Question 1 of 8: Signal flow, Routh–Hurwitz, error analysis (compulsory)
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 1 — Signal flow, Routh–Hurwitz, error analysis (compulsory) [20]
Figure Q1.1 — PI servo. Forward path $K(1+0.5/s)\!\cdot\!\frac{10}{s+3}\!\cdot\!\frac{1}{s+10}\!\cdot\!\frac1s$; $Y$ is returned to both summers (inner and outer unity feedback).
Find. $G_{cl}(s)$; stable range of $K$, $K_{crit}$, $\omega_{osc}$; $K_{op}$, stability at $K_{op}$ and the gain margin.
Approach. Reduce the two-loop graph algebraically to get $G_{cl}$, apply the Routh array to its denominator for the stable gain range, then use the ramp error constant to size $K$.
Item (1) — closed-loop transfer function
Reduce the two loops. Let $e_1=R-Y$ into the PI block and $e_2=C e_1-Y$ into the plant, so $Y=P\,e_2=P\big(C(R-Y)-Y\big)$. Collecting $Y$: $Y\left(1+P+PC\right)=PC\,R$, hence $$G_{cl}=\frac{Y}{R}=\frac{PC}{1+P+PC}.$$
Substitute and clear denominators. With $PC=\dfrac{10K(s+0.5)}{s^2(s+3)(s+10)}$ and $P=\dfrac{10}{s(s+3)(s+10)}$, multiply through by $s^2(s+3)(s+10)$. The numerator is $10K(s+0.5)$ and the denominator is $s^2(s+3)(s+10)+10s+10K(s+0.5)$.
Expand. $s^2(s+3)(s+10)=s^4+13s^3+30s^2$, so $$\boxed{G_{cl}(s)=\frac{10K\,(s+0.5)}{s^4+13s^3+30s^2+(10+10K)\,s+5K}.}$$
Item (2) — stable range by Routh–Hurwitz
Routh array of $D(s)=s^4+13s^3+30s^2+(10+10K)s+5K$: $$\begin{array}{c|ccc}s^4&1&30&5K\\ s^3&13&10+10K&0\\ s^2&b_1&5K&\\ s^1&c_1&&\\ s^0&5K&&\end{array}$$ with $b_1=\dfrac{13\cdot30-(10+10K)}{13}=\dfrac{380-10K}{13}$.
$s^1$ entry. $c_1=\dfrac{b_1(10+10K)-13\cdot5K}{b_1}$; its sign is set by the numerator $-100K^2+2855K+3800$ (after multiplying by 13). Setting it to zero, $100K^2-2855K-3800=0$.
Critical gain. The positive root is $$K_{crit}=\frac{2855+\sqrt{2855^2+4(100)(3800)}}{200}=\boxed{29.8}.$$ (The $b_1\gt0$ condition only requires $K\lt38$, so $c_1$ binds first; $s^0=5K\gt0$ needs $K\gt0$.)
Oscillation frequency. At $K=K_{crit}$ the $s^2$ auxiliary row gives $b_1 s^2+5K=0$, so with $b_1=\dfrac{380-10(29.8)}{13}=6.29$: $$\omega_{osc}=\sqrt{\frac{5K_{crit}}{b_1}}=\sqrt{\frac{149.1}{6.29}}=\boxed{4.87\ \text{rad/s}}.$$
Safe range. The first column stays positive for $$\boxed{0\lt K\lt 29.8}$$ — stable there, marginally stable at $K=29.8$.
Q1 root locus (locus gain $10K$). Open-loop poles $0,-0.40,-2.46,-10.14$ and zero $-0.5$; the branch pair crosses the jw-axis at $\pm j4.87$ when $K=29.8$, confirming Routh.
Item (3) — operating gain and gain margin
Error transfer function. The actuating error is $e_1=R-Y$, and $1-G_{cl}=\dfrac{s\,(s^3+13s^2+30s+10)}{D(s)}$ — a zero at the origin, so the step error is zero automatically (spec met for any stable $K$).
Ramp error. For $R=1/s^2$, $e_{ss(ramp)}=\lim_{s\to0}s\,E_1(s)=\lim_{s\to0}\dfrac{s^3+13s^2+30s+10}{D(s)}=\dfrac{10}{5K}=\dfrac{2}{K}.$ Setting $2/K=0.1$ gives $\boxed{K_{op}=20}$.
Stability at $K_{op}$. $K_{op}=20\lt K_{crit}=29.8$, so the loop is stable. The gain margin is the factor by which $K$ may rise before instability: $$G_m=\frac{K_{crit}}{K_{op}}=\frac{29.8}{20}=\boxed{1.49}\;(3.46\ \text{dB}).$$
