What does the reference vector $r(t)$ mean when it comes to LQG/LQR controllers? Because if I set $r(t) = 0$, then my controller will place the states at the equilibrium point. And if i changes my reference to $r(t) = 5$, the equilibrium point will move up or down.
But what does the reference vector $r(t)$ mean? Do I changes the eigenvalues of the system? Reference $r(t) = 5$ is just a number without any unit.
Assuming that $\hat{x}(t)$ converges to $x(t)$ in steady state, the system is asymptotically stable (so $A-L\,C$ and $A-B\,K$ are Hurwitz) and $r(t)$ is periodic with period $T$, and thus can be written as a Fourier series
$$ r(t) = \sum_{n=-\infty}^\infty c_n\,e^{\frac{2\,\pi\,i\,n\,t}{T}}, $$
with $c_n$ a complex valued vector with the same dimensions as $u$.
Under these assumptions when the transient of the system has died out the response of the system can be expressed as
$$ x(t) = \sum_{n=-\infty}^\infty \left(\frac{2\,\pi\,i\,n}{T}\,I - A + B\,K\right)^{-1} B\,c_n\,e^{\frac{2\,\pi\,i\,n\,t}{T}}. $$
For example when it is just a constant $r(t)=c_0$ then in steady state $x(t)=(B\,K-A)^{-1} B\,c_0$ and $y(t) = C\,(B\,K-A)^{-1} B\,c_0$. It can be noted that $A-B\,K$ is assumed to be Hurwitz, so does not have eigenvalues at zero and therefore that inverse should exist.
The input $r(t)$ can be seen as feedforward. So for example if you have some constant reference $y_r(t)$ for the output you would like to track, then you could scale this reference to get the feedforward $r(t) = \left(C\,(B\,K-A)^{-1} B\right)^{-1} y_r(t)$. If that matrix is not square or invertable, then you could use the right (pseudo) inverse.