Question 2 of 8: Closed-Loop Stability With a Non-Minimum-Phase Level Sensor

Problem 2: Closed-Loop Stability With a Non-Minimum-Phase Level Sensor (20%)

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. $G_p(s)=\dfrac1s$ (integrating tank level), proportional controller gain $K$, sensor $G_m(s)=\dfrac{s-1}{(s+1)^2}$ (note: numerator zero at $s=+1$ — a right-half-plane, non-minimum-phase sensor).

Find. (1) the feedback block diagram and closed-loop transfer function $Y(s)/R(s)$; (2) the range of $K$ for closed-loop stability.

Approach. Write the closed-loop transfer function for a non-unity-feedback loop, $Y/R=KG_p/(1+KG_pG_m)$, clear fractions to get a polynomial characteristic equation in $s$, and apply the Routh–Hurwitz array.

1. (1) Block diagram and closed-loop TF. See the figure. With $H(s)=G_m(s)$ in the feedback path, $$\boxed{\frac{Y(s)}{R(s)}=\frac{KG_p}{1+KG_pG_m}=\frac{K(s+1)^2}{s(s+1)^2+K(s-1)}.}$$
2. (2) Characteristic equation. Expanding the denominator, $s(s+1)^2+K(s-1)=s^3+2s^2+s+Ks-K$, so $$\boxed{s^3+2s^2+(1+K)s-K=0.}$$
3. Routh array. $$\begin{array}{c|cc}s^3&1&1+K\\ s^2&2&-K\\ s^1&\tfrac{2(1+K)-1(-K)}{2}=\tfrac{2+3K}{2}&0\\ s^0&-K&\end{array}$$ Stability requires every first-column entry positive: $2>0$ (always), $\tfrac{2+3K}{2}\gt0\Rightarrow K\gt-\tfrac23$, and $-K\gt0\Rightarrow K\lt0$.
4. Combine. $$\boxed{-\tfrac23 \lt K \lt 0.}$$ The stabilizing range is entirely negative because the sensor's right-half-plane zero at $s=+1$ inverts the effective sign of the measured signal at low frequency.
5. Cross-check. Direct root evaluation confirms $K=-0.3$ (inside the range) gives all closed-loop poles with negative real parts, while $K=0.5$ (outside) produces a positive real root; at the boundary $K=-2/3$ the dominant pole sits at the origin (marginal).