state and the disturbance similar to those used for noise-to-state stability analysis of stochastic systems (e.g., [@nunez2014; @zhang2016noise]). Specifically, in the next result, we provide an upper bound for the first moment of state by utilizing the second moment of disturbance.

\[TheoremDependentDisturbance\] Consider the closed-loop networked control system (\[eq:closed-loop-system\]). Suppose that the attacker’s interference power process $\{v(t)\in[0,\infty)\}_{t\in\mathbb{N}_{0}}$ satisfies Assumption \[SecondAssumption\]. Furthermore, suppose $\mathbb{E}[\|w(t)\|_{2}^{2}]<\infty$, $t\in\mathbb{N}_{0}$. If $$\begin{aligned}
 & (1-\hat{p}(\hat{v}))\|A+BK\|^{2}+\hat{p}(\hat{v})\|A\|^{2}<1,\label{eq:abkcond-difficult}\end{aligned}$$ then there exist $\hat{\mu},\hat{f}\geq0$, and $\hat{\theta}\in(0,1)$ such that for $t\in\mathbb{N}$, $$\begin{aligned}
\mathbb{E}[\|x(t)\|_{2}]\leq\hat{\mu}\hat{\theta}^{t}\|x_{0}\|_{2}+\hat{f}\max_{i\in\{0,\ldots,t-1\}}(\mathbb{E}[\|w(i)\|_{2}^{2}])^{\frac{1}{2}}.\label{eq:noise_to_state_three}\end{aligned}$$

The proof of this result relies on the following lemma.

\[LemmaSumTwo\] Suppose that the attacker’s interference power process $\{v(t)\in[0,\infty)\}_{t\in\mathbb{N}_{0}}$ satisfies Assumption \[SecondAssumption\]. Then for every $\gamma_{1}\geq0$, $\gamma_{0}\geq0$ that satisfy $$\begin{aligned}
(\gamma_{1}^{2}+2\gamma_{1}\gamma_{0})\hat{p}(\hat{v})+\gamma_{0}^{2} & <1,\label{eq:gamma_cond}\end{aligned}$$ there exists a scalar $f\geq0$ such that $$\begin{aligned}
\sum_{j=0}^{t-2}\Big(\mathbb{E}\Big[\big(\prod_{i=j+1}^{t-1}(\gamma_{1}l(i)+\gamma_{0})\big)^{2}\Big]\Big)^{\frac{1}{2}} & \leq f,\quad t\in\{2,3,\ldots\}.\label{eq:l_d1_ineq-1}\end{aligned}$$

For every $j\in\{0,1,\ldots,t-2\}$, $t\in\{2,3,\ldots\}$, we have $(\prod_{i=j+1}^{t-1}(\gamma_{1}l(i)+\gamma_{0})){}^{2}=\prod_{i=j+1}^{t-1}(\gamma_{1}l(i)+\gamma_{0})^{2}=\prod_{i=j+1}^{t-1}(\gamma_{1}^{2}l^{2}(i)+2\gamma_{1}\gamma_{0}l(i)+\gamma_{0}^{2}).$ Let $\alpha_{1}\triangleq\gamma_{1}^{2}+2\gamma_{1}\gamma_{0}$ and $\alpha_{0}\triangleq\gamma_{0}^{2}$. Since $l^{2}(i)=l(i)$, it follows that $$\mathbb{E}\Big[\big(\prod_{i=j+1}^{t-1}(\gamma_{1}l(i)+\gamma_{0})\big)^{2}\Big]=\mathbb{E}\Big[\prod_{i=j+1}^{t-1}(\alpha_{1}l(i)+\alpha_{0})\Big].\label{eq:gamma-alpha-eq}$$ By (\[eq:gamma\_cond\]), (\[eq:alpha\_ineq\]) holds. Therefore, it follows from Lemma \[LemmaDOne\] that $\mathbb{E}\Big[\prod_{i=j+1}^{t-1}(\alpha_{1}l(i)+\alpha_{0})\Big]\leq\mu\theta^{(t-j-1)}$, where $\mu\geq0$ and $\theta\in(0,1)$ are scalars that depend on $\alpha_{1}$ and $\alpha_{0}$. Letting $$\begin{aligned}
f & \triangleq\mu^{\frac{1}{2}}/(1-\theta^{\frac{1}{2}}),\label{eq:FDef}\end{aligned}$$ we obtain $$\begin{aligned}
 & \sum_{j=0}^{t-2}\Big(\mathbb{E}\Big[\big(\prod_{i=j+1}^{t-1}(\gamma_{1}l(i)+\gamma_{0})\big)^{2}\Big]\Big)^{\frac{1}{2}}\leq\sum_{j=0}^{t-2}\mu^{\frac{1}{2}}\theta^{\frac{1}{2}(t-j-1)}\nonumber \\
 & \quad=\mu^{\frac{1}{2}}\sum_{i=1}^{t-1}\theta^{\frac{1}{2}i}\leq\mu^{\frac{1}{2}}\sum_{i=0}^{\infty}\theta^{\frac{1}{2}i}=f,\label{eq:expineq-1}\end{aligned}$$ which completes the proof.

Lemma \[LemmaSumTwo\] enables us to deal with quadratic terms that involve the failure indicator $l(\cdot)$. We 