Laplace Transform of a Triangle Wave

Triangle wave with period T=2 and amplitude A=2

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Introduction

This article explains how to transform a periodic function (in this case a triangle wave). This is especially useful for analyzing circuits which contain triangle wave voltage sources.

Define F(t)

$m1={\frac {2+2}{.5+.5}}=4$

$m2={\frac {-2-2}{1.5-.5}}=-4$ So,

Using the theorem for the transform of a periodic function,

$L\left\{F\left(t\right)\right\}={\frac {1}{1-e^{-2s}}}\left[\int _{-.5}^{.5}{4te^{-st}dt}+\int _{.5}^{1.5}{\left(-4t+4.5\right)e^{-st}dt}\right]$

$L\left\{F\left(t\right)\right\}={\frac {1}{1-e^{-2s}}}\left[\int _{-.5}^{.5}{4te^{-st}dt}+\int _{.5}^{1.5}{-4te^{-st}dt}+\int _{.5}^{1.5}{4.5e^{-st}dt}\right]$

$\int _{-.5}^{.5}{4te^{-st}}=\;{\frac {4e^{.5s}-2se^{.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}}$

$\int _{.5}^{1.5}{-4te^{-st}}={\frac {6se^{-1.5s}+4e^{-1.5s}-2se^{-.5s}-4e^{-.5s}}{s^{2}}}$

$\int _{.5}^{1.5}{4.5e^{-st}}={\frac {4.5se^{-.5s}-4.5se^{-1.5s}}{s^{2}}}$

$F\left(s\right)={\frac {1}{1-e^{-2s}}}\left({\frac {-8e^{-.5s}+4e^{.5s}+4e^{-1.5s}}{s^{2}}}+{\frac {.5e^{-.5s}-2e^{.5s}-1.5e^{-1.5s}}{s}}\right)$

Author

• Michael Vier