State-space realisation, minimality conditions and sampled-data poles: 22-Elec-B2 Advanced Control Systems, May 2016, Q2

Question 2: State-space realisation, minimality conditions and sampled-data poles (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 third-order continuous plant with one free parameter in its numerator, to be realised in state space and then sampled behind a zero-order hold.

Given data
Item	Value
Plant	$P(s)=\dfrac{3(s+\alpha)}{s(s+4)^{2}}$
Poles	$s=0$, $s=-4$ (multiplicity 2)
Zero	$s=-\alpha$, with $\alpha$ a free real parameter
Denominator expanded	$s^{3}+8s^{2}+16s$
Sampling	uniform period $h$, ZOH on the input

Find. A state-space realisation $(A,b,c)$; the values of $\alpha$ for which that realisation is controllable and observable, with justification; and the poles of the sampled-data system as functions of $h$.

Pole–zero map. The zero is the only movable feature; the two values of α that place it on a pole are exactly the two values that destroy observability.

Approach. Put the transfer function in polynomial form and write the controller canonical realisation, whose controllability is automatic; then evaluate the observability determinant as a polynomial in $\alpha$ and interpret its roots as pole–zero cancellations; finally map each continuous pole through $z=e^{sh}$.

Part (a) — expand the transfer function into polynomial form. A canonical realisation needs the numerator and denominator as plain polynomials in $s$:$$P(s)=\frac{3(s+\alpha)}{s(s+4)^{2}}=\frac{3s+3\alpha}{s^{3}+8s^{2}+16s}$$The system is strictly proper (degree 1 over degree 3), so there is no direct feedthrough term and $D=0$.
Write the controller canonical realisation. With the denominator coefficients $a_{2}=8$, $a_{1}=16$, $a_{0}=0$ read off in reverse order along the bottom row, and the numerator coefficients placed in the output row,$$\boxed{\;A=\begin{bmatrix}0&1&0\\0&0&1\\0&-16&-8\end{bmatrix},\quad b=\begin{bmatrix}0\\0\\1\end{bmatrix},\quad c=\begin{bmatrix}3\alpha&3&0\end{bmatrix},\quad D=0\;}$$Recomputing $c(sI-A)^{-1}b$ symbolically returns $3(s+\alpha)/\!\left[s(s+4)^{2}\right]$ exactly, which is the only check this part needs.
Part (b) — test controllability. $$\mathcal{C}=\begin{bmatrix}b&Ab&A^{2}b\end{bmatrix}=\begin{bmatrix}0&0&1\\0&1&-8\\1&-8&48\end{bmatrix},\qquad \det\mathcal{C}=-1$$The determinant is a non-zero constant with no $\alpha$ in it, so this realisation is controllable for every value of $\alpha$. That is a structural property of the controller canonical form, not luck.
Test observability. Stacking $c$, $cA$ and $cA^{2}$,$$\mathcal{O}=\begin{bmatrix}3\alpha&3&0\\0&3\alpha&3\\0&-48&3\alpha-24\end{bmatrix},\qquad \det\mathcal{O}=27\alpha(\alpha-4)^{2}$$so the observability matrix loses rank exactly when$$\boxed{\alpha=0\quad\text{or}\quad\alpha=4}$$and the realisation is observable for all other $\alpha$.
Explain the two special values. They are not algebraic accidents: the zero of $P(s)$ sits at $s=-\alpha$ and the poles sit at $s=0$ and $s=-4$ (twice). At $\alpha=0$ the zero cancels the pole at the origin; at $\alpha=4$ it cancels one of the poles at $-4$. A pole–zero cancellation in a controllable realisation must show up as a loss of observability, because the cancelled mode is still excited by the input but no longer reaches the output. The realisation is minimal precisely when $\alpha\notin\{0,4\}$, and detectable only at $\alpha=4$ (the hidden mode at $-4$ decays), never at $\alpha=0$ (the hidden mode is the integrator).
Part (c) — map the continuous poles through the sampler. A zero-order hold followed by a continuous plant and an output sampler has the discrete dynamics matrix $\Phi=e^{Ah}$, whose eigenvalues are the exponentials of the eigenvalues of $A$. No convolution integral is needed, which is why the paper says detailed calculations are not necessary:$$z_{i}=e^{s_{i}h}\quad\Longrightarrow\quad \boxed{z=1,\qquad z=e^{-4h}\ \ (\text{double})}$$
Comment on how the poles move with $h$. The integrator maps to $z=1$ for every sample period — sampling never stabilises or destabilises a free integrator. The double pole starts at $z=1$ as $h\to 0$ and slides toward the origin as $h$ grows, reaching $e^{-4h}=0.670$ at $h=0.1$ s, $0.135$ at $h=0.5$ s and $0.018$ at $h=1$ s. Zeros are a different matter: the discrete zeros are not $e^{-\alpha h}$ and must be computed from the ZOH equivalent, which is why the question asks only about poles.

Final results
Quantity	Symbol	Value
State matrix	$A$	$[[0,1,0];[0,0,1];[0,-16,-8]]$
Input vector	$b$	$[0,\,0,\,1]^{T}$
Output vector	$c$	$[3\alpha,\,3,\,0]$
Controllability determinant	$\det\mathcal{C}$	$-1$ (controllable for all $\alpha$)
Observability determinant	$\det\mathcal{O}$	$27\alpha(\alpha-4)^{2}$
Observable / minimal	—	$\alpha\neq 0$ and $\alpha\neq 4$
Sampled-data poles	$z$	$1$ and $e^{-4h}$ (double)

Concept & Theory

Topic. Canonical realisations, minimality as the absence of pole–zero cancellation, and pole mapping under zero-order-hold sampling.

Key relations. Controller canonical form for $\dfrac{b_{2}s^{2}+b_{1}s+b_{0}}{s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}$; $\mathcal{C}=[b\;\;Ab\;\;A^{2}b]$; $\mathcal{O}=[c;\,cA;\,cA^{2}]$; discrete dynamics $\Phi=e^{Ah}$ so $z_{i}=e^{s_{i}h}$; minimal realisation $\Leftrightarrow$ controllable and observable $\Leftrightarrow$ no common root of numerator and denominator.

Why this works. A transfer function only determines a state-space model up to a similarity transformation and up to the modes it actually shows. The controller canonical form is built so that every mode is reachable from the input, which is why its controllability matrix is unimodular regardless of the numerator; all the numerator information lands in $c$, so the numerator is the only thing that can break observability. A shared root of numerator and denominator therefore appears as $\det\mathcal{O}=0$, and the cancelled mode becomes a hidden mode: still driven, no longer seen. Under sampling, the state-transition matrix over one period is the matrix exponential, so poles map by the scalar exponential and stability of a mode is preserved — a fact that does not extend to zeros.

Common pitfall. The classic error is to answer part (c) by mapping the ZERO as $e^{-\alpha h}$ as well. Sampled-data zeros come from the hold and the residues, not from the exponential map, and a ZOH equivalent can even acquire zeros outside the unit circle that the continuous plant never had. A second, subtler error is to declare the system “uncontrollable when $\alpha=0$”: it is the observability of this realisation that fails; had the observer canonical form been chosen instead, controllability would have been the casualty.

Source. Åström & Wittenmark, Computer-Controlled Systems, 3rd ed., §2.3–2.5 (ZOH equivalent, pole and zero mapping); Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed., §7.2–7.4 (canonical forms, controllability and observability).

Question 2 of 6: State-space realisation, minimality conditions and sampled-data poles

Question 2: State-space realisation, minimality conditions and sampled-data poles (25 marks)