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Laplace transform of periodic functions
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(Derivation)
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Let $f(t)$ be periodic with the positive period $p$ . Denote by $H(t)$ the Heaviside step function. If now $$g(t) \;:=\; f(t)H(t)\!-\!f(t\!-\!p)H(t\!-\!p),$$ then it follows
By the parent entry, the Laplace transform of $g$ is $$G(s) \;=\; F(s)\!-\!e^{-ps}F(s),$$ whence $$F(s) \;=\; \frac{G(s)}{1-e^{-ps}} \;=\; \frac{1}{1-e^{-ps}}\int_0^\infty\!e^{-st}g(t)\,dt \;=\; \frac{1}{1\!-\!e^{-ps}}\int_0^p\!e^{-st}f(t)\,dt.$$
Thus we have the rule
On the contrary, if $f(t)$ is antiperiodic with positive antiperiod $p$ , then the function $$g(t) \;:=\; f(t)H(t)\!+\!f(t\!-\!p)H(t\!-\!p)$$ also has the property (1). Analogically with the preceding procedure, one may derive the rule
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"Laplace transform of periodic functions" is owned by pahio.
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Cross-references: property, function, antiperiod, antiperiodic, Laplace transform, Heaviside step function, positive, periodic
This is version 2 of Laplace transform of periodic functions, born on 2009-07-11, modified 2009-07-11.
Object id is 11835, canonical name is LaplaceTransformOfPeriodicFunctions.
Accessed 468 times total.
Classification:
| AMS MSC: | 44A10 (Integral transforms, operational calculus :: Laplace transform) |
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Pending Errata and Addenda
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