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first order Bessel uses one inductor in series for the woofer and one
capacitor in series with the tweeter. lets say that for our example we are
using a 10mF cap and a 10mH coil. what if instead of using one of each we
put a 5mF cap on the negative lead and a 5mF cap on the positive lead of the
tweeter and then a 5mH coil on the negative lead and a 5mH coil on the
positive lead of the woofer. why doesn't this work to give the sum total of
a single 10 on each driver with the benefit of canceling out phase
distortion caused by having only one component on one side.

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

Troll wrote:
> first order Bessel uses one inductor in series for the woofer and one

> capacitor in series with the tweeter. lets say that for our example
we are
> using a 10mF cap and a 10mH coil. what if instead of using one of
each we
> put a 5mF cap on the negative lead and a 5mF cap on the positive lead
of the
> tweeter and then a 5mH coil on the negative lead and a 5mH coil on
the
> positive lead of the woofer. why doesn't this work to give the sum
total of
> a single 10 on each driver with the benefit of canceling out phase
> distortion caused by having only one component on one side.

Assuming you are hooking it up precisely as you descirbe, there is
NO difference between the two. None. There's NO diffierence in
"phase distortion" or anything of the sort. The total phase shift
in the stop band WILL be 90 degrees in both cases.

Consider the following (assuming you understand Thevenin and such):
Assuming you have a single inductor, a resistor representing the
load and a voltage source all hooked in series, can you show that
the transfer function across the resistor depends upon the order
of the inductor, resistor and voltage source? Hint: the "order"
changes as the polarity of the signal changes. Second hint: it
makes no difference.

Fine, assuming you accept the above (and if you don't, I'd ask
that you show why it's not true), we have established that the
transfer function is not dependent upon the order of the
components in a serial cricuit.

Next step. Instead of a single 10 mH inductor, consider 2 5 mH
inductors in series. Can you show that the transfer function is
different because you have two components in series equaling the
value of the first? (hint: no, you can't: two 5 mH inductors in
series is exactly the same as a single 10 mH inductor.)

Fine, so we have established that two series inductors of x/2 is
equivalent to a single inductor of x.

Now, combine the two: if the order in a series network is not
important, and two inductors in series of value x/2 is the same
as a single series inductor of x, then why would the transfer
function of a series circuit consisting of a voltage source,
a two 5 mH inductors and a resistor in ANY order be different?
(hint: it isn't).

I'll run a spice simulation later if you don't accept the results.

And, by the way, you'll find it interesting, I hope, that a
first-order Bessel filter has exactly the same response, phase
and amplitude, as a first order Butterworth.

The problems you are going to run into include the fact that
the driver does NOT present a resistive load, and that, due to
it's non-resistive behavior, you CANNOT implement a first-order
ANY filter, Bessel, Butterworth or otherwise.

Reply to Anonymous

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

On Sat, 1 Jan 2005 10:44:16 -0500, "Troll" <Troll@spam.net> wrote:

>first order Bessel uses one inductor in series for the woofer and one
>capacitor in series with the tweeter. lets say that for our example we are
>using a 10mF cap and a 10mH coil. what if instead of using one of each we
>put a 5mF cap on the negative lead and a 5mF cap on the positive lead of the
>tweeter and then a 5mH coil on the negative lead and a 5mH coil on the
>positive lead of the woofer. why doesn't this work to give the sum total of
>a single 10 on each driver with the benefit of canceling out phase
>distortion caused by having only one component on one side.

You are , in effect, putting the two 'halves' in series. This will
work for the inductances which will sum to 10. However, the series
capacitors will result in a 2.5mF capacitance.

Kal

Reply to Anonymous

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

On Sat, 1 Jan 2005 10:44:16 -0500, "Troll" <Troll@spam.net> wrote:

> with the benefit of canceling out phase
>distortion caused by having only one component on one side.

What phase distortion? Don't you realize that this is an AC circuit?

Reply to Anonymous

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

Troll wrote:
> first order Bessel uses one inductor in series for the woofer and one

> capacitor in series with the tweeter. lets say that for our example
we are
> using a 10mF cap and a 10mH coil. what if instead of using one of
each we
> put a 5mF cap on the negative lead and a 5mF cap on the positive lead
of the
> tweeter and then a 5mH coil on the negative lead and a 5mH coil on
the
> positive lead of the woofer. why doesn't this work to give the sum
total of
> a single 10 on each driver with the benefit of canceling out phase
> distortion caused by having only one component on one side.

As promised, here's the result of a spice simulation of the two
scenatios you've outlined. Column 1 is frequency, every third
octave, columns 2 and 3 are the magnitude and phase response across
resistor with a single 10 mH inductor in series with the resistor,
while columns 4 and 5 are the magnitude and phase response across
the resistor with a 5 mH inductor before the resistor, and a 5 mH
inductor after the resistor.

-------------------- --------------------
10 mH + 8 Ohm 5 mH + 8 Ohm + 5 mH
Frequency Mag Phase Mag Phase
--------- --------- --------- --------- ---------
20 Hz -0.106 dB -8.93 deg -0.106 dB -8.93 deg
25.2 -0.167 -11.2 -0.167 -11.2
31.7 -0.261 -14 -0.261 -14
39.9 -0.407 -17.4 -0.407 -17.4
50.2 -0.628 -21.5 -0.628 -21.5
63.2 -0.958 -26.4 -0.958 -26.4
79.6 -1.43 -32 -1.43 -32
100 -2.1 -38.2 -2.1 -38.2
126 -2.97 -44.7 -2.97 -44.7
159 -4.08 -51.3 -4.08 -51.3
200 -5.4 -57.5 -5.4 -57.5
252 -6.91 -63.2 -6.91 -63.2
317 -8.57 -68.1 -8.57 -68.1
399 -10.3 -72.3 -10.3 -72.3
502 -12.2 -75.8 -12.2 -75.8
632 -14.1 -78.6 -14.1 -78.6
796 -16 -80.9 -16 -80.9
1000 -18 -82.8 -18 -82.8

As, I hope, you can plainly see, both the magnituse AND the
phase response across the load are IDENTICAL in both topologies
(actually, according to Thevenin, the topologies are completely
equivalent: that's the point).

More to the point, what problem are you trying to solve and
how do you think your suggestion solves it?

For anyone interested, here are the two SPICE nets for the
above:

* Single 10 mH series inductor crossover
Vin 1 0 AC SIN 1.0 0.0
Lin 1 2 10MH
Rload 2 0 8
..AC DEC 10 20 1K
..PRINT AC VDB(2,0) VP(2,0)
..END

* Two split 5 mH series inductor crossover
Vin 1 0 AC SIN 1.0 0.0
Lin 1 2 5MH
Rload 2 3 8
Lout 3 0 5MH
..AC DEC 10 20 1K
.PRINT AC VDB(2,3) VP(2,3)
.END

Reply to Anonymous

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

On Sat, 1 Jan 2005 10:44:16 -0500, "Troll" <Troll@spam.net> wrote:

>first order Bessel uses one inductor in series for the woofer and one
>capacitor in series with the tweeter. lets say that for our example we are
>using a 10mF cap and a 10mH coil. what if instead of using one of each we
>put a 5mF cap on the negative lead and a 5mF cap on the positive lead of the
>tweeter and then a 5mH coil on the negative lead and a 5mH coil on the
>positive lead of the woofer. why doesn't this work to give the sum total of
>a single 10 on each driver with the benefit of canceling out phase
>distortion caused by having only one component on one side.

Two identical components in series are indeed exactly equivalent to
the equivalent single component - it's still the same first order
crossover, with all it's inherent problems, and now with higher cost
as well.

But two 5uF in series is equivalent to 2.5uF, NOT 10uF (I assume you
didn't really mean "10 milliFarads" )

Tony (remove the "_" to reply by email)

Reply to Tony

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