Passive high pass filter - Marchand XM446XLR-A

Great, thanks Al---Eric.

Thanks much Al!
The Claytons only have balanced inputs so I am confident the 100K input impedance is for the balanced condition.
Good point on matching the caps.  I plan to try something relatively high quality like Jupiter copper film caps.
Partsconnexion will match them for $1 per pair.
I plan to try 0.022uf caps for a cut-off frequency of about 72 Hz and then start by low-passing the subs at 80 Hz.
The next cap size up (0.047uf) gives a cut-off frequency that is a little low at 34Hz but may still be an option.
If I wanted something in-between, could I series the 0.022 with a 0.01 cap?
Caps are thankfully 25% off this month, which is good since I need four of them.

Tim (Mitch2), to be sure it’s clear, I was saying that in most cases manufacturers specify balanced input impedances as the sum of the input impedances of the two balanced signal legs. I have seen a few exceptions, however, in which the specification corresponds to the input impedance of each leg. The fact that an unbalanced input is not provided has no relevance to that.

If the 100K spec on your amp is defined in the usual manner, you would want to base your calculation for each of the two capacitors that are required (per channel) on a 50K input impedance, as I indicated earlier. The 0.022 uf cap would then result in 144 Hz, not 72 Hz. And the 0.047 uf cap would result in 68 Hz, not 34 Hz.

In the much less common situation where the input impedance is defined on a per leg basis, meaning that each leg is 100K, your calculations would be correct.

If I wanted something in-between, could I series the 0.022 with a 0.01 cap?

No, you would want to parallel them. The values of capacitors in parallel add. The value of two capacitors in series corresponds to their product (multiplication) divided by their sum, which will always be less than the lower of the two individual values. It’s the opposite of what happens when resistors or inductors are connected in series or in parallel.

Paralleling two resistors or two inductors or two capacitors results in a lower impedance than either of the two paralleled parts would present individually, at a given frequency. Since the impedance of a capacitor at a given frequency is inversely proportional to its capacitance, to have a lower impedance the capacitance of the paralleled capacitors must be greater than the capacitance of each of the two. Again, it’s the opposite for resistors and inductors. The impedance of a resistor is identical to its resistance (to the extent that it behaves in a theoretically ideal manner), and the impedance of an inductor is directly proportional to its inductance, rather than inversely proportional.

Best regards,
-- Al

Dang, my momma was right when she told me I shoulda finished high school.

Thanks again Al.  It seems the 0.047uf caps will be a good start.  If the cut-off is at 68Hz then that will actually be perfect and if it is at 34Hz then it still relieves the amplifier from trying to drive the lowest frequencies.  Another thing I could do is to try some less expensive caps first to determine the correct value, and then later move up to the caps I want to use.

I also appreciate the analysis and technical information because one of my favorite things about this site is learning about stuff I don't fully understand.

In summary, there doesn't seem to be anything close to a universal answer to this situation.  I have decided not to pursue a Marchand filter at this time because I don't want something in the path that may negatively affect the signal going to my main speakers.  The subs do have a (reportedly) high quality on-board high pass filter so I can give that a try.  The single cap in-line interests me and I may give that a try.  If it works well, I could actually wire them inside of my preamp, or make a pair of balanced cables with the capacitors in-line.  From what I have read, it doesn't seem to matter where the caps are placed as long as they are between the preamp output and the amplifier input.  Some put them inside the input of the amplifiers.

Thanks for everyone for the comments and help.