Question 2 of 7: Root-locus analysis of an unstable plant (compulsory)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 07-Elec-A2 Systems & Control, December 2013, 3 hours, closed book (approved calculator plus one signed 8.5 × 11" formula sheet). Questions 1 and 2 are compulsory; answer three of the remaining five (Q3–Q7). Each question is worth 20 marks. All seven are worked below so the set is complete.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — root locus (Ch. 8), frequency response / Bode and gain–phase margins (Ch. 10), stability and Routh–Hurwitz (Ch. 6), PID and cascade lead/lag compensation (Ch. 9–11), steady-state error and error constants (Ch. 7), state space and canonical forms (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of dominant-poles modelling, lead design and controllability; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci and Bode plots below are redrawn as inline figures.
Reading the exam figures. Two questions supply printed plots: Q1 gives an open-loop Bode plot (used for the gain-margin reading) and Q6 Part B gives a measured step-response curve (used to read $K_{dc},\zeta,\omega_n$). The transfer functions are given exactly in the question text, so every reading below is also confirmed analytically (Bode by direct evaluation, stability by Routh–Hurwitz). Where the printed plot annotation and the exact model differ slightly, the exact model governs.
Question 2 — Root-locus analysis of an unstable plant (compulsory) [10 + 10]
Given. Open-loop $L(s)=K_p\,G(s)$, $G(s)=\dfrac{1}{(s-5)(s^2+6s+40)}$. Poles: $+5$ (unstable) and $-3\pm j\sqrt{31}=-3\pm j5.57$; no finite zeros ($n-m=3$).
Find. Full root-locus geometry, the $j\omega$-axis crossings, $K_{crit}$/$\omega_{osc}$, and the safe gain range.
Open-loop poles: one in the right half-plane at $+5$, a lightly damped pair at $-3\pm j5.57$.
Part A — Root-locus construction [10]
Approach. Apply the standard construction rules (real-axis rule, asymptotes and centroid, break-away test, departure angle) to a three-pole, zero-free locus.
Real-axis segments. A real-axis point lies on the locus when the number of real poles/zeros to its right is odd. Only $+5$ is real, so the locus occupies the entire real axis to the left of $+5$: the segment $(-\infty,\,5]$. The branch that starts at $+5$ therefore travels left.
Asymptotes and centroid. With $n-m=3$ branches go to infinity along asymptotes at $\theta=\dfrac{(2k+1)180^\circ}{3}=\pm60^\circ,\,180^\circ$, meeting the real axis at the centroid $$\sigma_a=\frac{\sum p_i-\sum z_i}{n-m}=\frac{5+(-3)+(-3)}{3}=\boxed{-0.333}.$$
Break-away / break-in points. With $K_p=-(s-5)(s^2+6s+40)=-(s^3+s^2+10s-200)$, setting $\dfrac{dK_p}{ds}=-(3s^2+2s+10)=0$ gives $s=\dfrac{-2\pm\sqrt{4-120}}{6}$, which is complex — there are no real break-away/break-in points. The real branch runs straight to $-\infty$ without meeting another branch.
Angle of departure from the upper pole $p_1=-3+j5.57$: $\theta_d=180^\circ-\big[\angle(p_1-5)+\angle(p_1-\overline{p_1})\big]=180^\circ-[145.1^\circ+90^\circ]=\boxed{-55.2^\circ}$ (and $+55.2^\circ$ from the lower pole by symmetry).
Numerically computed root locus (red) with the $\pm60^\circ/180^\circ$ asymptotes (dashed) and centroid at $-0.333$. The real branch crosses the origin at $K_p=200$; the complex branches cross the imaginary axis at $\pm j3.16$ at $K_p=210$.
Part B — Imaginary-axis crossings, $K_{crit}$ and safe range [10]
Approach. Substitute $s=j\omega$ into the characteristic equation and split into real and imaginary parts; confirm with a Routh array to get the stability band.
Characteristic equation. $1+K_pG=0\Rightarrow (s-5)(s^2+6s+40)+K_p=0$, i.e. $s^3+s^2+10s+(K_p-200)=0$.
Imaginary-axis crossings. Put $s=j\omega$: $-j\omega^3-\omega^2+10j\omega+(K_p-200)=0$. Imaginary part $\omega(10-\omega^2)=0$ gives $\omega=0$ or $\omega=\sqrt{10}=3.16$ rad/s; the real part then fixes $K_p=\omega^2+200$. So the locus crosses at $$s=0\ (K_p=200)\quad\text{and}\quad s=\pm j\sqrt{10}=\pm j3.16\ (K_p=210).$$
Routh stability band. For $s^3+s^2+10s+(K_p-200)$: $$\begin{array}{c|cc}s^3&1&10\\ s^2&1&K_p-200\\ s^1&210-K_p&0\\ s^0&K_p-200&\end{array}$$ All first-column entries positive requires $K_p-200\gt0$ and $210-K_p\gt0$.
Critical gains and oscillation. The lower bound $K_p=200$ places a root at the origin ($\omega_{osc}=0$); the upper bound $K_p=210$ places the conjugate pair on the axis, giving $\boxed{K_{crit}=210,\ \omega_{osc}=\sqrt{10}=3.16\ \text{rad/s}}$ (the oscillatory marginal point).
Practical safe range. The closed loop is stable only inside the band $$\boxed{200\lt K_p\lt 210}.$$ Below $200$ the plant's RHP pole is not yet pulled into the LHP; above $210$ the complex branches enter the RHP.
