22-Elec-A2 Systems and Control · May 2018

Question 4 of 8: State-Space Model, Controllability, Observability and Pole Placement

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, 16-Elec-A2 Systems and Control, May 2018 — 3 hours, closed book (approved Casio or Sharp calculator plus one double-sided handwritten 8.5 × 11″ formula sheet). A short Laplace-transform table is printed on page 2, and the $\zeta$–overshoot, $\zeta$–resonant-peak and phase-margin design charts plus the second-order model, controllability/observability and state-space transfer-function definitions are on page 3. Questions 1 and 2 are compulsory; five questions constitute a complete paper, so a candidate chooses three of Q3–Q8. Each question carries 20 marks, for 100 in total. All eight questions are worked below so the set is a complete study resource.

Reference texts. N. S. Nise, Control Systems Engineering (7th ed., Wiley) — block-diagram and signal-flow reduction with Mason’s rule (Ch. 5), steady-state error and the static error constants (Ch. 7), Routh–Hurwitz (Ch. 6), root locus (Ch. 8), cascade compensator design (Ch. 9), frequency response and gain/phase margins (Ch. 10), frequency-domain lead design (Ch. 11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — second-order correlations, dominant-poles modelling and pole placement by state feedback. Every block diagram, root locus, pole–zero map, Bode plot, closed-loop magnitude curve and step response below is redrawn as an inline figure.

Reading the supplied design charts. Page 3 gives $PO = 100\,e^{-\zeta\pi/\sqrt{1-\zeta^{2}}}$ for percent overshoot, $M_r/K_{dc} = 1/\left(2\zeta\sqrt{1-\zeta^{2}}\right)$ for the closed-loop resonant peak, and the phase-margin correlation $\Phi_m \approx 100\zeta$ (equivalently $\zeta \approx 0.01\,\Phi_m$). Those three relations, together with $T_{settle(\pm 2\%)} \approx 4/(\zeta\omega_n)$ and $T_{rise(0-100\%)} = (\pi - \cos^{-1}\zeta)/\omega_d$, carry most of the numerical work in Questions 2, 5, 7 and 8.

Question 4: State-Space Model, Controllability, Observability and Pole Placement (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $A = \begin{bmatrix} -2 & 1 \\ 1 & 0 \end{bmatrix}$, $B = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $C = \begin{bmatrix} 1 & 2 \end{bmatrix}$, $D = 0$; control law $u = K(r - \mathbf{k}^{T}\mathbf{x})$; desired closed-loop poles at $-3$ and $-4$; zero steady-state error to a unit step.

Find. The open-loop eigenvalues and stability verdict; the open-loop transfer function; the controllability and observability verdicts; the gains $K$, $k_1$, $k_2$; and the closed-loop transfer function.

State-feedback configuration for Question 4. The full state vector $\mathbf{x}$ is fed back through the row vector $\mathbf{k}^{T}$ and the resulting signal is subtracted from the reference before the scalar gain $K$, giving $u = K(r - \mathbf{k}^{T}\mathbf{x})$.

Approach. Take the characteristic polynomial of $A$ for the eigenvalues, use $G(s)=C(sI-A)^{-1}B+D$ for the transfer function, test the ranks of the controllability and observability matrices from the page-3 definitions, then match the closed-loop characteristic polynomial to the desired one and set $K$ from the unity-DC-gain requirement.

Eigenvalues and open-loop stability. The characteristic polynomial is $$\det(sI - A) = \begin{vmatrix} s+2 & -1 \\ -1 & s \end{vmatrix} = s(s+2) - 1 = s^{2} + 2s - 1$$ so $s = -1 \pm \sqrt{2}$, i.e. $$\boxed{\;\lambda_1 = +0.4142, \qquad \lambda_2 = -2.4142\;}$$ One eigenvalue lies in the right half plane, therefore the open-loop system is unstable. Note that the constant term $-1$ is negative, which by itself already violates the necessary condition for stability — a useful one-line check before computing roots.
Open-loop transfer function. Using the page-3 definition $G(s) = C(sI-A)^{-1}B + D$ with $$(sI-A)^{-1} = \frac{1}{s^{2}+2s-1}\begin{bmatrix} s & 1 \\ 1 & s+2 \end{bmatrix}$$ we get $(sI-A)^{-1}B = \dfrac{1}{s^{2}+2s-1}\begin{bmatrix} s \\ 1 \end{bmatrix}$, and premultiplying by $C = [\,1\ \ 2\,]$, $$\boxed{\;G(s) = \frac{Y(s)}{U(s)} = \frac{s+2}{s^{2}+2s-1}\;}$$ The poles are of course the eigenvalues found above, and there is a finite zero at $s = -2$.
Controllability and observability. Building the page-3 matrices for this second-order system, $$M_c = \begin{bmatrix} B & AB \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}, \qquad M_o = \begin{bmatrix} C \\ CA \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$$ Both determinants equal $1 \ne 0$, so both matrices have full rank 2: $$\boxed{\;\text{the system is both controllable and observable}\;}$$ Controllability is the precondition for part 4 — arbitrary pole placement by state feedback is possible precisely because $M_c$ is non-singular.
Form the closed-loop state matrix. Substituting $u = K(r - \mathbf{k}^{T}\mathbf{x})$ into $\dot{\mathbf{x}} = A\mathbf{x}+Bu$ gives $\dot{\mathbf{x}} = (A - BK\mathbf{k}^{T})\mathbf{x} + BKr$, with $$A_{cl} = A - BK\mathbf{k}^{T} = \begin{bmatrix} -2-Kk_1 & 1-Kk_2 \\ 1 & 0 \end{bmatrix}$$ whose characteristic polynomial is $$\det(sI-A_{cl}) = s^{2} + (2+Kk_1)s - (1-Kk_2)$$
Match to the desired polynomial. The target poles $-3$ and $-4$ give $(s+3)(s+4) = s^{2}+7s+12$. Equating coefficients: $$2 + Kk_1 = 7 \;\Longrightarrow\; Kk_1 = 5, \qquad -(1 - Kk_2) = 12 \;\Longrightarrow\; Kk_2 = 13$$ Pole placement fixes only the products $Kk_1$ and $Kk_2$; the scalar $K$ is still free, and the steady-state requirement is what pins it down.
Use the zero-step-error condition to fix $K$. The closed-loop transfer function is $G_{cl}(s) = C(sI-A_{cl})^{-1}BK$. Since $A_{cl}$ differs from $A$ only in its first row, the same numerator survives and $$G_{cl}(s) = \frac{K(s+2)}{s^{2}+7s+12}$$ Zero steady-state error to a step demands unit DC gain, $G_{cl}(0)=1$: $$\frac{2K}{12} = 1 \;\Longrightarrow\; \boxed{\;K = 6\;}$$ and therefore $$\boxed{\;k_1 = \frac{5}{6} = 0.8333, \qquad k_2 = \frac{13}{6} = 2.1667\;}$$ A numerical eigenvalue check on $A_{cl}$ with these values returns exactly $-3$ and $-4$.
Closed-loop transfer function. Substituting $K=6$, $$\boxed{\;G_{cl}(s) = \frac{Y(s)}{R(s)} = \frac{6(s+2)}{(s+3)(s+4)} = \frac{6s+12}{s^{2}+7s+12}\;}$$ with $G_{cl}(0)=12/12=1$ as designed. State feedback has moved an unstable plant, with a pole at $+0.414$, to a well-damped pair of real poles at $-3$ and $-4$, and the extra scalar gain has removed the step error — all without adding a single dynamic element to the controller.

Quantity	Result
Eigenvalues of $A$	$+0.4142$, $-2.4142$
Open-loop stability	Unstable (one RHP eigenvalue)
Open-loop $G(s)$	$\dfrac{s+2}{s^{2}+2s-1}$
$\det M_c$ / controllable?	$1$ / yes
$\det M_o$ / observable?	$1$ / yes
Proportional gain $K$	$6$
State feedback gains $k_1$, $k_2$	$0.8333$, $2.1667$
Closed-loop $G_{cl}(s)$	$\dfrac{6(s+2)}{(s+3)(s+4)}$
Closed-loop DC gain	$1.000$ (zero step error)

Topic. State-space analysis of a second-order plant — eigenvalues, the transfer function from $(A,B,C,D)$, the controllability and observability tests, and pole placement by full state feedback.

Key relations. $\det(sI-A)=0$ for the eigenvalues; $G(s)=C(sI-A)^{-1}B+D$; $M_c=[\,B\ \ AB\ \ A^{2}B\,]$ and $M_o=[\,C;\ CA;\ CA^{2}\,]$ must have full rank; the closed-loop matrix $A_{cl}=A-BK\mathbf{k}^{T}$; and $G_{cl}(0)=1$ for zero step error.

Why this works. The eigenvalues of $A$ are the poles of the system because $(sI-A)^{-1}$ has $\det(sI-A)$ in its denominator, so the same polynomial governs both the free response and the transfer function. State feedback works because subtracting $BK\mathbf{k}^{T}\mathbf{x}$ modifies $A$ additively, and when the pair $(A,B)$ is controllable that modification can place the characteristic polynomial coefficients anywhere — here there are two free products for two coefficients, so the match is exact and unique. The reason a separate scalar $K$ is needed is that feedback alone changes only the denominator: it can make the plant fast and well damped but cannot by itself set the DC gain, so the loop would settle to the wrong value. Scaling the reference by $K$ fixes the amplitude without disturbing any pole, since $K$ multiplies the numerator only.

Common pitfall. Solving for $k_1$ and $k_2$ directly from the coefficient match, as if $K$ were already 1 — the match determines the products $Kk_1$ and $Kk_2$, and dividing by the wrong $K$ at the end gives gains that place the poles somewhere else entirely. A second trap is the sign in the $s^{0}$ coefficient: the open-loop constant term is $-1$, so the closed-loop term is $-(1-Kk_2)$, and dropping that minus sign yields $Kk_2 = -11$ instead of $+13$. Finally, do not conclude instability from a positive $k$ or a negative constant in $A$; always compute the eigenvalues.

Source. Nise, Control Systems Engineering 7th ed., Ch. 12 (controller design via pole placement, controllability and observability) and §3.6–3.7 (converting state space to a transfer function); Ogata 5th ed., Ch. 9 and §11.1–11.3.

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