23-Chem-A6 Process Dynamics and Control · December 2017
Question 1 of 8: Inverse Laplace Transform and Stability
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 2017 sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 1: Inverse Laplace Transform and Stability (20%)
Given. A Laplace-domain output whose quadratic factor completes to $s^2-6s+18=(s-3)^2+9$, so its roots are $s=3\pm 3j$; there is also a pole at the origin and a zero at $s=+3$.
Find. (a) $y(t)$ by partial-fraction inversion; (b) a justified stability verdict.
Pole–zero map. The complex pair $3\pm3j$ sits in the right-half plane, so the transient carries $e^{3t}$ — settle stability from the poles before inverting.
Approach. Read stability from the pole locations, then split $Y(s)$ by partial fractions and invert term by term with $\mathcal{L}^{-1}\{(s-a)/[(s-a)^2+\omega^2]\}=e^{at}\cos\omega t$ and $\mathcal{L}^{-1}\{\omega/[(s-a)^2+\omega^2]\}=e^{at}\sin\omega t$.
Locate the poles and rule on stability. The quadratic $s^2-6s+18$ has discriminant $36-72<0$, so its roots are $s=3\pm3j$ — real part $+3>0$, in the right-half plane. Hence the response contains a growing $e^{3t}$ envelope: $$\boxed{\text{the response is UNSTABLE.}}$$
Partial fractions. Write $\dfrac{s-3}{s(s^2-6s+18)}=\dfrac{A}{s}+\dfrac{Bs+C}{(s-3)^2+9}$. Evaluate at $s=0$: $A=\dfrac{-3}{18}=-\tfrac16$. Matching the $s^2$ coefficient $A+B=0\Rightarrow B=\tfrac16$; the $s^1$ coefficient $-6A+C=1\Rightarrow C=0$.
(a) Invert. The complex part is $\tfrac16\dfrac{s}{(s-3)^2+9}=\tfrac16\dfrac{(s-3)+3}{(s-3)^2+9}$, which supplies a cosine and a sine at $\omega=3$: $$\boxed{y(t)=-\tfrac16+\tfrac16\,e^{3t}\big(\cos 3t+\sin 3t\big).}$$
(b) Justify stability. A transform is BIBO-stable iff every pole has negative real part. Here the pair $3\pm3j$ lies in the RHP, so $y(t)$ grows without bound (the constant $-\tfrac16$ from the origin pole is bounded but does not offset the growing oscillation). The response is therefore unstable; no numerator could remove those poles.
