Stephane, I wrote this email right after our short discussion on slack.com but somehow I forgot to send it.
So I don't really know how to compare this filter with the existing filters forms in filters.lib, but please see the simple, stupid and "non-scientific" test below. Note! all 3 band-pass filters bp(fi.tf2s, F,Q), bp(fi.tf2snp,F,Q) and svf.bp(F,Q) have the same impulse responce for any F/Q. However they differ if F or/and Q is modulated, this is what this test tries to demonstrate, note that the 3rd output has a much more flat envelope. See http://people.redhat.com/onestero/svf/output.png Oleg. ------------------------------------------------------------------------------- import("stdfaust.lib"); F = 1000 + 200*os.oscrs(50); Q = 2; process = os.oscrs(F) <: bp(fi.tf2s, F,Q), bp(fi.tf2snp,F,Q), svf.bp(F,Q) // separate the outputs to make the difference more clear : par(i, 3, -(5*i)); bp(t2s, F,Q) = t2s(b2,b1,b0,a1,a0,wc) with { wc = 2*ma.PI*F; a1 = 1/Q; a0 = 1; b2 = 0; b1 = 1; b0 = 0; }; svf = environment { svf(T,F,Q,G) = tick ~ (_,_) : !,!,si.dot(3, mix) with { tick(ic1eq, ic2eq, v0) = 2*v1 - ic1eq, 2*v2 - ic2eq, v0, v1, v2 with { v1 = ic1eq + g *(v0-ic2eq) : /(1 + g*(g+k)); v2 = ic2eq + g * v1; }; A = pow(10.0, G / 40.0); g = tan(F * ma.PI / ma.SR) : case { (7) => /(sqrt(A)); (8) => *(sqrt(A)); (t) => _; } (T); k = case { (6) => 1/(Q*A); (t) => 1/Q; } (T); mix = case { (0) => 0, 0, 1; (1) => 0, 1, 0; (2) => 1, -k, -1; (3) => 1, -k, 0; (4) => 1, -k, -2; (5) => 1, -2*k, 0; (6) => 1, k*(A*A-1), 0; (7) => 1, k*(A-1), A*A-1; (8) => A*A, k*(1-A)*A, 1-A*A; } (T); }; lp(f,q) = svf(0, f,q,0); bp(f,q) = svf(1, f,q,0); hp(f,q) = svf(2, f,q,0); notch(f,q) = svf(3, f,q,0); peak(f,q) = svf(4, f,q,0); ap(f,q) = svf(5, f,q,0); bell(f,q,g) = svf(6, f,q,g); ls(f,q,g) = svf(7, f,q,g); hs(f,q,g) = svf(8, f,q,g); }; _______________________________________________ Faudiostream-users mailing list Faudiostream-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/faudiostream-users
