= Integral feedback
{title2=$\dot z=r-y$}
Integral feedback makes a <control signal> depend on the accumulated tracking error. With <reference signal> $r(t)$ and <measured output> $y(t)$, introduce
$$
e(t)=r(t)-y(t),\qquad \dot z(t)=e(t),\qquad u(t)=k_Iz(t).
$$
At any finite equilibrium of this controller, $\dot z=0$ forces zero tracking error. This explains why a stable integral-control loop can reject a constant <control disturbance> despite an imperfect model.
For the scalar <controlled plant> $\dot y=-ay+u+d$, with constants $a,k_I>0$ and constant $r,d$, equilibrium has $y=r$ and $u=ar-d$. Perturbations satisfy $\lambda^2+a\lambda+k_I=0$, whose roots have negative real parts, so this particular closed loop is stable. The conclusion depends on the plant and gain: adding integration does not guarantee <stability theory> for an arbitrary plant. Actuator limits can also let the accumulated error grow while the requested input cannot be delivered.
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