Question 4 of 8: Polar plot and Nyquist stability, RHP-pole plant
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 $PO$.
Given. $G_{open}/K=\dfrac{1+s}{s(s-1)}$: a pole at the origin, a right-half-plane pole at $s=+1$ ($P=1$) and a zero at $-1$. Find. the axis crossovers and the stable range of $K_p$.
Q4 open-loop singularities: integrator pole at the origin and an unstable pole at $+1$ (so $P=1$ RHP poles), with a zero at $-1$ — the loop cannot be stabilized by too small a gain.
Item (1) — polar plot
Real and imaginary parts. With $s=j\omega$, $j\omega(j\omega-1)=-\omega^2-j\omega$, so $\dfrac{1+j\omega}{-\omega^2-j\omega}$ rationalizes to $$\operatorname{Re}=\frac{-2}{\omega^2+1},\qquad \operatorname{Im}=\frac{1-\omega^2}{\omega(\omega^2+1)}.$$
Real-axis crossover. $\operatorname{Im}=0\Rightarrow \omega=1$ rad/s; there $\operatorname{Re}=\dfrac{-2}{2}=\boxed{-1}$ (so the actual plot crosses at $-K$).
Low-frequency asymptote. As $\omega\to0^{+}$, $\operatorname{Re}\to-2$ and $\operatorname{Im}\to+\infty$: the curve descends from $+j\infty$ along the vertical asymptote $\operatorname{Re}=-2$.
Shape. For $0\lt\omega\lt1$ the plot is in the second quadrant ($\operatorname{Re}\lt0,\operatorname{Im}\gt0$); it crosses the real axis at $-1$ at $\omega=1$, then for $\omega\gt1$ enters the third quadrant and spirals into the origin as $\omega\to\infty$.
Polar plot of $G_{open}(j\omega)/K$ ($\omega\gt0$ solid): down the $\operatorname{Re}=-2$ asymptote, through the real-axis crossing $-1$ at $\omega=1$, then into the origin. Arrow shows increasing $\omega$.
Item (2) — Nyquist criterion
Contour. Take the CW $\Gamma$-path enclosing the entire RHP, indented to the right around the origin pole. The full Nyquist contour is the $\omega\!:\!-\infty\to\infty$ image plus the large arc from the origin detour.
Stability count. With $P=1$ open-loop RHP pole, $Z=N+P$ needs $N=-1$: exactly one counter-clockwise encirclement of $-1$. That occurs only when the $-1$ point lies to the right of the $-K$ real-axis crossing, i.e. when $K\cdot(-1)\lt-1$, so $K\gt1$.
Confirm by Routh. The characteristic equation is $s(s-1)+K(s+1)=s^2+(K-1)s+K=0$; a 2nd-order polynomial is Hurwitz iff every coefficient is positive: $K-1\gt0$ and $K\gt0$. Hence $$\boxed{K_p\gt1}$$ with marginal oscillation at $K_p=1$. Too little gain leaves the RHP pole uncontrolled — a lower-gain stability limit, the opposite of the usual upper bound.
