simulate this circuit – Schematic created using CircuitLab

We can introduce the model of \$u(s)\$ into the controller \$K(s)\$ to realize high accuracy control.

  • \$u(t) = 1\$, i.e. \$u(s) = 1/s\$. We use \$K(s) = K_p + \frac{K_i}{s}\$ (PI controller) to realize zero static error.
  • \$u(t) = \sin(\omega t)\$, i.e. \$u(s) = \frac{\omega}{s^2+\omega^2}\$. We use \$K(s) = K_p + \frac{K}{s^2+\omega^2}\$ to realize zero static error.

Now my problem is that, does this work for any other \$u(t)\$? Such as \$u(t) = \exp(\alpha t)\$?

P.S. I learnt that the high accuracy of PI or proportional-resonance controller is high gain. I can understand this for DC input (\$u(t) = 1\$) or sine input. But if this kind of controller works for other input, how to understand the gain here?

  • \$\begingroup\$ Put a "\" before each instance of "$" to get Latex working properly. Why do you call a PI controller a "proportional-resonance" controller? \$\endgroup\$ – Andy aka Nov 29 '18 at 12:56
  • \$\begingroup\$ @Andyaka proportional-resonance controller (PR controller) is the controller shown in the 2nd bullet \$\endgroup\$ – Alexander Zhang Nov 29 '18 at 13:00

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