Laplace.2 - ut vt Input dv G dt dv dt RQ t GQ t GQ t t 0...

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u H t L = R Q L H t L + G Q H t L , t > 0 v H t L = G Q H t L = ' Output' voltage Input - Output Relation dv dt = G dQ dt = G u H t L R - v H t L R dv dt + kv H t L = ku H t L d k dt k = 1 fork = 0 d n v H t L dt n + 1 n - 1 a k d k v H t L dt k = m 0 b k d k u H t L dt k n th order differentialequation.Thecoefficientof thehhighestdegreeisnonzero. Comesfrom EQUILIBRIUM LAWSOF SCIENCE EXAMPLE: d 2 v H t L dt 2 + a 1 dv H t L dt + a 0 v H t L = du H t L dt + bu H t L CIRCUITEXAMPLE: L d 2 Q H t L dt 2 + R dQ H t L dt + GQ H t L = u H t L v H t L = GQ H t L Outputvolts u H t L InputVolts d 2 dt 2 Q H t L + R L d dt Q H t L + G L Q H t L = u H t L L d 2 dt 2 v H t L + R L d dt v H t L + G L v H t L = G L u H t L Thisisour IPOPRELATION ! Printed by Mathematica for Students
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HOWDOSOLVEFOR v H . L given u H . L ?? LAPLACE TRANSFORMS ** ** ** ** ** ** ** ** ** ** ** ** ** ** Recall v H t L = t W H t - s L u H s L ds whereW H t L is'generalises'e - kt ,t > 0 W H . L definedonthe" positive Half - line" Makethechangeofvariable Σ = t - s d Σ = - ds t W H t - s L u H s L ds = 0 ¥ W H Σ L u H t - Σ L d Σ CHANGE THE SUBJECT ..... COMPLEX PLANE z = x + iy - ¥ < x,y < ¥ RIGHTHALFPLANE Consistsofallpointszsuchthat Rez > Σ ‡ 0 PICTURE HERE ONTHE WHITE BOARD 2 Laplace.2.nb Printed by Mathematica for Students
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Weprefertousethelettersinsteadofz: RightHalfPlane = 8 s Res ‡ Σ ‡ 0 < Specifyingsigmaspecifiesthehalfplane. LaplaceTransform: of afunctionW H t L oftime t definedonthe HalfLine:t 0 F H s L = Limit T ¥ T 0 e - st W H t L dt = 0 ¥ e - st W H t L dt Res > Σ Σ iscalledthe"Abcissa of Convergence" H Forgetit ! L In Engineeringwejustcalculate - if theanswerdoesnotmakesense -- Wejustthrowitaway ! HEREANDBELOW F H s L sin righthalfplanewill denoteaLaplaceTransform: F H . L = L H W H . LL AfunctionintheTimeDomainbecomesafunctionintheLaplace DomainvizaRightHalfPlane. Example W H t L = e 4t t 0 Σ = 4 ForRes > 4, F H s L = Lim 0 T e - st e 4t dt = Lim 0 T e - H s - 4 L t - H s - 4 L = Lim 1 - e - H s - 4 L T s - 4 = 1 s - 4 For sNOTEQUALTO4 Blowsupifs = 4 ! s = 4 Re.s = 4isthebarrier ! W H t L = t k k t 0 k = 0 F H s L = Lim 0 T e - st 1ds = Lim I 1 - e - sT M 1 s 1 s with Σ = 0 Laplace.2.nb 3 Printed by Mathematica for Students
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ArbitrarykPositiveInteger 0 ¥ e - st t k dt Recall de - st ds = - t e - st d n e - st d n s = H - t L n e - st Hence 0 ¥ e - st t k dt = H - 1 L k d k ds k 0 ¥ e - st dt = H - 1 L k d k ds k 1 s k = 1 F H s L = 1 s 2 k = 2 F H s L = 2 s 3 k = 3 F H s L = 3.2 s 4 k F H s L = k ! s k + 1 Howabout NEGATIVEINTEGERS :W H t L = 1 t ?? Whatabout W H t L = t 1 4 4 Laplace.2.nb Printed by Mathematica for Students
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Learnaboutthe'GammaFunction" G H Α L = 0 ¥ e - t t Α- 1 dt 0 ¥ e - st t k dt make change of variable: s t This yields 0 ¥ e - st t k dt = 1 s k + 1 0 ¥ e - t t k dt = G H k + 1 L s k + 1 where we can now use noninteger values for k. Example k =- 1 2 0 ¥ e - st t k dt = G I 1 2 M s 1 2 Re.s > 0 Where in fact G H 1 2 L = Π More Examples: W H t L = Sin2 Π ft 0 ¥ e - st Sin2 Π ftdt = 0 ¥ e - st Ime i2 Π ft dt HerewecannotpulltheImout !
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