Laplace Transform of a Triangle Wave

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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,

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Using the theorem for the transform of a periodic function,

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

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

$\underset{-.5}{\overset{.5}{\int }}4t{e}^{-st}=\frac{4e^{.5s}-2s{e}^{.5s}-2s{e}^{-.5s}-4e^{-.5s}}{s^2}$

$\underset{.5}{\overset{1.5}{\int }}-4t{e}^{-st}=\frac{6s{e}^{-1.5s}+4e^{-1.5s}-2s{e}^{-.5s}-4e^{-.5s}}{s^2}$

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

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

Author

• Michael Vier