Inverse Laplace Transforms with Time Delay: 20-Bio-A2 Process Dynamics and Control, Undated paper, Q6

Problem 6: Inverse Laplace Transforms with Time Delay (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.

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The leading power of $s$ in part (a)'s denominator is unclear on the printed paper, with $s^2(s^2+4s+13)$ and $s(s^2+4s+13)$ both possible readings. The first reading is used below; note it cancels cleanly against the numerator's single factor of $s$, leaving exactly $Y(s)=(e^{-2s}-e^{-4s})/[s(s^2+4s+13)]$ either way the ambiguity is read, so the final answer is unaffected by which reading is correct.

Given. (a) $Y(s)=\dfrac{s(e^{-2s}-e^{-4s})}{s^2(s^2+4s+13)}$; (b) $Y(s)=\dfrac{s-3}{s^2-4s+5}$.

Find. $y(t)=\mathcal L^{-1}[Y(s)]$ for each.

Approach. (a) Cancel the common factor of $s$, complete the square in the denominator to expose a damped-sinusoid pair, partial-fraction the un-delayed part $F(s)$, then apply the real-translation (time-shift) theorem twice for the two delays. (b) Complete the square in the denominator directly and split the numerator into the two standard damped-sinusoid forms.

(a) Simplify and complete the square. $Y(s)=\dfrac{e^{-2s}-e^{-4s}}{s(s^2+4s+13)}=\left(e^{-2s}-e^{-4s}\right)F(s)$, where $F(s)=\dfrac{1}{s\left[(s+2)^2+9\right]}$ (using $s^2+4s+13=(s+2)^2+9$, poles at $s=-2\pm3j$).
(a) Partial-fraction $F(s)$. $F(s)=\dfrac As+\dfrac{Bs+C}{(s+2)^2+9}$: matching coefficients gives $A=1/13$, $B=-1/13$, $C=-4/13$, i.e. $$F(s)=\frac{1}{13}\left[\frac1s-\frac{(s+2)+2}{(s+2)^2+9}\right]=\frac{1}{13}\left[\frac1s-\frac{s+2}{(s+2)^2+9}-\frac{2}{(s+2)^2+9}\right].$$
(a) Invert $F(s)$. Using $\mathcal L^{-1}\!\left[\tfrac{s+2}{(s+2)^2+9}\right]=e^{-2t}\cos3t$ and $\mathcal L^{-1}\!\left[\tfrac{1}{(s+2)^2+9}\right]=\tfrac13e^{-2t}\sin3t$: $$f(t)=\frac{1}{13}\left[1-e^{-2t}\cos3t-\frac23e^{-2t}\sin3t\right]u(t).$$ (Check: $f(0)=\tfrac1{13}[1-1-0]=0$; $f(\infty)=\tfrac1{13}$, matching the final-value theorem $\lim_{s\to0}sF(s)=1/13$.)
(a) Apply the two time shifts. $$\boxed{y(t)=f(t-2)\,u(t-2)-f(t-4)\,u(t-4).}$$ The response is exactly zero for $t<2$, rises through $f(t-2)$ for $2\le t<4$ (e.g. $y(3)=f(1)=0.0862$), then for $t\ge4$ is the small difference of two nearly-settled values of $f$ (both close to $1/13=0.0769$), producing a lightly damped decay back toward zero (e.g. $y(5)=f(3)-f(1)=-0.0092$, $y(6)=0.0011$) — consistent with the underlying complex poles at $-2\pm3j$ and with the physical picture of a rectangular pulse (magnitude $1$, duration $2\ \text{s}$) applied to a lightly-damped second-order-plus-integrator system.
(b) Complete the square and split. $s^2-4s+5=(s-2)^2+1$ (poles at $s=2\pm j$), so $$Y(s)=\frac{s-3}{(s-2)^2+1}=\frac{(s-2)-1}{(s-2)^2+1}=\frac{s-2}{(s-2)^2+1}-\frac{1}{(s-2)^2+1}.$$
(b) Invert term by term. Using $\mathcal L^{-1}\!\left[\tfrac{s-2}{(s-2)^2+1}\right]=e^{2t}\cos t$ and $\mathcal L^{-1}\!\left[\tfrac{1}{(s-2)^2+1}\right]=e^{2t}\sin t$: $$\boxed{y(t)=e^{2t}\left(\cos t-\sin t\right).}$$ Check: $y(0)=1\times(1-0)=1$, matching the initial-value theorem $\lim_{s\to\infty}sY(s)=1$. The poles at $s=2\pm j$ are in the RHP, so this is (like Problem 5) an unbounded, growing oscillatory response — a legitimate inverse-transform exercise even though it does not correspond to a stable physical system.

Result	Value
(a) Un-delayed component	$f(t)=\tfrac1{13}\left[1-e^{-2t}\cos3t-\tfrac23e^{-2t}\sin3t\right]$
(a) $y(t)$	$f(t-2)u(t-2)-f(t-4)u(t-4)$
(b) $y(t)$	$e^{2t}(\cos t-\sin t)$

Concept & Theory

Topic. Inverse Laplace transforms of rational functions with complex-conjugate poles and pure time delays.

Key relations. Complete-the-square form $s^2+bs+c=(s+\beta)^2+\omega^2$; standard damped-sinusoid pairs $\mathcal L^{-1}[(s+\beta)/((s+\beta)^2+\omega^2)]=e^{-\beta t}\cos\omega t$, $\mathcal L^{-1}[1/((s+\beta)^2+\omega^2)]=\tfrac1\omega e^{-\beta t}\sin\omega t$; real-translation theorem $\mathcal L^{-1}[e^{-\theta s}F(s)]=f(t-\theta)u(t-\theta)$.

Why this works. Splitting a complex-pole-pair numerator into a "cosine part" (matching $s+\beta$) plus a "sine part" (the remainder) lets each be inverted by the standard pair directly, without ever working in complex arithmetic; two delayed copies of the same base function, subtracted, is exactly the transform of a finite-duration rectangular pulse passed through the system.

Common pitfall. Forgetting the $1/\omega$ scale factor on the sine term, or applying the time shift to the wrong sign (the delayed response is zero identically before $t=\theta$, not merely small).

Source. Seborg et al., Process Dynamics and Control, Ch. 3 (Laplace transforms, partial fractions, translation theorem).

Question 6 of 8: Inverse Laplace Transforms with Time Delay

Problem 6: Inverse Laplace Transforms with Time Delay (20%)