3-nd order SVF ?

Hello,

I've just started learning filter basics. If my question is dumb, I'm sorry in advance for it:

Here is 2-nd order SVF from Vadim Zavalishin's 'The Art of VA Filter Design':


How would the diargam of the 3rd order filter look like, based on same integrators?

Adding 1st order filter on the output definitely does a job, but resonance peak has to be narrower if we want the same amp value at cutoff frequency.
I built two 4th order filters: one based on two 2nd order SVFs from the above figure, and another one based on one 2nd order and two 1st order filters. Then I adjusted their R values to get equal amplitude values at cutoff point, and realized that resonance peak is much narrower in the filter containing 1st order filters, because we have to increase R of 2nd order filter to compensate the lack of resonance in 1st order filters.

I also tried to take an average of outputs of 2-order and 4-order filters to get a kind of 3-order one, but it causes slope distortion:

Is there any way to implement 3rd integrator into the scheme to get pure 3rd order filter with no ladder structure?

 

Check the footnote 1 below that picture. This should give you an idea how to build a 3-pole version of that filter, although I'm not sure whether anyone refers to it as an SVF (such filter design is not common anyway).

Edit: although you will end up with pretty much the same frequency response as if you simply put a 1-pole after the SVF. The trick is how you play around with the coefficients (like varying the cutoff of the 1-pole relatively to the SVF). As for the resonance peak, the resonance comes from the essentially complex poles of the filter, and those can be produced only by even filter orders. That is, the 3rd pole can not add any further resonance to the filter, because the respective pole will be purely real.

 

It is important to understand, that in the LTI (linear time-invariant) case (no nonlinearities, no audio-rate modulations), the properties of a filter are fully encoded in its transfer function. That is, any two filters with the same transfer function sound (almost) exactly the same. "Almost" covers things like floating-point precision issues and modulations which are still "sufficiently fast", whatever it means. Particularly you can get an exact match of a ladder filter sound by chaining two (properly tuned) 2-pole SVFs.

Therefore, the reasons to experiment with various filter topologies would be (in approximate order of importance):
- convenience of filter parameter control
- nonlinearities in the filter
- audio-rate modulation behavior (I'd expect that most analog-derived topologies will perform very similarly here, but I don't have even an experimental proof of that)
- precision issues

If those reasons do not apply, a chain of 2- and 1-pole SVFs could be good enough for most puproses.

 

Thanks for the clear and detailed explanation.
Now i see that the fun starts somethere close to this point:

But what does 'properly tuned' mean? Is just connecting 2 12db/oct filters with same cutoff frequency enough for making 24db/oct filter? Or there are tricks like arranging poles according some rules like this stuff?

For instance, is there a 'classic' place for pole of 1-pole filter to get 18db/oct?

 

Well, first we have to agree on what we understand as "classic filters". Such things as Butterworth/Chebyshev/Elliptic etc designs were use for long times in electronics, however these filters are not designed to be used in synthesizers. The biggest limitation of the analog filters was that it was not that easy to implement voltage-controlled cutoff (which is needed to apply envelopes, lfos and other control signals to the filter). The cutoff was controlled by manually turning a resistor or capacitor knob. AFAIK, Bob Moog was the first one to come up with a reasonable usable voltage-controlled filter circuit, by "abusing" the variation of the differential resistance of transistors in respect to the current passing through the transistor. Also, most of the filters didn't employ resonance and even for those which did, it was not designed to sound "musical".

Now, starting with Moog's filter, there were a number of various designs used in synthesizers. However, the practical goal of those designers was to implement a nice-sounding filter for a particular synth. The filters designed this way were 4-pole (based on ladder design) and 2-pole (including the SVF). I'm not aware of a 3-pole voltage-controlled filter used in any synths (it might well be that there was one, but it is not very known). For that reason the question of what is a classic 3-pole VCF might not have an answer.

As the "properly tuned" goes... First notice that a 2-pole LPF cannot have more than 2 degrees of freedom of control (you could also count the 3rd degree which simply controls the total gain, but it doesn't change the sound of the filter). Indeed, the denominator of the transfer function is a2*s^2+a1*s+a0. Firstly, we can factor out a2 into the total gain coefficient, thus we can assume a2=1. This leaves us with s^2+a1*s+a0, which has only two degrees of freedom corresponding to a1 and a0. Now a1 and a0 are not very intuitive control parameters, so we let a0=w^2 and we have s^2+(a1/w)*w*s+w^2. Now w would control the cutoff of the filter, simply shifting the amplitude and phase responses along the (logarithmic) frequency axis, as explained in the book. (a1/w) will control the direction, in which the poles are positioned relatively to the origin and thus will control the filter's resonance.

Notice that, even though there are 2 complex poles which should correspond to 4 degrees of freedom, the restriction that the denominator coefficients should be real leaves us only with 2 degrees. If the poles are real, then these two degrees correspond to real coordinates of the poles. If the poles are complex conjugate, the degrees correspond to the absolute magnitude and the argument of the poles.

As for the numerator of a 2-pole filter: b2*s^2+b1*s+b0. Firstly, it doesn't affect the positioning of the poles, hence it doesn't affect the resonance. The 3 terms of the numerator corresond to HP, BP and LP modes. A generic 2-pole filter's numerator will correspond to a mixture of these 3 modes. As for the LP filter, b0 coefficient affects only the overall gain of the filter and thus also can be ignored (in fact, b0 and a2 are usually chosen so that the gain of the filter at f=0 is 1). Thus, there are not too many options, how a 2-pole filter can be controlled. Any 2-pole is as good as another one as long as we stay in the LTI domain. The differences begin when we introduce nonlinearities or some other non-LTI effects come into play.

For a 4-pole filter the denominator a4*s^4+a3*s^3+a2*s^2+a1*s+a0 gives us 4 degrees of freedom (a4 can be factored into the total gain). These 4 degrees of freedom correspond to the 4-poles of the filter. So, in principle, a 4-pole lowpass filter could have 4 control parameters. However, Moog filter has only 2 control parameters, thus the two pairs of complex conjugate poles do not move independently. You can refer to the book for the formulas for the pole positions. Now, each pair of the poles can be simulated by a 2-pole filter where you can obtain the 2-pole coefficients from the respective positions of the poles. A generic 4-pole filter would have 5 different modes, corresponding to the 5 terms of the numerator (24dB LP, 6/18dB BP, 12/12dB BP, 18/6dB BP, 24dB HP). For the lowpass mode there is only the b0 coefficient, affecting the gain which can be obtained as the product of the numerators of the respective 2-pole lowpasses. The total gain of the Moog filters is 1/(1+k) and can be simulated by simply multiplying the input of the SVF LP series by 1/(1+k).

Of course you can choose any other settings for the two serial SVFs, resulting in some other 4-pole filter. Particularly you can emulate a 4-pole diode filter. However, there is no such thing as "classic" setting for a 4-pole, there is always some other analog design with a smaller than 4 number of controls, which will dictate how to tune the 2-poles. Because if there are 4 degrees of freedom, then you can obtain any combination of 2-pole SVFs in a 4-pole filter.

 

One important note. When replacing e.g. a ladder filter by its serial decomposition into two 2-poles, it's important that both 2-pole filters use a common BLT prewarping (which should be done for the same frequency as the ladder filter's prewarping). Otherwise you'll get a somewhat different response curve.