Laplace Transform of Half Wave Rectified Sine Curve

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Theorem

Consider the half wave rectified sine curve:

$\map f t = \begin {cases} \sin t & : 2 n \pi \le t \le \paren {2 n + 1} \pi \\ 0 & : \paren {2 n + 1} \pi \le t \le \paren {2 n + 2} \pi \end {cases}$


The Laplace transform of $\map f t$ is given by:

$\laptrans {\map f t} = \dfrac 1 {\paren {1 - e^{-\pi s} } \paren {s^2 + 1} }$


Proof

We have that $\map f t$ is periodic with period $2 \pi$:


Half-wave-rectified-sine-curve.png


Hence:

\(\displaystyle \laptrans {\map f t}\) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \int_0^{2 \pi} e^{-s t} \map f t \rd t\) Laplace Transform of Periodic Function
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \paren {\int_0^\pi e^{-s t} \sin t \rd t + \int_\pi^{2 \pi} e^{-s t} \times 0 \rd t}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \int_0^\pi e^{-s t} \sin t \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \intlimits {\dfrac {e^{-s t} \paren {-s \sin t - \cos t} } {s^2 + 1} } 0 \pi\) Primitive of $e^{a x} \sin b x$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-s \sin \pi - \cos \pi} } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {-s \sin 0 - \cos 0} } {s^2 + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\cos \pi} } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {- \cos 0} } {s^2 + 1} }\) Sine of Integer Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\paren {-1} } } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {-1} } {s^2 + 1} }\) Cosine of Integer Multiple of Pi
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\paren {-1} } } {s^2 + 1} - \dfrac {1 \times \paren {-1} } {s^2 + 1} }\) Exponential of Zero
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} + 1} {s^2 + 1} }\) simplification
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\paren {1 + e^{-\pi s} } \paren {1 - e^{-\pi s} } } \paren {\dfrac {e^{-s \pi} + 1} {s^2 + 1} }\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {\paren {1 - e^{-\pi s} } \paren {s^2 + 1} }\) simplification

$\blacksquare$


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