Question 5 of 8: Nyquist Stability of a Proportionally-Controlled Open-Loop-Unstable Process

Problem 5: Nyquist Stability of a Proportionally-Controlled Open-Loop-Unstable Process (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.

Given. $G_p(s)=\dfrac{20}{s-3}$ — an open-loop-unstable first-order process (pole at $s=+3$); proportional controller $K_c$; unity feedback assumed.

Find. (a) the qualitative Nyquist plot of $L(j\omega)=K_cG_p(j\omega)$ at $K_c=1$, key points, and the closed-loop stability verdict; (b) the range of $K_c$ for closed-loop stability by the Nyquist criterion.

Approach. Because $L(s)$ has a pole in the right half-plane ($P=1$), the standard "stable iff no encirclements" rule does not apply directly — use the general Nyquist criterion $Z=N+P$ (closed-loop RHP poles $=$ net clockwise encirclements of $-1$, plus open-loop RHP poles) and require $Z=0$. $L(j\omega)$ is a Möbius transform of $j\omega$, so its image is exactly a circle; locate that circle from its two easy points ($\omega=0$ and $\omega\to\infty$) and test whether $-1$ lies inside it.

1. (a) Frequency response and key points. $L(j\omega)=\dfrac{20K_c}{j\omega-3}=\dfrac{-60K_c-20K_cj\omega}{9+\omega^2}$. At $\omega=0$: $L=-\dfrac{20K_c}{3}$ (real, negative). As $\omega\to\pm\infty$: $L\to0$. Both real and imaginary parts are negative for all $\omega>0$, so the curve for $\omega:0\to\infty$ sweeps through the third quadrant from $(-20K_c/3,0)$ back to the origin; by symmetry ($L(-j\omega)=\overline{L(j\omega)}$) the $\omega:-\infty\to0$ branch sweeps the second quadrant. The full locus is the circle $$\left|L+\frac{10K_c}{3}\right|=\frac{10K_c}{3}\qquad(\text{centre }-\tfrac{10K_c}{3},\ \text{radius }\tfrac{10K_c}{3}),$$ found from its two real-axis crossings $(-\tfrac{20K_c}{3},0)$ and $(0,0)$.
2. (a) $K_c=1$ key points. Centre $(-3.333,0)$, radius $3.333$: $\omega=0\Rightarrow(-6.667,0)$; $\omega=3\Rightarrow L(j3)=20/(-3+3j)$, $|L|=4.714$ at $-135^{\circ}$, i.e. $(-3.333,-3.333)$; $\omega\to\infty\Rightarrow(0,0)$.
3. (a) Encirclement count and verdict. Distance from the circle's centre to $-1$ is $|-1-(-3.333)|=2.333<3.333=$ radius, so $-1$ lies inside the circle — encircled exactly once, and tracing the locus as $\omega$ increases through $0\to+\infty\to$(via $-\infty\to0$) shows the traversal is counter-clockwise (CCW), i.e. $N=-1$ (one CCW $=$ $-1$ clockwise encirclement in the $N$-counts-clockwise convention). With $P=1$ (one open-loop RHP pole at $s=+3$): $$\boxed{Z=N+P=-1+1=0\ \Rightarrow\ \text{closed loop is STABLE at }K_c=1.}$$ (Directly confirmed: char. eq. $s-3+20K_c=0\Rightarrow s=3-20(1)=-17<0$.)
4. (b) General $K_c$: circle scales with gain. For general $K_c>0$, the circle has centre $-\tfrac{10K_c}{3}$, radius $\tfrac{10K_c}{3}$ (same relation, scaled by $K_c$). The point $-1$ is inside this circle iff $\left|{-1}+\tfrac{10K_c}{3}\right|<\tfrac{10K_c}{3}$, which reduces (for $K_c>0$) to $K_c>\tfrac{3}{20}=0.15$: below this, the circle's left edge $-\tfrac{20K_c}{3}$ does not reach as far left as $-1$, so $-1$ is outside (not encircled), $N=0$, $Z=N+P=0+1=1$ — one unstable closed-loop pole.
5. (b) Direct confirmation via characteristic equation. $1+K_cG_p=0\Rightarrow s-3+20K_c=0\Rightarrow s=3-20K_c$. Requiring $s<0$: $$\boxed{K_c>\frac{3}{20}=0.15\ \text{for closed-loop stability.}}$$ This matches the Nyquist-criterion boundary exactly.