控制中的 LMI/pages/dt 混合 H2 Hinf 最優動態輸出反饋控制

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本部分展示如何在混合 ${\mathcal {H}}_{2}$ 和 ${\mathcal {H}}_{\infty }$ 意義上為離散時間設計動態輸出反饋控制。

問題

考慮離散時間廣義 LTI 工廠 ${\mathcal {P}}$ ，其最小狀態空間實現為

$x_{k+1}=A_{d}x_{k}+{\begin{bmatrix}B_{d1,1}&B_{d1,2}\end{bmatrix}}{\begin{bmatrix}w_{1,k}\\w_{2,k}\end{bmatrix}}+B_{d,2}u_{k},$

${\begin{bmatrix}z_{1,k}\\z_{2,k}\end{bmatrix}}={\begin{bmatrix}C_{d1,1}\\D_{d1,2}\end{bmatrix}}x_{k}+{\begin{bmatrix}D_{d11,11}&D_{d11,12}\\D_{d11,21}&D_{d11,22}\end{bmatrix}}{\begin{bmatrix}w_{1,k}\\w_{2,k}\end{bmatrix}}+{\begin{bmatrix}D_{12,1}\\D_{12,2}\end{bmatrix}}u_{k},$

$y_{k}=C_{d2}x_{k}+{\begin{bmatrix}D_{21,1}&D_{21,2}\end{bmatrix}}{\begin{bmatrix}w_{1,k}\\w_{2,k}\end{bmatrix}}+D_{d22}u_{k}$

定理

需要設計一個具有狀態空間實現 $(A_{dc},B_{dc},C_{dc},D_{dc})$ 的離散時間動態輸出反饋 LTI 控制器，以最小化閉環傳遞矩陣 $T_{11}(z)$ 的 ${\mathcal {H}}_{2}$ 範數，該矩陣從外源輸入 $w_{1},k$ 到效能輸出 $z_{1},k$ ，同時確保閉環傳遞矩陣 $T_{22}(z)$ 的 ${\mathcal {H}}_{\infty }$ 範數小於 $\gamma _{d}$ ，該矩陣從外源輸入 $w_{2},k$ 到效能輸出 $z_{2},k$ 。

$T_{11}(z)=C_{d_{CL}1,1}(zI-A_{d_{CL}})^{-1}B_{d_{CL}1,1},$

$T_{22}(z)=C_{d_{CL}1,2}(zI-A_{d_{CL}})^{-1}B_{d_{CL}1,2}+D_{d_{CL}11,22},$

$A_{d_{CL}}={\begin{bmatrix}A_{d}+B_{d2}D_{dc}{\tilde {D}}_{d}^{-1}C_{d2}&B_{d2}(I+D_{dc}{\tilde {D}}_{d}^{-1}D_{d22})C_{dc}\\B_{dc}{\tilde {D}}_{d}^{-1}C_{d2}&A_{dc}+B_{dc}{\tilde {D}}_{d}^{-1}D_{d22}C_{dc}\end{bmatrix}}$ ,

$B_{d_{CL}1,1}={\begin{bmatrix}B_{d1,1}+B_{d2}D_{dc}{\tilde {D}}_{d}^{-1}D_{d21,1}\\B_{dc}{\tilde {D}}_{d}^{-1}D_{d21,1}\end{bmatrix}}$ ,

$B_{d_{CL}1,2}={\begin{bmatrix}B_{d1,2}+B_{d2}D_{dc}{\tilde {D}}_{d}^{-1}D_{d21,2}\\B_{dc}{\tilde {D}}_{d}^{-1}D_{d21,2}\end{bmatrix}}$ ,

$C_{d_{CL}1,1}={\begin{bmatrix}C_{d1,1}+D_{d12,1}D_{dc}{\tilde {D}}_{d}^{-1}C_{d2,1}&D_{d12,1}(I+D_{dc}{\tilde {D}}_{d}^{-1}D_{d22})C_{dc}\end{bmatrix}}$ ,

$C_{d_{CL}1,2}={\begin{bmatrix}C_{d1,2}+D_{d12,2}D_{dc}{\tilde {D}}_{d}^{-1}C_{d2,2}&D_{d12,2}(I+D_{dc}{\tilde {D}}_{d}^{-1}D_{d22})C_{dc}\end{bmatrix}}$ ,

$D_{d_{CL}11,22}=D_{d11,22}+D_{d12,2}D_{dc}{\tilde {D}}_{d}^{-1}D_{d21,2}$ ,

以及 ${\tilde {D}}_{d}=I-D_{d22}D_{dc}$ .

綜合方法

求解 $A_{dn}\in \mathbb {R} ^{n_{x}\times n_{x}},B_{dn}\in \mathbb {R} ^{n_{x}\times n_{x}},C_{dn}\in \mathbb {R} ^{n_{u}\times n_{x}},D_{dn}\in \mathbb {R} ^{n_{u}\times n_{y}},X_{1},Y_{1}\in \mathbb {S} ^{n_{x}},Z\in \mathbb {S} ^{n_{Z_{1}}},$ 以及 $\mu \in \mathbb {R} _{>0}$ ，使 ${\mathcal {J}}(\mu )=\mu$ 最小化，滿足 $X_{1}>0,\ Y_{1}>0\ Z>0,$

${\begin{bmatrix}X_{1}&I&X_{1}A_{d}+B_{d_{n}}C_{d_{2}}&A_{d_{n}}&X_{1}B_{d1,1}+B{d_{n}}D_{d21,1}\\*&Y_{1}&A_{d}+B_{d2}C_{dn}D_{d2}&A_{d}Y_{1}+B_{d2}C_{dn}&B_{d1,1}+B_{d2}D_{dn}D_{d21,1}\\*&*&X_{1}&I&0\\*&*&*&Y_{1}&0\\*&*&*&*&I\end{bmatrix}}>0,$

${\begin{bmatrix}X_{1}&I&X_{1}A_{d}+B_{d_{n}}C_{d_{2}}&A_{d_{n}}&X_{1}B_{d1,2}+B{d_{n}}D_{d21,2}&0\\*&Y_{1}&A_{d}+B_{d2}C_{dn}D_{d2}&A_{d}Y_{1}+B_{d2}C_{dn}&B_{d1,2}+B_{d2}D_{dn}D_{d21,2}&0\\*&*&X_{1}&I&0&C_{d1,2}^{T}+C_{d2}^{T}D_{dn}^{T}D_{d12,2}^{T}\\*&*&*&Y_{1}&0&Y_{1}C_{d1,2}^{T}+C_{dn}^{T}D_{d12,2}^{T}\\*&*&*&*&-\gamma _{d}I&D_{d11,22}^{T}+D_{d21,2}^{T}D_{dn}^{T}D{d12,2}^{T}\\*&*&*&*&*&-\gamma _{d}I\end{bmatrix}}>0$

${\begin{bmatrix}Z&C_{d1,1}+D_{d12,1}D_{dn}C_{d2}&C_{d1,1,}Y_{1}^{T}+D_{d12,1}C_{dn}\\*&X_{1}&I\\*&*&Y_{1}\end{bmatrix}}>0,$

$D_{d11,11}+D_{d12,1}D_{dn}D_{d21,1}=0,$

${\begin{bmatrix}X_{1}&I\\*&Y_{1}\end{bmatrix}}>0,$

tr $Z<\mu .$

控制器透過以下公式恢復

$A_{dc}=A_{dK}-B{dc}(I-D_{d22}D_{dc})^{-1}D_{d22}C_{dc},$

$B_{dc}=B_{dK}(I-D_{dc}D_{d22}),$

$C_{dc}=(I-D_{dc}D_{d22})C_{dK},$

$D_{dc}=(I+D_{dK}D{d22})^{-1}D_{dK},where<math>{\begin{bmatrix}A_{dK}&B_{dK}\\C_{dK}&D_{dK}\end{bmatrix}}={\begin{bmatrix}X_{2}&X_{1}B_{d2}\\0&I\end{bmatrix}}^{-1}({\begin{bmatrix}A_{dn}&B_{dn}\\C_{dn}&D_{dn}\end{bmatrix}}-{\begin{bmatrix}X_{1}A_{d}Y_{1}&0\\0&0\end{bmatrix}}){\begin{bmatrix}Y_{2}^{T}&0\\C_{d2}Y_{1}&I\end{bmatrix}}^{-1}$ , 其中矩陣 $X_{2}$ 和 $Y_{2}$ 滿足 $X_{2}Y_{2}^{T}=I-X_{1}Y_{1}$ 。如果 $D_{22}=0$ , 那麼 $A_{dc}=A_{dK},B_{dc}=B{dK},C_{dc}=C_{dK}$ 以及 $D_{dc}=D_{dK}$ 。