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
- 1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Laplace Transforms of Special Functions: $11$