Hi, I'm trying to combine multiple LP and HP biquads to get higher order, but I still want to keep the -3db cut at the center frequency. On stage has 12db/oct, to get 24db/oct I can use one with Q=0.5 and the other with sqrt(2), but in reality I found better results with 1.27 and 0.55, but it's similar. Anyway I was unable to get higher slopes, mainly 36db/oct and 48db/oct. It is easy to get them with -6db at center as used in linkwitz-riley crossovers, but I have no idea how to make them with -3db. What's the trick? Thank you all. jungledmnc
How to combine multiple biquad LP/HP stages to get 36db/oct and higher order filters?
Started by ●September 5, 2011
Reply by ●September 5, 20112011-09-05
On 9/5/2011 8:44 AM, jungledmnc wrote:> Hi, > > I'm trying to combine multiple LP and HP biquads to get higher order, but I > still want to keep the -3db cut at the center frequency. > > On stage has 12db/oct, to get 24db/oct I can use one with Q=0.5 and the > other with sqrt(2), but in reality I found better results with 1.27 and > 0.55, but it's similar. Anyway I was unable to get higher slopes, mainly > 36db/oct and 48db/oct. It is easy to get them with -6db at center as used > in linkwitz-riley crossovers, but I have no idea how to make them with > -3db. What's the trick?It's not a trick, it's a method. Design the filter as a whole, then implement it as a cascade of quadratics, adding a linear section if the order is odd. The only "trick" is pairing the right numerator with the right denominator. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●September 5, 20112011-09-05
>It's not a trick, it's a method. Design the filter as a whole, then >implement it as a cascade of quadratics, adding a linear section if the >order is odd. The only "trick" is pairing the right numerator with the >right denominator. > >JerryAaaha, thanks, so it's quite more complicated than just setting nice Q values, right?
Reply by ●September 5, 20112011-09-05
Butterworth Filter Denominator Factors Order Quadratic Factors 1 s + 1.00000 2 s^2 + 1.41421s + 1.00000 (Q = 0.70711) 3 s^2 + 1.00000s + 1.00000 (Q = 1.00000) s + 1.00000 4 s^2 + 0.76536s + 1.00000 (Q = 1.30657) s^2 + 1.84776s + 1.00000 (Q = 0.54120) 6 s^2 + 0.51764s + 1.00000 (Q = 1.93184) s^2 + 1.41421s + 1.00000 (Q = 0.70711) s^2 + 1.93186s + 1.00000 (Q = 0.51764) 8 s^2 + 0.39018s + 1.00000 (Q = 2.56292) s^2 + 1.11114s + 1.00000 (Q = 0.89998) s^2 + 1.66294s + 1.00000 (Q = 0.60134) s^2 + 1.96158s + 1.00000 (Q = 0.50979) Filters other than Butterworth will have different factors. Greg
Reply by ●September 5, 20112011-09-05
>Butterworth Filter Denominator Factors > >Order Quadratic Factors > >1 s + 1.00000 > >2 s^2 + 1.41421s + 1.00000 (Q = 0.70711) > >3 s^2 + 1.00000s + 1.00000 (Q = 1.00000) > s + 1.00000 > >4 s^2 + 0.76536s + 1.00000 (Q = 1.30657) > s^2 + 1.84776s + 1.00000 (Q = 0.54120) > >6 s^2 + 0.51764s + 1.00000 (Q = 1.93184) > s^2 + 1.41421s + 1.00000 (Q = 0.70711) > s^2 + 1.93186s + 1.00000 (Q = 0.51764) > >8 s^2 + 0.39018s + 1.00000 (Q = 2.56292) > s^2 + 1.11114s + 1.00000 (Q = 0.89998) > s^2 + 1.66294s + 1.00000 (Q = 0.60134) > s^2 + 1.96158s + 1.00000 (Q = 0.50979) > >Filters other than Butterworth will have different factors. > >GregWhau! Thank you Greg!
Reply by ●September 5, 20112011-09-05
On 9/5/11 9:38 AM, jungledmnc wrote:>> It's not a trick, it's a method. Design the filter as a whole, then >> implement it as a cascade of quadratics, adding a linear section if the >> order is odd.i would call that a "first order section", Jerry. and, to save dmnc from confusion, i would stay the hell away from odd order. (36 dB/oct would be 6th order anyway.)>> The only "trick" is pairing the right numerator with the right denominator. > > Aaaha, thanks, so it's quite more complicated than just setting nice Q > values, right?actually, it *isn't* more complicated than just setting nice Q values. i am not sure what Jerry is trying to tell you, but the fact it, if you're doing it the butterworth way, each section is tuned to the same resonant frequency (the -3 dB corner frequency of the whole thing) but each section will have different a Q. if you're doing it the Tchebyshev way, then the different sections have somewhat different resonant frequencies (as well as different Q) and it's more complicated. look at Greg B's response. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●September 5, 20112011-09-05
On Mon, 05 Sep 2011 11:46:55 -0400, robert bristow-johnson <rbj@audioimagination.com> wrote:>if >you're doing it the butterworth way, each section is tuned to the same >resonant frequency (the -3 dB corner frequency of the whole thing) but >each section will have different a Q.Also note that, provided you _define_ the Q of a first-order section to be equal sqrt(2)/2, the product of the Qs of all sections is equal to sqrt(2)/2. Greg
Reply by ●September 5, 20112011-09-05
On 9/5/11 2:44 PM, Greg Berchin wrote:> On Mon, 05 Sep 2011 11:46:55 -0400, robert bristow-johnson > <rbj@audioimagination.com> wrote: > >> if you're doing it the butterworth way, each section is tuned to the >> same resonant frequency (the -3 dB corner frequency of the whole thing) >> but each section will have different a Q. > > Also note that, provided you _define_ the Q of a first-order section to be equal > sqrt(2)/2, the product of the Qs of all sections is equal to sqrt(2)/2.well, that's a curious factoid i hadn't known. i'm not doing that now, but i may (sometime in the future) try to justify that on the basis of the s-plane pole locations equally spaced on the left half-circle of radius 2*pi*f0. i'm sure that the Q of the 2nd-order section is related to the angle of the complex pole to the real axis, but i don't know what it is. does anyone have that formula handy? -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●September 5, 20112011-09-05
On Mon, 05 Sep 2011 16:58:14 -0400, robert bristow-johnson <rbj@audioimagination.com> wrote:>well, that's a curious factoid i hadn't known. i'm not doing that now, >but i may (sometime in the future) try to justify that on the basis of >the s-plane pole locations equally spaced on the left half-circle of >radius 2*pi*f0.You could also use RHP poles and still have a Butterworth filter, but it would be unstable.>i'm sure that the Q of the 2nd-order section is related to the angle of >the complex pole to the real axis, but i don't know what it is. does >anyone have that formula handy?I recall encountering this in control systems analysis, but I've long forgotten the details. According to http://www.edaboard.com/thread164378.html#post694250, "The Q-factor of a complex pole pair is defined using the angle between the negativ real axis of the complex s-plane and the vector directed to the pole: Q=1/(2�cos[angle])." I haven't attempted to verify its accuracy. Greg
Reply by ●September 6, 20112011-09-06
On Sep 5, 1:58�pm, robert bristow-johnson <r...@audioimagination.com> wrote:> On 9/5/11 2:44 PM, Greg Berchin wrote: > > > On Mon, 05 Sep 2011 11:46:55 -0400, robert bristow-johnson > > <r...@audioimagination.com> �wrote: > > >> if you're doing it the butterworth way, each section is tuned to the > >> same resonant frequency (the -3 dB corner frequency of the whole thing) > >> but each section will have different a Q. > > > Also note that, provided you _define_ the Q of a first-order section to be equal > > sqrt(2)/2, the product of the Qs of all sections is equal to sqrt(2)/2. > > well, that's a curious factoid i hadn't known. �i'm not doing that now, > but i may (sometime in the future) try to justify that on the basis of > the s-plane pole locations equally spaced on the left half-circle of > radius 2*pi*f0. > > i'm sure that the Q of the 2nd-order section is related to the angle of > the complex pole to the real axis, but i don't know what it is. �does > anyone have that formula handy? > > -- > > r b-j � � � � � � � � �r...@audioimagination.com > > "Imagination is more important than knowledge."I think that's: atan(sqrt(4*Q^2-1))
