Laplace Transform of Periodic Function/Proof 2
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Theorem
Let $f$ be periodic, that is:
- $\exists T \in \R_{\ne 0}: \forall x \in \R: \map f x = \map f {x + T}$
Then:
- $\laptrans {\map f t} = \dfrac 1 {1 - e^{-s T} } \ds \int_0^T e^{-s t} \map f t \rd t$
where $\laptrans {\map f t}$ denotes the Laplace transform.
Proof
\(\ds \laptrans {\map f t}\) | \(=\) | \(\ds \int_0^{\infty} e^{-s t} \map f t \rd t\) | Definition of Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + \int_T^{\infty} e^{-s t} \map f t \rd t\) | Sum of Integrals on Adjacent Intervals for Integrable Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s \paren {t + T} } \map f {t + T} \rd t\) | Change of Limits of Integration | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s \paren {t + T} } \map f t \rd t\) | Definition of Real Periodic Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s t - s T} \map f t \rd t\) | Real Multiplication Distributes over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s t} e^{-s T} \map f t \rd t\) | Product of Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + e^{-s T} \int_0^{\infty} e^{-s t} \map f t \rd t\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t + e^{-s T} \laptrans {\map f t}\) | Definition of Laplace Transform | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {1 - e^{-s T} } \laptrans {\map f t}\) | \(=\) | \(\ds \int_0^T e^{-s t} \map f t \rd t\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \laptrans {\map f t}\) | \(=\) | \(\ds \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t\) |
$\blacksquare$