Does the intrinsic low-end roll-off of a mid driver change the phase error?

In another thread here, is the statement that "if your driver has an intrinsic 6dB low-end roll-off then adding an 18dB HP filter (at the right frequency) will turn that low-end roll-off into a 24dB slope".

My question is:
* an 18dB HP filter has a defined phase response (from memory, it leads the corresponding LP filter?).
* if the driver's intrinsic roll-off is 6dB, if this changes the slope to 24dB has it also changed the phase behaviour of the HP filter to that of a 24dB HP filter?

Regards,

Andy
 
The short and perfectly correct answer to your question is "Yes".

A longer explanation would add that the phase response of almost any analogue filter (with some conditions attached, which I don't remember apart from excluding all-pass filters) and the amplitude response are intimately tied together so that one implies the other. In fact, given one in sufficient detail you can derive the other. It is something to do with where the poles and zeros of the filter transfer function are in the complex plane. Does that help?
 
So a filter suddenly loses its own phase response when included in a crossover? No, but it may be compensated by another driver and its crossover filter.

Low and high pass both depend on poles and zeros, so the same rule applies. I don't remember the details, but it is something like a filter or set of cascaded filters with only one path from input to output can be fully characterised by poles and zeros. These in turn determine both the phase and amplitude response.

You can get round this result by using an all-pass filter, which adds phase shift but no amplitude change - but it does this by having two paths from input to output so the theorem does not apply.

Digital filters can get round it too. For example, a sinc filter (naturally a part of most DACs, and the only filter in most NOS) gives a roll-off without changing phase (apart from a constant time delay).