Laplace Transforms — Question 8

Question 8 – Solution

Strategy. Sum delayed impulse kernels and distinguish a periodic limiting profile from a constant final value.

Step 1: Sum the transformed impulses. For $\operatorname{Re}s>0$, $$Y=\frac{e^{-sT}}{(1-e^{-sT})(s+1)^3},\qquad y(t)=\sum_{1\leq n\leq\lfloor t/T\rfloor}h(t-nT),\quad h(v)=\tfrac 12v^2e^{-v}.$$ The sum is empty before $T$; including the current impulse adds $h(0)=0$.

Step 2: Sum the limiting profile. At $t=NT+x$ the sum is $\sum_{k=0}^{N-1}h(x+kT)$. Geometric-series identities give $$\boxed{p(x)=\frac{e^{-x}}2\left[\frac{x^2}{1-q} +\frac{2xTq}{(1-q)^2}+\frac{T^2q(1+q)}{(1-q)^3}\right],\quad q=e^{-T}.}$$ The omitted tail is uniformly $O(N^2q^N)$ for $0\leq x\leq T$ and fixed $T$, so $y(NT+x)\to p(x)$ uniformly on a period.

Step 3: Check joins and mean. Since $(h,h',h'')(0)=(0,0,1)$, $y,y'$ are continuous and $y''$ jumps upward by one. Reindexing the convergent profile series gives the same periodic joins: $p(0)=p(T-)$, $p'(0)=p'(T-)$, $p''(0)-p''(T-)=1$. Also $\int_0^T p(x)\,dx=\int_0^\infty h(v)\,dv=1$, so its mean is $1/T$.

Step 4: Reject a constant final value. The profile is nonconstant: on each open gap $(D+1)^3p=0$, so a constant profile would be zero, contradicting its integral. Distinct phases therefore have distinct subsequential limits. Poles at $s=2\pi i k/T$, $k\ne 0$, remain in $sY$; the formal limit $\lim_{s\to 0}sY=1/T$ is a mean, not a final value.