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
Review of some clculus topics: In clculus most of the time you encountered proper integrls. Tht is, integrls f(t) dt where f is continuous function defined on the closed nd ounded intervl [, ]. 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 numer. 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.
Piecewise continuous functions: A function is clled piecewise continuous on n intervl if the intervl cn e roken into finite numer of suintervls on which the function is continuous on ech open suintervl (i.e. the suintervl without it s endpoints) nd hs finite limit t the endpoints of ech suintervl. Below is sketch of piecewise continuous function. In other words, piecewise continuous function is function tht hs finite numer of reks in it nd doesn t low up to infinity nywhere. If f is piecewise continuous on t for every >, then f is sid to e piecewise continuous on t.
Definitions: trnsform - to chnge the form of ( figure, expression, etc.) without chnging its vlue - mthemticl quntity otined from given quntity y n lgeric, geometric, or functionl trnsformtion trnsformtion - the ct or process of trnsforming In mthemtics trnsforms re used to tke n originl setting of prolem nd convert it (or trnsform it) into new setting in which it will e 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 e outlined s follows. Recll logrithmic differentition.
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 vrile t nd trnsforms it to n lger prolem in new vrile s. We then solve the resulting lger prolem, which is esier thn solving the DE. Finlly we use reverse process on the lgeric solution to otin the solution of the originl This procedure is often summrized y sying we go from the t-domin to the s-domin, solve the lger prolem in the s-domin, nd then invert to get the solution into the t-domin.
We strt y defining the Lplce trnsform of function. Definition Let f(t) e function with domin [, ). The Lplce trnsform of f is the function F defined y 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 symols we hve Linerity Property This property is just like integrl nd derivtive properties in clculus.
Oservtion: 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 sed on the exponentil function nd sinusoids, the Lplce trnsform is prticulrly useful for such equtions. It cn lso e 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 y 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 descried s piecewise continuous nd of exponentil order. (This is sttement out how fst the grph of f increses.)
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 elow 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 tle of Lplce trnsforms for functions tht we nticipte we will encounter. Oservtions: 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 y prts will most likely e used. The Lplce Trnsform is defined y n improper integrl, nd thus must e checked for convergence.
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. st st e L e dt lim e dt lim, s s s Let f (t) = e t for t. Determine the Lplce trnsform F(s) of f. t st t (s )t L e e e dt lim e dt (s)t e lim, s s s st Tle 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 y prts is not needed in either of these exmples. Find the following: L{8} st L{4-2e 5t } s e e lim lim since s s s s s L{e -7t }
Exmple: Let f (t) = sin(t) for t. Determine the Lplce trnsform F(s) of f. (We will need to use integrtion y prts twice.) st st F(s) L sin(t) e sin(t) dt lim e sin(t) dt st s st lim (e cos(t)) / e cos(t) dt s lim e st cos(t) dt Need second integrtion y prts here! First integrtion y prts. The limit of this term is /. The limit of this term is zero. 2 s st s st s 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 Tle 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
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. Oserve 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.
Tle 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
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 sutrct 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 2 2 2 2 2 s s s s s Definite integrl; compute it. From the integrl tle use #4.