State-space realisation and pole placement for a non-minimum-phase plant: 22-Elec-B2 Advanced Control Systems, May 2014, Q2

Question 2: State-space realisation and pole placement for a non-minimum-phase plant (25 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 second-order transfer function with a single parameter in the numerator, to be realised in state-space form with the output itself as a state.

Given data
Quantity	Symbol	Value
Transfer function	$G(s)$	$(\alpha - s)/(1+s)^{2}$
Denominator, expanded	—	$s^{2} + 2s + 1$ (double pole at $s=-1$)
Numerator, expanded	—	$-s + \alpha$ (zero at $s = +\alpha$)
Parameter for part (c)	$\alpha$	1
Desired closed-loop poles	—	$-3$ and $-2$

Find. A realisation $(A,B,C)$ with $x_1 = y$, the values of $\alpha$ for which the pair is controllable and observable, and the state-feedback gain vector that places the poles at $-3$ and $-2$ when $\alpha = 1$.

Question 2 with alpha = 1: the open-loop double pole at s = -1 (blue) and the right-half-plane zero at s = +1 (red). State feedback moves the poles to -3 and -2 (green); the zero is untouched, so the closed loop stays non-minimum phase.

Approach. The observer (observable) canonical form is the one realisation that makes the output a state variable by construction, so it answers (a) immediately; the Kalman rank tests then answer (b) symbolically in $\alpha$; and matching the characteristic polynomial of $A - BK$ against the desired one answers (c).

Put the transfer function in coefficient form. Expanding, $$G(s) = \frac{-s + \alpha}{s^{2} + 2s + 1} = \frac{b_1 s + b_0}{s^{2} + a_1 s + a_0}, \qquad b_1 = -1,\; b_0 = \alpha,\; a_1 = 2,\; a_0 = 1 .$$ The plant is strictly proper, so there is no feedthrough term and $D = 0$.
Write the observer canonical realisation. This form stacks the states so that $x_1$ is the output: $$\dot x_1 = -a_1 x_1 + x_2 + b_1 u, \qquad \dot x_2 = -a_0 x_1 + b_0 u, \qquad y = x_1 .$$ In matrix form, with the numbers above, $$A = \begin{bmatrix} -2 & 1\\ -1 & 0\end{bmatrix},\qquad B = \begin{bmatrix} -1\\ \alpha\end{bmatrix},\qquad C = \begin{bmatrix} 1 & 0\end{bmatrix},\qquad D = 0 .$$ Because $y = x_1$ by construction, the requirement “taking $y(t)$ as one of the state variables” is met exactly.
Verify the realisation. Evaluating $C(sI-A)^{-1}B$ symbolically returns $(\alpha - s)/(s^{2}+2s+1)$, and a numerical check at $s = 0.7,\,2.3,\,5.0$ reproduces $G(s)$ to machine precision. Any realisation that fails this check has a transposed $A$ or a mis-signed $b_1$.
Test controllability. The controllability matrix is $$\mathcal{C} = \begin{bmatrix} B & AB\end{bmatrix} = \begin{bmatrix} -1 & 2+\alpha\\ \alpha & 1 \end{bmatrix}, \qquad \det \mathcal{C} = -1 - \alpha(2+\alpha) = -(\alpha+1)^{2}.$$ so $$\boxed{\text{controllable} \iff \alpha \neq -1}.$$
Test observability. The observability matrix is $$\mathcal{O} = \begin{bmatrix} C\\ CA\end{bmatrix} = \begin{bmatrix} 1 & 0\\ -2 & 1\end{bmatrix}, \qquad \det \mathcal{O} = 1 \neq 0 \;\;\text{for every } \alpha,$$ so the realisation is $$\boxed{\text{observable for all } \alpha}.$$ That is no accident: the output is a state, so measuring $y$ hands us $x_1$ directly, and the coupling $\dot x_1 = -2x_1 + x_2 + b_1u$ then exposes $x_2$.
Interpret the loss of controllability. At $\alpha = -1$, $$G(s) = \frac{-1-s}{(1+s)^{2}} = \frac{-(s+1)}{(s+1)^{2}} = \frac{-1}{s+1},$$ a pole–zero cancellation at $s = -1$. The second-order realisation is then non-minimal: one mode is invisible to the input, which is precisely what $\det\mathcal{C} = 0$ reports. The hidden mode sits at $s=-1$ and is stable, so the system remains stabilisable; only exact pole placement is lost.
Set up the pole-placement equations for $\alpha = 1$. With $u = -Kx$ and $K = [\,k_1\;\;k_2\,]$, $$A - BK = \begin{bmatrix} -2 + k_1 & 1 + k_2\\ -1 - k_1 & -k_2 \end{bmatrix},$$ whose characteristic polynomial is $$\det(sI - A + BK) = s^{2} + (2 - k_1 + k_2)\,s + (1 + k_1 + 3k_2).$$
Match the desired polynomial and solve. The target is $(s+3)(s+2) = s^{2} + 5s + 6$, giving the linear pair $$2 - k_1 + k_2 = 5, \qquad 1 + k_1 + 3k_2 = 6 \;\Longrightarrow\; -k_1 + k_2 = 3,\quad k_1 + 3k_2 = 5 .$$ Adding the two equations gives $4k_2 = 8$, so $$\boxed{K = \begin{bmatrix} -1 & 2\end{bmatrix}}, \qquad u = -Kx = x_1 - 2x_2 .$$ Substituting back, the eigenvalues of $A - BK$ are exactly $-3$ and $-2$.
Note what feedback did not change. State feedback relocates poles but leaves transmission zeros where they are, so the closed loop still carries the zero at $s = +\alpha = +1$. The design is therefore still non-minimum phase, and its step response will start in the wrong direction — a limitation no choice of $K$ can remove. This is the same phenomenon Question 4 displays explicitly.

Question 2 — final results
Quantity	Result
State matrix (observer canonical, $x_1 = y$)	$A = \begin{bmatrix} -2 & 1\\ -1 & 0\end{bmatrix}$
Input and output matrices	$B = \begin{bmatrix} -1\\ \alpha\end{bmatrix}$, $C = \begin{bmatrix} 1 & 0\end{bmatrix}$, $D = 0$
Controllability determinant	$\det\mathcal{C} = -(\alpha+1)^{2}$ → controllable for $\alpha \neq -1$
Observability determinant	$\det\mathcal{O} = 1$ → observable for every $\alpha$
State-feedback gain ($\alpha = 1$, poles $-3,-2$)	$K = [\,-1\;\;\;2\,]$
Closed-loop zero	$s = +1$, unchanged by feedback

Topic. Canonical state-space realisations, the Kalman controllability and observability tests, and pole placement by characteristic-polynomial matching.

Key relations. Observer canonical form for $(b_1s+b_0)/(s^{2}+a_1s+a_0)$; $\mathcal{C} = [B\;\;AB]$ and $\mathcal{O} = [C;\,CA]$ must have full rank; $\det(sI - A + BK)$ matched term-by-term against $\prod(s - \lambda_i)$; $G(s) = C(sI-A)^{-1}B + D$.

Why this works. A transfer function fixes only the input–output behaviour, so infinitely many realisations exist; choosing the observer canonical form is what makes the output a state and turns part (a) into bookkeeping. The determinant $-(\alpha+1)^{2}$ is more informative than a bare rank condition: it is a perfect square, so controllability degrades only at the single value where the numerator zero collides with the double pole, and the algebra reproduces the pole–zero cancellation you can see by eye. Pole placement then works because a controllable single-input pair has exactly as many free gains as characteristic-polynomial coefficients.

Common pitfall. Assuming any realisation of a given $G(s)$ is minimal, and so never testing the rank conditions; and expecting state feedback to move the zero at $s = +\alpha$. A subtler slip is dropping the sign of $b_1 = -1$, which reverses the sign of $B$ and produces a gain vector that places the poles in the wrong place.

Source. Ogata, Modern Control Engineering, 5th ed., §9.2 (canonical realisations), §9.6 (controllability and observability) and §10.2 (pole placement).

Question 2 of 6: State-space realisation and pole placement for a non-minimum-phase plant

Question 2: State-space realisation and pole placement for a non-minimum-phase plant (25 marks)