mixed crossover phase question

Archived from groups: rec.audio.tech (More info?)

"Chris Berry" <christoforos@Notmail.com> wrote in message news:<c5jiu5$efh$00$1@news.t-online.com>...
> Can someone correct me if I've got this wrong please.
>
> Phase shift in crossovers works like this in 2-way systems:
> For every capacitor in series with the tweeter followed by an inductive load
> to earth you add 90 degrees to the phase shift.
> For every capacitor in parallel with the woofer you subtract 90 degrees from
> the woofer's phase (or add 90 degrees to the tweeter phase.
>
> Hence:
> first order high pass + 3rd order low pass = wire drivers out of phase.
>
> Is that right?

No, not exactly. You're confusing a couple of issues.

First, it is the order of the filter that, given the common use
in loudspeaker crossovers, that determines the ultimate STOP BAND
phase shift. Within the pass band, the phase shift approaches 0.

For example, well below the crossover point, the phase shift of
the woofer crossover is 0. Above it, it approaches -90 degrees
times the number of orders. On the other hand, well below the
crossover, the phase shift to the tweeter is 90 degrees times
the number of orders, while well above the crossover point, it's
approaching 0 degrees.

It's what's happen AT the crossover network that's interesting,
because it's at an around the crossover where the amplitudes of
the two are close enough for it to start to really matter. And
at that point, you're rule of thumb is wrong.

Let's take the simplest-to-understand case: assume that both the
woofer and tweeter crossover filters have Butterworth characteristics,
and that their -3dB points are the same frequency. Butterworth filters
conventiently have a simple to understand behavior: at their cutoff
frequency, they have half the passband amplitude, half the ultimate
stopband rolloff rate, and half the stopband phase shift.

THat means, for a simple Butterworth-aligned 2nd order network,
that while the ultimate stop-band pahse shift of the woofer is
-180 degrees and that of the tweeter is +180 degrees, at the
crossover point, the phase shifts are -90 and +90 respectively.
That's why, in 2nd order networks, you flip the phase of the
tweeter, because the woofer is at -90, the tweeter is at +90,
and the difference is 180 degrees, or complete cancellation.

In your case, you have a 3rd order low pass and a 1st order high
pass. The 3rd order low pass will have a stop-band phase shift
of -270 degrees, but at the crossover, it will be -135 degrees.
The 1st order high pass will have a stop-band phase shift of +90,
but will be +45. So the difference AT the crossover point will
be -135 - +45 or -180 degrees. On that basis alone, yes, you'd
want to flip the phase of the tweeter to prevent cancellation.

But because you're going with a non-symmetric topology, you run
into a difficulty. IN the symmetrical 2nd order case I illustrated
above, if you were to plot the phase difference vs frequency, you'd
find that the rate at which the phase of each changes with frequency
is the same: they are ALWAYS 180 degrees out of phase. This means
that flipping the phase will correct for the phase difference across
a wide bandwidth.

However, your situation is not so neat: the rate of change of phase
of the first-order high pass is 1/3 that of the 3rd-order low pass,
so that you meet the necessary condition at only one frequency. In
your particular situation, you may well find that the DIFFERENCE in
the rate of phase change is enough to cause problems above and below
the crossover point, ESPECIALLY considering that the 1st order low-
pass has poor stop-band attenuation for wuite a significant range of
frequency. Consider that an octave below the crossover, the tweeter
will be singing away only 6 dB down. With it's phase flipped 180
degrees, it will be in phase at the crossover, but will be approaching
-90 degrees only an octave down.

And, of course, you've neglected entirely the fact that the drivers
themselves add significantly to the total phase response of the system.
You need to concern yourself with the TOTAL phase response, not just
that of the crossover.

Archived from groups: rec.audio.tech (More info?)

"Dick Pierce" <dpierce@cartchunk.org> wrote in message
news c02c02f.0404141014.2252a762@posting.google.com...
> "Chris Berry" <christoforos@Notmail.com> wrote in message
news:<c5jiu5$efh$00$1@news.t-online.com>...
> > Can someone correct me if I've got this wrong please.
> >
> > Phase shift in crossovers works like this in 2-way systems:
> > For every capacitor in series with the tweeter followed by an inductive
load
> > to earth you add 90 degrees to the phase shift.
> > For every capacitor in parallel with the woofer you subtract 90 degrees
from
> > the woofer's phase (or add 90 degrees to the tweeter phase.
> >
> > Hence:
> > first order high pass + 3rd order low pass = wire drivers out of phase.
> >
> > Is that right?
>
> No, not exactly. You're confusing a couple of issues.
>
> First, it is the order of the filter that, given the common use
> in loudspeaker crossovers, that determines the ultimate STOP BAND
> phase shift. Within the pass band, the phase shift approaches 0.
>
> For example, well below the crossover point, the phase shift of
> the woofer crossover is 0. Above it, it approaches -90 degrees
> times the number of orders. On the other hand, well below the
> crossover, the phase shift to the tweeter is 90 degrees times
> the number of orders, while well above the crossover point, it's
> approaching 0 degrees.
>
> It's what's happen AT the crossover network that's interesting,
> because it's at an around the crossover where the amplitudes of
> the two are close enough for it to start to really matter. And
> at that point, you're rule of thumb is wrong.
>
> Let's take the simplest-to-understand case: assume that both the
> woofer and tweeter crossover filters have Butterworth characteristics,
> and that their -3dB points are the same frequency. Butterworth filters
> conventiently have a simple to understand behavior: at their cutoff
> frequency, they have half the passband amplitude, half the ultimate
> stopband rolloff rate, and half the stopband phase shift.
>
> THat means, for a simple Butterworth-aligned 2nd order network,
> that while the ultimate stop-band pahse shift of the woofer is
> -180 degrees and that of the tweeter is +180 degrees, at the
> crossover point, the phase shifts are -90 and +90 respectively.
> That's why, in 2nd order networks, you flip the phase of the
> tweeter, because the woofer is at -90, the tweeter is at +90,
> and the difference is 180 degrees, or complete cancellation.
>
> In your case, you have a 3rd order low pass and a 1st order high
> pass. The 3rd order low pass will have a stop-band phase shift
> of -270 degrees, but at the crossover, it will be -135 degrees.
> The 1st order high pass will have a stop-band phase shift of +90,
> but will be +45. So the difference AT the crossover point will
> be -135 - +45 or -180 degrees. On that basis alone, yes, you'd
> want to flip the phase of the tweeter to prevent cancellation.
>
> But because you're going with a non-symmetric topology, you run
> into a difficulty. IN the symmetrical 2nd order case I illustrated
> above, if you were to plot the phase difference vs frequency, you'd
> find that the rate at which the phase of each changes with frequency
> is the same: they are ALWAYS 180 degrees out of phase. This means
> that flipping the phase will correct for the phase difference across
> a wide bandwidth.
>
> However, your situation is not so neat: the rate of change of phase
> of the first-order high pass is 1/3 that of the 3rd-order low pass,
> so that you meet the necessary condition at only one frequency. In
> your particular situation, you may well find that the DIFFERENCE in
> the rate of phase change is enough to cause problems above and below
> the crossover point, ESPECIALLY considering that the 1st order low-
> pass has poor stop-band attenuation for wuite a significant range of
> frequency. Consider that an octave below the crossover, the tweeter
> will be singing away only 6 dB down. With it's phase flipped 180
> degrees, it will be in phase at the crossover, but will be approaching
> -90 degrees only an octave down.

My mistake... It's actually the other way round - 3rd order high pass, 1st
order low pass.
The theory would mean pretty much the same thing though - above the x-over
point.
Luckily, the phase plot of the woofer is between 0 and 10 degrees at the
x-over point but I have no phase data for the tweeter...


>
> And, of course, you've neglected entirely the fact that the drivers
> themselves add significantly to the total phase response of the system.
> You need to concern yourself with the TOTAL phase response, not just
> that of the crossover.

Just when I thought I had it sorted out, I find out that there's another
surprise round the bend...

I guess I'll just have to try using a 3rd order low and high pass and see if
it's worth the extra bit...

Thanks Dick. Appreciated.
cb

Archived from groups: rec.audio.tech (More info?)

Chris Berry wrote:

[Dick Pierce quoted]

>> And, of course, you've neglected entirely the fact that
>> the drivers themselves add significantly to the total
>> phase response of the system. You need to concern yourself
>> with the TOTAL phase response, not just that of the crossover.

> Just when I thought I had it sorted out, I find out that
> there's another surprise round the bend...

It may be easiest and best to determine the proper polarity of the
"higher frequency unit" by experiment.

> I guess I'll just have to try using a 3rd order low and high
> pass and see if it's worth the extra bit...

Yes, listening sometimes works. With a smooth rool-off LF unit with
extended response the 6/18 ploy may work very well, I used it on a pair
of Tannoys for some time way long time ago.

> cb


Kind regards

Peter Larsen

--
*******************************************
* My site is at: http://www.muyiovatki.dk *
*******************************************