Given a discrete-time LTI system as $$x_{i+1} = A x_{i} + B u_{i}$$ and suppose the feedback law $u_{i} = \mathcal{K}(x_{i})$ assymtotically stabilizes the closed-loop system. Now, consider the modified (i.e. delayed) input law $u_{i} = \tilde{\mathcal{K}}(x_{i-I})$ for some constant (but possibly unknown) $I \gg 1$ (e.g. 10). Given an initial condition $x_{0}$, what can I see about whether there exists some $i$ such that $\Vert x_{i} \Vert \geq \rho $? Does the state norm grow beyond some bound, given enough time? And does this depend on $A$ being (Schur) stable?
My intuition (and simulation) says that the state trajectory will eventually leave any finite set (regardless of $A$ being Schur ), as we are basically injecting an arbitrary input into the system, which is not related to the current state. Thus, the system is running in open loop. However, how do I prove this, or at least make this more mathematically rigorous? Any help would be appreciated!
P.S. This is related to replay attacks in control systems which I'm investigating.
If your controller is linear, then you can rewrite the closed-loop system into the form
$$z(k+1)=M_0z(k)+M_dz(k-d)$$
where $d$ is the constant delay. The state $z$ will contain here the state of the system and that of the controller.
The above system can be rewritten as
$$ \begin{bmatrix} z(k+1)\\ \hline z(k)\\ \vdots\\ z(k-d+2)\\ z(k-d+1)\\ \end{bmatrix}=\left[\begin{array}{c|cc} M_0 & 0_{n\times n(d-1)} & M_d\\ \hline I_n & 0_{n\times n(d-1)} & 0_{n\times n}\\ 0_{n(d-1)\times n} & I_{n(d-1)} & 0_{n(d-1)\times n} \end{array}\right]\begin{bmatrix} z(k)\\ \hline z(k-1)\\ \vdots\\ z(k-d+1)\\ z(k-d)\\ \end{bmatrix}. $$
The stability of the above system can be established by looking at the stability of the matrix, which is turn equivalent with the stability of the time-delay system.
When the time-delay $d(k)$ is time-varying with an upper bound of $\bar d$, then the system rewrites as
$$ \begin{bmatrix} z(k+1)\\ \hline z(k)\\ \vdots\\ z(k-\bar d+2)\\ z(k-\bar d+1)\\ \end{bmatrix}=\left[\begin{array}{c|c} Q_0(k) & Q_d(k)\\ \hline \begin{array}{c} I_n\\ 0_{n(\bar d-1)\times n} \end{array} & \begin{array}{cc} 0_{n\times n(\bar d-1)} & 0_{n\times n}\\ I_{n(\bar d-1)} & 0_{n(\bar d-1)\times n} \end{array} \end{array}\right]\begin{bmatrix} z(k)\\ \hline z(k-1)\\ \vdots\\ z(k-\bar d+1)\\ z(k-\bar d)\\ \end{bmatrix}, $$ where $Q_0(k)=M_0+M_d$ if $d(t)=0$, $M_0$ otherwise, and $Q_d(k) = e_{d(k)}^T\otimes M_d$ for $d(k)>0$, $0$ otherwise.
Then, this system can be analyzed using discrete-time LTV methods or switched systems methods. There are necessary and sufficient stability conditions for such systems, but they are difficul to apply. So, you will most likey have to turn to sufficient stabiliy conditions here.
Finally, when the delay is stochastic, time-varying, and evolves acording to a finite-state Markov chain described by $p(k+1)=p(k)\Pi$. The stability of the stochastic system can be anlyzed using the associated theory. Necessary and sufficient stability conditions exist.