Posted by GarMan [ 64.10.99.191 ] on May 27, 2004 at 18:32:57:

In Reply to: Re: NooBee Question posted by Dean Kukral on May 27, 2004 at 17:17:52:

In a straightforward L-pad with just two resistors, the attenuation is frequency independent. However, Wayne's compensation network incorporates a capacitor in parallel with the series resistor, and this in turn adjusts the level of attenuation by frequency.

Since you asked for mathematical models, everytime you reduce the voltage to a driver by half, you attenuate by 6dB. Level of attenuation is calulated by:

dB = 20*log(Vo/Vi)
where Vo/Vi is the percentage of the original voltage.

I think the easiest thing to do is to just use one of Wayne's crossover networks that he developed for his PI 18" speakers.

Gar.

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