[sdiy] Ladder filters and gain drop, that old chestnut

Richie Burnett rburnett at richieburnett.co.uk
Sun Aug 30 14:40:55 CEST 2015


> The gain needed for an
> n stage cascade is 1/(cos(pi/n)^n), so the gain for self oscillation
> is as follows:
> 
> n=3 gain=8
> n=4 gain=4
> n=5 gain=2.9
> n=6 gain=2.4
> n=7 gain=2.1
> n=8 gain=1.9

Thanks for posting the figures Andy. I was too lazy to work them all out!

To come back to the original enquiry about "gain drop" in the passband, you can calculate the passband gain drop from the feedback gain easily:

Passband gain is equal to 1 divided by (1 + feedback gain). So in the case of the moog ladder it is 1/(1+4) or 0.2, that's 14dB attenuation at the onset of self oscillation. It's easy to calculate the passband attenuation for other filters with different numbers of poles, using Andy's figures. 

As others have already said, it's easy to make up for all (or some) of this passband gain drop by feeding forward a little more of the drive signal to the input of the filter as the negative feedback around it is increased. 

> The TB-303 diode cascade is a 4 pole filter, but the stages are not
> buffered from each other, so the gain at the cutoff isn't -3dB, it's
> lower...

I know what you mean, but we need to be careful with terminology here. Most Electrical Engineer types will say the gain *IS* -3dB at the cutoff frequency, because the -3dB amplitude point is the definition of "cutoff frequency". It's just convention really.

You could say that the Tb-303 filter's resonance occurs at a frequency several octaves above its cutoff frequency...

From an EE perspective the cutoff frequency is where the amplitude of the filter with no feedback is 3dB down. The *crossover frequency* is where the phase shift goes through 180 degrees. The latter is the frequency at which the filter self oscillates if it is given enough feedback. These two frequencies don't have to be the same and are often separated. The Tb-303's unbuffered ladder is an extreme example of this.

-Richie,


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