# Laplace transform of periodic functions

Let $f(t)$ be periodic with the positive period (http://planetmath.org/PeriodicFunctions) $p$.  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

 $\displaystyle g(t)\;=\;\begin{cases}f(t)\quad\mbox{for}\;\;0 (1)

By the parent entry (http://planetmath.org/DelayTheorem), 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

 $\displaystyle\mathcal{L}\{f(t)\}\;=\;\frac{1}{1\!-\!e^{-ps}}\int_{0}^{p}\!e^{-% st}f(t)\,dt\qquad(\mbox{period }p).$ (2)

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

 $\displaystyle\mathcal{L}\{f(t)\}\;=\;\frac{1}{1\!+\!e^{-ps}}\int_{0}^{p}\!e^{-% st}f(t)\,dt\qquad(\mbox{antiperiod }p).$ (3)
Title Laplace transform of periodic functions LaplaceTransformOfPeriodicFunctions 2013-03-22 18:58:24 2013-03-22 18:58:24 pahio (2872) pahio (2872) 5 pahio (2872) Derivation msc 44A10 RectificationOfAntiperiodicFunction TableOfLaplaceTransforms