If I took LT : $L(u(t-1)\delta(t)) = u((0)-1)$ as Laplace of $\delta(t)=1$, and we substitute $t$ with zero don't we get a zero answer as $u(t-1) = 0$ when $t<1$?
but what if I did it the other way: $L(u(t-1)\delta(t)) = e^{-s} L(\delta(t+1)) = e^{-s} e^{s} = 1$
Am I missing something?
$\endgroup$ 02 Answers
$\begingroup$I suppose you could argue it like this.
$$L \{ f (t) U (t − a)\} = e^{−as}L \{ f (t+a)\}$$
so
$$L \{ \delta(t) U (t − 1)\} = e^{−s}L \{ \delta(t+1)\}$$
$$L\{ \delta(t+1)\}=\int_0^\infty e^{-st}\delta(t+1) dt=0$$
as $-1$ is not in the interval.
The other way
$$L\{ U (t − 1)\delta(t)\}=\int_0^\infty e^{-st}U (t − 1)\delta(t) dt=1\cdot U (0 − 1)=0$$
$\endgroup$ 1 $\begingroup$An alternate way
$$\mathcal{L}\left(u(t-1)\delta(t)\right)=\int_0^\infty u(t-1)\delta(t)e^{-st}dt=\int_1^\infty\delta(t)e^{-st}dt$$
So
$$\mathcal{L}\left(u(t-1)\delta(t)\right)=\mathcal{L}\left(\delta(t)\right)-\int_0^1\delta(t)e^{-st}dt=1-e^0=0$$
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