Transcription of Table of Laplace Transforms - Stanford University
1 S. BoydEE102 TableofLaplaceTransformsRemember thatwe considerall functions(signals)as de nedonlyont (t)F(s) =Z10f(t)e stdtf+gF+G f( 2R) FdfdtsF(s) f(0)dkfdtkskF(s) sk 1f(0) sk 2dfdt(0) dk 1fdtk 1(0)g(t) =Zt0f( )d G(s) =F(s)sf( t), >01 F(s= )eatf(t)F(s a)tf(t) dFdstkf(t)( 1)kdkF(s)dskf(t)tZ1sF(s)dsg(t) =(00 t < Tf(t T)t T,T 0G(s) =e sTF(s)1 Speci c11s 1 (k)skt1s2tkk!,k 01sk+1eat1s acos!tss2+!2=1=2s j!+1=2s+j!sin!t!s2+!2=1=2js j! 1=2js+j!cos(!t+ )scos !sin s2+!2e atcos!ts+a(s+a)2+!2e atsin!t!(s+a)2+!22 Notesonthederivative formulaatt= 0 TheformulaL(f0) =sF(s) f(0 ) must be interpretedvery carefullywhenfhas a discon-tinuity att= 0. We'll give two examplesof the ,suppose thatfis the constant 1, and has no discontinuity att= 0. In otherwords,fis the constant functionwithvalue1. Thenwe havef0= 0, andf(0 ) = 1 (sincethereisno jumpinfatt= 0).)
2 Now let'sapplythe derivative formula above. We haveF(s) = 1=s,so the formula readsL(f0) = 0 =sF(s) 1which is , let'ssuppose thatgis a unitstepfunction, ,g(t) = 1 fort >0, andg(0) = contrasttofabove,ghas a jumpatt= 0. In this case,g0= , andg(0 ) = 0. Now let'sapplythe derivative formula above. We haveG(s) = 1=s(exactlythe sameasF!), so theformula readsL(g0) = 1 =sG(s) 0which againis thesetwo examplesthe functionsfandgare the sameexceptatt= 0, so theyhavethe the rstcase,fhas no jumpatt= 0, whilein the secondcasegdoes. As a result,f0has no impulsive termatt= 0, whereasgdoes. As longas youkeep track of whetheryour functionhas,or doesn'thave, a jumpatt= 0, andapplytheformula consistently, everythingwill work
