Question 4 of 8: Root Locus, Gain Selection and Second-Order Model

Question 4: Root Locus, Gain Selection and Second-Order Model (20 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. $G(s)=10/[s(s^2+10s+36)]$, open-loop poles at $s=0$ and $s=-5\pm j3.317$ (from $s^2+10s+36=0$), in a unit-feedback loop with proportional gain $K_p$.

Find. Asymptote geometry, the marginal-stability gain/frequency, the gain giving $PO\approx5\%$ with its rise time, and which trace in Figure Q4.3 is the actual system.

Approach. Use the standard asymptote/centroid/departure-angle formulas on the three known open-loop poles; find $K_{crit}$ from the Routh array; search the constant-$\zeta$ ray implied by the $PO=5\%$ spec for the point where the angle (equivalently, the reciprocal-magnitude) condition gives a real positive $K$; and compare the resulting third-order closed-loop step response against its second-order dominant-pole model.

1. Part 1) — asymptotes, centroid, departure angles. With $n-m=3$ finite poles and no finite zeros, the asymptote angles are $\theta_k=\dfrac{(2k+1)180^{\circ}}{3}=\boxed{60^{\circ},180^{\circ},300^{\circ}}$ and the centroid is $$\sigma_a=\frac{\sum p_i-\sum z_i}{n-m}=\frac{0+(-5+j3.317)+(-5-j3.317)}{3}=\boxed{-3.33}.$$ The angle of departure from the upper complex pole $p=-5+j3.317$ uses $\sum(\text{angles from other poles to }p)-\theta_{dep}=180^{\circ}(2k+1)$: the angle from the pole at $0$ to $p$ is $180^{\circ}-33.7^{\circ}=146.3^{\circ}$ and from the conjugate pole at $-5-j3.317$ to $p$ is $90^{\circ}$, summing to $236.4^{\circ}$, so $\theta_{dep}=180^{\circ}-236.4^{\circ}=\boxed{-56.4^{\circ}}$ (i.e. $303.6^{\circ}$) — the branch initially curves down-and-right off the upper pole, matching the dip visible in Figure Q4.2 before the locus climbs back up toward the imaginary-axis crossing.
2. Part 2) — marginal stability. The characteristic equation is $s^3+10s^2+36s+10K=0$; Routh's $s^1$ row is $(360-10K)/10=36-K$, zero at $\boxed{K_{crit}=36}$. The $s^2$-row auxiliary equation $10s^2+360=0$ gives $s=\pm j6$, so $\boxed{\omega_{osc}=6\ \text{rad/s}}$ — confirmed by direct root-finding of the cubic at $K=36$, which returns poles at $-10$ and exactly $\pm j6.000$.
3. Part 3) — gain for $PO\approx5\%$ and rise time. $PO=5\%$ inverts to $\zeta=\dfrac{-\ln0.05}{\sqrt{\pi^2+\ln^20.05}}=\boxed{0.690}$. Searching the root locus along the ray $s=\omega_n(-\zeta+j\sqrt{1-\zeta^2})$ for the point where $K=-s(s^2+10s+36)/10$ is real and positive locates the dominant pole at $s=-2.305+j2.417$ with $\omega_n=\boxed{3.34\ \text{rad/s}}$ and $\boxed{K_{op}=6.01}$. The remaining (third) closed-loop pole is real, at $s=-5.39$ (from the coefficient sum $-10=-2(2.305)+s_3$). Using the exact underdamped rise-time formula $T_{rise(0-100\%)}=(\pi-\cos^{-1}\zeta)/\omega_d$ with $\omega_d=\omega_n\sqrt{1-\zeta^2}=2.417$ rad/s gives $\boxed{T_{rise(0-100\%)}\approx0.965\ \text{s}}$ for the dominant-pole model.
4. Part 4) — identifying the traces. Simulating the FULL third-order closed loop at $K_{op}=6.01$ (poles at $-2.305\pm j2.417$ and $-5.39$) against the pure second-order dominant-pole MODEL of Part 3 shows the model overshoots to almost exactly the designed $5.0\%$ and rises faster, while the actual system — slowed and additionally damped by the third, non-dominant real pole at $-5.39$ — overshoots only about $3.6\%$ and reaches its final value later. In Figure Q4.3 the SOLID trace (labelled Trace B in the legend) is the one that rises first and shows the larger, textbook overshoot, while the dash-dot trace (Trace A) lags slightly and settles with less overshoot. Matching this behaviour: $\boxed{\text{Trace A}=\text{the actual (third-order) system;}\quad\text{Trace B}=\text{the second-order model.}}$ The extra real pole is not far enough from the dominant pair (ratio $5.39/2.305\approx2.3$, well inside the usual “5–10$\times$” rule of thumb for negligibility) to be ignored, which is exactly why the two traces visibly differ.