Laplace Transform of a Triangle Wave: Difference between revisions

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<math>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]</math>
<math>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]</math>
<math>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]</math>
<math>\int_{}^{}{4te^{-st}}=\; L\left\{ 4t \right\}=\frac{4}{s^{2}}</math>
<math>\int_{}^{}{-4te^{-st}}=\; L\left\{ -4t \right\}=-\frac{4}{s^{2}}</math>
<math>\int_{}^{}{4.5e^{-st}}=\; L\left\{ 4.5 \right\}=\frac{4.5}{s}</math>
==Author==
==Author==



Revision as of 12:38, 25 January 2010

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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=2+2.5+.5=4

m2=221.5.5=4 So,

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Using the theorem for the transform of a periodic function,

L{F(t)}=11e2s[.5.54testdt+.51.5(4t+4.5)estdt]

L{F(t)}=11e2s[.5.54testdt+.51.54testdt+.51.54.5estdt]

4test=L{4t}=4s2

4test=L{4t}=4s2

4.5est=L{4.5}=4.5s

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