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Old 9th July 2007, 12:37 PM   #1
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Default Transient-"perfect" 2nd/1st order crossover

What I am going to present here is an old incarnation of my transient-„perfect“ crossover topology. I will try a more refined one in the future so I decided to let this one out to the public.

What I basically did is the electroacoustical approximation of the responses of the two branches of an asymmetrical subtractive crossover with a 2nd order highpass and a 1st order lowpass. Higher orders are theoretically possible but they don’t have advantages only.

The basic linear transfer function of a transient perfect crossover of the order N has the same numerator and denominator. That is actually the source of the infamous humps of subtractive crossovers. They don’t have the same cause as the humps of a transfer function with high Q-values – as it is often misunderstood.

But these humps are responsible for increased power needs, flat slopes around the crossover frequency (for very high orders they tend to get shallower than 6dB/octave !) and lobing.

So the lower the order, the less one runs into these problems. It is theoretically possible to cross any driver combination with such crossovers of arbitrary order - but doing it with a midwoofer/tweeter combination is not for the faint - hearted.

One of the most suitable topologies is therefore the fullrange (or wideband) + woofer topology because of 3 reasons:

1.) The wavelength at the crossover frequency is quite large relative to the physical dimensions involved (-> lobing !).
2.) The “tweeters” involved are more forgiving in terms of shallow crossover slopes.
3.) People using wideband/fullrange drivers often care more about transient accuracy than the “multiway fraction”.

So I thought it was a natural to post it here in the fullrange forum.

It is possible to do more accurate approximations than the one presented here but they would tend to be more complicated and need more accurate driver modelling. But everyone is free to do so. Since the circuit is quite simple it can easily be tried out on a breadboard if one has a second amp to use.

What we do to begin with is taking the response of the aforementioned subtractive crossover. As highpass function we take a 2nd order filter with a Q of 0.5. The actual crossover frequency wouldn’t be the pole frequency (where there is already a 6 dB drop compared to 3 dB for the Butterworth case) but about one octave higher. Both drivers are down by about two dB at the crossover frequency. For higher-order filters the relative level at the crossover point might be higher than 1 - just to mention one of the disadvantages of higher order subtractive crossovers.

So the original transfer function we are going to use will be:

H(s) = (1 + sT/0.5 + s^2*T^2) / (1 + sT/0.5 + s^2*T^2 )

This is then split into two parts:

H(s) = (1 + sT/0.5 ) / (1 + sT/0.5 + s^2*T^2 ) + s^2*T^2 / (1 + sT/0.5 + s^2*T^2 )

Giving the highpass function

Hh(s) = s^2*T^2 / (1 + sT/0.5 + s^2*T^2 )

And the lowpass function

Hl(s) = (1 + 2*sT) / (1 + sT/0.5 + s^2*T^2 )

As one can see the lowpass section is actually the sum of a 2nd order bandpass and a 2nd order lowpass. That’s where the 1st order slope and the hump of the lowpass come from. Another method to generate this transfer function is to multiply a 2nd order lowpass with the sum of unitiy gain and the output of a differentiator. It is the latter that we are going to do.

Enclosed you see the simulated responses of the topology we are going to approximate. It is shown for a highpass pole-frequency of 100 Hz.

More to come !!
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File Type: gif subtr_2_1_approx.gif (8.7 KB, 1521 views)
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Old 9th July 2007, 12:44 PM   #2
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The drivers one is going to use should be reasonably well behaved within their passband. Irregularities and baffle-step should be EQed out. This is shown as EQw and EQf. The woofer should at least be usable up to about 5 to 10 times the crossover frequency.

The 2nd order highpass function for the fullrange is achieved by simply using an LTF in order to push the pole-frequency upwards to the pole-frequency fx and achieve a Qtc of 0.5.
There are two problems that remain: 1) There is not much protection at the low end (though still a big improvement over real fullrange usage) and 2.) the phase-shift of the woofer can play havoc with proper SPL summation.
A simple and effective means for 2.) is the use of a highpass filter Hw in the fullrange branch that mimics the low-end response of the woofer and that also intrinsically improves 1.). For a closed box this would be a 2nd order highpass and for a reflex box a 4th order highpass respectively.
Though I am not sure if one wants to use a reflex box for a “transient-perfect” loudspeaker !

A little more complicated is the situation for the woofer.
As already mentioned we can achieve the required lowpass transfer function with a 2nd order lowpass, a summing stage and a differentiator.
But when doing this we have to take the woofer’s upper frequency limit Fu into consideration. We do therefore truncate the lowpass function at the upper frequency limit of the woofer by the use of the two resistors Rsl. This is of course only an approximation but it is better than doing nothing at all. The best thing would be to use a dual gang pot for these when experimenting. It can then be replaced by fixed resistors later on. Also R can be substituted by a dual gang pot during fine-tuning of the circuit.
The differentiator function does have to be truncated as well - otherwise we run the risk that this crossover branch has very high gain at high frequencies. This is done at a frequency between 10 and 30 times the crossover frequency by the use of Rsd (and therefore chose Rsd 10 to 30 times smaller than R).
The other reason not to try to achieve 1st order behaviour of the woofer/xover combination up to infinity: The fullrange driver’s response starts to fall off at a finite frequency and we would not want the woofer to try to compete with the HF response of the fullrange !!!

The differentiator’s feedback path can be used for gain setting with R1 and/or baffle step correction if needed.

The circuit presented may not give the flattest possible response (and the formulae are only coarse approximations) but it is one of the cheapest entry tickets to the world of transient-perfect loudspeakers.

Enclosed you’ll see the block diagram:
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File Type: gif tpc_trunc.gif (8.0 KB, 1500 views)
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Old 9th July 2007, 01:06 PM   #3
Hartono is offline Hartono  Indonesia
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Hi Phase_accurate,

I remember seeing article on AudioXpress on transient perfect second order x-over, is that article made by you ?

Hartono
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Old 9th July 2007, 01:46 PM   #4
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Hi Hartono

No it isn't - it is most likely by John Kreskovsky. I don't have this article but I assume it is dealing with this:

http://www.geocities.com/kreskovs/CrossoverdocN.html

regards

Charles
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Old 9th July 2007, 03:20 PM   #5
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Forgot to mention - here ist the measured step response of my Manger/Audiotechnology combination using this x-over toplology:

http://www.diyaudio.com/forums/showt...11#post1139711

And here is the QUAD ESL 63 for comparison:

http://stereophile.com/floorloudspea...6/index11.html

Regards

Charles
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Old 9th July 2007, 09:20 PM   #6
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Quote:
Originally posted by Hartono
Hi Phase_accurate,

I remember seeing article on AudioXpress on transient perfect second order x-over, is that article made by you ?
It may have been my "phase coherent crossovers", which you
can dowload from www.passlabs.com/np under the name "pcxvr"
which are tiff images of the Audio Amateur pages.

Funny you should mention this as I've been working with them
lately with Lowther drivers and woofers with good results



ps you will note from the 1982 vintage, there were no AP
systems or computer graphics - all those curves were generated
and drawn by hand.
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Old 9th July 2007, 09:31 PM   #7
Hartono is offline Hartono  Indonesia
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Hi Mr. Pass,

Your article seems to predate the AudioXpress article! the bottom of the page indicate SpeakerBuilder. Thanks for the information.

"all those curves were generated
and drawn by hand"

you must be very good at mathematics !!!

ps: on the last page there's classified advertising, 1 word cost 25 cents.......interesting, wonder how much it cost today on AX.

Hartono
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Old 9th July 2007, 09:51 PM   #8
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Quote:
Originally posted by Hartono
you must be very good at mathematics !!!
Unfortunately, no.

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Old 9th July 2007, 09:54 PM   #9
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I forgot to mention, there was also an article in the January 07
AudioXpress by Stokes on the "Yamanaka crossover" which
is an effort at transient perfect perfect performance, and which
uses a "filler" driver.

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Old 10th July 2007, 06:11 AM   #10
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N.P.'s article and the things written by John K and John Watkinson are indeed what sparked my interest in correct transient reproduction.

Quote:
there was also an article in the January 07 AudioXpress by Stokes
Is this "our" Sheldon Stokes or who ?

Regards

Charles
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