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

Question 7 — State-space model, controllability & 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.

[Figure not reproduced: Figure Q7.1 redrawn — integrator chain with feedback into $\dot x_1,\dot x_2$ and output taps $1,3,1$. See the official exam paper.]

Part A — State equations & structural tests [10]

Approach. Each integrator output is a state; read the incoming branch gains at every $\dot x_i$ node and the output taps.

1. State equations. From the graph, $$\dot x_1=-2x_1-x_2-x_3+u,\quad \dot x_2=x_1-2x_3,\quad \dot x_3=x_2,\quad y=x_1+3x_2+x_3.$$
2. Matrices. $A=\begin{bmatrix}-2&-1&-1\\1&0&-2\\0&1&0\end{bmatrix},\ B=\begin{bmatrix}1\\0\\0\end{bmatrix},\ C=\begin{bmatrix}1&3&1\end{bmatrix}.$
3. Controllability. $\mathcal{C}=[B\ AB\ A^2B]$ has $AB=\begin{bmatrix}-2\\1\\0\end{bmatrix},\ A^2B=\begin{bmatrix}3\\-2\\1\end{bmatrix}$; $\det\mathcal{C}=-1\ne0\Rightarrow\boxed{\text{controllable (rank 3)}}$.
4. Observability. $\mathcal{O}=[C;\ CA;\ CA^2]$ has rank 3 $\Rightarrow\boxed{\text{observable}}$. Full pole placement is therefore possible.

Part B — State-feedback pole placement [10]

Approach. Write the composite feedback row $\mathbf{f}=K\mathbf{k}^T$; match $\det(sI-A+B\mathbf{f})$ to the desired polynomial, then fix $K$ from the unity-DC-gain (zero step-error) condition.

1. Desired polynomial. $(s+3)(s^2+4s+13)=s^3+7s^2+25s+39$.
2. Coefficient match. With $\mathbf{f}=[f_1\ f_2\ f_3]$ acting on the first row only, $\det(sI-A+B\mathbf{f})=s^3+(2+f_1)s^2+(2+1+f_2\!-\!\ldots)$ — solving the three equations gives $\boxed{\mathbf{f}=[5\ \ 22\ \ 24]}$.
3. Scalar gain for zero step-error. $K=-\dfrac{1}{C\,A_{cl}^{-1}B}$ with $A_{cl}=A-B\mathbf{f}$ gives $C A_{cl}^{-1}B=-\tfrac{1}{13}$, hence $\boxed{K=13}$.
4. Feedback vector. $\mathbf{k}=\mathbf{f}/K=\Bigl[\tfrac{5}{13}\ \ \tfrac{22}{13}\ \ \tfrac{24}{13}\Bigr]=\boxed{[0.385\ \ 1.692\ \ 1.846]}$. This makes the closed-loop DC gain $-CA_{cl}^{-1}BK=1$, so $e_{ss}(\text{step})=0$.