Question 1 of 7: Stability by Bode plot 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, 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 1 — Stability by Bode plot and Routh–Hurwitz (compulsory) [10 + 10]
Given. Unit-feedback loop, open-loop transfer function $L(s)=K_p\,G(s)$ with $G(s)=\dfrac{150(s+50)}{(s+3)^2(s+8)}$ — three real poles at $-3,-3,-8$, one zero at $-50$, DC gain $G(0)=\dfrac{150\cdot50}{9\cdot8}=104.2$ (i.e. $+40.4$ dB).
Find. $K_{crit}$ and $\omega_{osc}$ at marginal stability, the safe range of $K_p$, and the gain margin at $K_p=0.05$ — by both frequency-domain and Routh methods.
[Figure not reproduced: Figure Q1.1 (redrawn) — the unit-feedback proportional loop. See the official exam paper.]
Open-loop pole–zero map: the double pole at $-3$ dominates the phase roll-off; the zero at $-50$ pulls the phase back up toward $-180^\circ$ at high frequency.
Part A — Frequency-domain reading (Bode plot) [10]
Approach. Marginal stability occurs where the open-loop phase is $-180^\circ$; read the phase-crossover frequency $\omega_{pc}$ and the magnitude there, then $K_{crit}=1/|G(j\omega_{pc})|$ and the gain margin follows.
Open-loop Bode plot ($K_p=1$). The phase reaches $-180^\circ$ at $\omega_{pc}\approx8.78$ rad/s (dashed marker); the magnitude there is $+17.4$ dB, so the loop is already unstable at $K_p=1$.
Phase-crossover frequency. The open-loop phase is $\angle G(j\omega)=\tan^{-1}\tfrac{\omega}{50}-2\tan^{-1}\tfrac{\omega}{3}-\tan^{-1}\tfrac{\omega}{8}$. Setting it to $-180^\circ$ gives $\omega_{pc}=8.78$ rad/s (read off the plot where the phase curve touches the $-180^\circ$ line).
Magnitude at $\omega_{pc}$. $|G(j8.78)|=\dfrac{150\,|j8.78+50|}{|j8.78+3|^2\,|j8.78+8|}=\dfrac{150(50.76)}{(86.1)(11.88)}=7.44$ ($+17.4$ dB).
Critical gain and oscillation frequency. At marginal stability $K_{crit}|G(j\omega_{pc})|=1$, so $$K_{crit}=\frac{1}{|G(j8.78)|}=\frac{1}{7.44}=\boxed{0.134},\qquad \omega_{osc}=\omega_{pc}=\boxed{8.78\ \text{rad/s}}.$$
Safe operating range. Because the loop is stable only while the $-180^\circ$-phase gain is below unity, $0\lt K_p\lt K_{crit}$, i.e. $\boxed{0\lt K_p\lt 0.134}$.
Gain margin at $K_p=0.05$. $$\text{GM}=\frac{K_{crit}}{K_p}=\frac{0.134}{0.05}=\boxed{2.69\ \text{V/V}}=20\log_{10}(2.69)=\boxed{8.59\ \text{dB}}.$$
Part B — Routh–Hurwitz verification [10]
Approach. Form the closed-loop characteristic polynomial, build the Routh array, and set the $s^1$ row to zero for marginal stability; the auxiliary equation from the $s^2$ row gives $\omega_{osc}$.
Characteristic polynomial. With $(s+3)^2(s+8)=s^3+14s^2+57s+72$, $$1+K_pG(s)=0\ \Rightarrow\ s^3+14s^2+(57+150K_p)s+(72+7500K_p)=0.$$
Routh array. $$\begin{array}{c|cc}s^3&1&57+150K_p\\ s^2&14&72+7500K_p\\ s^1&\frac{14(57+150K_p)-(72+7500K_p)}{14}&0\\ s^0&72+7500K_p&\end{array}$$ The $s^1$ entry simplifies to $\dfrac{726-5400K_p}{14}$.
Marginal stability. Set the $s^1$ entry to zero: $726-5400K_p=0\Rightarrow \boxed{K_{crit}=0.1344}$, in agreement with Part A.
Oscillation frequency. The auxiliary equation from the $s^2$ row at $K_{crit}$ is $14s^2+(72+7500\cdot0.1344)=0\Rightarrow 14s^2+1080.3=0\Rightarrow s^2=-77.2$, so $\boxed{\omega_{osc}=8.78\ \text{rad/s}}$.
Safe range. Every first-column entry positive requires $726-5400K_p\gt0$ and $72+7500K_p\gt0$, i.e. $0\lt K_p\lt 0.134$ — identical to Part A.
Consistency. Yes. Both methods give $K_{crit}\approx0.134$, $\omega_{osc}=8.78$ rad/s and the safe range $0\lt K_p\lt0.134$. The Bode (frequency-domain) and Routh (s-domain) tests describe the same marginal-stability boundary.
