Laplace Transforms — Question 5

Question 5 – Solution

Strategy. Inspect the residue of the positive pole before using a final-value theorem.

Step 1: Keep the initial-data numerator. Transformation yields $$P(s)Y=a(s+1)+b+\frac 1{s+1},\qquad P(s)=s(s-1)(s+2).$$ The candidate poles are $1,0,-1,-2$; $\operatorname{Re}s>1$ is a common convergence half-plane before any cancellation.

Step 2: Classify bounded responses. Residues give $$y(t)=-\frac{a+b+1}{2}+\frac{2a+b+1/2}{3}e^t +\frac 12e^{-t}+\frac{-a+b-1}{6}e^{-2t}.$$ Only $e^t$ is unbounded, and no other mode can cancel it asymptotically. Therefore $$\boxed{y\text{ is bounded}\ \Longleftrightarrow\ b=-2a-\tfrac 12.}$$ For this line of data, $\lim y=(a-1/2)/2$.

Step 3: Remove the constant mode as well. Zero limit requires $a=1/2$, $b=-3/2$, giving $$\boxed{y_*(t)=\tfrac 12(e^{-t}-e^{-2t}).}$$ Its first three initial values are $(0,1/2,-3/2)$. Since $P(-1)=2$ and $P(-2)=0$, $P(D)y_*=e^{-t}$, directly checking the equation. Its transform converges for $\operatorname{Re}s>-1$.

Step 4: Quantify sensitivity and theorem use. A change $b\mapsto b+\varepsilon$ gives $$\boxed{y-y_*=\varepsilon\left(-\frac 12+\frac{e^t}{3}+\frac{e^{-2t}}6\right).}$$ This error has initial state $(0,0,\varepsilon)$ and is asymptotic to $\varepsilon e^t/3$. Every nonzero perturbation is eventually unbounded. Only after the positive pole cancels do all poles of $sY$ lie strictly left of the imaginary axis; then $\lim y=\lim_{s\to 0}sY$ is valid. For unstable data, the finite algebraic value at $s=0$ does not represent a time limit.