Exponential stability of a class of networked control systems with disturbed controllers

Abstract

This paper studies the exponential stability problem for a class of networked control systems (NCSs) with time delays and packet dropouts. By considering the disturbed state-feedback controller, the closed-loop NCS is modeled as a new discrete-time switched system. A sufficient condition is established for the exponential stability of the NCS under a packet-dropout rate. The state-feedback controller gain is obtained through the cone complementarity linearization approach. A numerical example is provided to show the effectiveness of the proposed method.

1 Introduction

With the fast development of network technology, the network is being applied to the control field by researchers. Networked control systems (NCSs) whose control loops are connected via communication networks offer many advantages such as low cost of installation, ease of maintenance, high resource utilization, simple installation. Hence, NCSs have received increasing attention [1–3].

In the meantime, the introduction of network makes the analysis and design of system complexity. Due to the sampling data transmission through the network, the time delay and packet dropout are always inevitable, which often cause deterioration of system performance and instability of system. It is well known that the stability is one of the most important problems in the controller design. Therefore, the stability of systems has attracted much research interest [4–9]. It should be pointed out that the results in the aforesaid references do not apply to the stability of NCS.

Compared to the NCSs with only packet dropouts or time delays [10–12], it is more difficult to analyze the NCSs with both time delays and packet dropouts. When considering the packet-dropout problem, it is significant to establish the quantitative relation between the packet-dropout rate and the stability of the NCS. On this topic, only a few results have been presented [13–16]. In [13], the plant considered is a discrete-time one, therefore, the result of the paper does not be applied to the NCS when the plant is a continuous-time system. In [14, 15], the NCS with a continuous-time plant is studied. It should be pointed out that the delay is a constant or takes values in a finite set. However, the results are invalid when one encounters infinite possible values for the delay. In [16], the delay considered can be an arbitrary value in a finite interval. A sufficient condition is obtained for the exponential stability of the closed-loop NCS with the plant being a continuous-time one. It is worthwhile noting that the controller gain is unchanged.

In practice, due to interference (for example, machine aging and measurement error), there is a certain change in the controller parameters. This change may destroy the stable performance of the closed-loop systems. At this time, if considering the constant controller gain in the analysis of the system, the system may exhibit a high degree of vulnerability, which motivates the present research.

This paper studies a class of NCSs with disturbed controllers. The delay can arbitrarily take values in a finite interval which is smaller than the sampling period. A new discrete-time switched NCS model is proposed. A sufficient condition is obtained for the exponential stability of the closed-loop NCS under the maximum packet-dropout rate. The discrete-time feedback controller gain is derived by solving a set of linear matrix inequalities with inversion constraints. A numerical example verifies the developed theory.

2 Model for networked control system

The structure of the NCS is shown in Figure 1. The plant is a continuous-time linear system described by

$$ \dot{x} = A_{p}x(t)+B_{p}v(t), $$

(1)

where $x(t)\in R^{n}$ is the state, $v(t)\in R^{m}$ is the plant input, $A_{p}$ and $B_{p}$ are constant matrices of appropriate dimensions.

Figure 1

The structure of the NCS.

In the NCS, the discrete-time state-feedback controller is event-driven, the sensor is time-driven and the sampling period is T. The zeroth-order hold device does not update the output value until the new value arrives. The network-induced delay $\tau_{k}$ satisfies $0\leq\tau_{\min}\leq\tau_{k}\leq\tau_{\max}< T$. $x(k)$ is the value of $x(t)$ at the sampling instant kT.

The output value of the disturbed state-feedback controller corresponding to $x(k)$ is denoted by $u(k)$,

$$ \begin{aligned} &u(k):=\widetilde{K}(k) x(k), \\ &\widetilde{K}(k)=K+\triangle(k), \end{aligned} $$

(2)

where K is the designed feedback gain and $\triangle(k)$ is the controller gain perturbation that satisfies

$$\triangle^{\mathrm{T}}(k)\triangle(k)\leq\delta^{2} I. $$

By considering the network-induced delay, the plant input is

$$ v(t)=\textstyle\begin{cases} \hat{u}(k-1), & \mbox{if }kT< t\leq kT+\tau_{k},\\ \hat{u}(k), & \mbox{if }kT+\tau_{k}< t\leq(k+1)T, \end{cases} $$

(3)

where

$$\hat{u}(k)=\textstyle\begin{cases} u(k), & \mbox{if }u(k)\mbox{ and }x(k)\mbox{ is successfully transmitted,} \\ \hat{u}(k-1), & \mbox{if }u(k)\mbox{ or }x(k)\mbox{ is lost during transmission.} \end{cases} $$

From (3), system (1) with sampling period T is discretized to

$$ x(k+1)=A x(k)+ B_{1} \hat{u}(k-1)+B_{0} \hat{u}(k), $$

(4)

where

$$ A=\exp\{A_{p}T\},\qquad B_{1}= \int_{T-\tau_{k}} ^{T}\exp\{A_{p}s\} \,\mathrm{d}s B_{p}, \qquad B_{0}= \int_{0} ^{T-\tau_{k}}\exp\{A_{p}s\} \,\mathrm{d}s B_{p}. $$

(5)

During each sampling period, two cases may arise, which can be listed as follows:

• Packet dropout happens; (4) can be written as

  $$ \tilde{x}(k+1) = \tilde{A}_{0}(k) \tilde{x}(k), $$

  (6)

  where

  $$ \tilde{x}(k)=\left [ \textstyle\begin{array}{@{}c@{}} x(k) \\ \hat{u}(k-1) \end{array}\displaystyle \right ], \qquad \tilde{A}_{0}(k)=\left [ \textstyle\begin{array}{@{}cc@{}} A & B \\ 0 & I \end{array}\displaystyle \right ],\quad B= \int_{0}^{T}\exp\{A_{p}s\}\,\mathrm{d}s B_{p}. $$

  (7)

• No packet dropout happens; (4) can be written as

  $$ \tilde{x}(k+1) = \tilde{A}_{1}(k)\tilde{x}(k) , $$

  (8)

  where

  $$\begin{aligned} \tilde{A}_{1}(k) =& \left [ \textstyle\begin{array}{@{}cc@{}} A+B_{0}(\tau_{k})\widetilde{K}(k) & B_{1}(\tau_{k}) \\ \widetilde{K}(k) & 0 \end{array}\displaystyle \right ] \\ =&\left [ \textstyle\begin{array}{@{}cc@{}} A+(B_{00}+D F(\tau_{k})E)\widetilde{K}(k) & B_{11}-D F(\tau_{k})E \\ \widetilde{K}(k) & 0 \end{array}\displaystyle \right ], \end{aligned}$$

  (9)

$B_{00}$, $B_{11}$, D, and E are constant matrices, $F(\tau_{k})$ satisfies $F^{\mathrm{T}}(\tau_{k})F(\tau_{k})\leq I$.

If $A_{p}$ has non-zero mutually different eigenvalues $\lambda_{1}, \ldots, \lambda_{n}$ and Λ is the corresponding eigenvector matrix, choose $a_{1}, \ldots, a_{n}$ satisfying $\lambda _{1}(T-a_{1}-\tau_{k})<0$, …, $\lambda_{n}( T-a_{n}-\tau_{k})<0$, then $B_{00}$, $B_{11}$, D, E, and $F(\tau_{k})$ can be represented as

$$\begin{aligned}& E = \Lambda^{-1}B_{p}, \\& B_{00} = \Lambda \operatorname {diag}\biggl(-\frac{1}{\lambda_{1}} ,\ldots, - \frac{1}{\lambda_{n}} \biggr)\Lambda^{-1}B_{p}, \\& B_{11} = \Lambda \operatorname {diag}\biggl(\frac{1}{\lambda_{1}}\exp\{ \lambda_{1}T\} ,\ldots, \frac{1}{\lambda_{n}}\exp\{\lambda_{n}T\} \biggr)\Lambda^{-1}B_{p}, \\& D = \Lambda \operatorname {diag}\biggl(\frac{1}{\lambda_{1}}\exp\{\lambda _{1}a_{1} \}, \ldots, \frac{1}{\lambda_{n}}\exp\{\lambda_{n}a_{n}\} \biggr), \\& F(\tau_{k}) = \operatorname {diag}\bigl(\exp\bigl\{ \lambda_{1}(T-a_{1}- \tau_{k})\bigr\} , \ldots, \exp\bigl\{ \lambda_{n}(T-a_{n}- \tau_{k})\bigr\} \bigr). \end{aligned}$$

If $A_{p}$ has zero eigenvalues or multiple eigenvalues, for example, $A_{p}$ has one zero eigenvalue, one r multiple eigenvalue $\lambda _{*}$, and non-zero mutually different eigenvalues $\lambda_{2}, \ldots, \lambda_{n-r}$, in this case, $A_{p}$ can be represented as

$$A_{p}= \Lambda \operatorname {diag}(0, J_{1}, J_{2}) \Lambda^{-1}, $$

where $J_{1}$ is a diagonal matrix corresponding to $\lambda_{2}, \ldots, \lambda_{n-r}$, $J_{2}$ is the Jordan block corresponding to $\lambda_{*}$. Choose $a_{i}$, $i=1, 2, \ldots, n-r$, satisfying $a_{1}>\tau_{k}$, $\lambda _{l}(T-\tau_{k}-a_{l})<0$, $l=2, 3, \ldots, n-r$. Then we have

$$\begin{aligned}& B_{00} = \Lambda \operatorname {diag}(T, \tilde{J_{1}}, \tilde {J_{2}} )\Lambda^{-1}B_{p}, \\& B_{11} = \Lambda \operatorname {diag}(0, \hat{J_{1}}, \hat{J_{2}} )\Lambda^{-1}B_{p}, \\& D = \Lambda \operatorname {diag}\biggl(a_{1}, \frac{1}{\lambda_{2}}\exp\{ \lambda_{2}a_{2}\}, \ldots, \frac{1}{\lambda_{n-r}}\exp\{\lambda _{n-r}a_{n-r}\}, P_{2} \biggr), \\& F(\tau_{k}) = \operatorname {diag}\biggl(-\frac{\tau_{k}}{a_{1}}, \exp\bigl\{ \lambda _{2}(T-a_{2}-\tau_{k})\bigr\} , \ldots, \exp\bigl\{ \lambda_{n-r}(T-a_{n-r}-\tau _{k})\bigr\} , P_{2}^{-1}\bar{J_{2}} \biggr), \end{aligned}$$

where $P_{2}$ is a invertible diagonal matrix and satisfies $\|P_{2}^{-1}\bar{J}_{2}\|<1$, and

$$\begin{aligned}& \tilde{J_{1}} = \left [ \textstyle\begin{array}{@{}ccc@{}} -\frac{1}{\lambda_{2}} & & \\ & \ddots& \\ & & -\frac{1}{\lambda_{n-r}} \end{array}\displaystyle \right ], \\& \hat{J_{1}}=\left [ \textstyle\begin{array}{@{}ccc@{}} -\frac{1}{\lambda_{2}}\exp\{\lambda_{2}T\} & & \\ & \ddots& \\ & & -\frac{1}{\lambda_{n-r}}\exp\{\lambda_{n-r}T\} \end{array}\displaystyle \right ]. \end{aligned}$$

When $\lambda_{*}\neq0$,

$$\begin{aligned}& \hat{J_{2}} = \left [ \textstyle\begin{array}{@{}cccc@{}} \frac{1}{\lambda_{*}}\exp\{\lambda_{*}T\} & & & \\ \frac{1}{\lambda_{*}^{2}}(\lambda_{*}T-1)\exp\{\lambda_{*}T\} & \ddots & & \\ \vdots& \ddots& & \\ \frac{1}{\lambda_{*}^{r}}\sum_{k=1}^{r}(-1)^{k-1}\frac{(\lambda _{*}T)^{r-k}}{(r-k)!}\exp\{\lambda_{*}T\} & \cdots&\frac{1}{\lambda _{*}^{2}}(\lambda_{*}T-1)\exp\{\lambda_{*}T\} & \frac{1}{\lambda _{*}}\exp\{\lambda_{*}T\} \end{array}\displaystyle \right ], \\& \bar{J}_{2}= \left [ \textstyle\begin{array}{@{}cccc@{}} \frac{1}{\lambda_{*}}\exp\{x\} & & & \\ \frac{1}{\lambda_{*}^{2}}(x-1)\exp\{x\} & \ddots& & \\ \vdots& \ddots& & \\ \frac{1}{\lambda_{*}^{r}}\sum_{k=1}^{r}(-1)^{k-1}\frac {x^{r-k}}{(r-k)!}\exp\{x\} & \cdots&\frac{1}{\lambda_{*}^{2}}(x-1)\exp \{\lambda_{*}x\} & \frac{1}{\lambda_{*}}\exp\{x\} \end{array}\displaystyle \right ], \\& x=\lambda_{*}(T-\tau_{k}), \\& \tilde{J_{2}}= \left [ \textstyle\begin{array}{@{}cccc@{}} -\frac{1}{\lambda_{*}} & & & \\ -\frac{1}{\lambda_{*}^{2}} & \ddots& & \\ \vdots& \ddots& & \\ (-1)^{r}\frac{1}{\lambda_{*}^{r}} & \cdots&-\frac{1}{\lambda _{*}^{2}}& -\frac{1}{\lambda_{*}} \end{array}\displaystyle \right ]. \end{aligned}$$

When $\lambda_{*}=0$,

$$\begin{aligned}& \hat{J_{2}}=\left [ \textstyle\begin{array}{@{}cccc@{}} 0 & \cdots& & 0 \\ \frac{T^{2} }{2}& \ddots& & \vdots\\ \vdots& \ddots& \ddots& \\ \frac{T^{r} }{r!} &\cdots& \frac{T^{2} }{2}& 0 \end{array}\displaystyle \right ], \\& \bar{J}_{2} =\left [ \textstyle\begin{array}{@{}cccc@{}} -\tau_{k} & & & \\ \frac{(T-\tau_{k})^{2}}{2} & & \ddots& \\ \vdots& \ddots& & \\ \frac{(T-\tau_{k})^{r}}{r!} &\cdots& \frac{(T-\tau_{k})^{2}}{2} & -\tau_{k} \end{array}\displaystyle \right ], \\& \tilde{J}= \operatorname {diag}(T, T, \ldots, T). \end{aligned}$$

Combining (6) and (8) one obtains the following discrete-time switched system model:

$$ \tilde{x}(k+1) = \tilde{A}_{\sigma(k)}(k) \tilde{x}(k). $$

(10)

$\sigma(k)$ is called a switching signal. $\sigma(k)=1$ implies there is no packet dropout, while $\sigma(k)=0$ implies packet dropout.

3 Exponential stability analysis

Lemma 1

[17]

For constant matrices M, N, a symmetric matrix W, and scalar $\varepsilon>0$, the following inequality holds:

$$W+MFN+N^{\mathrm{T}}F^{\mathrm{T}}M^{\mathrm{T}}< 0, $$

where F satisfies $F^{\mathrm{T}}F \leq I$, if and only if that there exists a matrix $\varepsilon>0$

$$W+\varepsilon MM^{\mathrm{T}}+\varepsilon^{-1} N^{\mathrm{T}}N< 0. $$

Lemma 2

[18]

For the system (10), if there exist a Lyapunov function $V(\tilde{x}(k))=\tilde{x}(k)^{\mathrm{T}}\tilde{P}\tilde{x}(k)$ and positive scalars $\alpha_{0}$, $\alpha_{1}$, such that

$$\begin{aligned}& \alpha_{0}^{r_{0}}\alpha_{1}^{1-r_{0}}>1, \end{aligned}$$

(11)

$$\begin{aligned}& \left [ \textstyle\begin{array}{@{}cc@{}} -\tilde{P}^{-1} & \alpha_{0}\tilde{A}_{0} \\ {*} & -\tilde{P} \end{array}\displaystyle \right ]\leq0, \end{aligned}$$

(12)

$$\begin{aligned}& \left [ \textstyle\begin{array}{@{}cc@{}} -\tilde{P}^{-1} & \alpha_{1}\tilde{A}_{1} \\ {*} & -\tilde{P} \end{array}\displaystyle \right ]\leq0, \end{aligned}$$

(13)

then the system is exponentially stable with the packet-dropout rate $r\leq r_{0}$.

Theorem 3.1

For given positive scalars $r_{0}$, $\alpha_{0}$, $\alpha_{1}$, if there exist positive definite matrices P, Q, such that

$$\begin{aligned}& \alpha_{0}^{r_{0}}\alpha_{1}^{1-r_{0}}>1, \end{aligned}$$

(14)

$$\begin{aligned}& \left [ \textstyle\begin{array}{@{}cccc@{}} -P^{-1}& 0&\alpha_{0}A & \alpha_{0}B \\ {*}& -Q^{-1} & 0 & \alpha_{0}I \\ {*}& {*}& -P& 0\\ {*}& {*} &{*} & -Q \end{array}\displaystyle \right ]\leq0, \end{aligned}$$

(15)

$$\begin{aligned}& { \left [ \textstyle\begin{array}{@{}cccccccccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}B_{00}K& \alpha_{1}B_{11} & \varepsilon_{1}\alpha_{1}D & 0 & \varepsilon_{2}\sqrt{\delta}\alpha _{1}B_{00} & 0 & 0 & 0 \\ {*} & -Q^{-1} & \alpha_{1}K & 0 & 0 & 0 & \varepsilon_{2}\sqrt{\delta }\alpha_{1}I & 0 & 0 &0 \\ {*} & {*} &-P & 0& 0 & K^{\mathrm{T}}E^{\mathrm{T}} &0 & \sqrt{\delta }I & \varepsilon_{3}\sqrt{\delta}I&0 \\ {*} & {*} &{*} & -Q & 0 & -E^{\mathrm{T}} & 0 & 0 & 0& 0 \\ {*} &{*} & {*} & {*}&-\varepsilon_{1}I & 0 & 0 & 0 & 0 & 0\\ {*} & {*} & {*} & {*} & {*} & -\varepsilon_{1}I& 0 & 0 & 0& E \\ {*} & {*} & {*} & {*} & {*} & {*} & -\varepsilon_{2}I & 0& 0 & 0 \\ {*} & {*} & {*} & {*} & {*} & {*} & {*} & -\varepsilon_{2}I & 0 & 0 \\ {*} & {*} & {*} & {*} & {*} & {*} & {*} & {*} & -\varepsilon_{3}I & 0\\ {*} & {*} & {*} & {*} & {*} & {*} & {*} & {*} & {*} & -\varepsilon_{3}I \end{array}\displaystyle \right ]< 0,} \end{aligned}$$

(16)

then the system is exponentially stable with the packet-dropout rate $r\leq r_{0}$.

Proof

For the system (10), construct the Lyapunov function:

$$\begin{aligned}& V(k)=\tilde{x}^{\mathrm{T}}(k)\tilde{ P}\tilde{x}(k) \\& \tilde{P} =\left [ \textstyle\begin{array}{@{}cc@{}} P & 0\\ 0 & Q \end{array}\displaystyle \right ], \end{aligned}$$

where P, Q are positive definite matrices.

Then (12) can be written as (15) and (13) can be written as follows:

$$ \left [ \textstyle\begin{array}{@{}cccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}(B_{00}+DF(\tau_{k})E)\widetilde {K}(k)& \alpha_{1}B_{11}-\alpha_{1}DF(\tau_{k})E) \\ {*} & -Q^{-1} & \alpha_{1}\widetilde{K}(k) & 0\\ {*} & {*} &-P & 0 \\ {*} & {*} & {*} & -Q \end{array}\displaystyle \right ]\leq0, $$

(17)

which can be described as

$$\begin{aligned} & \left [ \textstyle\begin{array}{@{}cccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}B_{00}\widetilde{K}(k)& \alpha _{1}B_{11} \\ {*} & -Q^{-1} & \alpha_{1}\widetilde{K}(k) & 0\\ {*} & {*} &-P & 0 \\ {*} & {*} & {*} & -Q \end{array}\displaystyle \right ]+\left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}D \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]F(\tau_{k})\left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0\\ \widetilde{K}(k)^{\mathrm{T}}E^{\mathrm{T}} \\ -E^{\mathrm{T}} \end{array}\displaystyle \right ] ^{\mathrm{T}} \\ &\quad {} +\left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0\\ \widetilde{K}(k)^{\mathrm{T}}E^{\mathrm{T}} \\ -E^{\mathrm{T}} \end{array}\displaystyle \right ]F( \tau_{k})^{\mathrm{T}}\left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}D \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]^{\mathrm{T}} \leq0 . \end{aligned}$$

(18)

From Lemma 1, inequality (18) is true if and only if there exists a scalar $\varepsilon_{1} > 0$ such that the following inequality holds:

$$\begin{aligned} &\left [ \textstyle\begin{array}{@{}cccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}B_{00}\widetilde{K}(k)& \alpha _{1}B_{11} \\ {*} & -Q^{-1} & \alpha_{1}\widetilde{K}(k) & 0\\ {*} & {*} &-P & 0 \\ {*} & {*} & {*} & -Q \end{array}\displaystyle \right ]+\varepsilon_{1}\left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}D \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]\left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}D \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]^{\mathrm{T}} \\ &\quad {}+\varepsilon_{1}^{-1} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0\\ \widetilde{K}(k)^{\mathrm{T}}E^{\mathrm{T}} \\ -E^{\mathrm{T}} \end{array}\displaystyle \right ]\left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0\\ \widetilde{K}(k)^{\mathrm{T}}E^{\mathrm{T}} \\ -E^{\mathrm{T}} \end{array}\displaystyle \right ]^{\mathrm{T}}\leq0. \end{aligned}$$

(19)

It then follows from the Schur complement that inequality (19) is equivalent to the following matrix inequality:

$$\left [ \textstyle\begin{array}{@{}cccccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}B_{00}\widetilde{K}(k)& \alpha _{1}B_{11} & \varepsilon_{1}\alpha_{1}D & 0 \\ {*} & -Q^{-1} & \alpha_{1}\widetilde{K}(k) & 0 & 0 & 0 \\ {*} & {*} &-P & 0& 0 & \widetilde{K}(k)^{\mathrm{T}}E^{\mathrm{T}}\\ {*} & {*} &{*} & -Q & 0 & -E^{\mathrm{T}}\\ {*} &{*} & {*} & {*}&-\varepsilon_{1}I & 0\\ {*} & {*} & {*} & {*} & {*} & -\varepsilon_{1}I \end{array}\displaystyle \right ]< 0, $$

which can be written as

$$\begin{aligned} &\left [ \textstyle\begin{array}{@{}cccccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}B_{00}K& \alpha_{1}B_{11} & \varepsilon_{1}\alpha_{1}D & 0 \\ {*} & -Q^{-1} & \alpha_{1}K & 0 & 0 & 0 \\ {*} & {*} &-P & 0& 0 & K^{\mathrm{T}}E^{\mathrm{T}}\\ {*} & {*} &{*} & -Q & 0 & -E^{\mathrm{T}}\\ {*} &{*} & {*} & {*}&-\varepsilon_{1}I & 0\\ {*} & {*} & {*} & {*} & {*} & -\varepsilon_{1}I \end{array}\displaystyle \right ]+\left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}B_{00} \\ \alpha_{1}I \\ 0 \\ 0\\ 0 \\ 0 \end{array}\displaystyle \right ] \triangle(k) \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0\\ 0 \end{array}\displaystyle \right ]^{\mathrm{T}} \\ &\quad {}\times \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0\\ 0 \end{array}\displaystyle \right ]\triangle^{\mathrm{T}}(k)\left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}B_{00} \\ \alpha_{1}I \\ 0 \\ 0\\ 0 \\ 0 \end{array}\displaystyle \right ]^{\mathrm{T}} +\left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]\triangle^{\mathrm{T}}(k) \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ 0 \\ 0\\ 0 \\ E \end{array}\displaystyle \right ]^{\mathrm{T}}+\left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ 0 \\ 0\\ 0 \\ E \end{array}\displaystyle \right ]\triangle(k) \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ] \leq0. \end{aligned}$$

(20)

From Lemma 1, inequality (20) is true if and only if there exists a scalar $\varepsilon_{2} > 0$, $\varepsilon_{3} > 0$ such that the following inequality holds:

$$\begin{aligned} &\left [ \textstyle\begin{array}{@{}cccccc@{}} -P^{-1} & 0 & \alpha_{1}A+ \alpha_{1}B_{00}K& \alpha_{1}B_{11} & \varepsilon_{1}\alpha_{1}D & 0 \\ {*} & -Q^{-1} & \alpha_{1}K & 0 & 0 & 0 \\ {*} & {*} &-P & 0& 0 & K^{\mathrm{T}}E^{\mathrm{T}}\\ {*} & {*} &{*} & -Q & 0 & -E^{\mathrm{T}}\\ {*} &{*} & {*} & {*}&-\varepsilon_{1}I & 0\\ {*} & {*} & {*} & {*} & {*} & -\varepsilon_{1}I \end{array}\displaystyle \right ] \\ &\quad {}+\varepsilon_{2}\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}B_{00} \\ \alpha_{1}I \\ 0 \\ 0\\ 0 \\ 0 \end{array}\displaystyle \right ]\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} \alpha_{1}B_{00} \\ \alpha_{1}I \\ 0 \\ 0\\ 0 \\ 0 \end{array}\displaystyle \right ] ^{\mathrm{T}}+ \varepsilon_{2}^{-1}\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0\\ 0 \end{array}\displaystyle \right ]\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0\\ 0 \end{array}\displaystyle \right ]^{\mathrm{T}} \\ &\quad {}+\varepsilon_{3}\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ I \\ 0 \\ 0 \\ 0 \end{array}\displaystyle \right ]^{\mathrm{T}}+ \varepsilon_{3}^{-1}\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ 0 \\ 0\\ 0 \\ E \end{array}\displaystyle \right ]\sqrt{\delta} \left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0 \\ 0 \\ 0\\ 0 \\ E \end{array}\displaystyle \right ]^{\mathrm{T}}\leq0. \end{aligned}$$

(21)

From the Schur complement, we see that (21) is equivalent to (16).

From Lemma 2, we see that if (14), (15), and (16) hold, then the system (10) has exponential stability.

The conditions in Theorem 3.1 are a set of LMIs with inversion constraints. K can be solved by an iterative LMI approach which is called cone complementarity linearization [19, 20]. □

4 Numerical example

Consider the following system from [16]:

$$ \dot{x}(t) = \left [ \textstyle\begin{array}{@{}cc@{}} 0 & 1 \\ 0 & -0.1 \end{array}\displaystyle \right ]x(t)+\left [ \textstyle\begin{array}{@{}c@{}} 0 \\ 0.1 \end{array}\displaystyle \right ]v(t). $$

(22)

Choose the sampling period $T=0.3\mbox{ s}$, and suppose $0\leq\tau_{k}\leq 0.1\mbox{ s}$, $r_{0}=0.1$, $\delta=0.1$. Using Theorem 3.1, we get

$$K=[ \textstyle\begin{array}{@{}cc@{}} -2.8706 & -11.5619 \end{array}\displaystyle ]. $$

Suppose that the initial condition is $x^{\mathrm{T}}(0)=[0.5\ -0.5 ]$. Figure 2 shows the state trajectories of the NCS. We can see that the networked control system is exponentially stable.

Figure 2

State trajectories of the NCS.

In [16], the designed state-feedback gain is unchanged. When the feedback controller is disturbed, it cannot guarantee the system stability. However, from Theorem 3.1, the designed-state feedback gain subject to a certain additive interference still can make the system stable.

5 Conclusions

In this paper, a new discrete-time switched NCSs model that can deal simultaneously with packet dropout and time delay is presented. The criterion for the exponential stability of the system is derived. The gain of the disturbed state-feedback controller can be solved by the proposed method.