State-space realisation, controllability, observability and the sampled-data poles: 22-Elec-B2 Advanced Control Systems, December 2013, Q2

Question 2: State-space realisation, controllability, observability and the 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 type-1 plant with a single adjustable zero, to be realised in state-space form and then discretised.

Given data — Question 2
Quantity	Value
Plant	$P(s) = \dfrac{10(\beta s+1)}{s(0.4s+1)^{2}}$
Free parameter	$\beta$ (real, otherwise unrestricted)
Open-loop poles	$s=0$ and $s=-2.5$ (repeated)
Open-loop zero	$s=-1/\beta$ (absent when $\beta = 0$)
Discretisation	uniform sampling at period $h$, zero-order hold on the input

Find. (a) a state-space triple $(A,B,C)$ realising $P(s)$; (b) the values of $\beta$ for which that realisation is controllable and observable, with justification; (c) the poles of the ZOH-sampled system as functions of $h$.

Approach. Put the transfer function in monic polynomial form, write down the controllable canonical realisation, then argue controllability from the structure of that form and observability from pole–zero cancellation. Part (c) needs only the pole-mapping property of the zero-order-hold equivalent.

Expand the transfer function into polynomial form. The denominator is $$s(0.4s+1)^{2} = s(0.16s^{2}+0.8s+1) = 0.16s^{3}+0.8s^{2}+s.$$ Dividing numerator and denominator by the leading coefficient 0.16 to make the denominator monic, $$P(s) = \frac{62.5\beta\, s + 62.5}{s^{3}+5s^{2}+6.25s},$$ since $10/0.16 = 62.5$, $0.8/0.16 = 5$ and $1/0.16 = 6.25$.
Write the controllable canonical realisation. With denominator $s^{3}+a_{2}s^{2}+a_{1}s+a_{0}$ and numerator $b_{1}s+b_{0}$ the standard form is $$A = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ -a_{0} & -a_{1} & -a_{2}\end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & -6.25 & -5\end{bmatrix},\qquad B = \begin{bmatrix}0\\0\\1\end{bmatrix},$$ $$\boxed{C = \begin{bmatrix} 62.5 & 62.5\beta & 0\end{bmatrix},\qquad D = 0}$$ The states are the outputs of the three cascaded integrators, and $a_{0}=0$ records the free integrator in the plant.

Figure Q2.1 — controllable canonical realisation of P(s) = (62.5β s + 62.5)/(s³ + 5s² + 6.25s). The three integrator outputs are the states; the denominator coefficients return as input feedback, the numerator coefficients as output feedforward.

Test controllability. With $B$ the last unit vector, $$\mathcal{C} = \begin{bmatrix}B & AB & A^{2}B\end{bmatrix} = \begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & -5\\ 1 & -5 & 18.75\end{bmatrix}, \qquad |\det \mathcal{C}| = 1 .$$ The determinant is $\pm 1$ for any denominator coefficients, because the matrix is anti-triangular with unit anti-diagonal. Neither $A$ nor $B$ contains $\beta$, so $$\boxed{\text{the realisation is controllable for every value of }\beta}$$ This is the defining property of the controllable canonical form and is the reason it is the natural starting point for a pole-placement design.
Test observability. Only $C$ carries $\beta$, so observability is where the parameter can matter. A single-input single-output realisation of this order is observable exactly when the transfer function is irreducible — that is, when no zero cancels a pole. The zero sits at $s=-1/\beta$ and the poles at $s=0$ and $s=-2.5$ (twice). A zero can never land on the pole at the origin for finite $\beta$, so the only cancellation possible is $$-\frac{1}{\beta} = -2.5 \quad\Longrightarrow\quad \boxed{\beta = 0.4\ \text{loses observability; the system is observable for all }\beta \neq 0.4}$$ At $\beta = 0.4$ the plant collapses to $P(s)=10/[s(0.4s+1)]$, one of the two modes at $s=-2.5$ disappears from the output, and the observability matrix $\mathcal{O}=[\,C;\ CA;\ CA^{2}\,]$ drops to rank 2. Away from that value $\mathcal{O}$ has full rank 3. Note that $\beta = 0$ is perfectly acceptable: the numerator becomes the constant 62.5, there is no finite zero at all, and nothing cancels.
Map the poles through the zero-order hold. The ZOH equivalent of a continuous system is $G(z) = (1-z^{-1})\,\mathcal{Z}\{P(s)/s\}$, and the sampling operation maps every continuous pole $s_{i}$ to $z_{i}=e^{s_{i}h}$ — the hold and the $(1-z^{-1})$ factor change the zeros and the gain, never the poles. With $s = 0,\ -2.5,\ -2.5$, $$\boxed{z_{1}=e^{0\cdot h}=1,\qquad z_{2}=z_{3}=e^{-2.5h}\ \text{(repeated)}}$$ The integrator therefore stays marginally stable for every sample period — it sits on the unit circle at $z=1$ — while the repeated real pole slides in from the unit circle toward the origin as $h$ grows: $e^{-2.5h}=0.78$ at $h=0.1$ s, $0.47$ at $h=0.3$ s and $0.14$ at $h=0.8$ s.

Figure Q2.2 — z-plane image of the continuous poles. The integrator maps to z = 1 for every h; the repeated pole at s = −2.5 maps to z = e^(−2.5h), which slides from the unit circle toward the origin as the sample period grows.

Check: sampling does not change controllability or observability here, but it can in general. A ZOH-sampled system loses controllability or observability only when two continuous poles that differ solely in their imaginary parts alias onto the same $z$, which requires a complex pole pair and a pathological choice of $h$. All three poles here are real, so the discretised model inherits the continuous conclusions for every $h \gt 0$.

Final results — Question 2
Quantity	Result
Monic transfer function	$P(s)=\dfrac{62.5\beta s+62.5}{s^{3}+5s^{2}+6.25s}$
State matrix $A$	rows $[0\ 1\ 0]$, $[0\ 0\ 1]$, $[0\ -6.25\ -5]$
Input matrix $B$	$[0\ 0\ 1]^{\mathsf T}$
Output matrix $C$	$[62.5\ \ 62.5\beta\ \ 0]$, $D=0$
Controllability	full rank for every $\beta$ ($\|\det\mathcal{C}\|=1$)
Observability	full rank for every $\beta \neq 0.4$; rank 2 at $\beta=0.4$
Sampled-data poles	$z=1$ and $z=e^{-2.5h}$ (repeated)

Topic. Canonical state-space realisations, the structural meaning of controllability and observability, and pole mapping under zero-order-hold sampling.

Key relations. Controllable canonical form for $(b_{1}s+b_{0})/(s^{3}+a_{2}s^{2}+a_{1}s+a_{0})$; $\mathcal{C}=[B\ AB\ A^{2}B]$ and $\mathcal{O}=[C;\ CA;\ CA^{2}]$; a SISO realisation of minimal order is controllable and observable if and only if the transfer function is irreducible; $z_{i}=e^{s_{i}h}$ for the ZOH equivalent.

Why this works. The two canonical forms are duals: the controllable form buries the denominator in $A$ and $B$ and puts the numerator entirely in $C$, so any parameter appearing only in the numerator can affect observability but never controllability. That single observation answers part (b) without computing a $3\times3$ rank for every candidate $\beta$ — you only have to ask where the zero can cancel a pole. Part (c) rests on the fact that the ZOH equivalent is $e^{Ah}$ in state-space terms, whose eigenvalues are $e^{\lambda_{i}h}$; the hold affects how the input reaches the states, not the natural modes.

Common pitfall. Answering “the system is uncontrollable when $\beta=0.4$” — the cancellation costs observability in this realisation, not controllability, and swapping the two is the single most common error on this question. The dual statement would be true had the observable canonical form been written instead, which is why the answer must name the realisation it refers to. A second trap in part (c) is to try to compute $\mathcal{Z}\{P(s)/s\}$ in full; the question explicitly says detailed calculations are not necessary, because only the pole mapping is being tested.

Source. Ogata, Modern Control Engineering, 5th ed., §2.6–2.7 (canonical realisations) and Ch. 9 (controllability and observability); Franklin, Powell & Workman, Digital Control of Dynamic Systems, 3rd ed., Ch. 4 (discrete equivalents).

Question 2 of 6: State-space realisation, controllability, observability and the sampled-data poles

Question 2: State-space realisation, controllability, observability and the sampled-data poles (25 marks)