Steen Duelund Filter:
2vejs 12dB/okt
http://kbhhifi.dk/readarticle.php?article_id=140
Simulationer på Steen Duelunds 3 vejs Synkron filterhttp://tkhifi.homepage.dk/due3vejs/duesynkronsimu3vejs.html

The Duelund 3way crossover filter function D3
http://www.linkwitzlab.com/crossovers.htm
Recently I received an intriguing paper of the late Steen Duelund from one of his admiring followers. Steen, a "Danish Speaker Maniac", gives a fun derivation of known crossover filter functions for the mathematically inclined and he found some novel ones. To follow his text more easily I rewrite some of it here and develop it further to show a potentially useful 3way allpass crossover filter.
A) Transientperfect crossovers
1)  First order crossover with 6 dB/oct slopes for highpass and lowpass filters = Butterworth, B1+
H_{1}(s) = (s + 1) / (s + 1) = 1 where s = s +jw and the pole at s_{p} = 1 and zero at s_{z} = 1 cancel each other
H_{1} = [s / (s + 1)] + [1 / (s + 1)] = HP + LP ==> 2way
2)  Second order crossover by squaring H_{1}(s) ==> Baekgaard
(H_{1})^{2} = (s^{2} + 2s + 1) / (s + 1)^{2}
(H_{1})^{2 }= [s^{2} / (s + 1)^{2}] + [2s / (s + 1)^{2}] + [1 / (s +1)^{2}] = HP + BP + LP ==> 3way [1]
Highpass and lowpass filters have 12 dB/oct slopes, but the filler bandpass filter has only 6 dB/oct slopes and is difficult to realize. At the crossover frequencies w_{1} = 0.5 and w_{2} = 2 the three outputs are not in phase and the axis of the vertical polar pattern will tilt when the three separate drivers are aligned vertically. This can be avoided by using two filler drivers in a WFTFW arrangement instead of WFT.
More practical solutions of transientperfect crossovers are based on delay derived crossovers where HP(s) = e^{sT}  LP(s) [2] or they are constructed from lowpass and highpass filters which overlap in the crossover region in such a way that a minimum phase response is preserved [3]. Their summed response is not flat, but since it is minimum phase it can be equalized. Steeper slopes can be obtained from these two types of crossovers, but the lack of inphase addition in the crossover region tends to require symmetrical driver arrangements.
For further study I recommend:
[1] E. Baekgaard, " A novel approach to linearphase loudspeakers using passive crossover networks"
JAES, Vol. 25, pp. 284294, May 1977
[2] S. Lipshitz & J. Vanderkooy, "A family of linearphase crossover networks of high slope derived by time delay", JAES, Vol. 31, pp. 219, January/February 1983
[3] S. Lipshitz & J. Vanderkooy, "Use of frequency overlap and equalization to produce highslope linearphase loudspeaker crossover networks", JAES, Vol. 33, pp. 114126, March 1985
B) Allpass crossovers
3)  First order crossover with 6 dB/oct slopes for highpass and lowpass filters = Butterworth, B1
H_{2}(s) = (s  1) / (s + 1) = 1e^{j}^{f } when s = jw, i.e. the magnitude is one, but there is phase shift F(w)
The pole at s_{p} = 1 and zero at s_{z} = +1 do not cancel but form a 1st order allpass.
4)  2nd order LinkwitzRiley filter (12 dB/oct, LR2)
H_{2} H_{1} = (s  1) (s + 1) / (s +1) (s + 1) = (1  s^{2}) / (s + 1)^{2}
H_{2} H_{1} = [1 / (s + 1)^{2}]  [s^{2} / (s + 1)^{2}] = LP  HP ==> 2way and 1st order allpass as above.
HP and LP are 2 x 90 degrees out of phase at all frequencies and the polarity of one driver must be reversed to obtain allpass behavior.
5)  Squaring H2(s) and going to a more general expression by replacing '2' with 'a'
(H_{2})^{2} = (s^{2}  2s + 1) / (s^{2} + 2s +1)
H_{3} = (s^{2}  as + 1) / (s^{2} + as +1)
6)  4th order LinkwitzRiley filter (24 dB/oct, LR4) by squaring H_{3}
(H_{3})^{2} = (s^{2}  as +1)^{2} / (s^{2} + as +1)^{2} = [s^{4}  s^{2}(a^{2}  2) + 1] / (s^{2} + as + 1)^{2}
Let a = sqrt(2) = 1.41, then
H_{4} = (s^{4} + 1) / (s^{2} + 1.41s + 1)^{2} ==> 2way
which is a 2nd order allpass with a complex pole pair at 0.71 +/ j0.71
and a pair of complex zeros at +0.71 +/ j0.71 in the right half splane.
The individual filter functions are:
HP_{4} = s^{4} / (s^{2} + 1.41s + 1)^{2 } and LP_{4} = 1 / (s^{2} + 1.41s + 1)^{2 }
HP and LP are 4 x 90 degrees out of phase at all frequencies, i.e. they are inphase.
7)  Let a = 4 in (H_{3})^{2} to obtain a 3way allpass crossover
H_{5} = (s^{2}  4s +1)^{2} / (s^{2} + 4s +1)^{2} ==> 3way
This is an allpass with two real axis poles at 3.73 and two at 0.27. It has real axis zeros at +/3.73 and +/0.27 which combine to two 1st order allpasses.
The numerator of the polynomial is [s^{4} s^{2}(a^{2}  2) + 1] = s^{4}  14s^{2} + 1
The individual filter functions are:
HP_{5} = s^{4} / (s^{2} + 4s +1)^{2 }It is 4th order, but has in practice only 12 dB/oct filter slopes because the poles are widely spread apart.^{}BP_{5} = 14s^{2} / (s^{2} + 4s +1)^{2}
The bandpass filter slopes are 12 dB/oct.
LP_{5} = 1 / (s^{2} + 4s +1)^{2}Practical filter slopes are only 12 dB/oct
Crossover is at 6 dB points and the drivers are in phase so there is no tilt in the polar response. The filter slopes should be steeper though to make the crossover more realizable.
8)  Squaring (H_{3})^{2} once more leads to (H_{3})^{4} and selecting a = 3 gives the allpass
H_{6} = (s^{2}  3s +1)^{4} / (s^{2} + 3s +1)^{4}
There are 8 poles as calculated from (s^{2} + 3s +1)^{4} = 0, four at s_{p1} = 2.62 and four at s_{p2} = 0.38
Expanding the numerator yields (s^{8}  14s^{6} + 51s^{4}  14s^{2} + 1) and thus the numerator for the bandpass function becomes
N_{BP} = 14s^{2} [s^{4}  (51/14)s^{2} +1] with two zeros each at s_{z1} = 1.83, s_{z2} = 0.55 and s_{z3} = 0
The individual filter functions are:
LP_{6} = 1 / [(s + 2.62)^{4} (s +0.38)^{4}]
The lowpass is 8th order, but has in practice only 24 dB/oct filter slopes because the poles are widely spread apart.
HP_{6} = s^{8} / [(s + 2.62)^{4} (s +0.38)^{4}]^{}Highpass^{}filter slopes are 24 dB/oct as above.
BP_{6} = 14s^{2} (s + 1.83)^{2} (s + 0.55)^{2} / [(s + 2.62)^{4} (s +0.38)^{4}]
The bandpass filter slopes are 12 dB/oct. The filter is centered at w_{c} = 1 and the crossover frequencies are w_{1} = 0.24 and w_{2} = 4.1 at 6 dB as read from the graph. Thus a 3way loudspeaker with 120 Hz and 2.05 kHz crossover frequencies could be built that uses these filter functions. The gentle slopes of the bandpass filter function impose stringent requirements upon the midrange driver's volume displacement capability and its frequency range in order to obtain the targeted acoustic response.
9)  It is interesting to compare this Duelund 3way filter function to a crossover that uses LR4 filters.
The crossover frequencies must be at w_{1} = 0.24 and w_{2} = 4.1 and thus the LR4 poles are located at
0.24*(0.707 +/ j0.707) and at 4.1*(0.707 +/ j0.707)
The filter functions become
LP_{LR4} = 0.24^{4} / (s^{2} + 0.34s + 0.24^{2})^{2}
BP_{LR4} = 4.1^{4} s^{4} / [(s^{2} + 0.34s + 0.24^{2})^{2 }(s^{2} + 5.8s + 4.1^{2})^{2}]
HP_{LR4} = s^{4} / (s^{2} + 5.8s + 4.1^{2})^{2}
The sum of the three filters is approximately allpass since the two crossover frequencies are far (1:17) apart.
The Duelund filters roll off more gradually through the crossover region though highpass and lowpass reach 24 dB/oct slope. The bandpass covers a wider frequency range due to its more gradual 12 dB/oct rolloff. The group delay of the 3way also reflects this relative behavior. In particular the Duelund does not show any peaking because the allpass filter sections are 1st order squared. Based on my experimentation with 1st and 2nd order allpass crossovers this visually smoother response has no audible benefits since the amount of peaking is very small.
The frequency response curves and the group delay were determined in a spreadsheet using the methodology shown in 12dbhpf.gif.
10) The Duelund 3way is based on an 8th order polynomial yet the maximum slopes are only 4th order over the first 40 dB of attenuation. It seems likely that the same response can be approximated with 4th order functions. The good match, which was arrived at by an educated guess, can be seen in the graph below.
In the splane representation one can see the dominant poles taking over and cancellations of several adjacent polezero pairs.
Thus a practical Duelund 3way crossover would have parameters like this:
a) Crossover frequencies at 0.24 and 4.1 (e.g. at 120 Hz and 2050 Hz) Crossover points are at 6 dB. Filter outputs are in phase when the midrange bandpass has reversed polarity. Bandpass is centered at normalized frequency 1.0 (e.g. at 500 Hz). b) Woofer lowpass LP_{D3} = 0.38^{4} / ( s + 0.38)^{4} c) Midrange bandpass BP_{D3} = 15s^{2} / [(s + 0.28)^{2} (s + 3.6)^{2}] d) Tweeter highpass HP_{D3} = s^{4} / (s + 2.62)^{4} 
Note that all mathematical functions are meant to describe the acoustic response of a 3way loudspeaker. To the extend that woofer, midrange or tweeter drivers have a nonflat frequency response they must be either equalized to be flat, or their response must be made part of the filter function. In particular, the midrange driver's natural low frequency rolloff could be equalized with a "Linkwitz Transform" to obtain the desired LR2 highpass behavior of c) above. The tweeter can be treated similarly for two of the poles of d). The woofer's natural highpass behavior causes a phase lead which is probably far from zero at the first crossover frequency and thus affects the proper addition of woofer and midrange outputs. This can be corrected by placing a 1st order allpass in the midrange channel which simulates the highpass phase shift of the woofer.
The Duelund 3way crossover could provide a useful option for building a loudspeaker.
See also:
Rene Christensen, "Active AllPass Crossover Networks with Equal Resistors and Equal Capacitors", Journal of the Audio Engineering Society, Vol. 54, No. 1/2, 2006 January/February, pp. 4553.
LINKWITZ LAB