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
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
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.
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.
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
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
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?
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.
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.
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.
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.
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..
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..
OK I've been mulling over this for a few days now😀
What I *think* I was trying to ask is this:
I want my mid to be 'critically damped (Q=0.5)' , If I make the box to give a Q of 0.7, and add a high pass filter (2nd order, Q=0.7) that has it's F-3 the same as the box F-3, will I end up with a critically damped box(Q 0.5), or will it have the transient properties of the original Q=0.7 box? albeit with a bottom end roll of with a Q of 0.49 ?
😕
I don't know if that makes what I'm thinking about clearer or murkier🙂 please help!
Cheers
Rob
What I *think* I was trying to ask is this:
I want my mid to be 'critically damped (Q=0.5)' , If I make the box to give a Q of 0.7, and add a high pass filter (2nd order, Q=0.7) that has it's F-3 the same as the box F-3, will I end up with a critically damped box(Q 0.5), or will it have the transient properties of the original Q=0.7 box? albeit with a bottom end roll of with a Q of 0.49 ?
😕
I don't know if that makes what I'm thinking about clearer or murkier🙂 please help!
Cheers
Rob
I know what you are saying, Rob, and while I cannot speak from authority on these matters, it appears to me that you would like to have the transient response of a Qts=0.5 but the rolloff characteristics of Qts=0.7.
I don't know if that is possible.
The transient response of Qtc=0.7 is pretty close to that of Qtc=0.5, it seems to me. I am not sure if that is all that bad of a deal.
Here is Small's chart for the step response for various values of Qtc.
I don't know if that is possible.
The transient response of Qtc=0.7 is pretty close to that of Qtc=0.5, it seems to me. I am not sure if that is all that bad of a deal.
Here is Small's chart for the step response for various values of Qtc.
Attachments
If I understand the question, you'll get the 2nd order transient response of the box combined with the 2nd order electrical for a 4th order acoustic roll off.
You can use a L-R transform to electrically convert a higher Q cab to a lower Q one though: http://www.linkwitzlab.com/filters.htm and scroll down to '9 - 12 dB/oct highpass equalization ("Linkwitz Transform", Biquad)'.
For instance, 80Hz Fb = 0.7 Qtc, then 80*0.707 = 56.56Hz/0.5 Qtc. No free 'lunches' though, you take a 6dB hit on efficiency.
GM
You can use a L-R transform to electrically convert a higher Q cab to a lower Q one though: http://www.linkwitzlab.com/filters.htm and scroll down to '9 - 12 dB/oct highpass equalization ("Linkwitz Transform", Biquad)'.
For instance, 80Hz Fb = 0.7 Qtc, then 80*0.707 = 56.56Hz/0.5 Qtc. No free 'lunches' though, you take a 6dB hit on efficiency.
GM
Hi Kelticwizard,
Hey, I know those graphs😀
I actually want 'the roll of characteristics' of 24Db/oct with a Q of 0.49.
This is what I don't get - Does the Transient response of the box get affected by the high pass filter ?
I think I'd rather have a critically damped mid, with a slight Q mis match(?) on the bottom x/o than a mid with a Q of .7..
I've searched to see if anyone has posted about this, and double checked my books but I can't find an answer.
I believe that if I leave my box at 0.5 and add a 0.7 2nd order to it I'll get a Q of 0.35, but what about transient response ?- Does that become 0.35 too ?
Help😕
Cheers
Rob
Hey, I know those graphs😀
I actually want 'the roll of characteristics' of 24Db/oct with a Q of 0.49.
This is what I don't get - Does the Transient response of the box get affected by the high pass filter ?
I think I'd rather have a critically damped mid, with a slight Q mis match(?) on the bottom x/o than a mid with a Q of .7..
I've searched to see if anyone has posted about this, and double checked my books but I can't find an answer.
I believe that if I leave my box at 0.5 and add a 0.7 2nd order to it I'll get a Q of 0.35, but what about transient response ?- Does that become 0.35 too ?
Help😕
Cheers
Rob
Hi GM,
I've read about the Linkwitz transform, however it's this bit I don't get :
""you'll get the 2nd order transient response of the box combined with the 2nd order electrical ""
Does that mean with a box of 0.5, and a hi pass of 0.7 I'd get a Q of 0.35 ? Do I need a hi pass of Q=1.0 ?
Still😕
Rob
I've read about the Linkwitz transform, however it's this bit I don't get :
""you'll get the 2nd order transient response of the box combined with the 2nd order electrical ""
Does that mean with a box of 0.5, and a hi pass of 0.7 I'd get a Q of 0.35 ? Do I need a hi pass of Q=1.0 ?
Still😕
Rob
RobWells said:
The transient response of the enclosure with a Q of .7 is not altered by adding an electrical filter, whatever the Q of the filter might be. You still have a Q=.7 box. However, you need to look at the transfer function of the system.
A box Q of .7 cascaded with a filter Q of .7 gives a total Q of .5. The transient response is that of Q=.5.
Hi Bill,
I think I'm starting to sound really stupid here😱
You've just said "The transient response of the enclosure with a Q of .7 is not altered by adding an electrical filter, whatever the Q of the filter might be."
And then said "A box Q of .7 cascaded with a filter Q of .7 gives a total Q of .5. The transient response is that of Q=.5.""
So in point 1 you've said that my box of 0.7 is not altered by an electrical filter, then in point 2 you said that a box of Q0.7 with a filter of q0.7 = q0.5.....
Please explain.??
Basically, I want my mid to have a transient response of 0.5, but roll off nicely to join the subs.... (using a LR 24Db/oct x/o)
Cheers
Rob
I think I'm starting to sound really stupid here😱
You've just said "The transient response of the enclosure with a Q of .7 is not altered by adding an electrical filter, whatever the Q of the filter might be."
And then said "A box Q of .7 cascaded with a filter Q of .7 gives a total Q of .5. The transient response is that of Q=.5.""
So in point 1 you've said that my box of 0.7 is not altered by an electrical filter, then in point 2 you said that a box of Q0.7 with a filter of q0.7 = q0.5.....
Please explain.??
Basically, I want my mid to have a transient response of 0.5, but roll off nicely to join the subs.... (using a LR 24Db/oct x/o)
Cheers
Rob
Ok on re-reading your saying that the electrical and 'box' properties do sum, and give a final 'Q',
ie if I make my box have a Q of 0.71 and then high pass with a 12Db/oct Q=0.7 filter I will end up with a low end roll off of 24Db/oct (Q=0.49) and a transient response in my mid of Q=0.5.
Damn this chardonnay!
Cheers
Rob
ie if I make my box have a Q of 0.71 and then high pass with a 12Db/oct Q=0.7 filter I will end up with a low end roll off of 24Db/oct (Q=0.49) and a transient response in my mid of Q=0.5.
Damn this chardonnay!
Cheers
Rob
- Status
- Not open for further replies.