OK a bit of math. I'm using conventional operational mathematics notation where * is convolution and . is multiplication. d(w-w1) is a delta operator in frequency.
The normal representation is:
S1(w) = S0(w) . F1(w) . F2(w)
If F1 and F2 have the same BW but different center frequencies we can rewrite as:
S(1) = S0(w) . (F0*d(w-w1)) . (F0*d(w-w2))
which simplifies to
S(1(w) = S0(w) . (F0(w)**2 ) * d(w-w3)
This is the best explanation of why it works that I've been able to devise. It says that the pass band of the individual filters sets the time domain response, but the overlap sets the frequency domain response. The delta in frequency makes the time domain impedance complex.
Have Fun!
Reg
