Controllability, observability and pole placement

Why it matters

Before designing a state-feedback controller for a balancing robot or an active suspension you must know two things: can the actuators actually influence every mode of the system (controllability), and can the sensors see every mode (observability)? If both hold, pole placement lets you put every closed-loop pole exactly where you want it — something a single-gain root-locus design cannot do — and an observer can supply the states you do not measure.

Key ideas

Controllability. The LTI system ẋ = Ax + Bu is (completely state) controllable if some input can drive the state from any initial value to any final value in finite time. It is a property of the pair (A, B).

• Kalman test: the n × nm controllability matrix Qc = [B AB A²B … Aⁿ⁻¹B] must have rank n. For a single input, Qc is square and the test is det Qc ≠ 0.
• Physical meaning of failure: some mode is not connected to any actuator. Its eigenvalue cannot be changed by feedback. If that mode is stable the system is still stabilisable; if it is unstable, no controller can fix it.

Observability. The system is (completely) observable if the initial state x(0) can be found from the output y(t) and input u(t) over a finite interval. It is a property of the pair (A, C).

• Kalman test: Qo = [C; CA; CA²; …; CAⁿ⁻¹] (stacked rows) must have rank n.
• Failure means some mode never shows up in any measured output. If it is stable the system is detectable.

Duality. (A, B) is controllable if and only if (Aᵀ, Bᵀ) is observable. Every controller-design result has an observer-design twin.

Link to transfer functions. G(s) = C(sI − A)⁻¹B + D shows only the part of the system that is both controllable and observable. A pole–zero cancellation in G(s) signals an uncontrollable or unobservable mode. A realisation with no cancellations (both controllable and observable) is called minimal; its order equals the degree of the reduced denominator. Diagonal (modal) form gives an instant check: with distinct eigenvalues, a mode is uncontrollable if its row of B is zero and unobservable if its column of C is zero.

Pole placement by state feedback. Use u = −Kx + r, with K a 1 × n gain row for one input. The closed loop is ẋ = (A − BK)x + Br, with characteristic polynomial det(sI − A + BK). If and only if (A, B) is controllable, every eigenvalue of A − BK can be placed anywhere (complex ones in conjugate pairs).

• Direct method (n ≤ 3): equate det(sI − A + BK) with the desired polynomial (s − μ₁)(s − μ₂)… and match coefficients.
• Phase-variable form shortcut: if A has the companion structure with last row −a_n … −a₁, then kᵢ = (desired coefficient) − (open-loop coefficient) for each power of s.
• Ackermann's formula: K = [0 … 0 1]·Qc⁻¹·φ(A), where φ(s) is the desired characteristic polynomial.
• Choose desired poles from specifications (ζ, ωn, settling time). Very fast poles need large gains and actuator effort, and amplify noise.
• State feedback does not move the zeros, and it does not by itself give zero steady-state error; add a reference pre-gain or an integrator state.

Observers. When states are not measured, a Luenberger observer x̂̇ = Ax̂ + Bu + L(y − Cx̂) estimates them; its error dynamics are set by the eigenvalues of A − LC, placeable if and only if (A, C) is observable. By the separation principle, the controller poles (A − BK) and observer poles (A − LC) can be chosen independently; observer poles are usually 2–5 times faster.

Formulas

Qc = [B AB … Aⁿ⁻¹B], controllable ⇔ rank Qc = n Qo = [C; CA; …; CAⁿ⁻¹], observable ⇔ rank Qo = n

• n: number of states; A: n × n; B: n × m; C: p × n.

u = −K·x + r; ẋ = (A − B·K)·x + B·r det(sI − A + B·K) = desired characteristic polynomial

K = [0 … 0 1]·Qc⁻¹·φ(A) — Ackermann (single input).

x̂̇ = A·x̂ + B·u + L·(y − C·x̂); error dynamics ė = (A − L·C)·e

Worked examples

Example 1 (standard — tests and a hidden mode). For A = [[0, 1], [−2, −3]], B = [0; 1] and C = [1 1], check controllability and observability, and find G(s).

1. AB = [1; −3], so Qc = [[0, 1], [1, −3]]; det Qc = 0 × (−3) − 1 × 1 = −1 ≠ 0 → controllable.
2. CA = [1 1]·A = [1 × 0 + 1 × (−2), 1 × 1 + 1 × (−3)] = [−2, −2].
3. Qo = [[1, 1], [−2, −2]]; det Qo = 1 × (−2) − 1 × (−2) = 0 → rank 1 → not observable.
4. G(s) = C(sI − A)⁻¹B = [1 1]·[1; s]/((s + 1)(s + 2)) = (s + 1)/((s + 1)(s + 2)) = 1/(s + 2).
5. The cancelled pole at s = −1 is the unobservable mode. It is stable, so the system is detectable — but a transfer-function-only view would never show it.

Example 2 (GATE level — pole placement). For the same A and B, find K = [k₁ k₂] so that the closed-loop poles are at s = −2 ± j2. Comment on the resulting ζ and ωn.

1. Desired polynomial: (s + 2 − j2)(s + 2 + j2) = s² + 4s + 8.
2. A − BK = [[0, 1], [−2 − k₁, −3 − k₂]].
3. det(sI − A + BK) = s² + (3 + k₂)s + (2 + k₁).
4. Match: 3 + k₂ = 4 → k₂ = 1; 2 + k₁ = 8 → k₁ = 6. So K = [6 1].
5. ωn = √8 = 2.83 rad/s; ζ = 4/(2 × 2.83) = 0.707; settling time (2 %) ≈ 4/2 = 2 s, compared with 4 s for the open-loop slow pole at −1.
6. Ackermann check: φ(A) = A² + 4A + 8I = [[6, 1], [−2, 3]]; Qc⁻¹ = [[3, 1], [1, 0]]; [0 1]·Qc⁻¹ = [1, 0]; K = [1, 0]·φ(A) = [6 1]. ✔

Common mistakes

• Stacking Qc as rows or Qo as columns: Qc puts B, AB, … side by side; Qo stacks C, CA, … vertically.
• Computing CA as A·C (wrong order and wrong size).
• Saying "A is controllable" — controllability belongs to (A, B), observability to (A, C).
• Concluding a system is fine because its transfer function looks well behaved: a cancelled unstable mode is still there.
• Using A + BK instead of A − BK for u = −Kx.
• Expecting state feedback to give zero steady-state error automatically.

For GATE ME

Expect: rank or determinant of Qc or Qo for a 2 × 2 or 3 × 3 system; deciding controllability or observability from a diagonal form; recognising a pole–zero cancellation as a lost mode; computing K for given closed-loop poles in phase-variable form; and conceptual questions on duality and the separation principle. Practise matrix products quickly and keep track of row and column order.

Quick check

1. For A = [[0, 1], [−2, −3]] and B = [0; 1], what is the rank of Qc?
2. For A = [[1, 2], [3, 4]] and C = [1 0], is the system observable?
3. Which property must hold for arbitrary pole placement by state feedback?
4. What does a pole–zero cancellation in G(s) indicate?
5. For A = [[0, 1], [−5, −6]], what are the eigenvalues?

Answers: 1. 2. 2. Yes — Qo = [[1, 0], [1, 2]], determinant 2. 3. Controllability of (A, B). 4. A mode that is uncontrollable, unobservable or both. 5. −1 and −5.