Inference from Data and Models · §4.6 Controllability and Observability
Carl Wunsch · MIT OpenCourseWare · Course notes · MIT OCW
Read §4.6, printed p.249, equations 4.152–4.153 for the stacked observability matrix.
Open the sourceHidden states and observable shadows
Turn “cast down” into a map with measurable predictions.
Prerequisites: 09 · Rank and the information that disappears, 11 · Eigenvectors and the sign of a geometry, 24 · The Hodge bridge: signature → inequality
Understand
Objectives
- Predict the fiber of one reading before revealing another.
- Calculate the rank of a stacked observation map.
- Separate unique noiseless recovery from sensitivity to noise.
Start with something you can see
The recovery lab begins with x₀=(2,1). Before revealing the second reading, predict how many states share its first-coordinate reading 2. At θ=π/3, reveal the next reading. Can two starting points still give the same pair? Write the two equations before looking at the estimate.
Give the idea a precise name
The state x₀ is a column vector; A is a known rotation of the state, and C=[1,0] is the observation map that keeps only its first coordinate. A single observation has a whole line of possible starting states. Stack the observations to ask whether that line shrinks to one point.
xₜ₊₁=Axₜ, yₜ=Cxₜ; O=[C; CA]=[1 0; cosθ −sinθ]
For x₀=(a,b), y₀=a and y₁=a cosθ−b sinθ. Full column rank of O makes the mapping one-to-one: in a known exact linear model, the two noiseless readings uniquely determine the two coordinates. If sinθ=0, the second row is a multiple of the first. Full rank alone says nothing about how noise will be amplified.
Work one small world
At θ=π/3 and x₀=(2,1), the noiseless readings are y₀=2 and y₁=1−√3/2. The determinant is −√3/2, so a=2 and b=(cosθ·y₀−y₁)/sinθ=1. MIT’s stacked-observation construction gives the general rank test; this two-coordinate calculation is our own example.
Counterexample to “two readings always recover the state”: at θ=0, O has two identical rows. States (2,1) and (2,8) yield (2,2), however long we watch. No narrative about the hidden coordinate can distinguish them from these readings.
THE BRIDGEThe next lesson changes the rotation rate and asks what one scalar time series preserves.
WHERE THIS IDEA STOPSA single state model may fit data without being the only possible physical explanation.
Experiment
Select θ = π/3. Predict y₀=2 and y₁=1−√3/2, then use Reveal next step twice to see the constraint lines and estimate. Select θ = 0: no unique estimate appears. Add a noise pair and compare; Reset this lab restores the preset.
Do the dynamics ever move an invisible difference into a direction the sensor can see?
Check & explain
A quarter-turn observes (4,−7). What was the initial second coordinate?
At θ=0, what does another first-coordinate reading reveal about b?
- It determines b
- Nothing new; both rows coincide
- It reveals the sign of b
Teach back
Write C and CA for a quarter-turn and for θ=0. Give a pair of distinct states that collide in the latter case; explain what adding a noisy reading changes.
- A missing state/map/evolution distinction: name x₀, A and C separately.
- A missing rank argument: compute O, not just the observations.
- An overclaim about noise: full rank establishes exact uniqueness, not a small error.
Read deeper
Mathematics for Machine Learning
Deisenroth, Faisal & Ong · Freely readable textbook
Read §2.6, pp.44–48 for rank and basis before interpreting full column rank.
Open the source