CC BY-SA 4.0 |
added 299 characters in body
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| Oct 6, 2022 at 21:56 | comment | added | XYZT | Yes, that's what I used to compute the integral. I edited my question to show this is what I am doing. | |
| Oct 6, 2022 at 21:55 | history | edited | XYZT | CC BY-SA 4.0 |
added 299 characters in body
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| Oct 6, 2022 at 21:55 | comment | added | Ash | If $x(t)=\delta(t-t_0)$, then $x(t-\tau) = \delta(t-\tau-t_0)$. | |
| Oct 6, 2022 at 21:00 | history | edited | XYZT | CC BY-SA 4.0 |
added 336 characters in body
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| Oct 6, 2022 at 20:46 | history | asked | XYZT | CC BY-SA 4.0 |