[PD] [bp~] really equal to [vcf~]?
Claude Heiland-Allen
claude at mathr.co.uk
Fri Jul 25 17:24:15 CEST 2014
On 25/07/14 16:06, IOhannes m zmölnig wrote:
> On 07/25/2014 03:36 PM, Alexandre Torres Porres wrote:
>>> it's accurate to say that [biquad~] is a real-valued 2-pole/2-zero filter.
>>
>> but why, if the poles and zeros can have complex values?
>
> can they?
Yes, but they occur in conjugate pairs, as the transfer function has
real coefficients:
a s^2 + b s + c (s - z1) (s - z2)
H(s) = --------------- = k -----------------
e s^2 + f s + g (s - p1) (s - p2)
-b +/- sqrt(b^2 - 4 a c)
z1,z2 = ------------------------
2 a
-f +/- sqrt(f^2 - 4 e g)
p1,p2 = ------------------------
2 e
inside the sqrt is positive -> two real roots
inside the sqrt is negative -> complex conjugate pair
The imaginary parts of the complex conjugate "cancel out" to the
quadratic polynomial with all-real coefficients.
> afaict, [biquad~] only ever has real-valued coefficients and states.
sure, just different representations of the same thing
Claude
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