Timeline for Can I set a constraint on the first tap of an FIR filter such that its inverse is stable?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25 at 14:00 | answer | added | Matt L. | timeline score: 1 | |
| Oct 5, 2022 at 3:04 | history | tweeted | twitter.com/StackSignals/status/1577494944822050816 | ||
| Nov 4, 2021 at 17:18 | answer | added | Bob | timeline score: 1 | |
| May 22, 2020 at 21:21 | vote | accept | Condensation | ||
| May 22, 2020 at 3:24 | history | edited | robert bristow-johnson | CC BY-SA 4.0 | edited body |
| May 8, 2020 at 9:49 | comment | added | a concerned citizen | Just a thought: what do you do in the case of a half-band filter, with that $h_0$? | |
| May 5, 2020 at 22:49 | comment | added | TimWescott | Sufficently low $k$? Argh -- it's out there, in Mathmagic land. I can hear its mating cry, and I've found droppings, but I just can't see it yet. | |
| May 5, 2020 at 22:49 | answer | added | Dan Boschen | timeline score: 5 | |
| May 5, 2020 at 22:48 | comment | added | TimWescott | Yes. There is some minimum value for the first tap relative to the other taps (probably the sum of their absolute values) that would make your inverse stable. I base this on the Evans root locus of a feedback system $k H(z) / (1 + k H(z))$ where $H(z) = \sum_{k=0}^{n-1} a_k z^{-k}$. That will have a root of multiplicity $n$ at $z = 0$, so for sufficiently high $k$ the system will be guaranteed stable -- and the value of $k$ is related to the weight of the first tap of the filter relative to the weights of the rest. But I can't chase the proof now (and the filter would probably be useless). | |
| May 5, 2020 at 21:56 | review | First posts | |||
| May 6, 2020 at 0:32 | |||||
| May 5, 2020 at 21:54 | history | asked | Condensation | CC BY-SA 4.0 |