Timeline for Filter: IIR Filter
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2021 at 23:04 | history | edited | robert bristow-johnson | CC BY-SA 4.0 |
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| Apr 14, 2020 at 12:11 | vote | accept | ksi | ||
| Apr 14, 2020 at 12:11 | comment | added | ksi | Thanks a lot. got it now. | |
| Apr 14, 2020 at 11:36 | comment | added | Matt L. | @ksi: Mind the normalization. The $a_1$ given at the end of my answer is not equal to $2(1-k^2)$ but it is $2(1-k^2)/(k^2+\sqrt{2}k+1)$. Check again Eq. $(3)$, where I've added another step (to normalize $a_0$ to $1$). | |
| Apr 14, 2020 at 10:02 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Apr 13, 2020 at 16:22 | comment | added | Matt L. | @ksi: I've added a few steps to get you started. | |
| Apr 13, 2020 at 16:21 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Apr 13, 2020 at 15:22 | comment | added | Matt L. | @ksi: That's exactly what I tried to explain in my answer. You'll have to sit down yourself and use the formulae I've shown you. Have you actually tried replacing $s$ in $H(s)$ by $k(z-1)/(z+1)$? Once you've done that you just need to replace the constant $k$ (and $k^2$) by the expressions given in my answer, and you'll arrive at exactly the same formulae for the filter coefficients as I did. If you really arrive at a different result, please add the steps to your question and we can try to figure out where you went wrong. | |
| Apr 13, 2020 at 15:13 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Apr 13, 2020 at 14:24 | comment | added | Matt L. | @PeterK.: Oh yeah, for some reason I can never remember his name :) | |
| Apr 13, 2020 at 14:21 | comment | added | Peter K.♦ | That should be "robert bristow-johnson" :-) But I'll let it pass. | |
| Apr 13, 2020 at 14:15 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Apr 13, 2020 at 14:01 | history | edited | Matt L. | CC BY-SA 4.0 |
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| Apr 13, 2020 at 10:43 | history | answered | Matt L. | CC BY-SA 4.0 |