# Integrators and the mechanics of a state-variable filter

I'm trying to understand how this topology of state-variable filter with summing amplifier produces a band-pass filter at the second integrator's output, in the context a parametric audio equalizer. Full explanations I've read are based on equations that I'm not prepared to understand. With all my respect to those who can follow complex equations, I'm just trying to grasp the logic behind it, aware that it's not a simple linear process.

So the source signal goes into the non-inverting input of the summing amplifier (OPA1) and comes out into the input of the first integrator. Integration forms a sort of low-pass filter without (or extremely low) cutoff-frequency, as I understand, so its output is its full bandwidth input with a 90º phase shift. The second integrator produces another 90º shift, which would represent a 180º inversion in relation to the source signal, but for something to happen it should come out with an effective cutoff-frequency, determined by the values of R4 and C2, forming a low-pass filter. I don't know why this LPF is formed, if there is no resistor in the feedback loop of OPA3.

Then the low-passed signal comes into the inverting input of the summing amplifier which, subtracting it from the source signal, produces an effective high-pass filter. That output signal becomes again the input of the first integrator, which now in this second cycle forms a bandpass somehow. I suppose that R3 and C1 form this time a low-pass filter, which in combination with the input HPF gives a BP. But again, how is a low-pass filter formed if an integrator has almost no cutoff frequency–C1 has no parallel resistor?

The band-pass output is fed back into the non-inverting input of the summing amplifier, which I conceive as a way to control the bandpass influence on the source signal and therefore the overall Q. Not sure if that's correct.

Any effort of clarification will be highly appreciated and remembered. I've been dealing too long also in simulations trying to grasp the process without proper success.

UPDATE:

Here a reformulation of what puzzles me about this filter. There are two types of SVF, one with the input signal at the non-inverting (+) amplifier and another (more typical) where the signal comes into the inverting (-) input. This is of the the first type and here a simulation of each output.

As seen, In a non-inverting SVF the input is in phase with LP, 90º out of phase with BP and 180º out of phase with HP.

LP is the result of two consecutive 90º phase shifts carried out by the integrators, so it is logically 180º inverted in relation to the input signal. LP is summed into the inverting input of the differential amplifier, therefore brought back into phase to the input signal that comes in at the non-inverting input. During the inverting process, their common low frequencies are subtracted and an out-of-phase HP signal is brought out at the differential amplifier's output. That's my humble descriptive interpretation. But why is the HP out of phase in relation to the input signal? It is formed by the remaining high frequencies of the input signal, which comes at the non-inverting input of the differential amplifier, so in my limited understanding HP should be in phase with the input signal.

• @Domingo Figure 32(d) from here or Figure 5-55 from here are what I tend to imagine as the archtypes for state-variable. Anyway, good you found a youtube. The reason that the integrators work in this case is that their outputs are fed backwards towards what feeds them. So this stabilizes their behavior rather than causing them to rail. Dec 9, 2023 at 21:59
• @Domingo With the diagrams I show (not yours) the 2nd and 3rd stages are basically -1/(rcs) stages (relative to their inputs.) So the 3rd stage is now already then 1/(rcrcs^2) of the 1st stage's output. So the feedbacks to the 1st stage are -1/(rcs) from the band-pass output on the (+) and 1/(rcrcs^2) to the (-) input. So you get an s^2 numerator on the 1st stage (which is HP), then an s^2/s=s numerator on the 2nd stage (BP), then an s^2/s^2=s^0 numerator on the 3rd stage (LP). Dec 9, 2023 at 22:08
• @Domingo The '-1/rcs' stands for $\frac{-1}{R\,\cdot\, C\,\cdot\, s}$, where the $R$ is the resistor going to the (-) input and the $C$ is the capacitor connected between the output and the (-) input. The 's' has many facets to it but with a very powerful meaning that underlies all these facets. When multiplying by it then it may be seen as the $\frac{\text{d}}{\text{d}t}$ operator. When dividing by it, then it may be seen as the $\int \quad \text{d}t$ operator. The current in the capacitor depends on the time-derivative of the voltage across it. Basic behavior. Dec 10, 2023 at 3:18
• @Domingo There's a discussion of the filter here. If you skip over the math, you should be able to see from the rest of what's said there that the LP output is fed back (exactly 180-degrees out of phase because of the twice-integration that takes place in the 2nd and 3rd stages) and subtracted from the input. When you subtract out the low-pass from the signal, you get a high-pass result. So 1st stage's output is HP. Integrate that once to get back BP. Etc. But the text there is actually okay and better than I'd expected. Dec 10, 2023 at 4:09
• @Domingo Although I think the site I just linked may be adequate (but only you can say, for sure), if you feel the need for something more I may still give it a shot. The basic idea is to somehow subtract the low-pass of the input directly from the input signal located at the 1st stage. Then there's only the high-pass left over. The whole thing is a nice piece of work and especially good for high-Q, voltage-controlled tuning over wide ranges, quadrature outputs, and otherwise where all three outputs may be useful. Also, elliptical (cauer) filters are often built from these. Dec 10, 2023 at 8:48