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Finite observability identifies the initial state

Statement

Let $d \geq 0$ , $m, p \geq 1$ , and $q \geq 1$ be integers, and consider a real discrete input-state-output experiment with known matrices $A, B, C$ , known inputs $u_{0}, \dots, u_{q - 2}$ , and unknown initial state $x_{0}$ . For $0 \leq k < q$ , set $f_{0} : = 0, f_{k} : = \sum_{j = 0}^{k - 1} C A^{k - 1 - j} B u_{j} (k \geq 1),$ and define the finite observability matrix $O_{q} : = [\begin{array}{c} C \\ C A \\ ⋮ \\ C A^{q - 1} \end{array}] .$ The corrected noiseless output stack is $z : = [\begin{array}{c} y_{0} - f_{0} \\ y_{1} - f_{1} \\ ⋮ \\ y_{q - 1} - f_{q - 1} \end{array}] = O_{q} x_{0} .$ The map $x_{0} \mapsto z$ is injective if and only if $rank O_{q} = d$ . If $d = 0$ , both injectivity and the rank condition hold for the singleton state space. If $d \geq 1$ and $O_{q}$ has full column rank, let $σ_{\min} (O_{q}) > 0$ be its least singular value. For any additive stacked observation error $e \in R^{q p}$ , the minimum-norm least-squares estimate of $x_{0}$ from $z + e$ is ${\hat{x}}_{0} = O_{q}^{+} (z + e)$ , and it satisfies $∥ {\hat{x}}_{0} - x_{0} ∥_{2} \leq \frac{∥ e ∥_{2}}{σ_{\min} (O_{q})} .$

Statement references

Finite-horizon recurrence and exact observation convention

Rank as the dimension of the image

Rank-nullity for finite-dimensional domains

Finite-dimensional singular-value decomposition

Rank equals the number of positive singular values

Finite-matrix Moore--Penrose pseudoinverse

Existence and uniqueness of the pseudoinverse

Full-column-rank pseudoinverse identity

The product $O O^{+}$ is an orthogonal projection

The pseudoinverse gives the minimum-norm least-squares solution

Pythagoras for orthogonal vectors

Dahleh, Dahleh, and Verghese, Lectures on Dynamic Systems and Control, Chapter 25

Lall, State-Space Realization, Engr207c lecture notes

Facts & Assumptions

Given: A real discrete experiment of horizon $q$ with known matrices $A, B, C$ , known inputs $u_{0}, \dots, u_{q - 2}$ , unknown initial state $x_{0}$ , and a stacked additive error $e \in R^{q p}$ .

[F1]

Discrete linear input-state-output experiments: states obey $x_{k + 1} = A x_{k} + B u_{k}$ , outputs obey $y_{k} = C x_{k}$ , and observation errors are additive deterministic vectors.

[F2]

Matrix multiplication is associative, unital, distributive, and compatible with scalar multiplication: compatible matrix products are associative, identity matrices act on either side, and multiplication distributes over finite sums.

[F3]

Rank and nullity of a linear map with finite-dimensional domain: $rank T = \dim (im T)$ and $nullity T = \dim (\ker T)$ for finite-dimensional domains.

[F4]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map with finite-dimensional domain, $\dim V = nullity T + rank T$ .

[F5]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: a linear map between finite-dimensional real or complex inner-product spaces has orthonormal left and right singular bases, with its positive singular values ordered nonincreasingly.

[F6]

The rank of a linear map is the number of its nonzero singular values: the rank equals the number of positive singular values.

[F7]

The Moore--Penrose pseudoinverse $A^{+}$ as the solution of the four Penrose equations: $O^{+}$ satisfies $O O^{+} O = O$ , $O^{+} O O^{+} = O^{+}$ , and $O O^{+}$ and $O^{+} O$ are self-adjoint.

[F8]

Every finite real or complex matrix has a unique Moore--Penrose pseudoinverse: every finite real matrix has a unique Moore--Penrose pseudoinverse.

[F9]

If $A$ has full column rank, then $A^{+} = (A^{*} A)^{- 1} A^{*}$ : if $O$ has full column rank, then $O^{+} O = I_{d}$ .

[F10]

$A A^{+}$ and $A^{+} A$ are the orthogonal projections onto $im A$ and $im A^{*}$ : $O O^{+}$ is the orthogonal projection onto $im O$ .

[F11]

For every right-hand side $b$ , $A^{+} b$ is the unique least-squares solution of minimum Euclidean norm: for a finite real matrix $O$ and vector $b$ , $O^{+} b$ is the unique least-squares solution of minimum Euclidean norm.

[F12]

Pythagoras and finite orthogonal sums: if $a$ and $b$ are orthogonal, then $∥ a + b ∥_{2}^{2} = ∥ a ∥_{2}^{2} + ∥ b ∥_{2}^{2}$ .

Proof

technique · direct

1.1givenF1F2algebra

For each $0 \leq k < q$ , the recurrence gives $x_{k} = A^{k} x_{0} + \sum_{j = 0}^{k - 1} A^{k - 1 - j} B u_{j}$ , with the sum empty at $k = 0$ . At $k = 0$ this is $x_{0} = A^{0} x_{0}$ . If it holds at $k < q - 1$ , substitution into $x_{k + 1} = A x_{k} + B u_{k}$ and the power recursion give the formula at $k + 1$ , using associativity and distributivity from [F2]. Multiplying by $C$ and subtracting $f_{k}$ gives $y_{k} - f_{k} = C A^{k} x_{0}$ , so stacking yields $z = O_{q} x_{0}$ .

1.2givenF5F6algebra

Suppose $d \geq 1$ and $O_{q}$ has full column rank. By [F5] and [F6], an SVD has exactly $d$ positive singular values $s_{1} \geq \dots \geq s_{d} = σ_{\min} (O_{q}) > 0$ . Writing $w$ in the right singular basis gives $∥ O_{q} w ∥_{2}^{2} = \sum_{j = 1}^{d} s_{j}^{2} ∣ w_{j} ∣^{2} \geq σ_{\min} (O_{q})^{2} ∥ w ∥_{2}^{2}$ .

2.1givenF3F4step 1.1algebra

Two states $x, x^{'} \in R^{d}$ produce the same corrected stack exactly when $O_{q} (x - x^{'}) = 0$ . Thus injectivity is equivalent to $\ker O_{q} = {0}$ . By [F3] and [F4], this is equivalent to $rank O_{q} = d$ . When $d = 0$ , the map has a singleton domain and the empty matrix has rank zero, so the same equivalence holds.

2.2givenF7F8F9F10F11F12step 1.1algebra

By [F7] and [F8], $O_{q}^{+}$ is the unique Moore--Penrose inverse. Set ${\hat{x}}_{0} = O_{q}^{+} (z + e)$ ; [F11] identifies it as the minimum-norm least-squares estimate, and [F9] with $z = O_{q} x_{0}$ gives $w : = {\hat{x}}_{0} - x_{0} = O_{q}^{+} e$ . Hence $O_{q} w = O_{q} O_{q}^{+} e$ . By [F10], this is an orthogonal projection of $e$ ; the decomposition $e = O_{q} O_{q}^{+} e + (e - O_{q} O_{q}^{+} e)$ has orthogonal summands, so [F12] gives $∥ O_{q} w ∥_{2} \leq ∥ e ∥_{2}$ .

3.1step 1.2step 2.2step 2.1algebra∎

Combining steps 1.2 and 2.2 gives $σ_{\min} (O_{q}) ∥ {\hat{x}}_{0} - x_{0} ∥_{2} \leq ∥ O_{q} ({\hat{x}}_{0} - x_{0}) ∥_{2} \leq ∥ e ∥_{2}$ . Division by the positive singular value proves the claimed sensitivity bound; step 2.1 proves the injectivity assertion.

Depends on

Used by

An unobserved state mode Example

Dependency tree · two levels

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Sources

Mohammed Dahleh, Munther A. Dahleh, and George Verghese, Lectures on Dynamic Systems and Control, Chapter 25: Minimal State-Space Realization (standard reference, not scraped)
Sanjay Lall, Engr207c: Operator Theory and Linear Systems, State-Space Realization lecture notes (standard reference, not scraped)