Question 2 of 8: Nyquist stability, two loops (compulsory)

Question 2 — Nyquist stability, two loops (compulsory) [20]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part A — same plant as Q1

Given. $G_{open}(s)/K=\dfrac{10(s+10)}{(s+4)(s^2+4s+8)}$, open-loop poles $-4,\,-2\pm j2$ (all in the LHP, so $P=0$). Find. the polar-plot crossovers and the stable range of $K_p$.

1. Start and end. At $\omega=0$, $G/K=\dfrac{10(10)}{4\cdot 8}=3.125$ (positive real axis). As $\omega\to\infty$ the degree excess is $2$, so the locus tends to the origin at $-180^\circ$.
2. Real-axis crossover. Setting $\operatorname{Im}=0$ gives $\omega=10.2$ rad/s, where $G/K=-0.125=-1/8$. This is the phase-$-180^\circ$ point.
3. Imaginary-axis crossover. $\operatorname{Re}=0$ near $\omega\approx2.29$ rad/s, where $G/K\approx -j2.33$.
4. Nyquist criterion. With $P=0$ open-loop RHP poles, stability requires $N=0$ encirclements of $-1$ by $K_p\,(G/K)$. The critical point is enclosed once the real-axis crossing passes $-1$: $K_p\cdot(-0.125)=-1\Rightarrow K_{p,crit}=8$. Hence $$\boxed{0\lt K_p\lt 8}$$ — identical to Q1, as expected for the same loop.

Part B — open-loop-unstable plant $G_{open}(s)=K\dfrac{1+s}{s(s-2)}$

Given. One pole at the origin and one right-half-plane pole at $s=+2$, so $P=1$. One zero at $-1$. Find. the axis crossovers of $G_{open}(j\omega)/K$ and the stable range of $K_p$.

1. Split into real and imaginary parts. With $s=j\omega$, $\dfrac{1+j\omega}{j\omega(j\omega-2)}=\dfrac{-3\omega^2+j\,\omega(2-\omega^2)}{\omega^4+4\omega^2}$, so $$\operatorname{Re}=\frac{-3}{\omega^2+4},\qquad \operatorname{Im}=\frac{2-\omega^2}{\omega(\omega^2+4)}.$$
2. Real-axis crossover. $\operatorname{Im}=0\Rightarrow \omega^2=2$, $\omega=\sqrt2$ rad/s. There $\operatorname{Re}=-3/(2+4)=\boxed{-0.5}$.
3. Low-frequency asymptote. As $\omega\to0^{+}$, $\operatorname{Re}\to-0.75$ and $\operatorname{Im}\to+\infty$: the plot comes down from $+j\infty$ along the line $\operatorname{Re}=-0.75$, crosses the real axis at $-0.5$, and runs to the origin from the third quadrant.
Part B polar plot (solid $\omega\gt0$, dashed mirror). The one RHP pole requires one CCW encirclement of $-1$ for stability; that happens once $-1$ lies to the right of the $-0.5$ crossing, i.e. for $K_p\gt2$.
4. Nyquist criterion with $P=1$. For a CW $\Gamma$-path, closed-loop stability needs $Z=N+P=0$, i.e. $N=-1$ (one counter-clockwise encirclement of $-1$). The complete contour (adding the infinite arc from the origin pole) encircles $-1$ once CCW only when the scaled real-axis crossing $-0.5K_p$ has moved out past the critical point, i.e. lies to the left of $-1$: $K_p\cdot(-0.5)\lt-1\Rightarrow K_p\gt2$.
5. Cross-check by the characteristic equation. $s(s-2)+K_p(1+s)=s^2+(K_p-2)s+K_p=0$ is Hurwitz iff both $K_p-2\gt0$ and $K_p\gt0$, i.e. $$\boxed{K_p\gt2}$$ with marginal oscillation at $K_p=2$, $\omega_{osc}=\sqrt2$ rad/s. This is a lower-gain stability limit: too little gain cannot stabilize the RHP pole.