Question 1 of 8: State space vs. transfer function, controllability/observability, steady-state error (compulsory)

Question 1 — State space vs. transfer function, controllability/observability, steady-state error (compulsory) [5 + 5 + 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.

Given. State model of the Process:

Find. $L(s)=K_pG_p(s)H(s)$; controllability/observability; and the $K_p$ range for stability with $e_{ss(step)}\le3\%$.

[Figure not reproduced: Figure Q1.1 redrawn — proportional gain $K_p$ forward, the state-space Process, and feedback $H(s)=1/(s+1)$. See the official exam paper.]

Part 1 — Open-loop transfer function [5]

Approach. Convert the Process to a transfer function with $G_p(s)=C(sI-A)^{-1}B+D$, then multiply by the controller gain and the feedback element.

1. Resolvent. $sI-A=\begin{bmatrix}s&-1\\3&s+2\end{bmatrix}$, $\det(sI-A)=s(s+2)+3=s^2+2s+3$, so $(sI-A)^{-1}=\dfrac{1}{s^2+2s+3}\begin{bmatrix}s+2&1\\-3&s\end{bmatrix}$.
2. Process transfer function. $C(sI-A)^{-1}B=\dfrac{[\,1\ 1\,]}{s^2+2s+3}\begin{bmatrix}1\\s\end{bmatrix}=\dfrac{s+1}{s^2+2s+3}$; adding $D=1$: $$G_p(s)=\frac{s+1}{s^2+2s+3}+1=\boxed{\frac{s^2+3s+4}{s^2+2s+3}}.$$ Note $G_p(0)=\tfrac{4}{3}$ (the process is not strictly proper because $D=1$).
3. Open loop. Multiply the forward gain by the feedback path $H(s)=1/(s+1)$: $$L(s)=K_pG_p(s)H(s)=\boxed{\frac{K_p(s^2+3s+4)}{(s^2+2s+3)(s+1)}}=\frac{K_p(s^2+3s+4)}{s^3+3s^2+5s+3}.$$

Part 2 — Controllability and observability [5]

1. Controllability matrix. $AB=\begin{bmatrix}0&1\\-3&-2\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}1\\-2\end{bmatrix}$, so $\mathcal{C}=[\,B\ \ AB\,]=\begin{bmatrix}0&1\\1&-2\end{bmatrix}$, $\det\mathcal{C}=-1\ne0\Rightarrow\boxed{\text{controllable}}$.
2. Observability matrix. $CA=[\,1\ 1\,]\begin{bmatrix}0&1\\-3&-2\end{bmatrix}=[\,-3\ -1\,]$, so $\mathcal{O}=\begin{bmatrix}C\\CA\end{bmatrix}=\begin{bmatrix}1&1\\-3&-1\end{bmatrix}$, $\det\mathcal{O}=2\ne0\Rightarrow\boxed{\text{observable}}$.
3. Interpretation. Both matrices are full rank (2), so the realisation is minimal — no pole–zero cancellation, and the state-space and transfer-function descriptions carry the same information.

Part 3 — Gain range for stability and 3% step error [10]

Approach. Form the closed-loop characteristic polynomial $1+L(s)=0$, apply Routh–Hurwitz for stability, then use the position error constant ($H(0)=1$, so the loop is unity-feedback at DC) for the error spec.

1. Characteristic polynomial. $(s^3+3s^2+5s+3)+K_p(s^2+3s+4)=0$, i.e. $$s^3+(3+K_p)s^2+(5+3K_p)s+(3+4K_p)=0.$$
2. Routh array. $$\begin{array}{c|cc}s^3&1&5+3K_p\\s^2&3+K_p&3+4K_p\\s^1&\dfrac{(3+K_p)(5+3K_p)-(3+4K_p)}{3+K_p}&\\s^0&3+4K_p&\end{array}$$ The $s^1$ numerator is $3K_p^2+10K_p+12$, whose discriminant $100-144=-44\lt0$, so it is positive for every $K_p$. Stability therefore needs only $3+K_p\gt0$ and $3+4K_p\gt0$, i.e. $\boxed{K_p\gt-0.75}$ — satisfied by every positive gain.
3. Steady-state error (Type 0). $L(s)$ has no pole at the origin, so the loop is Type 0. Since $H(0)=1$ the DC feedback is unity, and the position error constant is $K_{pos}=\lim_{s\to0}L(s)=K_pG_p(0)H(0)=K_p\cdot\tfrac{4}{3}\cdot1=\dfrac{4K_p}{3}$.
4. Apply the 3% spec. $e_{ss(step)}=\dfrac{1}{1+K_{pos}}=\dfrac{1}{1+\tfrac{4K_p}{3}}=\dfrac{3}{3+4K_p}\le0.03$ gives $3+4K_p\ge100$, hence $\boxed{K_p\ge24.25}$.
5. Combined range. Stability holds for all $K_p\gt-0.75$, so the binding requirement is the error spec: $\boxed{K_p\ge24.25}$. (At $K_p=24.25$ the three closed-loop poles are all in the left half-plane, confirming stability.)