Question

find laplace transform of f(t)=d(t-4)*t*sin(10t+0.2)*e^2t where d(t) is an impulse function..
plzz help

Answer 1

The key property of the Dirac delta function that we're going to need to use here is
\[\large \int_{a-\epsilon}^{a+\epsilon}f(t)\delta(t-a)\,dt = f(a)\quad \text{ for $\epsilon>0$}\tag{1}\]This above result will be true for any integral evaluated over any interval containing \(a\); this will become the key needed to find the Laplace transform (since \(4\in(0,\infty)\)).

Therefore, by the integral definition of Laplace transform, we have that
\[\large\begin{aligned}\mathcal{L}\{te^{2t}\sin(10t+0.2)\delta(t-4)\} &= \int_0^{\infty}\underbrace{e^{-st}te^{2t}\sin(10t+0.2)}_{f(t)}\delta(t-4)\,dt\\ &= 4e^8\sin(40.2)e^{-4s}\qquad\qquad\text{by (1)}\end{aligned}\]Does this make sense?