LMIs in Control/pages/dt mixed H2 Hinf optimal output feedback control

WIP, Description in progress

This part shows how to design dynamic outpur feedback control in mixed $$\mathcal{H}_2$$ and $$\mathcal{H}_\infty$$ sense for the continuous time.

Problem
Consider the discrete-time generalized LTI plant $$\mathcal{P}$$ with minimal state-space realization

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

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

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

Theorem
A continuous-time dynamic output feedback LTI controllerwith state-space realization $$(A_c, B_c,C_c,D_c)$$ is to be designed to minimize the $$\mathcal{H}_2$$ norm of the closed-loop transfer matrix $$T_{11}(s)$$ from the exogenous input $$w_1$$ to the performance output $$z_1$$ while ensuring the H∞ norm of the closed-loop transfer matrix $$T_{22}(s)$$ from the exogenous input $$w_2$$ to the performance output $$z_2$$ is less than $$\gamma_d$$, where

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

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

$$A_{d_{CL}} = \begin{bmatrix} A+B_2D_c\tilde{D}^{-1}C_{2} & B_{2}(I+D_{c}\tilde{D}^{-1}D_{22})C_{c} \\ B_{c}\tilde{D}^{-1}C_{2} & A_{c}+B_{c}\tilde{D}^{-1}D_{22}C_{c} \end{bmatrix}$$,

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

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

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

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

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

and $$\tilde{D} = I - D_{22}D_{c}$$.

Synthesis Method
Solve for $$A_{n}\in \mathbb{R}^{n_x\times n_x}, B_{n} \in \mathbb{R}^{n_x \times n_x}, C_{n}\in \mathbb{R}^{n_u \times n_x}, D_{n} \in \mathbb{R}^{n_u\times n_y}, X_1, Y_1\in \mathbb{S}^{n_x}, Z \in \mathbb{S}^{n_{Z_1}},$$ and $$ \mu \in \mathbb{R}_{>0}$$ that minimizes $$\mathcal{J}(\mu) = \mu$$ subjects to $$X_1>0, \ Y_1>0 \ Z>0,$$

$$\begin{bmatrix}N_{11} & A+A_n^T + B_2D_nC_2 & B_{1,1}+B_2D_nD{21,1} \\
 * & X_1A+A^TX_1+B_nC_2+C_2^TB_n^T & X_1B_{1,1} + B_nD_{21,1} \\
 * & * & -I \end{bmatrix} < 0$$,

$$\begin{bmatrix}N_{11} & A+A_n^T + B_2D_nC_2 & B_{1,1}+B_2D_nD{21,1} & Y_1^TC_{1,2}^T+C_n^TD_{12,2}^T\\
 * & X_1A+A^TX_1+B_nC_2+C_2^TB_n^T & X_1B_{1,1} + B_nD_{21,1} & C_{1,2}^T+C_2^TD_n^TD_{12,2}^T\\
 * & * & -\gamma_d I & D_{11,22}^T+D_{21,2}^TD_n^TD_{12,2}^T \\
 * & * & * & -\gamma_d I\end{bmatrix} < 0$$,

$$\begin{bmatrix}Y_1 I Y_1C_{1,1}^T+C_n^TD_{12,1}^T \\
 * & X_1 & C_{1,1}^T+C_2^TD_n^TD_{12,1}^T \\
 * & * & Z \end{bmatrix} > 0$$,

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

$$\begin{bmatrix}X_1 & I \\
 * & Y_1\end{bmatrix}>0,$$

tr$$Z<\mu,$$

where $$N_{11} = AY_1+Y_1A^T+B_2C_n+C_n^TB_2^T$$.

The controller is recovered by

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

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

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

D_{c} = (I+D_{K}D{22})^{-1}D_{K},

where

$$\begin{bmatrix} A_{K} & B_{K} \\ C_{K} & D_{K} \end{bmatrix} = \begin{bmatrix} X_2 & X_1B_{2} \\ 0 & I \end{bmatrix}^{-1}(\begin{bmatrix} A_{n} & B_{n} \\ C_{n} & D_{n} \end{bmatrix}-\begin{bmatrix} X_1A_dY_1 & 0 \\ 0 & 0 \end{bmatrix})\begin{bmatrix} Y_2^T & 0 \\ C_{2}Y_1 & I \end{bmatrix}^{-1}$$, and the matrices $$X_2$$ and $$Y_2$$ satisfy $$X_2Y_2^T=I-X_1Y_1$$. If $$D_{22}=0$$, then $$A_{c}=A_{K}, B_{c}=B{K}, C_{c}=C_{K}$$ and $$D_{c}=D_{K}$$.

Given $$X_1$$ and $$Y_1$$, the matrices $$X_2$$ and $$Y_2$$ can be found using a matrix decomposition, such as a LU decomposition or a Cholesky decomposition.

If $$D_{11,11} = 0, D_{12,1} \neq 0, \text{ and } D_{21,1} \neq 0,$$ then it is often simplest to choose $$D_{n} = 0$$ in order to satisfy the equality constraint $$D_{11,11}+ D_{12,1}D_{n}D_{21,1}=0,$$.

WIP, additional references to be added