Inverse Laplace Transforms and Time-Response Stability: 23-Chem-A6 Process Dynamics and Control, December 2019, Q7

Problem 7: Inverse Laplace Transforms and Time-Response Stability (20%)

The two functions $Y(s)$ for Problem 7 used below are assumed. They are of the type this question tests: a proper rational $Y(s)$ with a simple pole (part a) or a double pole (part b) at the origin plus a complex-conjugate pair, requiring partial fractions, completing the square, and a stability judgement from the pole locations. Solve your own copy of the paper with its printed numerators; the method below transfers unchanged.

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 transforms sharing the quadratic $s^2-6s+13=(s-3)^2+2^2$, whose roots are the complex pair $s=3\pm2j$ — both in the right half-plane.

Find. (a) $y_a(t)$ and its stability; (b) $y_b(t)$ and its stability.

Problem 7(a): the inverse transform $y(t)=-\tfrac{3}{13}+e^{3t}\!\left(\tfrac{3}{13}\cos 2t+\tfrac{2}{13}\sin 2t\right)$ oscillates at $\omega=2$ inside an envelope that grows as $e^{3t}$ — the response is unbounded, so the system is unstable (poles $3\pm2j$ in the right half-plane).

Approach. Expand each $Y(s)$ by partial fractions, complete the square on the quadratic to expose the damped/undamped sinusoid $e^{3t}(\cos2t,\sin2t)$, invert term by term, then read stability directly from the pole locations (any pole with positive real part ⇒ unbounded response).

(a) Partial fractions. Write $\dfrac{s-3}{s(s^2-6s+13)}=\dfrac{A}{s}+\dfrac{Bs+C}{s^2-6s+13}$. Matching numerators, $A=-\tfrac{3}{13}$, $B=\tfrac{3}{13}$, $C=-\tfrac{5}{13}$.
(a) Complete the square and invert. With $s^2-6s+13=(s-3)^2+2^2$, regroup the numerator as $B(s-3)+(3B+C)$. Then $\dfrac{B(s-3)}{(s-3)^2+2^2}\!\rightarrow\!Be^{3t}\cos2t$ and $\dfrac{3B+C}{(s-3)^2+2^2}\!\rightarrow\!\dfrac{3B+C}{2}e^{3t}\sin2t$, giving $$\boxed{y_a(t)=-\frac{3}{13}+e^{3t}\!\left(\frac{3}{13}\cos2t+\frac{2}{13}\sin2t\right).}$$
(a) Stability. The oscillatory term carries the factor $e^{3t}$ (poles at $s=3\pm2j$, real part $+3>0$), so $|y_a|\to\infty$: the response is unstable (unbounded, growing oscillation).
(b) Partial fractions. Write $\dfrac{s-3}{s^2(s^2-6s+13)}=\dfrac{A}{s}+\dfrac{B}{s^2}+\dfrac{Cs+D}{s^2-6s+13}$. Matching numerators, $A=-\tfrac{5}{169}$, $B=-\tfrac{3}{13}$, $C=\tfrac{5}{169}$, $D=\tfrac{9}{169}$.
(b) Invert. The double pole at the origin contributes $A+Bt$; the complex pair contributes $e^{3t}$ terms as before (numerator regrouped as $C(s-3)+(3C+D)$, sine coefficient $(3C+D)/2=12/169$): $$\boxed{y_b(t)=-\frac{5}{169}-\frac{3}{13}\,t+e^{3t}\!\left(\frac{5}{169}\cos2t+\frac{12}{169}\sin2t\right).}$$
(b) Stability. Again the $e^{3t}$ envelope (poles $3\pm2j$ in the RHP) dominates the ramp and constant terms, so the response grows without bound — unstable.

Part	$y(t)$	Stability
(a)	$-\tfrac{3}{13}+e^{3t}\!\left(\tfrac{3}{13}\cos2t+\tfrac{2}{13}\sin2t\right)$	Unstable (poles $3\pm2j$)
(b)	$-\tfrac{5}{169}-\tfrac{3}{13}t+e^{3t}\!\left(\tfrac{5}{169}\cos2t+\tfrac{12}{169}\sin2t\right)$	Unstable (poles $3\pm2j$)

Concept & Theory

Topic. Inverse Laplace transforms by partial fractions; completing the square for complex-conjugate poles; the pole-location stability test.

Key relations. $\mathcal L^{-1}\{1/s\}=1$, $\mathcal L^{-1}\{1/s^2\}=t$; $\mathcal L^{-1}\!\left\{\dfrac{a\,\omega+b(s-a)}{(s-a)^2+\omega^2}\right\}=e^{at}(b\cos\omega t+a\!-\!\text{type }\sin\omega t)$; a time response is bounded (stable) iff every pole has $\mathrm{Re}(s)\le0$ with any imaginary-axis poles simple.

Why this works. Each partial-fraction term maps to an elementary time function whose growth is governed solely by the real part of its pole. Here the quadratic factors to poles at $3\pm2j$; the positive real part $+3$ makes the sinusoid’s amplitude grow as $e^{3t}$, overwhelming the bounded contributions from the poles at the origin, so both responses diverge.

Common pitfall. Reading $s^2-6s+13$ as having left-half-plane roots — the $-6s$ (not $+6s$) puts the real part at $+3$; and forgetting that a $1/s^2$ (double origin pole) term is already unbounded (a ramp), independent of the RHP pair.

Source. Coughanowr & LeBlanc, Process Systems Analysis and Control, Ch. 3–4 (inverse transforms, partial fractions); Seborg et al., Ch. 3 & 11 (transform inversion, stability from pole locations).

Question 7 of 8: Inverse Laplace Transforms and Time-Response Stability

Problem 7: Inverse Laplace Transforms and Time-Response Stability (20%)