補償器によるSM制御

補償器によるSM制御…Homework

[1] 制御対象

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

に対するSM制御則の線形制御部を、補償器

$\displaystyle{(2)\quad \boxed{\dot{x}_c(t)=Hx_c(t)+Dy(t)}\quad(x_c(t)\in{\rm\bf R}^q) }$

を用いて実現することを考えます。このときのスイッチング関数を

$\displaystyle{(3)\quad \boxed{s(t)=F_cx_c(t)+Fy(t)}\quad(s(t)\in{\rm\bf R}^m) }$

とします。

●出力FB型SM制御(p>m)の議論により、次のSM標準形を得ているものとします。

$\displaystyle{(18.1^*)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{c} \dot x'''_1(t)\\ \dot x'''_2(t) \end{array}\right] }_{\dot{x}'''(t)} = \underbrace{ \left[\begin{array}{ccc|c} A_{11}^o & A_{12}^o & A_{121}^m & A_{121}\\ 0_{n-p-r\times r} & A_{22}^o & A_{122}^m & A_{1221}\\ 0_{p-m\times r} & A_{21}^o & A_{22}^m & A_{1222}\\\hline A_{2120} & A_{2121} & A_{2122} & A_{22} \end{array}\right] }_{(T_aT_bT_c)A(T_aT_bT_c)^{-1}} \underbrace{ \left[\begin{array}{c} x'''_1(t)\\ x'''_2(t) \end{array}\right] }_{x'''(t)}\\ + \underbrace{ \left[\begin{array}{c} 0_{r\times m} \\ 0_{n-p-r\times m} \\ 0_{p-m\times m} \\\hline B_2 \end{array}\right] }_ {(T_aT_bT_c)B} \underbrace{(u(t)+\xi(t,x,u))}_{u'(t)}\\ (A_{11}^o\in{\rm\bf R}^{r\times r},A_{22}^o\in{\rm\bf R}^{n-p-r\times n-p-r},A_{22}^m\in{\rm\bf R}^{p-m\times p-m},A_{22}\in{\rm\bf R}^{m\times m}) \end{array} }$

$\displaystyle{(18.2^*)\quad \begin{array}{l} y(t) = \underbrace{ \left[\begin{array}{cc} 0_{p\times n-p} & T \\ \end{array}\right] }_{C(T_aT_bT_c)^{-1}} \underbrace{ \left[\begin{array}{c} x'''_1(t)\\ x'''_2(t) \end{array}\right] }_{x'''(t)} \end{array} }$

このとき、 $C_1=\left[\begin{array}{cc} 0_{p-m\times n-p} & I_{p-m} \end{array}\right]$ とおくと

$\displaystyle{(2')\quad \begin{array}{l} \dot{x}_c(t)=Hx_c(t)+Dy(t)\\ =Hx_c(t)+ \left[\begin{array}{cc} 0_{p\times n-p} & DT \\ \end{array}\right] %\underbrace{ \left[\begin{array}{c} x'''_1(t)\\ x'''_2(t) \end{array}\right] %}_{x'''(t)} \\ =Hx_c(t)+ \left[\begin{array}{cc} D_1C_1 & D_2 \\ \end{array}\right] %\underbrace{ \left[\begin{array}{c} x'''_1(t)\\ x'''_2(t) \end{array}\right] %}_{x'''(t)} \\ =Hx_c(t)+D_1C_1x'''_1(t)+D_2x'''_2(t)\\ (DT=\left[\begin{array}{cc} D_1 & D_2 \end{array}\right], D_1\in{\rm\bf R}^{m\times p-m}, D_2\in{\rm\bf R}^{m\times m}) \end{array} }$

および

$\displaystyle{(3')\quad \begin{array}{l} s(t)=F_cx_c(t)+Fy(t)\\ =Hx_c(t)+ \left[\begin{array}{cc} 0_{p\times n-p} & FT \\ \end{array}\right] %\underbrace{ \left[\begin{array}{c} x'''_1(t)\\ x'''_2(t) \end{array}\right] %}_{x'''(t)} \\ =Hx_c(t)+ \left[\begin{array}{cc} F_1C_1 & F_2 \\ \end{array}\right] %\underbrace{ \left[\begin{array}{c} x'''_1(t)\\ x'''_2(t) \end{array}\right] %}_{x'''(t)} \\ =Hx_c(t)+F_1C_1x'''_1(t)+F_2x'''_2(t)\\ (FT=\left[\begin{array}{cc} F_1 & F_2 \end{array}\right], F_1\in{\rm\bf R}^{m\times p-m}, F_2\in{\rm\bf R}^{m\times m}) \end{array} }$

●(18.1 $^*$ )と(2′)を結合し、 $\tilde{C}_1=\left[\begin{array}{cc} 0_{p-m\times n-p-r} & I_{p-m} \end{array}\right]$ を用いて、次式を得ます。

$\displaystyle{(4)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{c} \dot x'''_1(t)\\\hline \dot x'''_2(t)\\\hline\hline \dot x_c(t) \end{array}\right] }_{\dot{x}'''(t)} = \left[\begin{array}{c|c||c} \left.\begin{array}{ccc} A_{11}^o & A_{12}^o & A_{121}^m \\ 0_{n-p-r\times r} & A_{22}^o & A_{122}^m \\ 0_{p-m\times r} & A_{21}^o & A_{22}^m \\ \end{array}\right.& \left.\begin{array}{cc} A_{121}\\ A_{1221}\\ A_{1222} \end{array}\right.& \left.\begin{array}{cc} 0_{r\times q} \\ 0_{n-p-r\times q} \\ 0_{p-m\times q} \end{array}\right. \\\hline \left.\begin{array}{ccc} A_{211} & A_{2121} &A_{2122} \end{array}\right.& \left.\begin{array}{cc} A_{22} \end{array}\right.& 0_{m\times q} \\\hline\hline %D_1C_1 \left.\begin{array}{cc} 0_{q\times r} & D_1\tilde{C}_1 \end{array}\right. & D_2 & H \end{array}\right]\\ \times \underbrace{ \left[\begin{array}{c} x'''_1(t)\\\hline x'''_2(t)\\\hline\hline x_c(t) \end{array}\right] }_{x'''(t)} + \left[\begin{array}{c} 0_{r\times m} \\ 0_{n-p-r\times m} \\ 0_{p-m\times m} \\\hline B_2\\\hline\hline 0_{q\times m} \end{array}\right] u'(t) \end{array} }$

$x'''_2$ と $x_c$ を入れ替えて

$\displaystyle{(5)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{c} \dot x'''_1(t)\\\hline \dot x_c(t)\\\hline\hline \dot x'''_2(t) \end{array}\right] }_{\dot{x}'''(t)} = \left[\begin{array}{c|c||c} \left.\begin{array}{ccc} A_{11}^o & A_{12}^o & A_{121}^m \\ 0_{n-p-r\times r} & A_{22}^o & A_{122}^m \\ 0_{p-m\times r} & A_{21}^o & A_{22}^m \\ \end{array}\right.& \left.\begin{array}{cc} 0_{r\times q} \\ 0_{n-p-r\times q} \\ 0_{p-m\times q} \end{array}\right. & \left.\begin{array}{cc} A_{121}\\ A_{1221}\\ A_{1222} \end{array}\right. \\\hline %D_1C_1 \left.\begin{array}{ccc} 0_{q\times r} & D_1\tilde{C}_1 \end{array}\right. & \left.\begin{array}{cc} H \end{array}\right. & D_2 \\\hline\hline \left.\begin{array}{ccc} A_{211} & A_{2121} &A_{2122} \end{array}\right. & 0_{m\times q} & A_{22} \end{array}\right]\\ \times \underbrace{ \left[\begin{array}{c} x'''_1(t)\\\hline x_c(t)\\\hline\hline x'''_2(t) \end{array}\right] }_{x'''(t)} + \left[\begin{array}{c} 0_{r\times m} \\ 0_{n-p-r\times m} \\ 0_{p-m\times m} \\\hline 0_{q\times m}\\\hline\hline B_2\\ \end{array}\right] u'(t) \end{array} }$

ここで、 $x'''_1(t)$ を $x'''_{r}(t)\in{\rm\bf R}^{r}$ 、 $x'''_{11}(t)\in{\rm\bf R}^{n-p-r}$ 、 $x'''_{12}(t)\in{\rm\bf R}^{p-m}$ に分割し、座標変換

$\displaystyle{(6)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{c} x'''_r(t)\\ x'''_{11}(t)\\ x'''_{12}(t)\\\hline x_c(t)\\\hline\hline x''''_2(t)\\ \end{array}\right] }_{x''''(t)} = \underbrace{ \left[\begin{array}{ccc|c||c} I_{r} & 0 & 0 & 0& 0 \\ 0 & I_{n-p-r} & 0 & 0& 0 \\ 0 & 0 & I_{p-m} & 0 & 0\\\hline 0 & 0 & 0 & I_q &0 \\\hline\hline 0 & 0 & K & K_c & I_m \end{array}\right] }_{T_s} \underbrace{ \left[\begin{array}{c} x'''_r(t)\\ x'''_{11}(t)\\ x'''_{12}(t)\\\hline x_c(t)\\\hline\hline x'''_2(t)\\ \end{array}\right] }_{x'''(t)}\\ (K=F_2^{-1}F_1, K_c=F_2^{-1}F_c) \end{array} }$

を行って、次式を得ます。

$\displaystyle{(7)\quad \begin{array}{l} \underbrace{ \left[\begin{array}{c} \dot x'''_r(t)\\ \dot x'''_{11}(t)\\ \dot x'''_{12}(t)\\\hline \dot x_c(t)\\\hline \dot x''''_2(t) \end{array}\right] }_{\dot{x}''''(t)} = \left[\begin{array}{c|c} \left[\begin{array}{c|c} \left.\begin{array}{ccc} A_{11}^o & A_{12}^o & A_{121}^m \\ 0_{n-p-r\times r} & A_{22}^o & A_{122}^m \\ 0_{p-m\times r} & A_{21}^o & A_{22}^m \\ \end{array}\right. & \left.\begin{array}{cc} 0_{r\times q} \\ 0_{n-p-r\times q} \\ 0_{p-m\times q} \end{array}\right. \\\hline %D_1C_1 \left.\begin{array}{cc} 0_{q\times r} & D_1\tilde{C}_1 \end{array}\right. & H \end{array}\right]\\ - \left[\begin{array}{cc} A_{121}\\ A_{1221}\\ A_{1222}\\\hline D_2 \end{array}\right] \left[\begin{array}{ccccc} 0_{m\times r} & K\tilde{C}_1 & K_c \end{array}\right] & \left.\begin{array}{cc} A_{121}\\ A_{1221}\\ A_{1222}\\\hline D_2 \end{array}\right. \\\hline \left.\begin{array}{ccc|c} A_{211} & A_{2121} &A_{2122} & 0_{m\times q} \end{array}\right. & A_{22} \end{array}\right]\\ \times \underbrace{ \left[\begin{array}{c} x'''_r(t)\\ x'''_{11}(t)\\ x'''_{12}(t)\\\hline x_c(t)\\\hline x''''_2(t)\\ \end{array}\right] }_{x''''(t)} + \left[\begin{array}{c} 0_{r\times m} \\ 0_{n-p-r\times m} \\ 0_{p-m\times m} \\\hline 0_{q\times m}\\\hline B_2 \end{array}\right] u'(t)\\ = \underbrace{ \left[\begin{array}{c|c} \left.\begin{array}{c|c} \left.\begin{array}{ccc} A_{11}^o & A_{12}^o & A_{121}^m-A_{121}K \\ 0_{n-p-r\times r} & A_{22}^o & A_{122}^m-A_{1221}K \\ 0_{p-m\times r} & A_{21}^o & A_{22}^m-A_{1222}K \end{array}\right. & \left.\begin{array}{cc} -A_{121}K_c\\ -A_{1221}K_c\\ -A_{1222}K_c\\ \end{array}\right. \\\hline %(D_1-D_2K)C_1 \left.\begin{array}{cc} 0_{q\times r} & D(D_1-D_2K)\tilde{C}_1 \end{array}\right. & H-D_2K_c \end{array}\right. & \left.\begin{array}{cc} A_{121}\\ A_{1221}\\ A_{1222}\\\hline D_2 \end{array}\right.\\\hline \left.\begin{array}{ccc|c} A_{211} & A_{2121} &A_{2122} & 0_{m\times q} \end{array}\right. & A_{22} \end{array}\right] }_{ \left[\begin{array}{cc} \tilde{\cal A}_{11} & \tilde{\cal A}_{12} \\ \tilde{\cal A}_{21} & \tilde{\cal A}_{22} \end{array}\right]} \\ \times \underbrace{ \left[\begin{array}{c} x'''_r(t)\\ x'''_{11}(t)\\ x'''_{12}(t)\\\hline x_c(t)\\\hline x''''_2(t)\\ \end{array}\right] }_{x''''(t)} + \left[\begin{array}{c} 0_{r\times m} \\ 0_{n-p-r\times m} \\ 0_{p-m\times m} \\\hline 0_{q\times m}\\\hline B_2 \end{array}\right] u'(t) \end{array} }$

以前の議論と同様に、 $A_{11}^o$ が安定行列であることを前提として、 $\tilde{\cal A}_{11}$ を安定行列とする $H,D_1,D_2,K,K_c$ を求めます。

[2] 一つのアプローチとして

$\displaystyle{(8)\quad \begin{array}{l} \left[\begin{array}{c} \dot x'''_{11}(t)\\ \dot x'''_{12}(t) \end{array}\right]= \underbrace{ \left[\begin{array}{cc} A_{22}^o & A_{122}^m \\ A_{21}^o & A_{22}^m \\ \end{array}\right] }_{\tilde{A}_{11}} \left[\begin{array}{c} x'''_{11}(t)\\ x'''_{12}(t) \end{array}\right] + \underbrace{ \left[\begin{array}{cc} A_{1221} \\ A_{1222} \\ \end{array}\right] }_{A_{122}} \underbrace{x'''_{2}(t)}_{\tilde u(t)} \\ \underbrace{x'''_{12}(t)}_{\tilde y(t)} = \underbrace{ \left[\begin{array}{cc} 0_{p-m\times n-p-r} & I_{p-m} \end{array}\right] }_{\tilde{C}_1}} \left[\begin{array}{c} x'''_{11}(t)\\ x'''_{12}(t) \end{array}\right] \end{array} }$

に対する安定化状態フィードバック則を

$\displaystyle{(9)\quad \tilde u(t)= \left[\begin{array}{cc} K_1 & K_2 \end{array}\right] \left[\begin{array}{c} x'''_{11}(t)\\ x'''_{12}(t) \end{array}\right] }$

とします。ここで、 $x'''_{11}(t)$ のすべての要素は観測できないので、関数オブザーバ

$\displaystyle{(10)\quad \begin{array}{l} \dot{x}_{ob}(t)=\hat{A}x_{ob}(t)+\hat{B}\underbrace{x'''_{12}(t)}_{\tilde{y}(t)}+\hat{J}\underbrace{x'''_{2}(t)}_{\tilde{u}(t)}\\ z(t)=\hat{C}x_{ob}(t)+\hat{D}\underbrace{x'''_{12}(t)}_{\tilde{y}(t)} \end{array} }$

を考え、この出力 $z(t)$ に安定化状態フィードバック則 $\tilde{u}(t)$ を推定させます。すなわち

$\displaystyle{(11)\quad z(t)\rightarrow \tilde{u}(t)= K_1x'''_{11}(t)+K_2x'''_{12}(t) \quad(t\rightarrow\infty) }$

そのためには、関数オブザーバの状態 $x_{ob}(t)$ も適当な状態の線形関数を推定する必要があり、これを次のように表します。

$\displaystyle{(12)\quad x_{ob}(t)\rightarrow \underbrace{ \left[\begin{array}{cc} I_{n-p-r} & -L^o \end{array}\right] }_{U} \left[\begin{array}{c} x'''_{11}(t)\\ x'''_{12}(t) \end{array}\right]= x'''_{11}(t)-L^ox'''_{12}(t) \quad(t\rightarrow\infty) }$

いま $K_1,K_2,L^o$ を適当に与えたとき、関数オブザーバのパラメータは

$\displaystyle{(13)\quad \left[\begin{array}{c} U\tilde{A}_{11}\\ \left[\begin{array}{cc} {K}_1 & {K}_2 \end{array}\right] \end{array}\right] = \left[\begin{array}{cc} \hat{A} & \hat{B}\\ \hat{C} & \hat{D} \end{array}\right] \left[\begin{array}{c} U\\ \tilde{C}_1 \end{array}\right] }$

を解いて

$\displaystyle{(14)\quad \begin{array}{l} \left[\begin{array}{cc} \hat{A} & \hat{B}\\ \hat{C} & \hat{D} \end{array}\right] = \left[\begin{array}{c} U\tilde{A}_{11}\\ \left[\begin{array}{cc} {K}_1 & {K}_2 \end{array}\right] \end{array}\right] \left[\begin{array}{c} U\\ \tilde{C}_1 \end{array}\right]^{-1}\\ = \left[\begin{array}{c} \left[\begin{array}{cc} I_{n-p-r} & L^o \end{array}\right] \left[\begin{array}{cc} A_{22}^o & A_{122}^m \\ A_{21}^o & A_{22}^m \\ \end{array}\right]\\ \left[\begin{array}{cc} {K}_1 & {K}_2 \end{array}\right] \end{array}\right] \left[\begin{array}{cc} I_{n-p-r} & L^o\\ 0_{p-m\times n-p-r} & I_{p-m} \end{array}\right]^{-1}\\ = \left[\begin{array}{cc} A_{22}^o+L^oA_{21}^o & A_{122}^m+L^oA_{22}^m \\ {K}_1 & {K}_2 \end{array}\right] \left[\begin{array}{cc} I_{n-p-r} & -L^o\\ 0_{p-m\times n-p-r} & I_{p-m} \end{array}\right]\\ = \left[\begin{array}{cc} A_{22}^o+L^oA_{21}^o & A_{122}^m+L^oA_{22}^m-(A_{22}^o+L^oA_{21}^o)L^o \\ K_1 & K_2-K_1L^o \end{array}\right] \end{array} }$

ここで、 $L_o$ は $\hat{A}$ が安定行列となるように選んでおきます。また

$\displaystyle{(15)\quad \hat{J}=UA_{122}= \left[\begin{array}{cc} I_{n-p-r} & L^o \end{array}\right] \left[\begin{array}{cc} A_{1221} \\ A_{1222} \\ \end{array}\right] =A_{1221}+A_{1222}L^o }$

以上を踏まえて、 $K_c=K_1,K=K_2-K_1L^o$ と選び、線形関数オブザーバを

$\displaystyle{(16)\quad \begin{array}{l} \underbrace{\dot{x}_{ob}(t)}_{\dot{x}_c(t)} =\underbrace{\hat{A}}_{H}\underbrace{x_{ob}(t)}_{x_c(t)} +\underbrace{\hat{B}}_{D_1}\underbrace{x'''_{12}(t)}_{\tilde{y}(t)} +\underbrace{\hat{J}}_{D_2}\underbrace{x'''_{2}(t)}_{\tilde{u}(t)}\\ \underbrace{z(t)}_{\tilde{u}(t)} =\underbrace{\hat{C}}_{K_c}\underbrace{x_{ob}(t)}_{x_c(t)} +\underbrace{\hat{D}}_{K}\underbrace{x'''_{12}(t)}_{\tilde{y}(t)} \end{array} }$

とみなせば、第２式を第１式に代入して

$\displaystyle{(16')\quad \dot{x}_c(t)=(H-D_2K_c)x_c(t)+(D_1-D_2K)x'''_{12}(t) }$

ただし

$\displaystyle{(16'')\quad \boxed{\begin{array}{l} H=A_{22}^o+L^oA_{21}^o\\ D_1=A_{122}^m+L^oA_{22}^m-(A_{22}^o+L^oA_{21}^o)L^o\\ D_2=A_{1221}+A_{1222}L^o\\ K_c=K_1 \\ K=K_2-K_1L^o \end{array}} }$

となって、補償器を得ていることになります。以下では、この補償器を用いたSM制御則の構成法について検討します。

[3] 制御対象の状態方程式と観測方程式は、それぞれ(18.1 $^*$ )と(18.2 $^*$ )から次式で与えられます。

$\displaystyle{(17.1)\quad \begin{array}{l} \left[\begin{array}{c} \dot x'''_r(t)\\ \dot x'''_{11}(t)\\ \dot x'''_{12}(t)\\ \dot x'''_2(t) \end{array}\right]= \left[\begin{array}{ccccc} A_{11}^o & A_{12}^o & A_{121}^m & A_{121} \\ 0_{n-p-r\times r} & A_{22}^o & A_{122}^m & A_{1221} \\ 0_{p-m\times r} & A_{21}^o & A_{22}^m & A_{1222} \\ A_{211} & A_{212} & A_{213} & A_{22} \end{array}\right] \left[\begin{array}{c} x'''_r(t)\\ x'''_{11}(t)\\ x'''_{12}(t)\\ x'''_2(t) \end{array}\right]\\ + \left[\begin{array}{c} 0_{r\times m}\\ 0_{n-p-r\times m}\\ 0_{p-m\times m}\\ B_2\\ \end{array}\right] u(t)\\ (x'''_r(t)\in{\rm\bf R}^{r},x'''_{11}(t)\in{\rm\bf R}^{n-p-r},x'''_{12}(t)\in{\rm\bf R}^{p-m},x'''_{2}(t)\in{\rm\bf R}^{m}) \end{array} }$

$\displaystyle{(17.2)\quad \begin{array}{l} y(t) = \left[\begin{array}{cc} 0_{p\times n-p} & T \\ \end{array}\right] \left[\begin{array}{c} x'''_r(t)\\ x'''_{11}(t)\\ x'''_{12}(t)\\ x'''_2(t) \end{array}\right]=T \left[\begin{array}{c} x'''_{12}(t)\\ x'''_2(t) \end{array}\right] \quad(T\in{\rm\bf R}^{p\times p}) \end{array} }$

補償器は(16′)から次式で与えられます。

$\displaystyle{(18)\quad \begin{array}{l} \dot{x}_c(t) =Hx_c(t) +D_1x'''_{12}(t) +D_2}x'''_{2}(t)\\ (x'''_{2}(t)=K_cx_c(t)+Kx'''_{12}(t)) \end{array} }$

このとき補償器を合わせた制御対象のダイナミックスは次式のように表すことができます。

$\displaystyle{(19.1)\quad \underbrace{ \left[\begin{array}{c} \dot z_r(t)\\ \dot x_c(t)\\ \dot x'''_{12}(t)\\ \dot x'''_2(t) \end{array}\right] }_{\hat x(t)} = \underbrace{ \left[\begin{array}{ccccc} A_{11}^o & A_{12}^o & A_{121}^m-A_{12}^oL^o & A_{121} \\ 0_{q\times r} & H & D_1 & D_2 \\ 0_{p-m\times r} & A_{21}^o & A_{22}^m-A_{21}^oL^o & A_{1222} \\ A_{211} & A_{212} & A_{213}-A_{212}L^o & A_{22} \end{array}\right] }_{\hat A} \underbrace{ \left[\begin{array}{c} z_r(t)\\ x_c(t)\\ x'''_{12}(t)\\ x'''_2(t) \end{array}\right] }_{\hat x(t)} }$
$\displaystyle{ + \underbrace{ \left[\begin{array}{ccccc} 0_{r\times r} & 0_{r\times m} \\ 0_{q\times r} & 0_{q\times m} \\ 0_{p-m\times r} & A_{21}^o \\ A_{211} & A_{212} \end{array}\right] }_{\hat A_e} \underbrace{ \left[\begin{array}{c} e_r(t)\\ e_c(t) \end{array}\right] }_{\hat{e}(t)} + \underbrace{ \left[\begin{array}{c} 0_{r\times m}\\ 0_{q\times m}\\ 0_{p-m\times m}\\ B_2\\ \end{array}\right] }_{\hat B} u(t) }$

$\displaystyle{(19.2)\quad \underbrace{ \left[\begin{array}{c} \dot e_r(t)\\ \dot e_c(t) \end{array}\right] }_{\dot{\hat{e}}(t)} = \underbrace{ \left[\begin{array}{cc} A_{11}^o & A_{12}^o \\ 0 & H \end{array}\right] }_{\hat{H}} \underbrace{ \left[\begin{array}{c} e_r(t)\\ e_c(t) \end{array}\right] }_{\hat{e}(t)} }$

ここで、 $A_{11}^o$ が安定行列であることは前提条件であり、また $H$ は安定行列となるようにオブザーバゲイン $L_o$ を決めます。

●(12)における推定誤差を

$\displaystyle{(20)\quad e_c(t)=x_c(t) -(x'''_{11}(t)+L_ox'''_{12}(t)) }$

で定義します。補償器の状態方程式は

$\displaystyle{(21)\quad \begin{array}{l} \dot x_c(t)=Hx_c(t)+D_1x'''_{12}(t)+D_2x'''_{2}(t)}\\ =Hx_c(t)+(A_{122}^m+L^oA_{22}^m-(A_{22}^o+L^oA_{21}^o)L^o)x'''_{12}(t)\\ +(A_{1221}+A_{1222}L^o)x'''_{2}(t) \end{array} }$

となることから、次式すなわち(19.2)の下段式を得ます。

$\displaystyle{(22)\quad \begin{array}{l} \dot e_c(t)=Hx_c(t)\\ +(A_{122}^m+L^oA_{22}^m-(A_{22}^o+L^oA_{21}^o)L^o)x'''_{12}(t) +(A_{1221}+A_{1222}L^o)x'''_{2}(t)\\ -(A_{22}^ox'''_{11}(t) + A_{122}^mx'''_{12}(t) + A_{1221}x'''_2(t))\\ -L_o(A_{21}^ox'''_{11}(t) + A_{22}^mx'''_{12}(t) + A_{1222}x'''_2(t))\\ =H(x_c(t) -x'''_{11}(t)-L_ox'''_{12}(t))\\ =He_c(t) \end{array} }$

●(17.1)の第１式から

$\displaystyle{(23)\quad \begin{array}{l} \dot x'''_r(t)=A_{11}^ox'''_r (t)+ A_{12}^ox'''_{11}(t) + A_{121}^mx'''_{12}(t) + A_{121}x'''_2(t)\\ =A_{11}^ox'''_r(t) + A_{11}^o(-e_c(t) + x_c(t)-L_ox'''_{12}(t)) + A_{121}^mx'''_{12}(t) + A_{121}x'''_2(t)\\ = A_{11}^ox'''_r(t) + A_{11}^ox_c(t) + (A_{121}^m-A_{11}^oL_o)x'''_{12}(t) + A_{121}x'''_{2}(t) -A_{11}^oe_c(t) \end{array} }$

ここで、 $e_c\rightarrow 0$ の場合の目標となるダイナミックスを次式で表します。

$\displaystyle{(24)\quad \dot z_r(t)=A_{11}^oz_r(t) + A_{12}^ox_c(t) + (A_{121}^m-A_{12}^oL_o)x'''_{12}(t) + A_{121} x'''_2(t) }$

両者の状態の誤差を

$\displaystyle{(25)\quad e_r(t)=z_r(t)-x'''_r(t)\\ }$

とすると、次式すなわち(19.2)の上段式を得ます。

$\displaystyle{(26)\quad \begin{array}{l} \dot e_r(t)=A_{11}^oz_r(t) + A_{12}^ox_c(t) + (A_{121}^m-A_{12}^oL_o)x'''_{12}(t) + A_{121} x'''_2(t)\\ -(A_{11}^ox'''_r(t) + A_{12}^ox'''_{11}(t) + A_{121}^mx'''_{12}(t) + A_{121}x'''_2(t))\\ =A_{11}^oe_r(t) + A_{12}^o(x_c(t) -x'''_{11}(t)-L_ox'''_{12}(t))\\ =A_{11}^oe_r(t) + A_{12}^oe_c(t) \end{array} }$

●(19.1)の第３式と第４式は、それぞれ(17.1)の第３式と第４式から次式のように得られます。

$\displaystyle{(27)\quad \begin{array}{l} \dot x'''_{12}(t)= A_{21}^ox'''_{11}(t) + A_{22}^mx'''_{12}(t)+ A_{1222}x'''_{2}(t)\\ = A_{21}^o(-e_c(t) + x_c(t)-L_ox'''_{12}(t)) + A_{22}^mx'''_{12}(t)+ A_{1222}x'''_{2}(t)\\ = A_{21}^ox_c(t) + (A_{22}^m-A_{21}^oL_o)x'''_{12}(t)+ A_{1222}x'''_{2}(t) -A_{21}^oe_c(t) \end{array} }$

$\displaystyle{(28)\quad \begin{array}{l} \dot x'''_2(t)= A_{211}x'''_r(t) + A_{212}x'''_{11}(t) + A_{213}x'''_{12}(t) + A_{22}x'''_{2}(t) \\ = A_{211}(-e_r(t)+z_r(t)) + A_{212}(-e_c(t) + x_c(t)-L_ox'''_{12}(t))\\ + A_{213}x'''_{12}(t) + A_{22}x'''_{2}(t)+B_2u(t)\\ = A_{211}z_r(t) + A_{212}x_c(t) + (A_{213}-A_{212}L_o)x'''_{12}(t)\\ + A_{22}x'''_{2}(t) -A_{211}e_r(t)-A_{212}e_c(t)+B_2u(t) \end{array} }$

ちなみに(19.1)の第１式と第２式は、それぞれ(24)と(18)です。

●このときスイッチング関数(3′)は

$\displaystyle{(3'')\quad \begin{array}{l} s(t)=Hx_c(t)+F_1C_1x'''_1(t)+F_2x'''_2(t)\\ =\boxed{\underbrace{F_2\left[\begin{array}{cccc} 0_{m\times r} & K_c & K & I_m \end{array}\right]}_{S}} \underbrace{ \left[\begin{array}{c} z_r(t)\\ x_c(t)\\ x'''_{12}(t)\\ x'''_2(t) \end{array}\right] }_{\hat x(t)} \end{array} }$

と表すことができます。

[4] 以上の準備のもとで、補償器によるSM制御則を次のように構築します。これは次のように線形制御部とスイッチング部からなります。

$\displaystyle{(29)\quad u(t)=u_\ell(t)+u_n(t) }$

線形制御部を、SM標準形(19.1)に基づいて次式のように決めます。

$\displaystyle{(30)\quad \boxed{u_\ell(t)=-\underbrace{(S{\hat B})^{-1}(S{\hat A}-\Phi S)}_{L}{\hat x}(t)} }$

ここで、 ${\hat A}$ 、 ${\hat B}$ は(19.1)で定義されます。また ${\hat x}(t)$ は、(17.2)より

$\displaystyle{(31)\quad \left[\begin{array}{c} x'''_{12}(t)\\ x'''_2(t) \end{array}\right] =T'y(t) }$

と表せることに注意して、次の補償器の出力として得ます。

$\displaystyle{(32)\quad \boxed{\begin{array}{l} \left[\begin{array}{c} \dot z_r(t)\\ \dot x_c(t) \end{array}\right] = \underbrace{ \left[\begin{array}{ccccc} A_{11}^o & A_{12}^o \\ 0_{q\times r} & H \end{array}\right] }_{\hat H} \left[\begin{array}{c} z_r(t)\\ x_c(t) \end{array}\right] + \underbrace{ \left[\begin{array}{ccccc} A_{121}^m-A_{12}^oL^o & A_{121} \\ D_1 & D_2 \end{array}\right]T' }_{\hat D} y(t)\\ \underbrace{ \left[\begin{array}{c} z_r(t)\\ x_c(t)\\ x'''_{12}(t)\\ x'''_2(t) \end{array}\right] }_{\hat x(t)}= \underbrace{ \left[\begin{array}{ccccc} I_r & 0_{r\times q}\\ 0_{q\times r} & I_q\\ 0_{p-m\times r} & 0_{p-m\times q} \\ 0_{m\times r} & 0_{m\times q} \\ \end{array}\right]}_{\widehat{\cal C}} \left[\begin{array}{c} z_r(t)\\ x_c(t) \end{array}\right] + \underbrace{\left[\begin{array}{ccccc} 0_{r\times p-m} & 0_{r\times m} \\ 0_{q\times p-m} & 0_{q\times m} \\ I_{p-m} & 0_{p-m\times m}\\ 0_{m\times p-m} & I_m \end{array}\right]T' }_{\widehat{\cal D}} y(t) \end{array}} }$

スイッチング部を、次式のように決めます。

$\displaystyle{(33)\quad \boxed{u_n(t)=-\rho(t,y)\underbrace{(S{\hat B})^{-1}}_{L_n}\frac{Ps(t)}{||Ps(t)||}} }$

ここで、 $s(t)$ は(3”)で与えられます。また、 $P>0$ は適当な安定行列 $\Phi$ を与えて

$\displaystyle{(34)\quad P\Phi+\Phi^TP=-I_m }$

の解として求め、また $\rho(t,y)$ は

$\displaystyle{(35)\quad \rho(t,y)=\frac{k_1||S{\hat B}||||u_\ell(t)||+||S{\hat B}||\alpha(t,y)+\gamma_2}{1-k_1\kappa(S{\hat B})} }$

●以上に基づく設計手順を、数値例で示します。

MATLAB

%ex2_of_sm.m
%-----
 clear all, close all
%(A,B,C)
 A=[-2 1 0 0;
     0 0 4 1;
     0 1 0 0;
     1 -6 -9 -2];
 B=[0;0;0;1];
 C=[0 0 1 0;
    0 0 0 1];
 [nn,mm]=size(B);
 [pp,nn]=size(C); 
%-----
 [Af,Bf,Cf,r,Ta,Aa,Ba,Ca,Tb,T,Ac,Bc,Cc,Tc]=ca_form1(A,B,C)
%-----
%Assumes the triple (A,B,C) is in the canonical form of Lemma 5.3
%p1 is an nn-pp-r vector containing the desired poles of (A22o+Lo A21o)
%p2 is an nn-mm-r vector containing the desired poles of (A11tilde-A122 K)
%p3 is an mm vector representing the poles of the range space dynamics
 A11o=Af(1:r,1:r);
 A12o=Af(1:r,r+1:nn-pp);
 A22o=Af(r+1:nn-pp,r+1:nn-pp);
 A21o=Af(nn-pp+1:nn-mm,r+1:nn-pp);
 A121m=Af(1:r,nn-pp+1:nn-mm);
 A122m=Af(r+1:nn-pp,nn-pp+1:nn-mm);
 A22m=Af(nn-pp+1:nn-mm,nn-pp+1:nn-mm);
 A121=Af(1:r,nn-mm+1:nn);
 A122=Af(r+1:nn-mm,nn-mm+1:nn);
 A211=Af(nn-mm+1:nn,1:r);
 A212=Af(nn-mm+1:nn,r+1:nn-pp);
 A213=Af(nn-mm+1:nn,nn-pp+1:nn-mm);
 A22=Af(nn-mm+1:nn,nn-mm+1:nn);
 A11tilde=[A22o A122m;A21o A22m];
 A1221=A122(1:nn-pp-r,:);
 A1222=A122(nn-pp-r+1:nn-mm-r,:);
%-----
 p1=-2.5
 Lo=place(A22o',A21o',p1)
 Lo=Lo';
%-----
 p2=[-1,-1.5]
 calK=place(A11tilde,A122,p2)
 K1=calK(:,1:nn-pp-r)
 K2=calK(:,nn-pp-r+1:nn-mm-r)
%-----
 H=A22o+Lo*A21o
 D1=A122m+Lo*A22m-H*Lo
 D2=A1221+Lo*A1222
 K=K2-K1*Lo
 Kc=K1
 Hhat=[A11o A12o;
       zeros(nn-pp-r,r) H]
 Dhat=[A121m-A12o*Lo A121;
       D1 D2]*T'
%----- 
 S2=eye(mm);
 S=S2*[zeros(mm,r) Kc K eye(mm)]
 Ahat=[A11o A12o A121m-A12o*Lo A121;
       zeros(nn-pp-r,r) H D1 D2;
       zeros(pp-mm,r) A21o A22m-A21o*Lo A1222;
       A211 A212 A213-A212*Lo A22]
 p3=-5
 Phi=diag(p3)
 Lambda=S*Bf
 L=-inv(Lambda)*S*Ahat+inv(Lambda)*Phi*S
 P=lyap(Phi',eye(mm))
%-----
%eof

図1 補償器によるSM制御のシミュレーション例

制御系CAD

制御を意識したモノつくりを目指して

補償器によるSM制御