I am working on a project to design an optimal control setting. Now, I am considering the quadratic form based on the project's physical requirement.
$J(x,u) = \int_0^t (x^T Q x + u^T R u) dt$
However, I am not sure whether the cross term $x^T S u$ should be added to the above equation.
My question is what is the physical meaning of the cross term $x^T S u$ and are there real applications that should be considered the cross term based on the physical requirement?
thanks
Jie
In order to clarify I will define how I notate a standard state space model, namely using
\begin{align} \dot{x} &= A\,x + B\,u, \\ y &= C\,x + D\,u. \end{align}
Given a physical system it can be noted that its state space realization is not unique, since one can always do a non-singular linear transformation to redefine an alternate state representation. Therefore, one might argue that the chosen values for $Q$ might also not have physical meaning when $x$ doesn't. Though, usually the state $x$ is often chosen to have a physical meaning, but I believe certain models using such representation result in some numerical issues and instead one could for example use a balanced realization. Additionally, converting a transfer function into a state space model might also yield a model which doesn't have physically meaningful state.
Instead, one could penalize the output $y$, since it is not affected by linear transformations and hopefully it does have a physical meaning as well. So instead of $x^\top Q\,x$ you could use $y^\top M\,y$, which after substituting in the definition of $y$ yields
$$ y^\top M\,y = x^\top C^\top M\,C\,x + 2\,x^\top C^\top M\,D\,u + u^\top D^\top M\,D\,u. $$
So if $D \neq 0$ this would yield a cross term between $x$ and $u$. Many physical systems do have that $D = 0$, however one might also want to penalize $\dot{y}$ (or even higher order derivatives of the output) in the cost function and might therefore result in a cross term.
It can be noted that adding in the cross term can reduce assured stability margins of LQR without cross term, which is also a bit discussed in the comments of this question.
