four-pole state variable filters

[Toni Jovanovski]

At 15:12 19-11-98 -0500, Brian wrote:
>> Is this just a mistake ? The usual SVF has 2 poles.  You can get any
>> even order response by cascading such  2-poles.
>
>no mistake ... i realize casading two 2-pole filters could work.  however, i
>see some drawbacks to this method:
>
>1. the filters might have to be "tuned" together
>
>2. to get all the LP,HP,BP,notch functions, a selector switch would have to
>be used to select the correct output from the first stage to use in the
>second stage.  with a single, 4-pole filter, all the functions could be
>output at the same time.
>
>3. you can't change any of the filter coefficients independently in a casaded
>design - i really don't know if this is good or bad?
>
>here's a more general question, related to 3:
>
>- in a standard 2-pole SVF design, changing the resonance of the filter
>changes the coefficient of s ( right? ).

Well...yes.Generally the transfer function of a low pass filter is:

        H(s)=K/(s+a)(s+a)

Now , resonance is made when this second order pole splits to complex
conjugate:

            (s+b-j*c)(s+b+j*c)=s^2+2*b*s+b^2+c^2

>  what does the resonance do in a
>4-pole filter? does it just change the coefficient of s^3? what about s^2 and
>s?
>
>any ideas?
>

When two standard SVF are connected cascade with same control then :


               (s+b-j*c)^2*(s+b+j*c)^2=
               
               s^4+4*b*s^3+2*(3*b^2+c^2)*s^2+4*(b^3+b*c^2)*s+b^4+2*b^2*c^2+c^4


This is when you have same control.When the control is not same then:


                (s+b-j*c)*(s+b+j*c)*(s+d-j*e)*(s+d+j*e)=

                (s+2*b+b^2+c^2)(s+2*d+d^2+e^2)=.......

You will have varios tranfer functions...

                s^4+K1*s^3+K2*s^2+K3*s+K4

I hope this will help

TJ