最佳投影方程

降阶的LQG控制问题几乎和全阶的LQG控制问题相同。令${\displaystyle {\hat {\mathbf {x} }}_{r}(t)}$ 表示降阶LQG控制器的状态，唯一的差异是LQG控制器的状态维度${\displaystyle n_{r}=dim({\hat {\mathbf {x} }}_{r}(t))}$ 是事先定义好的值，比受控系统的状态维度${\displaystyle n=dim({\mathbf {x} }(t))}$ 要少。

降阶LQG控制器可以表示为下式：

${\displaystyle {\dot {\hat {\mathbf {x} }}}_{r}(t)=A_{r}(t){\hat {\mathbf {x} }}_{r}(t)+B_{r}(t){\mathbf {u} }(t)+K_{r}(t)\left({\mathbf {y} }(t)-C_{r}(t){\hat {\mathbf {x} }}_{r}(t)\right),{\hat {\mathbf {x} }}_{r}(0)={\mathbf {x} }_{r}(0),}$ 

${\displaystyle {\mathbf {u} }(t)=-L_{r}(t){\hat {\mathbf {x} }}_{r}(t).}$ 

上述公式刻意写的类似传统全阶LQG控制器的形式，降阶的LQG控制问题也可以改写为下式：

${\displaystyle {\dot {\hat {\mathbf {x} }}}_{r}(t)=F_{r}(t){\hat {\mathbf {x} }}_{r}(t)+K_{r}(t){\mathbf {y} }(t),{\hat {\mathbf {x} }}_{r}(0)={\mathbf {x} }_{r}(0),}$ 

${\displaystyle {\mathbf {u} }(t)=-L_{r}(t){\hat {\mathbf {x} }}_{r}(t),}$ 

其中

${\displaystyle F_{r}(t)=A_{r}(t)-B_{r}(t)L_{r}(t)-K_{r}(t)C_{r}(t).}$ 

降阶LQG控制器的矩阵${\displaystyle F_{r}(t),K_{r}(t),L_{r}(t)}$ 和${\displaystyle {\mathbf {x} }_{r}(0)}$ 是由所谓的最佳投影方程（optimal projection equations、OPE）来决定^[3]。

${\displaystyle n}$ 维的最佳投影方阵${\displaystyle \tau (t)}$ 是OPE的核心。此矩阵的秩在所有状态下几乎都等于${\displaystyle n_{r}}$ 。相关投影为斜投影（oblique projection）：${\displaystyle \tau ^{2}(t)=\tau (t)}$ 。最佳投影方程包括四个矩阵微分方程。前二个是LQG控制器对应的矩阵Riccati微分方程的扩展。在方程式中${\displaystyle \tau _{\perp }(t)}$ 表示${\displaystyle I_{n}-\tau (t)}$ ，而${\displaystyle I_{n}}$ 为${\displaystyle n}$ 维的单位矩阵

${\displaystyle {\begin{aligned}{\dot {P}}(t)={}&A(t)P(t)+P(t)A'(t)-P(t)C'(t)W^{-1}(t)C(t)P(t)+V(t)\\[6pt]&{}+\tau _{\perp }(t)P(t)C'(t)W^{-1}(t)C(t)P(t)\tau '_{\perp }(t),\\[6pt]P(0)={}&E\left({\mathbf {x} }(0){\mathbf {x} }'(0)\right),\\[6pt]&{}-{\dot {S}}(t)=A'(t)S(t)+S(t)A(t)-S(t)B(t)R^{-1}(t)B'(t)S(t)+Q(t)\\[6pt]&{}+\tau '_{\perp }(t)S(t)B(t)R^{-1}(t)B'(t)S(t)\tau _{\perp }(t),\end{aligned}}}$ 

${\displaystyle S(T)=F.}$ 

若LQG的维度没有减少，也就是${\displaystyle n=n_{r}}$ ，则${\displaystyle \tau (t)=I_{n},\tau _{\perp }(t)=0}$ ，上述二个方程就是二个没有耦合的矩阵Riccati微分方程，对应全阶的LQG控制器。若${\displaystyle n_{r}<n}$ ，则两个方程会有斜投影项${\displaystyle \tau (t).}$ 。这也是为何降阶的LQG控制器无法分离的原因，斜投影${\displaystyle \tau (t)}$ 是由另外二个矩阵微分方程所决定，其中也和秩的条件（rank conditions）有关。这四个矩阵微分方程组成了最佳投影方程。为了要列出另外二个矩阵微分方程，先定义以下二个矩阵：

${\displaystyle \Psi _{1}(t)=(A(t)-B(t)R^{-1}(t)B'(t)S(t)){\hat {P}}(t)+{\hat {P}}(t)(A(t)-B(t)R^{-1}(t)B'(t)S(t))'}$ 

${\displaystyle {}+P(t)C'(t)W^{-1}(t)C(t)P(t),}$ 

${\displaystyle \Psi _{2}(t)=(A(t)-P(t)C'(t)W^{-1}(t)C(t))'{\hat {S}}(t)+{\hat {S}}(t)(A(t)-P(t)C'(t)W^{-1}(t)C(t))}$ 

${\displaystyle {}+S(t)B(t)R^{-1}(t)B'(t)S(t).}$ 

则最后二个矩阵微分方程如下：

${\displaystyle {\dot {\hat {P}}}(t)=1/2\left(\tau (t)\Psi _{1}(t)+\Psi _{1}(t)\tau '(t)\right),{\hat {P}}(0)=E({\mathbf {x} }(0))E({\mathbf {x} }(0))',\operatorname {rank} ({\hat {P}}(t))=n_{r}}$  almost everywhere,

${\displaystyle -{\dot {\hat {S}}}(t)=1/2\left(\tau '(t)\Psi _{2}(t)+\Psi _{2}(t)\tau (t)\right),{\hat {S}}(T)=0,\operatorname {rank} ({\hat {S}}(t))=n_{r}}$  almost everywhere,

其中

${\displaystyle \tau (t)={\hat {P}}(t){\hat {S}}(t)\left({\hat {P}}(t){\hat {S}}(t)\right)^{*}.}$ 

此处的 * 表示群广义逆矩阵（group generalized inverse）或Drazin逆矩阵（英语：Drazin inverse），是唯一的，定义如下

${\displaystyle A^{*}=A(A^{3})^{+}A.}$ 

其中 + 是摩尔－彭若斯广义逆.

矩阵${\displaystyle P(t),S(t),{\hat {P}}(t),{\hat {S}}(t)}$ 都需要是非负对称矩阵。可以建构最佳投影方程的解，而此解可以决定降阶LQG控制器矩阵${\displaystyle F_{r}(t),K_{r}(t),L_{r}(t)}$ 和${\displaystyle {\mathbf {x} }_{r}(0)}$ ：

${\displaystyle F_{r}(t)=H(t)\left(A(t)-P(t)C'(t)W^{-1}(t)C(t)-B(t)R^{-1}(t)B'(t)S(t)\right)G(t)+{\dot {H}}(t)G'(t),}$ 

${\displaystyle K_{r}(t)=H(t)P(t)C'(t)W^{-1}(t),}$ 

${\displaystyle L_{r}(t)=R^{-1}(t)B'(t)S(t)G'(t),}$ 

${\displaystyle {\mathbf {x} }_{r}(0)=H(0)E({\mathbf {x} }(0)).}$ 

上式中的矩阵${\displaystyle G(t),H(t)}$ 是符合以下性质的矩阵：

${\displaystyle G'(t)H(t)=\tau (t),G(t)H'(t)=I_{n_{r}}}$ 几乎在所有状态下。

可以由${\displaystyle {\hat {P}}(t){\hat {S}}(t)}$ 的投影分解中得到^[4]：

若降阶LQG问题中的所有矩阵都是非时变的，且最终时间（horizon）${\displaystyle T}$ 趋近无限大，则最佳降阶LQG控制器和最佳投影方程也都会是非时变的^[1]。此情形下，最佳投影方程左侧的微分项会为零。

离散时间的情形类似连续时间的例子，要处理的是将${\displaystyle n}$ 阶传统离散时间全阶LQG问题转换为事先已知固定阶数的${\displaystyle n_{r}<n}$ 阶降阶LQG控制器。为了要表示离散时间的OPE，先引入以下二个矩阵：

${\displaystyle \Psi _{i}^{1}=\left(A_{i}-B_{i}(B'_{i}S_{i+1}B_{i}+R_{i})^{-1}B'_{i}S_{i+1}A_{i})\right){\hat {P}}_{i}\left(A_{i}-B_{i}(B'_{i}S_{i+1}B_{i}+R_{i})^{-1}B'_{i}S_{i+1}A_{i})\right)'}$ 

${\displaystyle {}+A_{i}P_{i}C'_{i}(C_{i}P_{i}C'_{i}+W_{i})^{-1}C_{i}P_{i}A'_{i}}$ 

${\displaystyle \Psi _{i+1}^{2}=\left(A_{i}-A_{i}P_{i}C'_{i}(C_{i}P_{i}C'_{i}+W_{i})^{-1}C_{i}\right)'{\hat {S}}_{i+1}\left(A_{i}-A_{i}P_{i}C'_{i}(C_{i}P_{i}C'_{i}+W_{i})^{-1}C_{i}\right)}$ 

${\displaystyle {}+A'_{i}S_{i+1}B_{i}(B'_{i}S_{i+1}B_{i}+R_{i})^{-1}B'_{i}S_{i+1}A_{i}}$ 

则离散时间OPE为

${\displaystyle P_{i+1}=A_{i}\left(P_{i}-P_{i}C'_{i}\left(C_{i}P_{i}C'_{i}+W_{i}\right)^{-1}C_{i}P_{i}\right)A'_{i}+V_{i}+\tau _{\perp i+1}\Psi _{i}^{1}\tau '_{\perp i+1},P_{0}=E\left({\mathbf {x} }_{0}{\mathbf {x'} }_{0}\right)}$ .

${\displaystyle S_{i}=A'_{i}\left(S_{i+1}-S_{i+1}B_{i}\left(B'_{i}S_{i+1}B_{i}+R_{i}\right)^{-1}B'_{i}S_{i+1}\right)A_{i}+Q_{i}+\tau '_{\perp i}\Psi _{i+1}^{2}\tau _{\perp i},S_{N}=F}$ .

${\displaystyle {\hat {P}}_{i+1}=1/2(\tau _{i+1}\Psi _{i}^{1}+\Psi _{i}^{1}\tau '_{i+1}),{\hat {P}}_{0}=E({\mathbf {x} }(0))E({\mathbf {x} }(0))',\operatorname {rank} ({\hat {P}}_{i})=n_{r}}$  almost everywhere,

${\displaystyle {\hat {S}}_{i}=1/2(\tau '_{i}\Psi _{i+1}^{2}+\Psi _{i+1}^{2}\tau _{i}),{\hat {S}}_{N}=0,\operatorname {rank} ({\hat {S}}_{i})=n_{r}}$  almost everywhere.

斜投影（oblique projection）矩阵为

${\displaystyle \tau _{i}={\hat {P}}_{i}{\hat {S}}_{i}\left({\hat {P}}_{i}{\hat {S}}_{i}\right)^{*}.}$ 

非负对称矩阵${\displaystyle P_{i},S_{i},{\hat {P}}_{i},{\hat {S}}_{i}}$ 是离散时间OPE的解，也决定了降阶LQG控制器的矩阵${\displaystyle F_{i}^{r},K_{i}^{r},L_{i}^{r}}$  and ${\displaystyle {\mathbf {x} }_{0}^{r}}$ ：

${\displaystyle F_{i}^{r}=H_{i+1}\left(A_{i}-P_{i}C'_{i}\left(C_{i}P_{i}C'_{i}+W_{i}\right)^{-1}C_{i}-B_{i}\left(B'_{i}S_{i+1}B_{i}+R_{i}\right)^{-1}B'_{i}S_{i+1}\right)G'_{i},}$ 

${\displaystyle K_{i}^{r}=H_{i+1}P_{i}C'_{i}\left(C_{i}P_{i}C'_{i}+W_{i}\right)^{-1},}$ 

${\displaystyle L_{i}^{r}=\left(B'_{i}S_{i+1}B_{i}+R_{i}\right)^{-1}B'_{i}S_{i+1}G'_{i},}$ 

${\displaystyle {\mathbf {x} }_{0}^{r}=H_{0}E({\mathbf {x} }_{0}).}$ 

在上述的方程中，矩阵${\displaystyle G_{i},H_{i}}$ 是有以下性质的矩阵：

${\displaystyle G'_{i}H_{i}=\tau _{i},G_{i}H'_{i}=I_{n_{r}}}$ 几乎在所有状态下。

这些矩阵可以从${\displaystyle {\hat {P}}_{i}{\hat {S}}_{i}}$ 的投影因式分解中求得^[4]。

如同在连续时间中的例子一样，若问题中所有的矩阵都是非时变，且且最终时间（horizon）${\displaystyle T}$ 趋近无限大，降阶LQG控制器就会是非时变的。因此离散时间OPE会收敛到稳态解，决定非时变的降阶LOG控制器^[2]。

离散时间OPE也可以应用在状态维度，输入维度或是输出维度可变的离散时间系统（具有时变维度的离散时间系统）^[6]。若在数位控制器中的取样是不同步的，就可能会出现这类的系统。