Laplace Transform of Heaviside Step Function/Proof 2

Theorem

Let $\map {u_c} t$ denote the Heaviside step function:

$\map {u_c} t = \begin{cases}

1 & : t > c \\ 0 & : t < c \end{cases}$

The Laplace transform of $\map {u_c} t$ is given by:

$\laptrans {\map {u_c} t} = \dfrac {e^{-s c} } s$

for $\map \Re s > c$.

Proof

\(\ds \laptrans 1\)

\(=\)

\(\ds \dfrac 1 s\)

Laplace Transform of 1

\(\ds \leadsto \ \ \)

\(\ds \laptrans {1 \times \map {u_c} t}\)

\(=\)

\(\ds \dfrac 1 s \times e^{-c s}\)

Laplace Transform of Function of t minus a

\(\ds \)

\(=\)

\(\ds \dfrac {e^{-s c} } s\)

simplification

Hence the result.

$\blacksquare$

Sources

• 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Impulse Functions. The Dirac Delta Function: $40$