Question 3 of 8: Root-locus analysis and gain selection for a RHP-pole plant
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 2015, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula/notes sheet; a Laplace-transform table and the standard ζ–overshoot / ζ–resonant-peak design plots are supplied). Questions 1 and 2 are compulsory; a complete paper is five questions, so candidates choose three of Q3–Q8. Each question is worth 20 marks. All eight are worked below so the set is a complete study resource.
Reference texts: N. S. Nise, Control Systems Engineering (7th ed., Wiley) — time response and second-order specifications (Ch. 4), block/signal-flow reduction and Mason’s rule (Ch. 5), Routh–Hurwitz stability (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID / lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain and phase margins (Ch. 10–11), state space, controllability/observability and canonical forms (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and Mason’s gain formula; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, the Nyquist plot, the network schematic and the signal-flow graph below are redrawn as inline figures.
Reading the exam figures. Every transfer function in this paper is stated exactly, so every boxed result is confirmed analytically and the exact model governs. Q2 supplies a Bode plot ($G_m=32.2$ dB at $3.9$ rad/s, $P_m=-145^\circ$ at $0.324$ rad/s) and a root locus, and Q7 supplies the uncompensated Bode plot; values read from those curves are flagged as reads and cross-checked against the Routh/analytic results.
Question 3 — Root-locus analysis and gain selection for a RHP-pole plant [10 + 10]
Given. $L(s)=K_p\dfrac{s+1}{s(s-3)}$: poles $0,+3$ (one RHP pole), zero $-1$; $n=2,m=1$, so one asymptote. Characteristic equation $s(s-3)+K_p(s+1)=0$, i.e. $s^2+(K_p-3)s+K_p=0$.
Find. Locus geometry; then $K_p$ at $\zeta=1,0,-1$.
Part A — Root-locus geometry [10]
Approach. Apply the real-axis, centroid/asymptote, break-point ($dK_p/ds=0$) and $j\omega$-crossing rules.
Root locus of $(s+1)/[s(s-3)]$: branches leave $0$ and $+3$, meet at the break-away $s=+1$ ($K_p=1$), circle through the complex plane, cross $j\omega$ at $\pm j\sqrt3$ ($K_p=3$), and re-enter the real axis at the break-in $s=-3$ ($K_p=9$).
Real-axis segments. The singularities on the axis are the zero $-1$ and poles $0,+3$. A point is on the locus if the number of singularities to its right is odd: the segment $[0,3]$ (one pole, $+3$, to the right) and the segment $(-\infty,-1]$ (three singularities to the right) are on the locus; $(-1,0)$ is not.
Centroid & asymptote. $\sigma_a=\dfrac{(0+3)-(-1)}{2-1}=\boxed{4}$; with $n-m=1$ there is a single asymptote at $\boxed{180^\circ}$ (the branch that ends at the break-in runs left along the real axis to $-\infty$).
Break points. $K_p=-\dfrac{s(s-3)}{s+1}$; setting $dK_p/ds=0$ gives $(s+3)(s-1)=0$. $s=+1$ lies on segment $[0,3]$ — the $\boxed{\text{break-away at }s=1\ (K_p=1)}$; $s=-3$ lies on $(-\infty,-1]$ — the $\boxed{\text{break-in at }s=-3\ (K_p=9)}$.
Imaginary-axis crossing. In $s^2+(K_p-3)s+K_p=0$ set $s=j\omega$: the real part gives $K_p-\omega^2=0$ and the imaginary part $(K_p-3)\omega=0$. The non-trivial solution is $K_p=3$, giving $\omega^2=3$, i.e. $\boxed{s=\pm j\sqrt3=\pm j1.73}$ at $K_p=3$.
Stability. The $s^1$ Routh term is $K_p-3$, so the closed loop is stable only for $\boxed{K_p\gt3}$ — the RHP open-loop pole must be pulled into the LHP by enough gain.
Part B — Gain for each damping ratio [10]
Approach. Match $s^2+(K_p-3)s+K_p$ to $s^2+2\zeta\omega_ns+\omega_n^2$: $\omega_n^2=K_p$ and $2\zeta\omega_n=K_p-3$, so $\zeta=\dfrac{K_p-3}{2\sqrt{K_p}}$. Solve for each target.
(a) $\zeta=1$ (critically damped). $K_p-3=2\sqrt{K_p}$; with $x=\sqrt{K_p}$, $x^2-2x-3=0\Rightarrow x=3$, so $\boxed{K_p=9}$. Then $s^2+6s+9=(s+3)^2$ — the double root at $-3$ is exactly the break-in point.
(b) $\zeta=0$ (marginally stable). $2\zeta\omega_n=K_p-3=0$, so $\boxed{K_p=3}$, with poles $\pm j\sqrt3$ — the imaginary-axis crossing.
(c) $\zeta=-1$ (unstable). $K_p-3=-2\sqrt{K_p}$; with $x=\sqrt{K_p}$, $x^2+2x-3=0\Rightarrow x=1$, so $\boxed{K_p=1}$. Then $s^2-2s+1=(s-1)^2$ — a double root at $+1$ (RHP), the break-away point.
Note: the two break points are the two critically-damped gains. The break-away ($s=+1$, $K_p=1$) is a double RHP root ($\zeta=-1$); the break-in ($s=-3$, $K_p=9$) is a double LHP root ($\zeta=+1$). Between them ($1\lt K_p\lt9$) the poles are complex, passing through the $j\omega$-axis at $K_p=3$ ($\zeta=0$).
