Question 1 of 8: Stability by root locus, frequency response and Routh (compulsory)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, December 2017, 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 $\zeta$–overshoot, $\zeta$–resonant-peak and phase-margin design plots are supplied on pages 2–3). Questions 1 and 2 are compulsory; a complete paper is five questions, so a candidate chooses 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) — Routh–Hurwitz (Ch. 6), root locus (Ch. 8), steady-state error and static error constants (Ch. 7), frequency response, Nyquist, gain/phase margins (Ch. 10), lead/lag and PID design (Ch. 9–11), state space and controllability/observability (Ch. 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations, pole placement and frequency-domain compensator design. All block diagrams, root loci, pole–zero maps, Bode and Nyquist plots, closed-loop magnitude curves and step responses below are redrawn as inline figures.
Reading the supplied design charts. The percent-overshoot chart uses $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}$, the phase-margin chart uses $\Phi_m\approx100\,\zeta$, and the second-order model is $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Every result here is derived analytically in the $s$-domain from the exact transfer functions given in the paper; the supplied plots are used only to corroborate $\zeta$, $\Phi_m$ and the crossover frequencies. Where a printed figure’s hand-read tick differs from the exact value, the exact value governs and the gap is flagged.
Question 1 — Stability by root locus, frequency response and Routh (compulsory) [20]
Figure Q1.1 — unit-feedback loop, proportional gain $K_p$ ahead of $G(s)=100(s+20)(s+10)/[(s+1)^3(s+100)]$.
Given.
Process
$G(s)=\dfrac{100(s+20)(s+10)}{(s+1)^3(s+100)}$
Open-loop poles / zeros
poles $-1$ (triple), $-100$; zeros $-10,-20$
jw crossings (from RL)
$s_1=j2.25$, $s_2=j12.5$ rad/s
Controller
proportional, gain $K_p$
Find. $K_{crit}$ at each crossing, the safe gain range(s), a Bode cross-check, and Routh confirmation.
Approach. Apply the magnitude criterion at the two given jw points, read the same gains off the Bode phase-crossover frequencies, then build the Routh array of the closed-loop denominator and solve its $s^1$ condition.
Item (1) — magnitude criterion and safe range
Lower crossing $s_1=j2.25$. Evaluate the open-loop magnitude: $|G(j2.25)|=13.82$, so $$K_{crit,1}=\frac{1}{|G(j2.25)|}=\boxed{0.0724}.$$ (The angle $\angle G(j2.25)=-180^\circ$ confirms the point lies on the locus.)
Upper crossing $s_2=j12.5$. $|G(j12.5)|=0.190$, so $$K_{crit,2}=\frac{1}{|G(j12.5)|}=\boxed{5.26}.$$
Interpret the locus. With all four open-loop poles in the left half-plane, the loop is stable at low gain. As $K_p$ rises, the branches leaving the triple pole at $-1$ bow out and cross into the right half-plane at $\omega=2.25$ (gain $0.072$), making the loop unstable; at the higher gain $5.26$ they cross back at $\omega=12.5$ and the loop restabilizes. Hence there are two disjoint safe bands: $$\boxed{0\lt K_p\lt 0.072\quad\cup\quad K_p\gt 5.26.}$$ This conditional-stability behaviour is the signature of a plant whose two finite zeros pull the branches back across the axis.
[Figure not reproduced: Q1.1 redrawn: branches from the triple pole $-1$ cross the jw-axis at $\pm j2.24$ ($K_p=0.072$) and again at $\pm j12.5$ ($K_p=5.26$); the far pole $-100$ and zeros $-10,-20$ close the map. See the official exam paper.]
Item (2) — Bode verification
Phase-crossover frequencies. On the open-loop Bode plot the phase reaches $-180^\circ$ at exactly the two locus crossing frequencies, $\omega_{pc1}=2.24$ and $\omega_{pc2}=12.5$ rad/s.
Gains there. The magnitude readings are $|G(j\omega_{pc1})|=+22.8$ dB $=13.8$ and $|G(j\omega_{pc2})|=-14.4$ dB $=0.190$, giving $K_{crit}=1/|G|=0.072$ and $5.26$ — identical to Item 1.
[Figure not reproduced: Q1.2 redrawn (exact): the phase curve touches $-180^\circ$ at $2.24$ and $12.5$ rad/s (dashed marker at $2.24$); the corresponding magnitudes are $+22.8$ dB and $-14.4$ dB. See the official exam paper.]
Reading Fig Q1.2. The printed plot (a landscape page) matches the exact $G(s)$: flat at $+46$ dB ($=20\log_{10}200$) with $0^\circ$ phase at low frequency, and the phase dips below $-180^\circ$ between about $2.2$ and $12.5$ rad/s (minimum $\approx-198^\circ$ near $5$ rad/s). Reading the magnitude at the two phase crossovers gives about $+23$ dB and $-14$ dB, i.e. $K_{crit}\approx0.07$ and $5.3$, the same values Items 1 and 3 give.
Item (3) — Routh–Hurwitz
Closed-loop denominator. $1+K_pG=0$ gives $(s+1)^3(s+100)+100K_p(s+20)(s+10)=0$, i.e. $$D(s)=s^4+103s^3+(303+100K_p)s^2+(301+3000K_p)s+(100+20000K_p).$$
Routh array. $$\begin{array}{c|ccc}s^4&1&303+100K_p&100+20000K_p\\ s^3&103&301+3000K_p&0\\ s^2&b_1&100+20000K_p&\\ s^1&c_1&&\\ s^0&100+20000K_p&&\end{array}$$ with $b_1=\dfrac{103(303+100K_p)-(301+3000K_p)}{103}$.
$s^1$ condition. Setting the $s^1$ entry $c_1=0$ reduces (after clearing $b_1$) to the quadratic $$5\,475\,000\,K_p^2-29\,314\,675\,K_p+2\,060\,602=0,$$ whose roots are $$\boxed{K_{crit}=0.0712\ \text{and}\ 5.283.}$$ Because the leading coefficient is positive, $c_1\gt0$ (stable) for $K_p\lt0.0712$ and for $K_p\gt5.283$, and $c_1\lt0$ (unstable) between — exactly the two safe bands of Item 1.
Oscillation frequencies. At each $K_{crit}$ the auxiliary row $b_1s^2+(100+20000K_p)=0$ gives $\omega_{osc}=\sqrt{(100+20000K_p)/b_1}$: $\boxed{\omega_{osc1}=2.24}$ and $\boxed{\omega_{osc2}=12.5}$ rad/s, matching the stated crossings.
