Laplace Transforms — Question 10

Question 10 – Solution

Strategy. Use partial fractions for a closed form and convolution for a stable coalescing-pole limit.

Step 1: Invert distinct repeated poles. For $\varepsilon>0$, $H_\varepsilon=1/[z^2(z+\varepsilon)^2]$ and $$H_\varepsilon=\frac 1{\varepsilon^2}\left(\frac 1{z^2}+\frac 1{(z+\varepsilon)^2}\right) +\frac 2{\varepsilon^3}\left(\frac 1{z+\varepsilon}-\frac 1z\right).$$ Hence $h_\varepsilon=e^{-t}\{t(1+e^{-\varepsilon t})/\varepsilon^2 -2(1-e^{-\varepsilon t})/\varepsilon^3\}$.

Step 2: Take the limit through convolution. Convolving $te^{-t}$ with $te^{-(1+\varepsilon)t}$ gives $$\boxed{h_\varepsilon(t)=e^{-t}\int_0^t v(t-v)e^{-\varepsilon v}\,dv,\qquad h_0(t)=\frac{t^3e^{-t}}6.}$$ For fixed $t$, dominated convergence justifies the limit, including $t=0$.

Step 3: Prove global error bounds. Bounding the integrand and using $1-e^{-\varepsilon v}\leq\varepsilon v$ yields $$e^{-(1+\varepsilon)t}\frac{t^3}6\leq h_\varepsilon(t)\leq e^{-t}\frac{t^3}6, \qquad 0\leq h_0-h_\varepsilon\leq\frac{\varepsilon t^4e^{-t}}{12}.$$ The maximum of $t^4e^{-t}$ is $256e^{-4}$ at $t=4$. Thus $\boxed{\sup_{t\geq 0}|h_\varepsilon-h_0|\leq 64\varepsilon/(3e^4)}$.

Step 4: Check poles, impulses and evaluation. Two double poles merge into a quadruple pole at $-1$; absolute convergence holds for $\operatorname{Re}s>-1$. Convolution gives $h_\varepsilon=t^3/6+O(t^4)$, so its right-hand jet is $(0,0,0,1)$, producing exactly a unit $\delta$. For small $\varepsilon t$, use the integral or its exponential series; the closed form subtracts large nearly equal terms and can lose significant digits.