The Laplace Transform of a unit step function $u_{a} (t)$ , defined as
$u_{a} (t) = {\begin{matrix} 0 f o r t < a 1 f o r t > a \end{matrix}$ is

\frac{e^{- a s}}{s}

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s e^{- a s}

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s - u(0)

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s e^{- a s} - 1

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Solution

The correct option is A $\frac{e^{- a s}}{s}$
$u_{a} (t) = {\begin{matrix} 0, t < a 1, t > a \end{matrix}$
$L [u_{a} (t)] = \int_{a}^{\infty} e^{- s t} (1) d t$
$= \frac{- 1}{s} [e^{- s t}]_{a}^{\infty}$
$= \frac{- 1}{s} [0 - e^{- a s}] = \frac{e^{- a s}}{s}$

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a

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