Question 1 of 8: Proportional + rate-feedback design: transfer function, error and overshoot (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, May 2015, 3 hours, closed book (approved Casio/Sharp calculator plus one signed, double-sided 8.5 × 11" formula/notes sheet; a Laplace-transform table and design plots are supplied with the paper). Questions 1 and 2 are compulsory; a complete paper is five questions, so candidates choose 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) — time response and second-order specifications (Ch. 4), block/signal-flow reduction and Mason’s rule (Ch. 5), Routh–Hurwitz stability (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID / lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain and phase margins (Ch. 10–11), state space, controllability/observability and pole placement (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling and Mason’s gain formula; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, Bode plots, signal-flow graphs and the Nyquist plot below are redrawn as inline figures.
Reading the exam figures. Where a transfer function is stated exactly (Q1–Q5, Q7, Q8A) every result is confirmed analytically and the exact model governs. Q4 supplies a Bode plot ($G_m=16.1$ dB at $2.05$ rad/s, $P_m=-85.1^\circ$ at $0.589$ rad/s) and a root locus; Q5 Parts A–B and Q6 use printed frequency-response plots, and Q8B a printed Nyquist diagram — values read from those curves are flagged as reads and cross-checked against the analytics wherever a transfer function is available.
Question 1 — Proportional + rate-feedback design: transfer function, error and overshoot (compulsory) [8 + 5 + 7]
Given. Forward path $K_pG(s)$ with $G(s)=\dfrac{0.1(s+50)}{(s+0.5)(s+1)}$; return path $H(s)=K_ds+1$ (rate feedback plus unity). $G(0)=\dfrac{0.1(50)}{0.5(1)}=10$, $H(0)=1$.
Find. $T(s)=Y/R$ in $K_p,K_d$; then $K_p$ for $e_{ss}=5\%$; then an approximate $K_d$ for $PO=15\%$.
[Figure not reproduced: Figure Q1.1 redrawn — proportional gain $K_p$ in the forward path, rate feedback $K_ds+1$ in the return path; $G(s)=0.1(s+50)/[(s+0.5)(s+1)]$. See the official exam paper.]
Part 1 — Closed-loop transfer function [8]
Approach. Standard single-loop reduction $T=\dfrac{K_pG}{1+K_pGH}$; clear the denominator $(s+0.5)(s+1)$.
Forward and loop gains. $K_pG=\dfrac{0.1K_p(s+50)}{(s+0.5)(s+1)}$ and $K_pGH=\dfrac{0.1K_p(s+50)(K_ds+1)}{(s+0.5)(s+1)}$.
Assemble $T=\dfrac{K_pG}{1+K_pGH}$. Multiplying through by $(s+0.5)(s+1)=s^2+1.5s+0.5$ and using $0.1K_p(s+50)(K_ds+1)=0.1K_pK_ds^2+0.1K_p(1+50K_d)s+5K_p$:$$T(s)=\frac{0.1K_p(s+50)}{(1+0.1K_pK_d)s^2+\bigl(1.5+0.1K_p+5K_pK_d\bigr)s+(0.5+5K_p)}.$$
Structure. Two poles, one zero at $s=-50$; the rate term $K_d$ appears in both the $s^2$ and $s^1$ coefficients (it adds damping without moving the DC gain, since $H(0)=1$).
Part 2 — Proportional gain for 5% steady-state error [5]
Approach. The loop is Type 0 with unity feedback at DC ($H(0)=1$); use the position error constant $K_{pos}=K_pG(0)H(0)$.
Position error constant. $K_{pos}=K_pG(0)H(0)=K_p(10)(1)=10K_p$.
Substitute $K_p=1.9$. With $0.1K_p=0.19$, $5K_p=9.5$:$$T(s)=\frac{0.19(s+50)}{(1+0.19K_d)s^2+(1.69+9.5K_d)s+10}.$$ Only $K_d$ is unknown.
Part 3 — Approximate rate gain for 15% overshoot [7]
Simplifying assumption (stated clearly). Two reductions are available and both are used: (i) the closed-loop zero at $s=-50$ is far in the left half-plane, so it barely affects the dominant response and the system is treated as a standard second order; (ii) because $K_d$ is expected small, $0.19K_d\ll1$, so the $s^2$ coefficient $\approx1$. The denominator then reads $s^2+(1.69+9.5K_d)s+10$.
Damping for 15% overshoot. $\zeta=\dfrac{-\ln(0.15)}{\sqrt{\pi^2+\ln^2(0.15)}}=\boxed{0.517}$.
Natural frequency. With the $s^2$ coefficient $\approx1$, $\omega_n=\sqrt{10}=3.162$ rad/s, so $2\zeta\omega_n=2(0.517)(3.162)=3.269$.
Match the $s^1$ coefficient. $1.69+9.5K_d=2\zeta\omega_n=3.269$ gives $9.5K_d=1.579$, so $\boxed{K_d\approx0.166}$.
Refinement (optional). Retaining the exact $s^2$ coefficient and solving $1.69+9.5K_d=2\zeta\sqrt{10/(1+0.19K_d)}$ gives $K_d=0.161$ — within 3% of the approximate value, confirming the assumption.
