Nyquist Stability of an Open-Loop-Unstable First-Order Process: 23-Chem-A6 Process Dynamics and Control, December 2015, Q7

Problem 7: Nyquist Stability of an Open-Loop-Unstable First-Order 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. $G_p=\dfrac{20}{s-3}$ — a single right-half-plane pole at $s=+3$, so the open loop has $P=1$ unstable pole; $L(s)=k_cG_p=\dfrac{20k_c}{s-3}$.

Find. (a) the Nyquist verdict at $k_c=1$; (b) the stabilising range of $k_c$.

Approach. Plot a few points of $L(j\omega)$ to fix the locus, count encirclements of $-1$, apply $Z=P+N$ (with $N$ the clockwise encirclements), and cross-check against the characteristic equation.

Key points ($k_c=1$). $L(j\omega)=\dfrac{20}{j\omega-3}=\dfrac{20(-3-j\omega)}{9+\omega^2}$. At $\omega=0$: $L=-\tfrac{20}{3}=-6.67$ (negative real axis). At $\omega=3$: $L=\dfrac{20}{-3+3j}=-3.33-3.33j$. As $\omega\to\infty$: $L\to0$ from the third quadrant. The full locus ($\omega:-\infty\to\infty$) is a circle centred at $(-3.33,0)$ of radius $3.33$, passing through $-6.67$ and the origin.
(a) Encirclement and stability. The critical point $-1$ lies inside that circle (between $-6.67$ and $0$), so the locus encircles $-1$ once counter-clockwise: $N=-1$. With $P=1$ open-loop RHP pole, $$Z=P+N=1+(-1)=0,$$ so there are no closed-loop RHP poles — the loop is stable at $k_c=1$. (The single RHP pole demands exactly one CCW encirclement, which the locus supplies.)
Direct check. Characteristic equation $1+L=0\Rightarrow s-3+20k_c=0\Rightarrow s=3-20k_c$. For $k_c=1$, $s=-17<0$ — stable, confirming $Z=0$. ✓
(b) Range of $k_c$. The real-axis crossing at $\omega=0$ is $-20k_c/3$. The locus encircles $-1$ (giving $Z=0$) only when this crossing lies to the left of $-1$: $\dfrac{20k_c}{3}>1$. Equivalently the closed-loop pole $s=3-20k_c$ is in the LHP when $$\boxed{k_c>\frac{3}{20}=0.15}.$$ For $0can stabilise this plant — unlike a process with two mismatched RHP contributions — because there is only one pole to drive into the LHP.

Problem 7(a): Nyquist plot of $L=20/(s-3)$ at $k_c=1$ — a circle from $-6.67$ ($\omega=0$) through $(-3.33,-3.33)$ to the origin. It encircles $-1$ once counter-clockwise ($N=-1$); with $P=1$, $Z=P+N=0$, so the closed loop is stable.

Item	Result
Open-loop RHP poles	$P=1$ (at $s=+3$)
$L(0)$ at $k_c=1$	$-6.67$ (encircles $-1$)
Encirclements / $Z$ at $k_c=1$	$N=-1$, $Z=P+N=0$ → stable
Stabilising range	$k_c>3/20=0.15$

Concept & Theory

Topic. Nyquist criterion for an open-loop-unstable process; encirclement counting; when proportional gain alone stabilises.

Key relations. $Z=P+N$ ($N=$ clockwise encirclements of $-1$; CCW counts negative); stability $\Leftrightarrow Z=0$; $L(0)=-20k_c/3$; closed-loop pole $s=3-20k_c$.

Why this works. The Nyquist map turns $1+L=0$ into an encirclement count; an RHP open-loop pole ($P=1$) must be cancelled by one CCW encirclement for closed-loop stability. Increasing $k_c$ scales the locus radially outward, and once the $\omega=0$ crossing passes $-1$ the required encirclement appears — exactly the $k_c>0.15$ boundary the algebra gives.

Common pitfall. Forgetting $P=1$ and calling the encircling locus “unstable”; mixing the sign convention for $N$; or assuming an unstable open-loop plant can never be fixed by proportional control (a single RHP pole often can).

Source. Stephanopoulos, Ch. 17–18 (Nyquist criterion, unstable open-loop systems); Seborg et al., Ch. 14 (frequency-response stability).

Question 7 of 8: Nyquist Stability of an Open-Loop-Unstable First-Order Process

Problem 7: Nyquist Stability of an Open-Loop-Unstable First-Order Process (20%)