On Sun, Feb 19, 2012 at 1:04 PM, Mathieu Bouchard <ma...@artengine.ca> wrote: > Le 2012-02-02 à 19:13:00, Mike Moser-Booth a écrit : >> I don't think this is entirely accurate. I think a and c should be >> switched here, though of course when finding b²-4ac that doesn't >> really matter. > > > It depends whether you write ax²+bx+c or a+bx+cx². Both forms are > convenient, and the latter expands better in cases of variable degrees > (letters don't get renamed when adding a term), but the former is more > common for cases that have only a degree fixed at 2 or 3.
Ah, okay. >> Also, when applying a gain to a recursive filter, it's not really the same >> as scaling all the coefficients. If you were to scale them first, then the >> gain would affect the feedback portions of the filter. Applying the gain >> after means the feedback samples are not scaled by the gain. > > > Those filters are all linear. This means that you can effectively commute > them with a constant gain [*~] without any difference. However, it will make > a difference when the gain of [*~] changes quickly while the main input > changes too. Right. I was just saying that, if you separate the gain from the filter, the gain has to be applied at the input or the output. It's a single operation. So if you incorporate it into the filter, scaling the feedback coefficients has the effect of making it multiple operations. Or at least that's how I understand it, anyway. At any rate, the only reason I picked up on that in your previous email is because I wanted to plot the frequency response of [bp~], and at first I just applied the gain to all the coefficients without thinking about. It didn't work. Then I remembered the gain should just be applied to the FIR part of the filter, and it worked fine. >> If you think of it in terms of its transfer function, it would look >> more like this: H(z) = g*(1 / (1 - 2r*cos(ω)*(z^-1) + r^2 * z(^-2) )) > > > Well, I was thinking of it in terms of 1/H(z) or 1/gH(z). > > But are you sure that you got the signs right in the denominator ? Pretty sure, though I mess that up all the time. Aren't you supposed change the signs of the feedback coefficients when z-transforming the difference equation? .mmb -- Mike Moser-Booth - mmoserbo...@gmail.com Master's Student in Music Technology Schulich School of Music, McGill University Centre for Interdisciplinary Research in Music Media and Technology "If you think education is expensive, try ignorance" -Derek Bok _______________________________________________ Pd-list@iem.at mailing list UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list
