I designed a 4th-order active Linkwitz-Riley crossover filter by cascading two 2nd-order Linkwitz-Riley filters.

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Below are the simulation results. The red and purple waveforms are taken at the output of the first stage of op-amps (i.e. 2nd order), while the cyan and green waveforms are taken at the output of the second stage of op-amps (i.e. 4th order).

enter image description here

For 1st-order filters, the crossover frequency is defined as the -3 dB point. For 2nd-order filters, I believe it is normal to define the crossover point as -6 dB. However, I haven't seen many references online where people define higher-order filter cut-offs as the -9 dB, -12 dB point. It seems they still stick to the -6 dB point. Obviously, cascading these 2nd-order stages drops has caused my designed crossover point of 3.2 kHz to drop from -6 dB to -12 dB.

In general, are higher-order filter cut-off frequencies still defined by their -6 dB point? And also, do I need to "shift" my high-pass and low-pass filters so that they meet at their -6 dB points and not at their -12 dB points?

  • \$\begingroup\$ As James writes, a 2nd order Butterworth is \$-3.01\:\text{dB}\$ at \$\omega_{_\text{c}}\$. But a 2nd order Bessel is \$-4.77\:\text{dB}\$ at \$\omega_{_\text{c}}\$. You can compute all this by just looking at \$20\cdot\log_{10}\left(\sqrt{H\; H^{^{*}}}\right)\$ where \$\omega=\omega_{_\text{c}}\$. \$\endgroup\$
    – jonk
    Sep 14, 2022 at 7:43
  • \$\begingroup\$ Cascading filters will never preserve the type of response. E.g. if you cascade two Butterworth filters (designed so that they are two, distinct Butterworth filters) then the result will no longer be a Butterworth. But it's not necessary that they will become an LR filter. E.g.2nd + 4th =/= 3rd + 3rd. \$\endgroup\$ Sep 14, 2022 at 11:46
  • \$\begingroup\$ Closed. This question is opinion-based. It is not currently accepting answers. Update the question so it can be answered with facts and citations. This will help others answer the question. You can edit the question. Closed just now by Andy aka, Russell McMahon♦.**SEP 14 2022** VTC cleared as understandable technical question which now has good answers. \$\endgroup\$
    – Russell McMahon
    Sep 15, 2022 at 1:26

2 Answers 2


Linkwitz-Riley filters are designed with 4 poles but each 2 pole stage should be designed to be 3dB down at the cross-over frequency so that each 4 pole stage is 6dB down at the cross-over frequency. This then gives an overall flat response across the frequency band because at the cut-off frequency both the low pass filter and the high pass filter will be at half the pass band signal level.

Linkwitz-Riley filters are constructed from Butterworth filters in order to get each 2 pole stage -3 dB down at the cut-off frequency.

At the moment each of your 2 pole stages is -6 dB down at the cut-off frequency giving -12 dB down for 4 poles. This will cause a dip in your frequency response at the cross-over frequency which is undesirable.

So, you need to convert your filters to Butterworth in order to get each 2 pole stage down -3 dB at the cross-over frequency.

To convert your low-pass filters to Butterworth change C5, C6, C7 & C8 to 2 nF. You can do this by adding another 1 nF capacitor in parallel to the existing 1 nF capacitors.

To convert your high pass filters to Butterworth change R2, R4, R15 & R16 to 100 k ohms (double them).

Each 4 pole stage should now be down 6 dB at the cross-over frequency giving an overall flat frequency response.

The cut-off frequency of both the high and the low pass filters is given by the following equation:-

$$f_c = \frac{1}{\sqrt{2} \times 2\pi RC}$$

Where R = 50 k and C = 1 nF

So the distinguishing feature of Linkwitz-Riley filters is that they are designed with two separate cascaded 2 pole filters, each 2 pole stage having a -3 dB drop at the cut-off frequency giving -6 dB at the cut-off frequency from the 4 poles. If you were to design a "normal" 4 pole Butterworth filter it would be designed to have an overall -3 dB drop at the cut-off frequency instead of -6 dB. Using these "normal" -3dB 4 pole Butterworth filters in an active cross-over network would give a "hump" in the overall frequency response, at the cross-over frequency, of 1.414 times the passband signal level which wouldn't be good.

Obviously, looking at my equation, making those changes to the component values has shifted the cross-over frequency. You can play around with the component values to get it back where it was but make sure that, to get the Butterworth response, C7 is twice the value of C9 for the low pass filters and R2 is twice the value of R1 for the high pass filters (this is so for all the stages).

The 4 pole low pass filter will have -180 degrees of lag at its -6 dB cut-off frequency and the 4 pole high pass filter will have 180 degrees of lead at its -6 dB cut-off frequency. This means that, at the cross-over frequency, the two outputs will be in phase which is a desirable feature.

The active cross-over network has performance advantages over the passive cross-over network. One disadvantage of the active type is that a power amplifier is required for each cross-over network output. So, if you have an active cross-over network with stereo treble, mid-range and base outputs then you will need 6 power amplifiers (compared to just two for the passive cross-over network). Although, you can set each amplifier's power output level to the actual required level for each driver and so there will be less power required overall compared to the passive cross-over network which can waste power in the high frequency section of the cross-over network in the effort to get the base driver's power to a high enough level.

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    \$\begingroup\$ Just one remark: LR filters are not restricted to a multiple of 4th order -- their characteristic is the squared Butterworth response that gives them the -6 dB crossover which allows for summing a lowpass and a highpass (and even a bandass in-between) with a perfectly flat response. A 2nd order LR would be a critically damped 2nd order (2 x 1st order). A 6th order would be 2 x 3rd order Butterworth, etc. So, from this perspective, they are made from two Butterworth halves (except 1st order), but their characteristic is the flatness when summed. \$\endgroup\$ Sep 14, 2022 at 7:31
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    \$\begingroup\$ Thank you, this was extremely helpful! \$\endgroup\$
    – zigzidane
    Sep 14, 2022 at 17:51

Cascading filters will never preserve the type of response, if the filters are designed so that they are distinct. E.g. if you cascade two Butterworth filters then the result will no longer be a Butterworth. But it's not necessary that they will become an Linkwitz-Riley filter (also see the references at the bottom). For example a 2nd and a 4th, cascaded, will not be the same as two 3rd orders (I'll get to this later on).

For 1st-order filters, the crossover frequency is defined as the -3 dB point. For 2nd-order filters, I believe it is normal to define the crossover point as -6 dB.

There's a slight misunderstanding: the order of the filter has nothing to do with the chosen corner frequency (fc) -- that is part of either the design, or the transfer function, itself. What I mean is that, usually, the -3 dB point is considered for 1st order filters (they have no name, no Butterworth, or Chebyshev, or...; they are universal) and for Butterworth filters, because it's only for them that the -3 dB fc coincides with the phase being half the end value (asymptotically). That's not to say that it is a universal point, it can be whatever the designer chooses to be: you can make a Butterworth to have a -1 dB fc, you can make a 0.1 dB ripple Chebyshev to have a -3 dB fc, etc; that's why frequency scaling exists. It's only the fact that this 3 dB is used in the vast majority of the cases that made it so prevalent and, sometimes, misunderstood.

However, I haven't seen many references online where people define higher-order filter cut-offs as the -9 dB, -12 dB point.

If you do it's either part of special requirements (custom attenuation at fc), or SF literature.

It seems they still stick to the -6 dB point.

If you mean for Linkwitz-Riley filters then yes, that is their defined point, and it is so because they were born with the idea that audio channels (bass and treble) can be split and, their summed response will be flat. And, to reach this, the simplicity turned out to be cascading two Butterworth filters, which means LR filters always have an even order:

  • A 2nd order LR is two, identical, cascaded 1st order filters which, when summed up, they result in the critically damped 2nd order transfer function. Each 1st order has a normal -3 dB fc which means that the output of two identical stages, cascaded, will be -6 dB -- or half value for the signal (\$10^{-6.0206/20}=0.5\$, or \$20\log_{10}(0.5)=6.0206...\$).
  • The rest, N=4th/6th/8th/etc, are two, identical, cascaded N/2 order Butterworth filters. If you do the math you will get the same -6 dB @ fc: $$\begin{align} B_2(s)&=\dfrac{1}{s^2+\sqrt2 s+1} \\ LR_4(s)&=B_2(s)^2 \\ &=\dfrac{1}{s^4+2\sqrt2 s^3+4s^2+2\sqrt2 s+1} \tag{1} \\ |LR_4(j\cdot 1)|&=\dfrac12 \\ \\ B_3(s)&=\dfrac{1}{s^3+2s^2+2s+1} \\ LR_6(s)&=B_4(s)^2 \\ &=\dfrac{1}{s^6+4s^5+8s^4+10s^3+8s^2+4s+1} \tag{2} \\ |LR_6(j\cdot 1)|&=\dfrac12 \\ \\ ... \end{align}$$

Note that it's not enough to have two Butterworth filters, they have to be identical, which is half the LR order. For example, a 6th order LR can't be made with a 2nd and a 4th:

$$\begin{align} H_4(s)&=\dfrac{1}{\left(s^2+2\sin\left(\dfrac18\pi\right)s+1\right)\left(s^2+2\sin\left(\dfrac38\pi\right)s+1\right)} \\ &=\dfrac{1}{(s^2+\sqrt{2-\sqrt2}s+1)(s^2+\sqrt{2+\sqrt2}s+1)} \\ &=\dfrac{1}{s^4+\sqrt{4+2\sqrt2}s^3+(4+\sqrt2)s^2+\sqrt{4+2\sqrt2}s+1} \\ &=\dfrac{1}{s^4+2.6131s^3+3.4142s^2+2.6131s+1} \tag{3} \\ H_2(s)H_4(s)&=\dfrac{1}{s^6+4.0273s^5+8.1097s^4+10.055s^3+8.1097s^2+4.0273s+1} \tag{4} \end{align}$$

If you compare (2) with (4), they differ.

In general, are higher-order filter cut-off frequencies still defined by their -6 dB point? And also, do I need to "shift" my high-pass and low-pass filters so that they meet at their -6 dB points and not at their -12 dB points?

Yes, as explained above, LR filters are defined to have -6 dB @ fc, so you don't have to "shift" anything. If you feel you have to then you need to recalculate something. And, to exemplify what I meant earlier, here's how two, 4th order lowpass and highpass LR are separately, vs summed up:

LR 4th LP + HP

Some final notes:

  • LR filters are, usually, considered to be 2nd, 4th, or 8th order filters, but it is possible to have a 6th order, as long as the flatness of the summed lowpass and highpass is preserved.
  • It may be possible to squeeze in a three-way cross-over (bass, mid, high), provided you can tolerate the slightly less-than-flat response:

3-way "LR" cross-over

The peak is ~60 mdB (milli dB), or ~1.007, and the bandpass is made of two 4th order Butterworth (the t.f/ is 4th order, not the lowpass prototype). You could tweak it to have symmetrical errors relative to 1, which will give you ~±20 mV error (~17 mdB). If you manage to make a practical filter that has this error stand out above the flatness of your response, I will tip my hat to you and call you "Sir!". In short: it's not mathematical, but maybe practical.


This came as a surprise for me so I thought I'd mention it.

Normally, when converting a lowpass to a bandpass, \$s\$ is replaced with \$(s^2+f_c^2)/(BW\cdot s)\$, resulting in twice the original order and a filter that accurately matches the specifications. That's because using a lowpass and a highpass, without conversion, will not match the requirements due to the cascading effect (the input will influence the output). The resulting bandpass filter may also be a combination of a lowpass and a highpass but, with they will have modified frequencies, to account for the cascading effect.

That is what I did in my example about the LR bandpass. The problem is that -- as mentioned both above and in the 1st comment below -- in order to get a fairly good "flatness" you need the corner frequencies to be at least one decade apart.

However, if, instead of a lowpass→bandpass conversion you use the lowpass and its highpass equivalent, similar to a bandpass but without conversion, it turns out you get much better results!

To exemplify, here's a 3-way LR with 2nd orders (top) and one with 4th orders (bottom), with this proposed method, calculated for a fairly typical 300 Hz ... 3 kHz band edges:

2nd & 4th LR with "brute" bandpass

The center frequency is \$\sqrt{300\cdot 3000}\$ and the ratio is \$\sqrt{3000/300}\approx 3.16\$, which is about one and a half octaves. And look how they compare to the version with the "by-the-book" bandpass:


There is no mistake: the top plot shows an overall gain of 2nd orders LP+HP+(LP+HP) (V(x2)) vs LP+HP+BP (V(w2)), while the bottom plot shows the same but for 4th orders (y1 = BP). The "brute-force" LP+HP tops a BP! And, to take things a bit further, here's how the same configurations respond when the band edges are chosen to be one octave apart, 500 Hz ... 2 kHz:

one octave apart

True, the droops are now in the dB range, they are not acceptable, but even now the 3-way with the BP has worse response than the one with the LP+HP! This was a complete surprise for me.

To conclude, I'll add one more thing: my assertion in the comment below (that you would need a 1-LowPass(s)-HighPass(s) as an equivalent) is not exactly true -- instead of 1 you need an allpass with the same frequency as the bandpass. That's because 1 has zero phase and the difference will cause very unwanted peakings. It's a minor detail but which matters.

Anyway, considering a normalized version (\$\omega=1\$), the allpass is made from a 2nd order Butterworth, squared to have a LR phase response:

$$AP(s)=\left(\dfrac{s^2-\sqrt2 s+1}{s^2+\sqrt2 s+1}\right)^2 \tag{5}$$

The lowpass is \$B_2(s/f)\$ and the highpass is \$B_2(1/f/s)\$, where \$f\$ is the band edge relative to the center frequency. The resulting transfer function is this:

monster bandpass

Now, if the 3-way is made with active filters then there's no problem: split the monster into 2nd order sections and use whichever one/two/n opamp biquad topology. But for a passive 3-way this transfer function is rather prohibitive, not as much due to the number of elements but, because the numerator has damped zeroes (they have real part) -- which means "pure" LC is no longer available and you need to get crafty with lossy resistors. Simply put: even if it is possible (I don't know and, frankly, I don't care to find out, at least not today), it's not worth it, even if the response will be mathematically flat.

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    \$\begingroup\$ Forgot to say that the trick with the bandpass is only valid for suficiently low center_frequency/bandwidth ratio. But the problem can still be solved by solving for 1-LowPass(s)-HighPass(s) (which has the potential to result in a rather un-appealing, but possible transfer function). \$\endgroup\$ Sep 14, 2022 at 13:33
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    \$\begingroup\$ Thank you! This was very helpful, and your explanation with the transfer functions was very useful. \$\endgroup\$
    – zigzidane
    Sep 14, 2022 at 17:52
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    \$\begingroup\$ @zigzidane It may not matter now but, I've updated my answer to include some new information regarding the "nonconformist" 3-way. \$\endgroup\$ Sep 15, 2022 at 21:10
  • \$\begingroup\$ Thanks for the additional info! \$\endgroup\$
    – zigzidane
    Sep 16, 2022 at 16:48

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