Asymmetric Crossover Filters
The top priority when designing crossover filters is generally to achieve a flat amplitude response when the two or more outputs are summed acoustically. In the real world our drive units have their own response variations, and there are other effects including cabinet diffraction and also path length differences from the drive units to the listening position, all of which need to be taken into account. Here the drive units will be assumed to be perfect, time aligned and unaffected by cabinets and surroundings, but then the crossovers derived must be regarded as just a starting point for a real world design, and considerable adjustment may be needed in practice.
Apart from flat amplitude response we may decide we would like flat phase response, or at least linear phase response so that phase shift is proportional to frequency, equivalent to a constant time delay. There is general agreement that with music signals under normal listening conditions including room reverberation typical levels of 'phase distortion' are almost entirely inaudible, only under other conditions, e.g. in an anechoic chamber, a difference may be reliably heard. This can therefore be given a relatively low priority
The simple first-order crossover, giving just a 6dB/octave slope, gives a summed output with both flat amplitude and zero phase shift, so in that respect it is ideal. The responses for the high-pass (HP) and low-pass (LP) are shown in Fig.1 below, and this shows that even 3 octaves away from the crossover frequency the attenuation is only 18dB, and so we would need to use drive units with well maintained response far beyond the chosen crossover frequency. This is possible, but is far from ideal, it is easier to achieve a good performance over a more limited frequency range.
There are other crossover filters with flat amplitude and phase, and an easy way to achieve this is shown in Fig.2 where a higher slope low-pass filter has its output subtracted from the original input signal to give a high-pass result. This guarantees that LP + HP is equal to the original input, so again we have the ideal flat responses. Unfortunately when we do this we get the result shown in Fig.3 where we see the high-pass response rises to a peak at the crossover frequency and then rolls off at only 6dB/octave again. The attenuation at 3 octaves below crossover is now only 14dB down, worse than even the basic first-order filter. For this diagram I used a second-order low-pass, but it makes little difference using higher-orders, the resulting high-pass is always poor. This is an example of an 'asymmetric crossover filter' where high-pass and low-pass are of different order, in this case a 2nd order low-pass and 1st order high-pass.
It isn't entirely clear who first thought of a solution to this, the most widely quoted paper was presented at the AES Convention in May 1981, 'A family of Linear-Phase Crossover Networks Of High Slope Derived By Time Delay' by Stanley Lipshitz and John Vanderkooy. Their own references include Tanaka and Iwahara, 1979, and R.M.Golden, 1973, and there was also a paper by Ng and Rothenberg published in 1982. I did some analysis of the problem myself, but I can only say for certain that was some time prior to 1983, so it's quite possible I picked up the idea somewhere without remembering the source, but anyway most of the diagrams on this page date back to that period.
The solution is shown in Fig.4 where we again subtract the low-pass output, but this time from a time delayed version of the input. Fig.5 shows an example of what is then possible, in this case we have the same low-pass as in Fig.3 but now the resulting high-pass has far better attenuation. The HP response in this example is actually 3rd order, though this is not obvious from the graph, but the eventual rate of roll off at very low frequencies is 18dB/octave. There is a small rise in the high-pass response reaching 0.5dB two octaves above the crossover frequency, but the summed HP + LP is flat.
Time delays are rather inconvenient, almost impossible for passive filters and not easy for active filters, although they have become easier to implement with digital processing. There is however an alternative possibility if we can relax the requirement for linear phase. Instead of a time delay we can use a filter known as an 'all-pass' which has flat gain and a phase response deviating from linearity at high frequencies. An example is shown in Fig.6. This has a second-order Butterworth low-pass filter, the same as the two previous examples, and its output is subtracted from a first-order all-pass. There are alternative all-pass circuits, and it is possible to save one op-amp with some of these. The resulting high-pass response is again 3rd-order. The crossover frequency is around 3kHz for this example.
The phase as a function of frequency is shown in Fig.7 for a first and second-order all-pass and for a constant time delay. It can be seen that at very low frequencies these all converge, so an all-pass can be regarded as a good approximation to a time delay over a limited low frequency range. A general first-order all-pass is (a-s)/(a+s) and the reason why a = square root of 2 in the diagram can be found in the explanation of the calculation method in a footnote at the end of this page.
To bring things up to date a little, here in Fig.8 is a Spice simulation showing the responses for the circuit in Fig.6 with a 3kHz crossover frequency. (To get an accurate 3kHz crossover frequency I had to adjust the capacitor values a little, they became 46.3nF and 92.6nF).The HP and LP are in green and red. The attenuation at the crossover frequency is only 5dB, a perfect time delay instead of the all-pass would achieve 6dB, but on the plus side there is no longer the 0.5dB rise found using a time delay. The sum of the outputs is again a flat line shown in light blue. For comparison a second-order Linkwitz-Riley (L-R) response is shown in blue and purple, this is I believe the only second-order crossover with flat response, and the HP and LP are in phase at all frequencies, giving 6dB attenuation at the crossover frequency. The asymmetric filter gives slightly better attenuation than the L-R with 6dB more attenuation at 1kHz, but otherwise there is little benefit.
The asymmetric second plus third-order filters are not confined to active or subtractive filters, the same result can be achieved using a passive filter, and an example is shown next, the speakers being rather unrealistically represented by 8R resistors. The frequency responses are almost exactly identical to those shown in red and green above in the Spice simulation of the active version, and again the sum is flat, but note that the bass and treble drivers must be connected in opposite polarity.
There are low-pass responses for which the corresponding time delayed high-pass is one order higher. I worked that out myself for 2nd, 3rd and 4th order low-pass. The calculation is easy enough, just start with a first order all-pass (a-s)/(a+s) and subtract a low-pass e.g. 1/(1+bs+s2) and the result can then be made into a third order high-pass by choosing a and b to make all the terms in the numerator zero except the -s3, which makes both a and b equal the square-root of 2. Generally starting with a nth order low-pass we get n equations for n unknowns, which at least for n= 2, 3 or 4 is solvable. If we get a higher order high-pass using the all-pass function then that will also be true for a time delay chosen such that the all-pass and time delay become identical as frequency is reduced towards zero, which is also where the ultimate slope of the high-pass is reached. Using this method of analysis only the final slope at low frequencies is predicted, actually plotting the complete responses may then reveal problems.
It is said in the Lipshitz and Vanderkooy paper that the 5th or higher order low-pass results are 'unstable'. The 4th order low-pass and resulting 5th order high-pass will work, but the responses are plotted in their fig 7b, and clearly have significant peaks in both high and low-pass, so they are not of any practical use. The mathematical method used in that paper is a little different to my own method starting with an all-pass response, but the end results are identical.