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23-Chem-A6 Process Dynamics and Control · December 2014

Question 4 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

National Exams / EGBC — December 2014 — 04-Chem-A6 Process Dynamics & Control. Three-hour open-book examination; any non-communicating calculator is permitted. Eight problems are printed and any five constitute a complete paper (each worth 20%); all eight are solved below for completeness. The parts are quantitative throughout — two-capacitance thermal modelling, state-space transfer functions, IMC design with a right-half-plane zero, Nyquist stability of an open-loop-unstable plant, ultimate gain with sensor dead time, second-order damping regimes, a non-linear CSTR, and Bode gain-margin design — and every requested plot (step response, IMC servo response, Nyquist locus, damping family, Bode diagram) is drawn as a real figure.

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, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Nyquist and Bode stability, dead-time systems and controller design; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first- and second-order dynamics, linearisation of non-linear balances. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout.

Problem 4: Nyquist Stability of an 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. $G_p=\dfrac{100}{s-10}$ — a single right-half-plane pole at $s=+10$ ($P=1$), proportional control $K_c$.

Find. (a) stability at $K_c=1$ and $0.01$; (b) the limiting stabilising gain.

Approach. Check the closed-loop pole directly, then interpret with the Nyquist criterion $Z=N+P$ (need $Z=0$, so one CCW encirclement of $-1$ since $P=1$).

  1. Closed-loop pole (direct check). $1+K_c\dfrac{100}{s-10}=0\Rightarrow s=10-100K_c$. So $K_c=1\Rightarrow s=-90$ (stable); $K_c=0.01\Rightarrow s=+9$ (unstable).
  2. (a) Nyquist interpretation. $L(j\omega)=\dfrac{100K_c}{j\omega-10}$ has $P=1$ RHP pole, so stability needs exactly one CCW encirclement of $-1$ ($N=-1$). The locus crosses the real axis at $\omega=0$, where $L=-10K_c$. For $K_c=1$ the crossing is $-10$ (left of $-1$): the locus encircles $-1$ once CCW $\Rightarrow$ stable. For $K_c=0.01$ the crossing is $-0.1$ (right of $-1$): no encirclement $\Rightarrow$ unstable.
  3. (b) Limiting gain. Encirclement requires the $\omega=0$ crossing to lie left of $-1$: $-10K_c<-1$, i.e. $$\boxed{K_c>0.1.}$$ This is a minimum stabilising gain, not an upper limit — an unstable plant needs enough loop gain before feedback can pull the RHP pole into the left half plane.
−1 ω=0: -10 Re L(jω) Im L(jω)P4: Nyquist locus, L=100/(s−10), K_c=1 (−1 encircled → stable)
Problem 4(a): Nyquist locus of $L=100/(s-10)$ at $K_c=1$. The $\omega=0$ real-axis crossing at $-10$ lies left of the critical point $-1$, giving the single counter-clockwise encirclement an unstable-plant loop needs for stability.
CaseResult
$K_c=1$pole $-90$ → stable (encircles $-1$)
$K_c=0.01$pole $+9$ → unstable (no encirclement)
(b) Limiting gain$K_c>0.1$ (minimum, not maximum)