Observability — when KF can estimate state: Worked examples

Recall the one tool we use everywhere: $O = C C A C A^{2} ⋮ C A^{n - 1}, observable ⟺ rank (O) = n .$ Here $n$ = number of states, $O$ has one block-row per power of $A$ , and "rank" counts how many rows are linearly independent (i.e. not copies or scaled sums of each other).

The scenario matrix

Every observability problem falls into one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	What makes it different	Covered by
A. Coupling saves you	unmeasured state drives a measured one → observable	Ex 1
B. Decoupled & blind	unmeasured state connected to nothing → NOT observable	Ex 2
C. Degenerate zero dynamics	$A = 0$ (pure integrators), single sensor	Ex 3
D. Multi-output redundancy	extra sensor rows add no rank	Ex 4
E. 3-state chain	need the full $A^{2}$ block; picture of information flow	Ex 5
F. Weak/limiting observability	full rank but nearly singular as a parameter $\to 0$	Ex 6
G. Real-world word problem	attitude + gyro bias, sensor choice matters	Ex 7
H. Exam twist	"add a sensor to fix it" / duality with controllability	Ex 8

The examples together touch every cell.

Example 1 — Coupling saves you (Cell A)

Worked example Position–velocity, measure position only

$A = [0010], C = [10], n = 2.$ State is $x = (position, velocity)$ . Sensor reads position.

Forecast: we only measure position — can a filter still recover velocity? Guess yes/no before reading.

Bottom power first: $C = [10]$ . Why this step? The first block-row of $O$ is always $C$ — it says "what do we see with zero derivatives."
$C A = [10] [0010] = [01]$ . Why this step? $C A$ is the first derivative $\overset{y}{˙} (0) = C A x (0)$ ; it tells us what the rate of change of the measurement reveals. Here it exposes the velocity component.
Stack and rank: $O = [1001], rank = 2 = n \Rightarrow observable .$ Why this step? Two independent rows spanning both coordinates ⇒ null space is ${0}$ ⇒ unique $x (0)$ .

Verify: the two rows are the standard basis vectors — clearly independent, determinant $= 1 \neq = 0$ . Because $A_{12} = 1$ (velocity drives position), how fast the position moves is literally the velocity. The window shows position; the motion of that window leaks velocity.

Example 3 — Degenerate zero dynamics (Cell C)

Example 4 — Multi-output redundancy (Cell D)

Worked example A second sensor that adds nothing

$A = [- 1 0 0 - 2], C = [1200] .$ We added a second sensor measuring $2 x_{1}$ . Same system as Example 2 but with "more data".

Forecast: more sensors — does that finally make $x_{2}$ observable? (This is the classic trap.)

$C$ has two rows $[10]$ and $[20]$ . Why? two sensors ⇒ two block-rows of $C$ .
$C A = [1200] [- 1 0 0 - 2] = [- 1 - 2 00]$ . Why this step? Every derivative still has a zero in the $x_{2}$ slot.
Stack (4 rows): $O = 12 - 1 - 2 0000, rank = 1 \neq = 2 \Rightarrow NOT observable .$ Why? All four rows are multiples of $[10]$ . Four rows, one independent direction.

Verify: the second sensor reads $2 x_{1}$ — a scalar multiple of what sensor 1 already gives. It is a linear combination already visible. Rank counts independence, not row count. Adding sensors that see the same thing never buys observability.

Example 5 — 3-state chain (Cell E)

Worked example A cascade

$x_{3} \to x_{2} \to x_{1}$ , measure only $x_{1}$ $A = 000100010, C = [100], n = 3.$ This is position–velocity–acceleration: $\overset{x}{˙}_{1} = x_{2}, \overset{x}{˙}_{2} = x_{3}, \overset{x}{˙}_{3} = 0$ . We measure position $x_{1}$ only. This is the case where you must go all the way to the $A^{2}$ block.

Forecast: one sensor at the end of a three-link chain. Observable? Look at the figure of information flow.

$C = [100]$ . Why? we see position (derivative order 0).
$C A = [010]$ . Why this step? first derivative $\overset{y}{˙} = x_{2}$ — velocity leaks in through $A_{12} = 1$ .
$C A^{2} = [001]$ . Why this step? second derivative $\overset{y}{¨} = x_{3}$ — acceleration leaks in through the chain $A_{12} A_{23}$ . This is why we needed the $A^{2}$ block: the deepest state only shows up two differentiations later.
Stack: $O = 100010001 = I_{3}, rank = 3 = n \Rightarrow observable .$

Verify: $O = I$ , determinant $1$ — perfectly observable. In the figure, the red arrows show each state pushing into the next until it reaches the window at $x_{1}$ . Lesson: a full chain into the measured state is observable even from a single sensor, but you must climb to $A^{n - 1}$ to prove it.

Example 6 — Weak / limiting observability (Cell F)

Worked example Full rank, but barely — the near-degenerate limit

$A = [- 1 0 0 - 1 - ε], C = [11], ε > 0 small .$ Two decay modes with almost equal rates, measured through their sum.

Forecast: the two modes are nearly identical. Technically observable? And what happens as $ε \to 0$ ?

$C = [11]$ . Why? we measure $x_{1} + x_{2}$ .
$C A = [- 1 - 1 - ε]$ . Why this step? the derivative weights the two modes differently — this difference is the only thing separating them.
Stack and compute the determinant: $O = [1 - 1 1 - 1 - ε], det O = (1) (- 1 - ε) - (1) (- 1) = - ε .$ Why this step? For a $2 \times 2$ square $O$ , rank $= 2 ⟺ det \neq = 0$ .
Case split on $ε$ :
- $ε > 0$ : $det = - ε \neq = 0 \Rightarrow observable$ .
- $ε = 0$ : $det = 0 \Rightarrow NOT observable$ (both modes decay identically, only their sum is ever visible — exactly Example 3's "difference is hidden").

Verify: as $ε \to 0^{+}$ the determinant $\to 0$ , so $O$ becomes ill-conditioned. Technically full rank, but the Gramian has a tiny eigenvalue: the difference mode is weakly observable and noise swamps it. Lesson: "observable" is a yes/no, but the degree matters — near-degenerate systems give confident-looking but fragile estimates.

Example 7 — Real-world word problem (Cell G)

Worked example Spacecraft attitude + drifting gyro bias

A gyro reports rate $ω_{meas} = ω_{true} + b$ , where $b$ is a slowly-drifting bias. Integrating the gyro gives angle. A star tracker measures the true angle $θ$ . States $x = (θ, b)$ : $\dot{θ} = ω_{meas} - b, \dot{b} = 0 \Rightarrow A = [00 - 1 0], C = [10] .$

Forecast: we measure angle, but really want to know the bias we can't sense directly. Can the filter calibrate the gyro?

$C = [10]$ . Why? star tracker sees angle, not bias.
$C A = [10] [00 - 1 0] = [0 - 1]$ . Why this step? The derivative of the angle measurement is influenced by $b$ (through $A_{12} = - 1$ ): a nonzero bias makes the measured angle drift relative to the true angle, and that drift is visible.
Stack: $O = [10 0 - 1], rank = 2 = n \Rightarrow observable .$

Verify: $det O = - 1 \neq = 0$ . The bias reveals itself as a creeping mismatch between the integrated gyro angle and the star-tracker angle. This is exactly why real spacecraft use star-trackers to calibrate gyro bias — the bias is unmeasured yet fully observable. (Contrast Example 2: there the hidden state had no pathway to the output; here $A_{12} = - 1$ opens one.)

Example 8 — Exam twist: fix it by adding the right sensor (Cell H)

Worked example Repair an unobservable system, and check the duality

Start from the blind system of Example 2: $A = [- 1 0 0 - 2] .$ We showed $C_{old} = [10]$ gives rank 1. Twist: which single new sensor row makes it observable — $C_{a} = [20]$ or $C_{b} = [01]$ ?

Forecast: guess which sensor actually helps before computing.

Try $C_{a} = [20]$ (redundant, like Ex 4). $C_{a} A = [- 20]$ . $O_{a} = [2 - 2 00], rank = 1 \Rightarrow still blind .$ Why? $[20]$ is a multiple of $[10]$ — no new direction.
Try $C_{b} = [01]$ (measures $x_{2}$ ). $C_{b} A = [0 - 2]$ . $O_{b} = [00 1 - 2], rank = 1.$ Why? Alone it sees only $x_{2}$ ! Rank still 1 — this sensor is blind to $x_{1}$ .
Combine both sensors $C = [1001]$ . $O = 10 - 1 0 010 - 2, rank = 2 = n \Rightarrow observable .$ Why this step? Now the two independent sensor directions together span $R^{2}$ .
Duality cross-check (controllability link): $(A, C)$ observable $⟺ (A^{⊤}, C^{⊤})$ controllable. With $A^{⊤} = A$ (diagonal) and $B = C^{⊤} = [1001] = I$ , the controllability matrix $[B A^{⊤} B] = [1001 - 1 0 0 - 2]$ has rank 2 — controllable, consistent.

Verify: decoupled diagonal dynamics need one independent sensor per decoupled mode. Redundant sensors ( $C_{a}$ ) fail; a genuinely new direction succeeds. And the Cayley–Hamilton-guaranteed dual test agrees.

Recall

Recall Which cell is which — quick self-test

Coupling into a measured state (position–velocity) — observable? ::: Yes (Cell A) — velocity drives position. Two decoupled modes, one measured — observable? ::: No (Cell B) — hidden mode has no output pathway. $A = 0$ with a single summing sensor — observable? ::: No (Cell C) — only the sum is seen, difference hidden. Extra sensor = scalar multiple of an existing one — does it help? ::: No (Cell D) — rows without rank. Nearly-equal decay rates through their sum — observable and how strongly? ::: Full rank for $ε > 0$ but weakly ( $det = - ε \to 0$ ) (Cell F). Why can a star tracker calibrate gyro bias? ::: Bias creates a visible drift in the measured angle ( $A_{12} = - 1$ ), so it is observable (Cell G).

Worked examples — Observability — when KF can estimate state