Q6Control System

Controllability and observability are two fundamental structural properties of a state-space system ẋ = Ax + Bu, y = Cx. A system is completely state controllable if there exists some control input u(t) capable of driving the state from any initial value to any desired final value in finite time; a system is completely state observable if the entire initial state x(0) can be uniquely determined from observation of the output y(t) (and known input) over a finite time interval. Both properties are checked algebraically using the Kalman rank test, without needing to solve the state equations explicitly: the system is controllable if and only if the controllability matrix Qc = [B, AB, A²B, …, Aⁿ⁻¹B] has full rank n, and observable if and only if the observability matrix Qo (formed from stacked rows C, CA, CA², …, CAⁿ⁻¹) has full rank n. For a system with a single input and single output, as here (n=3), full rank is equivalent to the corresponding matrix being non-singular, i.e. having a non-zero determinant, which is the fastest way to check the condition by hand.

Given:

Controllability test: the controllability matrix is Qc = [B, AB, A²B]. Compute AB:

Compute A²B = A(AB):

So:

Expanding det(Qc) along the first row: det = 1×(1×65 - (-12)×(-12)) - 0 + 1×(0×(-12)-1×1) = 1×(65-144) + 1×(-1) = -79 - 1 = -80. Since det(Qc) = -80 ≠ 0, Qc is full rank (rank 3), so the system is completely state controllable.

Observability test: the observability matrix is Qo = [Cᵀ; CᵀA; CᵀA²] (rows). With C = [10, 5, 1] (as a row vector, from the given Cᵀ column):

Expanding det(Qo) along the first row: det = 10×[(-1)(5)-(-1)(5)] - 5×[(-6)(5)-(-1)(6)] + 1×[(-6)(5)-(-1)(6)] = 10×[-5+5] - 5×[-30+6] + 1×[-30+6] = 10×0 - 5×(-24) + 1×(-24) = 0 + 120 - 24 = 96.

Re-checking this computation carefully: the first minor is (-1×5 - (-1)×5) = -5+5 = 0, so the first term is 10×0 = 0. The second minor is (-6×5 - (-1)×6) = -30+6 = -24, so the second term is -5×(-24) = 120. The third minor is (-6×5 - (-1)×6) = -30+6 = -24, so the third term is 1×(-24) = -24. Total det(Qo) = 0 + 120 - 24 = 96 ≠ 0. Since det(Qo) = 96 ≠ 0, Qo is full rank, and the system is completely state observable.

Conclusion: with det(Qc) = -80 ≠ 0 and det(Qo) = 96 ≠ 0, both matrices are of full rank 3, so the given system is both completely controllable and completely observable.