Question 1 of 8: Stability of a lag-compensated loop (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 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 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, Bode, Nyquist, gain/phase margins (Ch. 10), lead–lag and PID design (Ch. 9–11), state space and Mason’s rule (Ch. 3, 5); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and second-order correlations. All block diagrams, root loci, pole–zero maps, Bode sketches, Nyquist plots and the realization diagrams 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Questions 4, 6 and 7 are read from printed Bode/response plots, so those values carry normal chart-reading tolerance; the s-domain results (Q1, Q2, Q3, Q5, Q8) are exact.
Question 1 — Stability of a lag-compensated loop (compulsory) [20]
Figure Q1.1 — lag controller $C(s)=K(s+10)/(s+4)$ in series with the process $G(s)=10/(s^2+4s+8)$, unit feedback.
Given. $C(s)=K(s+10)/(s+4)$, $K\gt 0$; $G(s)=10/(s^2+4s+8)$; unit negative feedback. Find. $K_{crit}$, $\omega_{osc}$ and the stable range of $K$, by Routh–Hurwitz, the root locus and the Bode plot.
Approach. Form the closed-loop characteristic polynomial $1+C(s)G(s)=0$, build the Routh array, set the $s^1$ row to zero for marginal stability, then read the auxiliary equation for $\omega_{osc}$; confirm on the root locus and Bode.
Item (1) — Routh–Hurwitz
Characteristic polynomial. The open loop is $L(s)=\dfrac{10K(s+10)}{(s+4)(s^2+4s+8)}$, so $1+L=0$ gives $(s+4)(s^2+4s+8)+10K(s+10)=0$. Expanding $(s+4)(s^2+4s+8)=s^3+8s^2+24s+32$: $$s^3+8s^2+(24+10K)s+(32+100K)=0.$$
Routh array. $$\begin{array}{c|cc}s^3&1&24+10K\\ s^2&8&32+100K\\ s^1&\frac{8(24+10K)-(32+100K)}{8}&0\\ s^0&32+100K&\end{array}$$ The $s^1$ entry simplifies to $\dfrac{160-20K}{8}=20-2.5K$.
Marginal stability. Set the $s^1$ row to zero: $20-2.5K=0\Rightarrow \boxed{K_{crit}=8}$. The $s^0$ term $32+100K\gt 0$ for all $K\gt 0$, so the only sign change comes from the $s^1$ row.
Oscillation frequency. At $K=8$ the auxiliary equation is the $s^2$ row: $8s^2+(32+800)=0\Rightarrow s^2=-104$, hence $$\omega_{osc}=\sqrt{104}=\boxed{10.2\ \text{rad/s}}.$$
Safe range. The first Routh column stays positive for $0\lt K\lt 8$; the loop is stable there and marginally stable at $K=8$. $$\boxed{0\lt K\lt 8}.$$
Item (2) — root locus and the magnitude criterion
[Figure not reproduced: Q1.2 redrawn: three branches leave the poles $-4,\,-2\pm j2$ toward the zero $-10$ and infinity; two branches cross the jw-axis at $\pm j10.2$. See the official exam paper.]
jw-axis crossover. The locus intersects the imaginary axis at $s^\ast=\pm j10.2$, i.e. $\omega_{osc}=10.2$ rad/s — identical to the Routh result.
Magnitude criterion. With the open loop written as $K\cdot G_{ol}(s)$, $G_{ol}(s)=\dfrac{10(s+10)}{(s+4)(s^2+4s+8)}$, marginal stability needs $|K\,G_{ol}(s^\ast)|=1$. Evaluating $|G_{ol}(j10.2)|=0.125$ gives $$K_{crit}=\frac{1}{|G_{ol}(j10.2)|}=\frac{1}{0.125}=\boxed{8},$$ confirming item (1).
Interpretation. For $K\lt 8$ the dominant branches sit in the left half plane (stable); at $K=8$ they touch the axis; for $K\gt 8$ they cross into the right half plane (unstable). Safe range $0\lt K\lt 8$.
Item (3) — Bode verification
Bode of the normalized open loop $L(s)/K$: the phase reaches $-180^\circ$ at $\omega=10.2$ rad/s, where $|L/K|=0.125$ ($-18.1$ dB), so $K_{crit}=1/0.125=8$.
On the open-loop frequency response the phase crosses $-180^\circ$ at $\omega_{pc}=10.2$ rad/s. There the magnitude of the normalized loop is $|L(j\omega_{pc})/K|=0.125$, i.e. $-18.1$ dB; the gain that lifts this to $0$ dB is $K_{crit}=1/0.125=8$. All three methods agree.
