[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 ) 系統的狀態方程求解。

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。

NOTE: 若上述對 ${\bf{A}}\left( t \right)$ 時變矩陣的連續性假設成立，則對任意初始狀態 ${\bf{x}}\left( {{t_0}} \right)$ 與任意輸入 ${{\bf{u}}\left( t \right)}$， 狀態方程有唯一解。
(Proof ommitted)

在我們進行求解之前，我們首先求解
\[
{{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)}
\] 其中 ${\bf{A}}\left( t \right)$ 為 $n \times n$ 且每一個 entry 都為 對時間 $t$ 連續的函數。故對任意初始狀態 ${\bf{x}}_i\left( {{t_0}} \right)$  狀態方程存在唯一解 $ {{\bf{x}}_i}\left( t \right),\forall i = 1,2,...,n$ 。

我們可以將這些 $n$ 個解蒐集起來寫作矩陣形式如下：
\[{\bf{X}}\left( t \right): = \left[ {\begin{array}{*{20}{c}}
{{{\bf{x}}_1}\left( t \right)}&{{{\bf{x}}_2}\left( t \right)}& \cdots &{{{\bf{x}}_n}\left( t \right)}
\end{array}} \right]
\]由於 每一個 $ {{\bf{x}}_i}\left( t \right)$ 都滿足 ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)}$ 故我們有
\[{\bf{\dot X}}\left( t \right) = {\bf{A}}\left( t \right){\bf{X}}\left( t \right)
\]

現在我們給出下面的定義：
====================
Definition: (Fundamental Matrix)
若 ${\bf{X}}\left( {{t_0}} \right)$ 為 nonsingular 或者 $n$ 個初始狀態彼此之間為線性獨立，則時變矩陣 ${\bf{X}}\left( t \right)$ 稱作 ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)}$ 的 Fundamental matrix 。
====================

Comment:
Fundamental matrix 並無唯一表示式 (因為初始狀態可以任選).

接著我們定義 狀態轉移矩陣 (State Transition Matrix)

====================
Definition: (State Transition Matrix)
令 ${\bf{X}}\left( {{t}} \right)$ 為 ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)}$ 的 Fundamental matrix，則我們定義其對應的 狀態轉移矩陣 (State Transition Matrix) ${\bf{\Phi }}\left( {t,{t_0}} \right)$ 如下：
\[
{\bf{\Phi }}\left( {t,{t_0}} \right): = {\bf{X}}\left( t \right){{\bf{X}}^{ - 1}}\left( {{t_0}} \right)
\] 且 此狀態轉移矩陣 ${\bf{\Phi }}\left( {t,{t_0}} \right)$ 為 下列狀態方程的唯一解
\[\frac{\partial }{{\partial t}}{\bf{\Phi }}\left( {t,{t_0}} \right) = {\bf{A}}\left( t \right){\bf{\Phi }}\left( {t,{t_0}} \right)\]且 初始條件為 ${\bf{\Phi }}\left( {{t_0},{t_0}} \right) = {\bf{I}}$。
====================

====================
Theorem:
給定任意初始狀態 $t_0$，狀態方程 ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)}$  的解為
\[{\bf{x}}\left( t \right) = {\bf{\Phi }}\left( {{t},{t_0}} \right){\bf{x}}\left( {{t_0}} \right)
\]====================

Proof: Omitted.

下面我們看個例子看看給定狀態方程  ${{\bf{\dot x}}\left( t \right) = {\bf{A}}\left( t \right){\bf{x}}\left( t \right)}$  如何求出對應的 Fundamental matrix。以及 State Transition Matrix。

====================
Example
考慮下列狀態方程
\[{\bf{\dot x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]{\bf{x}}\left( t \right)
\]試求對應的 Fundamental matrix，State Transition Matrix，與 ${\bf{x}}\left( t \right)$。
====================

Solution:
由於 時變矩陣 ${\bf{A}}\left( t \right)$ 符合連續性假設，故我們有對任意初始狀態 ${\bf{x}}\left( {{t_0}} \right)$，存在唯一解。

現在我們觀察
\[{\bf{\dot x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]{\bf{x}}\left( t \right) \Rightarrow \left[ {\begin{array}{*{20}{c}}
{{{\dot x}_1}\left( t \right)}\\
{{{\dot x}_2}\left( t \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{x_2}\left( t \right)}
\end{array}} \right]
\]亦即
\[\left\{ \begin{array}{l}
{{\dot x}_1}\left( t \right) = 0\\
{{\dot x}_2}\left( t \right) = t{x_1}\left( t \right)
\end{array} \right.
\]故給定初始時間 $t_0 =0$ 我們可求解 $x_1(t)$ 與 $x_2(t)$ 如下
\[\begin{array}{l}
{{\dot x}_1}\left( t \right) = 0\\
 \Rightarrow \int_0^t {d{x_1}\left( \tau  \right)}  = 0\\
 \Rightarrow {x_1}\left( t \right) = {x_1}\left( 0 \right)
\end{array}\]與
\[\begin{array}{l}
{{\dot x}_2}\left( t \right) = t{x_1}\left( t \right)\\
 \Rightarrow \int_0^t {d{x_2}\left( \tau  \right)}  = \int_0^t {\tau {x_1}\left( 0 \right)d\tau } \\
 \Rightarrow {x_2}\left( t \right) = {x_1}\left( 0 \right)\frac{{{t^2}}}{2} + {x_2}\left( 0 \right)
\end{array}
\]為了建構 Fundamental matrix，我們可任意選定 等同時變矩陣階數數目的初始狀態，在此例中由於 $\bf{A}$ 為 $2 \times 2$ 時變矩陣，故我們可任選兩個 線性獨立的 初始狀態 來建構 Fundamental matrix ${\bf{X}}\left( t \right)$，比如說選 ${\bf{x}}\left( 0 \right)=[1 \; 0]^T$ 與 ${\bf{x}}\left( 0 \right)=[0 \; 1]^T$，則我們有
\[\begin{array}{l}
{\bf{x}}\left( 0 \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( 0 \right)}\\
{{x_2}\left( 0 \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right]\\
 \Rightarrow {\bf{x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{x_2}\left( t \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( 0 \right)}\\
{{x_1}\left( 0 \right)\frac{{{t^2}}}{2} + {x_2}\left( 0 \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
{\frac{{{t^2}}}{2}}
\end{array}} \right]
\end{array}
\]與
\[\begin{array}{l}
{\bf{x}}\left( 0 \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( 0 \right)}\\
{{x_2}\left( 0 \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right]\\
 \Rightarrow {\bf{x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{x_2}\left( t \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( 0 \right)}\\
{{x_1}\left( 0 \right)\frac{{{t^2}}}{2} + {x_2}\left( 0 \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right]
\end{array}
\] 由於 $[1 \; 0]^T$ 與 $[0 \; 1]^T$ 彼此線性獨立，故由 Fundamental matrix 的定義，我們確實得到了 一組 (不唯一) Fundamental Matrix 如下：
\[{\bf{X}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2}}}{2}}&1
\end{array}} \right]
\] 故由 State Transition Matrix 定義，我們可知
\[
{\bf{\Phi }}\left( {t,{t_0}} \right): = {\bf{X}}\left( t \right){{\bf{X}}^{ - 1}}\left( {{t_0}} \right)
\]其中
\[{{\bf{X}}^{ - 1}}\left( {{t_0}} \right) = {\left. {\frac{1}{1} \cdot \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{ - {t^2}}}{2}}&1
\end{array}} \right]} \right|_{t = {t_0}}} = \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{ - {t_0}^2}}{2}}&1
\end{array}} \right]
\]故 State Transition Matrix 為
\[\begin{array}{l}
{\bf{\Phi }}\left( {{t},{t_0}} \right) = {\bf{X}}\left( t \right){{\bf{X}}^{ - 1}}\left( {{t_0}} \right)\\
\begin{array}{*{20}{c}}
{}&{}&{}&{}
\end{array} = \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2}}}{2}}&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{ - {t_0}^2}}{2}}&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2} - {t_0}^2}}{2}}&1
\end{array}} \right]
\end{array}
\] 由前述 Theorem 可知，狀態方程 ${\bf{\dot x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]{\bf{x}}\left( t \right)$ 的解為
\[
{\bf{x}}\left( t \right) = {\bf{\Phi }}\left( {t,{t_0}} \right){\bf{x}}\left( {{t_0}} \right) = \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2} - {t_0}^2}}{2}}&1
\end{array}} \right]{\bf{x}}\left( {{t_0}} \right) = \left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2}}}{2}}&1
\end{array}} \right]{\bf{x}}\left( 0 \right)
\]現在我們帶回驗證 上式 ${\bf{\Phi }}\left( {t,{t_0}} \right)$ 確實為 ，亦即對其微分
\[\begin{array}{l}
\frac{d}{{dt}}{\bf{x}}\left( t \right) = \frac{d}{{dt}}\left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2}}}{2}}&1
\end{array}} \right]{\bf{x}}\left( 0 \right) = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]\underbrace {{{\left[ {\begin{array}{*{20}{c}}
1&0\\
{\frac{{{t^2}}}{2}}&1
\end{array}} \right]}^{ - 1}}{\bf{x}}\left( t \right)}_{{\bf{x}}\left( 0 \right)}\\
 \Rightarrow {\bf{\dot x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&0\\
{ - \frac{{{t^2}}}{2}}&1
\end{array}} \right]{\bf{x}}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0&0\\
t&0
\end{array}} \right]{\bf{x}}\left( t \right)
\end{array}
\]故得證。