Question 1 of 8: Servo-Positioning System under PI Control — Closed-Loop TF, Stability Range, Operating Gain

Question 1: Servo-Positioning System under PI Control — Closed-Loop TF, Stability Range, Operating Gain (20 marks, compulsory)

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. Plant $P(s)=\dfrac{10}{s(s+3)(s+10)}$ inside its own inner unity negative-feedback loop; PI controller $C(s)=K\!\left(1+\dfrac{0.5}{s}\right)=\dfrac{K(s+0.5)}{s}$ around the OUTER loop; design targets $e_{ss(step)}=0$ and $e_{ss(ramp)}=0.1$ V/V.

Find. 1) $G_{cl}(s)=Y(s)/R(s)$ as a polynomial ratio in $K$. 2) The range of $K$ for closed-loop stability, $K_{crit}$, $\omega_{osc}$. 3) $K_{op}$ meeting the SSE specification, its stability, and the Gain Margin.

Approach. Reduce the inner loop first ($Y/u=P/(1+P)$ for a signal $u$ injected at the second summing junction), then close the outer loop around the PI controller; collect the resulting quartic characteristic polynomial and apply the Routh–Hurwitz array (parts 1–2); use the ramp-error specification to solve for $K_{op}$, then re-use the same characteristic polynomial's frequency response for the Gain Margin (part 3).

1. Part 1) — reduce the inner loop, then close the outer loop. With $u(s)$ the PI-controller output feeding the second summing junction, $Y=P(u-Y)\Rightarrow Y/u=P/(1+P)$. The outer loop then gives $Y=\dfrac{P}{1+P}\,C\,(R-Y)$, which rearranges to $$G_{cl}(s)=\frac{Y(s)}{R(s)}=\frac{PC}{1+P+PC}.$$ Substituting $P(s)=10/[s(s+3)(s+10)]$ and $C(s)=K(s+0.5)/s$ and clearing denominators: $$\boxed{G_{cl}(s)=\frac{10Ks+5K}{s^4+13s^3+30s^2+(10+10K)s+5K}}$$
2. Part 2) — Routh–Hurwitz on the closed-loop characteristic polynomial. $Q(s)=s^4+13s^3+30s^2+(10+10K)s+5K=0$. The array: $$\begin{array}{c|cc} s^4 & 1 & 30 & 5K\\ s^3 & 13 & 10{+}10K &\\ s^2 & \tfrac{380-10K}{13} & 5K &\\ s^1 & \tfrac{20K^2-571K-760}{2(K-38)} & &\\ s^0 & 5K & & \end{array}$$ All four left-column entries must stay positive for $K>0$: row $s^2$ needs $K\lt38$; row $s^0$ needs $K>0$; row $s^1$'s numerator $20K^2-571K-760=0$ has roots $K=\tfrac{571\pm13\sqrt{2289}}{40}$, i.e. $K\approx-1.27$ or $K\approx29.82$, and since the row's denominator $2(K-38)$ is negative throughout $0\lt K\lt38$, the row stays positive only for $0\lt K\lt29.82$. Combining every row: $\boxed{0\lt K\lt K_{crit}=29.82}$ (exactly $\tfrac{571+13\sqrt{2289}}{40}$). At $K=K_{crit}$ the $s^1$ row vanishes; the auxiliary equation from the $s^2$ row, $\tfrac{380-10K_{crit}}{13}s^2+5K_{crit}=0$, gives $s^2=-23.71\Rightarrow\boxed{\omega_{osc}=\sqrt{23.71}\approx4.87\ \text{rad/s}}$.
3. Part 3) — operating gain from the ramp-error spec, stability check, Gain Margin. For this Type-1 loop $e_{ss(ramp)}=1/K_v$ with $K_v=\lim_{s\to0}sG_{open}(s)$ (Question 2 derives $K_v=0.5K$); setting $e_{ss(ramp)}=1/(0.5K)=0.1$ gives $\boxed{K_{op}=20}$. Substituting $K_{op}=20$ into $G_{cl}(s)$ reproduces the paper's own printed factored form $G_{cl}(s)=\dfrac{200(s+0.5)}{(s+11.9)(s+0.505)(s^2+0.5925s+16.64)}$ to within rounding — a strong cross-check that $K_{op}=20$ is right. Since $K_{op}=20\lt K_{crit}=29.82$, the system is $\boxed{\text{still stable}}$ at $K_{op}$. The phase-crossover frequency (where $\text{Im}\,G_{open}(j\omega)=0$) is set entirely by the SHAPE of $G_{open}(s)$, not by $K$, so it equals $\omega_{osc}=4.87$ rad/s found in Part 2 — a second independent check. At $K_{op}=20$, $|G_{open}(j4.87)|=0.671$, so $$G_m=\frac{1}{|G_{open}(j\omega_{osc})|}=\frac{1}{0.671}\approx\boxed{1.49\ \text{V/V}\ (3.47\ \text{dB})},$$ and indeed $G_m\times K_{op}=1.49\times20=29.8\approx K_{crit}$, confirming the margin is consistent with how far $K_{op}$ sits below $K_{crit}$.