Question 1 of 8: Basic definitions and concepts of control (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 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 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 specs (Ch. 4), block reduction (Ch. 5), Routh–Hurwitz (Ch. 6), steady-state error and static error constants (Ch. 7), root locus (Ch. 8), PID and lead–lag design (Ch. 9–11), frequency response, Bode, Nyquist, gain/phase margins (Ch. 10–11), state space, controllability/observability (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling. All step responses, block diagrams, pole–zero maps, root loci, the Bode sketches and the Nyquist plot 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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$; the figures on the exam paper (step responses, root loci, Bode plots) are read graphically, so the values below carry normal chart-reading tolerance.
Question 1 — Basic definitions and concepts of control (compulsory) [20]
[Figure not reproduced: Q1(1)–(2): unit-step response redrawn; peak $\approx1.23$, steady state $\approx0.90$, $\pm5\%$ band shown. See the official exam paper.]
Items (1) and (2) — reading the step response
Percent overshoot. Peak $\approx1.23$ over steady value $y_{ss}\approx0.90$: $PO=\dfrac{1.23-0.90}{0.90}\times100\approx \boxed{37\%}$.
Steady-state error. Unit reference, $y_{ss}\approx0.90$, so $e_{ss(step)}=1-0.90=\boxed{10\%}$.
Settling time (5%). The response last leaves the $0.855$–$0.945$ band near the second peak: $T_{settle(5\%)}\approx \boxed{1.0\text{ s}}$.
Damping ratio. Inverting the $PO$ chart at $37\%$: $\zeta=\dfrac{-\ln(PO/100)}{\sqrt{\pi^2+\ln^2(PO/100)}}\approx \boxed{0.30}$.
Natural frequency. Peak time $T_p\approx0.33$ s gives $\omega_d=\pi/T_p\approx9.5$, so $\omega_n=\omega_d/\sqrt{1-\zeta^2}\approx \boxed{10\text{ rad/s}}$; check $T_{settle}=3/(\zeta\omega_n)\approx1.0$ s. DC gain $K_{dc}=y_{ss}/r=\boxed{0.90}$.
Items (3) and (4) — $G(s)=22.5/(s^2+s+25)$
Standard form. $s^2+2\zeta\omega_n s+\omega_n^2$ with $\omega_n^2=25\Rightarrow\omega_n=5$, $2\zeta\omega_n=1\Rightarrow \boxed{\zeta=0.10}$, and $K_{dc}=22.5/25=\boxed{0.90}$; $\omega_n=\boxed{5\text{ rad/s}}$.
Q1(5): Type-0 unit-feedback loop with the disturbance injected between $G_1$ and $G_2$.
Item (5) — steady-state output with reference and disturbance
DC gains. $G_1(0)=\tfrac{3}{2}=1.5$, $G_2(0)=\tfrac{3}{1}=3$, so the forward DC gain is $G_1G_2(0)=4.5$.
Superpose the two inputs. Reference: $\dfrac{G_1G_2}{1+G_1G_2}\Big|_0=\dfrac{4.5}{5.5}=\tfrac{9}{11}$. Disturbance (enters after $G_1$): $\dfrac{G_2}{1+G_1G_2}\Big|_0=\dfrac{3}{5.5}=\tfrac{6}{11}$.
(6) Match $G(s)$ (no calculation). The response contains a ramp term $0.0833\,t$ and a constant (so $Y(s)$ has a $1/s^2$ term $\Rightarrow$ $G$ has a pole at the origin), plus $t\,e^{-2t},e^{-2t}$ (double pole at $-2$) and $t\,e^{-3t},e^{-3t}$ (double pole at $-3$). Only $\boxed{G(s)=\dfrac{3}{s(s+2)^2(s+3)^2}}$ has that pole pattern.
(7) Ideal tracking under P-control. Zero tracking error needs the loop gain $\to\infty$: the open-loop gain should be as close to infinity as possible.
(8) Type and error constants. Recover the loop from the unit-feedback closed loop: $L=\dfrac{T}{1-T}$. Here $1-T=\dfrac{s^4+6s^3+7s^2}{\text{den}}=\dfrac{s^2(s^2+6s+7)}{\text{den}}$, so $L(s)=\dfrac{5s^2+10s+3}{s^2(s^2+6s+7)}$ — two poles at the origin, $\boxed{N=2\text{ (Type 2)}}$. Then $K_{pos}=\infty$, $K_{vel}=\infty$, $K_{acc}=\lim_{s\to0}s^2L=\tfrac{3}{7}=\boxed{0.43}$; $e_{ss(step)}=0$, $e_{ss(ramp)}=0$, $e_{ss(parab)}=1/K_{acc}=\tfrac{7}{3}=\boxed{2.33}$.
(9) Ideal disturbance rejection — open-loop gain. High loop gain divides down the disturbance: the open-loop gain should be as close to infinity as possible.
(10) Ideal disturbance rejection — closed-loop gain. The disturbance-to-output closed-loop gain should be driven to zero: the closed-loop gain should be as close to zero as possible.
