Laplace Transform of Heaviside Step Function

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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 1

\(\ds \laptrans {\map {u_c} t}\) \(=\) \(\ds \int_0^{\to +\infty} \map {u_c} t e^{-s t} \rd t\) Definition of Laplace Transform
\(\ds \) \(=\) \(\ds \int_0^c \map {u_c} t e^{-s t} \rd t + \int_c^{\to +\infty} \map {u_c} t e^{-s t} \rd t\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \int_0^c 0 \times e^{-s t} \rd t + \int_c^{\to +\infty} 1 \times e^{-s t} \rd t\) Definition of Heaviside Step Function
\(\ds \) \(=\) \(\ds \int_c^{\to +\infty} e^{-s t} \rd t\)
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to +\infty} \int_c^L e^{-s t} \rd t\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to +\infty} \intlimits {\dfrac {e^{-s t} } {-s} } c L\) Primitive of $e^{a x}$
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to +\infty} \dfrac {e^{-s L} } {-s} - \dfrac {e^{-s c} } {-s}\)
\(\ds \) \(=\) \(\ds 0 + \dfrac {e^{-s c} } s\) simplification

Hence the result.

$\blacksquare$


Proof 2

\(\ds \laptrans 1\) \(=\) \(\ds \dfrac 1 s\) 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