Question 1 of 8: Nyquist stability of a proportional loop (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 2014, 3 hours, closed book (approved calculator plus one signed 8.5 × 11" formula sheet). Questions 1 and 2 are compulsory; answer three of the remaining six (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 and stability (Ch. 6), steady-state error and error constants (Ch. 7), root locus (Ch. 8), PID and lag/lead design (Ch. 9–11), frequency response, Nyquist and gain/phase margins (Ch. 10–11), state space, controllability/observability and pole placement (Ch. 3, 12); K. Ogata, Modern Control Engineering (5th ed., Prentice Hall) — companion treatment of dominant-poles modelling and Mason’s rule; G. F. Franklin, J. D. Powell & A. Emami-Naeini, Feedback Control of Dynamic Systems. All block diagrams, pole–zero maps, root loci, Bode/Nyquist plots and signal-flow graphs below are redrawn as inline figures.
Reading the exam figures. Q1 supplies an open-loop Bode plot and Q6 supplies open-loop Bode plots of the process. Where a transfer function is given exactly in the text (Q1’s system is the same plant as Q2, stated exactly there), every graphical reading is also confirmed analytically and the exact model governs. Q6 is a pure graph-reading problem (no closed form given); its numeric reads are approximate and are flagged accordingly. Note the printed exam mislabels the header of the last two pages “May 2013” — this is the May 2014 sitting throughout.
Question 1 — Nyquist stability of a proportional loop (compulsory) [20]
Given. Unit-feedback loop, open-loop $L(s)=K_p\,G(s)$. The DC gain fixes $G(0)=83.33$. The plant is stated exactly in Question 2 (the two questions share one system): $G(s)=\dfrac{(s+40)(s+50)}{(s+2)^2(s+6)}$ — three poles at $-2,-2,-6$, two zeros at $-40,-50$; check $G(0)=\tfrac{40\cdot50}{2\cdot2\cdot6}=\tfrac{2000}{24}=83.33$ ✓.
Find. $K_{crit}$ and $\omega_{osc}$ at marginal stability, and the safe range(s) of $K_p$, by the Nyquist criterion (confirmed by Routh–Hurwitz in Q2).
[Figure not reproduced: Figure Q1.1 (redrawn) — the unit-feedback proportional loop. See the official exam paper.]
[Figure not reproduced: Figure Q1.2 (redrawn) — open-loop Bode plot of $G(s)$ at $K_p=1$. The double pole at $-2$ drives the phase below $-180^\circ$; the zeros at $-40,-50$ pull it back up so the phase crosses $-180^\circ$ twice before settling at $-90^\circ$ (relative degree $1$). See the official exam paper.]
Approach. Convert the frequency response into a polar plot; each crossing of the negative real axis (open-loop phase $=-180^\circ$) is a candidate marginal-stability point with $K_{crit}=1/|G(j\omega_{pc})|$. Because the open loop has no RHP poles ($P=0$), the closed loop is stable whenever the Nyquist contour makes zero net encirclements of the $-1/K_p$ point.
Polar plot of $G(j\omega)$ at $K_p=1$ (not to scale). It starts on the positive real axis at the DC gain $+83.33$ and ends at the origin approaching along $-90^\circ$ (relative degree 1).
Zoom near the origin: the locus crosses the negative real axis at $-4.10$ (at $\omega=7.07$ rad/s) and at $-0.0857$ (at $\omega=32.8$ rad/s).
Phase-crossover frequencies (real-axis crossings). Evaluating $G(j\omega)$, its imaginary part vanishes at two frequencies: $$\omega_{pc,1}=7.07\ \text{rad/s}\ \ (G=-4.10),\qquad \omega_{pc,2}=32.8\ \text{rad/s}\ \ (G=-0.0857).$$ These are the two points where the polar plot cuts the negative real axis.
Critical gains. Marginal stability occurs when $K_p\,|G(j\omega_{pc})|=1$, i.e. the crossing lands exactly on $-1$: $$K_{crit,1}=\frac{1}{4.10}=\boxed{0.244},\qquad K_{crit,2}=\frac{1}{0.0857}=\boxed{11.67}.$$ Correspondingly $\omega_{osc,1}=7.07$ rad/s and $\omega_{osc,2}=32.8$ rad/s.
Encirclement / safe-range reasoning. With $P=0$ open-loop RHP poles, we need zero encirclements of $-1/K_p$. The two real-axis crossings sit at $-4.10$ and $-0.0857$, so the segment $[-4.10,\,-0.0857]$ of the negative real axis is “inside” the locus. The point $-1/K_p$ avoids that segment when $-1/K_p\lt-4.10$ or $-1/K_p\gt-0.0857$:
Safe operating ranges. $-1/K_p\lt-4.10\Rightarrow K_p\lt0.244$, and $-1/K_p\gt-0.0857\Rightarrow K_p\gt11.67$. Hence $$\boxed{0\lt K_p\lt0.244\quad\text{or}\quad K_p\gt11.67.}$$ The loop is unstable in between ($0.244\lt K_p\lt11.67$) — an unusual “stable–unstable–stable” gain map produced by the phase dipping below and then recovering past $-180^\circ$.
