0
$\begingroup$

I'm trying to find the equation of a shelving filter of the second order. I easily found the equation for a first order filter on wikipedia with a transition region which have a 6db per octave slope. The equations are :

  • coef^2 * (1 + (f / (coef * f0))^2) / (1 + (f/f0)^2) for the lowshelf

  • (1 + (f / (coef * f0))^2) / (1 + (f/f0)^2) for the highshelf

With :

coef = multiplicator coef (gain = 10 log10 (coef))

f = frequency

f0 = cutoff frequency

But now, how can I use a kind of Q factor to modify the slope of the transition region ?

Of course, as I'm using already a frequency signal and not a temporal signal, I can't use the equations made for the z transform...

Thank you for all your answers

Improve this question
$\endgroup$
6
  • $\begingroup$ You're using a frequency signal? What do you mean? $\endgroup$ – Ben Jul 11 '18 at 17:46
  • $\begingroup$ I mean I'm using a spectral signal and not a temporal signal $\endgroup$ – Dr_Click Jul 11 '18 at 17:49
  • $\begingroup$ I fail to see why you would use an IIR or an analog filter in this case... $\endgroup$ – Ben Jul 11 '18 at 18:09
  • $\begingroup$ I'd like to increase the "slope coefficient" in the transition region. WIth that 1st order filter, I have a 6 db per octave slope, but I'd like to increase it to 12 or more. $\endgroup$ – Dr_Click Jul 11 '18 at 18:51
  • 1
    $\begingroup$ check out the Audio EQ Cookbook. $\endgroup$ – robert bristow-johnson Jul 11 '18 at 20:02
0
$\begingroup$

Ok, I finally found a solution after few search about sigmoïd charts and few try. What I wanted is to modify the slope factor in the transition area without modifying the high and low border values. So, the equations are becoming : 

coef^2 * (1 + (f / f0)^(SFactor+2) * (1/coef)^2) / (1 + (f/f0)^(SFactor+2)) for the lowshelf

(1 + (f / f0)^(SFactor+2)) * coef^2 / (1 + (f/f0)^(SFactor+2)) for the highshelf

With :

coef = multiplicator coef (gain = 10 log10 (coef))

f = frequency

f0 = cutoff frequency

SFactor = slope factor. If SFactor = 0, it acts like a classical first order shelving filter, if SFactor increases, the slope in the transition area is steeper.

Thank you for all the previous comments and the Audio EQ CookBook is also a very interesting source I'm going to study...

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.