控制中的 LMI / 頁面 / 離散時間混合 H2 Hinf 最優輸出反饋控制

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本部分介紹瞭如何設計混合 ${\mathcal {H}}_{2}$ 和 ${\mathcal {H}}_{\infty }$ 意義上的動態輸出反饋控制，用於連續時間。

問題

考慮具有最小狀態空間實現的離散時間廣義 LTI 系統 ${\mathcal {P}}$

${\dot {x}}=Ax+{\begin{bmatrix}B_{1,1}&B_{1,2}\end{bmatrix}}{\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix}}+B_{2}u,$

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

$y=C_{d2}x+{\begin{bmatrix}D_{21,1}&D_{21,2}\end{bmatrix}}{\begin{bmatrix}w_{1}\\w_{2}\end{bmatrix}}+D_{d22}u$

定理

需要設計一個具有狀態空間實現 $(A_{c},B_{c},C_{c},D_{c})$ 的連續時間動態輸出反饋 LTI 控制器，以最小化閉環傳遞矩陣 $T_{11}(s)$ 的 ${\mathcal {H}}_{2}$ 範數，從外源輸入 $w_{1}$ 到效能輸出 $z_{1}$ ，同時確保從外源輸入 $w_{2}$ 到效能輸出 $z_{2}$ 的閉環傳遞矩陣 $T_{22}(s)$ 的 H∞ 範數小於 $\gamma _{d}$ ，其中

$T_{11}(s)=C_{CL1,1}(sI-A_{CL})^{-1}B_{CL1,1},$

$T_{22}(s)=C_{CL1,2}(sI-A_{CL})^{-1}B_{CL1,2}+D_{CL11,22},$

$A_{d_{CL}}={\begin{bmatrix}A+B_{2}D_{c}{\tilde {D}}^{-1}C_{2}&B_{2}(I+D_{c}{\tilde {D}}^{-1}D_{22})C_{c}\\B_{c}{\tilde {D}}^{-1}C_{2}&A_{c}+B_{c}{\tilde {D}}^{-1}D_{22}C_{c}\end{bmatrix}}$ ,

$B_{CL1,1}={\begin{bmatrix}B_{1,1}+B_{2}D_{c}{\tilde {D}}^{-1}D_{21,1}\\B_{c}{\tilde {D}}^{-1}D_{21,1}\end{bmatrix}}$ ,

$B_{CL1,2}={\begin{bmatrix}B_{1,2}+B_{2}D_{c}{\tilde {D}}^{-1}D_{21,2}\\B_{c}{\tilde {D}}^{-1}D_{21,2}\end{bmatrix}}$ ,

$C_{CL1,1}={\begin{bmatrix}C_{1,1}+D_{12,1}D_{c}{\tilde {D}}^{-1}C_{2,1}&D_{12,1}(I+D_{c}{\tilde {D}}^{-1}D_{22})C_{c}\end{bmatrix}}$ ,

$C_{CL1,2}={\begin{bmatrix}C_{1,2}+D_{12,2}D_{c}{\tilde {D}}^{-1}C_{2,2}&D_{12,2}(I+D_{c}{\tilde {D}}^{-1}D_{22})C_{c}\end{bmatrix}}$ ,

$D_{CL11,22}=D_{11,22}+D_{12,2}D_{c}{\tilde {D}}^{-1}D_{21,2}$ ,

以及 ${\tilde {D}}=I-D_{22}D_{c}$ .

合成方法

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

${\begin{bmatrix}N_{11}&A+A_{n}^{T}+B_{2}D_{n}C_{2}&B_{1,1}+B_{2}D_{n}D{21,1}\\*&X_{1}A+A^{T}X_{1}+B_{n}C_{2}+C_{2}^{T}B_{n}^{T}&X_{1}B_{1,1}+B_{n}D_{21,1}\\*&*&-I\end{bmatrix}}<0$ ,

${\begin{bmatrix}N_{11}&A+A_{n}^{T}+B_{2}D_{n}C_{2}&B_{1,1}+B_{2}D_{n}D{21,1}&Y_{1}^{T}C_{1,2}^{T}+C_{n}^{T}D_{12,2}^{T}\\*&X_{1}A+A^{T}X_{1}+B_{n}C_{2}+C_{2}^{T}B_{n}^{T}&X_{1}B_{1,1}+B_{n}D_{21,1}&C_{1,2}^{T}+C_{2}^{T}D_{n}^{T}D_{12,2}^{T}\\*&*&-\gamma _{d}I&D_{11,22}^{T}+D_{21,2}^{T}D_{n}^{T}D_{12,2}^{T}\\*&*&*&-\gamma _{d}I\end{bmatrix}}<0$ ,

${\begin{bmatrix}Y_{1}IY_{1}C_{1,1}^{T}+C_{n}^{T}D_{12,1}^{T}\\*&X_{1}&C_{1,1}^{T}+C_{2}^{T}D_{n}^{T}D_{12,1}^{T}\\*&*&Z\end{bmatrix}}>0$ ,

$D_{11,11}+D_{12,1}D_{n}D_{21,1}=0,$

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

tr $Z<\mu ,$

其中 $N_{11}=AY_{1}+Y_{1}A^{T}+B_{2}C_{n}+C_{n}^{T}B_{2}^{T}$ .

控制器可由下式恢復

$A_{c}=A_{K}-B{c}(I-D_{22}D_{c})^{-1}D_{22}C_{c},$

$B_{c}=B_{K}(I-D_{c}D_{22}),$

$C_{c}=(I-D_{c}D_{22})C_{K},$

$D_{c}=(I+D_{K}D{22})^{-1}D_{K},where<math>{\begin{bmatrix}A_{K}&B_{K}\\C_{K}&D_{K}\end{bmatrix}}={\begin{bmatrix}X_{2}&X_{1}B_{2}\\0&I\end{bmatrix}}^{-1}({\begin{bmatrix}A_{n}&B_{n}\\C_{n}&D_{n}\end{bmatrix}}-{\begin{bmatrix}X_{1}A_{d}Y_{1}&0\\0&0\end{bmatrix}}){\begin{bmatrix}Y_{2}^{T}&0\\C_{2}Y_{1}&I\end{bmatrix}}^{-1}$ , 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 $A_{c}=A_{K},B_{c}=B{K},C_{c}=C_{K}$ 並且 $D_{c}=D_{K}$ .