Butterfield Audio: Fun with Math

Suppose we had a 2nd order high pass filter for a tweeter of impedance Rt. The filter would have a crossover frequency fc, and a Q. We might want to mate a woofer to this tweeter in such a way as to avoid all frequency and phase distortion. This would be a transient perfect design. The woofer might have impedance Rw. There would have to be an output level discrepency favoring the woofer in the following amount:

WooferSPL = TweeterSPL + 20*log10(1/(1-Q^2))
(Note that Q should be less than 1.)

Next we would want to put an inductor in series with our woofer to supress the high frequency output. The equation for this inductor is as follows:

L = Q*Rw/(2*pi*fc*(1-Q^2))

Next we would put a capacitor (C) and resistor (Rc) in parallel with eachother, and place that unit in series with our woofer and inductor. We would do this according to the following equations:

Rc = Rw/(1/Q^2 - 1)
C = (1-Q^2)/(wc*Rw*Q^3)

So what we have in the end is a first order band-pass filter, with the high pass capacitor shorted out with a parallel resistor. Depending on the values of fc and Q, the output would look something like the picture above. Note that the woofer and tweeter output do not cross at fc. Also, these represent target slopes, not real life crossover networks. In real life, speakers have very complex impedances, and their own response characteristics which must be taken into account.