Hallo, Charles Henry hat gesagt: // Charles Henry wrote:
On Mon, Sep 15, 2008 at 7:49 AM, Damian Stewart damian@frey.co.nz wrote:
dude, that's not 'easy'. i can barely remember how the quadratic formula works with pen and paper, let alone in C, and let alone to the point where i could confidently transfer from C to Pd. i'd have no idea where the bugs were... and indeed, no idea
Use pen and paper. Then, take your results and make implementations in Pd (or C, your choice). This is a good exercise, so I'll leave it to you to do the derivation. I'm more than happy to help. I do this stuff all the time.
- Solve for x in
a*x^2 + b*x + c =0 (use "completing the square") This is how you derive the quadratic formula.
Okay, that's easy (or can be looked up e.g. here: http://www.purplemath.com/modules/sqrquad2.htm )
- Then, solve
ff1*z^2 + ff2*z^1 + ff3 = 0 and 1*z^2 - fb1*z^1 - fb2 = 0 using the result from (1)
Easy now as well.
- Then, solve for fb1, fb2, ff1, ff2, ff3 in terms of the poles/zero
locations from (2)
After (2) I have something like this:
x = (-b +- sqrt(b*b - 4*a*c)) / 2*a
zf = (-ff2 +- sqrt(ff2*ff2 - 4*ff1*ff3)) / 2*ff1 zb = (-fb1 +- sqrt(fb1*fb1 - 4*fb2)) / 2
This gives me four complex numbers zf+, zf-, zb+ and zb-, right?
Now in a biquad~-clone with czero~/cpole~, the values for fb1, fb2, ff1, ff2, ff3 would be known so I also could calculate zf and zb now.
But what I still don't get is how to feed this into a patch as below as coefficients to make a biquad~ clone:
[*~] ? | | ? ? [czero~] | | | | ? ? [czero~] | | | | ? ? [cpole~] | | | | ? ? [cpole~]
This still has several question marks. :(
Ciao
Frank