Hi, I've been working on trying to implement a 4rth order butterworth from 2 biquads. I know the coefficients for these two filters will not be the same because then I get Linkwitz-Riley which I do not want. I looked through some old threads and someone pointed out that the Fc of each of the two necessary biquads will be the same, however the Q factor of each will be different. Someone else has told me this as well, but how can this be since the pole locations are different for each filter(because each biquad has one complex conjugate pair)? Also, if this is true, then the a0,a2 coefficients of each biquad should be the only ones that change in each filter, since they seem to be the only ones that depend on the Q factor, but in going through the math so far, this doesn't seem to be the case either? Obviously I am going astray somewhere but can someone point out where I am in error? Thanks, Ethan

# damping in biquads for butterworth filter?

Started by ●April 8, 2009

Reply by ●April 8, 20092009-04-08

*update* Okay, I found my mistake in the math, so I can see now that a0 and a2 are indeed the only coefficients that change. Which seems to confirm what others have said, namely that the fc will be the same for each filter and only the Q will differ. Is there an easier way to do this? My design procedure was: 1)figure out what the pre-warped pole locations in the s-domain are 2)take each complex conjugate pole pair to create two second order sections 3)apply BLT to each biquad, with results that the filters are the same except for the Q factor. Since the Q factor is the only thing that changes, is there a way to figure out what the Q for each of these biquads will need to be to get a fourth-order butterworth, without going through all the tedium described above???? Thanks again, Ethan

Reply by ●April 8, 20092009-04-08

Oh, what a lazy idiot. Enjoy: Q0 = 0.541196 Q1 = 1.306563 VLV esfield wrote:> *update* > Okay, I found my mistake in the math, so I can see now that a0 and a2 are > indeed the only coefficients that change. Which seems to confirm what > others have said, namely that the fc will be the same for each filter and > only the Q will differ. > > Is there an easier way to do this? My design procedure was: > > 1)figure out what the pre-warped pole locations in the s-domain are > 2)take each complex conjugate pole pair to create two second order > sections > 3)apply BLT to each biquad, with results that the filters are the same > except for the Q factor. > > Since the Q factor is the only thing that changes, is there a way to > figure out what the Q for each of these biquads will need to be to get a > fourth-order butterworth, without going through all the tedium described > above???? > > Thanks again, > Ethan

Reply by ●April 8, 20092009-04-08

Thanks for the help, Vladimir! Wow. So all I had to do was find values of Q such that when multiplied together, they yeild 1/sqrt(2)?????? It seems so obvious now. So painfully, painfully obvious. Ethan

Reply by ●April 8, 20092009-04-08

esfield wrote:> Thanks for the help, Vladimir! > > Wow. > > So all I had to do was find values of Q such that when multiplied > together, they yeild 1/sqrt(2)??????There an infinite number of such Q pairs. Vlad cave you the only ones that work right.> It seems so obvious now. > > > So painfully, painfully obvious.Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●April 8, 20092009-04-08

Yeah, I just realized that.... so what are the other constraints on these values that force them to be the ones Vlad gave? Thanks, Ethan

Reply by ●April 8, 20092009-04-08

esfield wrote:> Yeah, I just realized that.... so what are the other constraints on these > values that force them to be the ones Vlad gave?Math. All low-pass Butterworth filters have all derivatives zero at f=0. That gives a 4th-order filter the response 1 + x^4 where x = w/w0. Solve for the roots. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●April 8, 20092009-04-08

On Apr 8, 12:34�pm, "esfield" <ethanstubblefi...@hotmail.com> wrote:> *update* > Okay, I found my mistake in the math, so I can see now that a0 and a2 are > indeed the only coefficients that change. �Which seems to confirm what > others have said, namely that the fc will be the same for each filter and > only the Q will differ. > > Is there an easier way to do this? �My design procedure was: > > 1)figure out what the pre-warped pole locations in the s-domain are > 2)take each complex conjugate pole pair to create two second order > sections > 3)apply BLT to each biquad, with results that the filters are the same > except for the Q factor. > > Since the Q factor is the only thing that changes, is there a way to > figure out what the Q for each of these biquads will need to be to get a > fourth-order butterworth, without going through all the tedium described > above???? > > Thanks again, > EthanEthan, You may like this: http://www.claysturner.com/dsp/Butterworth%20Filter%20Formulae.pdf Clay

Reply by ●April 8, 20092009-04-08

Thanks again Jerry. Please bear with me a bit longer! That gives pole locations at sqrt(j), sqrt(-j), -1/sqrt(j), -1/sqrt(-j). But, aren't these the pole locations for a second order butterworth? There are four poles but we only use the left half plane, so this only gives two poles so this is a second order filter, right? Also, it's not obvious to me how the Q values come from the pole locations. Can you please enlighten me? Clay, thanks for the link. I wish I'd had that reference when I started! Just a quick question: to transform the lpf into hpf, all I do is change the '1' in each numerator to 's^2', right?

Reply by ●April 9, 20092009-04-09

On Apr 8, 3:33�pm, "esfield" <ethanstubblefi...@hotmail.com> wrote:> Thanks again Jerry. �Please bear with me a bit longer! > > That gives pole locations at sqrt(j), sqrt(-j), -1/sqrt(j), -1/sqrt(-j). > > But, aren't these the pole locations for a second order butterworth? > There are four poles but we only use the left half plane, so this only > gives two poles so this is a second order filter, right? > > Also, it's not obvious to me how the Q values come from the pole > locations. �Can you please enlighten me? > > Clay, thanks for the link. �I wish I'd had that reference when I started! > Just a quick question: to transform the lpf into hpf, all I do is change > the '1' in each numerator to 's^2', right?I look at it as replacing each "s" by a 1/s - that inverts the frewquency response. For example 1/(1+s) -> transforms to 1/(1+1/s) now mult top and bot by s to clean up the expresion and you get s/(s+1) which is now your 1st order Butterworth highpass filter. IHTH, Clay