Solution of state equations, controllability and observability

Why it matters

Once a plant is written as ẋ = Ax + Bu, two practical questions follow: how does it actually move in time from a given starting state, and can we steer every state with the actuators we have and see every state with the sensors we have? The state transition matrix answers the first. Controllability and observability answer the second — and they tell you, before any design work, whether state feedback or an observer is even possible.

Key ideas

Solution of the state equation

• Homogeneous (zero-input) case ẋ = Ax: x(t) = e^(At)·x(0). The matrix Φ(t) = e^(At) is the state transition matrix (STM) — it carries the state from time 0 to time t.
• Forced case: x(t) = Φ(t)·x(0) + ∫₀ᵗ Φ(t − τ)·B·u(τ) dτ — zero-input response plus a convolution (zero-state response).
• Output: y(t) = C·x(t) + D·u(t).

Ways to find Φ(t)

1. Laplace: Φ(t) = L⁻¹[(sI − A)⁻¹] — the usual hand method.
2. Series: e^(At) = I + At + A²t²/2! + … — exact in a few terms if A is nilpotent (e.g. [[0, 1], [0, 0]] gives [[1, t], [0, 1]]).
3. Diagonalisation: if A = P·Λ·P⁻¹, then e^(At) = P·e^(Λt)·P⁻¹, and e^(Λt) is diagonal with entries e^(λi·t).
4. Cayley–Hamilton: e^(At) = α0(t)·I + α1(t)·A + …, with the αi found from the eigenvalues.

Properties of Φ(t): Φ(0) = I; Φ⁻¹(t) = Φ(−t); Φ(t1 + t2) = Φ(t1)·Φ(t2); dΦ/dt = A·Φ. So A = dΦ/dt at t = 0 — useful for checking whether a proposed matrix can be a valid STM.

Controllability: the system is (completely state) controllable if some input u(t) can transfer any initial state to any final state in finite time.

• Kalman test: the n×nm matrix Qc = [B AB A²B … A^(n−1)B] must have rank n. For a single input, det Qc ≠ 0.
• Diagonal form test (distinct eigenvalues): controllable if no row of the transformed B is zero.

Observability: the system is (completely) observable if the initial state x(0) can be found from the output y(t) (and known input) over a finite interval.

• Kalman test: Qo = [C; CA; CA²; …; CA^(n−1)] (stacked, pn×n) must have rank n. For a single output, det Qo ≠ 0.
• Diagonal form test: observable if no column of the transformed C is zero.

Duality: (A, B) is controllable exactly when (Aᵀ, Bᵀ) is observable.

Link to transfer functions: a pole-zero cancellation in C(sI − A)⁻¹B means the cancelled mode is uncontrollable, unobservable, or both. The transfer function is a minimal description only if the realisation is both controllable and observable.

Why it matters for design: with full controllability, state feedback u = −Kx can place the closed-loop poles anywhere (pole placement). With full observability, an observer can estimate all states from measured outputs. Weaker notions — stabilisability (only stable modes are uncontrollable) and detectability (only stable modes are unobservable) — are enough for a working stable design.

Formulas

• Φ(t) = e^(At) = L⁻¹[(sI − A)⁻¹]
• x(t) = Φ(t)·x(0) + ∫₀ᵗ Φ(t − τ)·B·u(τ) dτ
• X(s) = (sI − A)⁻¹·x(0) + (sI − A)⁻¹·B·U(s)
• Qc = [B AB … A^(n−1)B]; controllable ⇔ rank Qc = n
• Qo = [C; CA; …; CA^(n−1)]; observable ⇔ rank Qo = n
• Φ(0) = I, Φ(−t) = Φ⁻¹(t), A = dΦ/dt |t=0

Symbols: n number of states, m inputs, p outputs; Φ is dimensionless when all states have the same unit; t, τ in s; eigenvalues λ in s⁻¹.

Worked examples

Example 1 — state transition matrix and response Given: A = [[0, 1], [−2, −3]], B = [0; 1], x(0) = [1; 0].

1. (sI − A)⁻¹ = [[s + 3, 1], [−2, s]]/[(s + 1)(s + 2)].
2. Partial fractions, element by element: Φ11 = 2e^(−t) − e^(−2t), Φ12 = e^(−t) − e^(−2t), Φ21 = −2e^(−t) + 2e^(−2t), Φ22 = −e^(−t) + 2e^(−2t).
3. Check: at t = 0, Φ = [[1, 0], [0, 1]].
4. Zero-input response: x(t) = Φ(t)·[1; 0] = [2e^(−t) − e^(−2t); −2e^(−t) + 2e^(−2t)].
5. At t = 1 s: x1 = 2(0.3679) − 0.1353 = 0.600, x2 = −2(0.3679) + 2(0.1353) = −0.465.
6. Unit-step zero-state response: X(s) = [1; s]/[s(s + 1)(s + 2)] → x1(t) = 0.5 − e^(−t) + 0.5e^(−2t), x2(t) = e^(−t) − e^(−2t).

Result: x(1) = [0.600; −0.465] from x(0) = [1; 0]; the step drives x1 to a final value of 0.5.

Example 2 — controllable but not observable (GATE level) Given: A = [[−1, 0], [0, −2]], B = [1; k], C = [1, 0].

1. AB = [−1; −2k], so Qc = [[1, −1], [k, −2k]], det Qc = −2k + k = −k. Controllable for every k ≠ 0.
2. CA = [−1, 0], so Qo = [[1, 0], [−1, 0]], det Qo = 0, rank 1 — not observable for any k.
3. The second state (mode e^(−2t)) never reaches the output because the second column of C is zero.
4. Transfer function: C(sI − A)⁻¹B = [1, 0]·[1/(s + 1); k/(s + 2)] = 1/(s + 1).

Result: controllable for k ≠ 0, never observable; the transfer function 1/(s + 1) hides the mode at −2. Because that hidden mode is stable, the system is still detectable.

Common mistakes

• Writing Φ(t) = (sI − A)⁻¹ and forgetting the inverse Laplace transform.
• Taking e^(At) element by element (e^(aij·t)); that is wrong unless A is diagonal.
• Building Qo as [C, AC, …] instead of [C; CA; …].
• Multiplying AB in the wrong order (BA is usually not even defined).
• Concluding "observable" from a transfer function with no visible cancellation, without checking the state model.
• Checking only det Qc for multi-input systems, where Qc is not square — use rank.

For GATE IN

Expect questions that ask for e^(At) or one of its elements, the zero-input or step response at a given time, whether a given (A, B, C) is controllable or observable, the parameter value that destroys controllability or observability, and whether a given matrix can be a valid STM. Practise 2×2 partial fractions and the Kalman tests until they take a minute each.

Quick check

1. What is e^(At) for A = [[0, 1], [0, 0]]?
2. Can [[e^(−t), 0], [0, 2e^(−2t)]] be a state transition matrix?
3. For A = [[0, 1], [−2, −3]], B = [0; 1], is the system controllable?
4. For A = [[1, 2], [3, 4]], C = [1, 0], what is the rank of Qo?

Answers: 1. [[1, t], [0, 1]]; 2. No — at t = 0 it gives [[1, 0], [0, 2]] ≠ I; 3. Yes — Qc = [[0, 1], [1, −3]], det = −1; 4. Qo = [[1, 0], [1, 2]], rank 2.