State space, canonical forms, eigenvalues and stability: 22-Elec-A2 Systems and Control, December 2013, Q7

Question 7 — State space, canonical forms, eigenvalues and stability [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.

Part A — Transfer function and eigenvalues [10]

Approach. The first column of $A$ carries the (negated) characteristic coefficients — this is observable canonical form — so the denominator is read off directly and the numerator from $B$.

Characteristic polynomial by inspection. The first column $[-9,-23,-15]^{T}$ gives $\det(sI-A)=s^3+9s^2+23s+15$.
Remaining eigenvalues. Given the root $-5$, factor: $s^3+9s^2+23s+15=(s+5)(s^2+4s+3)=(s+5)(s+1)(s+3)$, so the eigenvalues are $\boxed{-5,\ -3,\ -1}$ — all in the left half-plane, so the system is stable.
Transfer function by inspection. In observable canonical form $C(sI-A)^{-1}B$ has numerator built from $B=[0,1,3]^{T}$, i.e. $0\,s^2+1\,s+3$: $$G(s)=\frac{s+3}{s^3+9s^2+23s+15}=\frac{s+3}{(s+1)(s+3)(s+5)}=\boxed{\frac{1}{(s+1)(s+5)}}.$$
Interpretation. The numerator zero at $-3$ cancels the pole at $-3$: that mode is hidden (uncontrollable/unobservable) from the input–output map, leaving a second-order $G(s)=\dfrac{1}{s^2+6s+5}$ even though the state model is third-order.

Eigenvalues $-5,-3,-1$ (all stable); the input–output zero at $-3$ cancels the pole there.

Part B — Stability and controllable canonical form [10]

Approach. Apply Routh to the denominator for stability, then write the standard controllable canonical (phase-variable) matrices from the coefficients.

(a) Stability by Routh. For $s^3+7s^2+2s+15$: $$\begin{array}{c|cc}s^3&1&2\\ s^2&7&15\\ s^1&\frac{7(2)-15}{7}=-0.14&0\\ s^0&15&\end{array}$$ The first column changes sign twice ($7\to-0.14\to15$), so there are two right-half-plane poles — the system is unstable.
(b) Controllable canonical form. With denominator coefficients $a_2=7,a_1=2,a_0=15$ and numerator $3s^2+2s+1$ ($b_2=3,b_1=2,b_0=1$), $D=0$: $$A=\begin{bmatrix}0&1&0\\0&0&1\\-15&-2&-7\end{bmatrix},\quad B=\begin{bmatrix}0\\0\\1\end{bmatrix},\quad C=\begin{bmatrix}1&2&3\end{bmatrix},\quad D=0.$$
Check. $\det(sI-A)=s^3+7s^2+2s+15$ reproduces the denominator, and $C(sI-A)^{-1}B=\dfrac{3s^2+2s+1}{s^3+7s^2+2s+15}$ recovers $G(s)$.

Quantity	Result
Part A eigenvalues	$-5,-3,-1$ → stable
Part A transfer function	$\dfrac{1}{(s+1)(s+5)}$ (pole–zero cancellation at $-3$)
Part B stability	2 RHP poles → unstable
Part B $(A,B,C,D)$	controllable canonical form above

Concept & Theory

Topic. Duality of state-space canonical forms and modal cancellation.

Key relations. Observable canonical form → denominator from the leading column, numerator from $B$; controllable canonical form → companion $A$ from denominator, $C$ from numerator; eigenvalues $=$ poles $=$ roots of $\det(sI-A)$ (Nise Ch. 3, 6; Ogata Ch. on canonical forms).

Why this works. Canonical forms encode the transfer-function coefficients directly in the matrices, so $G(s)$ is written by inspection without inverting $sI-A$. A numerator zero coinciding with an eigenvalue signals a mode that is uncontrollable or unobservable and cancels in the I/O transfer function.

Common pitfall. Reading a sign change in the Routh first column as a single unstable pole. Each sign change counts one RHP root, so two changes mean two RHP poles — still unstable, but the count matters for the root distribution.

Question 7 of 7: State space, canonical forms, eigenvalues and stability

Question 7 — State space, canonical forms, eigenvalues and stability [10 + 10]

Part A — Transfer function and eigenvalues [10]

Part B — Stability and controllable canonical form [10]