Question 2 of 8: Root-locus design of a PD controller (compulsory)

Question 2 — Root-locus design of a PD controller (compulsory) [10 + 7 + 3]

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. $G_c(s)=K_p(1+2s)=2K_p(s+0.5)$, $G(s)=\dfrac{2}{s^2(s+5)}$, so the loop gain in root-locus form is $L(s)=\dfrac{4K_p(s+0.5)}{s^2(s+5)}$ (locus gain $K=4K_p$).

Find. Locus geometry; the two $\zeta=0.707$ gains; the chosen $K_p$ and its transient specifications.

Part A — Root-locus geometry [10]

Approach. Poles $0,0,-5$; one zero $-0.5$; $n-m=2$ branches to infinity. Apply the centroid, asymptote, real-axis and break-point rules.

1. Real-axis segment. An odd count of real singularities to the right places the locus on $[-5,-0.5]$; the origin segment $(-0.5,0)$ has an even count and is not on the locus.
2. Centroid & asymptotes. $\sigma_a=\dfrac{(0+0-5)-(-0.5)}{2}=\boxed{-2.25}$; asymptote angles $\dfrac{(2k+1)180^\circ}{2}=\boxed{\pm90^\circ}$.
3. Break points. With $K_p(s)=-\dfrac{s^3+5s^2}{2(2s+1)}$, $dK_p/ds=0$ gives $-2s(4s^2+13s+10)=0$, i.e. $s=0,\,-1.25,\,-2$. The double pole departs vertically at $s=0$; on the real segment there is a $\boxed{\text{break-in at }s=-1.25\ (K_p\approx1.95)}$ and a $\boxed{\text{break-away at }s=-2\ (K_p=2.0)}$.
4. Imaginary-axis crossings. The Routh $s^1$ entry is $18K_p/5\gt0$ for every $K_p\gt0$ (characteristic $s^3+5s^2+4K_ps+2K_p$), so the locus never crosses the $j\omega$-axis — the loop is stable for all $K_p\gt0$.

Part B — Gain selection for $\zeta=0.707$ [7]

Approach. Intersect the locus with the $\zeta=0.707$ ray ($135^\circ$ from the positive real axis, $s=-p+jp$); solving the characteristic equation on that ray yields two positive gains.

1. Low-gain intersection. $K_p=1.27$ places the complex pair at $-0.575\pm j0.575$ ($\omega_n=0.813$ rad/s) with the third pole far away at $-3.85$ — the pair is genuinely dominant.
2. High-gain intersection. $K_p=3.07$ places the pair at $-2.175\pm j2.175$ ($\omega_n=3.08$ rad/s), but the third pole moves in to $-0.65$, closer to the imaginary axis than the pair — the real pole now dominates and the $\zeta=0.707$ pair no longer governs the response.
3. Choice. Select the $\boxed{\text{low-gain point, }K_p\approx1.27}$. Although its $\omega_n$ is smaller, the far third pole guarantees a clean, predictable second-order response — exactly what the $\zeta=0.707$ specification is meant to deliver. The high-gain point would be sabotaged by its slow dominant real pole.

Part C — Transient specifications [3]

1. Overshoot. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-\pi}=\boxed{4.3\%}$.
2. Settling time. $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{0.707(0.813)}=\boxed{6.96\ \text{s}}$.
3. Steady-state errors. The plant has two free integrators $\Rightarrow$ Type 2: $e_{ss}(\text{step})=e_{ss}(\text{ramp})=0$; $K_a=\lim_{s\to0}s^2L=0.4K_p=0.508$, so $e_{ss}(\text{parabola})=\dfrac{1}{K_a}=\boxed{1.97}$.