出力FB用変数変換

出力FB用変数変換…Homework

[1]　 $n$ 次系

$\displaystyle{(1)\quad \left\{\begin{array}{l} \dot{x}=Ax+Bu\\ y=Cx \end{array}\right. }$

に対する出力フィードバック

$\displaystyle{(2)\quad \left\{\begin{array}{l} \dot{x}_K=A_Kx_K+B_Ky \\ u=C_Kx_K+D_Ky \end{array}\right. }$

による閉ループ系

$\displaystyle{(3)\quad \left[\begin{array}{c} \dot{x} \\ \dot{x}_K \end{array}\right] = \underbrace{ \left[\begin{array}{ccc} A+BD_KC & BC_K \\ B_KC & A_K \end{array}\right] }_{A_{CL}} \left[\begin{array}{c} x \\ x_K \end{array}\right] }$

を考えます。

出力フィードバックに関する変数変換として、以下の議論が知られています。

$\displaystyle{(4)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{cc} S & N \\ N^T & S' \end{array}\right] }_{X_{CL}} \underbrace{ \left[\begin{array}{cc} R & M \\ M^T & R' \end{array}\right] }_{Y_{CL}=X^{-1}_{CL}} = \left[\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right] \quad(SR+NM^T=I)\\ \underbrace{ \left[\begin{array}{cc} R & M \\ M^T & R' \end{array}\right] }_{Y_{CL}} \underbrace{ \left[\begin{array}{cc} S & N \\ N^T & S' \end{array}\right] }_{X_{CL}=Y^{-1}_{CL}} = \left[\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right] \quad(RS+MN^T=I) \end{array} }$

が与えられるとき、次の命題が成立します（Note B41参照）。

$\displaystyle{(5)\quad X_{CL}>0 \Leftrightarrow Y_{CL}>0 \Leftrightarrow \left[\begin{array}{cc} R & I \\ I & S \end{array}\right] >0 }$

このとき、次が成り立ちます。

$\displaystyle{(6)\quad X_{CL}= \underbrace{ \left[\begin{array}{cc} I & S \\ 0 & N^T \end{array}\right] }_{\Pi_2} \underbrace{ \left[\begin{array}{cc} R & I \\ M^T & 0 \end{array}\right]^{-1} }_{\Pi_1^{-1}} = \underbrace{ \left[\begin{array}{cc} R & M \\ I & 0 \end{array}\right]^{-1} }_{\Pi_1^{-T}} \underbrace{ \left[\begin{array}{cc} I & 0 \\ S & N \end{array}\right] }_{\Pi_2^T} }$

$\displaystyle{(7)\quad Y_{CL}= \underbrace{ \left[\begin{array}{cc} R & I \\ M^T & 0 \end{array}\right] }_{\Pi_1} \underbrace{ \left[\begin{array}{cc} I & S \\ 0 & N^T \end{array}\right]^{-1} }_{\Pi_2^{-1}} = \underbrace{ \left[\begin{array}{cc} I & 0 \\ S & N \end{array}\right]^{-1} }_{\Pi_2^{-T}} \underbrace{ \left[\begin{array}{cc} R & M \\ I & 0 \end{array}\right] }_{\Pi_1^T} }$

$\displaystyle{(8)\quad \left[\begin{array}{cc} R & I \\ I & S \end{array}\right] = \underbrace{ \left[\begin{array}{cc} R & M \\ I & 0 \end{array}\right] }_{\Pi_1^T} \underbrace{ \left[\begin{array}{cc} I & S \\ 0 & N^T \end{array}\right] }_{\Pi_2} = \underbrace{ \left[\begin{array}{cc} I & 0 \\ S & N \end{array}\right] }_{\Pi_2^T} \underbrace{ \left[\begin{array}{cc} R & I \\ M^T & 0 \end{array}\right] }_{\Pi_1} }$

以下では、次の変換を用います。

$\displaystyle{(9)\quad \begin{array}{l} \Pi_2^TA_{CL}\Pi_1= \left[\begin{array}{cc} I & 0 \\ S & N \end{array}\right] \left[\begin{array}{ccc} A+BD_KC & BC_K \\ B_KC & A_K \end{array}\right] \left[\begin{array}{cc} R & I \\ M^T & 0 \end{array}\right] \\ =\left[\begin{array}{cc} I & 0 \\ S & N \end{array}\right] \left[\begin{array}{ccc} (A+BD_KC)R+BC_KM^T & A+BD_KC \\ B_KCR+A_KM^T & B_KC \end{array}\right] \\ =\left[\begin{array}{ccc} (A+BD_KC)R+BC_KM^T & \\ S(A+BD_KC)R+SBC_KM^T+NB_KCR+NA_KM^T & \end{array}\right. \\ \left.\begin{array}{ccc} & A+BD_KC \\ & S(A+BD_KC)+NB_KC \end{array}\right] \\ =\left[\begin{array}{cc} AR+B{\cal C}_K & A+BD_KC \\ {\cal A}_K & SA+{\cal B}_KC \end{array}\right] \end{array} }$

ただし

$\displaystyle{(10)\quad \begin{array}{lll} {\cal A}_K&=&NA_KM^T+NB_KCR+SBC_KM^T+S(A+BD_KC)R \nonumber\\ {\cal B}_K&=&NB_K+SBD_K \nonumber\\ {\cal C}_K&=&C_KM^T+D_KCR \nonumber \end{array} }$

[2] $\lambda(A)\subset{\rm\bf LHP}$ となるLMI条件は、次の通りでした。

$\displaystyle{(11) \begin{array}{lll} &&\lambda(A)\subset {\rm\bf LHP}=\{s=x+jy\in{\rm\bf C}: s+s^*<0 \}\\ &\Leftrightarrow& \exists X>0:\ XA+A^TX<0\\ &\Leftrightarrow& \exists Y>0:\ AY+YA^T<0 \end{array} }$

したがって、 $\lambda(A_{CL})\subset{\rm\bf LHP}$ となるLMI条件は、次のようになります。

$\displaystyle{(12)\quad \exists Y_{CL}>0: A_{CL}Y_{CL}+Y_{CL}A_{CL}^T<0 }$

$Y_{CL}=\Pi_1\Pi_2^{-1}=\Pi_2^{-T}\Pi_1^T$ を代入して

$\displaystyle{(13)\quad \exists \Pi_1\Pi_2^{-1}>0: A_{CL}\Pi_1\Pi_2^{-1}+\Pi_2^{-T}\Pi_1^TA_{CL}^T<0 }$

左から $\Pi_2^T$ 、右から $\Pi_2$ をかけると、次のようなLMIとなります。

$\displaystyle{(14)\quad \exists \Pi_2^T\Pi_1>0: \Pi_2^TA_{CL}\Pi_1+(\Pi_2^TA_{CL}\Pi_1)^T<0 }$

すなわち

$\displaystyle{(15)\quad \exists \left[\begin{array}{cc} R & I \\ I & S \end{array}\right]>0: }$
$\displaystyle{ \left[\begin{array}{cc} AR+B{\cal C}_K & A+BD_KC \\ {\cal A}_K & SA+{\cal B}_KC \end{array}\right] + \left[\begin{array}{cc} AR+B{\cal C}_K & A+BD_KC \\ {\cal A}_K & SA+{\cal B}_KC \end{array}\right]^T <0 }$

これを解いて $R=R^T,S=S^T,{\cal A}_K,{\cal B}_K,{\cal C}_K,D_K$ を求め、次式によって出力フィードバックのゲインを決定します。

$\displaystyle{(16)\quad \begin{array}{lll} A_K=N^{-1}({\cal A}_K-S(A-BD_KC)R-{\cal B}_KCR-SB{\cal C}_K)M^{-T} \nonumber\\ B_K=N^{-1}({\cal B}_K-SBD_K) \nonumber\\ C_K=({\cal C}_K-D_KCR)M^{-T} \nonumberr \end{array} }$

ただし、 $N^{-1}$ と $M^{-T}$ は次の分解を行って求めます。

$\displaystyle{(17)\quad I-SR=U\Sigma V^T=\underbrace{U\Sigma^{1/2}}_{N}\underbrace{\Sigma^{1/2}V^T}_{M^T} \Rightarrow N^{-1}=\Sigma^{-1/2}U^T,\ M^{-T}=V\Sigma^{-1/2} }$

または、 $I-SR=(I-RS)^T=NM^T$ に注意して

$\displaystyle{(18)\quad \begin{array}{l} I-RS=V\Sigma U^T=\underbrace{V\Sigma^{1/2}}_{M}\underbrace{\Sigma^{1/2}U^T}_{N^T}\\ \Rightarrow M^{-T}=(M^{-1})^T=(\Sigma^{-1/2}V^T)^T=V\Sigma^{-1/2},\\ N^{-1}=(N^{-T}))^T=(U\Sigma^{-1/2})^T=\Sigma^{-1/2}U^T \end{array} }$

演習B41…Flipped Classroom
$1^\circ$ 次のコードを参考にして、安定化出力FBを求める関数を作成せよ。

MATLAB

%of_syn_lmi0.m 
%-----
 clear all, close all
 A=[0 1;0 0]; B=[0;1]; C=[1 0];
%-----
 setlmis([]); 
 [R,xxx,Rdec]=lmivar(1,[2 1]); 
 [S,xxx,Sdec]=lmivar(1,[2 1]); 
 Ak=lmivar(2,[2 2]); 
 Bk=lmivar(2,[2 1]); 
 Ck=lmivar(2,[1 2]); 
 Dk=lmivar(2,[1 1]); 
%-----
 lmiPL=newlmi; 
 lmiterm([lmiPL 1 1 R],A,1,'s');      %#1:R*A'+AR 
 lmiterm([lmiPL 1 1 Ck],B,1,'s');     %#1:Ck'*B'+B*Ck 
 lmiterm([lmiPL 2 1 Ak],1,1);         %#1:Ak 
 lmiterm([lmiPL 1 2 0],A);            %#1:A 
 lmiterm([lmiPL 1 2 Dk],B,C);         %#1:B*Dk*C 
 lmiterm([lmiPL 2 2 S],1,A,'s');      %#1:A'*S+S*A 
 lmiterm([lmiPL 2 2 Bk],1,C,'s');     %#1:C'*Bk'+Bk*C 
%-----
 posX=-newlmi;
 lmiterm([posX 1 1 R],1,1);           %#2:R 
 lmiterm([posX 2 1 0],1);             %#2:I 
 lmiterm([posX 2 2 S],1,1);           %#2:S 
%-----
 LMIs=getlmis; 
 [tmin,xfeas]=feasp(LMIs);   
 R=dec2mat(LMIs,xfeas,R); 
 S=dec2mat(LMIs,xfeas,S); 
 ak=dec2mat(LMIs,xfeas,Ak); 
 bk=dec2mat(LMIs,xfeas,Bk); 
 ck=dec2mat(LMIs,xfeas,Ck); 
 dk=dec2mat(LMIs,xfeas,Dk); 
 [u,sd,v]=svd(eye(size(A,1)-S*R);  
 sd=diag(sqrt(1./diag(sd))); 
 Ni=sd*u'; Mti=v*sd; 
 AK=Ni*(ak-S*(A-B*dk*C)*R-bk*C*R-S*B*ck)*Mti; 
 BK=Ni*(bk-S*B*dk); 
 CK=(ck-dk*C*R)*Mti; 
 DK=dk; 
%-----        
 pl=eig(A)          
 ACL=[A+B*DK*C B*CK;          
      BK*C AK];          
 plCL=eig(ACL)          
 close all,figure(1)          
 dregion(0,0,5,pi/2,5*[-1,1,-1,1])            
 plot(real(pl),imag(pl),'x',real(plCL),imag(plCL),'*')          
%-----
%eof

Note B41　補題

●　 ${\bf Projection\ Lemma}$

$\displaystyle{(1)\quad \exists \Theta:\ \Psi+P^T\Theta^TQ+Q^T\Theta P < 0 \Leftrightarrow \left\{\begin{array}{l} W_P^T\Psi W_P<0 \quad (PW_P=0)\\ W_Q^T\Psi W_Q<0 \quad (QW_Q=0) \end{array}\right. }$

実際

$\displaystyle{(2)\quad \exists U_{PQ}:\ \left[\begin{array}{c} P \\ Q \end{array}\right]U_{PQ}=0 \quad\Rightarrow\quad \left\{\begin{array}{l} \exists U_{P}:\ P\underbrace{ \left[\begin{array}{cc} U_{PQ} & U_P \end{array}\right] }_{W_P}=0\\ \exists U_{Q}:\ Q\underbrace{ \left[\begin{array}{cc} U_{PQ} & U_Q \end{array}\right] }_{W_Q}=0\\ \end{array}\right. }$

を満足する $U_{PQ},U_P,U_Q$ を定めますと、次が成り立ちます。

$\displaystyle{(3)\quad T= \left[\begin{array}{cccc} U_{PQ} & U_P & U_Q & V \end{array}\right] \quad\Rightarrow\quad \left\{\begin{array}{l} PT= \left[\begin{array}{cccc} 0 & 0 & P_3 & P_4 \end{array}\right]\\ QT= \left[\begin{array}{cccc} 0 & Q_2 & 0 & Q_4 \end{array}\right] \end{array}\right. }$

この $T$ を用いて次が成り立ちます。

$\displaystyle{(4)\quad \Psi+P^T\Theta^TQ+Q^T\Theta P < 0 \ \Leftrightarrow\ T^T\Psi T+(PT)^T\Theta^T(QT)+(QT)^T\Theta (PT) < 0 }$
$\displaystyle{ \Leftrightarrow \left[\begin{array}{cccc} \Psi_{11} & \Psi_{12} & \Psi_{13} & \Psi_{14} \\ \Psi_{12}^T & \Psi_{22} & \Psi_{23}& \Psi_{24}\\ \Psi_{13}^T & \Psi_{23}^T & \Psi_{33} & \Psi_{34}\\ \Psi_{14}^T & \Psi_{24}^T & \Psi_{34}^T & \Psi_{44} \end{array}\right] +\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & \Theta_{11}^T & 0 & \Theta_{12}^T \\ 0 & \Theta_{21}^T & 0 & \Theta_{22}^T \end{array}\right] +\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & \Theta_{11} & \Theta_{21} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \Theta_{12} & \Theta_{22} \end{array}\right] < 0 }$
$\displaystyle{ \Leftrightarrow \left[\begin{array}{ccc|c} \Psi_{11} & \Psi_{12} & \Psi_{13} & \Psi_{14} \\ \Psi_{12}^T & \Psi_{22} & \Psi_{23}+\Theta_{11} & \Psi_{24}+\Theta_{21} \\ \Psi_{13}^T & \Psi_{23}^T+\Theta_{11}^T & \Psi_{33} & \Psi_{34}+\Theta_{12} \\ \hline \Psi_{14}^T & \Psi_{24}^T+\Theta_{21}^T & \Psi_{34}^T+\Theta_{12}^T & \Psi_{44}+\Theta_{22}+\Theta_{22}^T \end{array}\right]<0}$
$\displaystyle{\Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{cccc} \Psi_{11} & \Psi_{12} & \Psi_{13} \\ \Psi_{12}^T & \Psi_{22} & \Psi_{23}+\Theta_{11} \\ \Psi_{13}^T & \Psi_{23}^T+\Theta_{11}^T & \Psi_{33} \end{array}\right] < 0 \\ \quad\\ \Psi_{44}+\Theta_{22}+\Theta_{22}^T-\\ \left[\begin{array}{c} \Psi_{14} \\ \Psi_{24}+\Theta_{21} \\ \Psi_{34}+\Theta_{12} \end{array}\right]^T \left[\begin{array}{cccc} \Psi_{11} & \Psi_{12} & \Psi_{13} \\ \Psi_{12}^T & \Psi_{22} & \Psi_{23}+\Theta_{11} \\ \Psi_{13}^T & \Psi_{23}^T+\Theta_{11}^T & \Psi_{33} \end{array}\right]^{-1} \left[\begin{array}{c} \Psi_{14} \\ \Psi_{24}+\Theta_{21}^T \\ \Psi_{34}+\Theta_{12}^T \end{array}\right] < 0 \end{array}\right. }$

この第２式は、与えられた $\Theta_{11},\Theta_{12},\Theta_{21}$ に対して、 $\Theta_{22}=Q_4^T\Theta P_4$ を適当に選んで、常に満足させることができます。一方、第１式は

$\displaystyle{(5)\quad \begin{array}{l} \left[\begin{array}{ccc} I & 0 & 0 \\ -\Psi_{12}^T\Psi_{11}^{-1} & I & 0 \\ -\Psi_{13}^T\Psi_{11}^{-1} & 0 & I \end{array}\right] \left[\begin{array}{cccc} \Psi_{11} & \Psi_{12} & \Psi_{13} \\ \Psi_{12}^T & \Psi_{22} & \Psi_{23}+\Theta_{11} \\ \Psi_{13}^T & \Psi_{23}^T+\Theta_{11}^T & \Psi_{33} \end{array}\right]\\ \times \left[\begin{array}{ccc} I & -\Psi_{11}^{-1}\Psi_{12} & -\Psi_{11}^{-1}\Psi_{13} \\ 0 & I & 0 \\ 0 & 0 & I \end{array}\right]<0 \end{array} }$
$\displaystyle{ \Leftrightarrow \left[\begin{array}{ccc} \Psi_{11} & 0 & 0 \\ 0 & \Psi_{22}-\Psi_{12}^T\Psi_{11}^{-1}\Psi_{12} & \Theta_{11}+\Psi_{23}-\Psi_{12}^T\Psi_{11}^{-1}\Psi_{13} \\ 0 & \Theta_{11}^T+\Psi_{23}^T-\Psi_{13}^T\Psi_{11}^{-1}\Psi_{12} & \Psi_{33}-\Psi_{13}^T\Psi_{11}^{-1}\Psi_{13} \end{array}\right] < 0 }$

ここで、 $\Theta_{11}=Q_2^T\Theta P_3$ を

$\displaystyle{(6)\quad \Theta_{11}+\Psi_{23}-\Psi_{12}^T\Psi_{11}^{-1}\Psi_{13}=0 }$

を満足するように選ぶと、次の条件を得ます。

$\displaystyle{(7)\quad \underbrace{ \left[\begin{array}{cc} \Psi_{11} & \Psi_{12} \\ \Psi_{12}^T & \Psi_{22} \end{array}\right] }_{W_P^T\Psi W_P} < 0,\ \underbrace{ \left[\begin{array}{cc} \Psi_{11} & \Psi_{13} \\ \Psi_{13}^T & \Psi_{33} \end{array}\right] }_{W_Q^T\Psi W_Q}< 0 }$

●　 ${\bf Completion\ Lemma}$

$\displaystyle{(8)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{cc} S & N \\ X & S' \end{array}\right] }_{W} \underbrace{ \left[\begin{array}{cc} R & M \\ Y & R' \end{array}\right] }_{V=W^{-1}} = \left[\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right] \quad(I=SR+NY=SR-NS'^{-1}XR)\\ \underbrace{ \left[\begin{array}{cc} R & M \\ Y & R' \end{array}\right] }_{V} \underbrace{ \left[\begin{array}{cc} S & N \\ X & S' \end{array}\right] }_{W=V^{-1}} = \left[\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right] \quad(I=RS+MX=RS-MR'^{-1}YS) \end{array} }$

が与えられるとき、次が成り立ちます。

$\displaystyle{(9)\quad W+W^T>0 \Leftrightarrow V+V^T>0 \Leftrightarrow \left\{\begin{array}{l} R+R^T>0\\ S+S^T>0 \end{array}\right. }$

実際

$\displaystyle{(10)\quad \left\{\begin{array}{l} I=SR-NS'^{-1}XR\\ I=RS-MR'^{-1}YS \end{array}\right. \Rightarrow \left\{\begin{array}{l} S-R^{-1}=NS'^{-1}X\\ R-S^{-1}=MR'^{-1}Y \end{array}\right. }$

に注意して

$\displaystyle{(11)\quad S-R^{-1}=L_{\ell}L_r }$

のように分解し、 $N=L_{\ell}S', X=L_r$ と選ぶと

$\displaystyle{(12)\quad 0< \underbrace{ \left[\begin{array}{cc} S & L_{\ell}S' \\ L_r & S' \end{array}\right] }_W + \underbrace{ \left[\begin{array}{cc} S^T & L_r^T \\ S'\,^TL_{\ell}^T & S'\,^T \end{array}\right] }_{W^T} }$
$\displaystyle{ = \underbrace{ \left[\begin{array}{cc} S+S^T & L_r^T \\ L_r & 0 \end{array}\right] }_{\Psi} + \underbrace{ \left[\begin{array}{c} 0 \\ I \end{array}\right] }_{P^T} \underbrace{ S'\,^T }_{\Theta^T} \underbrace{ \left[\begin{array}{cc} L_\ell^T & I \end{array}\right] }_{Q} + \underbrace{ \left[\begin{array}{c} L_{\ell} \\ I \end{array}\right] }_{Q^T} \underbrace{ S' }_{\Theta} \underbrace{ \left[\begin{array}{cc} 0 & I \end{array}\right] }_{P}$
$\Leftrightarrow \left\{\begin{array}{l} \underbrace{ \left[\begin{array}{cc} I & 0 \end{array}\right] }_{W_P^T} \underbrace{ \left[\begin{array}{cc} S+S^T & L_r^T \\ L_r & 0 \end{array}\right] }_{\Psi} \underbrace{ \left[\begin{array}{c} I \\ 0 \end{array}\right] }_{W_P}<0 \quad (\underbrace{ \left[\begin{array}{cc} 0 & I \end{array}\right] }_{P} \underbrace{ \left[\begin{array}{c} I \\ 0 \end{array}\right] }_{W_P}=0)\\ \underbrace{ \left[\begin{array}{cc} I & -L_\ell \end{array}\right] }_{W_Q^T} \underbrace{ \left[\begin{array}{cc} S+S^T & L_r^T \\ L_r & 0 \end{array}\right] }_{\Psi} \underbrace{ \left[\begin{array}{c} I \\ -L_\ell^T \end{array}\right] }_{W_Q}<0 \quad (\underbrace{ \left[\begin{array}{cc} L_\ell^T & I \end{array}\right] }_{Q} \underbrace{ \left[\begin{array}{c} I \\ -L_\ell^T \end{array}\right] }_{W_Q}=0)\\ \end{array}\right. }$
$\Leftrightarrow \left\{\begin{array}{l} S+S^T<0 \\ \underbrace{S-L_\ell L_r}_{R^{-1}}+ \underbrace{S^T-L_r^TL_\ell^T}_{R^{-T}}<0 \Leftrightarrow R+R^T<0 \end{array}\right. }$

一方、

$\displaystyle{(13)\quad R-S^{-1}=L_{\ell}L_r }$

のように分解し、 $M=L_{\ell}R', Y=L_r$ と選ぶと

$\displaystyle{(14)\quad 0< \underbrace{ \left[\begin{array}{cc} R & L_{\ell}R' \\ L_r & R' \end{array}\right] }_V + \underbrace{ \left[\begin{array}{cc} R^T & L_r^T \\ R'\,^TL_{\ell}^T & R'\,^T \end{array}\right] }_{V^T} }$
$\displaystyle{ = \underbrace{ \left[\begin{array}{cc} R+R^T & L_r^T \\ L_r & 0 \end{array}\right] }_{\Psi} + \underbrace{ \left[\begin{array}{c} 0 \\ I \end{array}\right] }_{P^T} \underbrace{ R'\,^T }_{\Theta^T} \underbrace{ \left[\begin{array}{cc} L_\ell^T & I \end{array}\right] }_{Q} + \underbrace{ \left[\begin{array}{c} L_{\ell} \\ I \end{array}\right] }_{Q^T} \underbrace{ R' }_{\Theta} \underbrace{ \left[\begin{array}{cc} 0 & I \end{array}\right] }_{P}$
$\Leftrightarrow \left\{\begin{array}{l} \underbrace{ \left[\begin{array}{cc} I & 0 \end{array}\right] }_{W_P^T} \underbrace{ \left[\begin{array}{cc} R+R^T & L_r^T \\ L_r & 0 \end{array}\right] }_{\Psi} \underbrace{ \left[\begin{array}{c} I \\ 0 \end{array}\right] }_{W_P}<0 \quad (\underbrace{ \left[\begin{array}{cc} 0 & I \end{array}\right] }_{P} \underbrace{ \left[\begin{array}{c} I \\ 0 \end{array}\right] }_{W_P}=0)\\ \underbrace{ \left[\begin{array}{cc} I & -L_\ell \end{array}\right] }_{W_Q^T} \underbrace{ \left[\begin{array}{cc} R+R^T & L_r^T \\ L_r & 0 \end{array}\right] }_{\Psi} \underbrace{ \left[\begin{array}{c} I \\ -L_\ell^T \end{array}\right] }_{W_Q}<0 \quad (\underbrace{ \left[\begin{array}{cc} L_\ell^T & I \end{array}\right] }_{Q} \underbrace{ \left[\begin{array}{c} I \\ -L_\ell^T \end{array}\right] }_{W_Q}=0)\\ \end{array}\right. }$
$\Leftrightarrow \left\{\begin{array}{l} R+R^T<0 \\ \underbrace{S-L_\ell L_r}_{S^{-1}}+ \underbrace{R^T-L_r^TL_\ell^T}_{S^{-T}}<0 \Leftrightarrow S+S^T<0 \end{array}\right. }$

●特に、 $S=S^T,R=R^T,S'=S'\,^T,R'=R'\,^T$ 　かつ

$\displaystyle{(15)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{cc} S & N \\ N^T & S' \end{array}\right] }_W \underbrace{ \left[\begin{array}{cc} R & M \\ M^T & R' \end{array}\right] }_{V=W^{-1}} = \left[\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right] \quad(SR+NM^T=I) \\ \underbrace{ \left[\begin{array}{cc} R & M \\ M^T & R' \end{array}\right] }_{V} \underbrace{ \left[\begin{array}{cc} S & N \\ N^T & S' \end{array}\right] }_{W=V^{-1}} = \left[\begin{array}{cc} I & 0 \\ 0 & I \end{array}\right] \quad(RS+MN^T=I) \end{array} }$

のときは

$\displaystyle{(16)\quad W>0 \quad\Leftrightarrow\quad V>0 \quad\Leftrightarrow\quad R>0,\ S>0 }$

となりますが、 $R$ と $S$ は

$\displaystyle{(17)\quad \left\{\begin{array}{l} S-R^{-1}=NS'^{-1}N^T>0\\ R-S^{-1}=MR'^{-1}M^T>0 \end{array}\right. }$

を考慮して、次の命題を得ます。

$\displaystyle{(15)\quad W>0 \quad\Leftrightarrow\quad V>0 \quad\Leftrightarrow\quad \left[\begin{array}{cc} R & I \\ I & S \end{array}\right]>0 }$

●以上では、次のシュール補元に関する公式を多用しました。

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

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

制御系CAD

制御を意識したモノつくりを目指して

出力FB用変数変換