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I have noticed that when you create a crossover using even-order Butterworth filters, you need to use two filters instead of one (Linkwitz-Riley) in order to obtain a flat amplitude response (two low-pass and two high-pass), whereas using odd-order you can only use one (one low-pass and one high-pass). Both even-order and odd-order have an amplitude response of -3 dB at the crossover frequency and you need -6 dB in order to have a perfectly flat sum (after adding up the low-pass and the high-pass). So why odd-order filters don't require two filters to have a flat sum? Thanks.

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I have noticed that when you create a crossover using even-order Butterworth filters, you need to use two filters instead of one (Linkwitz-Riley)

That's not really correct. A LR (Linkwitz-Riley) is defined as the cascade of two Butterworth (BW) filter. A LR filter is simply the cascade of two equal BW filters. That's why the order comes out to be even. Two BW filters and one LR filters are the same thing.

Both even-order and odd-order have an amplitude response of -3 dB in the passband

Incorrect. Both have 0 dB amplitude in the passband

and you need -6 dB in order to have a perfectly flat sum (after adding up the low-pass and the high-pass).

Incorrect again. You just need to add them with 0 dB of gain.

So why odd-order filters don't require two filters to have a flat sum?

Odd order BW high and low pass filters have a phase difference of 90 degrees. Hence the gain of each is -3dB at the cross over frequency and they sum to flat.

Even order or LR filters have either 0 or 180 degree phase difference and so the gain of each is -6dB at the cross over so the sum (suing the proper sign) is 1.

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    $\begingroup$ "Both even-order and odd-order have an amplitude response of -3 dB in the passband" .... "Incorrect. Both have 0 dB amplitude in the passband" ---- Uhm, is that after adding them together, Hil? $\endgroup$ Oct 9, 2022 at 17:34
  • $\begingroup$ 1. That's exactly what I said, so not sure what is "not really correct" there. Linkwitz-Riley = two identical Butterworth. I was talking about the order of the Butterworth there, not the Linkwitz-Riley. If you build a LR filter using even-order BW filters, it will sum to a flat amplitude, but not when you are using odd-order BW filters. That's basically the essence of my question. $\endgroup$
    – Golitan11
    Oct 9, 2022 at 20:24
  • $\begingroup$ 2. Huh no. BW filters have an amplitude response of -3 dB in the passband, and LR filters have an amplitude response of -6 dB in the passband. They only sum up to 0 dB after adding them together (i.e. crossover). $\endgroup$
    – Golitan11
    Oct 9, 2022 at 20:25
  • $\begingroup$ 3. If you know your dB, a +6 dBFS increase means that the amplitude of the signal has doubled. When adding two "crossovered" signals together, you are effectively doubling their amplitude in the passband. So, the passband of the underlying filters must be -6 dB in order to obtain a flat sum. That's the essence of Linkwitz-Riley filters. $\endgroup$
    – Golitan11
    Oct 9, 2022 at 20:29
  • $\begingroup$ I think the confusion comes from the term "passband". By passband, I meant the crossover frequency, which is wrong. Sorry for that. I edited my original question. $\endgroup$
    – Golitan11
    Oct 9, 2022 at 20:33

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