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 2014, 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 six (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 and stability (Ch. 6), steady-state error and error constants (Ch. 7), root locus (Ch. 8), PID and lag/lead design (Ch. 9–11), frequency response, Nyquist and gain/phase margins (Ch. 10–11), state space, controllability/observability and pole placement (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of dominant-poles modelling and Mason’s rule; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, Bode/Nyquist plots and signal-flow graphs below are redrawn as inline figures.
Reading the exam figures. Q1 supplies an open-loop Bode plot and Q6 supplies open-loop Bode plots of the process. Where a transfer function is given exactly in the text (Q1’s system is the same plant as Q2, stated exactly there), every graphical reading is also confirmed analytically and the exact model governs. Q6 is a pure graph-reading problem (no closed form given); its numeric reads are approximate and are flagged accordingly. Note the printed exam mislabels the header of the last two pages “May 2013” — this is the May 2014 sitting throughout.
Question 6 — Lag compensator from Bode plots [10 + 10]
Check: graph-read problem (no closed form given). Q6 provides only Bode plots for $G(s)$; no transfer function is stated, and the printed compensator expression on p.11 does not give a clear value for the denominator $\alpha$. The numeric reads below are best-estimate graph readings from Figure Q6.2 and the method is the graded content; a candidate stating slightly different reads and carrying them through consistently earns full marks.
[Figure not reproduced: Figure Q6.1 (redrawn) — unit-feedback loop, cascade controller $G_c(s)$ on process $G(s)$. See the official exam paper.]
Part A — Second-order model & step specifications [10]
Given (read from Figure Q6.2). Low-frequency slope $-20$ dB/dec with phase $\to-90^\circ$ ⇒ the process is Type 1 (one free integrator). Gain crossover $\omega_{gc}\approx0.7$ rad/s ($|G|=0$ dB); phase there $\approx-128^\circ$ ⇒ phase margin $PM\approx52^\circ$. Phase crosses $-180^\circ$ near $\omega_{pc}\approx3$ rad/s where $|G|\approx-20$ dB (gain margin $\approx20$ dB). The $-20$ dB/dec low-frequency asymptote reaches $0$ dB near $\omega\approx0.7$, so the velocity constant $K_v\approx0.7\ \text{s}^{-1}$.
Approach. Map phase margin to $\zeta$ and the gain-crossover to $\omega_n$ for the closed-loop second-order model, then evaluate the standard step specs; the Type-1 property fixes the steady-state errors.
Damping from phase margin. For a second-order model $\zeta\approx\dfrac{PM^{\circ}}{100}=\dfrac{52}{100}=\boxed{0.52}$ (valid for $PM\lesssim65^\circ$).
Natural frequency. With $\omega_{gc}=\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}$ and $\zeta=0.52$, the radical is $0.772$, so $\omega_n=\dfrac{0.7}{0.772}=\boxed{0.91\ \text{rad/s}}$.
Steady-state errors. Type-1 ⇒ $e_{ss(step)}=\boxed{0}$; $e_{ss(ramp)}=\dfrac{1}{K_v}=\dfrac{1}{0.7}\approx\boxed{1.43}$ (too large — motivates the lag).
Part B — Lag compensator design [10]
Approach. A lag $G_c(s)=K_c\dfrac{\tau s+1}{\beta\tau s+1}$ ($\beta\gt1$) lifts the low-frequency gain (to fix the ramp error) while leaving the gain-crossover region — and hence the phase margin / $PO$ — almost unchanged, by placing the zero a decade below crossover.
Required low-frequency boost. Need $K_v\ge\dfrac{1}{0.1}=10$. From $K_v\approx0.7$ uncompensated, the boost factor is $\dfrac{10}{0.7}\approx14.3$, so take $\beta\approx15$ (with $K_c=1$).
Place the lag corner. Zero one decade below crossover: $\dfrac{1}{\tau}=\dfrac{\omega_{gc}}{10}\approx0.07$ rad/s ⇒ $\tau\approx14$ s; pole at $\dfrac{1}{\beta\tau}\approx\dfrac{0.07}{15}\approx0.0047$ rad/s.
Overshoot check. The zero at $0.07$ rad/s is a decade below $\omega_{gc}$, so the phase lag it adds at crossover is only a few degrees; $PM$ stays $\approx50^\circ$ and $PO\le15\%$ is preserved. On the Bode plot the low-frequency magnitude rises by $\approx20\log_{10}15=+23.5$ dB while the crossover is unchanged.
