(1) Then we could wave our hands over this and it would become:

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Preston Butler
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1 MAT* K285 Spring 28 Anthony Bnoit 4/17/28 Wk 12: Laplac Tranform Rading: Kohlr & Johnon, Chaptr 5 to p. 35 HW: 5.1: 3, 7, 1*, : 1, 5*, 13*, 19, 45* 5.3: 1, 11*, 19 * Pla writ-up th problm natly and turn thm in on 4/24. Not: Your grad will b hlpd if you turn thm in and may b harmd if you don t. What i a Laplac tranform and why? Imagin if w had a way of rmoving diffrntial (and intgral) from diffrntial (or intgral) quation. Suppo for xampl that ach drivativ wa rplacd by a impl algbraic xprion. Thn, olving th diffrntial quation would rquir only algbraic manipulation. For xampl, w could tart with a diffrntial quation uch a: y () t + ay() t = g() t (1) Thn w could wav our hand ovr thi and it would bcom: ( + a) Y C = G (2) Whr i a nw indpndnt variabl and Y and G ar tranformd function. C i jut a contant (dpndnt on initial condition). You can aily that: Y + ( + a) G = C (3) W could thn tranform Y() back into y(t) and w d b don. (Don t gt caught up in th dtail hr; thi i jut a hand-waving xampl.) In fact thr i a proc for tranforming function that w can u in jut thi way, on that convrt drivativ and intgral into algbraic xprion. Th Laplac tranform rplac a function of t with a tranformd function of. (Th variabl nam ar not important, jut that thy chang.) Th tranformation i vry impl: t { } () F = L f t = f t dt (4)

2 Laplac Tranform, Pag 2 of 8 Th ymbol L{} i ud to indicat th Laplac tranformation opration. You can u thi formula to prov om uful fatur of th Laplac tranform, but in practic, w can jut look up th tranform of common function (uch a in th tabl inid th front covr of th txt.) Baic proprti of th Laplac tranform Th firt proprty w want i xitnc. Th Laplac tranform xit for many function of intrt, but not all. But thr i a impl tt. A function which i at lat picwi continuou and which i xponntially boundd i guarantd to hav a Laplac tranform. Unboundd function ar a problm bcau th intgral can t convrg. Similarly, function with gap or infinit dicontinuiti (within a finit intrval) cau problm with intgration. Th Laplac tranform i a linar oprator (lik diffrntiation or intgration), which man that addition i ditributd ovr it: { () ()} () { } { ()} { ()} { ()} L a f t + a f t = L a f t + L a f t = al f t + a L f t (5) Anothr way to ay thi i that a linar combination can b movd in or out of th opration. Th proof of thi proprty follow traight from th dfinition of th Laplac tranform. W can t mov multiplication in and out o aily, but w can b confidnt that if f 1 and f 2 ar both picwi continuou and xponntially boundd, thn w can NOT ay that th tranform of th product i th product of th tranform. But, w can b confidnt that th product function i alo p.c. and.b., and o it too ha a Laplac tranform. A advrtid, drivativ and intgral vaporat in th tranformd domain. If a function f(t) i continuou for all non-ngativ t and f'(t) i at lat picwi continuou and i xponntially boundd. Thn: { ()} () { } ( ) L f t = L f t f (6) Thi rul can b applid quntially to gt highr ordr drivativ. And it can b rtatd to tll u how tranformd intgral bhav. Firt, rcall that: d t g( u) du dt = () g t (7) Now think carfully about how th pic of thi would fit into th prviou quation. Th right hand id i th drivativ, o it would go into th lft id of quation (6). What i inid th drivativ i th original function, o it would go into th Laplac oprator on th right id of quation (6). Think about what f() would b. Think carfully.

3 Laplac Tranform, Pag 3 of 8 Right, it would b th intgral from to. So, f() =. Thu: t 1 { f ( u) du} f () t F L = L { } = (8) So, in a n, whn w mov to th tranform domain, diffrntiation i quivalnt to multiplying by and intgration i quivalnt to dividing by. How do w gt back to th tim domain? Thr i mor than on formula that will rturn an original (tim domain) function whn fd a Laplac tranform. Th mot common u a lin intgral in th complx plan. I will how it blow, but jut covr your y if you don t want to think about it now: () γ + i t { } f t = L F = F d 2π i (9) γ i 1 1 For now at lat, w will u th tabular mthod of invr Laplac tranformation. That i, w will look for tranform domain function that look familiar and rad th original function out of a tabl (uch a 5.1 in th book or th on inid th front covr). A t of two function, on original in th tim domain, th othr it Laplac tranform, i calld a Laplac pair. Onc w hav th tranform domain function in a rcognizabl format, thi proc i fairly ay. But aftr carrying out th algbraic manipulation uch a lad to quation (3) abov, w may hav a my xprion. Th ar typically dcompod uing th mthod of partial fraction. A fw pair Bfor working an xampl, lt actually tranform a fw function. Firt of all, what happn to f(t) = 1? L 1 1 {} 1 = dt = = (1) How about f(t) = t? In thi ca, w nd to do a littl intgration by part: In fact, w could how that: L 1 1 {} t = t dt t dt = = 2 (11) n! L (12) n { t } = n 1 How about xponntial function, of th form f(t) = αt? Apply th tranform: +

4 Laplac Tranform, Pag 4 of 8 L αt t t ( ) t 1 { } = α α dt = dt for all α = > α (13) To gt th tranform of coin and in, w nd to u intgration by part twic. That rturn our original intgral and w can olv for it: L t coωt ω { coωt} = coωtdt = + 2 ( inωt ω coωtdt ) But th intgral on th far right i th on w tartd with: (14) 2 coωt ω inωt ω L{ coωt} = + L { coωt} (15) If w rtrict to valu gratr than zro, thn th bracktd xprion bcom 1/. Solving for th tranform: Th am raoning lad to: Hr an intrting thing: L { co ωt} = for all + ω > (16) ω L { in ωt} = for all > (17) + ω ω L co + ( L in ) = ω + ω = (18) ( { ωt} ) { ωt} But don t b mild. Product of tranform ar not ncarily th tranform of th product. 1 L{ co ωt} + L{ in ωt} = L{ co ωt + in ωt} = L{} 1 = ( L{ coωt} ) L{ inωt} + (19) Th hift Rcall that w can tranform picwi continuou function providd that thy ar xponntially boundd. On uch function i th tp function, which i zro for all ngativ valu but qual to on for any valu gratr than or qual to zro:

5 Laplac Tranform, Pag 5 of 8 () ht for t < = 1 for t (2) Thi function i picwi continuou, o w can tranform it. In fact inc w intgrat only ovr t whn tranforming a function, th tp function tranform a if it wr f(t) = 1. So, 1 L { ht ()} = (21) But om intrting thing happn whn w look at th hiftd tp function. Thi jump from zro to on at t qual to om valu α: ( α ) ht for t < α = 1 for t α (22) W can tranform thi radily by hand: L α ht α = ht α dt= dt for all and α = = > (23) α { } Notic that thi i conitnt with quation (21). In fact, th rlationhip btwn (21) and (23) hold for othr function for which tranform xit. In gnral, a hiftd function [f(t - α)] ha th am tranform a th non-hiftd function but multiplid by -α. Mor prcily: α { f ( t α) h( t α) } = F a L (24) So, hifting a function in th tim domain cau it tranform to b multiplid by an xponntial. It i fairly ay to tablih a complmntary proprty: Multiplying a function by an xponntial in th tim domain hift th tranform: OK, an xampl { α t f () t } = F ( α ) L (25) U Laplac tranform to olv th following initial valu problm: whn t < 4 y () t y() t = g() t whr g() t = with y 3t ( ) = 1 whn 4 t < (26) Th forcing function can b rwrittn uing th tp function: () ( t ) g t h t h t ( t ) = 4 = 4 (27)

6 Laplac Tranform, Pag 6 of 8 Tak th Laplac tranform of both id of quation (26): So: 12 3( t 4) { } { () ()} = ( 4) L y t y t L h t (28) Y y ( ) Y = (29) Applying th initial condition: = 3 ( ) Y (3) So: Y = (31) W can u partial fraction to dcompo th firt trm on th right: A B = + A 3A+ B B = (32) So: A + B = A = B 1 1 3A+ B = 1 2B = 1 B = A= (33) Writing out Y() rady for invr tranforming: Y = (34) Not that: 4 4 L 1 3( t 4) 1 t 4 = ht ( 4 ) and = ht ( 4) 3 L 1 (35) So:

7 Laplac Tranform, Pag 7 of ( t 4) t 4 t y () t = h( t 4) h( t 4) + (36) Or: ( 4) 3 8 ht t t+ t y () t = + 2 (37)

8 Laplac Tranform, Pag 8 of 8 Rfrnc Laplac tranform. 12 April 29. In Wikipdia: Th Fr Encyclopdia. Rtrivd 17 April 28 from Kohlr, W. and L. Johnon. 26. Elmntary Diffrntial Equation, 2 nd d. Boton: Paron/Addion-Wly.

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