Hello.
I have started to model an active filter design using Boxsim to sim the acoustic output and i have arrived at a 5 pole lowpass and 4pole highpass for my loudspeakers.
I have used each 2order stage Q to account for driver response vagaries, and in on case i have Q=1.5.
This filter has similar output to what i am using passively at the moment, in the active design im working on i was just a little concerned about ringing in the underdamped stages, and with that im seeking advice as im just starting out in active filter and a noob
Can someone here help me? Im concerned that the transient effect of this will be audible. Or am I concerned over nothing?
Thanks for reading
I have started to model an active filter design using Boxsim to sim the acoustic output and i have arrived at a 5 pole lowpass and 4pole highpass for my loudspeakers.
I have used each 2order stage Q to account for driver response vagaries, and in on case i have Q=1.5.
This filter has similar output to what i am using passively at the moment, in the active design im working on i was just a little concerned about ringing in the underdamped stages, and with that im seeking advice as im just starting out in active filter and a noob
Can someone here help me? Im concerned that the transient effect of this will be audible. Or am I concerned over nothing?
Thanks for reading
Last edited:
I havent worked out the combined Q Andrew.
In boxsim i can add a single pole, then add a 2pole 2nd stage in series and so on to reach 5 poles cascaded.
Each 2 pole stage has Fc and Q, with the single pole stage fixed at 0.7 as you would expect.
This is a LP btw.
1st stage 1st order at 400hz,
2nd stage 2nd order Q = 1.2 at 3500hz,
3rd stage 2nd order at 3300hz Q=1
IF memory serves. Ill check my numbers and post them a little later (I shouldve started there really lol )
In boxsim i can add a single pole, then add a 2pole 2nd stage in series and so on to reach 5 poles cascaded.
Each 2 pole stage has Fc and Q, with the single pole stage fixed at 0.7 as you would expect.
This is a LP btw.
1st stage 1st order at 400hz,
2nd stage 2nd order Q = 1.2 at 3500hz,
3rd stage 2nd order at 3300hz Q=1
IF memory serves. Ill check my numbers and post them a little later (I shouldve started there really lol )
Last edited:
thank you!
thanks Andrew. You filled in a gap in my uni (read basic) filter theory. I didnt realise you multiplied Qs for the total filter Q. They stopped at 2nd order at uni...
One of the 2 pole stages has a Q of between 1.2 and 1.5 for a good acoustic response, and i was just concerned that it may not be a great idea. But then im a noob and this is my 1st serious attempt at making a sallen key filter of practical use for my hifi.
thanks Andrew. You filled in a gap in my uni (read basic) filter theory. I didnt realise you multiplied Qs for the total filter Q. They stopped at 2nd order at uni...
One of the 2 pole stages has a Q of between 1.2 and 1.5 for a good acoustic response, and i was just concerned that it may not be a great idea. But then im a noob and this is my 1st serious attempt at making a sallen key filter of practical use for my hifi.
Last edited:
The standard RC is a Butterworth roll off.
That Butterworth roll off implies a Q = sqrt(2) (~0.7)
When you go active, you use component values to arrange the Q you want, Butterworth, L-R, Bessel, or any of the many rippled and anywhere in the infinite series that is available.
When you create a 3pole, using a passive RC, plus an active 2pole, to create a Butterworth, you assume Q=0.7 for the passive and set the active Q=1 for a combined Q = 0.7 for the 3pole. It's the same for all the other odd order Butterworths.
The same assumption would apply if your desired Q for the cascaded filters were not Butterworth.
If anybody sees this as wrong, please correct me with references to where I can learn the correct way.
I have been in previous Threads where some Members simply stated I was wrong without any attempt to help me to learn. I am open to learning and to correcting my mistakes.
That Butterworth roll off implies a Q = sqrt(2) (~0.7)
When you go active, you use component values to arrange the Q you want, Butterworth, L-R, Bessel, or any of the many rippled and anywhere in the infinite series that is available.
When you create a 3pole, using a passive RC, plus an active 2pole, to create a Butterworth, you assume Q=0.7 for the passive and set the active Q=1 for a combined Q = 0.7 for the 3pole. It's the same for all the other odd order Butterworths.
The same assumption would apply if your desired Q for the cascaded filters were not Butterworth.
If anybody sees this as wrong, please correct me with references to where I can learn the correct way.
I have been in previous Threads where some Members simply stated I was wrong without any attempt to help me to learn. I am open to learning and to correcting my mistakes.
Thanks again.
Im familiar with how to structure the filter, but the individual stage Q just foxed me a bit. I could also do the RC with opamp to match impedance or gain. A system Q of ~0.8 isnt too bad then. I had visions of making an oscillator rather than a filter
.
Im not designing to cheby 3db dips or a bessel curve per se, just adjusting Q and Fc in boxsims active editor to smooth the response and align phase as best as i can. In doing that i required a stage Q higher than 1 and didnt know if it was a problem. After further reading i found the LR filter Q of
.55 and 1.3 ish so now im not worried.
Im familiar with how to structure the filter, but the individual stage Q just foxed me a bit. I could also do the RC with opamp to match impedance or gain. A system Q of ~0.8 isnt too bad then. I had visions of making an oscillator rather than a filter
.Im not designing to cheby 3db dips or a bessel curve per se, just adjusting Q and Fc in boxsims active editor to smooth the response and align phase as best as i can. In doing that i required a stage Q higher than 1 and didnt know if it was a problem. After further reading i found the LR filter Q of
.55 and 1.3 ish so now im not worried.
Last edited:
Q=1/sqrt(2) is for a second-order Butterworth, where some choice is possible. A first-order is just a first order: the most flat passband and the most linear phase are both the same. As I said, I believe Q is only defined and meaningful for a second-order system. It does not appear in the equations for a first-order filter. For a third-order it only appears in the second-order factor, when the third-order polynomial is expressed in factored form. For a 5th order there are two different Qs, in the two second-order factors.AndrewT said:
That gives you the right answer by the wrong method for third-order.
For 5th order Butterworth you need Q=0.618 and Q=1.618 in the two second-order sections. This is according to 'Network Analysis and Synthesis' by Franklin F. Kuo (2ed., Wiley 1966) - Table 13.1 on page 372. The two Q's are found from (sqrt(5)+1)/2 and (sqrt(5)-1)/2. Their product is 1, as you expect.
yes i recall that from somewhere. Mainly due to the golden ratio like Qs. But using different Qs in order to align phase and iron out bumps in the acoustic output, that isnt an issue is it? Like i said my goals are to align phase and smooth the acoustic freqency response, not to build a text book Bessel, BW, Cauer, Chebychev curve.
There's something I am not following.
This 5pole implies that the Q=0.7 is being assumed as coming from the passive RC section.
Why then should the 3pole combining the passive RC and a 2pole active not be the same? i.e. the RC provides the Q=0.7 and the 2pole active provides the Q=1 giving a combined Q=0.7 for the 3pole circuit?
I am aware of the different Q's in 4th order filters.
Duncan and Self handled explanation of that way back in the 80's where I followed their writings in Wireless World.
That's the way I build 4th orders with a Butterworth rolloff to this day.
Whereas two of those 2pole Butterworths for an LR4 crossover, require the the single Q=0.7 in each part of the cascaded filter for a combined Q=0.5
You say that the third order has the Q set in the 2pole active part, then show a 5pole example where the two 2pole actives arriving at a combined Q=1 (Q=0.618 and Q=1.618, combined Q=0.618*1.618=1)
This 5pole implies that the Q=0.7 is being assumed as coming from the passive RC section.
Why then should the 3pole combining the passive RC and a 2pole active not be the same? i.e. the RC provides the Q=0.7 and the 2pole active provides the Q=1 giving a combined Q=0.7 for the 3pole circuit?
I am aware of the different Q's in 4th order filters.
Duncan and Self handled explanation of that way back in the 80's where I followed their writings in Wireless World.
That's the way I build 4th orders with a Butterworth rolloff to this day.
Whereas two of those 2pole Butterworths for an LR4 crossover, require the the single Q=0.7 in each part of the cascaded filter for a combined Q=0.5
thanks Andrew. I do understand a fair bit of what you both have posted, indeed much of it prodded my brain into recalling earlier learning which i thought had been displaced by other info. Much like Homer Simpson. On Linkwitz site he shows a LR filter Q 0.55 and 1.3 so yet another ratio to get you to Q 0.7 for 2nd order rather than 1. Ill give the textbook 0.618 and 1.618 a go and post my curves as i have them now. Q>1 being passband gain (ok >0.7 in truth) and clipping a possibility. Makes perfect sense. Some days I reckon i need a cranking handle to turn this brain over!
Sometimes a picture speaks a thousand words...
my current passive setup sims like this:
and next my rubbish attempt at active (making up the Qs as i went along....)
Using:
And then Using the suggested Qs (0.618 and 1.618):
Using:
Also using a DC blocking cap of 33uF in the HP cct.
Complex......
Worth the effort to switch from passive? Considering stop band rejection only then YES.
my current passive setup sims like this:
and next my rubbish attempt at active (making up the Qs as i went along....)
Using:
And then Using the suggested Qs (0.618 and 1.618):
Using:
Also using a DC blocking cap of 33uF in the HP cct.
Complex......
Worth the effort to switch from passive? Considering stop band rejection only then YES.
Last edited:
You are continuing to argue from two false premises:AndrewT said:
1. that Q=0.7 for first order. For first order there is no Q!
2. that 'combined Q'=0.7 for all odd-order Butterworth. For 3rd, 4th etc. order there is no Q!
Now it is possible that we are arguing at cross-purposes. It may be true to say that for a Butterworth characteristic the product of second-order Qs equals 1 (for odd-order) or 0.7 (for even-order). Thus if you pretend that a first order has Q=0.7 (it doesn't - Q is undefined) and you pretend that for Nth order the Q is the product of individual Qs (it isn't - total Q is undefined) then you will get your result - although it doesn't help you find the individual Qs for each second-order factor. Maybe it is the difference between a technician/engineer vs. a scientist/mathematician.
The 4-pole coefficients are something like sqrt(2-sqrt(2)) and sqrt(2+sqrt(2)), or possibly the reciprocal of those.
Basically you are looking for a set of polynomials P(s) in s such that the product of the polynomial and its complex conjugate PP^* gives just (1+ s^2N) - the Butterworth response. The first is 1+s: (1+s)(1-s)=1+s^2. Second is 1+sqrt(2)s+s^2: (1+sqrt(2)s+s^2)(1-sqrt(2)s+s^2)=1+s^4 etc. Above second you factorise into seconds (plus a first for odd-order). The book I have gives factors up to 8th-order.
The TI datasheet Ive just got gives a table for up to 10th order, butterworth bessel and chebychev 3dB
http://www.google.co.uk/url?sa=t&rc...s6-KHrmoplqWdIZGonErkYw&bvm=bv.51495398,d.ZGU
Im also a little familiar with the s-plane format of TFs but my math isnt so good, and im not all that sure the TF would help me (or that I would be able to use it to my advantage)
1st: 1 / (ns+1)
2nd: 1/ (ns^2 + ms + 1)
thats where id finish higher orders I could probably derive, but I hate such exercises in Math....
Id have to study Linkwitz' site further to learn more about his curves, as it stands im stuck with 4th order plus HP cap for the highpass, and 5th order (RC+4th order) for lowpass.
I need that RC for shelving the filter.
I think the sim looks promising. Opinions on what ive done?
http://www.google.co.uk/url?sa=t&rc...s6-KHrmoplqWdIZGonErkYw&bvm=bv.51495398,d.ZGU
Im also a little familiar with the s-plane format of TFs but my math isnt so good, and im not all that sure the TF would help me (or that I would be able to use it to my advantage)
1st: 1 / (ns+1)
2nd: 1/ (ns^2 + ms + 1)
thats where id finish
higher orders I could probably derive, but I hate such exercises in Math....Id have to study Linkwitz' site further to learn more about his curves, as it stands im stuck with 4th order plus HP cap for the highpass, and 5th order (RC+4th order) for lowpass.
I need that RC for shelving the filter.
I think the sim looks promising. Opinions on what ive done?
Last edited:
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.