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30th August 2013, 02:40 PM  #11  
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Join Date: May 2007

Quote:
Quote:
For 5th order Butterworth you need Q=0.618 and Q=1.618 in the two secondorder 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. 

30th August 2013, 02:51 PM  #12 
diyAudio Member

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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Every new piece of knowledge pushes something else out of my brain  Homer.....................Simpson 
30th August 2013, 02:55 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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Every new piece of knowledge pushes something else out of my brain  Homer.....................Simpson 
30th August 2013, 03:00 PM  #14  
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Join Date: Jul 2004
Location: Scottish Borders

There's something I am not following.
Quote:
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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regards Andrew T. 

30th August 2013, 03:16 PM  #15 
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Join Date: Jul 2004
Location: Scottish Borders

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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regards Andrew T. 
30th August 2013, 03:30 PM  #16  
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Join Date: Jul 2004
Location: Scottish Borders

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 oops !
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regards Andrew T. 

30th August 2013, 04:11 PM  #17 
diyAudio Member

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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Every new piece of knowledge pushes something else out of my brain  Homer.....................Simpson 
30th August 2013, 05:20 PM  #18 
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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.
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Every new piece of knowledge pushes something else out of my brain  Homer.....................Simpson Last edited by mondogenerator; 30th August 2013 at 05:31 PM. Reason: picturemania 
30th August 2013, 06:16 PM  #19  
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Join Date: May 2007

Quote:
1. that Q=0.7 for first order. For first order there is no Q! 2. that 'combined Q'=0.7 for all oddorder Butterworth. For 3rd, 4th etc. order there is no Q! Now it is possible that we are arguing at crosspurposes. It may be true to say that for a Butterworth characteristic the product of secondorder Qs equals 1 (for oddorder) or 0.7 (for evenorder). 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 secondorder factor. Maybe it is the difference between a technician/engineer vs. a scientist/mathematician. Quote:
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)(1s)=1+s^2. Second is 1+sqrt(2)s+s^2: (1+sqrt(2)s+s^2)(1sqrt(2)s+s^2)=1+s^4 etc. Above second you factorise into seconds (plus a first for oddorder). The book I have gives factors up to 8thorder. 

30th August 2013, 07:42 PM  #20 
diyAudio Member

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 splane 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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Every new piece of knowledge pushes something else out of my brain  Homer.....................Simpson Last edited by mondogenerator; 30th August 2013 at 07:49 PM. 
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