Question 4 of 8: State-Space Model — Eigenvalues, TF, Controllability/Observability, Pole Placement

Question 4: State-Space Model — Eigenvalues, TF, Controllability/Observability, 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. $A=\begin{bmatrix}-2&1\\1&0\end{bmatrix}$, $B=\begin{bmatrix}1\\0\end{bmatrix}$, $C=\begin{bmatrix}1&2\end{bmatrix}$, $D=0$; state feedback law $u=K(r-k^Tx)$.

Find. 1) Eigenvalues, stability. 2) $G(s)$. 3) Controllability/observability. 4) $K,k$ for poles at $-3,-4$ and zero step SSE. 5) $G_{cl}(s)$.

Approach. Eigenvalues from $\det(sI-A)=0$; $G(s)=C(sI-A)^{-1}B$; controllability/observability from the standard $2\times2$ matrices $M_c=[B\ AB]$, $M_o=[C;CA]$; pole placement by matching $\det(sI-A_{cl})$ to $(s+3)(s+4)$, with the zero-SSE condition supplying the third equation needed to fix $K$ uniquely.

1. Part 1) — eigenvalues and stability. $$\det(sI-A)=\det\begin{bmatrix}s+2&-1\\-1&s\end{bmatrix}=s^2+2s-1=0\ \Rightarrow\ s=-1\pm\sqrt2.$$ $$\boxed{\lambda_1=-1+\sqrt2\approx+0.414,\qquad \lambda_2=-1-\sqrt2\approx-2.414}$$ Since $\lambda_1>0$ lies in the right half-plane, the OPEN-LOOP SYSTEM IS UNSTABLE.
2. Part 2) — transfer function. $G(s)=C(sI-A)^{-1}B$, with $(sI-A)^{-1}=\dfrac{1}{s^2+2s-1}\begin{bmatrix}s&1\\1&s+2\end{bmatrix}$, so $C(sI-A)^{-1}B$ picks out the first column dotted with $C$: $$\boxed{G(s)=\frac{s+2}{s^2+2s-1}}$$ (poles at $-1\pm\sqrt2$, matching Part 1 exactly, as they must.)
3. Part 3) — controllability and observability. $$M_c=[B\ \ AB]=\begin{bmatrix}1&-2\\0&1\end{bmatrix},\ \det M_c=1\neq0\ \Rightarrow\ \boxed{\text{fully controllable}}.$$ $$M_o=\begin{bmatrix}C\\CA\end{bmatrix}=\begin{bmatrix}1&2\\0&1\end{bmatrix},\ \det M_o=1\neq0\ \Rightarrow\ \boxed{\text{fully observable}}.$$
4. Part 4) — pole placement with zero step SSE. With $u=K(r-k^Tx)=Kr-Kk^Tx$, the closed-loop state matrix is $A_{cl}=A-BKk^T$, giving $$\det(sI-A_{cl})=s^2+(2+Kk_1)s+(Kk_2-1).$$ Matching to $(s+3)(s+4)=s^2+7s+12$: $\ Kk_1+2=7\Rightarrow Kk_1=5$, and $Kk_2-1=12\Rightarrow Kk_2=13$ — two equations, three unknowns. The third condition is zero steady-state error to a step, i.e. $G_{cl}(0)=1$ for $G_{cl}(s)=C(sI-A_{cl})^{-1}BK=\dfrac{K(s+2)}{s^2+7s+12}$: $$G_{cl}(0)=\frac{2K}{Kk_2-1}=\frac{2K}{12}=1\ \Rightarrow\ K=6.$$ Then $k_1=5/6$ and $k_2=13/6$ from the two equations above: $$\boxed{K=6,\qquad k=\begin{bmatrix}5/6\\13/6\end{bmatrix}}$$ (check: $A_{cl}=\begin{bmatrix}-7&-12\\1&0\end{bmatrix}$ has eigenvalues exactly $-3,-4$.)
5. Part 5) — closed-loop transfer function. Substituting $K,k$, $$\boxed{G_{cl}(s)=\frac{6(s+2)}{(s+3)(s+4)}=\frac{6(s+2)}{s^2+7s+12}}$$ (and indeed $G_{cl}(0)=6\times2/12=1$, confirming zero step SSE as designed.)