Questions: Controllability and Observability

Question 1

You test controllability for a 3rd-order system and find rank(𝒞) = 2. Which statement is most accurate?

• A. The system has one unstable mode that feedback cannot fix
• B. There exists one direction in state space that the input cannot influence, regardless of the control law applied
• C. The system cannot be stabilized under any circumstances
• D. The B matrix must be incorrect since rank should equal n for any physical system

Answer: (B)

A rank deficiency of 1 means one direction in state space lies outside the column span of [B, AB, …, Aⁿ⁻¹B]. The input simply has no effect on that mode. This does not by itself mean the system is unstable — an uncontrollable mode that is also stable (negative eigenvalue) will decay on its own. Only an uncontrollable *unstable* mode is catastrophic. Option A conflates uncontrollability with instability, which is a common and consequential error.

Question 2

A control engineer moves an actuator to a different location on the plant, keeping the same A matrix but changing B. The controllability matrix rank drops from n to n−1. What is the most precise explanation?

• A. The plant dynamics changed, making one eigenvalue uncontrollable
• B. One mode of the system is now uncontrollable because B changed, even though A is identical
• C. The system is now unstable due to the actuator placement
• D. The PBH test is no longer valid; the Kalman test must be used instead

Answer: (B)

Controllability depends on the pair (A, B), not on A alone. Moving an actuator changes the B matrix — which directions in state space the input directly enters — without altering the plant's intrinsic dynamics (A). A poor placement can make some modes unreachable from the new actuator position. This is exactly why controllability analysis must be re-run whenever sensor or actuator placement changes, not just when the plant dynamics are modified.

Question 3

Moving a sensor to a different location on a plant with the same A matrix can change whether the system is observable, even though the plant dynamics are unchanged.

• T. True
• F. False

Answer: True

Observability depends on the pair (A, C), and C encodes sensor placement — it maps the state to the measured output. Changing the sensor location changes C, which changes the observability matrix 𝒪 = [C; CA; …; CAⁿ⁻¹]. A new sensor position may fail to 'see' certain modes, destroying observability even though the plant's A matrix is identical. This is the observability analogue of the controllability dependence on B.

Question 4

An uncontrollable mode in a state-space system is generally unstable and is expected to be addressed before feedback can stabilize the system.

• T. True
• F. False

Answer: False

An uncontrollable mode is simply one the input cannot influence — it evolves according to its own eigenvalue regardless of what control is applied. If that eigenvalue has a negative real part (stable mode), the mode decays to zero on its own; it is benign. Only an uncontrollable *unstable* mode (positive real-part eigenvalue) is catastrophic, because it grows and no feedback can move its eigenvalue. The dangerous distinction is uncontrollable-and-unstable, not merely uncontrollable.

Question 5

Why does the Kalman controllability matrix stop at Aⁿ⁻¹B rather than including AⁿB and higher powers, and what theorem justifies this?

Model answer: The Cayley-Hamilton theorem states that every matrix satisfies its own characteristic polynomial, so Aⁿ can be expressed as a linear combination of I, A, …, Aⁿ⁻¹. This means AⁿB lies in the column span of [B, AB, …, Aⁿ⁻¹B], adding no new reachable directions. The same applies to all higher powers. Therefore the first n columns of the infinite power series fully characterize the reachable subspace, and rank([B AB … Aⁿ⁻¹B]) is the complete test.

This result is important because it makes the Kalman rank test finite and computable. Without Cayley-Hamilton, you might worry that longer sequences of input-propagation steps could reach new state-space directions — but the theorem guarantees they cannot. The controllability matrix is exactly as wide as it needs to be.