Suppose we have a real-valued square matrix $A$ that is Hurwitz, i.e., all eigenvalues of
$A$ have strictly negative real parts.
I want to show that given a scalar $h>0$, if A is Hurtiwz, then $e^{Ah}$ is Schur, i.e., all the eigenvalues of $e^{Ah}$ are inside the open unit disc.
This statement is based on the fact that if $\lambda$ is an eigenvalue of A, then $e^{\lambda h}$ is the eigenvalue of $e^{Ah}$. If the real part of $\lambda$ is negative, the magnitude of $e^\lambda h$ is less than. Is my argument correct?
A related question is about the sampling period of a linear stochastic control system. Suppose we have a continuous-time linear stochastic system:
$$
dx = Ax(t)dt + Bu(t)dt + Cdw(t),\ \ \ x(0) = x_0,
$$
where $\{w(t),t \geq 0\}$ is a real-valued standard Wiener process. The state is only measured periodically with sampling period $h$ and the control takes the form of $K\hat{x}$, where between two sampling instances $[kh,(k+1)h)$, the estimate $\hat{x}(t)$ evolves according to
$$
d\hat{x}(t) = (A+BK)d\hat{x}(t),
$$
for $t\in[kh,(k+1)h)$ for $k=1,2,\cdots$.
Whenever the state is measured, the estimate is set to be the actual state $\hat{x}(kh)= x((k+1)h)$.
We can write
$$
\begin{bmatrix}
dx(t) \\ d\hat{x}(t)
\end{bmatrix}= \begin{bmatrix}
A & BK\\
0 & A+BK
\end{bmatrix}\begin{bmatrix}
dx(t) \\ d\hat{x}(t)
\end{bmatrix} + \begin{bmatrix}
Cdw\\0
\end{bmatrix},
$$
for $t\in[kh,(k+1)h)$.
Then, the discretized system is
$$
z((k+1)h) = e^{\tilde{A}h}z(kh) + \begin{bmatrix}
v_k\\0
\end{bmatrix},
$$
where
$$
z(t)= \begin{bmatrix}
x(t)\\ \hat{x}(t)
\end{bmatrix},\ \ \
\tilde{A} = \begin{bmatrix}
A & BK\\
0 & A+BK
\end{bmatrix},\ \ \ v_k = \int_{kh}^{(k+1)h} e^{A[(k+1)h-\tau]}Cdw(\tau).
$$
Using the discretized system, we can show that if $A+BK$ is Hurwitz and $h$ is finite, $\mathbb{E}x(t)\rightarrow 0$ and $\sup_{t\geq 0} \mathbb{E}\Vert x(t) \Vert^2\leq \infty$.
What if we fix the control $u(t) = Kx(kh)$ for $t\in[kh,(k+1)h)$ for every $k$. What is the upper bound on the sampling period $h$ such that the system is stable in the sense that $\mathbb{E}x(t)\rightarrow 0$ and $\sup_{t\geq 0} \mathbb{E}\Vert x(t) \Vert^2\leq \infty$? Any textbook or reference that discuss this problem?
Best Answer
It is a well-known fact that if $A$ is Hurwitz stable, then $\exp(Ah)$ is Schur stable for all $h>0$.
This can be seen from the definition of the exponential. Assume for simplicity that $A$ is diagonalizable. Then, there exists an invertible matrix $P$ such that $A=PDP^{-1}$ where $D$ is a diagonal matrix that contains the eigenvalues of $A$ on the diagonal.
Then, we have that
$$\exp(Ah)=\sum_{i=0}^\infty\dfrac{A^ih^i}{i!}=\sum_{i=0}^\infty\dfrac{(PDP^{-1})^ih^i}{i!}=P\sum_{i=0}^\infty\dfrac{D^ih^i}{i!}P^{-1}=P\exp(Dh)P^{-1}.$$
We can conclude that the eigenvalues of $\exp(Ah)$ are simply the exponentiated eigenvalues of $A$ scaled with $h>0$. Therefore, if all the eigenvalues of $A$ have negative real part, then all the eigenvalues of $\exp(Ah)$ will lie within the unit disc. The general case can be proven analogously.
For your second question. You should model your sampled-data systems using a hybrid formulation so that you do not need to discretize and handle the difficult integral term that you have. This may seem more complex but it is not, and the hybrid systems formulation actually circumvents a lot of problems.
For instance, you can express your system as an impulsive system of the form $$\begin{array}{rcll} dz(t)&=&\bar A z(t)dt + Edw(t)&,\ t\ne t_k\\ z(t_k^+)&=&Jz(t_k)&,\ t=t_k \end{array}$$
where $t_k=kh$, $z=(x,u)$, $\bar{A}=\begin{bmatrix}A & B\\0 & 0\end{bmatrix}$, $E=\begin{bmatrix}C\\0\end{bmatrix}$, and $J=\begin{bmatrix}I & 0\\K & 0\end{bmatrix}$.
If you want to look at the synamics of the moments, then one can see that $$\begin{array}{rcll} \dfrac{d}{dt}\mathbb{E}[z(t)]&=&\bar A \mathbb{E}[z(t)]&,\ t\ne t_k\\ \mathbb{E}[z(t_k^+)]&=&J\mathbb{E}[z(t_k)]&,\ t=t_k \end{array}$$
which is going to be exponentially stable if and only if $$\exp(\bar AT)J=\begin{bmatrix}\exp(Ah) & \Psi(h)\\0 & I \end{bmatrix}\begin{bmatrix}I & 0\\K & 0 \end{bmatrix}=\begin{bmatrix}\exp(Ah)+\Psi(h)K & 0\\K & 0 \end{bmatrix}$$ is Schur stable where $$\Psi(h)=\int_0^h\exp(As)Bds.$$
This will be the case if and only if $\exp(Ah)+\Psi(h)K$ is Schur stable. If we assume that $A+BK$ is Hurwitz, then $\exp(Ah)+\Psi(h)K$ will be Schur stable for any sufficiently small $h>0$. To see this, just observe that a Taylor expansion at $h=0$ of $\exp(Ah)+\Psi(h)K$ yields
$$\exp(Ah)+\Psi(h)K=I+h(A+BK)+o(h)$$ from which the result follows.
For the second-order moments, we have that
$$\begin{array}{rcll} \dfrac{d}{dt}\mathbb{E}[z(t)z(t)^T]&=&\bar A [z(t)z(t)^T]+[z(t)z(t)^T]\bar A^T+EE^T&,\ t\ne t_k\\ \mathbb{E}[z(t_k^+)z(t_k^+)^T]&=&J\mathbb{E}[z(t_k)z(t_k^+)^T]J^T&,\ t=t_k \end{array}$$
and the system is stable provided that the operator $L:\mathbb{S}_{\succeq0}^n\mapsto\mathbb{S}_{\succeq0}^n$ defined as
$$L(X)=J\exp(\bar Ah)X\exp(\bar A^Th)J^T$$
has eigenvalues in the unit disc. Assuming that $J\exp(\bar Ah)$ is diagonalizable, it turns out that the eigenvalues of this operator are given by $|\lambda_i|^2$ where the $\lambda_i$'s are the eigenvalues of $J\exp(\bar Ah)$.
Of course, it is possible to to some continuous-time analysis using looped-functionals, Lyapunov functionals, clock-/timer-dependent Lyapunov functions, etc. Those methods are used to design controllers that can guarantee that the closed-loop system will be stable for some range of values for the sampling period or even deal with the aperiodic case where sampling is not periodic. They can also be used to find the largest admissible sampling period or even find the controller that maximizes the largest value of the sampling period.
For resources, you may look at the book by Goebel, Sanfelice and Teel on hybrid systems.
There is also an issue with your formulation is that it will not work when $A$ is unstable while it is possible to stabilize an unstable system with sampled-data control law.