# RBJ Audio-EQ-Cookbook¶

**Author or source:**Robert Bristow-Johnson**Type:**EQ filter kookbook**Created:**2005-05-04 20:31:18**Linked files:**`Audio-EQ-Cookbook.txt`

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Equations for creating different equalization filters.
see linked file
```

## Comments¶

**Date**: 2006-08-30 22:14:22**By**: ude.odu@grebnesie.nitram

```
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.
```

**Date**: 2006-08-31 17:08:27**By**: ude.odu@grebnesie.nitram

```
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.
```

**Date**: 2013-10-05 18:06:20**By**: moc.liamg@56rekojbm

```
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.
```