Discussion
The Laplace transform of a unit impulse function is :
Answer: D
The Laplace transform of a unit impulse is 1.
By the definition of the transform and the sifting property of the impulse:
\[ \begin{aligned} \mathcal{L}\{\delta(t)\} &= \int_{0^-}^{\infty} \delta(t)e^{-st}\,dt \\ &= e^{-s \cdot 0} = 1 \end{aligned} \]
1/s is the transform of the unit step, not the impulse. \(e^{-st}\) is the kernel of the transform, and \(e^{-sT}\) would be the transform of an impulse delayed by T. \(\delta(t)\) is the impulse function itself in the time domain, not its transform.
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