Q of box re: hi-pass x/o

[RobWells]

Hi everyone,

I originally planned to make my mid box with a Q of .51. However, I want to add a 12Db/oct filter to this to make it a 24Db/oct roll off.(high pass). I read here:

http://www.snippets.org/filters/crossover.htm

That if I want to add a 12Db to the natural roll off, then both slopes shopuld have a Q of .71, to give a combined Q of 0.5

What do you guys think ? - I'm halfway through constructing my boxes, and can adapt what I've done already.

Cheers

Rob

[sreten]

You have a cascaded filter, with the Q of the drivers 2nd order
roll-off and the Q of the electrical filter.

There is no reason why the two Q's have to be the same.
However they will combine to form a 4th order alignment.

A passive crossover combining with the units Q is very, very
difficult due to the units impedance rise at resonance.

Using an active crossover its simply a question of reading up
on cascaded 2nd order alignments to form 4th order alignments.

Your mid unit must a bass/midrange with long throw to utilise its natural in box roll-off.
If it has a short throw nearly always it must be crossed over at a higher frequency.

sreten.

[RobWells]

Hi Sreten,

It's an active filter. The reason I am asking is the Q of the bass lowpass is 0.49, so I want to match that to get a 4th order linkwitz-reilly x/o. If I add a 12Db/oct high pass with a Q of .7 to the mids natural roll of, I'd need it to be .7 to match.

If I left the box at Q0.5 then I'd have to make my active x/o a Q of 1 ? Would this be feasible ?

The mid will be at -3Db at 83Hz (simulated - I've yet to measure it)

It's a s-speak 8530K-00

Cheers

Rob

[kelticwizard]

How much will the mid enclosure's F3 rise if you make the box smaller to get a Qtc of 0.7? Are you willing to adjust the crossover point?

[sreten]

I'm sure you can cascade an 2nd order active filter to get your
target alignment using the speaker with a Q of 0.5.

I don't know the exact numbers but the Q will be near 1,
if depends on the F (if lower Q higher) to get the right phase
response for the LR alignment.

My maths isn't what it used to be so I can't work out the exact
values for you. But I'm sure with the box alignment you can get
a LR alignment at one particular frequency by using a 2nd order
active filter with a different Q and F.

sreten.

[kelticwizard]

I'll answer my own question. I plugged the Scan Speak 15W 8530 K's numbers into Win ISD.

The red line is 4.8 liters, Qtc=0.71

The blue line is 11 liters, Qtc=0.51

It looks to me that all you have to do is to make the box smaller.

[sreten]

Reducing the box volume is the easiest way to do it.

sreten.

[kelticwizard]

Will that move the crossover frequency upward?

Will what is now the F3 become the F6?

[sreten]

For the cascaded Butterworths the Butterworths F3 becomes
the F6 of the Linkwitz/Riley alignment.

I'm not sure what F6 you'd get using the 0.5 box alignment
with a different 2nd order filter, suspect it would be lower.

sreten.

[RobWells]

Thanks Sreten, Kelticwizard.

Reducing the vol. lowers the x/o freq a bit - the tricky bit is making the -3Db of the x/o match the -3Db of the box. The LR x/o sums at the -6Db point so I'll probably have to experiment a bit to get it bang on. I've just e-mailed John Pomann about a kit to breadboard up my x/o. I've got the ESP boards here to make a pair of 3-ways, but these don't have provision for RF filters or baffle step compensation.

As I was building up my mids in 'layers' it means they're practically finished rather than halfway through which is nice

cheers

Rob

edit - sreten replied with the above while I was typing this..