Hello Federico,
Le 20/07/2022 à 22:47, Federico Miyara a écrit :
Dear All,
I have problems with this script:
Fs = 44100;
fo = 500;
B = 2^(1/6)-2^(-1/6);
// Low-pass Butterworth
hLPs = analpf(3,"butt",[],1);
// Band-pass Butterworth centered at fo
hBPs = horner(hLPs, (2*%pi*fo/%s + %s/(2*%pi*fo))/B)
// Apply bilinear transform
hBPz = horner(hBPs, 2*Fs*(%z - 1)/(%z + 1))
// Attempt to get the same using iir
hBPz1 = iir(3, "bp", "butt", fo/Fs*[2^(1/6),2^(-1/6)], [0 0])
I attempt to design a discrete Butterworth band-pass filter using two
equivalent methods:
1) Manually applying a bilinear transform to the analog filter
designed with analpf()
2) Applying the function iir()
For some resaon I couldn't figure out, the first method yields an
unexpected result:
--> hBPz
hBPz =
0.0000005 +0.0000016z +0.0000016z² +0.0000005z³
-----------------------------------------------
-0.9671578 +2.9195957z -2.9518237z² +z³
--> hBPz1
hBPz1 =
0.0000006 -0.0000017z² +0.0000017z⁴ -0.0000006z⁶
-----------------------------------------------------------------------------
1.0335428 -6.1515162z +15.271063z² -20.239437z³ +15.104038z⁴
-6.0176897z⁵ +z⁶
The official function iir() is correct. The manual procedure
unexpectedly reduces the order of the denominator.
Maybe someone can find out what's going on...
Sorry, but nothing is clear to me about your statements:
1) what shows you that iir() is correct while the analpf + horner way is
not?
2) With the analpf + horner method, assuming that it is not correct,
what shows you that horner is not correct, while analpf is correct,
instead of the opposite or both incorrect?
3) do you have a reference about the equivalence?
4) have you tried after simp_mode(%f)?
Samuel
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