Main Archive Specials Wiki | FAQ Links Submit Forum


RBJ Audio-EQ-Cookbook

Type : EQ filter kookbook
References : Posted by Robert Bristow-Johnson
Linked file : Audio-EQ-Cookbook.txt

Notes :
Equations for creating different equalization filters.
see linked file





Comments


Added on : 30/08/06 by martin[ DOT ]eisenberg[ AT ]udo[ DOT ]edu
Comment :
rbj writes with regard to shelving filters:

> _or_ S, a "shelf slope" parameter (for shelving EQ only).  When S = 1,
> the shelf slope is as steep as it can be and remain monotonically
> increasing or decreasing gain with frequency.  The shelf slope, in
> dB/octave, remains proportional to S for all other values for a
> fixed f0/Fs and dBgain.

The precise relation for both low and high shelf filters is

  S = -s * log_2(10)/40 * sin(w0)/w0 * (A^2+1)/(A^2-1)
  
where s is the true shelf midpoint slope in dB/oct and w0, A are defined in
the Cookbook just below the quoted paragraph. It's your responsibility to keep
the overshoots in check by using sensible s values. Also make sure that s has
the right sign -- negative for low boost or high cut, positive otherwise.

To find the relation I first differentiated the dB magnitude response of the
general transfer function in eq. 1 with regard to log frequency, inserted the
low shelf coefficient expressions, and evaluated at w0. Second, I equated this
derivative to s and solved for alpha. Third, I equated the result to rbj's
expression for alpha and solved for S yielding the above formula. Finally
I checked it with the high shelf filter.




Added on : 31/08/06 by martin[ DOT ]eisenberg[ AT ]udo[ DOT ]edu
Comment :
Sorry, a slight correction: rewrite the formula as

  S = s * log_2(10)/40 * sin(w0)/w0 * (A^2+1)/abs(A^2-1)

nad make s always positive.




Added on : 05/10/13 by mbjoker65[ AT ]gmail[ DOT ]com
Comment :
This is a very famous article. I saw many are asking what is the relationship between "Q" and the resonance in low-pass and hi-pass filters.

By experimenting, I found that Q should always be >= 1/2. Value < 1/2 seems to alter f0 "wherever it's happenin', man", cutting off frequencies not where it was planned. In fact Q = 1/2 is the value for which H(s) = 1 / (s^2 + s/Q + 1) gets two poles, real and coincident. In other words the filter becomes like two 1st order filters in cascade, with no resonance at all.

When Q tends to infinite the poles get close to the unit circle, the gain around the cutoff frequency increases, creating resonance.




Add your own comment
Comments are displayed in fixed width, no HTML code allowed!
Email:

Comment:

Are you human?



Site created and maintained by Bram
Graphic design by line.out | Server sponsered by fxpansion