Exponential stabilization of aero-engine T-S fuzzy system with decentralized dynamic event-triggered mechanism

Abstract

This paper proposes a decentralized dynamic event-triggered control method of aero-engine T-S fuzzy system, aiming to achieve both exponential stabilization of aero-engine networked control systems (NCSs) and reduction in network communication loads. Firstly, a novel decentralized dynamic event-triggered mechanism (DETM) is developed to regulate data transmissions in each communication channel independently. The closed-loop model is then established, by considering network-induced delays, dynamic quantization effects, external disturbance, and parameter perturbation. Stability criteria are derived using the Lyapunov–Krasovskii method, and a collaborative design method based on linear matrix inequalities (LMIs) is proposed for controller, quantizers, and DETM. Lastly, a parameter tuning method based on the iL-SHADE algorithm is proposed for better feasibility of the obtained LMIs. Simulation results show that the proposed method has good robustness to multi-uncertainty conditions and can effectively reduce communication resource wastage while ensuring well control performance across a wide range of flight envelopes.

Appendices

Appendix 1

Proof of Theorem 1

Choose the following Lyapunov–Krasovskii function as

$$ V\left( t \right) = \sum\limits_{l = 1}^{4} {V_{l} \left( t \right)} $$

(30)

where $V_{1} \left( t \right) = {\varvec{x}}^{T} \left( t \right){\varvec{Px}}\left( t \right)$ with ${\varvec{P}} > 0$, $V_{2} \left( t \right) = \int_{{t - \tau_{{\text{M}}} }}^{t} {e^{{2\lambda \left( {s - t} \right)}} {\varvec{x}}^{T} \left( s \right){\varvec{Qx}}\left( s \right)ds}$ with ${\varvec{Q}} > 0$, $V_{3} \left( t \right) = \int_{{ - \tau_{{\text{M}}} }}^{0} {\int_{t + s}^{t} {e^{{2\lambda \left( {v - t} \right)}} {\dot{\varvec{x}}}^{T} \left( v \right){\varvec{R}}{\dot{\varvec{x}}}\left( v \right)dv} ds}$ with ${\varvec{R}} > 0$, $V_{4} \left( t \right) = \sum\limits_{i = 1}^{2} {\eta_{i} \left( t \right)}$.

This together with Lemma 1 yields

$$ \lambda_{\min } \left( {\varvec{P}} \right)\left\| {{\varvec{x}}\left( t \right)} \right\|^{2} + \sum\limits_{i = 1}^{2} {\eta_{0{\text{i}}} e^{{ - \left( {1 + \rho_{i} } \right)t}} } \le V\left( t \right) \le \lambda_{{\text{M}}} \left\| {{\varvec{x}}\left( t \right)} \right\|^{2} + \sum\limits_{i = 1}^{2} {\eta_{i} \left( t \right)} $$

(31)

where $\lambda_{{\text{M}}} = \lambda_{\max } \left( {\varvec{P}} \right) + \tau_{{\text{M}}} \lambda_{\max } \left( {\varvec{Q}} \right) + \frac{1}{2}\tau_{{\text{M}}}^{2} \lambda_{\max } \left( {\varvec{R}} \right)$.

Calculating the derivative of $V_{l} \left( t \right)\left( {l = 1,2,3,4} \right)$ along the trajectory of system (18) yields

$$ \dot{V}\left( t \right) + 2\lambda V\left( t \right) = \sum\limits_{l = 1}^{4} {\left[ {\dot{V}_{l} \left( t \right) + 2\lambda V_{l} \left( t \right)} \right]} $$

(32)

$$ \begin{aligned} &\dot{V}_{1} \left( t \right) + 2\lambda V_{1} \left( t \right) = 2{\varvec{x}}^{T} \left( t \right){\varvec{P}}{\dot{\varvec{x}}}\left( t \right) + 2\lambda {\varvec{x}}^{T} \left( t \right){\varvec{Px}}\left( t \right) \hfill \\ & = {{\boldsymbol{\xi}}}^{T} \left( t \right)\left[ {\sum\limits_{m = 1}^{r} {\sum\limits_{n = 1}^{r} {\mu_{m} \mu_{n} \left\{ {2{\varvec{e}}_{1}^{T} {\varvec{P}}{\boldsymbol{\varLambda}}_{mn} } \right\}} } + 2\lambda {\varvec{e}}_{1}^{T} {\varvec{Pe}}_{1} } \right]{{\boldsymbol{\xi}}}^{T} \left( t \right) \end{aligned} $$

$$ \begin{aligned} &\dot{V}_{1} \left( t \right) + 2\lambda V_{1} \left( t \right) = 2{\varvec{x}}^{T} \left( t \right){\varvec{P}}{\dot{\varvec{x}}}\left( t \right) + 2\lambda {\varvec{x}}^{T} \left( t \right){\varvec{Px}}\left( t \right) \hfill \\ & = {{\boldsymbol{\xi}}}^{T} \left( t \right)\left[ {\sum\limits_{m = 1}^{r} {\sum\limits_{n = 1}^{r} {\mu_{m} \mu_{n} \left\{ {2{\varvec{e}}_{1}^{T} {\varvec{P}}{\boldsymbol{\varLambda}}_{mn} } \right\}} } + 2\lambda {\varvec{e}}_{1}^{T} {\varvec{Pe}}_{1} } \right]{{\boldsymbol{\xi}}}^{T} \left( t \right) \end{aligned} $$

$$ \begin{gathered} \dot{V}_{3} \left( t \right) + 2\lambda V_{3} \left( t \right) = \tau_{{\rm{M}}}^{2} {\dot{\varvec{x}}}^{T} \left( t \right){\varvec{R}}{\dot{\varvec{x}}}\left( t \right) + {\Gamma }_{1} + {\Gamma }_{2} \hfill \\ \, = {{\boldsymbol{\xi}}}^{T} \left( t \right)\left[ {\sum\limits_{m = 1}^{r} {\sum\limits_{n = 1}^{r} {\sum\limits_{p = 1}^{r} {\sum\limits_{q = 1}^{r} {\mu_{m} \mu_{n} } } \mu_{p} \mu_{q} \left\{ {\tau_{{\rm{M}}}^{2} {{\boldsymbol{\varLambda}}}_{mn}^{T} {\varvec{R}}{\boldsymbol{\varLambda }}_{pq} } \right\}} } + {\Gamma }_{1} + {\Gamma }_{2} } \right]{{\boldsymbol{\xi}}}^{T} \left( t \right) \hfill \\ \, \le {{\boldsymbol{\xi}}}^{T} \left( t \right)\left[ {\sum\limits_{m = 1}^{r} {\sum\limits_{n = 1}^{r} {\mu_{m} \mu_{n} \left\{ {\tau_{{\rm{M}}}^{2} {{\boldsymbol{\varLambda}}}_{mn}^{T} {\varvec{R}}{\boldsymbol{\varLambda }}_{mn} } \right\}} } + {\Gamma }_{1} + {\Gamma }_{2} } \right]{{\boldsymbol{\xi}}}^{T} \left( t \right) \hfill \\ \end{gathered} $$

$$ \dot{V}_{4} \left( t \right) + 2\lambda V_{4} \left( t \right) = \sum\limits_{i = 1}^{2} {\left[ {\dot{\eta }_{i} \left( t \right) + 2\lambda \eta_{i} \left( t \right)} \right]} $$

where $\Gamma_{1} = - \tau_{{\text{M}}} e^{{ - 2\lambda \tau_{{\text{M}}} }} \int_{{t - \tau_{{\text{M}}} }}^{t - \tau \left( t \right)} {{\dot{\varvec{x}}}^{T} \left( s \right){\varvec{R}}{\dot{\varvec{x}}}\left( s \right)ds}$, $\Gamma_{2} = - \tau_{{\text{M}}} e^{{ - 2\lambda \tau_{{\text{M}}} }} \int_{t - \tau \left( t \right)}^{t} {{\dot{\varvec{x}}}^{T} \left( s \right){\varvec{R}}{\dot{\varvec{x}}}\left( s \right)ds}$.

To handle the nonlinear integral terms in $\dot{V}_{3} \left( t \right) + 2\lambda V_{3} \left( t \right)$, we utilize Lemma 2 and reciprocally convex combination inequality [45] to obtain

$$ \Gamma_{1} + \Gamma_{2} \le - {{\boldsymbol{\xi}}}^{T} \left( t \right){\boldsymbol{\varTheta \xi }}\left( t \right) $$

(33)

Follow to $\dot{V}_{4} \left( t \right) + 2\lambda V_{4} \left( t \right)$, owing to $\rho_{i} > 2\lambda$, combining (19) and (20), it can be deduced that

$$ \begin{gathered} \dot{\eta }_{i} \left( t \right) + 2\lambda \eta_{i} \left( t \right) < - v_{i} {\varvec{e}}_{i}^{T} \left( t \right){{\boldsymbol{\varPhi}}}_{i} {\varvec{e}}_{i} \left( t \right) + v_{i} \sigma_{i} \left( t \right)\left[ {{\varvec{x}}_{i} \left( {t - \tau \left( t \right)} \right) + {\varvec{e}}_{i} \left( t \right)} \right]^{T} {{\boldsymbol{\varPhi}}}_{i} \left[ {{\varvec{x}}_{i} \left( {t - \tau \left( t \right)} \right) + {\varvec{e}}_{i} \left( t \right)} \right] \end{gathered} $$

(34)

where $v_{i} = 1 + \rho_{i} - 2\lambda$. It follows that

$$ \dot{V}_{4} \left( t \right) + 2\lambda V_{4} \left( t \right) \le {{\boldsymbol{\xi}}}^{T} \left( t \right)\left\{ {\sigma \left[ {{\varvec{e}}_{2}^{T} + {\varvec{e}}_{8}^{T} } \right]{\boldsymbol{\nu \varPhi }}\left[ {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right] - {\varvec{e}}_{8}^{T} {\boldsymbol{\nu \varPhi e}}_{8} } \right\}{{\boldsymbol{\xi}}}\left( t \right) $$

(35)

Define $J = \dot{V}\left( t \right) + 2\lambda V\left( t \right) - \gamma^{2} {\varvec{w}}^{T} \left( t \right){\varvec{w}}\left( t \right) + {\varvec{y}}^{T} \left( t \right){\varvec{y}}\left( t \right)$. Combining (32)–(35), one obtains

$$ J \le {{\boldsymbol{\xi}}}^{T} \left( t \right){{\boldsymbol{\varSigma}}}_{0} {{\boldsymbol{\xi}}}\left( t \right) $$

(36)

where

${{\boldsymbol{\varSigma}}}_{0} = \sum\limits_{m = 1}^{r} {\sum\limits_{n = 1}^{r} {\mu_{m} \mu_{n} \left\{ \begin{gathered} {\varvec{e}}_{1}^{T} \left[ {2\lambda {\varvec{P}} + {\text{sym}}\left\{ {{\varvec{P}}\left( {{\varvec{A}}_{m} + \Delta {\varvec{A}}_{m} } \right)} \right\} + {\varvec{Q}}} \right]{\varvec{e}}_{1} + \sigma {\varvec{e}}_{2}^{T} {\boldsymbol{\nu \varPhi e}}_{2} - e^{{ - 2\lambda \tau_{{\text{M}}} }} {\varvec{e}}_{3}^{T} {\varvec{Qe}}_{3} \hfill \\ - \left( {1 - \sigma } \right){\varvec{e}}_{8}^{T} {\boldsymbol{\nu \varPhi e}}_{8} - \gamma^{2} {\varvec{e}}_{9}^{T} {\varvec{e}}_{9} + {\text{sym}}\left\{ {{\varvec{e}}_{1}^{T} {\varvec{PB}}_{m} {\varvec{K}}_{n} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} + {\varvec{e}}_{10} } \right) + {\varvec{e}}_{1}^{T} {\varvec{Pe}}_{9} } \right\} \hfill \\ + {\text{sym}}\left\{ {{\varvec{e}}_{1}^{T} {\varvec{PB}}_{m} {\varvec{e}}_{11} + \sigma {\varvec{e}}_{2}^{T} {\boldsymbol{\nu \varPhi e}}_{8} } \right\} - {{\boldsymbol{\varTheta}}} + \tau_{{\text{M}}}^{2} {{\boldsymbol{\varLambda}}}_{mn}^{T} {\varvec{R\varLambda }}_{mn} + {\varvec{e}}_{1}^{T} {\varvec{C}}_{m}^{T} {\varvec{C}}_{m} {\varvec{e}}_{1} \hfill \\ \end{gathered} \right\}} }$.

From (12), we can get whenever $\left\| {\frac{{{\varvec{x}}_{i} \left( {\tilde{t}_{k}^{i} h} \right)}}{{\mu_{x}^{i} }}} \right\| \le M_{x}^{i}$ and $\left\| {\frac{{{\varvec{u}}_{\text{c}} \left( t \right)}}{{\mu_{u} }}} \right\| \le M_{u}$, then

$$ \left\| {q\left( {\frac{{{\varvec{x}}_{i} \left( {\tilde{t}_{k}^{i} h} \right)}}{{\mu_{x}^{i} }}} \right) - \frac{{{\varvec{x}}_{i} \left( {\tilde{t}_{k}^{i} h} \right)}}{{\mu_{x}^{i} }}} \right\| \le \Delta_{x}^{i} $$

(37)

$$ \left\| {q\left( {\frac{{{\varvec{u}}_{\text{c}} \left( t \right)}}{{\mu_{u} }}} \right) - \frac{{{\varvec{u}}_{\text{c}} \left( t \right)}}{{\mu_{u} }}} \right\| \le \Delta_{u} $$

(38)

Without loss of generality, the following variables are defined

$$ \mu_{x}^{i} = \frac{{\kappa_{x}^{i} }}{{M_{x}^{i} }}\left\| {{\varvec{x}}_{i} \left( {\tilde{t}_{k}^{i} h} \right)} \right\| $$

(39)

$$ \mu_{u} = \frac{1}{{\sqrt {\kappa_{u} } M_{u} }}\left\| {{\varvec{u}}_{\text{c}} \left( t \right)} \right\| $$

(40)

where $\kappa_{x}^{i} \ge 1$, $0 < \kappa_{u} \le 1$ are scalars to be designed. It is easy to verify that (39) can guarantee the precondition of (37), and (40) can guarantee that of (38).

By combining (13), (37) and (39), we can get

$$ \phi_{x}^{T} \left( t \right)\phi_{x} \left( t \right) \le \frac{{\kappa_{x}^{2} {\Delta }_{x}^{2} }}{{M_{x}^{2} }}{{\boldsymbol{\xi}}}^{T} \left( t \right)\left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right)^{T} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right){{\boldsymbol{\xi}}}\left( t \right) $$

(41)

where $\kappa_{x} = \max \left\{ {\kappa_{x}^{1} ,\kappa_{x}^{2} } \right\}$, $\Delta_{x} = \max \left\{ {\Delta_{x}^{1} ,\Delta_{x}^{2} } \right\}$, $M_{x} = \min \left\{ {M_{x}^{1} ,M_{x}^{2} } \right\}$.

Then, the inequality (41) can be rewritten as

$$ {{\boldsymbol{\xi}}}^{T} \left( t \right){{\boldsymbol{\varSigma}}}_{1} {{\boldsymbol{\xi}}}\left( t \right) \ge 0 $$

(42)

where ${{\boldsymbol{\varSigma}}}_{1} = \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right)^{T} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right) - \frac{{M_{x}^{2} }}{{\kappa_{x}^{2} {\Delta }_{x}^{2} }}{\varvec{e}}_{10}^{T} {\varvec{e}}_{10}$.

Considering the property of Euclidean norm, we have

$$ \left\| {\left[ {\begin{array}{*{20}c} {{\overline{\varvec{x}}}_{1}^{T} \left( {\tilde{t}_{k}^{1} h} \right)} \\ {{\overline{\varvec{x}}}_{2}^{T} \left( {\tilde{t}_{k}^{2} h} \right)} \\ \end{array} } \right]} \right\| \le \left\| {\left[ {\begin{array}{*{20}c} {{\varvec{x}}_{1}^{T} \left( {\tilde{t}_{k}^{1} h} \right)} \\ {{\varvec{x}}_{2}^{T} \left( {\tilde{t}_{k}^{2} h} \right)} \\ \end{array} } \right]} \right\| + \left\| {\phi_{x} \left( t \right)} \right\| \le \left( {1 + \frac{{\kappa_{x} {\Delta }_{x} }}{{M_{x} }}} \right)\left\| {\left[ {\begin{array}{*{20}c} {{\varvec{x}}_{1}^{T} \left( {\tilde{t}_{k}^{1} h} \right)} \\ {{\varvec{x}}_{2}^{T} \left( {\tilde{t}_{k}^{2} h} \right)} \\ \end{array} } \right]} \right\| $$

(43)

From (14), (22), (38), (40), and (43), one has

$$ \phi_{u}^{T} \left( t \right)\phi_{u} \left( t \right) \le \frac{{\Delta_{u}^{2} }}{{s\kappa_{u} M_{u}^{2} }}\left( {1 + \frac{{\kappa_{x} \Delta_{x} }}{{M_{x} }}} \right)^{2} {{\boldsymbol{\xi}}}^{T} \left( t \right)\left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right)^{T} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right){{\boldsymbol{\xi}}}\left( t \right) $$

(44)

Further, we can rewrite the inequality (44) to

$$ {{{\xi}}}^{T} \left( t \right){{{\boldsymbol{\varSigma }}}}_{2} {{\boldsymbol{\xi}}}\left( t \right) \ge 0 $$

(45)

where ${{\boldsymbol{\varSigma}}}_{2} = \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right)^{T} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} } \right) - \frac{{s\kappa_{u} M_{u}^{2} }}{{\Delta_{u}^{2} }}\left( {1 + \frac{{\kappa_{x} \Delta_{x} }}{{M_{x} }}} \right)^{ - 2} {\varvec{e}}_{11}^{T} {\varvec{e}}_{11}$.

Define $\kappa_{1} = \frac{1}{{\kappa_{x}^{2} }}$, $\kappa_{2} = \kappa_{u} \left( {1 + \frac{{\kappa_{x} \Delta_{x} }}{{M_{x} }}} \right)^{ - 2}$. Since $\kappa_{x}^{i} \ge 1$, $0 < \kappa_{u} \le 1$, it is easy to obtain

$0 < \kappa_{1} \le 1$

$$0 < \kappa_{2} \le \left( {1 + \frac{{\Delta_{x} }}{{M_{x} }}} \right)^{ - 2}$$

(46)

Then, according to Lemma 3, it can be deduced that ${{\boldsymbol{\xi}}}^{T} \left( t \right){{\boldsymbol{\Sigma}}}_{0} {{\boldsymbol{\xi}}}\left( t \right) < 0$ holds if

$$ {{\boldsymbol{\varSigma}}}_{0} + \frac{1}{{\kappa_{1} }}{{\boldsymbol{\varSigma}}}_{1} + \frac{1}{{\kappa_{2} }}{{\boldsymbol{\varSigma}}}_{2} < 0 $$

(47)

Noting that $\sum\nolimits_{{m = 1}}^{r} {\sum\nolimits_{{m = 1}}^{r} {\mu _{m} \mu _{n} {\text{ }}\left\{ {{\boldsymbol{\varOmega }}_{{mn}} } \right\}} } = \sum\nolimits_{{m = 1}}^{r} {\mu _{m}^{2} } \left\{ {{\boldsymbol{\varOmega }}_{{mm}} } \right\} + \sum\nolimits_{{m = 1}}^{r} {\sum\nolimits_{{n = m + 1}}^{r} {\mu _{m} \mu _{n} } } \left\{ {{\boldsymbol{\varOmega }}_{{mn}} + {\boldsymbol{\varOmega }}_{{nm}} } \right\}$, we can obtain that (47) can be guaranteed by

$$ \begin{gathered} \sum\limits_{m = 1}^{r} {\mu_{m}^{2} } \left\{ {{{\boldsymbol{\varPsi}}}_{mm} - {{\boldsymbol{\varTheta}}} + \tau_{{\text{M}}}^{2} {{\boldsymbol{\varLambda}}}_{mm}^{T} {\varvec{R\varLambda }}_{mm} - {{\boldsymbol{\varPi}}}_{13} {{\boldsymbol{\varPi}}}_{33}^{ - 1} {{\boldsymbol{\varPi}}}_{13}^{T} } \right\} + \sum\limits_{m = 1}^{r} {\sum\limits_{n = m + 1}^{r} {\mu_{m} \mu_{n} } } \hfill \\ \left\{ {{{\boldsymbol{\varPsi}}}_{mn} + {{\boldsymbol{\varPsi}}}_{nm} - {{\boldsymbol{\varTheta}}} + \frac{1}{2}\tau_{{\text{M}}}^{2} \left( {{{\boldsymbol{\varLambda}}}_{mn} + {{\boldsymbol{\varLambda}}}_{nm} } \right)^{T} {\varvec{R}}\left( {{{\boldsymbol{\varLambda}}}_{mn} + {{\boldsymbol{\varLambda}}}_{nm} } \right) - {{\boldsymbol{\varPi}}}_{13} {{\boldsymbol{\varPi}}}_{33}^{ - 1} {{\boldsymbol{\varPi}}}_{13}^{T} } \right\} < 0 \hfill \\ \end{gathered} $$

(48)

Based on Schur complement, (23) and (24) is equivalent to

$$ {{\boldsymbol{\varPsi}}}_{mm} - {{\boldsymbol{\varTheta}}} + \tau_{{\text{M}}}^{2} {{\boldsymbol{\varLambda}}}_{mm}^{T} {\varvec{R\varLambda }}_{mm} - {{\boldsymbol{\varPi}}}_{13} {{\boldsymbol{\varPi}}}_{33}^{ - 1} {{\boldsymbol{\varPi}}}_{13}^{T} < 0 $$

(49)

$$ {{\boldsymbol{\varPsi}}}_{mn} + {{\boldsymbol{\varPsi}}}_{nm} - {{\boldsymbol{\varTheta}}} + \frac{1}{2}\tau_{{\text{M}}}^{2} \left( {{{\boldsymbol{\varLambda}}}_{mn} + {{\boldsymbol{\varLambda}}}_{nm} } \right)^{T} {\varvec{R}}\left( {{{\boldsymbol{\varLambda}}}_{mn} + {{\boldsymbol{\varLambda}}}_{nm} } \right) - {{\boldsymbol{\varPi}}}_{13} {{\boldsymbol{\varPi}}}_{33}^{ - 1} {{\boldsymbol{\varPi}}}_{13}^{T} < 0 $$

(50)

respectively. Therefore, combining (36), (47), and (48), we can obtain that if (23) and (24) hold, then $J \le 0$. It implies

$$ \dot{V}\left( t \right) \le \dot{V}\left( t \right) + 2\lambda V\left( t \right) \le \gamma^{2} {\varvec{w}}^{T} \left( t \right){\varvec{w}}\left( t \right) - {\varvec{y}}^{T} \left( t \right){\varvec{y}}\left( t \right) $$

(51)

Under zero conditions, the integration of both sides of (51) from $0$ to $+ \infty$ yields

$$ \left\| {{\varvec{y}}\left( t \right)} \right\|_{2} \le \gamma \left\| {{\varvec{w}}\left( t \right)} \right\|_{2} $$

(52)

With condition $w\left( t \right) \equiv 0$, we can get $\dot{V}\left( t \right) \le - 2\lambda V\left( t \right), t \in \Omega_{{t_{k} }}$ from (51). Thus, we have

$$ V\left( t \right) \le e^{{ - 2\lambda \left( {t - t_{k} h - \tau_{{t_{k} }} } \right)}} V\left( {t_{k} h + \tau_{{t_{k} }} } \right) \le \cdots \le e^{ - 2\lambda t} V\left( {0} \right) $$

(53)

From (31) and (53), it follows that

$$ \lambda_{\min } \left( {\varvec{P}} \right)\left\| {{\varvec{x}}\left( t \right)} \right\|^{2} \le e^{ - 2\lambda t} \lambda_{{\text{M}}} \left\| {{\varvec{x}}\left( 0 \right)} \right\|^{2} + \sum\limits_{i = 1}^{2} {\eta_{0{\text{i}}} \left[ {e^{ - 2\lambda t} - e^{{ - \left( {1 + \rho_{i} } \right)t}} } \right]} \le e^{ - 2\lambda t} \lambda_{{\text{M}}} \left\| {{\varvec{x}}\left( 0 \right)} \right\|^{2} $$

(54)

Letting $\beta = \sqrt {{{\lambda_{{\text{M}}} } \mathord{\left/ {\vphantom {{\lambda_{{\text{M}}} } {\lambda_{\min } \left( {\varvec{P}} \right)}}} \right. \kern-0pt} {\lambda_{\min } \left( {\varvec{P}} \right)}}}$, (54) follows that

$$ \left\| {{\varvec{x}}\left( t \right)} \right\| \le \beta e^{ - \lambda t} \left\| {{\varvec{x}}\left( 0 \right)} \right\| $$

(55)

Then from (52), (55) and Definition 1, we naturally obtain that system (18) is exponentially stable while having $H\infty$ performance level $\gamma$. This completes the proof.

Appendix 2

Proof of Theorem 2

Using the Schur complement to (22) leads to

$$ \left[ {\begin{array}{*{20}c} { - \frac{1}{s}{\varvec{I}}} & {{\varvec{K}}_{m}^{T} } \\ * & { - {\varvec{I}}} \\ \end{array} } \right] < 0 $$

(56)

with ${\varvec{Y}}_{m} = {\varvec{K}}_{m} {\varvec{X}}$, performing congruence transformation to (56) with ${\text{diag}}\left\{ {{\varvec{X}},\,{\varvec{I}}} \right\}$ yields

$$ \left[ {\begin{array}{*{20}c} { - {\varvec{X}}\frac{1}{s}{\varvec{IX}}} & {{\varvec{Y}}_{m}^{T} } \\ * & { - {\varvec{I}}} \\ \end{array} } \right] < 0 $$

(57)

It is known that $\left( {{{\boldsymbol{\Delta}}} - \frac{1}{\vartheta }{\varvec{X}}} \right){{\boldsymbol{\Delta}}}^{ - 1} \left( {{{\boldsymbol{\Delta}}} - \frac{1}{\vartheta }{\varvec{X}}} \right) \ge 0$ always holds for any matrix ${{\boldsymbol{\Delta}}} > 0$ and scalar $\vartheta > 0$, which implies that

$$ - {\varvec{X\Delta }}^{ - 1} {\varvec{X}} \le \vartheta^{2} {{\boldsymbol{\Delta}}} - 2\vartheta {\varvec{X}} $$

(58)

Let ${{\boldsymbol{\Delta}}}$ stands for $s{\varvec{I}}$, (58) becomes $- {\varvec{X}}\frac{1}{s}{\varvec{X}} \le \vartheta_{1}^{2} s{\varvec{I}} - 2\vartheta_{1} {\varvec{X}}$. Then, the inequality (57) can be guaranteed by (26).

Note that the inequalities (23) and (24) can be equivalently expressed as (59) and (60), respectively.

$$ \left[ {\begin{array}{*{20}c} {{\tilde{\varvec{\varPsi }}}_{{mm}} - {\varvec{\varTheta }}} & {\tau _{{\rm{M}}} {\tilde{\varvec{\varLambda }}}_{{mm}} ^{T} \user2{R}} & {{\varvec{\varPi }}_{{13}} } \\ * & { - \user2{R}} & {\varvec{0}} \\ * & * & {{\varvec{\varPi }}_{{33}} } \\ \end{array} } \right] + {\text{sym}}\left\{ {\user2{H}_{{G_{m} }} \user2{F}_{m} \left( t \right)\user2{H}_{{E_{m} }} } \right\} < 0 $$

(59)

$$ \left[ {\begin{array}{*{20}c} {{\tilde{\boldsymbol{\varPsi }}}_{mn} + {\tilde{\boldsymbol{\varPsi }}}_{nm} - {{\boldsymbol{\varTheta}}}} &{\tau_{{\text{M}}} \left( {{\tilde{\boldsymbol{\varLambda }}}_{mn} + {\tilde{\boldsymbol{\varLambda }}}_{nm} } \right)^{T} {\varvec{R}}} & {{{\boldsymbol{\varPi}}}_{13} } \\ * & { - 2{\varvec{R}}} & {\varvec{0}} \\ * & * & {{{\boldsymbol{\varPi}}}_{33} } \\ \end{array} } \right] \, + {\text{sym}}\left\{ {{\varvec{H}}_{{G_{m} }} {\varvec{F}}_{m} \left( t \right){\varvec{H}}_{{E_{m} }} } \right\} + {\text{sym}}\left\{ {{\varvec{H}}_{{G_{n} }} {\varvec{F}}_{n} \left( t \right){\varvec{H}}_{{E_{n} }} } \right\} < 0 $$

(60)

where ${\varvec{H}}_{{G_{m} }} = \left[ {\begin{array}{*{20}c} {{\varvec{G}}_{m}^{T} {\varvec{Pe}}_{1} } & {\tau_{{\text{M}}} {\varvec{G}}_{m}^{T} {\varvec{R}}} & {\varvec{0}} \\ \end{array} } \right]^{T}$, ${\varvec{H}}_{{E_{m} }} = \left[ {\begin{array}{*{20}c} {{\varvec{E}}_{m} {\varvec{e}}_{1} } & {\varvec{0}} & {\varvec{0}} \\ \end{array} } \right]$, $\begin{gathered} {\tilde{\boldsymbol{\varPsi }}}_{mn} = {\varvec{e}}_{1}^{T} \left[ {2\lambda {\varvec{P}} + {\varvec{A}}_{m}^{T} {\varvec{P}} + {\varvec{PA}}_{m} + {\varvec{Q}}} \right]{\varvec{e}}_{1} + \sigma {\varvec{e}}_{2}^{T} {\boldsymbol{\nu \varPhi e}}_{2} - e^{{ - 2\lambda \tau_{{\text{M}}} }} {\varvec{e}}_{3}^{T} {\varvec{Qe}}_{3} - \left( {1 - \sigma } \right){\varvec{e}}_{8}^{T} {\boldsymbol{\nu \varPhi e}}_{8} - \gamma^{2} {\varvec{e}}_{9}^{T} {\varvec{e}}_{9} \hfill \\ \quad \quad \quad - \frac{{M_{x}^{2} }}{{{\Delta }_{x}^{2} }}{\varvec{e}}_{10}^{T} {\varvec{e}}_{10} - \frac{{sM_{u}^{2} }}{{\Delta_{u}^{2} }}{\varvec{e}}_{11}^{T} {\varvec{e}}_{11} + {\text{sym}}\left\{ {{\varvec{e}}_{1}^{T} {\varvec{PB}}_{m} {\varvec{K}}_{n} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} + {\varvec{e}}_{10} } \right) + {\varvec{e}}_{1}^{T} {\varvec{Pe}}_{9} + {\varvec{e}}_{1}^{T} {\varvec{PB}}_{m} {\varvec{e}}_{11} + \sigma {\varvec{e}}_{2}^{T} {\boldsymbol{\nu \varPhi e}}_{8} } \right\},\; \hfill \\ \quad \quad \quad {\tilde{\boldsymbol{\varLambda }}}_{mn} = {\varvec{A}}_{m} {\varvec{e}}_{1} + {\varvec{B}}_{m} {\varvec{K}}_{n} \left( {{\varvec{e}}_{2} + {\varvec{e}}_{8} + {\varvec{e}}_{10} } \right) + {\varvec{e}}_{9} + {\varvec{B}}_{m} {\varvec{e}}_{11} . \hfill \\ \end{gathered}$

Using Lemma 4, there exist positive scalar $\varepsilon_{m}$, $\varepsilon_{n}$$\left( {1 \le m < n \le r} \right)$ such that

$$ \begin{gathered} {\text{sym}}\left\{ {{\varvec{H}}_{{G_{m} }} {\varvec{F}}_{m} \left( t \right){\varvec{H}}_{{E_{m} }} } \right\} \le \varepsilon_{m} {\varvec{H}}_{{G_{m} }} {\varvec{H}}_{{G_{m} }}^{T} + \varepsilon_{m}^{ - 1} {\varvec{H}}_{{E_{m} }}^{T} {\varvec{H}}_{{E_{m} }} \hfill \\ {\text{sym}}\left\{ {{\varvec{H}}_{{G_{n} }} {\varvec{F}}_{n} \left( t \right){\varvec{H}}_{{E_{n} }} } \right\} \le \varepsilon_{n} {\varvec{H}}_{{G_{n} }} {\varvec{H}}_{{G_{n} }}^{T} + \varepsilon_{n}^{ - 1} {\varvec{H}}_{{E_{n} }}^{T} {\varvec{H}}_{{E_{n} }} \hfill \\ \end{gathered} $$

(61)

Therefore, (59) and (60) are equivalent to

$$ \left[ {\begin{array}{*{20}l} {{\tilde{\boldsymbol{\varPsi }}}_{mm} - {{\boldsymbol{\varTheta}}}} & {\tau_{{\text{M}}} {\tilde{\boldsymbol{\varLambda }}}_{mm}^{T} {\varvec{R}}} & {{{\boldsymbol{\varPi}}}_{13} } & {{{\boldsymbol{\varPi}}}_{14} } \\ * & { - {\varvec{R}}} & {\varvec{0}} & {\varvec{0}} \\ * & * & {{{\boldsymbol{\varPi}}}_{33} } & {\varvec{0}} \\ * & * & * & {{{\boldsymbol{\varPi}}}_{44} } \\ \end{array} } \right] < 0 $$

(62)

$$ \left[ {\begin{array}{*{20}l} {{\tilde{\boldsymbol{\varPsi }}}_{mn} + {\tilde{\boldsymbol{\varPsi }}}_{nm} - {{\boldsymbol{\varTheta}}}} & {\tau_{{\text{M}}} \left( {{\tilde{\boldsymbol{\varLambda }}}_{mn} + {\tilde{\boldsymbol{\varLambda }}}_{nm} } \right)^{T} {\varvec{R}}} & {{{\boldsymbol{\varPi}}}_{13} } & {{\tilde{\boldsymbol{\varPi }}}_{14} } \\ * & { - 2{\varvec{R}}} & {\varvec{0}} & {\varvec{0}} \\ * & * & {{{\boldsymbol{\varPi}}}_{33} } & {\varvec{0}} \\ * & * & * & {{\boldsymbol{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{\Pi } }}_{44} } \\ \end{array} } \right] < 0 $$

(63)

where $\begin{aligned}&{{\boldsymbol{\varPi}}}_{14} = \left[ {\begin{array}{*{20}l} {\varepsilon_{m} {\varvec{e}}_{1}^{T} {\varvec{PG}}_{m} } & {{\varvec{e}}_{1}^{T} {\varvec{E}}_{m}^{T} } \\ {\tau_{{\text{M}}} \varepsilon_{m} {\varvec{RG}}_{m} } & {\varvec{0}} \\ {\varvec{0}} & {\varvec{0}} \\ \end{array} } \right],\\&{\tilde{\boldsymbol{\varPi }}}_{14} = \left[ {\begin{array}{*{20}c} {\varepsilon_{m} {\varvec{e}}_{1}^{T} {\varvec{PG}}_{m} } & {{\varvec{e}}_{1}^{T} {\varvec{E}}_{m}^{T} } & {\varepsilon_{n} {\varvec{e}}_{1}^{T} {\varvec{PG}}_{n} } & {{\varvec{e}}_{1}^{T} {\varvec{E}}_{n}^{T} } \\ {\tau_{{\text{M}}} \varepsilon_{m} {\varvec{RG}}_{m} } & {\varvec{0}} & {\tau_{{\text{M}}} \varepsilon_{n} {\varvec{RG}}_{n} } & {\varvec{0}} \\ {\varvec{0}} & {\varvec{0}} & {\varvec{0}} & {\varvec{0}} \\ \end{array} } \right].\end{aligned}$

Define ${\varvec{X}} = {\varvec{P}}^{ - 1}$. Multiplying the left- and right-hand sides of (21) by ${\text{diag}}\left\{ {{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},} \right.$$\left. {\varvec{X}} \right\}$, one can obtain (25). Similarly, multiplying the left- and right-hand sides of inequality (62) by ${\text{diag}}\left\{ {{\varvec{X}},\,{\varvec{X}},\,{\varvec{X}},\,{\varvec{X}},\,{\varvec{X}},\,{\varvec{X}},\,{\varvec{X}},\,{\varvec{X}},\,{\varvec{I}},\,{\varvec{X}},\,{\varvec{I}},\,{\varvec{R}}^{ - 1} ,\,{\varvec{I}},\,{\varvec{I}},\,{\varvec{I}},\,{\varvec{I}},\,{\varvec{I}}} \right\}$ and inequality (63) by ${\text{diag}}\left\{ {{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{X}},{\varvec{I}},{\varvec{X}},{\varvec{I}},{\varvec{R}}^{ - 1} ,{\varvec{I}},{\varvec{I}},{\varvec{I}},{\varvec{I}},{\varvec{I}},{\varvec{I}},{\varvec{I}}} \right\}$. Define new matrix variables ${\overline{\varvec{Q}}} = {\varvec{XQX}}$, ${\overline{\varvec{R}}} = {\varvec{XRX}}$, ${\overline{\boldsymbol{\varPhi }}} = \left( {{\boldsymbol{\nu \varPhi }}} \right)^{ - 1}$, ${\overline{\varvec{U}}} = {\text{diag}}\left\{ {{\varvec{X}},{\varvec{X}},{\varvec{X}}} \right\} \cdot {\varvec{U}} \cdot {\text{diag}}\left\{ {{\varvec{X}},{\varvec{X}},{\varvec{X}}} \right\}$. Let ${{\boldsymbol{\Delta}}}$ stands for ${\overline{\varvec{R}}}$, ${\overline{\boldsymbol{\varPhi }}}$, and ${\varvec{I}}$, (58) becomes $- {\varvec{R}}^{ - 1} = - {\varvec{X\overline{R}}}^{ - 1} {\varvec{X}} \le \vartheta_{2}^{2} {\overline{\varvec{R}}} - 2\vartheta_{2} {\varvec{X}}$, $- {\varvec{X}}{\overline{\boldsymbol{\varPhi }}}^{ - 1} {\varvec{X}} \le$$\vartheta_{3}^{2} {\overline{\boldsymbol{\varPhi }}} - 2\vartheta_{3} {\varvec{X}}$, and $- {\varvec{XX}} \le \vartheta_{4}^{2} {\varvec{I}} - 2\vartheta_{4} {\varvec{X}}$, respectively. Using Schur complement, (27) is equivalent to ${\varvec{X}}{\overline{\boldsymbol{\varPhi }}}^{ - 1} {\varvec{X}} < \varepsilon_{0} {\varvec{I}}$. Then, by the Schur complement, (28) and (29) can be obtained easily. This completes the proof.