[最佳估計] 狀態估測器的收斂性

考慮離散時間 LTI 系統
$$\begin{array}{l} x(k + 1) = Ax(k) + w(k)\\ y(k) = Cx\left( k \right) + v(k) \end{array}$$ 其中 $x \in \mathbb{R}^n, u \in \mathbb{R}^m, y \in \mathbb{R}^p$ 且 $x(0) = x_0$；

基本狀態估計問題：給定初始估計誤差 ( i.e., $\bar x(0) \neq x_0$) 且 不考慮 雜訊 無外部干擾的情況 ($w(k) = v(k)=0$)，我們想問 $\hat x(k) \to x(k)$ as $k \to \infty$ ?

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Theorem: (Convergence of Estimator Cost)
給定無 noise 量測輸出 ${\bf y}( T) = \{Cx(0), CAx(0),..., CA^T x(0)\}$ 則最佳估測器的 cost $V_T^*({\bf y}(T))$ 在 $T \to \infty$ 時收斂 。
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Proof:
由於
$$V_T^{} = \frac{1}{2}\left( \begin{array}{l} \left| {\hat x\left( 0 \right) - \bar x\left( 0 \right)} \right|_{{{\left( {{P^ - }\left( 0 \right)} \right)}^{ - 1}}}^2\\ \begin{array}{*{20}{c}} {}&{} \end{array} + \sum\limits_{k = 0}^{T - 1} {|\hat x\left( {k + 1} \right) - A\hat x\left( k \right)|_{{Q^{ - 1}}}^2}  + \sum\limits_{k = 0}^T {\left| {y\left( k \right) - C\hat x\left( k \right)} \right|_{{R^{ - 1}}}^2} \end{array} \right)$$ 故我們分三個步驟證明 $V_T^*({\bf y}(T))$ 在 $T \to \infty$ 時收斂 ：
首先證明 sequence $\{V_T^*\}$ 有上界(bounded above)；接著我們證明 $\{V_T^*\}$ 為 nondecreasing。則由步驟一與步驟二可推論 $\{V_T^*\}$ 必定收斂。

現在我們開始證明 $V_T^*({\bf y}(T))$ 有界：
由於我們的目標為 $\min_{\hat{ {\bf x}}(k)} V_k$ 但我們並知道 怎樣的 $\hat x$ 可以幫助我們達成此目標，故我們先暫取 $\hat x(0) :=  x_0$ (並不一定為最佳解!) 則我們有
$$\begin{array}{l} \left\{ \begin{array}{l} \hat x(1) = A\hat x(0)\\ \hat x(2) = A\hat x(1) = {A^2}\hat x(0)\\  \vdots \end{array} \right.\\  \Rightarrow y\left( k \right) = C{A^k}{x_0} \end{array}$$ 故若將此結果帶入我們的 cost 可得
$$\bar{V}_T^{} = \frac{1}{2}\left| {\hat x\left( 0 \right) - \bar x\left( 0 \right)} \right|_{{{\left( {{P^ - }\left( 0 \right)} \right)}^{ - 1}}}^2 < \infty$$ 但注意到此並非最佳 cost，若代入最佳解 則我們必有
$$V_T^* \le \bar V_T$$ 故可推論 sequence $\{V_T^*\}$ 必定 bounded above。

接著我們證明 optimal cost sequence $\{V_T^*\}$ 為 nondecreasing。
給定量測輸出 ${\bf y}( T) = \{Cx(0), CAx(0),..., CA^T x(0)\}$，定義 在時間 $T$ 的最佳狀態 sequence 為
$$\left\{ {\hat x\left( {0} \right),\hat x\left( {1} \right),...,\hat x\left( {T} \right)} \right\}$$ 現在我們比對 $T$ 時刻的 optimal cost 與 $T-1$ 時刻的 optimal cost 可知
$$V_T^* - \frac{1}{2}\left( {\left| {y\left( {T - 1} \right) - C\hat x\left( {T - 1} \right)} \right|_{{R^{ - 1}}}^2 + |\hat x\left( T \right) - A\hat x\left( k \right)|_{{R^{ - 1}}}^2} \right) \ge V_{T - 1}^*$$ 此說明了 sequence $\{V_T^*\}$ 為 nondecreasing。

故 由  optimal cost sequence $\{V_T^*\}$  bounded above 與 $\{V_T^*\}$ 為 nondecreasing，我們可推論當 $T \to \infty$  $\{V_T^*\}$ 必定收斂。 $\square$


注意到 optimal estimator cost $V_T^*$ 的收斂性與 系統可觀測性無關，但若我們要求我們的估計狀態 $\hat x \to x$ 則系統的觀測性將扮演重要腳色，我們將此結果記做以下定理：

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Theorem: Estimator Convergence
考慮控制系統 $(A,C)$ 可觀測 且 $Q,R >0$ 為正定矩陣 且 給定一組無雜訊量測輸出
$${\bf y}(T) := \{Cx(0), CAx(0), ..., CA^T x(0)\}$$ 則 最佳狀態估測 收斂到原本系統狀態；亦即
$$\hat x(T) \to x(T) \;\; \text{ as $T \to \infty$}$$ ==========================

Proof:

利用 在時刻 $T+n-1$ 的最佳解作為在時刻 $T-1$的 decision variables ，則前述 Theorem 告訴我們可寫

$$\small V_{T + n - 1}^* - \frac{1}{2}\left[ {\sum\limits_{k = T - 1}^{T + n - 2} {|\hat x\left( {k + 1} \right) - A\hat x\left( k \right)|_{{Q^{ - 1}}}^2}  + \sum\limits_{k = T}^{T + n - 1} {\left| {y\left( k \right) - C\hat x\left( k \right)} \right|_{{R^{ - 1}}}^2} } \right] \ge V_{T - 1}^*$$ 做變數變換 令 $j = k - T$ 我們可得

$$\small V_{T + n - 1}^* - \frac{1}{2}\left[ {\sum\limits_{j =  - 1}^{n - 2} {|\hat x\left( {T + j + 1} \right) - A\hat x\left( {T + j} \right)|_{{Q^{ - 1}}}^2}  + \sum\limits_{j = 0}^{n - 1} {\left| {y\left( {j + T} \right) - C\hat x\left( {j + T} \right)} \right|_{{R^{ - 1}}}^2} } \right] \ge V_{T - 1}^*$$ 注意到當 $T \to \infty$時， $\{V_T^*\}$ 收斂 且 $Q^{-1}, R^{-1} >0$ 故上式

$$\begin{array}{l} \underbrace {V_{T + n - 1}^* - V_{T - 1}^*}_{ \to 0} \ge \frac{1}{2}\left[ \begin{array}{l} \sum\limits_{j =  - 1}^{n - 2} {|\hat x\left( {T + j + 1} \right) - A\hat x\left( {T + j} \right)|_{{Q^{ - 1}}}^2} \\ \begin{array}{*{20}{c}} {}&{} \end{array} + \sum\limits_{j = 0}^{n - 1} {\left| {y\left( {j + T} \right) - C\hat x\left( {j + T} \right)} \right|_{{R^{ - 1}}}^2} \end{array} \right]\\  \Rightarrow \sum\limits_{j =  - 1}^{n - 2} {|\hat x\left( {T + j + 1} \right) - A\hat x\left( {T + j} \right)|_{{Q^{ - 1}}}^2}  + \sum\limits_{j = 0}^{n - 1} {\left| {y\left( {j + T} \right) - C\hat x\left( {j + T} \right)} \right|_{{R^{ - 1}}}^2}  \to 0\\  \Rightarrow \left\{ \begin{array}{l} \hat x\left( {T + j + 1} \right) - A\hat x\left( {T + j} \right) \to 0,\begin{array}{*{20}{c}} {} \end{array}\forall j =  - 1,...,n - 2\\ y\left( {j + T} \right) - C\hat x\left( {j + T} \right) \to 0,\begin{array}{*{20}{c}} {} \end{array}\forall j = 0,...,n - 1 \end{array} \right. \ \ \ \ \ \ (**) \end{array}$$ 令 $\hat w_T(j) := \hat x(T+j+1|T+n-1) - A \hat x(T+j|T+n-1)$ 並且透過系統方程 $x(k+1) = Ax(k) + w(k)$ 我們有
$$\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {\hat x\left( {T|T + n - {\rm{1}}} \right)}\\ {\hat x\left( {T{\rm{ + 1}}|T + n - {\rm{1}}} \right)}\\  \vdots \\ {\hat x\left( {T + n - 1|T + n - {\rm{1}}} \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} I\\ A\\  \vdots \\ {{A^{n - 1}}} \end{array}} \right]\hat x\left( {T|T + n - {\rm{1}}} \right)\\ \begin{array}{*{20}{c}} {}&{}&{}&{} \end{array} + \left[ {\begin{array}{*{20}{c}} 0&{}&{}&{}\\ I&0&{}&{}\\  \vdots & \vdots & \ddots &{}\\ {{A^{n - 2}}}&{{A^{n - 3}}}& \cdots &I \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\hat w}_T}\left( 0 \right)}\\ {{{\hat w}_T}\left( 1 \right)}\\  \vdots \\ {{{\hat w}_T}\left( {n - 2} \right)} \end{array}} \right]\ \ \ \ (*) \end{array}$$ 且由於我們的量測輸出 滿足
$$\left\{ \begin{array}{l} y\left( T \right) = Cx\left( T \right)\\ y\left( {T + 1} \right) = CAx\left( T \right)\\  \vdots \\ y\left( {T + n - 1} \right) = C{A^{n - 1}}x\left( T \right) \end{array} \right. \Rightarrow \left[ {\begin{array}{*{20}{c}} {y\left( T \right)}\\ {y\left( {T + 1} \right)}\\  \vdots \\ {y\left( {T + n - 1} \right)} \end{array}} \right] = Ox\left( T \right)$$ 其中 $O$ 為 observability matrix。現在用上式減去 同乘 $C$ 矩陣 後的 $(*)$ 可得
$$\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {y\left( T \right)}\\ {y\left( {T + 1} \right)}\\  \vdots \\ {y\left( {T + n - 1} \right)} \end{array}} \right] - C\left[ {\begin{array}{*{20}{c}} {\hat x\left( {T|T + n - {\rm{1}}} \right)}\\ {\hat x\left( {T{\rm{ + 1}}|T + n - {\rm{1}}} \right)}\\  \vdots \\ {\hat x\left( {T + n - 1|T + n - {\rm{1}}} \right)} \end{array}} \right] = O\left[ {x\left( T \right) - \hat x\left( {T|T + n - {\rm{1}}} \right)} \right]\\ \begin{array}{*{20}{c}} {}&{}&{}&{} \end{array} + \left[ {\begin{array}{*{20}{c}} 0&{}&{}&{}\\ C&0&{}&{}\\  \vdots & \vdots & \ddots &{}\\ {C{A^{n - 2}}}&{C{A^{n - 3}}}& \cdots &C \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\hat w}_T}\left( 0 \right)}\\ {{{\hat w}_T}\left( 1 \right)}\\  \vdots \\ {{{\hat w}_T}\left( {n - 2} \right)} \end{array}} \right] \end{array}$$ 現在用 $(**)$ 可知
$$\begin{array}{l} \underbrace {\left[ {\begin{array}{*{20}{c}} {y\left( T \right)}\\ {y\left( {T + 1} \right)}\\  \vdots \\ {y\left( {T + n - 1} \right)} \end{array}} \right] - C\left[ {\begin{array}{*{20}{c}} {\hat x\left( {T|T + n - {\rm{1}}} \right)}\\ {\hat x\left( {T{\rm{ + 1}}|T + n - {\rm{1}}} \right)}\\  \vdots \\ {\hat x\left( {T + n - 1|T + n - {\rm{1}}} \right)} \end{array}} \right]}_{ \to 0} = O\left[ {x\left( T \right) - \hat x\left( {T|T + n - {\rm{1}}} \right)} \right]\\ \begin{array}{*{20}{c}} {}&{}&{}&{} \end{array} + \underbrace {\left[ {\begin{array}{*{20}{c}} 0&{}&{}&{}\\ C&0&{}&{}\\  \vdots & \vdots & \ddots &{}\\ {C{A^{n - 2}}}&{C{A^{n - 3}}}& \cdots &C \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{{\hat w}_T}\left( 0 \right)}\\ {{{\hat w}_T}\left( 1 \right)}\\  \vdots \\ {{{\hat w}_T}\left( {n - 2} \right)} \end{array}} \right]}_{ \to 0}\\  \Rightarrow O\left[ {x\left( T \right) - \hat x\left( {T|T + n - {\rm{1}}} \right)} \right] \to 0 \end{array}$$ 又因為 observability matrix $O$ 有 linear independent columns，故我們可推知
$$x\left( T \right) - \hat x\left( {T|T + n - {\rm{1}}} \right) \to 0 \;\; \text{as $T \to \infty$}$$ 亦即
$$\hat x\left( {T|T + n - {\rm{1}}} \right) \to x\left( T \right)$$ 再者如果我們觀察 $(*)$ 可以發現因為 $\hat w_T (j) \to 0$ 當 $T \to \infty$，故
$$\hat x\left( {T + n - 1|T + n - {\rm{1}}} \right) \to {A^{n - 1}}\hat x\left( {T|T + n - {\rm{1}}} \right) \;\;\;\; \text{ as $T \to \infty$}$$ 又因為 $A^{n-1} x(T) = x(T + n -1)$ ，我們有
$$\begin{array}{l} x\left( {T + n - 1} \right) - \hat x\left( {T + n - 1|T + n - {\rm{1}}} \right) \to {A^{n - 1}}x\left( T \right) - {A^{n - 1}}\hat x\left( {T|T + n - {\rm{1}}} \right)\\  \Rightarrow x\left( {T + n - 1} \right) - \hat x\left( {T + n - 1|T + n - {\rm{1}}} \right) \to {A^{n - 1}}\underbrace {\left[ {x\left( T \right) - \hat x\left( {T|T + n - {\rm{1}}} \right)} \right]}_{ \to 0}\\  \Rightarrow x\left( {T + n - 1} \right) \to \hat x\left( {T + n - 1|T + n - {\rm{1}}} \right)\\  \Rightarrow x\left( {T + n - 1} \right) \to \hat x\left( {T + n - 1} \right)\\  \Rightarrow x\left( j \right) \to \hat x\left( j \right),\begin{array}{*{20}{c}} {} \end{array}\forall j \to \infty \end{array}$$ 至此得證