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Old 30th August 2013, 01:40 PM   #11
DF96 is offline DF96  England
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Quote:
Originally Posted by AndrewT
The standard RC is a Butterworth roll off.
That Butterworth roll off implies a Q = sqrt(2) (~0.7)
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.

Quote:
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.
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.
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Old 30th August 2013, 01:51 PM   #12
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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.
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Old 30th August 2013, 01:55 PM   #13
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my question should be: Is there a point when a stage Q>1 becomes a problem? I know Sallen Key with Q =>3 will be prone to oscillate, but is that the only limit?
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Old 30th August 2013, 02:00 PM   #14
AndrewT is offline AndrewT  Scotland
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There's something I am not following.
Quote:
That gives you the right answer by the wrong method for third-order.
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
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Old 30th August 2013, 02:16 PM   #15
AndrewT is offline AndrewT  Scotland
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When an active stage has Q>1 you must take care to ensure that the stage does not voltage clip.
Precede the high Q stage with the low Q stage to avoid clipping. But do check voltage levels in each stage.
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Old 30th August 2013, 02:30 PM   #16
AndrewT is offline AndrewT  Scotland
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Quote:
The two Q's are found from (sqrt(5)+1)/2 and (sqrt(5)-1)/2.
I don't recall this relationship when I have read numerous filter papers.
I tried inserting sqrt(4) for sqrt(5) hoping to get the 4pole coefficients, but it came out as 1.5 & 0.5 = 0.75
oops !
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Old 30th August 2013, 03:11 PM   #17
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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!
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Old 30th August 2013, 04:20 PM   #18
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Default Sometimes a picture speaks a thousand words...

my current passive setup sims like this:

Click the image to open in full size.Click the image to open in full size.

and next my rubbish attempt at active (making up the Qs as i went along....)

Click the image to open in full size.Click the image to open in full size.

Using:

Click the image to open in full size.Click the image to open in full size.

And then Using the suggested Qs (0.618 and 1.618):

Click the image to open in full size.Click the image to open in full size.

Using:

Click the image to open in full size.Click the image to open in full size.



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.
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Last edited by mondogenerator; 30th August 2013 at 04:31 PM. Reason: picture-mania
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Old 30th August 2013, 05:16 PM   #19
DF96 is offline DF96  England
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Quote:
Originally Posted by AndrewT
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?
You are continuing to argue from two false premises:
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.

Quote:
I tried inserting sqrt(4) for sqrt(5) hoping to get the 4pole coefficients, but it came out as 1.5 & 0.5 = 0.75
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.
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Old 30th August 2013, 06:42 PM   #20
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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&rct...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?
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Last edited by mondogenerator; 30th August 2013 at 06:49 PM.
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