追従SMI-OBS制御 | 制御系CAD

SMオブザーバベース・追従SMI制御…Homework

[0] まず準備の復習から始めます。

制御対象の状態空間表現として次式を考えます。

$\displaystyle{ \begin{array}{ll} (1.1) & \dot{x}(t)=Ax(t)+Bu(t)+B\xi(t,x,u)\\ (1.2) & y(t)=Cx(t)\\ & (x(t)\in{\rm\bf R}^n, u(t)\in{\rm\bf R}^m, y(t)\in{\rm\bf R}^p, 1\le m= p < n) \end{array} }$

これに対して、積分動作

$\displaystyle{ \begin{array}{ll} (2.1) &\dot{x}_r(t)=r(t)-y(t)\quad (x_r(t)\in{\rm\bf R}^m)\\ (2.2) &\underbrace{\frac{d}{dt}(r(t)-r_c)}_{\dot{e}_r(t)}=\Gamma\underbrace{(r(t)-r_c)}_{e_r(t)} \end{array} }$

とSM状態オブザーバ

$\displaystyle{ \begin{array}{ll} (3.1) &\dot{z}(t)=Az(t)+Bu(t)-G \overbrace{C\underbrace{(z(t)-x(t))}_{e(t)}}^{e_y(t)=Cz(t)-y(t)}+B\nu_o\quad(z(t)\in{\rm\bf R}^n)\\ (3.2) &\displaystyle{\nu_o=-\rho_o(u_\ell,y)\frac{Fe_y(t)}{||Fe_y(t)||}} \end{array} }$

を考えます。

●以下では、 $x(t)$ の代わりに $z(t)$ を用いた制御則 $u(t)$ を $\hat{u}(t)$ で表します。

$\displaystyle{(4)\quad \begin{array}{l} \hat{u}(t)=\underbrace{-L\left[\begin{array}{c} x_r(t)\\ z(t) \end{array}\right]-L_rr(t)+L_{\dot r}\dot{r}(t)}_{\hat {u}_L}+\hat{\nu}_c\\ \displaystyle{\hat{\nu}_c=L_n\frac{P_2(S\left[\begin{array}{c} x_r(t)\\ z(t) \end{array}\right]-S_rr(t))}{||P_2(S\left[\begin{array}{c} x_r(t)\\ z(t) \end{array}\right]-S_rr(t))||}}\\ L=(S\tilde{B})^{-1}(S\tilde{A}-\Phi S)\\ L_r=(S\tilde{B})^{-1}(\Phi S_r+S_2MB_r)\\ L_{\dot r}=(S\tilde{B})^{-1}S_r\\ L_n=(S\tilde{B})^{-1}\rho_c(\hat{u}_L,y) \end{array} }$

ここでスイッチング関数は、 $\left[\begin{array}{c} x_r(t)\\ z(t) \end{array}\right]\in{\rm\bf R}^{m+n}$ を $\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]\in{\rm\bf R}^{n+m}$ と分割して、次式で表されるとしています。

$\displaystyle{(5)\quad s(t)= \underbrace{ \left[\begin{array}{cc} S_1 & S_2 \\ \end{array}\right] }_{S} %\underbrace{ \left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right] %}_{x(t)} = \underbrace{S_2 \left[\begin{array}{cc} M & I_m \\ \end{array}\right] }_{S} %\underbrace{ \left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right] %}_{x(t)} \ (M=S_2^{-1}S_1) }$

[1] (3.1)から(1.1)を辺々引き算して、誤差方程式

$\displaystyle{(6)\quad \dot{e}(t)=(A-GC)e(t)+B\nu_o-B\xi(t,x,\hat{u}) }$

を得ます。 $x(t)=z(t)-e(t)$ を用いて、(2.1)は

$\displaystyle{(7)\quad \dot{x}_r(t)=r(t)-y(t)=r(t)-Cx(t)=r(t)-Cz(t)+\underbrace{Ce(t)}_{e_y(t)} }$

となります。また(3.1)は、 $u(t)$ の代わりに $\hat{u}(t)$ を用いることにすれば

$\displaystyle{(8)\quad \dot{z}(t)=Az(t)-Ge_y(t)+B(\hat{u}(t)+\nu_o) }$

と表せます。

●(7)と(8)をまとめて

$\displaystyle{(9)\quad \boxed{\begin{array}{l} \left[\begin{array}{c} \dot x_r(t)\\ \dot z(t) \end{array}\right] = \left[\begin{array}{cc} 0_{m\times m} & -C \\ 0_{n\times m} & A \\ \end{array}\right] \left[\begin{array}{c} x_r(t)\\ z(t) \end{array}\right] + \left[\begin{array}{c} I_p\\ -G \end{array}\right] e_y(t)\\ + \left[\begin{array}{c} r(t)\\ B(\hat{u}(t)+\nu_o) \end{array}\right] \end{array}} }$

ここで座標変換

$\displaystyle{(10)\quad \begin{array}{l} \left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right]= \underbrace{\left[\begin{array}{cc} I_{n} & 0_{n\times m} \\ S_1 & S_2 \\ \end{array}\right]}_{\bar T} \left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]\\ \Leftrightarrow \left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]= \underbrace{\left[\begin{array}{cc} I_{n} & 0_{n\times m} \\ -M & S_2^{-1} \\ \end{array}\right]}_{\bar{T}^{-1}} \left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right]\quad(M=S_2^{-1}S_1) \end{array} }$

を行うと

$\displaystyle{(11)\quad \begin{array}{l} \bar{T}^{-1} \left[\begin{array}{c} \dot{\tilde{z}}_1(t)\\ \dot{s}(t) \end{array}\right] = \underbrace{\left[\begin{array}{cc|c} 0_{m\times m} & -C_1 & -C_2 \\ 0_{n-m\times m} & A_{11} & A_{12} \\\hline 0_{m\times m} & A_{21} & A_{22} \end{array}\right]}_{\left[\begin{array}{cc} \tilde{A}_{11} & \tilde{A}_{12} \\ \tilde{A}_{21} & \tilde{A}_{22} \end{array}\right]} \left[\begin{array}{cc} I_{n} & 0_{n\times m} \\ -M & S_2^{-1} \\ \end{array}\right] \left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right]\\ + \left[\begin{array}{c} I_p\\ -G_1\\\hline -G_2 \end{array}\right] e_y(t) + \left[\begin{array}{c} r(t)\\ 0\\\hline B_2(\hat{u}(t)+\nu_o) \end{array}\right]\\ = \underbrace{\left[\begin{array}{cc} \tilde{A}_{11}-\tilde{A}_{12}M & \tilde{A}_{12}S_2^{-1}\\ \tilde{A}_{21}-\tilde{A}_{22}M & \tilde{A}_{22}S_2^{-1} \end{array}\right]}_{\tilde{A}\bar{T}^{-1}} \left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right] - \left[\begin{array}{c} \bar{G}_1\\ G_2 \end{array}\right] e_y(t)\\ + \left[\begin{array}{c} B_rr(t)\\ B_2(\hat{u}(t)+\nu_o) \end{array}\right]\\ (\left[\begin{array}{c} G_1\\ G_2 \end{array}\right] =\left[\begin{array}{c} A_{12}C_2^{-1} \\ A_{22}C_2^{-1}-C_2^{-1}A_{22}^S \end{array}\right], \quad\bar{G}_1= \left[\begin{array}{c} -I_p\\ G_1 \end{array}\right]) \end{array} }$

したがって

$\displaystyle{(12)\quad \begin{array}{l} \left[\begin{array}{c} \dot{\tilde{z}}_1(t)\\ \dot{s}(t) \end{array}\right] = \underbrace{\left[\begin{array}{cc} I_{n} & 0_{n\times m} \\ S_1 & S_2 \\ \end{array}\right] \left[\begin{array}{cc} \tilde{A}_{11}-\tilde{A}_{12}M & \tilde{A}_{12}S_2^{-1}\\ \tilde{A}_{21}-\tilde{A}_{22}M & \tilde{A}_{22}S_2^{-1} \end{array}\right]}_{\bar{T}\tilde{A}\bar{T}^{-1}} \left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right]\\ - \left[\begin{array}{cc} I_{n} & 0_{n\times m} \\ S_1 & S_2 \\ \end{array}\right] \left[\begin{array}{c} \bar{G}_1\\ G_2 \end{array}\right] e_y(t) + \left[\begin{array}{cc} I_{n} & 0_{n\times m} \\ S_1 & S_2 \\ \end{array}\right] \left[\begin{array}{c} B_rr(t)\\ B_2(\hat{u}(t)+\nu_o) \end{array}\right]\\ \end{array} }$

すなわち、第１ブロック行は

$\displaystyle{(13)\quad \begin{array}{l} \dot{\tilde{z}}_1(t)= \underbrace{(\tilde{A}_{11}-\tilde{A}_{12}M)}_{\bar{A}_{11}}\tilde{z}_1(t) +\underbrace{\tilde{A}_{12}S_2^{-1}}_{\bar{A}_{12}}\bar{z}_2(t)-\bar{G}_1e_y(t)+B_rr(t)\\ =\bar{A}_{11}\tilde{z}_1(t) +\bar{A}_{12}(\bar{z}_2(t)-S_rr(t))+(B_r+\bar{A}_{12}S_r)r(t)-\bar{G}_1e_y(t) \end{array} }$

第２ブロック行は(5)を用いて、 $\Lambda=S\bar{B}$ とおいて

$\displaystyle{(14)\quad \begin{array}{l} \dot{s}(t)= \underbrace{((S_1\bar{A}_{11}+S_2(\tilde{A}_{21}-\tilde{A}_{22}M))\tilde{z}_1(t)+(S_1\bar{A}_{12}+S_2\tilde{A}_{22}S_2^{-1})s(t)}_{S\tilde{A}\bar{T}^{-1}\left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right]=S\tilde{A}\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]}\\ -\underbrace{(S_1\bar{G}_1+S_2G_2)}_{\bar G_2}e_y(t)+S_1B_rr(t)\\ +\underbrace{S_2B_2}_{\Lambda}(L\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]+L_rr(t)+S_r\dot{r}(t)+\hat{\nu}_c+\nu_o)\\ =S\tilde{A}\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]-\bar{G}_2e_y(t)+S_1B_rr(t)\\ +\Lambda(-\Lambda^{-1}(S\tilde{A}-\Phi S)\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]-\Lambda^{-1}(\Phi S_r+S_2MB_r)r(t)+\Lambda^{-1}S_r\dot{r}(t))\\ +\Lambda(\hat{\nu}_c+\nu_o)\\ =S\tilde{A}\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]-\bar{G}_2e_y(t)+S_1B_rr(t)\\ -(S\tilde{A}-\Phi S)\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]-(\Phi S_r+S_2MB_r)r(t)+S_r\dot{r}(t)+\Lambda(\hat{\nu}_c+\nu_o)\\ =\Phi (\underbrace{S\left[\begin{array}{c} \tilde z_1(t)\\ \tilde z_2(t) \end{array}\right]}_{s(t)}-S_rr(t))+S_r\dot{r}(t)-\bar{G}_2e_y(t)+\Lambda(\hat{\nu}_c+\nu_o) \end{array} }$

(13)と(14)の結果は、次のようにまとめられます。

$\displaystyle{ \begin{array}{ll} (15.1) &\dot{\tilde{z}}_1(t) =\bar{A}_{11}\tilde{z}_1(t) +\bar{A}_{12}(\bar{z}_2(t)-S_rr(t))+(B_r+\bar{A}_{12}S_r)r(t)-\bar{G}_1e_y(t)\\ (15.2) &\dot{s}(t)=\Phi (s(t)-S_rr(t))+S_r\dot{r}(t)-\bar{G}_2e_y(t)+\Lambda(\hat{\nu}_c+\nu_o) \end{array} }$

ここで

$\displaystyle{(16)\quad \begin{array}{l} \underbrace{\left[\begin{array}{c} \zeta_1(t)\\ \zeta_2(t) \end{array}\right]}_{\zeta(t)}= \left[\begin{array}{c} \tilde z_1(t)\\ s(t) \end{array}\right]+ \left[\begin{array}{c} \bar{A}_{11}^{-1}(\bar{A}_{12}S_r+B_r)r_c\\ -S_rr(t) \end{array}\right] \end{array} }$

を定義すると

$\displaystyle{\quad \begin{array}{ll} (17.1) &\dot{\zeta}_1(t) =\bar{A}_{11}\underbrace{(\tilde{z}_1(t)+\bar{A}_{11}^{-1}(\bar{A}_{12}S_r+B_r)r_c)}_{\zeta_1(t)} +\bar{A}_{12}\zeta_2(t)\\ &+(B_r+\bar{A}_{12}S_r)\underbrace{(r(t)-r_c)}_{e_r(t)}-\bar{G}_1e_y(t)\\ (17.2) &\underbrace{\dot{s}(t)-S_r\dot{r}(t)}_{\dot{\zeta}_2(t)}=\Phi \underbrace{(s(t)-S_rr(t))}_{\zeta_2(t)}-\bar{G}_2e_y(t)+\Lambda(\hat{\nu}_c+\nu_o) \end{array} }$

すなわち

$\displaystyle{(18)\quad \begin{array}{l} \underbrace{\left[\begin{array}{c} \dot \zeta_1(t)\\ \dot \zeta_2(t) \end{array}\right]}_{\dot{\zeta}(t)}= \underbrace{\left[\begin{array}{cc} \bar{A}_{11} & \bar{A}_{12}\\ 0 & \Phi \end{array}\right]}_{A_c} \underbrace{\left[\begin{array}{c} \zeta_1(t)\\ \zeta_2(t) \end{array}\right]}_{\zeta(t)}\\ - \underbrace{\left[\begin{array}{c} \bar{G}_1\\ \bar{G}_2 \end{array}\right]}_{\bar{G}} \underbrace{e_y(t)}_{Ce(t)} + \underbrace{\left[\begin{array}{c} B_r+\bar{A}_{12}S_r\\ 0 \end{array}\right]}_{\bar{G}_r}e_r(t) + \underbrace{\left[\begin{array}{c} 0\\ \Lambda \end{array}\right]}_{\bar\Lambda}(\hat{\nu}_c+\nu_o) \end{array} }$

を得ます。これを誤差方程式(6)すなわち

$\displaystyle{(6')\quad \dot{e}(t)=\underbrace{(A-GC)}_{A_o}e(t)+B(\nu_o-\xi(t,x,\hat{u})) }$

と合わせて、閉ループ系は次式で表されます。

$\displaystyle{(19)\quad \begin{array}{l} \left[\begin{array}{c} \dot{e}(t)\\ \dot \zeta(t) \end{array}\right]= \underbrace{\left[\begin{array}{cc} A_o & 0\\ -\bar{G}C & A_c \end{array}\right]}_{A_G} \left[\begin{array}{c} e(t)\\ \zeta(t) \end{array}\right]\\ + \left[\begin{array}{c} 0\\ \bar{G}_r \end{array}\right]e_r(t) + \left[\begin{array}{c} B\\ \bar\Lambda \end{array}\right]\nu_o + \left[\begin{array}{c} 0\\ \bar\Lambda \end{array}\right]\hat{\nu}_c - \left[\begin{array}{c} B\\ 0 \end{array}\right]\xi(t,x,\hat{u}) \end{array} }$

ここまでの手順を、関数ob_smiとしてプログラムすることにします。

●以上に基づく設計手順を、数値例で示します。

MATLAB

%ex8_ob_smi.m
%-----
 clear all, close all
 A=[0 1 0;
    0 0 1;
   -0.0001 -0.0082 -0.1029];
 B=[0;0;1];
 C=[0.0001  0.0022  0.0053];
 Gamma=0;
%-----
 po=-0.2; 
 [G,Gn,F]=smobs3(A,B,C,po)
 rhoo=10;
%----- 
 p1=[-0.025 -0.03+j*0.025 -0.03-j*0.025];
 p2=-0.1;
 [L,Lr,Ldr,Ln,SS,Sr,P2]=ob_smi(A,B,C,p1,p2)
 rhoc=20;
%-----
%eof

図1 SMオブザーバベースSMI追従制御系のシミュレーション例

Note C92 閉ループ系の安定性

●閉ループ系は(19)と(3.2)を合わせて

$\displaystyle{ \begin{array}{ll} (1.1) & \dot{e}(t)=A_oe(t)+B(\nu_o-\xi(t,x,u))\\ (1.2) & \dot{\zeta}(t)=A_c\zeta(t)-\bar{G}Ce(t)+\bar{G}_re_r(t)+\bar{\Lambda}(\nu_o+\hat{\nu}_c)\\ (1.3) & \dot{e}_r(t)=\Gamma e_r(t) \end{array} }$

で表されます。これに対して次のリャプノフ関数を考えます。

$\displaystyle{(2)\quad \boxed{\begin{array}{l} V(e,\zeta,e_r) =e^T(t)Pe(t)+\zeta^T(t)\bar{P}\zeta(t)+e_r^T(t)P_re_r(t) \end{array}} }$

ここで、正定行列 $P$ 、 $\bar{P}$ 、 $P_r$ は、それぞれ安定行列 $A_o$ 、 $A_c$ 、 $\Gamma$ のリャプノフ行列とし、次のリャプノフ方程式の解とします。

$\displaystyle{ \begin{array}{ll} (3.1) & PA_o+A_o^TP=-Q\\ (3.2) & \bar{P}A_c+A_c^T\bar{P}=-\bar{P}\bar{G}CQ^{-1}C^T\bar{G}^T\bar{P}-\underline{\bar{P}\bar{Q}\bar{P}}\\ (3.3) & P_r\Gamma+\Gamma^TP_r=-Q_r-\underline{\bar{G}_r^T\bar{Q}^{-1}\bar{G}_r} \end{array} }$

● $\bar{P}$ は、リャプノフ方程式

$\displaystyle{(4)\quad \underbrace{\left[\begin{array}{cc} \bar{P}_1 & 0\\ 0 & \bar{P}_2 \end{array}\right]}_{\bar P} \underbrace{\left[\begin{array}{cc} \bar{A}_{11} & \bar{A}_{12}\\ 0 & \Phi \end{array}\right]}_{A_c} +\underbrace{\left[\begin{array}{cc} \bar{A}_{11} & \bar{A}_{12}\\ 0 & \Phi \end{array}\right]^T}_{A_c^T} \underbrace{\left[\begin{array}{cc} \bar{P}_1 & 0\\ 0 & \bar{P}_2 \end{array}\right]}_{\bar P}<0 }$

を満足するブロック対角行列とします。ここで、正定行列 $\bar{P}_1$ 、 $\bar{P}_2$ は、それぞれ安定行列 $\bar{A}_{11}$ 、 $\Phi$ に対する次のリャプノフ不等式の解です。

$\displaystyle{(5)\quad \bar{P}_1\bar{A}_{11}+\bar{A}_{11}^T\bar{P}_1<0 \Leftrightarrow \bar{A}_{11}\bar{P}_1^{-1}+\bar{P}_1^{-1}\bar{A}_{11}^T<0 }$
$\displaystyle{(6)\quad \bar{P}_2\Phi+\Phi^T\bar{P}_2<0 \Leftrightarrow \Phi\bar{P}_2^{-1}+\bar{P}_2^{-1}\Phi^T<0 }$

このとき、 $P$ と $\bar P$ は、次のリャプノフ不等式の解を構成するとします。

$\displaystyle{(7)\quad \underbrace{\left[\begin{array}{cc} P & 0\\ 0 & \bar{P} \end{array}\right]}_{P_G} \underbrace{\left[\begin{array}{cc} A_o & 0\\ -\bar{G}C & A_c \end{array}\right]}_{A_G} +\underbrace{\left[\begin{array}{cc} A_o & 0\\ -\bar{G}C & A_c \end{array}\right]^T}_{A_G^T} \underbrace{\left[\begin{array}{cc} P & 0\\ 0 & \bar{P} \end{array}\right]}_{P_G}<0 }$

これは次式と等価です。

$\displaystyle{(8)\quad \begin{array}{l} \underbrace{\left[\begin{array}{cc} PA_o & 0\\ -\bar{P}\bar{G}C & \bar{P}A_c \end{array}\right]}_{P_GA_G}+ \underbrace{\left[\begin{array}{cc} PA_o & 0\\ -\bar{P}\bar{G}C & \bar{P}A_c \end{array}\right]^T}_{(P_GA_G)^T} =\left[\begin{array}{cc} -Q & -C^T\bar{G}^T\bar{P}\\ -\bar{P}\bar{G}C & \bar{P}A_c+A_c^T\bar{P} \end{array}\right]<0\\ \Leftrightarrow \bar{P}A_c+A_c^T\bar{P}+\bar{P}\bar{G}CQ^{-1}C^T\bar{G}^T\bar{P}<0\\ \Leftrightarrow A_c\bar{P}^{-1}+\bar{P}^{-1}A_c^T+\bar{G}CQ^{-1}C^T\bar{G}^T<0\\ \Leftrightarrow A_c\bar{P}^{-1}+\bar{P}^{-1}A_c^T+\bar{G}CQ^{-1}C^T\bar{G}^T=-\bar{Q}\quad(\bar{Q}>0) \end{array} }$

ここで、新しいパラメータ $\bar{Q}$ を導入し、次の公式を用いています。

$\displaystyle{ \begin{array}{lll} && \left[\begin{array}{cc} P & M \\ M^T & Q \end{array}\right]<0\\ &\Leftrightarrow& P-MQ^{-1}M^T<0,\ Q<0\\ &\Leftrightarrow& P<0,\ Q-M^TP^{-1}M<0 \end{array} }$

いま適当な $\hat{Q}_1$ と $\hat{Q}_2$ を与えて、

$\displaystyle{(9)\quad \bar{A}_{11}\bar{P}_1^{-1}+\bar{P}_1^{-1}\bar{A}_{11}^T=-\hat{Q}_1 \quad(\hat{Q}_1>0)}$
$\displaystyle{(10)\quad \Phi\bar{P}_2^{-1}+\bar{P}_2^{-1}\Phi^T=-\hat{Q}_2\quad(\hat{Q}_2>0) }$

を解いて、 $\bar{P}_1$ と $\bar{P}_2$ を定めるものとします。このとき(7)は

$\displaystyle{(11)\quad \begin{array}{l} \underbrace{\left[\begin{array}{cc} \bar{A}_{11} & \bar{A}_{12}\\ 0 & \Phi \end{array}\right]}_{A_c} \underbrace{\left[\begin{array}{cc} \bar{P}_1^{-1} & 0\\ 0 & \bar{P}_2^{-1} \end{array}\right]}_{\bar{P}^{-1}} + \underbrace{\left[\begin{array}{cc} \bar{P}_1^{-1} & 0\\ 0 & \bar{P}_2^{-1} \end{array}\right]}_{\bar{P}^{-1}} \underbrace{\left[\begin{array}{cc} \bar{A}_{11} & \bar{A}_{12}\\ 0 & \Phi \end{array}\right]^T}_{A_c^T}\\ +\underbrace{\left[\begin{array}{c} \bar{G}_1\\ \bar{G}_2 \end{array}\right]}_{\bar{G}} \underbrace{CQ^{-1}C^T}_{Q_{22}} \underbrace{\left[\begin{array}{c} \bar{G}_1\\ \bar{G}_2 \end{array}\right]^T}_{\bar{G}^T}\\ = \underbrace{\left[\begin{array}{cc} \bar{A}_{11}\bar{P}_1^{-1} & \bar{A}_{12}\bar{P}_2^{-1}\\ 0 & \Phi \bar{P}_2^{-1} \end{array}\right]}_{A_c\bar{P}^{-1}} + \underbrace{\left[\begin{array}{cc} \bar{A}_{11}\bar{P}_1^{-1} & \bar{A}_{12}\bar{P}_2^{-1}\\ 0 & \Phi \bar{P}_2^{-1} \end{array}\right]^T}_{(A_c\bar{P}^{-1})^T}\\ +\underbrace{\left[\begin{array}{cc} \bar{G}_1Q_{22}\bar{G}_1^T &\bar{G}_1Q_{22}\bar{G}_2^T\\ \bar{G}_2Q_{22}\bar{G}_1^T &\bar{G}_2Q_{22}\bar{G}_2^T \end{array}\right]}_{\bar{G}CQ^{-1}C^T\bar{G}^T}\\ = \left[\begin{array}{cc} \bar{A}_{11}\bar{P}_1^{-1}+\bar{P}_1^{-1}\bar{A}_{11}^T+\bar{G}_1Q_{22}\bar{G}_1^T & \bar{A}_{12}\bar{P}_2^{-1}+\bar{G}_1Q_{22}\bar{G}_2^T\\ \bar{P}_2^{-1}\bar{A}_{12}^T+\bar{G}_2Q_{22}\bar{G}_1^T & \Phi\bar{P}_2^{-1}+\bar{P}_2^{-1}\Phi^T+\bar{G}_2Q_{22}\bar{G}_2^T \end{array}\right]\\ = \left[\begin{array}{cc} -\hat{Q}_1+\bar{G}_1Q_{22}\bar{G}_1^T & \bar{A}_{12}\bar{P}_2^{-1}+\bar{G}_1Q_{22}\bar{G}_2^T\\ \bar{P}_2^{-1}\bar{A}_{12}^T+\bar{G}_2Q_{22}\bar{G}_1^T & -\hat{Q}_2+\bar{G}_2Q_{22}\bar{G}_2^T \end{array}\right]<0\\ \Leftrightarrow \left\{\begin{array}{l} \bar{G}_1Q_{22}\bar{G}_1^T +(\bar{A}_{12}\bar{P}_2^{-1}+\bar{G}_1Q_{22}\bar{G}_2^T)\\ \times(\bar{G}_2Q_{22}\bar{G}_2^T -\hat{Q}_2)^{-1}(\bar{P}_2^{-1}\bar{A}_{12}^T+\bar{G}_2Q_{22}\bar{G}_1^T) <\hat{Q}_1\\ \bar{G}_2Q_{22}\bar{G}_2^T <\hat{Q}_2 \end{array}\right. \end{array} }$

となります。したがって、この制約を(8)の $\hat{Q}_1$ 、(9)の $\hat{Q}_2$ に付けておきます。

●以上の準備の下で次式が示され、閉ループ系のリャプノフ安定性が成り立ちます。

$\displaystyle{(12)\quad \begin{array}{l} \dot{V}(e,\zeta,e_r)=2e^T(t)P\dot{e}(t) +2\zeta^T(t)\bar{P}\dot{\zeta}(t) +2e_r^T(t)P_r\dot{e}_r(t)\\\\ =2e^T(t)P(A_oe(t)+B(\nu_o-\xi(t,x,u))\\ +2\zeta^T(t)\bar{P}(A_c\zeta(t)-\bar{G}Ce(t)+\bar{G}_re_r(t)+\bar{\Lambda}(\nu_o+\hat{\nu}_c))\\ +2e_r^T(t)P_r\Gamma e_r(t)\\\\ =e^T(t)\underbrace{(PA_o+A_o^TP)}_{-Q}e(t)+2e^T(t)PB(\nu_o-\xi(t,x,u))\\ +\zeta^T(t)\underbrace{(\bar{P}A_c+A_c^T\bar{P})}_{-\bar{P}\bar{G}CQ^{-1}C^T\bar{G}^T\bar{P}-\bar{P}\bar{Q}\bar{P}}\zeta(t)\\ -2\zeta^T(t)\bar{P}\bar{G}Ce(t)+2\zeta^T(t)\bar{P}\bar{G}_re_r(t)+2\zeta^T(t)\bar{P}\bar{\Lambda}(\nu_o+\hat{\nu}_c)\\ +e_r^T(t)\underbrace{(P_r\Gamma+\Gamma^TP_r)}_{-Q_r-\bar{G}_r^T\bar{Q}^{-1}\bar{G}_r}e_r(t)\\\\ =-e^T(t)Qe(t)+2e^T(t)PB(\nu_o-\xi(t,x,u))\\ -\zeta^T(t)\bar{P}\bar{G}CQ^{-1}C^T\bar{G}^T\bar{P}\zeta(t)-\underline{\zeta^T(t)\bar{P}\bar{Q}\bar{P}\zeta(t)}\\ -2\zeta^T(t)\bar{P}\bar{G}Ce(t)+2\zeta^T(t)\bar{P}\bar{G}_re_r(t)+2\zeta^T(t)\bar{P}\bar{\Lambda}(\nu_o+\hat{\nu}_c)\\ -e_r^T(t)Q_re_r(t)-\underline{e_r^T(t)\bar{G}_r^T\bar{Q}^{-1}\bar{G}_re_r(t)}\\\\ =\underbrace{2e^T(t)PB(\nu_o-\xi(t,x,u))}_{\le -2\gamma_o||FCe(t)||}+\underbrace{2\zeta^T(t)\bar{P}\bar{\Lambda}(\nu_o+\hat{\nu}_c)}_{\le -2\gamma_c||\bar{P}_2\zeta_2(t)||}\\ -\underbrace{(e^T(t)Qe(t)+2\zeta^T(t)\bar{P}\bar{G}Ce(t)+\zeta^T(t)\bar{P}\bar{G}CQ^{-1}C^T\bar{G}^T\bar{P}\zeta(t))}_{(e(t)+Q^{-1}C^T\bar{G}^T\bar{P}\zeta(t))^TQ(e(t)+Q^{-1}C^T\bar{G}^T\bar{P}\zeta(t))}\\ -\underbrace{\underline{\zeta^T(t)\bar{P}\bar{Q}\bar{P}\zeta(t)}-2\zeta^T(t)\bar{P}\bar{G}_re_r(t)+\underline{e_r^T(t)\bar{G}_r^T\bar{Q}^{-1}\bar{G}_re_r(t)}}_{(\zeta(t)-\bar{P}^{-1}\bar{Q}^{-1}\bar{G}_re_r(t))^T \bar{P}\bar{Q}\bar{P} (\zeta(t)-\bar{P}^{-1}\bar{Q}^{-1}\bar{G}_re_r(t))}\\ -e_r^T(t)Q_re_r(t)\\\\ \le -2\gamma_o||FCe(t)||-2\gamma_c||\bar{P}_2\zeta_2(t)||\\ -(e(t)+Q^{-1}C^T\bar{G}^T\bar{P}\zeta(t))^TQ(e(t)+Q^{-1}C^T\bar{G}^T\bar{P}\zeta(t))\\ -(\zeta(t)-\bar{P}^{-1}\bar{Q}^{-1}\bar{G}_re_r(t))^T \bar{P}\bar{Q}\bar{P} (\zeta(t)-\bar{P}^{-1}\bar{Q}^{-1}\bar{G}_re_r(t))\\ -e_r^T(t)Q_re_r(t)<0 \end{array} }$

ここで、次の平方完成を行っています。

$\displaystyle{ \begin{array}{ll} (13.1) &(e(t)+Q^{-1}C^T\bar{G}^T\bar{P}\zeta(t))^TQ(e(t)+Q^{-1}C^T\bar{G}^T\bar{P}\zeta(t))\\ &=e^T(t)Qe(t)+2\zeta^T(t)\bar{P}\bar{G}Ce(t)+\zeta^T(t)\bar{P}\bar{G}CQ^{-1}C^T\bar{G}^T\bar{P}\zeta(t)\\ (13.2) &(\zeta(t)-\bar{P}^{-1}\bar{Q}^{-1}\bar{G}_re_r(t))^T \bar{P}\bar{Q}\bar{P} (\zeta(t)-\bar{P}^{-1}\bar{Q}^{-1}\bar{G}_re_r(t))\\ &=\underline{\zeta^T(t)\bar{P}\bar{Q}\bar{P}\zeta(t)}-2\zeta^T(t)\bar{P}\bar{G}_re_r(t)+\underline{e_r^T(t)\bar{G}_r^T\bar{Q}^{-1}\bar{G}_re_r(t)} \end{array} }$

また、次が成り立つことを用いています。

$\displaystyle{ \begin{array}{ll} (14.1) &e^T(t)PB(\nu_o-\xi(t,x,u))\le -\gamma_o||FCe(t)||\\ (14.2) &\zeta^T(t)\bar{P}\bar{\Lambda}(\nu_o+\hat{\nu}_c)\le -\gamma_c||\bar{P}_2\zeta_2(t)|| \end{array} }$

まず、(14.1)は

$\displaystyle{(15)\quad \rho_o(\hat{u}_L,y)=\frac{k_1||\hat{u}_L||+\alpha(t,y)+k_1\gamma_c||\Lambda^{-1}||+\gamma_o}_{1-k_1\kappa(\Lambda)} }$

を変形して得られる

$\displaystyle{(16)\quad \begin{array}{l} \rho_o(\hat{u}_L,y)= k_1||\hat{u}_L||+\alpha(t,y)+k_1\gamma_c||\Lambda^{-1}||+\gamma_o+k_1\underbrace{||\Lambda||||\Lambda^{-1}||}_{\kappa(\Lambda)}\rho_o(\hat{u}_L,y)\\ =k_1(||\hat{u}_L||+||\Lambda^{-1}||\underbrace{(\rho_o(\hat{u}_L,y)||\Lambda|| +\gamma_c)}_{\rho_c(\hat{u}_L,y)})+\alpha(t,y)+\gamma_o\\ >k_1(||\hat{u}_L||+||\hat{\nu}_c||)+\alpha(t,y)+\gamma_o\\ \ge k_1(||\hat{u}_L+\hat{\nu}_c||)+\alpha(t,y)+\gamma_o\\ = k_1||\hat{u}||+\alpha(t,y)+\gamma_o \end{array} }$

を用いて

$\displaystyle{(17)\quad \begin{array}{l} 2e^T(t)PB(\nu_o-\xi(t,x,u))\\ =2e^T(t)C^TF^T(-\rho_o(\hat{u}_L,y)\frac{FCe(t)}{||FCe(t)||}-\xi(t,x,u))\\ =-2||FCe(t)||\rho_o(\hat{u}_L,y)-2e^T(t)C^TF^T\xi(t,x,u)\\ \le -2||FCe(t)||\rho_o(\hat{u}_L,y)+2||FCe(t)||||\xi(t,x,u)||\\ \le 2||FCe(t)||(k_1||\hat{u}||+\alpha(t,y)-\rho_o(\hat{u}_L,y))\\ \le -2\gamma_o||FCe(t)|| \end{array} }$

次に、(14.2)は

$\displaystyle{(18)\quad \begin{array}{l} 2\zeta^T(t)\bar{P}\bar{\Lambda}(\nu_o+\hat{\nu}_c)\\ =2\underbrace{\left[\begin{array}{c} \zeta_1(t)\\ \zeta_2(t) \end{array}\right]^T}_{\zeta^T(t)} \underbrace{\left[\begin{array}{cc} \bar{P}_1 & 0\\ 0 & \bar{P}_2 \end{array}\right]}_{\bar P} \underbrace{\left[\begin{array}{c} 0\\ \Lambda \end{array}\right]}_{\bar\Lambda}(\nu_o+\hat{\nu}_c)\\ =-2\zeta_2^T(t)\bar{P}_2\Lambda\rho_o(\hat{u}_L,y)\frac{FCe(t)}{||FCe(t)||}-2\zeta_2^T(t)\bar{P}_2\Lambda\rho_c(\hat{u}_L,y)\Lambda^{-1} \frac{\bar{P}_2\zeta_2(t)}{||\bar{P}_2\zeta_2(t)||}\\ \le 2||\bar{P}_2\zeta_2(t)||||\Lambda||\rho_o(\hat{u}_L,y)-2\rho_c(\hat{u}_L,y)||\bar{P}_2\zeta_2(t)||\\ =2||\bar{P}_2\zeta_2(t)||(\rho_o(\hat{u}_L,y)||\Lambda||-\rho_c(\hat{u}_L,y))\\ =2||\bar{P}_2\zeta_2(t)||(\rho_c(\hat{u}_L,y)-\gamma_c-\rho_c(\hat{u}_L,y))\\ =-2\gamma_c||\bar{P}_2\zeta_2(t)|| \end{array} }$

Note C92-2 スライディングモードの検討

このSM制御系では、超平面

$\displaystyle{(1)\quad {\cal S}_c=\left\{\left[\begin{array}{c} e\\ \zeta_1\\ \zeta_2 \end{array}\right]\in{\bf R}^{2n+p}: \left\{\begin{array}{l} e_y(t)=Ce(t)=Cz(t)-y(t)=0\\ \zeta_2(t)=s(t)-S_rr(t)=0 \end{array}\right. \right\} }$

においてスライディングモードが生じることを示します。そのために

$\displaystyle{ \begin{array}{ll} (2.1) & \dot{e}(t)=A_oe(t)+B(\nu_o-\xi(t,x,\hat{u}))\\ (2.2) & \dot{\zeta}_2(t)=\Phi\zeta_2(t)-\bar{G}_2\underbrace{e_y(t)}_{Ce(t)} +\Lambda(\hat{\nu}_c+\nu_o) \end{array} }$

に対して次のリャプノフ関数を考えます。

$\displaystyle{(3)\quad \boxed{V_c(e,\zeta_2)=e_y^T(t)P_2e_y(t)+\zeta_2(t)^T\bar{P}_2\zeta_2(t)} }$

この時間微分は次のように計算できます。

$\displaystyle{(4)\quad \dot{V}_c=2e_y^T(t)P_2\dot{e}_y(t)+2\zeta_2(t)^T\bar{P}_2\dot{\zeta}_2(t) }$

(4)の第１項は、 $P_2=F^T(CB)^{-1}$ より

$\displaystyle{(5)\quad \begin{array}{l} P_2\dot{e}_y(t)=F^T(CB)^{-1}C\dot{e}(t)\\ =F^T(CB)^{-1}CA_oe(t)+F^T(CB)^{-1}CB(\nu_o-\xi(t,x,\hat{u}))\\ =F^T(CB)^{-1}CA_oe(t)+F^T(\nu_o-\xi(t,x,\hat{u})) \end{array} }$

また、(14.1)は $PB=C^TF^T$ を適用して

$\displaystyle{(6)\quad \underbrace{e^T(t)C^T}_{e_y^T(t)}F^T(\nu_o-\xi(t,x,u))\le -\gamma_o||Fe_y(t)|| }$

したがって

$\displaystyle{(7)\quad \begin{array}{l} e_y^T(t)P_2\dot{e}_y(t)\\ =e_y^T(t)F^T(CB)^{-1}CA_oe(t)+e_y^T(t)F^T(\nu_o-\xi(t,x,\hat{u}))\\ \le ||Fe_y(t)||||(CB)^{-1}CA_oe(t)||-\gamma_o||Fe_y(t)||\\ =-||Fe_y(t)||(\gamma_o-||(CB)^{-1}CA_oe(t)||) \end{array} }$

(4)の第２項は、

$\displaystyle{(8)\quad \begin{array}{l} 2\zeta_2(t)^T\bar{P}_2\dot{\zeta}_2(t) =2\zeta_2(t)^T\bar{P}_2(\Phi\zeta_2(t)-\bar{G}_2e_y(t) +\Lambda(\hat{\nu}_c+\nu_o))\\ =-\zeta_2(t)^T(\bar{P}_2\Phi+\Phi^T\bar{P}_2)\zeta_2(t)-2\zeta_2(t)^T\bar{P}_2\bar{G}_2e_y(t)+2\zeta_2(t)^T\bar{P}_2\Lambda(\hat{\nu}_c+\nu_o)\\ \le 2||\bar{P}_2\zeta_2(t)||||\bar{G}_2e_y(t)||-2\gamma_c||\bar{P}_2\zeta_2(t)||\\ =-2||\bar{P}_2\zeta_2(t)||(\gamma_c-||\bar{G}_2e_y(t)||) \end{array} }$

したがって

$\displaystyle{(9)\quad \Omega_\eta=\left\{\left[\begin{array}{c} e\\ \zeta_1\\ \zeta_2 \end{array}\right]\in{\bf R}^{2n+p}: \left\{\begin{array}{l} ||(CB)^{-1}CA_oe(t)||<\gamma_o-\eta\\ ||\bar{G}_2e_y(t)||<\gamma_c-\eta \end{array}\right. \right\} }$

においては

$\displaystyle{(10)\quad \begin{array}{l} \dot{V}_c=2e_y^T(t)P_2\dot{e}_y(t) +2\zeta_2(t)^T\bar{P}_2\dot{\zeta}_2(t)\\ \le -2||Fe_y(t)||(\gamma_o-||(CB)^{-1}CA_oe(t)||) -2||\bar{P}_2\zeta_2(t)||(\gamma_c-||\bar{G}_2e_y(t)||)\\ \le -2\eta(||Fe_y(t)||+||\bar{P}_2\zeta_2(t)||)\\ \le -2\eta \kappa \sqrt{V_c}\\ (\kappa=\min\{\underline{\sigma}(F),\underline{\sigma}(\bar{P}_2)\} \times\sqrt{\max\{\bar{\sigma}(P_2),\bar{\sigma}(\bar{P}_2)\}}) \end{array} }$

が成り立ち、これより有限時間で $V_c=0$ となり、超平面 ${\cal S}_c$ に突入することが分かります（ $\kappa$ の表現式はEdwards and Spurgeonによるものです）。