Observability matrix — rank test — Physics notes

WHAT is observability?

WHY only $A$ and $C$ ? Because the inputs $u$ and $D$ produce a known contribution to $y$ ; we can subtract it off. What remains is the free response $y_{\text{free}}=C e^{At}x(0)$ . Observability is purely about whether $C$ and $A$ let us "see" every direction of $x(0)$ .

HOW to derive the test (from scratch)

Set $u=0$ (legitimate, since its effect is known). The output is $y(t) = C e^{At} x(0).$

WHY differentiate? A single instant $y(0)=Cx(0)$ only tells us $x(0)$ 's projection onto $C$ . To grab more directions, take derivatives at $t=0$ :

\begin{aligned} y(0) &= C x(0)\\ \dot{y}(0) &= C A x(0)\\ \ddot{y}(0) &= C A^2 x(0)\\ &\;\;\vdots\\ y^{(k)}(0) &= C A^{k} x(0) \end{aligned}

Why stop at $k=n-1$ ? By the Cayley–Hamilton theorem, $A^n$ is a linear combination of $I, A, \dots, A^{n-1}$ . So $CA^n, CA^{n+1}, \dots$ give no new rows — every higher derivative is redundant. Stacking derivatives $0$ through $n-1$ :

\underbrace{\begin{bmatrix} y(0)\\ \dot y(0)\\ \vdots \\ y^{(n-1)}(0)\end{bmatrix}}_{\text{known}} = \underbrace{\begin{bmatrix} C\\ CA\\ \vdots \\ CA^{n-1}\end{bmatrix}}_{\mathcal{O}} x(0).

We can solve uniquely for $x(0)$ iff the matrix $\mathcal{O}$ has full column rank $n$ (so its null space is only $\{0\}$ ).

Figure — Observability matrix — rank test

Worked examples

Common mistakes

Flashcards

Define observability of an LTI system

(A,C)

is observable if

x(0)

can be uniquely determined from

u(t),y(t)

on any finite interval.

State the observability matrix

\mathcal{O}=[\,C;\ CA;\ CA^2;\dots;\ CA^{n-1}\,]

, size

pn\times n

State the rank test

Observable

\iff \operatorname{rank}(\mathcal{O})=n

Why stop at power

n-1

Cayley–Hamilton:

A^n

is a combination of lower powers, so

CA^k

for

k\ge n

adds no new rows.

What is an unobservable state?

A nonzero

x_0\in\ker(\mathcal{O})

; it gives

Ce^{At}x_0\equiv0

, invisible to sensors.

Why can we ignore

B,D,u

in observability?

Their contribution to

y

is known and subtractable; observability concerns only the free response

Ce^{At}x(0)

Duality statement

(A,C)

observable

\iff (A^\top,C^\top)

controllable.

Can a single output observe two identical decoupled modes?

No — a scalar

C

cannot separate a repeated eigenvalue with 2 eigenvectors; need

p\ge2

Recall Feynman: explain to a 12-year-old

Imagine a box with hidden gears (the states). You can only see one dial on the outside (the output). Observability asks: by watching that dial — its value and how fast it changes — can you figure out how every gear was spinning at the start? If some gear never moves the dial no matter what, it's "hidden," and the box is not observable. The rank test is just a way to count how many gears the dial can actually reveal; if that count equals the total number of gears, you can see everything.

Connections

Controllability matrix — rank test (dual concept)
Cayley–Hamilton theorem (why we stop at $A^{n-1}$ )
Kalman Filter (observability is required for state estimation to converge)
Kalman decomposition (splits system into observable/unobservable subsystems)
State-space representation
Rank and null space of a matrix (linear algebra core)
PBH test (eigenvalue-based alternative to the rank test)

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho tumhare paas ek system hai jiske andar kuch chhupe hue "states" hain (jaise position, velocity, temperature), par tum sirf bahar wala sensor ka output $y$ dekh sakte ho. Observability ka sawaal simple hai: kya sirf output dekh ke tum initial state $x(0)$ ko poori tarah nikal sakte ho? Agar haan, to system observable hai. Yeh cheez bahut important hai kyunki Kalman filter jaisa estimator tabhi kaam karta hai jab system observable ho.

Test kaise banate hain? Output $y=Ce^{At}x(0)$ hota hai. Ek time pe sirf ek projection milta hai, isliye hum derivatives lete hain: $y, \dot y, \ddot y \dots$ Har derivative ek nayi row deta hai — $C, CA, CA^2 \dots$ Cayley–Hamilton theorem kehta hai ki $A^n$ wala term purane powers ka combination hota hai, isliye $n-1$ power pe ruk jao. In sab rows ko upar-neeche stack karo — yeh hai observability matrix $\mathcal{O}$ .

Ab rule ekdum clean hai: agar $\operatorname{rank}(\mathcal{O})=n$ (total states), to observable. Agar rank kam hai, to koi na koi state "chhupa" hai — us direction ka $x_0$ output mein kabhi dikhta hi nahi ( $y\equiv 0$ ). Diagram mein red curve dekho: ek state aisi hai jo sensor ko bilkul touch nahi karti.

Do common galtiyan yaad rakho: (1) matrix ka bada size matlab zyada observable nahi hota — sirf rank matter karta hai. (2) Stability aur observability alag cheezein hain; system stable ho sakta hai par phir bhi unobservable. Aur ek pyaari duality: $(A,C)$ observable hona $=(A^T, C^T)$ controllable hona. Ek seekh lo, doosra free.

Observability matrix — rank test

WHAT is observability?

HOW to derive the test (from scratch)

Worked examples

Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Guidance, Navigation & Control (GNC)

Connections