In message , Old Fart at Play
writes

Chris, you are still confused but nearly there.

As you say, a 4th order L-R is two 2nd order Butterworths.
At the crossover frequency a B2 filter is -3dB and 90 degrees phase
shift.
Therefore the 4LR is -6dB and 180 degrees.
The LP and HP outputs are in phase at all frequencies
and the voltage sum is constant.

HTH,

Roger.

You are quite correct, I'll go and hang my head in shame now :-(

For a 4-th order L-R, you do indeed cross over at the -6dB point and the
outputs of the LP and HP sections are always in phase. (And the
magnitude sums to be flat).

(I even went and SPICEd it to check!)
--
Chris Morriss

In article , Chris Morriss
wrote:
In message , Jim Lesurf
writes
However you may not actually want that. :-)

Personally, for active filtering, I'd tend to prefer using a LPF, then
creating a HPF output by subtracting the LPF output from the input. The
result if you keep the levels matched is a LP and HP pair of signals
whose vector sum always equals the input. Thus the combined result
shows no phase errors due to the filtering.

For the actual filters I tend to lift the basic designs from the Active

Oh yes, I quite agree, a complex phase-compensated crossover has only
one advantage: it does help keep down vertical lobing problems.

I would put this slightly differently. The 'lobing' problem arises as a
result of having an 'array' of speakers in operation in the crossover
frequency region. There will always tend to be a frequency region where the
two units are radiating similar powers. If the speakers are not very close
(in wavelength terms) lobing is then inevitable. The phasing in this region
won't prevent lobing, it will just displace the maxima and minima in
angular terms w.r.t. the line through the speakers and the speaker plane.

As Arnie has also said, it does also depend on the inherent amplitude
and phase response of the drivers.

The key point here for me is the phase responses of the two drivers in the
frequency region where they are tending to radiate similar powers.

If they are 'in phase' at this point, then ensuring the vector sum is
unchanged should mean that the 'far field' power sent normal to the line
through the speaker units (i.e. towards the nominal listener) will be
correct.

However the above makes assumptions about what is the case. So, for
example, if the speakers have phase delays that differ when they are
radiating similar amounts, you'd need to change what you are giving them.
We also have to worry about where the listener may be and the room
acoustic.

All of this is another reason why I'm not really a fan of 'dynamic'
speakers. :-)

The advantage of the method I prefer is that it ensures both constant
amplitude sum (for the correct unit phase behaviour) and constant total
power. Does this by ensuring the vector sum gain from the filtering is
frequency independent. However this may not be what a specific speaker
requires.

Above said, for electronic crossovers, I'd tend to do it this way, then add
any required modifiers to 'pre-correct' the split signals before delivering
them to the power amps and units...

I use constant-voltage subtraction crossovers, but without any phase
compensation they do force one of the outputs to only roll off at 6db
per octave.

The advantage of higher orders is they can cut down to size of the region
where we have an (unwanted) array effect. However you can do this using my
approach, and it saves money as you only need one high-order LPF and then
get the HPF that matches it 'for free'. :-)

Slainte,

Jim

--
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html

In message , Jim Lesurf
writes
The advantage of higher orders is they can cut down to size of the region
where we have an (unwanted) array effect. However you can do this using my
approach, and it saves money as you only need one high-order LPF and then
get the HPF that matches it 'for free'. :-)

Slainte,

Jim


But even if your HPF (say) is a 4-th order, the LPF you get by
subtraction is still only a first order. The phase shift of the summed
output is zero, and it does sum flat of course, so it still is a good
thing.

(It took me ages to work this out, but it is correct, and a quick SPICE
simulation shows it)

As I said in an earlier post, I do this on a homemade two-way (M-T-M)
that I use. A second order HPF for the tweeter, and the subtractive LPF
for the Bass units. I put the main filter as the HPF to roll-off the
tweeter reasonably quickly, whereas I wasn't too worried about the
low-order roll-of of the subtractive LP output.


--
Chris Morriss

Chris Morriss wrote:

In message , Old Fart at Play
writes

Chris, you are still confused but nearly there.

As you say, a 4th order L-R is two 2nd order Butterworths.
At the crossover frequency a B2 filter is -3dB and 90 degrees phase
shift.
Therefore the 4LR is -6dB and 180 degrees.
The LP and HP outputs are in phase at all frequencies
and the voltage sum is constant.

HTH,

Roger.

You are quite correct, I'll go and hang my head in shame now :-(

For a 4-th order L-R, you do indeed cross over at the -6dB point and the
outputs of the LP and HP sections are always in phase. (And the
magnitude sums to be flat).

Thanks.

Of course it only works if the woofer and tweeter work well
for an octave or so beyond the crossover frequency, otherwise
as Arny said, you have to consider the acoustical output.

--
Roger.

In article , Chris Morriss
wrote:

This is the crux of the issue. To get a flat amplitude response from an
even order filter, the crossover frequency should be at the -6dB point,

Not quite. The only general requirement would be that the vector sum adds
up to the 0dB level. If you allow the components to be in quadrature, for
example, this can occur if they cross at -3dB.

A problem here is that there is often an ambiguity in discussions due to
the coherent effects of the way the radiation pattern is affected by phase,
and the mean (space-averaged) power around a room, and on axis.

Ideally, you'd want to know the directional and amplitude/phase properties
of the speakers, and the acoustics of the room, and the choice of listening
location. Lacking these, you end up with having to choose a set of
assumptions.

Slainte,

Jim

--
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html

In article , Chris Morriss
wrote:
In message , Jim Lesurf
writes
The advantage of higher orders is they can cut down to size of the
region where we have an (unwanted) array effect. However you can do
this using my approach, and it saves money as you only need one
high-order LPF and then get the HPF that matches it 'for free'. :-)

Slainte,

Jim


But even if your HPF (say) is a 4-th order, the LPF you get by
subtraction is still only a first order.

I've not done this for a while, but that strikes me as rather odd (apology
for the pun! :-) ) as a general claim. I think you may find it depends upon
the details of the filter shape of the LPF filter, not just the order.

IIRC when I did some filtering like this a while ago for analysis of the
effects of HF on tweeters the HP anf LP sections (done this way) were of
the same sort or roll-off slopes. May be mis-remembering, though...

The phase shift of the summed output is zero, and it does sum flat of
course, so it still is a good thing.

(It took me ages to work this out, but it is correct, and a quick SPICE
simulation shows it)

I think this may depend upon some specific assumptions you may have made.
However I'll be interested to hear what you can report on this.

Consider designing a LPF that is approaching a 'brick wall' with a flat top
near 0dB and a fall-off to, say, -60dB that occurs over a narrow range.

Where the LPF is near 0dB the output from the 'HPF' must be very small.
However as with transit the turnover region it rises to near 0dB. The
narrower the transition region, the steeper the slope of the turnover of
both the HP and LP outputs.

OTOH if you choose a high order filter that has an inband 'droop' whose
size scales up significantly with the order, you will, indeed reduce the
slope of the HP output in the 'droop' region. But what matters here is the
filter shape, not just the order.

I've forgotten the latin for 'taking an absurd example' for the sake of
extreme illustration. However I won't let my lack of decent classical
education deter me from the following... :-)

Imagine building an analogue version of one of the 96th order low pass
filters used for digital. These can have an inband ripple that is very
close to 0dB, yet die the death in the space of about 2kHz. If you were to
apply the above methods to get the HP difference I doubt it would show just
a first order rolloff as you went down into the low frequency range. I
think the response would change very rapidly over the same 2kHz-ish band.

Or am I missing something?

Slainte,

Jim

--
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html

In message , Jim Lesurf
writes
In article , Chris Morriss
wrote:
In message , Jim Lesurf
writes
The advantage of higher orders is they can cut down to size of the
region where we have an (unwanted) array effect. However you can do
this using my approach, and it saves money as you only need one
high-order LPF and then get the HPF that matches it 'for free'. :-)

Slainte,

Jim


But even if your HPF (say) is a 4-th order, the LPF you get by
subtraction is still only a first order.

I've not done this for a while, but that strikes me as rather odd (apology
for the pun! :-) ) as a general claim. I think you may find it depends upon
the details of the filter shape of the LPF filter, not just the order.

IIRC when I did some filtering like this a while ago for analysis of the
effects of HF on tweeters the HP anf LP sections (done this way) were of
the same sort or roll-off slopes. May be mis-remembering, though...

The phase shift of the summed output is zero, and it does sum flat of
course, so it still is a good thing.

(It took me ages to work this out, but it is correct, and a quick SPICE
simulation shows it)

I think this may depend upon some specific assumptions you may have made.
However I'll be interested to hear what you can report on this.


It struck me as rather odd as well when it was first put to me. Do you
have SPICE available to you? It's easy enough to make an arbitary LP or
HP filter and use an "ideal subtractor" element to produce the other
output. It's easy enough to verify it.

I'll try it with a high order filter, (6-th order or so) and run the
simulation. I guess your email address is valid, so I'll email you the
gain/phase plot of the sim run.

Regards,

--
Chris Morriss

In message , Jim Lesurf
writes
I think this may depend upon some specific assumptions you may have made.
However I'll be interested to hear what you can report on this.


I've sent you an email with the SPICE results. They are in .wmf format.
The email is 216k long so I hope you allow emails that size through!

--
Chris Morriss

In article , Chris Morriss
wrote:
In message , Jim Lesurf
writes
I think this may depend upon some specific assumptions you may have
made. However I'll be interested to hear what you can report on this.


I've sent you an email with the SPICE results. They are in .wmf format.
The email is 216k long so I hope you allow emails that size through!

The size is no problem. The snag is that I don't use 'doze, so now will
have to translate the WMFs and hope the translations comes out OK.
;-)

Don't send me a different format (yet!), though. I'll have a bash when I
can and reply in detail by email.

Slainte,

Jim

--
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html

In article , Chris Morriss
wrote:
In message , Jim Lesurf
writes


I think this may depend upon some specific assumptions you may have
made. However I'll be interested to hear what you can report on this.


It struck me as rather odd as well when it was first put to me. Do you
have SPICE available to you?

I don't use SPICE much. I tend to write my own programs, or analyse things
like filters using the response equations given in the books like the
cookbook I mentioned. I don't think I have a copy of SPICE to hand that is
32-bit 'clean' for the OS/CPU I use. I'll investigate when I get a chance,
though.


I'll try it with a high order filter, (6-th order or so) and run the
simulation. I guess your email address is valid, so I'll email you the
gain/phase plot of the sim run.

The files have arrived OK. :-)

Slainte,

Jim

--
Electronics http://www.st-and.ac.uk/~www_pa/Scot...o/electron.htm
Audio Misc http://www.st-and.demon.co.uk/AudioMisc/index.html
Armstrong Audio http://www.st-and.demon.co.uk/Audio/armstrong.html
Barbirolli Soc. http://www.st-and.demon.co.uk/JBSoc/JBSoc.html