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Is a PI-controller considered an LTI system?

Intuitively it seems that the integral part would break the time-invariant requirement requirement, because the output depends on how wound up the integrator is.

On the other hand, the output of an RC circuit also depends on the capacitor charge, so this would also seem to be non-time-invariant, but this is the most common example of an LTI-system.

K0ICHI
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    Integration and differentiation are perfectly LTI. Every system that can be described by a set of homogeneous differential equations is. – Hilmar Nov 15 '23 at 12:17

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A wound-up integrator is an indication of non-zero initial conditions. Any linear, time-invariant system stops being so when one considers non-zero initial conditions.

In your example, if the capacitor was charged before application of a sinusoidal voltage, then it would not be considered LTI.

In terms of differential equation solutions, there are usually two: the homogeneous (unforced) solution and the particular (forced) solution. The homogeneous solution is generally what causes "transients" when there are non-zero initial conditions.

Peter K.
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