Inverse Laplace Transforms and Stability: 23-Chem-A6 Process Dynamics and Control, December 2016, Q1

Problem 1: Inverse Laplace Transforms and Stability (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. Two Laplace-domain outputs sharing the quadratic factor $s^2-6s+18=(s-3)^2+9$, whose roots are $s=3\pm 3j$.

Find. $y_a(t)$ and $y_b(t)$, and a stability verdict for each.

Pole–zero map. Both transforms carry the complex pair $3\pm 3j$ in the right-half plane; part (a) adds a pole at the origin, part (b) a double pole there. Any right-half-plane pole forces an $e^{3t}$ term, so both responses are unstable — settle this from the poles before inverting.

Approach. Read stability directly from the pole locations, then split each transform by partial fractions and invert term by term using $\mathcal{L}^{-1}\{(s-a)/[(s-a)^2+\omega^2]\}=e^{at}\cos\omega t$ and $\mathcal{L}^{-1}\{\omega/[(s-a)^2+\omega^2]\}=e^{at}\sin\omega t$.

Locate the poles & rule on stability. The quadratic $s^2-6s+18$ has discriminant $36-72<0$, so its roots are $s=3\pm3j$. Their real part $+3>0$ lies in the right-half plane, hence the time response contains $e^{3t}$ and grows without bound: $$\boxed{\text{both (a) and (b) are UNSTABLE.}}$$
(a) Partial fractions. Write $\dfrac{s-3}{s(s^2-6s+18)}=\dfrac{A}{s}+\dfrac{Bs+C}{(s-3)^2+9}$. Matching: $s=0\Rightarrow-3=18A\Rightarrow A=-\tfrac16$; $s^2$-coeff $A+B=0\Rightarrow B=\tfrac16$; $s^1$-coeff $-6A+C=1\Rightarrow C=0$.
(a) Invert. The complex part is $\tfrac16\dfrac{s}{(s-3)^2+9}=\tfrac16\dfrac{(s-3)+3}{(s-3)^2+9}$, giving cosine and sine at $\omega=3$: $$\boxed{y_a(t)=-\tfrac16+\tfrac16 e^{3t}\big(\cos 3t+\sin 3t\big).}$$ The $e^{3t}$ envelope confirms instability.
(b) Partial fractions. $\dfrac{s-3}{s^2(s^2-6s+18)}=\dfrac{A}{s}+\dfrac{B}{s^2}+\dfrac{Cs+D}{(s-3)^2+9}$. Matching: $s=0\Rightarrow-3=18B\Rightarrow B=-\tfrac16$; $s^1\Rightarrow18A-6B=1\Rightarrow A=0$; $s^3\Rightarrow A+C=0\Rightarrow C=0$; $s^2\Rightarrow-6A+B+D=0\Rightarrow D=\tfrac16$.
(b) Invert. With $A=C=0$, $B=-\tfrac16$, $D=\tfrac16$ the sine term is $\tfrac16\dfrac{1}{(s-3)^2+9}=\tfrac{1}{18}\dfrac{3}{(s-3)^2+9}$: $$\boxed{y_b(t)=-\tfrac{t}{6}+\tfrac{1}{18}e^{3t}\sin 3t.}$$ Again the growing exponential makes the response unstable.

Part	Inverse transform	Stability
(a)	$y_a(t)=-\tfrac16+\tfrac16 e^{3t}(\cos3t+\sin3t)$	Unstable (poles $3\pm3j$)
(b)	$y_b(t)=-\tfrac{t}{6}+\tfrac{1}{18}e^{3t}\sin3t$	Unstable (poles $3\pm3j$)

Concept & Theory

Topic. Inverse Laplace by partial fractions; stability from pole location.

Key relations. A quadratic $s^2+bs+c$ completes to $(s-a)^2+\omega^2$ with $a=-b/2$; a system is BIBO-stable iff every pole has negative real part.

Why this works. The denominator roots are the natural modes of the response. Because $s^2-6s+18=(s-3)^2+9$ has positive real part $+3$, each mode is a growing oscillation $e^{3t}\{\cos,\sin\}3t$; no choice of numerator can remove a pole, so the transient cannot decay.

Common pitfall. Do not be fooled by the finite algebra — students often invert first and forget the verdict, or mis-sign the completed square. Check the sign of the real part before computing residues.

Source. Seborg, Edgar, Mellichamp & Doyle, Process Dynamics and Control (4th ed., Wiley) Ch. 3 (Laplace transforms) and Ch. 11 (stability); Coughanowr & LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) Ch. 2.

Question 1 of 8: Inverse Laplace Transforms and Stability

Problem 1: Inverse Laplace Transforms and Stability (20%)