NivaarExam PrepOfficial exam papers ↗

22-Elec-A2 Systems and Control · May 2016

Question 2 of 8: Marginal stability by root locus, Bode and Routh–Hurwitz (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, May 2016, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula sheet; a Laplace-transform table and the standard ζ–overshoot and 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 specs (Ch. 4), block reduction (Ch. 5), Routh–Hurwitz (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID and lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10–11), state space, controllability/observability (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling. All step responses, block diagrams, pole–zero maps, root loci, the Bode sketches and the Nyquist plot below are redrawn as inline figures.

Reading the supplied design charts. The percent overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$ and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$; the figures on the exam paper (step responses, root loci, Bode plots) are read graphically, so the values below carry normal chart-reading tolerance.

Question 2 — Marginal stability by root locus, Bode and Routh–Hurwitz (compulsory) [6 + 4 + 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. $G(s)=\dfrac{(s+50)^2}{(s+4)^3}$, DC gain $G(0)=50^2/4^3=39.1$ ($31.8$ dB).

Find. all $(K_{crit},\omega_{osc})$ pairs where the locus crosses the $j\omega$-axis, and the ranges of $K_p$ giving a stable closed loop.

Q2 — Root locus of G(s)=(s+50)^2/(s+4)^3-150-145-140-135-130-125-120-115-110-105-100-95-90-85-80-75-70-65-60-55-50-45-40-35-30-25-20-15-10-50510152025-90-85-80-75-70-65-60-55-50-45-40-35-30-25-20-15-10-551015202530354045505560657075808590ReImw=9.5 (Kp=0.42)w=35.5 (Kp=12.1)
Q2.1: root locus of $G(s)$; the locus crosses the imaginary axis twice.

Part (1) — root-locus magnitude criterion

  1. Read the crossovers. The locus leaves the triple pole at $-4$ and bends into the right half-plane, crossing $j\omega$ near $\omega\approx9.5$ and again near $\omega\approx35.5$ rad/s before returning left toward the double zero at $-50$.
  2. Magnitude criterion at the low crossing. $|G(j9.5)|=\dfrac{|j9.5+50|^2}{|j9.5+4|^3}=\dfrac{50.9^2}{10.31^3}=2.36$, so $K_{crit}=1/|G|=\boxed{0.42}$ at $\omega_{osc}=9.5$ rad/s.
  3. Magnitude criterion at the high crossing. $|G(j35.5)|=\dfrac{61.3^2}{35.72^3}=0.0825$, so $K_{crit}=1/0.0825=\boxed{12.1}$ at $\omega_{osc}=35.5$ rad/s.
  4. Interpret the safe ranges. The locus is in the left half-plane for small gains, enters the RHP between the two crossings, and returns to the LHP for large gains (excess of one pole $\to$ one asymptote at $180^\circ$). Hence stable for $0\lt K_p\lt0.42$ and for $K_p\gt12.1$.
Q2 — Bode of G(s); two -180 deg crossings-60-40-2002040Magnitude (dB)-225-180-135-90-45010^010^110^210^3Phase (deg)Frequency (rad/s)
Q2.2: Bode of $G(s)$; the phase dips through $-180^\circ$ then recovers, giving two phase-crossover frequencies.

Part (2) — Bode verification

  1. Two $-180^\circ$ crossings. The triple pole drives the phase toward $-270^\circ$, but the double zero at $-50$ pulls it back up, so the phase crosses $-180^\circ$ at $\omega\approx9.5$ (descending) and $\omega\approx35.5$ (ascending).
  2. Gains at those frequencies. $|G(j9.5)|=+7.5$ dB $\Rightarrow K_{crit}=10^{-7.5/20}=0.42$; $|G(j35.5)|=-21.7$ dB $\Rightarrow K_{crit}=10^{+21.7/20}=12.1$ — identical to Part (1). The two gain margins are $\boxed{-7.5\text{ dB and }+21.7\text{ dB}}$ relative to unity gain.

Part (3) — Routh–Hurwitz

  1. Characteristic equation. $(s+4)^3+K_p(s+50)^2=0\Rightarrow s^3+(12+K_p)s^2+(48+100K_p)s+(64+2500K_p)=0$.
  2. Routh array. The $s^1$ row is $\dfrac{(12+K_p)(48+100K_p)-(64+2500K_p)}{12+K_p}$; its numerator is $100K_p^2-1252K_p+512$.
  3. Marginal gains. $100K_p^2-1252K_p+512=0\Rightarrow K_p=\boxed{0.42}\text{ or }\boxed{12.10}$ — matching both graphical reads.
  4. Oscillation frequencies. From the auxiliary polynomial $(12+K_p)s^2+(64+2500K_p)=0$: $\omega_{osc}=\sqrt{\tfrac{64+2500K_p}{12+K_p}}$ gives $\boxed{9.5}$ rad/s at $K_p=0.42$ and $\boxed{35.5}$ rad/s at $K_p=12.10$.
  5. Safe range. The $s^1$ quadratic is positive outside its roots, so the closed loop is stable for $0\lt K_p\lt0.42$ or $K_p\gt12.10$ (unstable in between).
QuantityValue
Low critical gain$K_{crit}=0.42$ at $\omega_{osc}=9.5$ rad/s
High critical gain$K_{crit}=12.10$ at $\omega_{osc}=35.5$ rad/s
Stable operating ranges$0\lt K_p\lt0.42$ and $K_p\gt12.10$