cascading Q

[ocool_15]

I understand that I can use 2 12db/oct butterworth High pass filters to make a 24dB/oct linkwitz-riley. I have a theory below but have never seen it done.

sealed .7 q box + butterworth active filter at same frequency
.707 *.707= .500

linkwitz should be .49 but above is my understanding.

I believe I could use 2 filters like above but not matching
smaller box *modified filter = same q?
.831*.60=.49


The main advantage I see is to use a smaller box but be able to get the same frequency response. I have never seen this done but am hoping it is possible.

[Ron E]

Q is only defined for 2nd order systems.

[ocool_15]

Perhaps I am using the term Q incorrectly but what I am intending to do is end up with a 24db/octave Linkwitz-Riley using the enclosure as part of the High pass.



here is an example website on a quick search that uses Q
4th order still refers to Q here


Often 4th order are referred to as 2 cascaded 2nd order filters like used here
Rod Elliott


At this point I think I have made a mistake. In short is there a way through an active filter I can use my sealed enclosure with a Q of ~.707(Butterworth) to end up with a 24dB/oct Linkwitz-Riley

To end up with a 24db/oct Linkwitz_Riley should I have built my sealed enclosure to a q of .49 and used a 12dB/oct HP linkwitz riley also?

[Dag]

I do't think you can do that because to get a real 4th order L-R filter you have to get the poles at the right places. But you can first do a "Linkwitz Transform" and then a 2nd order Butterworth. Best done with an active filter.

http://www.linkwitzlab.com/filters.htm#9
9 - 12 dB/oct highpass equalization ("Linkwitz Transform", Biquad)

D

[john k...]

Originally posted by ocool_15
I understand that I can use 2 12db/oct butterworth High pass filters to make a 24dB/oct linkwitz-riley. I have a theory below but have never seen it done.

sealed .7 q box + butterworth active filter at same frequency
.707 *.707= .500

linkwitz should be .49 but above is my understanding.

I believe I could use 2 filters like above but not matching
smaller box *modified filter = same q?
.831*.60=.49


The main advantage I see is to use a smaller box but be able to get the same frequency response. I have never seen this done but am hoping it is possible.

(JPK) For and LR4 Q1 and Q2 = 0,707. If Q1 and Q2 are not equal but Q1 x Q2 = 0,5 the filter will not be a true LR4 but if they are not too far from 0,707 the difference for audio use will not be important. Even with Q1 =1,0 and Q2 = 0.5 there is only a fraction of a dB difference at worst.

[planet10]

That we have pigeon-holed certain alignments, filters etc with names doesn't mean that the ones that don't coincide with the named ones aren't useful... the named ones are primarily an artifact of when all the math was done with pencil & paper... use what works and don't be afraid to explore....

dave

[catapult]

In short is there a way through an active filter I can use my sealed enclosure with a Q of ~.707(Butterworth) to end up with a 24dB/oct Linkwitz-Riley

Yes, that's the theory behind the so-called THX crossover used in most AV receivers and prepros. The speaker is designed with a sealed Q=.7 (B2) alignment and the receiver applies a B2 electrical filter to end up with an LR4 acoustical response. On the low end, an LR4 electrical filter is applied to the sub giving, in theory, a nice LR4 acoustical crossover between the sub and the main.

Linkwitz says if the box's Q is between .6 and .8, that's close enough to use a standard Q=.7 electrical filter. If it's outside that range or you want to shift the frequency of the poles, he recommends adding a Linkwitz Transform circuit in addition to the Q=.7 filter.

http://www.linkwitzlab.com/thor-eq.htm -- Other applications.

[cuibono]

I'd like to resurrect this thread. I've been using the computer to develop active crossovers. In particular, I've been using Cocko's Reaper and EQ, hooked up directly to my speaker/measurement system. It allows real time adjustment and accurate measurements to hit whatever acoustic target I want. It is by far the easiest way I've done it so far (still not perfect though).

One thing though, is that the LP/HP sections of the equalizers I've tried all use 2nd order filters. You have to adjust the Q factor to get the roll-off you want. So I have some questions wrt Q and cascading filters...

Originally Posted by john k...

(JPK) For and LR4 Q1 and Q2 = 0,707. If Q1 and Q2 are not equal but Q1 x Q2 = 0,5 the filter will not be a true LR4 but if they are not too far from 0,707 the difference for audio use will not be important. Even with Q1 =1,0 and Q2 = 0.5 there is only a fraction of a dB difference at worst.

Is this true for B4 (fourth order butterworth)? When simulating them, it seemed necessary to use different Q values to hit the target. When is this true? LR4 didn't.

Originally Posted by planet10

That we have pigeon-holed certain alignments, filters etc with names doesn't mean that the ones that don't coincide with the named ones aren't useful... the named ones are primarily an artifact of when all the math was done with pencil & paper... use what works and don't be afraid to explore....

dave

I agree, particularly with DSP in the computer, you have really a ton of control over the acoustic response of the driver. BUT, if you don't use one of the 'named' alignments, you can't guess what the phase response will be beforehand, correct?

[john k...]

Originally Posted by cuibono

I'd like to resurrect this thread. I've been using the computer to develop active crossovers. In particular, I've been using Cocko's Reaper and EQ, hooked up directly to my speaker/measurement system. It allows real time adjustment and accurate measurements to hit whatever acoustic target I want. It is by far the easiest way I've done it so far (still not perfect though).

One thing though, is that the LP/HP sections of the equalizers I've tried all use 2nd order filters. You have to adjust the Q factor to get the roll-off you want. So I have some questions wrt Q and cascading filters...



Is this true for B4 (fourth order butterworth)? When simulating them, it seemed necessary to use different Q values to hit the target. When is this true? LR4 didn't.





I agree, particularly with DSP in the computer, you have really a ton of control over the acoustic response of the driver. BUT, if you don't use one of the 'named' alignments, you can't guess what the phase response will be beforehand, correct?

All filters of order greater than 2 can be expressed as a cascade of 1st and 2nd order filters. Mathematically it is called factored form. A B3, for example is a 1st order cascaded with a Q = 1, 2nd order. A B4 is two 2nd order; a Q = 0.541 cascaded with a Q = 1.306. The produce of the Qs does indicate the amplitude at the corner frequency, but it is incorrect to refer to the produce as the Q of the filter. For example, 0.541 x 1.306 = 0.707, then amplitude is -3dB at Fc, but there are an infinited number of Q1 x Q2 products that yield 0.707, -3dB at fc, but all have different behavior above and below fc.

[cuibono]

Thanks John, that's exactly the info I was looking for. Is there any place on the internet that shows those values for various filter combinations?