LR4 Phase shift and driver polarity

when we use LR4 xover then there will be exact 180 degree phase shift happens right on the center of the midrange frequency. Is there anything to consider in flipping the midrange polarity? When LR 4 is 360 degree phase shift crossover and quiet surprised why this 180 deg phase shift happening?
 

Attachments

  • image_2020-10-07_003000.png
    image_2020-10-07_003000.png
    22.9 KB · Views: 171
The output signals of a fourth-order Linkwitz-Riley low-pass and a fourth-order Linkwitz-Riley high-pass are exactly in phase(*) at all frequencies, so swapping the polarity of one driver with respect to the other will result in cancellation at the crossover frequency. I'm not sure if that answers the question, as I don't understand what you are asking.

The sum of the high-pass and low-pass responses are a second-order all-pass response, and a second-order all-pass goes from 0 degrees to -360 degrees phase shift, with -180 degrees at the centre frequency.

(*): Or exactly 360 degrees out of phase, depending on how you look at it, but that boils down to the same thing.
 
Last edited:
360deg out of phase means time alignment is 1 wavelength delayed at XO frequency. For 2kHz that is 17cm, not an insignificant distance to compensate for if one wanted to get back to 0deg.

Hi xrk, i see where you are coming from. If it's truly 360 deg out of phase, then yes one time period.

I think what MarcelvdG was saying is that in the question/graph from the OP, whether you look at it as 0 degrees or 360 degrees makes no difference.

All is in phase, and in time, with nothing to fix. (assuming the LR4 is at about 1.8kHz)
 
Of course, because it is an all-pass filter. The high-pass and low-pass have exactly equally crappy phase responses, so their outputs are always in phase and the sum of their output signals is an all-pass filtered version of the input signal.