Question 3 of 8: Analytical step response by partial fractions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, 16-Elec-A2 Systems & Control, May 2017, 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 $\zeta$–overshoot and $\zeta$–resonant-peak design plots are supplied on pages 2–3). 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, Nyquist, gain/phase margins (Ch. 10), PID and compensator design (Ch. 9–11), state space, controllability/observability and Mason’s rule (Ch. 3, 5, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — dominant-poles modelling, second-order correlations and pole-placement. All block diagrams, root loci, pole–zero maps, Nyquist plots, step responses and the state-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 $G_m(s)=K_{dc}\,\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Every result here is derived analytically in the $s$-domain and cross-checked; the design charts are used only to corroborate $\zeta$ from $PO$.
Question 3 — Analytical step response by partial fractions [20]
Given. $G(s)=\dfrac{7s(s+3)}{(s+0.5)(s^2+1.6s+16)}$, $R(s)=1/s$. Find. $y(t)$. Approach. form $Y=GR$, cancel the input pole against the numerator $s$, expand in partial fractions and invert term by term.
Q3 pole–zero map of $Y(s)$: a real pole $-0.5$ and the complex pair $-0.8\pm j3.92$; the zero at $-3$ shapes the transient (the input pole at the origin cancels the plant zero at the origin).
Form and simplify $Y(s)$. $Y=\dfrac{7s(s+3)}{s(s+0.5)(s^2+1.6s+16)}=\dfrac{7(s+3)}{(s+0.5)(s^2+1.6s+16)}$ — the origin pole of the step cancels the plant zero at the origin.
Partial fractions. Write $Y=\dfrac{A}{s+0.5}+\dfrac{Bs+C}{s^2+1.6s+16}$. Residue at $-0.5$: $A=\dfrac{7(2.5)}{(-0.5)^2+1.6(-0.5)+16}=\dfrac{17.5}{15.45}=\boxed{1.133}$.
Remaining coefficients. Matching $s^2$: $0=A+B\Rightarrow B=-1.133$; matching the constant: $21=16A+0.5C\Rightarrow C=\boxed{5.754}$ (the $s^1$ identity $7=1.6A+0.5B+C$ checks).
Complete the square. $s^2+1.6s+16=(s+0.8)^2+15.36$, so $a=0.8$, $\omega_d=\sqrt{15.36}=3.92$ rad/s. Rewrite the quadratic term as $-1.133\dfrac{s+0.8}{(s+0.8)^2+15.36}+\dfrac{6.660}{(s+0.8)^2+15.36}$, where the sine coefficient over $\omega_d$ is $6.660/3.92=1.699$.
Invert. $$\boxed{y(t)=1.133\,e^{-0.5t}+e^{-0.8t}\big[-1.133\cos(3.92t)+1.699\sin(3.92t)\big].}$$ The oscillatory part has amplitude $\sqrt{1.133^2+1.699^2}=2.04$, so $y(t)=1.133e^{-0.5t}+2.04\,e^{-0.8t}\sin(3.92t-\phi)$, $\phi=\tan^{-1}(1.133/1.699)$.
Sanity checks. $y(0)=1.133-1.133=0$ (strictly proper $Y$), and $y(\infty)=0$ because $G$ has a zero at the origin (zero DC gain) — the step produces a decaying transient that returns to zero.
Analytical unit-step response $y(t)$: rises to a peak near $t\approx0.4$ s then rings out to zero (the two decays $e^{-0.5t}$ and $e^{-0.8t}$ set the envelope).
