**An easy way, using normalisation**

Filter order 1, resistive load

~~~~~~~~~~~~~~~~~~

f = 1 Hz, R = 1 Ohm ==> C' = 160000 µF, L' = 160000 µH

. for other frequencies, divide C' et L' by the value of the required frequency

. for other resistive loads, divide C' et L' by the value of the required load

At wo = 2.Pi.fo (fo being the -3 dB frequency),

the impedances of C' and L' are equal to R

ZC' =1/(wo*C') = R

ZL' = wo*L' = R

Filter order 2, resistive load

~~~~~~~~~~~~~~~~~~

f = 1 hz, R = 1 Ohm and Q = 1.00 ==> C" = 160000 µF, L" = 160000 µH

. for other frequencies, divide C" et L" by the value of the required frequency

. for other resistive loads, divide C" et L" by the value of the required load

. for other Qs :

multiply C" by the value of the required Q

divide L" by the value of the required Q

When wo² = 1/L".C" = (2.Pi.fo)² (fo being the resonance frequency resulting from the combination of reactive components C" and L")

ZC" = 1 / (Q*wo*C")

ZL" = (wo*L1") / Q

For Q = 1,

ZC" = ZL" = R, the value of C" and L" are the same as C' and L' of the order 1 filter if the load is the same.

Note

the exact value of 160000 is 159236, the difference is less than 0.5%. It is much more precise than the usual components of passive crossovers and than the driver impedance and frequency (often 5% variation for this last one) .