Question 2 of 6: State-space realisation and pole placement with reference scaling

Question 2: State-space realisation and pole placement with reference scaling (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, strictly proper, Type-1 plant with one finite zero, to be realised in state-space form and then placed by full state feedback with a scalar reference gain.

Find. A realisation $(A,B,C,D)$ of $P(s)$, the row vector $K$ that places the closed-loop eigenvalues at $-1,-2,-3$, and the scalar $F$ that makes the closed-loop DC gain unity.

Approach. Write the controllable canonical realisation, in which the characteristic coefficients appear explicitly in the last row of $A$ so pole placement is a subtraction, then choose $F$ as the reciprocal of the closed-loop DC gain.

1. Part (a) — expand the denominator and identify the coefficients.$$P(s)=\frac{2s+1}{s^{3}+s^{2}+4s}=\frac{b_{2}s^{2}+b_{1}s+b_{0}}{s^{3}+a_{2}s^{2}+a_{1}s+a_{0}}$$with $a_{2}=1$, $a_{1}=4$, $a_{0}=0$ and $b_{2}=0$, $b_{1}=2$, $b_{0}=1$.
2. Part (a) — write the controllable canonical form. Taking $x_{1}$ as the output of the innermost integrator,$$\dot{x}=\begin{pmatrix}0&1&0\\0&0&1\\-a_{0}&-a_{1}&-a_{2}\end{pmatrix}x+\begin{pmatrix}0\\0\\1\end{pmatrix}u=\begin{pmatrix}0&1&0\\0&0&1\\0&-4&-1\end{pmatrix}x+\begin{pmatrix}0\\0\\1\end{pmatrix}u$$$$y=\begin{pmatrix}b_{0}&b_{1}&b_{2}\end{pmatrix}x=\boxed{\begin{pmatrix}1&2&0\end{pmatrix}x},\qquad D=0$$The realisation is minimal here (numerator and denominator share no root: the zero at $-0.5$ is not a pole), so it is simultaneously controllable and observable. Any similarity transform of it is an equally valid answer to part (a); the canonical form is chosen because it makes part (b) trivial.
3. Part (a) — verify the realisation reproduces $P(s)$. Evaluating $C\left(sI-A\right)^{-1}B$ symbolically returns $\left(b_{0}+b_{1}s+b_{2}s^{2}\right)/\left(s^{3}+a_{2}s^{2}+a_{1}s+a_{0}\right)$, and numerically at $s=2$ it gives $5/22=0.22727$, which matches $P(2)=(2\cdot2+1)/[2(4+2+4)]=5/22$. This one-line substitution is worth doing in the exam: a transposed $B$ or a mis-ordered $C$ is the commonest realisation error and it shows up immediately.
4. Part (b) — form the desired characteristic polynomial.$$\alpha_{d}(s)=(s+1)(s+2)(s+3)=s^{3}+6s^{2}+11s+6$$
5. Part (b) — place the poles by inspection. In controllable canonical form the last row of $A-BK$ is $\left(-(a_{0}+k_{1}),\,-(a_{1}+k_{2}),\,-(a_{2}+k_{3})\right)$, so matching it to $\alpha_{d}$ term by term gives$$k_{1}=6-a_{0}=6,\qquad k_{2}=11-a_{1}=11-4=7,\qquad k_{3}=6-a_{2}=6-1=5$$$$\boxed{K=\begin{pmatrix}6&7&5\end{pmatrix}}$$Checking directly, the eigenvalues of$$A-BK=\begin{pmatrix}0&1&0\\0&0&1\\-6&-11&-6\end{pmatrix}$$are $-1$, $-2$ and $-3$ as required. Note that this works only because the pair $(A,B)$ is controllable — the canonical form guarantees it, and Question 4 shows what happens when it is not.
6. Part (b) — choose $F$ from the closed-loop DC gain. State feedback rescales but does not move the numerator, so$$\frac{Y(s)}{R(s)}=F\,C\left(sI-A+BK\right)^{-1}B=\frac{F\left(2s+1\right)}{s^{3}+6s^{2}+11s+6}$$Setting $s=0$ for a step command,$$\left.\frac{Y}{R}\right|_{s=0}=\frac{F\times1}{6}=1\;\Rightarrow\;\boxed{F=6}$$so $u(t)=-\begin{pmatrix}6&7&5\end{pmatrix}x(t)+6\,r(t)$ places the poles and makes $y(\infty)=r$.
7. Part (b) — state the limitation of the $F$ trick. $F$ is pure feed-forward: it inverts the nominal DC gain and provides no integral action. If the plant gain drifts by ten percent, or a constant load disturbance appears, the tracking error reappears in proportion. The exam asks only for tracking “in steady state” for the given model, so $F=6$ is the complete answer, but the robust alternative — augmenting the state with $\dot{x}_{I}=r-y$ and placing four poles — is worth one sentence.