State Space Model, Pole Placement by State Feedback, Steady State Errors (20 marks): 22-Elec-A2 Systems and Control, Undated paper, Q8

Question 8: State Space Model, Pole Placement by State Feedback, Steady State Errors (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. The open-loop state-space model

$$\begin{bmatrix}\dot{x}_1\\\dot{x}_2\end{bmatrix}=\begin{bmatrix}-8&1\\1&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}+\begin{bmatrix}1\\0\end{bmatrix}u,\qquad y=\begin{bmatrix}1&8\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}+[0]u.$$

Find. (1) eigenvalues and open-loop stability; (2) $G_{open}(s)$; (3) controllability and observability; (4) $K$ and $k$ placing the closed-loop poles at $-10$ and $-30$ with zero step error; (5) $G_{cl}(s)$, its poles, zeros and DC gain.

Figure S8.1 - State-feedback configuration u = K(r - k^T x) for Question 8.

Approach. Work directly with the $2\times2$ matrices: characteristic polynomial for the eigenvalues, the resolvent for the transfer function, the Kalman rank tests, and coefficient matching for the state-feedback gains.

Eigenvalues (part 1).$$\det(sI-A)=\det\begin{bmatrix}s+8&-1\\-1&s\end{bmatrix}=s(s+8)-1=s^2+8s-1.$$Solving, $s=\dfrac{-8\pm\sqrt{64+4}}{2}$, so$$\boxed{\lambda_1=-8.123,\qquad\lambda_2=+0.1231}$$One eigenvalue lies in the right half-plane, therefore the open-loop system is unstable. The tell-tale sign is the negative constant term in $s^2+8s-1$: a Hurwitz polynomial needs all coefficients positive.
Open-loop transfer function (part 2). Using $(sI-A)^{-1}=\dfrac{1}{\Delta}\begin{bmatrix}s&1\\1&s+8\end{bmatrix}$ with $\Delta=s^2+8s-1$, we get $(sI-A)^{-1}B=\dfrac{1}{\Delta}\begin{bmatrix}s\\1\end{bmatrix}$, so with $D=0$,$$G_{open}(s)=C(sI-A)^{-1}B=\frac{1\cdot s+8\cdot1}{s^2+8s-1}=\boxed{\frac{s+8}{s^2+8s-1}}$$
Controllability and observability (part 3). The system is second order, so the test matrices are $2\times2$:$$M_c=\begin{bmatrix}B&AB\end{bmatrix}=\begin{bmatrix}1&-8\\0&1\end{bmatrix},\qquad\det M_c=1\neq0,$$$$M_o=\begin{bmatrix}C\\CA\end{bmatrix}=\begin{bmatrix}1&8\\0&1\end{bmatrix},\qquad\det M_o=1\neq0.$$Both have full rank 2, so the system is completely controllable and completely observable. Pole placement by state feedback is therefore guaranteed to be possible — which is what part 4 relies on.
Form the closed-loop state matrix (part 4). With $u=K(r-k^Tx)$ and $k^T=[k_1\ k_2]$,$$\dot{x}=(A-BKk^T)x+BKr,\qquad A-BKk^T=\begin{bmatrix}-8-Kk_1&1-Kk_2\\1&0\end{bmatrix}.$$Its characteristic polynomial is$$\det(sI-A_{cl})=s\left(s+8+Kk_1\right)-\left(1-Kk_2\right)=s^2+(8+Kk_1)s-(1-Kk_2).$$
Match the desired pole polynomial. Poles at $-10$ and $-30$ require $(s+10)(s+30)=s^2+40s+300$, so$$8+Kk_1=40\;\Rightarrow\;Kk_1=32,\qquad -(1-Kk_2)=300\;\Rightarrow\;Kk_2=301.$$
Use the zero-error condition to split $K$ from $k$. The pole placement fixes only the products $Kk_1$ and $Kk_2$; the scalar $K$ is set by requiring zero steady-state error to a step, i.e. $G_{cl}(0)=1$. Since $(sI-A_{cl})^{-1}B$ has the same $[s\ \ 1]^T/\Delta_{cl}$ structure as before,$$G_{cl}(s)=\frac{K(s+8)}{s^2+40s+300}\;\Rightarrow\;G_{cl}(0)=\frac{8K}{300}=1,$$so$$\boxed{K=37.5,\qquad k_1=\frac{32}{37.5}=0.8533,\qquad k_2=\frac{301}{37.5}=8.0267}$$
Closed-loop transfer function (part 5). Substituting $K$,$$\boxed{G_{cl}(s)=\frac{37.5(s+8)}{s^2+40s+300}=\frac{37.5s+300}{s^2+40s+300}}$$with poles at $s=-10$ and $s=-30$ (as designed), a single zero at $s=-8$ inherited from $C$ and $B$, and DC gain $300/300=1$. State feedback has moved the unstable eigenvalue at $+0.123$ deep into the left half-plane while leaving the zero untouched — state feedback relocates poles only.

Result	Value
Open-loop characteristic polynomial	$s^2+8s-1$
Eigenvalues	$-8.123$, $+0.1231$ — unstable
Open-loop transfer function	$\dfrac{s+8}{s^2+8s-1}$
$\det M_c$ / $\det M_o$	1 / 1 — controllable and observable
Gain products	$Kk_1=32$, $Kk_2=301$
Proportional gain	$K=37.5$
State feedback vector	$k^T=[0.8533\ \ 8.0267]$
Closed-loop transfer function	$\dfrac{37.5s+300}{s^2+40s+300}$
Closed-loop poles / zero / DC gain	$-10,\ -30$ / $-8$ / 1

Topic: State-space analysis and pole placement by state feedback.

Key relations: $G(s)=C(sI-A)^{-1}B+D$; $M_c=[B\ \ AB]$, $M_o=[C;\ CA]$ (full rank = controllable / observable); $A_{cl}=A-BKk^T$; DC gain $G_{cl}(0)=-CA_{cl}^{-1}BK$.

Why this works: If the pair $(A,B)$ is controllable, the state feedback gains can place the closed-loop eigenvalues anywhere. Matching the coefficients of $\det(sI-A+BKk^T)$ against the desired polynomial gives as many linear equations as there are states, and the extra scalar $K$ is then available to normalise the DC gain.

Common pitfall: Building $3\times3$ $M_c$ and $M_o$ because the formula sheet prints them that way — the test matrices must be $n\times n$, and here $n=2$. Also, expecting state feedback to move the zero at $-8$: it cannot, so the DC gain must be corrected with the scalar $K$ instead.

Source: Ogata, Modern Control Engineering, 5th ed., Ch. 9 and Sec. 10–2; Nise, Control Systems Engineering, 8th ed., Ch. 12 (Design via State Space).

Question 8 of 8: State Space Model, Pole Placement by State Feedback, Steady State Errors (20 marks)

Question 8: State Space Model, Pole Placement by State Feedback, Steady State Errors (20 marks)