We again consider the linear system of ODE

where M ∈ M^n×n.

In this section we address the observability problem, modeled as follows. We suppose that we can observe

(O)                                             y(t) := Nx(t)               (t ≥ 0),

for a given matrix N ∈ M^m×n. Consequently, y(t) ∈ R^m. The interesting situation is when m << n and we interpret y(.) as low-dimensional “observations” or  “measurements” of the high-dimensional dynamics x(.).

OBSERVABILITY QUESTION: Given the observations y(.), can we in principle reconstruct x(.)? In particular, do observations of y(.) provide enough information for us to deduce the initial value x⁰ for (ODE)?

DEFINITION. The pair (ODE),(O) is called observable if the knowledge of y(.) on any time interval [0, t] allows us to compute x⁰.

More precisely, (ODE),(O) is observable if for all solutions x₁(.), x₂(.), Nx₁(.) ≡Nx₂(.) on a time interval [0, t] implies x₁(0) = x₂(0).

TWO SIMPLE EXAMPLES. (i) If N ≡ 0, then clearly the system is not observable.

(ii) On the other hand, if m = n and N is invertible, then clearly x(t) = N−1y(t)

is observable.

The interesting cases lie between these extremes.

THEOREM 1.1 (OBSERVABILITY AND CONTROLLABILITY). The system

(1.11)

is observable if and only if the system

(1.12)

is controllable, meaning that C = Rⁿ.

INTERPRETATION. This theorem asserts that somehow “observability and controllability are dual concepts” for linear systems.

Proof. 1. Suppose (1.11) is not observable. Then there exist points x¹ ≠ x² ∈Rⁿ, such that

Let t = 0, to find Nx⁰ = 0. Then differentiate this expression k times in t and let t = 0, to discover as well that

                                               NM^kx⁰= 0

for k = 0, 1, 2, . . . . Hence (x⁰)^T (M^k)^TN^T = 0, and hence (x⁰)^T (M^T )kN^T = 0.

This implies

                               (x⁰) ^T [N^T ,M^TN^T , . . . , (M^T ) ⁿ⁻¹N^T ] = 0.

Since x⁰ = 0, rank[N^T , . . . , (M^T )ⁿ⁻¹N^T ] < n. Thus problem (1.12) is not controllable. Consequently, (1.12) controllable implies (1.11) is observable.

2. Assume now (1.12) not controllable. Then rank[N^T , . . . , (M^T )ⁿ⁻¹N^T ] < n,  and consequently according to Theorem 2.3 there exists x0 = 0 such that

(x⁰) ^T [N^T , . . . , (M^T ) ⁿ⁻¹N^T ] = 0.

That is, NM^kx⁰ = 0 for all k = 0, 1, 2, . . . , n − 1.

We want to show that y(t) = Nx(t) ≡ 0, where

According to the Cayley–Hamilton Theorem, we can write

for appropriate constants. Consequently NMⁿx⁰ = 0. Likewise,

and so NMⁿ⁺¹x⁰ = 0. Similarly, NM^kx⁰ = 0 for all k.

Now

and therefore

References

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