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Old 29th September 2012, 12:50 PM   #1
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Default linkwitz riley crossover

I have designed an active crossover using two 2nd order linkwitz riley filters Q= 0.5 cutoff freq. 2.5kHz. Now the signals after the crossover go into two separate amplifiers and finally to the speakers where they combine into the air. So if we sum the two output we get a deep notch at the cut off frequency, but if we invert one of the signals and sum them we get a completely flat response. The question is: if I use an inverting circuit for the time delay of the bass then the bass will be 180 out of phase and the treble will be in phase so then they will sum flat. is it right?
an other way will be make the outputs in phase with the input and then connect one of the speakers with the polarity reversed but i would prefer the first even though one of the output is not in phase with the input.
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Old 29th September 2012, 01:04 PM   #2
AndrewT is online now AndrewT  Scotland
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It would seem that the inverting delay will get your phases back in line.
That would be the simplest and it allows for correct phase matching at the input and correct phase matching at the output.
You don't need to remember to swap phases when you swap gear.

I hope I have got that right, can anyone else confirm?
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Old 29th September 2012, 01:22 PM   #3
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The crossover you describe, (two cascaded second order Butterworth filters giving a Q of .5) is the normal Linkwitz Riley fourth order crossover. This is called an "in phase" crossover, meaning that no phase inversion is required to bring the HP and LP outputs into phase.

What you seem to be describing are second order crossovers where the HP and LP are 180 degrees apart requiring phase inversion of one output to avoid the cancellation null.

Keith

Edited PS, are you building a 12dB/octave crossover or a 24 dB/octave?

Last edited by Keith Taylor; 29th September 2012 at 01:34 PM.
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Old 29th September 2012, 01:22 PM   #4
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With just an inverting circuit on the woofer side should bring both outputs into a state that adds to get a flat response. However if the delay is a RC type circuit shouldn't it affect the phase vs frequency around the crossover point. Would it now sum flat ? I guess a simple sim of this should give the answer.
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Old 29th September 2012, 01:23 PM   #5
puppet is offline puppet  United States
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An LR2 active circuit doesn't guaranty LR2 acoustic output. Have you measured the driver(s) response?
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Old 29th September 2012, 01:26 PM   #6
AndrewT is online now AndrewT  Scotland
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two 2nd order LR would be, one 2nd order LR for the low pass and one 2nd order LR for the high pass. This would give the inverted phases at the crossover frequency.

A 4th order LR crossover would require four 2nd order Butterworth (Q=1/sqrt(2)).
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Old 29th September 2012, 01:40 PM   #7
Dissi is online now Dissi  Switzerland
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I confirm. The two outputs of the linkwitz-riley filter have opposite phase and you will have to invert one to get a flat sum. Which one and where is not really important.

The problem you will encounter is, that the transfer function of the drivers will add to the electronic filter and the acoustical result will be anything, but not the desired linkwitz-riley filter...
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Old 29th September 2012, 01:44 PM   #8
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I have two 2nd order filter separate one for the bass the oder for the treble 12dB/octave.
so is right to do the first way i described?
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Old 29th September 2012, 01:52 PM   #9
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man.. change the wires on the speaker...
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Old 29th September 2012, 02:19 PM   #10
tinitus is offline tinitus  Europe
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are all your amps exactly the same ?

but its like puppet said, varying acoustic output/response plays the major part
and actual speaker design
you may have to trust your ears, or measure
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