Section 6.1 INTRO to LAPLACE TRANSFORMS

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1 Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform (n integrl trnsform) Linerity

2 History of the Lplce trnsform: The Lplce trnsform is widely used integrl trnsform with mny pplictions in physics nd engineering. The Lplce trnsform is nmed fter mthemticin nd stronomer Pierre-Simon Lplce, who used similr trnsform (now clled z trnsform) in his work on probbility theory. The current widespred use of the Lplce trnsform cme bout soon fter World Wr II lthough it hd been used in the 9th century. The older history of similr trnsforms is s follows. From 744, Leonhrd Euler investigted integrls of the form s solutions of differentil equtions but did not pursue the mtter very fr. Joseph Louis Lgrnge ws n dmirer of Euler nd studied integrls of similr type. These types of integrls seem first to hve ttrcted Lplce's ttention in 782 where he ws following in the spirit of Euler in using the integrls themselves s solutions of equtions. Source:

3 Review of some clculus topics: In clculus most of the time you encountered proper integrls. Tht is, integrls b f(t) dt where f is continuous function defined on the closed nd bounded intervl [, b]. A prticulr type of improper integrl hs n infinite intervl of integrtion like f(t)dt Such n integrl is defined s limit of integrls over finite intervls; thus A A f(t)dt lim f(t)dt where A is positive rel number. If the integrl from to A exists for ech A >, nd if the limit s A exists, (tht is, hs rel numericl vlue) then the improper integrl is sid to converge to tht limiting numericl vlue. Otherwise the integrl is sid to diverge, or to fil to exist.

4 Piecewise continuous functions: A function is clled piecewise continuous on n intervl if the intervl cn be broken into finite number of subintervls on which the function is continuous on ech open subintervl (i.e. the subintervl without it s endpoints) nd hs finite limit t the endpoints of ech subintervl. Below is sketch of piecewise continuous function. In other words, piecewise continuous function is function tht hs finite number of breks in it nd doesn t blow up to infinity nywhere. If f is piecewise continuous on t b for every b >, then f is sid to be piecewise continuous on t.

5 Definitions: trnsform - to chnge the form of ( figure, expression, etc.) without chnging its vlue - mthemticl quntity obtined from given quntity by n lgebric, geometric, or functionl trnsformtion trnsformtion - the ct or process of trnsforming In mthemtics trnsforms re used to tke n originl setting of problem nd convert it (or trnsform it) into new setting in which it will be esier to solve. Exmple: The use of logrithms is n exmple of trnsformtion. We hve the following rules. log(ab) = log(a) + log(b) log(a/b) = log(a) - log(b) log(a B ) = B log(a) The process of using logrithms cn be outlined s follows. Recll logrithmic differentition.

6 The Lplce Trnsform Previously we hd differentil opertors which took function nd trnsformed it into nother function vi differentition. The Lplce trnsform is n integrl opertor which performs nother type of trnsformtion. The Lplce Trnsform is widely used in engineering differentil eqution pplictions (mechnicl nd electronic), especilly where the driving force (the nonhomogeneous term) is not continuous. The Lplce trnsform tkes single IVP or system of IVPs in the independent vrible t nd trnsforms it to n lgebr problem in new vrible s. We then solve the resulting lgebr problem, which is esier thn solving the DE. Finlly we use reverse process on the lgebric solution to obtin the solution of the originl This procedure is often summrized by sying we go from the t-domin to the s-domin, solve the lgebr problem in the s-domin, nd then invert to get the solution into the t-domin.

7 We strt by defining the Lplce trnsform of function. Definition Let f(t) be function with domin [, ). The Lplce trnsform of f is the function F defined by the integrl L st f F(s) e f(t)dt The domin of F(s) is ll vlues of s for which the integrl exists. The Lplce trnsform of f is denoted F or L{f} or L{f}. We hve st N st F(s) e f(t)dt lim e f(t)dt N whenever the integrl exists. NOTE: The Lplce trnsform involves n improper integrl nd since it involves n integrl it is liner opertor. Tht is, the Lplce trnsform of sum of functions is the sum of the individul Lplce trnsforms nd the Lplce trnsform of constnt times function is the constnt times the Lplce trnsform of the function. In symbols we hve Linerity Property This property is just like integrl nd derivtive properties in clculus.

8 Observtion: The Lplce trnsform of function f is involves the exponentil function. L st f L{f} F(s) e f(t)dt hence it Since the solutions of liner differentil equtions with constnt coefficients re bsed on the exponentil function nd sinusoids, the Lplce trnsform is prticulrly useful for such equtions. It cn lso be used when the coefficients re not constnts. The Lplce trnsform F of function f exists if f stisfies certin conditions, such s those stted in the following theorem. Theorem: Suppose tht. f is piecewise continuous on the intervl t A for ny positive A. 2. f(t) Ke t when t M. In this inequlity, K,, nd M re rel constnts, K nd M necessrily positive. Then the Lplce trnsform L{f(t)} = F(s), defined by st L f F(s) e f(t)dt exists for s >. (This restriction ensures tht the improper integrl exists. See P.39.) Functions which stisfy the condition f(t) Ke t re described s piecewise continuous nd of exponentil order. (This is sttement bout how fst the grph of f increses.)

9 The ide tht f(t) is of exponentil order mens tht f(t) does not grow fster thn constnt multiple of n exponentil function. Tht is, for some constnts K nd the grph of f(t) s t will sty below tht of Ke t. Not ll functions re of exponentil order; for exmple e t 2. Since we do not wnt to repet the sme clcultion over nd over gin, we will construct tble of Lplce trnsforms for functions tht we nticipte we will encounter. Observtions: Computtion of L st f F(s) e f(t)dt will involve limit since st N st F(s) e f(t)dt lim e f(t)dt N nd the ntiderivtive of product. Thus integrtion by prts will most likely be used. The Lplce Trnsform is defined by n improper integrl, nd thus must be checked for convergence.

10 Exmples: Insted of the script L, the stndrd L is sometimes used. Let f (t) = for t. Determine the Lplce trnsform F(s) of f. b st st e L e dt lim e dt lim, s s s b b Let f (t) = e t for t. Determine the Lplce trnsform F(s) of f. t st t b (s )t L e e e dt lim e dt (s)t e lim, s s s b b b st b Tble of Lplce Trnsforms L, s s t L e, s s We need s > so tht the improper integrl exists; tht is, the limit is finite. Integrtion by prts is not needed in either of these exmples. Find the following: L{8} L{e -7t } L{4-2e 5t } st b e e lim lim since s b s b s s s bs

11 Exmple: Let f (t) = sin(t) for t. Determine the Lplce trnsform F(s) of f. (We will need to use integrtion by prts twice.) b st b st F(s) L sin(t) e sin(t) dt lim e sin(t) dt b st s st lim (e cos(t)) / e cos(t) dt b s lim e st cos(t) dt b b b Need second integrtion by prts here! First integrtion by prts. The limit of this term is /. The limit of this term is zero. 2 s st b s b st s b 2 2 lim (e sin(t)) / e sin(t) dt F(s) Note tht we hve F(s) s F(s) 2 We get F(s) L sin(t), s 2 2 s Tble of Lplce L, s t Trnsforms s (so fr) L e, s s so we need to solve for F(s). 2 2 L sin(t) s, s

12 Exmple: Determine the Lplce trnsform F(s) of, if t 3 f(t) 5, if 3 t 6, if 6 t 25 st 3 st 6 st 25 st L f(t) f(t)e dt e dt 5e dt e dt 3 6 Determine the Lplce trnsform F(s) of This function is often clled the unit pulse. 6 st 6 6s 3s 6s 3s st 5e 5e 5e 5e 5e 5 e dt 3 s s s s s f(t), if t k, if t, if t st s st st e e e s 3 L f(t) f(t)e dt e dt,s s s s s Grph this function! Grph this function! where k is constnt. Observe tht L{f(t)} does not depend on k, the function vlue t the point of discontinuity. Even if f(t) is not defined t this point, the Lplce trnsform of f remins the sme. Thus there re mny functions, differing only in their vlue t single point, tht hve the sme Lplce trnsform.

13 Tble of Lplce L, s t Trnsforms s (so fr) L e, s s = -2 Find the Lplce trnsform of f(t) = 5e -2t -3 sin(4t), t >. 2 2 L sin(t) s = 4, s Recll tht the Lplce trnsform is liner opertor. 5 2 s 2 s 6 Tht mens L{5e -2t -3 sin(4t)} = 5 L{e -2t } - 3 L{sin(4t)} = 2

14 Exmple: Let f(t), t / 2 cos(t), t / 2 Find L(f(t)). Let s tke s fct tht Then s st L(cos(t)) e cos(t)dt 2 s st st So this is not 2 /2 L(f(t)) e f(t)dt e cos(t)dt Wht do we? s s Hint: We re only missing chunk - nmely /2 st e Lets dd zero in disguise; dd nd subtrct the chunk! st st /2 st L(f(t)) e f(t)dt e cos(t)dt e cos(t)dt Then we hve f(t) dt st s /2 st L(f(t)) e f(t)dt e cos(t)dt 2 s st st s e L(f(t)) e f(t) dt s cos(t) sin(t) 2 2 s s /2 s /2 s e e s s /2 ( s) e s s s s s s Definite integrl; compute it. From the integrl tble use #4.

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