Lyapunov criteria for discrete linear system with noise

Question

Consider constant model matrix $A$ and $B$, the Lyapunov criteria for system $x_{k+1}=Ax_k+Bu_k$ with state feedback input $u_k=Kx_k$ (K is designed matrix) is $P-(A+BK)P(A+BK)^\top>0$, where $P$ is positive definite matrix.

When the system is affected by bounded noise $w_k$, it becomes $x_{k+1}=Ax_k+Bu_k+w_k$. In this case, can $P-(A+BK)P(A+BK)^\top>0$ still guarantee the stability of closed-loop system?

Answer 1

The question is not very well-posed. Asymptotic stability is usually considered for unforced systems, so for systems with no inputs.

A useful concept is that of input-to-state-stability (ISS) that aims at dealing exactly at the properties of systems with inputs. It can be proven that for the system you consider that the system is ISS if and only if the system with no input is asymptotically stable; i.e. $A+BK$ is Schur stable. In this regard, the system can be considered as stable.

Another issue with the question is that $w_k$ is said to be noise, making the system stochastic while the above discussion involves purely deterministic arguments.

In this regard, we can assume now that the noise term $w_k$ is independent of $x_k$ and that it is stationary with mean $\mathbb{E}[w_k]=\bar w$ and second-order moment $\mathbb{E}[w_kw_k^T]=\Sigma$.

Then, one can see that

$$\mathbb{E}[x_{k+1}]=(A+BK)\mathbb{E}[x_k]+\bar w$$ which implies that $$\mathbb{E}[x_k]\to (I-A-BK)^{-1}\bar w\ \mathrm{as}\ k\to\infty$$ provided that $A+BK$ is Schur stable. In the particular case $\bar w=0$, the state converges to the origin in average.

Still assuming that $\bar w=0$, one can see that $$\mathbb{E}[x_{k+1}x_{k+1}^T]=(A+BK)\mathbb{E}[x_kx_k^T](A+BK)^T+\Sigma.$$ Whenever $A+BK$ is Schur stable, then we have that

$$\mathbb{E}[w_kw_k^T]\longrightarrow\sum_{k=0}^\infty (A+BK)^k\Sigma(A^T+K^TB^T)^k\ \mathrm{as\ }k\to\infty,$$

which shows that the system remains stable in mean-square, but not asymptotically stable in mean-square (as $\mathbb{E}[w_kw_k^T]$ does not go to zero).

Answer 2

The short answer is: no, it does not guarantee stability.

You can see it intuitively by considering what happens close to the origin. If stability was guaranteed, you would have that $x_k\rightarrow 0$, but this means that at some point the state update would be: $$ x_{k+1} = (A+BK)x_k + w_k \approx w_k \neq 0, $$ meaning that the system would be driven by the noise, preventing convergence.

More formally, Lyapunov criterion says that the quadratic energy function defined as $V(x)=x^TPx$ decreases over time, i.e., $V(x_{k+1})-V(x_k)<0$ (check that this is indeed equivalent to $P-(A+BK)^TP(A+BK)>0$ in the non-perturbed case). If you add noise, you get: $$ V(x_{k+1})-V(x_k) = \left[(A+BK)x_k+w_k\right]^TP\left[(A+BK)x_k+w_k\right]-x_k^TPx_k, $$ which is not negative in general.