State-space model, controllability/observability, pole placement: 22-Elec-A2 Systems and Control, May 2014, Q7

Question 7 — State-space model, controllability/observability, pole placement [10 + 10]

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 (from Figure Q7.1). Paths give $\dot x_1=u-2x_1-x_3$, $\dot x_2=x_1-2x_3$, $\dot x_3=x_2$, $y=x_1+3x_2+x_3$, i.e. $$A=\begin{bmatrix}-2&0&-1\\1&0&-2\\0&1&0\end{bmatrix},\ B=\begin{bmatrix}1\\0\\0\end{bmatrix},\ C=\begin{bmatrix}1&3&1\end{bmatrix}.$$

Check: figure typo resolved. The extraction shows two branches from $x_2$ into $\dot x_3$ (gains $+1$ and $-1$), which would cancel and make $x_3$ uncontrollable/unobservable — flagged in the source as a likely diagram error. Taking the single integrator path $\dot x_3=x_2$ (drop the spurious $-1$) yields a controllable and observable system, consistent with the pole-placement task, so that reading is adopted.

Part A — State equations, controllability, observability [10]

Controllability. $\mathcal{C}=[B\ AB\ A^2B]=\begin{bmatrix}1&-2&4\\0&1&-2\\0&0&1\end{bmatrix}$, $\det\mathcal{C}=1\ne0$ ⇒ $\boxed{\text{controllable}}$.
Observability. $\mathcal{O}=\begin{bmatrix}C\\CA\\CA^2\end{bmatrix}=\begin{bmatrix}1&3&1\\1&1&-7\\-1&-7&-3\end{bmatrix}$, $\det\mathcal{O}=-28\ne0$ ⇒ $\boxed{\text{observable}}$.

Target closed-loop poles: $-3$ and $-2\pm j3$ (all in the left half-plane).

Part B — State feedback: $K$ and $k$ [10]

Approach. The control law $u=K(r-k^Tx)=Kr-g^Tx$ with $g^T=Kk^T$ places the poles; $g$ is fixed by matching characteristic polynomials, then the scalar $K$ is set by the zero-step-error (unity DC gain) condition.

Desired characteristic polynomial. $(s+3)(s^2+4s+13)=s^3+7s^2+25s+39$.
Closed-loop polynomial. With $A_{cl}=A-Bg^T$ (feedback enters row 1 only), $$\det(sI-A_{cl})=s^3+(2+g_1)s^2+(2+g_2)s+(5+2g_1+g_3).$$
Match → $g=Kk$. $2+g_1=7,\ 2+g_2=25,\ 5+2g_1+g_3=39$ give $$g=Kk=\begin{bmatrix}5&23&24\end{bmatrix}^T.$$
Fix $K$ from zero step error. Closed-loop DC gain $C(-A_{cl})^{-1}BK=\dfrac{K}{13}$ must equal $1$ (unity reference tracking), so $$\boxed{K=13},\qquad k=\frac{g}{K}=\begin{bmatrix}\tfrac{5}{13}&\tfrac{23}{13}&\tfrac{24}{13}\end{bmatrix}^T=\begin{bmatrix}0.385&1.769&1.846\end{bmatrix}^T.$$
Check. Eigenvalues of $A_{cl}=A-B(Kk)^T=\begin{bmatrix}-7&-23&-25\\1&0&-2\\0&1&0\end{bmatrix}$ are $-3,\,-2\pm j3$ ✓, and the DC gain is exactly $1$ (zero step error) ✓.

Quantity	Value
Controllable / observable	Yes / Yes ($\det\mathcal{C}=1$, $\det\mathcal{O}=-28$)
State-feedback product $g=Kk$	$[5,\ 23,\ 24]^T$
Scalar gain $K$	$13$
Feedback vector $k$	$\boxed{[0.385,\ 1.769,\ 1.846]^T}$

Question 7 of 8: State-space model, controllability/observability, pole placement

Question 7 — State-space model, controllability/observability, pole placement [10 + 10]

Part A — State equations, controllability, observability [10]

Part B — State feedback: $K$ and $k$ [10]