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Old 11th March 2012, 10:43 PM   #471
system7 is offline system7  United Kingdom
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Quote:
Originally Posted by AllenB View Post
@System7, how about we settle on the topology? Will the final shunt pair remain independent or joined?
I'm not sure it makes much difference actually, except to wiring the driver polarity. But tweeter and bass shunted together seems to be most people's convention. As in Tony Gee's diagrams.

I'm hoping there is a flat impedance third order series lurking at some set of values at 1000Hz. Just not getting it...

@speakerdave. I am using an 8 ohm source resistor as a convenience to easily measure the overall impedance, which is usually flat for paired butterworth filters. This is nice because such a load won't be affected by the type of amp you have. You can see the impedance is flat 8 ohms here for 2nd order butterworth series filter. Usual 12dB per octave and -3dB crosover points.
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File Type: png Constant_resistance_series_crossover.PNG (13.2 KB, 80 views)
File Type: png Constant-resistance-series.PNG (24.5 KB, 73 views)
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Old 11th March 2012, 11:44 PM   #472
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So the 8 ohms in front is for modeling but wouldn't be used in a built up network. (also explains the 6dB drop)

Which is the impedance curve? The phase curves are of network voltage? 180 degrees apart means the tweeter polarity should be flipped? And are R1 and R2 real added resistors or are they the simulated drivers?

David

Last edited by speaker dave; 11th March 2012 at 11:46 PM.
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Old 11th March 2012, 11:55 PM   #473
system7 is offline system7  United Kingdom
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Originally Posted by speaker dave View Post
So the 8 ohms in front is for modeling but wouldn't be used in a built up network. (also explains the 6dB drop)

Which is the impedance curve? The phase curves are of network voltage? 180 degrees apart means the tweeter polarity should be flipped? And are R1 and R2 real added resistors or are they the simulated drivers?

David
VM1, VM2 and VM3 voltage sensors give the three curves. VM3 is the brown one, and measures voltage relative to earth. It would have a 3dB peak in it for a Linkwitz-Riley filter which is not constant resistance. The subsequent curves ARE affected for the L-R filter, but not the butterworth. 2nd order butterworth has a 3dB amplitude peak. I quite like modelling the 8 ohm resistor, because I have a cunning scheme to replace it with a 8 ohm LR bafflestep correction. R! and R2 are simulated drivers. I'm not too worried about that, because you can simulate a bass as a coil and a resistor, so it will be just a question of replacing an inductor with the speakers inductance. The tweeter is easily corrected with a zobel or some padding. Polarity is not an issue at this stage. Get that right later.

In fact the third-order parallel Butterworth is also flat impedance, and IIRC, flat amplitude. I'm just hoping AllenB can do better than me at some component values for the series 3rd order. My 0.5, 4/3 and 1.5 ratio really didn't hack it.
Attached Images
File Type: png Third_order_butterworth_1000Hz.PNG (10.8 KB, 72 views)
File Type: png Third_order_butterworth_Freq_resp.PNG (26.6 KB, 72 views)
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Last edited by system7; 12th March 2012 at 12:04 AM.
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Old 12th March 2012, 06:44 AM   #474
VaNarn is offline VaNarn  Australia
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System 7,the ratios that suit a third order constant resistance series circuit are 0.67R, 1.33R and 2R. The second order case in theory has a 3dB rise,but with real world speakers, where we measure the parameters and design the xover accordingly; the problem is non-existant.Labouring the point a little further,to achieve the 3dB gain would require the drivers to be co-incidently mounted and have identical amplitude velocity characteristics in the xover region .Peaks an dips occur by virtue of spacing and phase effects even with perfectly flat(freq response) drivers.Text book explanations are not helpful in the real world and the current fad concerning baffle step compensation is yet another example of "heading off madly in all directions."
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Old 12th March 2012, 10:13 AM   #475
AllenB is offline AllenB  Australia
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Originally Posted by VaNarn View Post
System 7,the ratios that suit a third order constant resistance series circuit are 0.67R, 1.33R and 2R.
That's helpful, VaNarn.

System7. I notice that on-axis the drivers are in quadrature, does this count?

Quote:
the current fad concerning baffle step compensation is yet another example of "heading off madly in all directions."
Yup.
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Old 12th March 2012, 10:33 AM   #476
sreten is offline sreten  United Kingdom
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Originally Posted by system7 View Post
Though I'm not entirely sure what a Solen Split first order does...
Hi,

Neither am I and like assymetric 1st order parallel I don't think a series
version of the solen split (-6dB x/o point instead of -3dB) is possible.

It seems to me the purpose of the solen split is where your crossing
over two drivers at a point where they both begin to roll-off, the mid
above that point and the tweeter below.

If your lucky the solen split values will end up approximating 2nd order
L/R acoustic, if your unlucky the standard butterworth arrangement
might give a better acoustic result.

I can't find any information describing what the the problem
is that the Solen Split is supposed to fix and how it does it.

rgds, sreten.

BTW your filters will behave differently with a 8 ohm source impedance
and with zero source impedance, the two cases are not the same.
(TinaTi has an impedance meter function, replaces the voltage source.)
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Last edited by sreten; 12th March 2012 at 10:43 AM.
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Old 12th March 2012, 10:46 AM   #477
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Quote:
Originally Posted by AllenB View Post

System7. I notice that on-axis the drivers are in quadrature, does this count?
Drivers in quadrature will force non-symmetric polars with a peak in one direction and a dip in the other. Symmetry only comes with LR or LR-like in-phase networks. I assume that Butterworth targets aren't the only option? What about unequal corner shape (high Q on one side, low Q on the other)?

In my oppinion, aiming for constant resistance input Z will be at odds with achieving flat response. As such it is a very poor design choice.

David
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Old 12th March 2012, 09:44 PM   #478
VaNarn is offline VaNarn  Australia
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I overlooked a query from system 7(#473) in which he sought confirmation of the // third order Butterworth xover constants.My old sources say ,that they are 1.5R, 0.75R and 0.5R.
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Old 12th March 2012, 09:48 PM   #479
system7 is offline system7  United Kingdom
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Quote:
Originally Posted by VaNarn View Post
System 7,the ratios that suit a third order constant resistance series circuit are 0.67R, 1.33R and 2R. The second order case in theory has a 3dB rise,but with real world speakers, where we measure the parameters and design the xover accordingly; the problem is non-existant.Labouring the point a little further,to achieve the 3dB gain would require the drivers to be co-incidently mounted and have identical amplitude velocity characteristics in the xover region .Peaks an dips occur by virtue of spacing and phase effects even with perfectly flat(freq response) drivers.Text book explanations are not helpful in the real world and the current fad concerning baffle step compensation is yet another example of "heading off madly in all directions."
Flippin' excellent post, VaNarn. At last someone is talking sense!

I did actually finally find a good link for the 3rd. order series crossover values @ 3kHz which were published by Jeff Bagby years back:
Passive Crossover Design Calculator

The inductor ratios were as you said 3:2:1, the capacitor ratios were a bit more subtle, eh? 3:3/2:1, and the L/C ratios and Q altogether deeper which is why my measly brain got confused...

Very nice work, AllenB.

We are there now on the constant resistance networks, which have a certain elegance. Anybody got any idea what the 4th. order series values are?
Attached Images
File Type: png Third_order_butterworth_1000Hz.PNG (10.8 KB, 18 views)
File Type: png Third_order_butterworth_Freq_resp.PNG (26.6 KB, 13 views)
File Type: png 3rd_order_series_butterworth_schematic.PNG (12.3 KB, 14 views)
File Type: png 3rd-order_series_butterworth.PNG (42.9 KB, 13 views)
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Old 12th March 2012, 10:41 PM   #480
AllenB is offline AllenB  Australia
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Quote:
Originally Posted by speaker dave View Post
In my oppinion, aiming for constant resistance input Z will be at odds with achieving flat response. As such it is a very poor design choice.
Quote:
Originally Posted by system7 View Post
constant resistance networks, which have a certain elegance.
I assume that in a simple series network where the loads behave themselves and the drivers are of equal sensitivity, then a flat impedance curve would be indicative of a flat acoustic power response.

It seems to come down to the usual flat power/symmetrical lobing issue, of which the second option is normally preferred, so why not allow the impedance to peak? You could always conjugate it.

Quote:
Originally Posted by sreten View Post
I don't think a series version of the solen split (-6dB x/o point instead of -3dB) is possible.
You can approximate it using only damping. The effective slope near the crossover will be less than first order, but otherwise the benefits are the same.

Quote:
I can't find any information describing what the the problem
is that the Solen Split is supposed to fix and how it does it.
The problem it is supposed to reduce is the assymetrical lobing, whilst retaining a flat on-axis response. This means there should be no phase difference between the drivers on-axis, and they should be 6dB down at the crossover point.

By pulling the crossover points apart, the 6dB points can be matched easily enough. The phase, which is already 90 degrees separated will separate further. Reversing one of the drivers will bring them closer than before.

However the phase responses will still not meet up and so there will still be some lobing. The drivers will not sum entirely on-axis and there will be a droop of a couple of dB. The full response will come in above or below. I have typically been of the notion that a 4.5dB compromise split makes sense.
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