[線性系統] 動態方程式的求解(3) - LTV state equation- Total Solution

延續前篇文章 [線性系統] 動態方程式的求解(2) - LTV state equation- Homogeneous solution，這次要介紹線性時變 (Linear Time Varying, LTV ) 系統的狀態方程的全解。


考慮下列 LTV 動態系統
\[\left\{ {\begin{array}{*{20}{l}}
{{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right) + {\bf{B}}\left( t \right){\bf{u}}\left( t \right)}\\
{{\bf{y}}\left( t \right) = {\bf{C}}\left( t \right){\bf{x}}\left( t \right) + {\bf{D}}\left( t \right){\bf{u}}\left( t \right)}
\end{array}} \right.
\] 且假設  ${\bf{A}}\left( t \right)$ 為 $n \times n$ 且矩陣中每一項元素 都為對時間 $t$ 連續函數。

Comment:
1. 上式中 ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right) + {\bf{B}}\left( t \right){\bf{u}}\left( t \right)}$ 稱為狀態方程 (State equation)
2. ${{\bf{y}}\left( t \right) = {\bf{C}}\left( t \right){\bf{x}}\left( t \right) + {\bf{D}}\left( t \right){\bf{u}}\left( t \right)}$ 稱為 輸出方程 (Output equation)

===================
Claim:
給定初始狀態 ${\bf{x}}\left( {{t_0}} \right)$ 與 輸入 ${{\bf{u}}\left( t \right)}$，則狀態方程 ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right) + {\bf{B}}\left( t \right){\bf{u}}\left( t \right)}$ 的解為
\[
{\bf{x}}\left( t \right) = {\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } \ \ \ \ (*)
\]其中  ${\bf{\Phi }}\left( {t,\tau } \right): = {\bf{X}}\left( t \right){{\bf{X}}^{ - 1}}\left( \tau  \right)$ 為 ${\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)$ 的 State Transition matrix 滿足\[\frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,{t_0}} \right) = {\bf{A}}\left( t \right){\bf{\Phi }}\left( {t,{t_0}} \right)\]且 初始條件為 ${\bf{\Phi }}\left( {{t_0},{t_0}} \right) = {\bf{I}}$。
===================

Proof:
首先證明 $(*)$ 滿足初始條件：
\[\begin{array}{l}
{\bf{x}}\left( t \right) = {\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } \\
 \Rightarrow {\bf{x}}\left( {{t_0}} \right) = {\bf{\Phi }}\left( {{t_0},{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \underbrace {\int_{{t_0}}^{{t_0}} {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } }_{ = 0}\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = {\bf{X}}\left( {{t_0}} \right){{\bf{X}}^{ - 1}}\left( {{t_0}} \right){\bf{x}}\left( {{t_0}} \right) = {\bf{Ix}}\left( {{t_0}} \right) = {\bf{x}}\left( {{t_0}} \right)
\end{array}\]接著我們證明 $(*)$ 確實滿足狀態方程。
\[\begin{array}{l}
{\bf{x}}\left( t \right) = {\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } \\
\frac{d}{{dt}}{\bf{x}}\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } } \right]\\
 \Rightarrow {\bf{\dot x}}\left( t \right) = \frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \frac{\partial }{{\partial t}}\left[ {\int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } } \right]
\end{array}
\] 利用 Fundamental Theorem of Calculus:
\[\frac{\partial }{{\partial t}}\int_{{t_0}}^t {f\left( {t,\tau } \right)d\tau }  = \left. {f\left( {t,\tau } \right)} \right|_{\tau  = t}^{} + \int_{{t_0}}^t {\left( {\frac{\partial }{{\partial t}}f\left( {t,\tau } \right)} \right)d\tau }
\] 我們得知
\[\begin{array}{l}
{\bf{\dot x}}\left( t \right) = \frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) \\
\ \ \ \ \ \ \ \ \ \ \ + \left[ {{\bf{\Phi }}\left( {t,t} \right){\bf{B}}\left( t \right){\bf{u}}\left( t \right) + \int_{{t_0}}^t {\left( {\frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)} \right)d\tau } } \right]\\
 \Rightarrow {\bf{\dot x}}\left( t \right) = \frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + {\bf{\Phi }}\left( {t,t} \right){\bf{B}}\left( t \right){\bf{u}}\left( t \right) \\
\ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  \ \ \  + \int_{{t_0}}^t {\left( {\frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)} \right)d\tau }
\end{array}
\]再由 State Transition Matrix 定義 $\frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,{t_0}} \right) = {\bf{A}}\left( t \right){\bf{\Phi }}\left( {t,{t_0}} \right)$我們知道
\[\begin{array}{l}
{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + {\bf{\Phi }}\left( {t,t} \right){\bf{B}}\left( t \right){\bf{u}}\left( t \right) \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ + \int_{{t_0}}^t {\left( {\frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,\tau } \right)} \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } \\
 \Rightarrow {\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + {\bf{\Phi }}\left( {t,t} \right){\bf{B}}\left( t \right){\bf{u}}\left( t \right)\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  + \int_{{t_0}}^t {{\bf{A}}\left( t \right){\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } \\
 \Rightarrow {\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right)\underbrace {\left[ {{\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right)  + \int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } } \right]}_{ = {\bf{x}}\left( t \right)} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + \underbrace {{\bf{X}}\left( t \right){{\bf{X}}^{ - 1}}\left( t \right)}_{ = {\bf{I}}}{\bf{B}}\left( t \right){\bf{u}}\left( t \right)\\
 \Rightarrow {\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right) + {\bf{B}}\left( t \right){\bf{u}}\left( t \right)
\end{array}
\]

有了上述結果之後，我們便可以進一步求得 輸入輸出之間關係，將
\[
{\bf{x}}\left( t \right) = {\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) + \int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau } \] 帶回輸出方程 ${{\bf{y}}\left( t \right) = {\bf{C}}\left( t \right){\bf{x}}\left( t \right) + {\bf{D}}\left( t \right){\bf{u}}\left( t \right)}$，可得
\[\begin{array}{l}
 \Rightarrow {\bf{y}}\left( t \right) = {\bf{C}}\left( t \right){\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} + {\bf{C}}\left( t \right)\int_{{t_0}}^t {{\bf{\Phi }}\left( {t,\tau } \right){\bf{B}}\left( \tau  \right){\bf{u}}\left( \tau  \right)d\tau }  + {\bf{D}}\left( t \right){\bf{u}}\left( t \right)
\end{array}\]