laplace transform of impulse function
By Emma Terry •
The laplace tranform of the following function impulse function is
$$\int_{0}^{\infty}\delta(t).e^{-st}\,dt$$ $$=\int_{0}^{\infty}\delta(t)\,dt=1$$ (area under unit impulse is always 1) $$=1.\int_{0}^{\infty}e^{-st}\,dt$$ $$=1.\frac{1}{S}$$ $$=\frac{1}{S}$$ but the correct answer is 1, I don't know why.
$\endgroup$ 21 Answer
$\begingroup$Hint. Note that by the sampling property of the delta function (which is actually a distribution) $$\int_{-\infty}^{+\infty}f(t)\delta(t-t_0)dt=f(t_0).$$
$\endgroup$ 2
