November 2, 2001ece309-09-1 November 2, 2001 12.5 Operational Transform a. Linearty (Multiplication by a constant) If {}()()ftFsL =, Then {}()()Kf tKF sL = b. Addition (Subtraction) If {}11()()ftFsL =, {}22()()ftFsL =, {}33()()ftFsL = Then {}123 1 2 3() () ()() () ()ft ft ftFs Fs FsL ++ =++ Similar {}123 1 2 3() () ()() () ()ft ft ftFs Fs FsL −− =−− c. Differentiation {}()() (0)ftsF s fdtL−=− d. Integration 0()()FsftdtsL−∞=∫ e. Translation in the time Domain Translation in the time domain corresponds to multiplication by an exponential in the frequency domain {}()()()for 0asft aut aeFs aL−−−=> f. Translation in the Frequency Domain Translation in the frequency domain corresponds to multiplication by an exponential in the time domain {}()()ateftFs aL−=+ g. Scale Changing If {}()()ftFsL =, Then {}()1( ), for 0fatsFasaL =>ece309-09-2 Properties of Laplace Transform Property ()ft ()Fs Linearty 11 2 2() ()Kf t Kf t+ 11 2 2() ()KF s KF s+ Scaling (), 0fat a> 1()sFaa Time shifting ()(),0ft aut a a−−> ()aseFs− Frequency shifting ()ateft− ()Fs a+ Time differentiation ()df tdt () (0)sF s f−− SecondTime differentiation 22()dftdt 2(0 )() (0)dfsFs sfdt−−−− nth time differentiation ()nndftdt 1121(0 ) (0 )( ) (0 ) ...nnn nndf d fsFs s f sdt dt−−−−− −−−− −− Time integration 0()tftdt∫ 1()Fss First derivative(s) ()tf t ()dF sds− nth derivative(s) ()ntft ()(1)nnndFsds− s integral ()ftt ()sFsds∞∫ Time preiodicity () ( )ft ft nT=+ ()1sTFse−− Initial value (0 )f+ lim ( )ssF s→∞ Final value ()f∞ 0lim ( )ssF s→ Convolution 12()* ()ft ft 12() ()FsFsece309-09-3 Examples: 1. Find the Laplace Transform of 2() () 4 () 2tft tuteδ−=+ − {} {}{}2() () 4 () 2tFstuteLLLδ−=+ − 211 2482() 1 4 22(2)sss sFsss ss+++−=+ − =++ 248()(2)ssFsss++=+ 2. Find the Laplace Transform of []() 4 ( 1) ( 4)ft ut ut=−−− {}{ }()4() 4 ( 1) ( 4) 4sseeFs ut utssLL−−=−−−=− 4() 4sseeFss−−−== 3. Find the Laplace Transform of 2() cos4ft t t= {}22222 2() cos4 (1)4dsFs Lt tds s==−+ {}24222 2316 2 ( 256)() cos4( 16) ( 16)ds ssFs Lt tds s s−+ −== =++ 4. Find the Laplace Transform of the following figure []( ) 10 ( 2) ( 4)f t ut ut=−−− {}{}()24() 5 ( 2) ( 4) 5sseeFs ut utssLL−−=−−−=− 24() 5sseeFss−−−= t f(t) 5 2
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