Question 7 of 8: State-Space Model, Controllability, Observability and Pole Placement by State Feedback
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, December 2018 — 3 hours, CLOSED BOOK (approved calculator plus one double-sided handwritten aid sheet). Eight questions of 20 marks each; Questions 1 and 2 are compulsory and the candidate chooses three of the remaining six, so five answers make a complete paper (100 marks). All eight questions are solved here, because the set is intended as a study resource. A short Laplace-transform table and the standard second-order design charts are supplied with the paper.
Reference texts. N. S. Nise, Control Systems Engineering, 8th ed. (root locus, frequency-domain design, steady-state error); K. Ogata, Modern Control Engineering, 5th ed. (Routh–Hurwitz, state-space methods); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (dominant-pole modelling, lag/lead compensator design); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed.
Question 7: State-Space Model, Controllability, Observability and Pole Placement by State Feedback (20 marks)
Find. The open-loop eigenvalues and stability verdict; $G_{open}(s)$; the controllability and observability verdicts; the scalar gain $K$ and the feedback vector $k$; and the closed-loop transfer function.
State-feedback configuration: the plant (A, B, C) inside a loop closed through the state-feedback vector, with the scalar gain K setting the DC gain.
Approach. Take the characteristic polynomial of $A$ for the eigenvalues, use the resolvent formula for the transfer function, test the rank of the two structural matrices, match the closed-loop characteristic polynomial coefficient-by-coefficient for $Kk_1$ and $Kk_2$, and finally fix $K$ from the unity-DC-gain requirement.
Part 1 — eigenvalues and stability. The characteristic polynomial is $$\det(sI - A) = \det\begin{bmatrix} s+4 & -1 \\ -1 & s \end{bmatrix} = s(s+4) - 1 = s^2 + 4s - 1.$$ Its roots are $s = (-4 \pm \sqrt{16 + 4})/2 = -2 \pm \sqrt{5}$, that is $$\boxed{\lambda_1 = -4.236, \qquad \lambda_2 = +0.236 \;\Rightarrow\; \text{the open loop is UNSTABLE.}}$$ One eigenvalue sits in the right half-plane, so any non-zero initial state grows without bound. Note the giveaway in the polynomial itself: the constant term is negative, so a sign change is guaranteed.
Part 2 — the open-loop transfer function. Using $G(s) = C(sI - A)^{-1}B + D$ and inverting the $2\times2$ resolvent, $$(sI - A)^{-1} = \frac{1}{s^2 + 4s - 1}\begin{bmatrix} s & 1 \\ 1 & s+4 \end{bmatrix}.$$ Post-multiplying by $B$ selects the first column, $[\,s \;\; 1\,]^{T}$ divided by the determinant, and pre-multiplying by $C = [1\ \ 1]$ sums its entries: $$\boxed{G_{open}(s) = \frac{s + 1}{s^2 + 4s - 1}.}$$ The right-half-plane pole at $+0.236$ is visible here too, and there is no pole–zero cancellation, so the model is a minimal realisation.
Part 3 — controllability. With $n = 2$ the controllability matrix is $M_c = [\,B \;\; AB\,]$. Since $AB = \begin{bmatrix} -4 \\ 1 \end{bmatrix}$, $$M_c = \begin{bmatrix} 1 & -4 \\ 0 & 1 \end{bmatrix}, \qquad \det M_c = 1 \ne 0,$$ so $M_c$ has full rank 2 and the system is controllable — which is precisely what makes part 4 solvable.
Observability. The observability matrix is $M_o = [\,C \;;\; CA\,]$, and $CA = [-3 \;\; 1]$, so $$M_o = \begin{bmatrix} 1 & 1 \\ -3 & 1 \end{bmatrix}, \qquad \det M_o = 1 + 3 = 4 \ne 0.$$ Full rank again, so the system is also observable. Both properties holding is consistent with the minimal (no cancellation) transfer function found in part 2.
Part 4 — form the closed-loop state matrix. Substituting $u = K(r - k^{T}x)$ into $\dot{x} = Ax + Bu$ gives $\dot{x} = (A - BKk^{T})x + BKr$. With $B = [1\ \ 0]^{T}$ the product $BKk^{T}$ modifies only the first row, so $$A_{cl} = \begin{bmatrix} -4 - Kk_1 & 1 - Kk_2 \\ 1 & 0 \end{bmatrix}.$$ Its characteristic polynomial is $\det(sI - A_{cl}) = s\,(s + 4 + Kk_1) - (1 - Kk_2) = s^2 + (4 + Kk_1)s + (Kk_2 - 1)$.
Match the desired polynomial. The requested poles give $(s+5)(s+6) = s^2 + 11s + 30$. Comparing coefficients, $$4 + Kk_1 = 11 \;\Rightarrow\; Kk_1 = 7, \qquad Kk_2 - 1 = 30 \;\Rightarrow\; Kk_2 = 31.$$ Pole placement alone fixes only these two products; the scalar $K$ is still free, which is exactly what the zero-error requirement will consume.
Fix $K$ from the zero-steady-state-error requirement. The closed-loop DC gain must be unity for a step to be tracked exactly. Since $G_{cl}(s) = C(sI - A_{cl})^{-1}BK$, the same resolvent structure as in part 2 gives $G_{cl}(s) = K(s+1)/(s^2 + 11s + 30)$, so $$G_{cl}(0) = \frac{K}{30} = 1 \;\Longrightarrow\; K = 30.$$ Dividing the two products by this value, $$\boxed{K = 30, \qquad k_1 = \tfrac{7}{30} = 0.2333, \qquad k_2 = \tfrac{31}{30} = 1.0333.}$$
Part 5 — the closed-loop transfer function. Collecting the result just derived, $$\boxed{G_{cl}(s) = \frac{30\,(s+1)}{s^2 + 11 s + 30} = \frac{30\,(s+1)}{(s+5)(s+6)}.}$$ A numerical check confirms $\det(sI - A_{cl})$ has roots exactly $-5$ and $-6$, that $G_{cl}(0) = 1$, and that the open-loop zero at $-1$ survives unchanged — state feedback relocates poles but never moves zeros, which is worth remembering when a design specification is written in terms of overshoot.
