Timeline for How to derive Q from r in this filter definition?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| S Nov 27, 2015 at 6:50 | history | suggested | user11906 | CC BY-SA 3.0 |
removed irrelevant parts, corrected case
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| Nov 27, 2015 at 6:37 | review | Suggested edits | |||
| S Nov 27, 2015 at 6:50 | |||||
| Nov 12, 2015 at 0:34 | vote | accept | danfuzz | ||
| Nov 11, 2015 at 20:41 | comment | added | Matt L. | I know it's a band pass, but even there pole and resonant frequency are generally not equal. Check (your own) cookbook for the $a_1$ coefficient and you'll see what I mean. | |
| Nov 11, 2015 at 18:52 | comment | added | robert bristow-johnson | take a look at $b_0$, $b_1$, and $b_2$. it's a bandpass filter. the peak frequency and resonant frequency are the same thing. but i take your point, in the Impulse Invariant transformation, i may have $2\pi f_0 T$ as the angle the pole has against the $\Re\{z\}$ axis. | |
| Nov 11, 2015 at 8:12 | comment | added | Matt L. | Robert, your $f_0$ is not the same as the OP's $f_0$. This will cause misunderstandings. Your $f_0$ is the peak frequency, whereas the OP's $f_0$ is the pole frequency. I also left a comment for the OP to clear this up. | |
| Nov 11, 2015 at 6:19 | history | edited | robert bristow-johnson | CC BY-SA 3.0 |
added 2 characters in body
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| Nov 11, 2015 at 6:13 | history | answered | robert bristow-johnson | CC BY-SA 3.0 |