Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. 16-Elec-B2 Advanced Control Systems, May 2017 — a three-hour open-book examination. The cover page states “Any four questions constitute a complete paper. Only the first four questions as they appear in your answer book will be marked” and “All questions are of equal value (25%)”. This sitting prints five questions, so each carries 25 marks and a candidate answers four; Question 1 is a seventeen-item multiple-choice block whose per-item weights are printed in the margin as [1] or [2] and total exactly 25. Tables of inverse Laplace transforms and of Laplace/z transforms are appended to the paper. All five questions are worked below, because this set is a study resource rather than a timed sitting.
Reference texts. N. S. Nise, Control Systems Engineering, 8th ed. (Ch. 4 time response and the rise-time/peak-time/settling-time relations, Ch. 6 Routh–Hurwitz stability, Ch. 7 steady-state error and system type, Ch. 8 root-locus sketching rules, Ch. 9 root-locus design of cascade compensators, Ch. 10 frequency response and stability margins); R. C. Dorf and R. H. Bishop, Modern Control Systems, 13th ed. (Ch. 5, 6, 7, 9); K. Ogata, Modern Control Engineering, 5th ed. (Ch. 5 transient response, Ch. 6 root-locus design, Ch. 7 frequency-response design); G. F. Franklin, J. D. Powell and A. Emami-Naeini, Feedback Control of Dynamic Systems, 8th ed. (Ch. 3, 5, 6). These are the references listed for exam code 16-Elec-B2 in the Engineers Canada syllabus, and this paper’s vocabulary (percent overshoot, settling time, asymptote intercept, “summation of the angles”) follows Nise closely.
Check — two source readings that were checked against the printed paper.
(1) Question 1, item 5. The paper prints the \(s^2\) first-column entry of the Routh table as \((1-4\varepsilon)/\varepsilon\), which is exactly what the Routh–Hurwitz recursion gives, and which is confirmed independently by the printed \(s^1\) entry. The table on the paper is correct and internally consistent. The answer is unaffected by the alternative form \(\varepsilon/(1-4\varepsilon)\), because both are positive for small \(\varepsilon>0\).
(2) Question 4. The block diagram prints the plant as \(10/[s(s+2)(s+10)]\) with no \(K\) shown, while part (a) says “the open-loop transfer function when \(K = 20\)”. This solution takes the direct reading, \(G(s)=K/[s(s+2)(s+10)]\) with \(K = 20\), so the printed 10 is one sample value of the same numerator gain. The alternative reading — a fixed plant gain of 10 multiplied by \(K\), i.e. \(G(s)=10K/[s(s+2)(s+10)]\) — is carried through in a callout under part (b) so that both mark schemes are covered.
