Question 2 of 8: Stabilising an unstable plant with PID: Bode, root locus, Routh–Hurwitz (compulsory)

Question 2 — Stabilising an unstable plant with PID: Bode, root locus, Routh–Hurwitz (compulsory) [10 + 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. The PID term simplifies with $T_i=1,\ T_d=0.5$: $\left(1+\tfrac{1}{s}\right)(1+0.5s)=\dfrac{0.5s^2+1.5s+1}{s}$, and $\times2$ over the plant gives forward numerator $2(0.5s^2+1.5s+1)$. Grouping the constant into $K_p$, the open loop is $$G_{open}(s)=\frac{K_p(s^2+2s+2)}{s^2(s+10)(s-2)},$$ poles $0,0,-10,+2$ (a double integrator and one RHP pole at $+2$), zeros $-1\pm j$.

Find. The stable $K_p$ range and $\omega_{osc}$, twice (figures then Routh).

Part 1 — Stability range from the figures [10]

Approach. With an open-loop RHP pole the closed loop is stable only for sufficient gain; the Bode gain margin measures the distance from the plotted gain to the marginal boundary, and the root-locus $j\omega$ crossing gives the same boundary and its frequency.

1. Read the Bode margin. The plot (drawn at $K_p=1$) shows $G_m=32.2$ dB at the phase-crossover $\omega\approx3.9$ rad/s. The gain must be raised by $32.2$ dB, a factor $10^{32.2/20}\approx40.7$, to reach marginal stability, so from the figure $\boxed{K_{p,\min}\approx41}$ and $\omega_{osc}\approx3.9$ rad/s.
2. Read the root locus. The branches leaving the origin and the RHP pole $+2$ migrate left as $K_p$ increases and cross the imaginary axis at $\pm j\omega_{osc}$; for gains above the crossing all closed-loop poles lie in the LHP, so the loop is stable only above that gain (no finite upper limit).
3. Negative phase margin explained. The reported $P_m=-145^\circ$ at $K_p=1$ simply confirms the loop is unstable at unit gain — expected for an open-loop-unstable plant, not an error.

Part 2 — Routh–Hurwitz verification [10]

Approach. Form $s^2(s+10)(s-2)+K_p(s^2+2s+2)=0$ and require a positive Routh first column.

1. Characteristic polynomial. $s^2(s+10)(s-2)=s^4+8s^3-20s^2$, so $$s^4+8s^3+(K_p-20)s^2+2K_p\,s+2K_p=0.$$
2. Routh array. $$\begin{array}{c|ccc}s^4&1&K_p-20&2K_p\\s^3&8&2K_p&\\s^2&0.75K_p-20&2K_p&\\s^1&\dfrac{2K_p(0.75K_p-28)}{0.75K_p-20}&&\\s^0&2K_p&\end{array}$$ where the $s^2$ entry is $\dfrac{8(K_p-20)-2K_p}{8}=0.75K_p-20$.
3. First-column conditions. $2K_p\gt0$ for all $K_p\gt0$; the $s^2$ entry needs $K_p\gt26.7$; the $s^1$ entry needs $0.75K_p-28\gt0$, i.e. $K_p\gt37.33$. The binding condition is $\boxed{K_p\gt37.33=\tfrac{112}{3}}$ (with no upper bound).
4. Oscillation frequency. At $K_p=37.33$ the $s^2$ entry is $0.75(37.33)-20=8$, and the auxiliary equation $8s^2+2K_p=0$ gives $s^2=-\dfrac{2(37.33)}{8}=-\dfrac{28}{3}$, so $\boxed{\omega_{osc}=\sqrt{28/3}=3.06\ \text{rad/s}}$.