I have this ordinary Sallen Key bandpass filter: -


simulate this circuit – Schematic created using CircuitLab

I'm being asked, if the system is linear and/or time-invariant and/or causal. Furthermore, if the decomposition property holds for this system.

My answers

  • The system is linear, because the superposition principle holds.

  • The system is time-invariant, because a time-shifted input yields a correspondingly time-shifted output.

  • The system is causal, because the output depends only on present and past inputs.

  • The decomposition property is based on the principle of superposition. Since superposition holds, the system has the decomposition property.

As you can see, my answers doesn't have much argumentation, and I'm not 100% sure if what I've written is correct. Is there a reliable way to check if these properties, for example through simulation?

  • \$\begingroup\$ The property of linearity depends on the parts values - in particular the ratio R4/R5. \$\endgroup\$ – LvW Feb 9 at 9:15
  • \$\begingroup\$ @LvW I will add component values now. \$\endgroup\$ – Carl Feb 9 at 9:16
  • \$\begingroup\$ Background of my comment: When the bandpass works correctly within the quasi-linear range of the opamp, the circuit can be regarded as "linear". \$\endgroup\$ – LvW Feb 9 at 9:55
  • \$\begingroup\$ The decomposition property is not widely referenced in circuit analysis texts. This filter circuit without source Vin can be decomposed in many ways. One way is to divide it to two 1st order subsystems. One of them has has 3 ports. It contains parts R1, C2 and R3. The rest belong to a 2-port. What kind of decomposition did you think when writing your question? Maybe something which talks about the effect to the signal, not the structure of the circuit? \$\endgroup\$ – user287001 Feb 9 at 10:41
  • \$\begingroup\$ The decomposition property I'm talking about involves the differential equation for the system. If we have our differential equation \$Q(D)y(t) = P(D)x(t) \$. If the decomposition property holds, then we can write it like this: \$Q(D)y_{zi}(t)+Q(D)y_{zs}(t) = P(D)x(t) \$, where \$y_{zi} \$ is the zero input response and \$y_{zs} \$ is the zero state response. Can you follow me @user287001 ? \$\endgroup\$ – Carl Feb 9 at 10:55

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