2nd Order Xover question

johninCR · 6th August 2004, 10:37 PM

In a basic 2nd order crossover, what happens if I use a different cap value than what is called for? Do I just end up with a stepped xover slope, 6db/oct then 12db/oct ?

Svante · 7th August 2004, 12:03 PM

Either that or a peak in the frequency response, depending on if the value is smaller or bigger. Also the cutoff frequency (which will be a bit hard to define) will move.

Volenti · 7th August 2004, 01:06 PM

In my practical experience with passive x-overs, incorrect component values often result in "inbetween" x-over slopes rather then stepped, ie; 9db rather than 12db per oct. It's only when the values are really far out that a step becomes noticable.

I have an older RTA that can electrically measure a passive x-over in circuit, very handy tool.

Svante · 7th August 2004, 08:34 PM

Yes, of course, the "knees" in the "theoretical" response is just a way to make the physics behind the actual response curve. For example this means that in a 2nd order butterworth filter the slope gradually increases from 0 to 12 dB/oct, it is just in some simplified graphs that there are sharp knees in the response.

Swedish Chef · 8th August 2004, 08:03 PM

Be sure to use a Zobel compensation network if doing a 2nd order LP. Otherwise you might end up with something else than you intended...

Kittle · 9th August 2004, 04:13 PM

Quote:

Originally posted by Swedish Chef
Be sure to use a Zobel compensation network if doing a 2nd order LP. Otherwise you might end up with something else than you intended...

Like what?

Im trying to keep the # of components in my signal path down to a minimum. and had not planned on using one of them.
(im working on 2nd order xo as well)

Volenti · 10th August 2004, 07:38 AM

Quote:

Originally posted by Kittle

Like what?

Like a big hump in the response at or just after the intended x-over point, caused by the drivers impedance rising with frequency, a zobel corrects this (as far as the x-over is concerned)

AGGEMAM · 10th August 2004, 08:15 AM

Or instead of a Zobel, you could experiment with a low pass that is 5%-10% lower in frequency than the high pass filter. That will often give the same result, plus the filter slope will be 15/db octave instead of 12/db octave with a 2nd order XO.

6th August 2004, 10:37 PM	#1
johninCR diyAudio Member Join Date: Apr 2004 Location: Costa Rica	2nd Order Xover question In a basic 2nd order crossover, what happens if I use a different cap value than what is called for? Do I just end up with a stepped xover slope, 6db/oct then 12db/oct ? __________________ Everyone has a photographic memory. It's just that most are out of film.

7th August 2004, 12:03 PM	#2
Svante diyAudio Member Join Date: Feb 2004 Location: Stockholm	Either that or a peak in the frequency response, depending on if the value is smaller or bigger. Also the cutoff frequency (which will be a bit hard to define) will move. __________________ Simulate loudspeakers: Basta! Simulate the baffle step: The Edge

7th August 2004, 08:34 PM	#4
Svante diyAudio Member Join Date: Feb 2004 Location: Stockholm	Yes, of course, the "knees" in the "theoretical" response is just a way to make the physics behind the actual response curve. For example this means that in a 2nd order butterworth filter the slope gradually increases from 0 to 12 dB/oct, it is just in some simplified graphs that there are sharp knees in the response. __________________ Simulate loudspeakers: Basta! Simulate the baffle step: The Edge

8th August 2004, 08:03 PM	#5
Swedish Chef diyAudio Member Join Date: Jun 2004 Location: Stockholm	Be sure to use a Zobel compensation network if doing a 2nd order LP. Otherwise you might end up with something else than you intended... __________________ "Knowing what to do but not why is no use in a changing world" - The Art of Sound Reproduction

10th August 2004, 08:15 AM	#8
AGGEMAM diyAudio Member Join Date: Jul 2002 Location: Copenhagen	Or instead of a Zobel, you could experiment with a low pass that is 5%-10% lower in frequency than the high pass filter. That will often give the same result, plus the filter slope will be 15/db octave instead of 12/db octave with a 2nd order XO. __________________ Hmm .. no .. I really haven't got anythig cool to say ..


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7th August 2004, 01:06 PM	#3
Volenti diyAudio Member Join Date: Aug 2003 Location: Australia	In my practical experience with passive x-overs, incorrect component values often result in "inbetween" x-over slopes rather then stepped, ie; 9db rather than 12db per oct. It's only when the values are really far out that a step becomes noticable. I have an older RTA that can electrically measure a passive x-over in circuit, very handy tool.

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