LQG制御 | 制御系CAD

LQG制御…Homework

[1]　次のようなオブザーバベース・コントローラによる閉ループ系を考えます。ここで，新しい入力 $w$ と $v$ がそれぞれ $W>0$ と $V>0$ の平方根行列（ $X>0(\ge0)$ に対し $YY=X$ を満足する行列 $Y>0(\ge0)$ を $X^{\frac{1}{2}}$ で表す）により重み付けられて， $n$ 次系の入力側（ $B'$ を介して）と出力側に設置されています。また新しい出力 $z=C'x$ と入力 $u$ が取り出されており，それぞれ $Q>0$ と $R>0$ の平方根行列により重み付けられています。

図１　LQG制御系設計の枠組み

可制御かつ可観測な制御対象

$\displaystyle{(1)\quad \left\{\begin{array}{ll} \dot{x}(t)=Ax(t)+Bu(t)+B'W^{\frac{1}{2}}w(t)&(x(t)\in{\bf R}^n,u(t),w(t)\in{\bf R}^m)\\ y(t)=Cx(t)+V^{\frac{1}{2}}v(t)&(y(t),v(t)\in{\bf R}^p)\\ z(t)=C'x(t)&(z(t)\in{\bf R}^p) \end{array}\right. }$

を安定化するオブザーバベース・コントローラ

$\displaystyle{(2)\quad \left\{\begin{array}{l} \dot{\hat{x}}(t)=(A-HC-BF)\hat{x}(t)+Hy(t)\\ u(t)=-F\hat{x}(t) \end{array}\right. }$

を２次形式評価関数

$\displaystyle{(3)\quad J=\int_0^\infty(z^T(t)Qz(t)+u^T(t)Ru(t))\,dt\quad (Q>0,R>0) }$

を最小にするように、 $F$ と $H$ を決定する問題を考えます。これらは

$\displaystyle{(4)\quad \boxed{{\bf CARE}:\Pi A+A^T\Pi-\Pi BR^{-1}B^T\Pi+C'^TQC'=0} }$

$\displaystyle{(5)\quad \boxed{{\bf FARE}:\Gamma A^T+A\Gamma-\Gamma C^TV^{-1}C\Gamma+B'WB'^T=0} }$

を満足する $\Pi>0$ と $\Gamma>0$ を用いて、次のように与えられます。

$\displaystyle{(6)\quad \boxed{F=R^{-1}B^T\Pi} }$

$\displaystyle{(7)\quad \boxed{H=(V^{-1}C\Gamma)^T} }$

このような制御方式をLQG制御と呼びます。

[2]　上の安定な閉ループ系は次式で表されます。

$\displaystyle{(8)\quad \boxed{\begin{array}{l} \left[\begin{array}{cc} \dot{x}(t) \\ \dot{\hat{x}}(t) \end{array}\right]= \underbrace{ \left[\begin{array}{cc} A & -BF \\ HC&A-HC-BF \end{array}\right] }_{A_{CL1}} \left[\begin{array}{cc} {x}(t) \\ \hat{x}(t) \end{array}\right] + \underbrace{ \left[\begin{array}{cc} B'W^{\frac{1}{2}} & 0 \\ 0 & HV^{\frac{1}{2}} \end{array}\right] }_{B_{CL1}} \left[\begin{array}{cc} {w}(t) \\ {v}(t) \end{array}\right]\\ \left[\begin{array}{cc} {y'}(t) \\ {u'}(t) \end{array}\right] = \underbrace{ \left[\begin{array}{cc} Q^{\frac{1}{2}}C' & 0 \\ 0 & -R^{\frac{1}{2}}F \end{array}\right] }_{C_{CL1}} \left[\begin{array}{cc} {x}(t) \\ \hat{x}(t) \end{array}\right] \end{array}} }$

これに座標変換

$\displaystyle{(9)\quad \left[\begin{array}{cc} {x}(t) \\ {e}(t) \end{array}\right] = \left[\begin{array}{cc} I_n & 0 \\ -I_n & I_n \end{array}\right] \left[\begin{array}{cc} {x}(t) \\ \hat{x}(t) \end{array}\right],\quad \left[\begin{array}{cc} {x}(t) \\ \hat{x}(t) \end{array}\right] = \left[\begin{array}{cc} I_n & 0 \\ I_n & I_n \end{array}\right] \left[\begin{array}{cc} {x}(t) \\ {e}(t) \end{array}\right] }$

を行うと

$\displaystyle{(10)\quad \boxed{\begin{array}{l} \left[\begin{array}{cc} \dot{x}(t) \\ \dot{e}(t) \end{array}\right]= \underbrace{ \left[\begin{array}{cc} A-BF & -BF \\ 0 & A-HC \end{array}\right] }_{A_{CL2}} \left[\begin{array}{cc} {x}(t) \\ {e}(t) \end{array}\right] +\underbrace{ \left[\begin{array}{cc} B'W^{\frac{1}{2}} & 0 \\ -B'W^{\frac{1}{2}} & HV^{\frac{1}{2}} \end{array}\right] }_{B_{CL2}} \left[\begin{array}{cc} {w}(t) \\ {v}(t) \end{array}\right]\\ \left[\begin{array}{cc} {y'}(t) \\ {u'}(t) \end{array}\right] = \underbrace{ \left[\begin{array}{cc} Q^{\frac{1}{2}}C' & 0 \\ -R^{\frac{1}{2}}F & -R^{\frac{1}{2}}F \end{array}\right] }_{C_{CL2}} \left[\begin{array}{cc} {x}(t) \\ {e}(t) \end{array}\right] \end{array}} }$

となります。このとき閉ループ系のインパルス応答は次式で与えられます。

$\displaystyle{(11)\quad G_{CL}(t) =C_{CL1} \exp(A_{CL1}t) B_{CL1} =C_{CL2} \exp(A_{CL2}t) B_{CL2} }$

● $i=1,2$ に対して、評価関数 $J$ の総和は次式で与えられます。

$(12)\quad \begin{array}{l} \displaystyle{\int_0^\infty{\rm tr}(G_{CL}^T(t)G_{CL}(t))\,dt\quad (Q>0,R>0)}\\ \displaystyle{={\rm tr}(\int_0^\infty B_{CLi}^T\exp(A_{CLi}^Tt)C_{CLi}^TC_{CLi}\exp(A_{CLi}t)B_{CLi}\,dt)}\\ \displaystyle{={\rm tr}(B_{CLi}^T\int_0^\infty \exp(A_{CLi}^Tt)\underbrace{C_{CLi}^TC_{CLi}}_{Q_{CLi}}\exp(A_{CLi}t)\,dt\,B_{CLi})}\\ \displaystyle{={\rm tr}(B_{CLi}^T\underbrace{\int_0^\infty \exp(A_{CLi}^Tt)Q_{CLi}\exp(A_{CLi}t)\,dt}_{\Pi}\,B_{CLi})}\\ \displaystyle{={\rm tr}(\Pi B_{CLi}B_{CLi}^T)} \end{array} }$

ここで $\Pi>0$ は次式を満足します。

$\displaystyle{(13)\quad \Pi A_{CLi}+A_{CLi}^T\Pi+Q_{CLi}=0 }$

ラグランジュの未定定数法を適用するために、次の評価関数を考えます。

$\displaystyle{(14)\quad J'={\rm tr}(\Pi B_{CLi}B_{CLi}^T)+{\rm tr}((\Pi A_{CLi}+A_{CLi}^T\Pi+Q_{CLi})\Gamma) }$

これを最小化する場合の必要条件は、次式となります。

$\displaystyle{(15)\quad \frac{\partial J'}{\partial\Gamma}=0,\ \frac{\partial J'}{\partial\Pi}=0,\ \frac{\partial J'}{\partial F}=0,\ \frac{\partial J'}{\partial H}=0 }$

以下では、次の分割を考えます。

$\displaystyle{(16)\quad \Pi=\left[\begin{array}{cc} \Pi_{1}&\Pi_{3} \\ \Pi_{3}^T&\Pi_{2} \end{array}\right],\ \Gamma=\left[\begin{array}{cc} \Gamma_{1}&\Gamma_{3} \\ \Gamma_{3}^T&\Gamma_{2} \end{array}\right] }$

[3]　 $i=1$ の場合について、必要条件を一つ一つ調べていきます。

● $\frac{\partial J'}{\partial\Gamma}=0$

$\displaystyle{(17)\quad \frac{\partial J}{\partial\Gamma}=\Pi A_{CL1}+A_{CL1}^T\Pi+Q_{CL1}=0 }$

において

$\displaystyle{(18)\quad \begin{array}{l} \Pi A_{CL1}= \left[\begin{array}{cc} \Pi_{1}&\Pi_{3} \\ \Pi_{3}^T&\Pi_{2} \end{array}\right] \left[\begin{array}{cc} A & -BF \\ HC&A-HC-BF \end{array}\right]\\ = \left[\begin{array}{cc} \Pi_{1}A+\Pi_{3}HC & -\Pi_{1}BF+\Pi_{3}(A-HC-BF)\\ \Pi_{3}^TA+\Pi_{2}HC & -\Pi_{3}^TBF+\Pi_{2}(A-HC-BF) \end{array}\right] \end{array} }$

$\displaystyle{(19)\quad \begin{array}{l} A_{CL1}^T\Pi =(\Pi A_{CL1})^T\\ =\left[\begin{array}{cc} A^T\Pi_{1}+C^TH^T\Pi_{3}^T&\\ -F^TB^T\Pi_{1}+(A-HC-BF)^T\Pi_{3}^T & \end{array}\right.\\ \left.\begin{array}{cc} A^T\Pi_{3}+C^TH^T\Pi_{2}\\ -F^TB^T\Pi_{3}+(A-HC-BF)^T\Pi_{2} \end{array}\right] \end{array} }$

$\displaystyle{(20)\quad Q_{CL1}= \left[\begin{array}{cc} C'^TQC' & 0 \\ 0 & F^TRF \end{array}\right] }$

を代入して、次を得ます。

$\displaystyle{(21)\quad \Pi_{1}A+\Pi_{3}HC+A^T\Pi_{1}+C^TH^T\Pi_{3}^T+C'^TQC'=0 }$

$\displaystyle{(22)\quad -\Pi_{1}BF+\Pi_{3}(A-HC-BF)+A^T\Pi_{3}+C^TH^T\Pi_{2}=0 }$

$\displaystyle{(23)\quad \begin{array}{l} -\Pi_{3}^TBF+\Pi_{2}(A-HC-BF) \\ -F^TB^T\Pi_{3}+(A-HC-BF)^T\Pi_{2}+F^TRF=0 \end{array} }$

(22)+(23)より

$\displaystyle{(24)\quad \begin{array}{l} (\Pi_{3}+\Pi_{2})(A-HC)+(A-BF)^T(\Pi_{3}+\Pi_{2})\\ +\underbrace{F^TRF-(\Pi_{1}+\Pi_{3}+\Pi_{3}^T+\Pi_{2})BF}_{0\ (assumed)}=0\\ \Rightarrow \Pi_{3}+\Pi_{2}=0\Rightarrow \Pi_{3}=-\Pi_{2}\Rightarrow \Pi_{3}^T=\Pi_{3} \end{array} }$

(23)より

$\displaystyle{(25)\quad \begin{array}{l} \Pi_{2}(A-HC)+(A-HC)^T\Pi_{2}+F^TRF\\ \underbrace{-(\Pi_{3}^T+\Pi_{2})BF-F^TB^T(\Pi_{3}+\Pi_{2})}_{0}=0\\ \Rightarrow \Pi_{2}>0 \end{array} }$

(21)+(22)より

$\displaystyle{(26)\quad \begin{array}{l} (\Pi_{1}+\Pi_{3})(A-BF)+A^T(\Pi_{1}+\Pi_{3})\\ +\underbrace{C^TH^T(\Pi_{3}^T+\Pi_{2})}_{0}+C'^TQC'+F^TRF-F^TB^T(\Pi_{1}+\Pi_{3})=0 \end{array} }$

$\displaystyle{(27)\quad \begin{array}{l} (\Pi_{1}+\Pi_{3})(A-BF)+(A-BF)^T(\Pi_{1}+\Pi_{3})+C'^TQC'+F^TRF=0\\ \Rightarrow \Pi_{1}+\Pi_{3}=\Pi_{1}-\Pi_{2}>0 \end{array} }$

● $\frac{\partial J'}{\partial\Pi}=0$

$\displaystyle{(28)\quad \frac{\partial J}{\partial\Pi}=B_{CL1}B_{CL1}^T+\Gamma A_{CL1}^T+A_{CL1}\Gamma=0 }$

において

$\displaystyle{(29)\quad \begin{array}{l} A_{CL1}\Gamma= \left[\begin{array}{cc} A & -BF \\ HC&A-HC-BF \end{array}\right] \left[\begin{array}{cc} \Gamma_{1}&\Gamma_{3} \\ \Gamma_{3}^T&\Gamma_{2} \end{array}\right]\\ =\left[\begin{array}{cc} A\Gamma_{1} -BF\Gamma_{3}^T &A\Gamma_{3} -BF\Gamma_{2}\\ HC\Gamma_{1}+(A-HC-BF)\Gamma_{3}^T &HC\Gamma_{3}+(A-HC-BF)\Gamma_{2} \end{array}\right] \end{array} }$

$\displaystyle{(30)\quad \begin{array}{l} \Gamma A_{CL1}^T =(A_{CL1}\Gamma)^T\\ =\left[\begin{array}{cc} \Gamma_{1}A^T -\Gamma_{3}F^TB^T & \Gamma_{1}C^TH^T+\Gamma_{3}(A-HC-BF)^T\\ \Gamma_{3}^TA^T -\Gamma_{2}F^TB^T & \Gamma_{3}^TC^TH^T+\Gamma_{2}(A-HC-BF)^T \end{array}\right] \end{array} }$

$\displaystyle{(31)\quad B_{CL1}B_{CL1}^T =\left[\begin{array}{cc} B'WB'^T & 0 \\ 0 & HVH^T \end{array}\right] }$

を代入して、次を得ます。

$\displaystyle{(32)\quad A\Gamma_{1} -BF\Gamma_{3}^T+\Gamma_{1}A^T -\Gamma_{3}F^TB^T+B'WB'^T=0 }$

$\displaystyle{(33)\quad A\Gamma_{3} -BF\Gamma_{2}+\Gamma_{1}C^TH^T+\Gamma_{3}(A-HC-BF)^T=0 }$

$\displaystyle{(34)\quad \begin{array}{l} HC\Gamma_{3}+(A-HC-BF)\Gamma_{2}\\ +\Gamma_{3}^TC^TH^T+\Gamma_{2}(A-HC-BF)^T+HVH^T=0 \end{array} }$

(33)-(34)より

$\displaystyle{(35)\quad \begin{array}{l} (A-HC)(\Gamma_{3}-\Gamma_{2})+(\Gamma_{3}-\Gamma_{2})(A-BF)^T\\ +\underbrace{(\Gamma_{1}-\Gamma_{3}-\Gamma_{3}^T+\Gamma_{2})C^TH^T-HVH^T}_{0\ (assumed)}=0\\ \Rightarrow \Gamma_{3}-\Gamma_{2}=0\Rightarrow \Gamma_{3}=\Gamma_{2}\Rightarrow \Gamma_{3}^T=\Gamma_{3} \end{array} }$

(34)より

$\displaystyle{(36)\quad \begin{array}{l} (A-BF)\Gamma_{2}+\Gamma_{2}(A-BF)^T+H^TVH\\ +\underbrace{HC(\Gamma_{3}-\Gamma_{2})+(\Gamma^T_{3}-\Gamma_{2})C^TH^T}_{0}=0\Rightarrow \Gamma_{2}>0 \end{array} }$

(32)-(33)より

$\displaystyle{(37)\quad \begin{array}{l} (\Gamma_{1}-\Gamma_{3})(A-HC)^T+A(\Gamma_{1}-\Gamma_{3})\\ +\underbrace{BF(\Gamma_{3}^T-\Gamma_{2})}_{0}+B'WB'^T+HVH^T-HC(\Gamma_{1}-\Gamma_{3})=0 \end{array} }$

$\displaystyle{(38)\quad \begin{array}{l} (\Gamma_{1}-\Gamma_{3})(A-HC)^T+(A-HC)(\Gamma_{1}-\Gamma_{3})+B'WB'^T+HVH^T=0\\ \Rightarrow \Gamma_{1}-\Gamma_{3}=\Gamma_{1}-\Gamma_{2}>0 \end{array} }$

●準備１

$\displaystyle{(39)\quad \Pi B_{CL1}B_{CL1}^T= \left[\begin{array}{cc} \Pi_{1}&\Pi_{3} \\ \Pi_{3}^T&\Pi_{2} \end{array}\right] \left[\begin{array}{cc} B'WB'^T & 0 \\ 0 & HVH^T \end{array}\right] }$

(1) $\begin{eqnarray*} \end{eqnarray*}$

$\displaystyle{(40)\quad {\rm tr}(\Pi B_{CL1}B_{CL1}^T)={\rm tr}(\Pi_{1}B'WB'^T+\Pi_{2}HVH^T) }$

$\displaystyle{(41)\quad \frac{\partial {\rm tr}(\Pi B_{CL1}B_{CL1}^T)}{\partial H}=2\Pi_{2}HV }$

●準備２

$\displaystyle{(42)\quad \begin{array}{l} \Pi A_{CL1}\Gamma= \left[\begin{array}{cc} \Pi_{1}&\Pi_{3} \\ \Pi_{3}^T&\Pi_{2} \end{array}\right] \left[\begin{array}{cc} A & -BF \\ HC&A-HC-BF \end{array}\right] \Gamma\\ =\left[\begin{array}{cc} \Pi_{1}A+\Pi_{3}HC & -\Pi_{1}BF+\Pi_{3}(A-HC-BF)\\ \Pi_{3}^TA+\Pi_{2}HC & -\Pi_{3}^TBF+\Pi_{2}(A-HC-BF) \end{array}\right] \left[\begin{array}{cc} \Gamma_{1}&\Gamma_{3} \\ \Gamma_{3}^T&\Gamma_{2} \end{array}\right] \end{array} }$

$\displaystyle{(43)\quad \begin{array}{l} {\rm tr}(\Pi A_{CL1}\Gamma)\\ ={\rm tr}(\Pi_{1}A\Gamma_{1}+\Pi_{3}HC\Gamma_{1} -\Pi_{1}BF\Gamma_{3}^T+\Pi_{3}(A-HC-BF)\Gamma_{3}^T)\\ +{\rm tr}(\Pi_{3}^TA\Gamma_{3}+\Pi_{2}HC\Gamma_{3} -\Pi_{3}^TBF\Gamma_{2}+\Pi_{2}(A-HC-BF)\Gamma_{2}) \end{array} }$

$\displaystyle{(44)\quad \frac{\partial {\rm tr}(\Pi A_{CL1}\Gamma)}{\partial F} =-B^T\Pi_1\Gamma_{3}-B^T\Pi_3^T\Gamma_3-B^T\Pi_{3}\Gamma_{2}-B^T\Pi_{2}\Gamma_{2} }$

$\displaystyle{(45)\quad \frac{\partial {\rm tr}(\Pi A_{CL1}\Gamma)}{\partial H}=\Pi_{3}^T\Gamma_{1}C^T-\Pi_{3}^T\Gamma_{3}C^T+\Pi_{2}\Gamma_{3}^TC^T-\Pi_{2}\Gamma_{2}C^T }$

●準備３

$\displaystyle{(46)\quad \begin{array}{l} A_{CL1}^T\Pi\Gamma=(\Pi A_{CL1})^T\Gamma=\\ \left[\begin{array}{cc} A^T\Pi_{1}+C^TH^T\Pi_{3}^T&\\ -F^TB^T\Pi_{1}+(A-HC-BF)^T\Pi_{3}^T & \end{array}\right.\\ \left.\begin{array}{cc} A^T\Pi_{3}+C^TH^T\Pi_{2}\\ -F^TB^T\Pi_{3}+(A-HC-BF)^T\Pi_{2} \end{array}\right] \left[\begin{array}{cc} \Gamma_{1}&\Gamma_{3} \\ \Gamma_{3}^T&\Gamma_{2} \end{array}\right] \end{array} }$

$\displaystyle{(47)\quad \begin{array}{l} {\rm tr}(A_{CL1}^T\Pi\Gamma)\\ ={\rm tr}(A^T\Pi_{1}\Gamma_{1}+C^TH^T\Pi_{3}^T\Gamma_{1}+A^T\Pi_{3}\Gamma_{3}^T+C^TH^T\Pi_{2}\Gamma_{3}^T)\\ +{\rm tr}( -F^TB^T\Pi_{1}\Gamma_{3}+(A-HC-BF)^T\Pi_{3}^T\Gamma_{3} \\ -F^TB^T\Pi_{3}\Gamma_{2}+(A-HC-BF)^T\Pi_{2}\Gamma_{2}) \end{array} }$

$\displaystyle{(48)\quad \frac{\partial {\rm tr}(\Pi A_{CL1}\Gamma)}{\partial F}=-B^T\Pi_1\Gamma_{3}-B^T\Pi_3^T\Gamma_3-B^T\Pi_{3}\Gamma_{2}-B^T\Pi_{2}\Gamma_{2} }$

$\displaystyle{(49)\quad \frac{\partial {\rm tr}(\Pi A_{CL1}\Gamma)}{\partial H}=\Pi_{3}^T\Gamma_{1}C^T-\Pi_{3}^T\Gamma_{3}C^T+\Pi_{2}\Gamma_{3}^TC^T-\Pi_{2}\Gamma_{2}C^T }$

●準備４

$\displaystyle{(50)\quad Q_{CL1}^T\Gamma= \left[\begin{array}{cc} C^TQC & 0 \\ 0 & F^TRF \end{array}\right] \left[\begin{array}{cc} \Gamma_{1}&\Gamma_{3} \\ \Gamma_{3}^T&\Gamma_{2} \end{array}\right] }$

$\displaystyle{(51)\quad {\rm tr}(Q_{CL1}\Gamma)={\rm tr}(C^TQC\Gamma_{1}+F^TRF\Gamma_{2}) }$

$\displaystyle{(52)\quad \frac{\partial {\rm tr}(Q_{CL1}\Gamma)}{\partial F}=2RF\Gamma_{2} }$

● $\frac{\partial J'}{\partial F}=0$

$\displaystyle{(53)\quad \begin{array}{l} \frac{\partial J}{\partial F}=2RF\Gamma_{2}-2B^T(\Pi_1+\underbrace{\Pi_3^T}_{-\Pi_2})\underbrace{\Gamma_3}_{\Gamma_2}-2B^T\underbrace{(\Pi_{3}+\Pi_{2})}_{0}\Gamma_{2}=0\\ \qquad\Downarrow\Pi=\Pi_1-\Pi_2>0\\ 2(RF-B^T\Pi)\Gamma_{2}=0\\ \qquad\Downarrow\Gamma_2>0\\ \boxed{F=R^{-1}B^T\Pi}\\ \qquad\Downarrow\Pi(A-BF)+(A-BF)^T\Pi+C'^TQC'+F^TRF=0\\ \boxed{\Pi A+A^T\Pi-\Pi BR^{-1}B^T\Pi+C'^TQC'=0} \end{array} }$

● $\frac{\partial J'}{\partial H}=0$

$\displaystyle{(54)\quad \begin{array}{l} \frac{\partial J}{\partial H}=2\Pi_{2}HV+2\underbrace{\Pi_{3}^T}_{-\Pi_{2}}(\Gamma_{1}-\underbrace{\Gamma_{3}}_{\Gamma_{2}})C^T+2\Pi_{2}\underbrace{(\Gamma_{3}^T-\Gamma_{2})}_{0}C^T=0\\ \qquad\Downarrow\Gamma=\Gamma_1-\Gamma_2>0\\ 2\Pi_{2}(HV-\Gamma C^T)=0\\ \qquad\Downarrow\Pi_2>0\\ \boxed{H=(V^{-1}C\Gamma)^T}\\ \qquad\Downarrow(A-HC)\Gamma+\Gamma(A-HC)^T+B'WB'^T+HVH^T=0\\ \boxed{\Gamma A^T+A\Gamma-\Gamma C^TV^{-1}C\Gamma+B'WB'^T=0} \end{array} }$

補遺　上述の議論では、次についての検討が必要です。

検討事項１　(13)の妥当性
検討事項２　(24),(35)における仮定の妥当性
検討事項３　 $i=2$ の場合の導出
検討事項４　十分性の証明