23-Chem-A6 Process Dynamics and Control · December 2016
Question 8 of 8: Nyquist Stability of an Open-Loop-Unstable Process
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order thermal/level dynamics, step/ramp/pulse response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This December 2016 sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 8: Nyquist Stability of an Open-Loop-Unstable Process (20%)
Given. $G_p=10/(s-5)$ — a single open-loop-unstable pole at $s=+5$ ($P=1$ RHP pole), proportional control $L=k_cG_p$.
Find. (a) Nyquist sketch and stability at $k_c=1$; (b) the stabilising range of $k_c$.
Nyquist plot of $L=10/(s-5)$ at $k_c=1$: a circle (centre $-1$, radius $1$) lying entirely in the left half-plane, from $L(0)=-2$ through $L(j5)=-1-j$ toward the origin as $\omega\to\infty$. Because the plot encircles the critical point $-1$ once counter-clockwise, it cancels the one RHP open-loop pole and the closed loop is stable.
Approach. With one RHP open-loop pole ($P=1$), the Nyquist criterion requires exactly one counter-clockwise encirclement of $-1$ ($N=1$) so that $Z=P-N=0$. Locate the real-axis crossing and set the condition for encirclement.
Key points at $k_c=1$. $L(j\omega)=\dfrac{10}{j\omega-5}$. $\omega=0\Rightarrow L=-2$; $\omega=5\Rightarrow L=\dfrac{10}{5(j-1)}=-1-j$; $\omega\to\infty\Rightarrow L\to0$. The locus is a circle of centre $-1$ and radius $1$ passing through $(-2,0)$ and the origin, traced through the third quadrant for $\omega>0$ (and its mirror image in the second quadrant for $\omega<0$).
(a) Encirclement & verdict. The plot passes to the left of $-1$ (real crossing at $-2$), so it encircles $-1$ once counter-clockwise: $N=1$. With $P=1$, $Z=P-N=0$ — no closed-loop RHP poles, so $$\boxed{\text{the system is STABLE at }k_c=1.}$$ (Check: characteristic equation $s-5+10k_c=0\Rightarrow s=-5<0$.)
(b) Real-axis crossing vs $k_c$. The locus meets the real axis at $\omega=0$ with value $L(0)=-2k_c$. To enclose $-1$ (the one CCW encirclement needed) requires $-2k_c<-1$.
(b) Stability range. $-2k_c<-1\Rightarrow$ $$\boxed{k_c>0.5.}$$ For $k_c<0.5$ the crossing lies to the right of $-1$, giving $N=0$, $Z=1$ — the closed loop keeps the RHP pole and is unstable. Equivalently the characteristic root $s=5-10k_c<0$ requires $k_c>0.5$.
