Question 2 of 8: Nyquist Stability of a Proportionally-Controlled Open-Loop-Unstable Process

Problem 2: Nyquist Stability of a Proportionally-Controlled Open-Loop-Unstable Process (20% total)

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. $G_p(s)=\dfrac{20}{s-2}$ — an open-loop-unstable first-order process (pole at $s=+2$); proportional controller $K_c$; unity feedback assumed.

Find. (a) the qualitative Nyquist plot of $L(j\omega)=K_cG_p(j\omega)$ at $K_c=0.2$, its general shape/direction, 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 one right-half-plane pole ($P=1$), the general Nyquist criterion $Z=N+P$ (closed-loop RHP poles $=$ net clockwise encirclements of $-1$ plus open-loop RHP poles) must be used instead of the "stable iff zero encirclements" shortcut; $L(j\omega)$ is a Möbius transform of $j\omega$ so its locus is exactly a circle, located from its real-axis crossings at $\omega=0$ and $\omega\to\infty$.

1. (a) Frequency response and circle geometry. $L(j\omega)=\dfrac{20K_c}{j\omega-2}=\dfrac{-40K_c-20K_cj\omega}{4+\omega^2}$, which traces the circle $$\left|L+\frac{10K_c}{2}\right|=\frac{10K_c}{2}\ \Longrightarrow\ \text{centre }-5K_c,\ \text{radius }5K_c$$ (found from the two real-axis crossings $(-10K_c,0)$ at $\omega=0$ and $(0,0)$ as $\omega\to\infty$).
2. (a) $K_c=0.2$ key points. Centre $(-1,0)$, radius $1$: $\omega=0\Rightarrow(-2,0)$; $\omega=2\Rightarrow L(j2)=4/(j2-2)$, giving $(-1,-1)$; $\omega\to\infty\Rightarrow(0,0)$. The locus is traced clockwise for $\omega:0\to+\infty$ (moving from $(-2,0)$ down through $(-1,-1)$ toward the origin), with the mirror-image branch for $\omega:-\infty\to0$ tracing the upper half — net one counter-clockwise encirclement of $-1$ when the full contour ($\omega:-\infty\to+\infty$) is read in the standard order.
3. (a) Encirclement count and verdict. The circle is centred exactly on $-1$ with radius $1>0$, so $-1$ lies strictly inside it and is encircled once: $N=-1$ (one CCW $=-1$ clockwise encirclement). With $P=1$ open-loop RHP pole ($s=+2$): $$\boxed{Z=N+P=-1+1=0\ \Rightarrow\ \text{closed loop is STABLE at }K_c=0.2.}$$ (Direct check: characteristic equation $s-2+20K_c=0\Rightarrow s=2-20(0.2)=-2<0$ — confirmed stable.)
4. (b) General $K_c$: circle scales with gain. For general $K_c>0$ the circle has centre $-5K_c$, radius $5K_c$ (same relation, scaled by $K_c$). The point $-1$ is inside this circle iff $|-1+5K_c|<5K_c$.
5. (b) Solve the encirclement condition, confirm via characteristic equation. For $5K_c\ge1$ the condition $|5K_c-1|<5K_c$ holds automatically; for $5K_c<1$ it reduces to $1-5K_c<5K_c\Rightarrow K_c>0.1$. So $-1$ is encircled ($N=-1$, $Z=N+P=0$, stable) exactly when $K_c>0.1$. Direct confirmation: $s=2-20K_c<0\Rightarrow$ $$\boxed{K_c>\frac{2}{20}=0.1\ \text{for closed-loop stability.}}$$ This matches the Nyquist-geometry boundary exactly, and confirms part (a): $K_c=0.2>0.1$ is comfortably inside the stable range.