Question 4 of 8: Nyquist Stability of an Open-Loop-Unstable First-Order Process
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2017 — 04-Bio-A2 Process Dynamics and Control. Three-hour open-book examination; any non-communicating calculator is permitted. The cover page states that only the first five questions in the answer book are marked, yet all eight problems are printed on the paper — all eight are solved below for completeness, since the full set is a study resource. Content spans sampled-data (digital) control, state-space-to-transfer-function modelling, Internal Model Control (IMC) of a non-minimum-phase dead-time process, Nyquist and Routh stability, frequency response (Bode/gain-margin) design and nonlinear-reactor linearisation.
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace/z-domain modelling, transfer functions, Routh and Jury stability, Nyquist/Bode frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Nyquist criterion, dead-time systems and sampled-data control. Standard control conventions (deviation variables; unity sensor/valve gains unless stated) are used throughout.
Problem 4: Nyquist Stability of an Open-Loop-Unstable First-Order Process (20% total)
Given. $G_p(s)=\dfrac{100}{s-10}$ — one open-loop right-half-plane pole at $s=+10$ (so $P=1$); proportional gain $k_c$; loop transfer $L(s)=k_cG_p(s)=\dfrac{100k_c}{s-10}$.
Find. (a) the Nyquist verdict at $k_c=1$ and $k_c=0.01$; (b) the limiting (stabilising) range of $k_c$.
Problem 4(a): Nyquist loci of $L=100k_c/(s-10)$. Each locus is a circle of diameter $[0,-10k_c]$ on the real axis (endpoints at $\omega=0$ and $\omega=\pm\infty$). For $k_c=1$ the circle spans $0$ to $-10$ and encloses $-1$ (one CCW encirclement); for $k_c=0.01$ it spans $0$ to $-0.1$ and does not reach $-1$.
Approach. Because $L(s)$ is a bilinear map of the imaginary axis, its Nyquist locus is exactly a circle with real-axis endpoints at $L(0)=-10k_c$ and $L(\pm\infty)=0$; whether $-1$ lies inside that circle is decided by a single inequality. Count encirclements with $Z=P+N$ ($P=1$ RHP pole; $N$ = clockwise encirclements, negative for counter-clockwise), then cross-check against the closed-loop characteristic equation directly.
Locus shape. $L(j\omega)=\dfrac{100k_c}{j\omega-10}=\dfrac{100k_c(-10-j\omega)}{100+\omega^2}$ has real part always negative and traces (as $\omega:-\infty\to\infty$) a circle of diameter $[0,-10k_c]$, centred at $(-5k_c,0)$ with radius $5k_c$, symmetric about the real axis.
(a) $k_c=1$. Circle centre $-5$, radius $5$, spanning $0$ to $-10$. The point $-1$ satisfies $|-1-(-5)|=4<5$, so it lies inside the circle: the locus encircles $-1$ once, counter-clockwise ($N=-1$). With $P=1$, $$Z=P+N=1+(-1)=0\ \Rightarrow\ \boxed{\text{stable at }k_c=1.}$$
(a) $k_c=0.01$. Circle centre $-0.05$, radius $0.05$, spanning $0$ to $-0.1$ — far too small to reach $-1$. No encirclement ($N=0$), so $$Z=P+N=1+0=1\ \Rightarrow\ \boxed{\text{one closed-loop RHP pole; unstable at }k_c=0.01.}$$
Direct check. The characteristic equation $1+L(s)=0$ gives $(s-10)+100k_c=0$, i.e. a single closed-loop pole at $s=10-100k_c$. At $k_c=1$: $s=-90<0$ (stable). At $k_c=0.01$: $s=+9>0$ (unstable). Both agree exactly with the Nyquist verdicts above.
(b) Limiting $k_c$. The point $-1$ is inside the circle of diameter $[0,-10k_c]$ (i.e. $|-1-(-5k_c)|<5k_c$) exactly when $10k_c>1$, matching the pole condition $10-100k_c<0$: $$\boxed{k_c>\frac{1}{10}=0.1.}$$ There is no upper bound — any gain above the critical value $k_c=0.1$ keeps the single closed-loop pole in the left-half plane.
