I don't understand one thing about 3 way EQ based on Linkwitz-Riley crossover. We split the signal to 3 bands - low, mid and high. Why is mid band phase inverted? It doesn't make sense to me, but people do it for some fundamental reason.

The most confusing thing is that in LTspice simulation inverting the "mids" phase ruins the frequency response. The response is flat when 3 bands have equal, non inverted phase. The last picture shows the response of 3 bands of circuit which doesn't invert the phase of any band.

More info about inverting here: https://sound-au.com/articles/eq.htm#s9

Thanks for help :)

ISO eq


non inverted


  • 1
    \$\begingroup\$ The three filters on the right are of the Sallen-Key type, the left one isn't. Did you simulate that actual schematic or used filter building blocks? \$\endgroup\$
    – Janka
    Commented Dec 5, 2017 at 18:33
  • \$\begingroup\$ Hey, the problem is solved, but thanks ! :) \$\endgroup\$
    – cubix
    Commented Dec 5, 2017 at 18:37
  • \$\begingroup\$ I didn't simulate the circuit which is in description - that's why i got confused. \$\endgroup\$
    – cubix
    Commented Dec 5, 2017 at 18:37

1 Answer 1


Any good discussion of the L-R crossovers should cover this so I'll just outline it quickly.

Plot the phase shift of the 2nd order LF section (low pass filter) and you'll find it is 0 at DC, 90 degrees at the crossover, and 180 degrees at HF.

Now the mid section has a high pass filter at the same crossover frequency; do the same for it and you'll find it is 0 degrees at HF, -90 degrees at crossover and -180 degrees at LF.

If you sum their outputs, this leaves them 180 degrees apart at all frequencies. (The L-R filter characteristics are designed to achieve this). Which doesn't matter where one or the other attenuates the signal (and the other passes it) but at the crossover frequency, both signals are -3dB but 180 degrees apart in phase ... thus they cancel out leaving a deep notch in the frequency response.

By inverting the mid section, to cancel this phase shift, they are in phase at all frequencies, so sum constructively, and maintain a flat frequency response.

The same trick applies between MF and HF sections, where leaving the HF section unchanged provides the phase inversion required between sections.

Although your example is used as some sort of equaliser, the classic use for this filter is frequency splitting in active loudspeaker systems; instead of VR1-3, each section feeds its own power amplifier driving LF (woofer), mid-range, and HF (tweeter) loudspeaker drive units.

Think through the same argument for 4th order L-R crossover and you'll see that two 180 degree inversions would be required to maintain phase coherence - i.e. it is naturally coherent with no phase correction. This makes 4th order Linkwitz-Riley quite a popular choice for active loudspeaker systems.

EDIT : From the updated plots in the question, that last paragraph was prescient; they show a 4th order L-R which requires a 360 degree phase shift, i.e. no phase shift at all. (If you could drop the HF (green) phase plot by 360 degrees, it would very closely match the mid phase plot at HF, just as the LF plot does at lower frequencies.

Your linked article doesn't discuss the phase inversion necessary in the 2nd order version, nor does it then explain that the 4th order filter shouldn't have it.

  • \$\begingroup\$ Thanks for your input. I've updated the question, and the simulations say just the opposite... \$\endgroup\$
    – cubix
    Commented Dec 5, 2017 at 18:03
  • \$\begingroup\$ Something odd there ... can you plot gain/phase of each section independently? (U3/U4 outputs at least) \$\endgroup\$
    – user16324
    Commented Dec 5, 2017 at 18:10
  • \$\begingroup\$ i've done the update - it shows the response of 3 bands of circuit which use only non inverting stages \$\endgroup\$
    – cubix
    Commented Dec 5, 2017 at 18:22
  • \$\begingroup\$ And all is clear ... the ultimate rolloff rates are about 24dB/octave, which suggests you are simulating 4th order L-R sections (despite showing us a 2nd order schematic in the question). Now see the last paragraph of my answer. \$\endgroup\$
    – user16324
    Commented Dec 5, 2017 at 18:29
  • \$\begingroup\$ Dayum, ok. So 24dB/octave crossover doesn't suffer of this issue and all is just right? \$\endgroup\$
    – cubix
    Commented Dec 5, 2017 at 18:31

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