23-Chem-A6 Process Dynamics and Control · May 2018
Question 7 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, state-space models and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order tank/thermal dynamics, step/ramp response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 7: Nyquist Stability of an Open-Loop-Unstable Process (20% total)
Given. $G_p=\dfrac{1}{s^2-s-2}=\dfrac{1}{(s-2)(s+1)}$ — open-loop poles at $s=+2$ (right-half-plane) and $s=-1$; proportional control $G_c=k_c$.
Find. (a) the qualitative Nyquist locus of $L=k_cG_p$ at $k_c=1$ and the stability verdict; (b) the range of $k_c$ (if any) that stabilises the loop.
Problem 7(a): the locus is a closed loop lying entirely in the left half-plane, from $-0.5$ ($\omega=0$) to the origin ($\omega\to\infty$), reaching only $\mathrm{Re}=-0.5$. It never encircles the $-1$ point, so $N=0$.
Approach. Plot $L(j\omega)=k_c/((j\omega-2)(j\omega+1))$, count encirclements $N$ of $-1$, and apply $Z=N+P$ with $P=1$ (one open-loop RHP pole); closed-loop stability needs $Z=0$. Then confirm algebraically with the characteristic equation.
(a) Key points of the locus ($k_c=1$). Write $L(j\omega)=\dfrac{1}{(j\omega)^2-j\omega-2}=\dfrac{1}{-(\omega^2+2)-j\omega}$. At $\omega=0$: $L=1/(-2)=-0.5$ (negative real axis). As $\omega\to\infty$: $|L|\to0$, so the locus ends at the origin. The real part $-(\omega^2+2)/|D|^2$ is negative for all $\omega$, so the entire curve stays in the left half-plane (it bulges to $\mathrm{Im}\approx\pm0.105$ near $\omega=\pm0.77$).
(a) Encirclements. The closed locus reaches only to $\mathrm{Re}=-0.5$; the critical point $-1$ lies to its left and is therefore not encircled: $N=0$.
(a) Nyquist verdict. With one RHP open-loop pole, $P=1$, the closed-loop RHP-pole count is $$Z=N+P=0+1=1\neq0.$$ The closed loop is unstable at $k_c=1$ (one right-half-plane pole).
(b) Range of $k_c$. Scaling $k_c$ only stretches the locus radially along the same directions; since the locus lies wholly on the negative-real / imaginary side and the point $-1$ sits on the negative real axis at the level the curve only reaches for $k_c$ large, examine the characteristic equation directly: $1+k_cG_p=0\Rightarrow s^2-s+(k_c-2)=0$. The coefficient of $s$ is fixed at $-1<0$, so by the Routh (or Hurwitz) necessary condition at least one root has a positive real part for every $k_c$ (the roots sum to $+1$).
(b) Conclusion. $$\boxed{\text{No finite }k_c\text{ stabilises the loop.}}$$ Proportional control cannot move the fixed root-sum $+1$ into the left half-plane; e.g. at $k_c=1$ the roots are $\tfrac{1\pm\sqrt5}{2}=\{-0.618,\,+1.618\}$, always one in the RHP. Stabilising this plant requires dynamic (e.g. PD/lead) compensation, consistent with the Nyquist result $Z\ge1$.
