技術(shù)特征：

1.一種非線性網(wǎng)絡(luò)化控制系統(tǒng)的非脆弱耗散濾波方法，其特征在于，具體包括以下步驟：

1)建立非線性網(wǎng)絡(luò)化濾波誤差系統(tǒng)模型：

$\left\{\begin{array}{c}x(k+1)=Ax\left(k\right)+Bw\left(k\right)+Gf(k,x(k\left)\right)\\ e\left(k\right)=Cx\left(k\right)+Dw\left(k\right)\end{array}\right.$

其中：x(k+1)＝[x(k+1)^T x_f(k+1)^T Y(k)^T y(k)^T]^T，x(k)∈Rⁿ是系統(tǒng)的狀態(tài)向量，x_f(k)∈Rⁿ是狀態(tài)估計(jì)，y(k)∈R^r是濾波器接收到的測(cè)量輸出，w(k)∈R^p是外部干擾信號(hào)，f(k,x(k))滿足Lipschitz條件非線性向量項(xiàng)||f(k,x(k))||≤||Wx(k)||，w(k)∈l₂[0,∞)為擾動(dòng)輸入，e(k)＝z(k)-z_f(k)是濾波誤差，z(k)∈R^m是被估計(jì)信號(hào)，z_f(k)∈R^m是濾波器的輸出；

$A=\stackrel{\‾}{A}+\widetilde{A_1}+\widetilde{A_2},B=\stackrel{\‾}{B}+\widetilde{B_1}+\widetilde{B_2}$

$\widetilde{A_2}=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ \left(\mathrm{\ξ}\right(k)-\stackrel{\‾}{\mathrm{\ξ}})C& 0& -\left(\mathrm{\ξ}\right(k)-\stackrel{\‾}{\mathrm{\ξ}})I& 0\\ 0& 0& 0& 0\end{array}\right]$

A_fd＝A_f+ΔA_f，B_fd＝B_f+ΔB_f，C_fd＝C_f+ΔC_f

A∈R^n×n，B∈R^n×p，C∈R^r×n，D∈R^r×p，L₁∈R^m×n，L₂∈R^m×p為系統(tǒng)參數(shù)矩陣；

A_f∈R^n×n，B_f∈R^n×r，C_f∈R^m×n為濾波器參數(shù)矩陣，ΔA_f＝H₁F₁(k)E₁，ΔB_f＝H₂F₂(k)E₂，ΔC_f＝H₃F₃(k)E₃為濾波器參數(shù)攝動(dòng)矩陣；H₁∈R^n×d，H₂∈R^n×h，H₃∈R^m×a，E₁∈R^d×n，E₂∈R^h×r，E₃∈R^a×n，F(xiàn)₁(k)∈R^d×d，F(xiàn)₂(k)∈R^h×h，F(xiàn)₃(k)∈R^a×a；

0和I是為零矩陣和單位陣；

Y(k)＝ξ(k)y(k)+(1-ξ(k))y(k)，ξ(k)＝(1-θ(k))δ(k+1)，有如下統(tǒng)計(jì)特性，E表示數(shù)學(xué)期望：

$E\[\mathrm{\θ}\left(k\right)\]=\stackrel{\‾}{\mathrm{\θ}},$

$E\[\mathrm{\δ}\left(k\right)\]=\stackrel{\‾}{\mathrm{\δ}},$

$E\[\mathrm{\ξ}\left(k\right)\]=(1-\stackrel{\‾}{\mathrm{\θ}})\stackrel{\‾}{\mathrm{\δ}}=\stackrel{\‾}{\mathrm{\ξ}},$

$E\[{\left(\mathrm{\θ}\right(k)-\stackrel{\‾}{\mathrm{\θ}})}^2\]=\stackrel{\‾}{\mathrm{\θ}}(1-\stackrel{\‾}{\mathrm{\θ}})={\mathrm{\σ}}_1^2,$

$E\[{\left(\mathrm{\ξ}\right(k)-\stackrel{\‾}{\mathrm{\ξ}})}^2\]=\stackrel{\‾}{\mathrm{\ξ}}(1-\stackrel{\‾}{\mathrm{\ξ}})={\mathrm{\σ}}_2^2,$

$E\[\left(\mathrm{\ξ}\right(k)-\stackrel{\‾}{\mathrm{\ξ}})\left(\mathrm{\θ}\right(k)-\stackrel{\‾}{\mathrm{\θ}})\]=-\stackrel{\‾}{\mathrm{\θ}}\stackrel{\‾}{\mathrm{\ξ}}=-{\mathrm{\σ}}_3^2,$

其中：δ_k和θ_k是不相關(guān)的隨機(jī)變量，由模型

y(k)＝θ_ky(k)+(1-θ_k)(1-θ_k-1)δ_ky(k-1)+(1-θ_k)[1-(1-θ_k-1)δ_k)]y(k-1)

其中：y(k)∈R^r是測(cè)量輸出；

時(shí)延發(fā)生概率為

丟包概率為

數(shù)據(jù)按時(shí)接收概率為

2)構(gòu)造Lyapunov函數(shù)；

$V\left(\tilde{x}\right(k\left)\right)=\tilde{x}{\left(k\right)}^{T}P\tilde{x}\left(k\right)$

其中P是正定對(duì)稱矩陣；

3)計(jì)算非脆弱耗散濾波器參數(shù)矩陣A_f，B_f，C_f和系統(tǒng)性能指標(biāo)γ，系統(tǒng)均方指數(shù)穩(wěn)定和非脆弱耗散控制器存在的充分條件為：

針對(duì)下列線性矩陣不等式：

$\left[\begin{array}{cc}{\mathrm{\Ψ}}_1& {\mathrm{\Ψ}}_2^T\\ {\mathrm{\Ψ}}_2& {\mathrm{\Ψ}}_3\end{array}\right]\<0$

其中：

${\mathrm{\Ψ}}_1=\left[\begin{array}{ccccccc}-\mathrm{\Π}+{\mathrm{\Π}}_{11}& *& *& *& *& *& *\\ 0& -P_2& *& *& *& *& *\\ 0& 0& -\mathrm{\τ}I& *& *& *& *\\ -{\mathrm{\Π}}_1& 0& 0& {\mathrm{\Π}}_2& *& *& *\\ {\mathrm{\Θ}}_{11}& {\mathrm{\Θ}}_{12}& {\mathrm{\Θ}}_{13}& {\mathrm{\Θ}}_{14}& {\mathrm{\Π}}_3& *& *\\ {\mathrm{\Θ}}_{21}& {\mathrm{\Θ}}_{22}& {\mathrm{\Θ}}_{23}& {\mathrm{\Θ}}_{24}& 0& {\mathrm{\Π}}_4& *\\ {\mathrm{\Θ}}_c& 0& 0& {\mathrm{\Θ}}_d& 0& 0& Q^{-1}\end{array}\right]$

${\mathrm{\Ψ}}_2=\left[\begin{array}{ccccccccccccccccccccccc}0& 0& 0& 0& 0& 0& {\hat{H}}_1^T& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& {\mathrm{\ϵ}}_1{\hat{E}}_1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& {\hat{H}}_3^{T}S& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& -{\hat{H}}_3^T\\ 0& {\mathrm{\ϵ}}_2{\hat{E}}_3& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& \stackrel{\‾}{\mathrm{\θ}}{\hat{H}}_2^T& 0& {\mathrm{\β}}_1{\hat{H}}_2^T& 0& 0& 0& {\mathrm{\σ}}_3{\hat{H}}_2^T& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {\mathrm{\ϵ}}_3{\hat{E}}_{2}C& {\mathrm{\ϵ}}_3{\hat{E}}_{2}C& 0& 0& 0& {\mathrm{\ϵ}}_3{\hat{E}}_{2}D& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& \left(1-\stackrel{\‾}{\mathrm{\θ}}\right){\hat{H}}_2^T& 0& -{\mathrm{\β}}_1{\hat{H}}_2^T& 0& 0& 0& -{\mathrm{\σ}}_3{\hat{H}}_2^T& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& {\mathrm{\ϵ}}_4{\hat{E}}_2& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right]$

Ψ₃＝diag{-ε₁I,-ε₁I,-ε₂I,-ε₂I,-ε₃I,-ε₃I,-ε₄I,-ε₄I}

$\mathrm{\Π}=\left[\begin{array}{cc}X& Z\\ Z& Z\end{array}\right],P_2=\left[\begin{array}{cc}{Y}_1& Y_3\\ Y_3& Y_2\end{array}\right],{\mathrm{\Π}}_{11}=\left[\begin{array}{cc}{\mathrm{\τW}}^{T}W& {\mathrm{\τW}}^{T}W\\ {\mathrm{\τW}}^{T}W& {\mathrm{\τW}}^{T}W\end{array}\right]$

${\mathrm{\Π}}_1=\left[\begin{array}{cc}-S^T{L}_1& -S^T{L}_1+S^T{\hat{C}}_f\end{array}\right],{\mathrm{\Π}}_2=-D^{T}S-S^{T}D-R$

Π₃＝diag{-Π,-Π,-Π,-Π}，Π₄＝diag{-P₂,-P₂,-P₂,-P₂}

${\mathrm{\Θ}}_{11}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{A_1}^T& {\mathrm{\Θ}}_{{\hat{M}}_1}^T& {\mathrm{\Θ}}_{{\hat{M}}_3}^T& {\mathrm{\Θ}}_{{\hat{M}}_{1,3}}^T\end{array}\right]}^T,{\mathrm{\Θ}}_{12}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{A_2}^T& {\mathrm{\Θ}}_{{\hat{M}}_2}^T& {\mathrm{\Θ}}_{{\hat{M}}_4}^T& {\mathrm{\Θ}}_{{\hat{M}}_{2,4}}^T\end{array}\right]}^T,$

${\mathrm{\Θ}}_{14}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{B_1}^T& {\mathrm{\Θ}}_{{\hat{N}}_1}^T& {\mathrm{\Θ}}_{{\hat{N}}_2}^T& {\mathrm{\Θ}}_{{\hat{N}}_{1,2}}^T\end{array}\right]}^T,{\mathrm{\Θ}}_{21}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{A_3}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_1}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_3}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_{1,3}}^T\end{array}\right]}^T$

${\mathrm{\Θ}}_{22}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{A_4}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_2}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_4}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_{2,4}}^T\end{array}\right]}^T,{\mathrm{\Θ}}_{24}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{B_2}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{N}}_1}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{N}}_2}^T& {\mathrm{\Θ}}_{{\stackrel{\‾}{N}}_{1,2}}^T\end{array}\right]}^T$

${\mathrm{\Θ}}_{13}={\left[\begin{array}{cccc}{\mathrm{\Θ}}_{c_1}^T& 0& 0& 0\end{array}\right]}^T,{\mathrm{\Θ}}_{23}={\left[\begin{array}{cccc}0& 0& 0& 0\end{array}\right]}^T$

${\mathrm{\Θ}}_c=\left[\begin{array}{cc}{L}_1& L_1-{\hat{C}}_f\end{array}\right],{\mathrm{\Θ}}_d=L_2,{\mathrm{\Θ}}_{c_1}={\left[\begin{array}{cc}{X}^T& Z^T\end{array}\right]}^T$

${\mathrm{\Θ}}_{A_1}=\left[\begin{array}{cc}XA+\stackrel{\‾}{\mathrm{\θ}}{\hat{B}}_{f}C& XA+\stackrel{\‾}{\mathrm{\θ}}{\hat{B}}_{f}C+{\hat{A}}_f\\ ZA& ZA\end{array}\right],{\mathrm{\Θ}}_{A_2}=\left[\begin{array}{cc}\left(1-\stackrel{\‾}{\mathrm{\θ}}\right){\hat{B}}_f& 0\\ 0& 0\end{array}\right]$

${\mathrm{\Θ}}_{A_3}=P_2\left[\begin{array}{cc}(\stackrel{\‾}{\mathrm{\θ}}+\stackrel{\‾}{\mathrm{\ξ}})C& (\stackrel{\‾}{\mathrm{\θ}}+\stackrel{\‾}{\mathrm{\ξ}})C\\ \stackrel{\‾}{\mathrm{\θ}}C& \stackrel{\‾}{\mathrm{\θ}}C\end{array}\right],{\mathrm{\Θ}}_{A_4}=P_2\left[\begin{array}{cc}(1-\stackrel{\‾}{\mathrm{\θ}}-\stackrel{\‾}{\mathrm{\ξ}})I& 0\\ (1-\stackrel{\‾}{\mathrm{\θ}})I& 0\end{array}\right]$

${\mathrm{\Θ}}_{{\hat{M}}_1}={\mathrm{\β}}_1\left[\begin{array}{cc}{\hat{B}}_{f}C& {\hat{B}}_{f}C\\ 0& 0\end{array}\right],{\mathrm{\Θ}}_{{\hat{M}}_3}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right],{\mathrm{\Θ}}_{{\hat{M}}_{1,3}}={\mathrm{\σ}}_3\left[\begin{array}{cc}{\hat{B}}_{f}C& {\hat{B}}_{f}C\\ 0& 0\end{array}\right]$

${\mathrm{\Θ}}_{{\hat{M}}_2}={\mathrm{\β}}_1\left[\begin{array}{cc}-{\hat{B}}_f& 0\\ 0& 0\end{array}\right],{\mathrm{\Θ}}_{{\hat{M}}_4}=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right],{\mathrm{\Θ}}_{{\hat{M}}_{2,4}}={\mathrm{\σ}}_3\left[\begin{array}{cc}-{\hat{B}}_f& 0\\ 0& 0\end{array}\right]$

${\mathrm{\Θ}}_{{\hat{N}}_1}={\mathrm{\β}}_1\left[\begin{array}{c}{\hat{B}}_{f}D\\ 0\end{array}\right],{\mathrm{\Θ}}_{{\hat{N}}_2}={\mathrm{\β}}_1\left[\begin{array}{c}0\\ 0\end{array}\right],{\mathrm{\Θ}}_{{\hat{N}}_{1,2}}={\mathrm{\σ}}_3\left[\begin{array}{c}{\hat{B}}_{f}D\\ 0\end{array}\right]$

${\mathrm{\Θ}}_{B_1}={\mathrm{\β}}_1\left[\begin{array}{c}XB+\stackrel{\‾}{\mathrm{\θ}}{\hat{B}}_{f}D\\ ZB\end{array}\right],{\mathrm{\Θ}}_{B_2}=P_2\left[\begin{array}{c}\left(\stackrel{\‾}{\mathrm{\θ}}+\stackrel{\‾}{\mathrm{\ξ}}\right)D\\ \stackrel{\‾}{\mathrm{\θ}}D\end{array}\right]$

${\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_1}={\mathrm{\β}}_1{P}_2\left[\begin{array}{cc}C& C\\ C& C\end{array}\right],{\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_3}={\mathrm{\β}}_1{P}_2\left[\begin{array}{cc}C& C\\ 0& 0\end{array}\right],{\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_{1,3}}={\mathrm{\σ}}_3{P}_2\left[\begin{array}{cc}0& 0\\ C& C\end{array}\right]$

${\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_2}={\mathrm{\β}}_1{P}_2\left[\begin{array}{cc}-I& 0\\ -I& 0\end{array}\right],{\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_4}={\mathrm{\β}}_2{P}_2\left[\begin{array}{cc}-I& 0\\ 0& 0\end{array}\right],{\mathrm{\Θ}}_{{\stackrel{\‾}{M}}_{2,4}}={\mathrm{\σ}}_3{P}_2\left[\begin{array}{cc}0& 0\\ -I& 0\end{array}\right]$

${\mathrm{\Θ}}_{{\stackrel{\‾}{N}}_1}={\mathrm{\β}}_1{P}_2\left[\begin{array}{c}D\\ D\end{array}\right],{\mathrm{\Θ}}_{{\stackrel{\‾}{N}}_2}={\mathrm{\β}}_2{P}_2\left[\begin{array}{c}D\\ 0\end{array}\right],{\mathrm{\Θ}}_{{\stackrel{\‾}{N}}_{1,2}}={\mathrm{\σ}}_3{P}_2\left[\begin{array}{c}0\\ D\end{array}\right]$

$\begin{array}{cccccc}{A}_f=W^{-1}{\hat{A}}_f{Z}^{-1}{V}^{-T},& B_f=W^{-1}{\hat{B}}_f,& C_f={\hat{C}}_f{Z}^{-1}{V}^{-T},& H_1=W^{-1}{\hat{H}}_1,& E_1={\hat{E}}_1{Z}^{-1}{V}^{-T}& H_2=W^{-1}{\hat{H}}_2\end{array},$

$E_3={\hat{E}}_3{Z}^{-1}{V}^{-T},H_3={\hat{H}}_3,E_2={\hat{E}}_2;$

其中：W,V均是非奇異常數(shù)矩陣，滿足，WV^T＝I-XZ^-1；

X∈R^n×n，Z∈R^n×n，P₂∈R^2r×2r，ε_i＞0(i＝1,2,3,4)均為未知變量，其它變量均是已知的，可以根據(jù)系統(tǒng)參數(shù)得出或直接給定，利用Matlab LMI工具箱進(jìn)行求解，如果存在對(duì)稱正定矩陣X,Z,P₂和矩陣和標(biāo)量ε_i＞0(i＝1,2,3,4)，則網(wǎng)絡(luò)化濾波誤差系統(tǒng)是均方指數(shù)穩(wěn)定的且具有嚴(yán)格耗散性，非脆弱濾波器參數(shù)矩陣為γ＝Σ(||e(k)||)/Σ(||w(k)||)，且可以繼續(xù)進(jìn)行步驟4)；如果上述未知變量無解，則網(wǎng)絡(luò)化濾波誤差系統(tǒng)不是均方指數(shù)穩(wěn)定且不滿足嚴(yán)格耗散性，不能得到非脆弱濾波器的參數(shù)矩陣，也不可以進(jìn)行步驟4)；

4)計(jì)算非脆弱H_∞濾波器參數(shù)矩陣A_f，B_f，C_f，各矩陣參數(shù)取為：Q＝-I，R＝γ²I，S＝0，H_∞濾波下最優(yōu)擾動(dòng)抑制比γ_opt優(yōu)化的條件為：

令e＝γ²，如果以下優(yōu)化問題成立：

$\begin{array}{ccc}\min e& s.t.& \left[\begin{array}{cc}{\mathrm{\Ψ}}_1& {\mathrm{\Ψ}}_2^T\\ {\mathrm{\Ψ}}_2& {\mathrm{\Ψ}}_3\end{array}\right]\<0\end{array}$

X＝X^T＞0，Z＝Z^T＞0，X-Z＞0,ε_i＞0(i＝1,2,3,4)

系統(tǒng)的最小擾動(dòng)抑制率同時(shí)非脆弱耗散濾波器的參數(shù)矩陣被優(yōu)化為