Chapter 2 Further Properties of the Laplace Transform

Transcription

1 Chapter 2 Further Propertie of the Laplace Tranform 2.1 Real Function Sometime, a function F(t) repreent a natural or engineering proce that ha no obviou tarting value. Statitician call thi a time erie. Although we hall not be conidering F(t) a tochatic, it i neverthele worth introducing a way of witching on a function. Let u tart by finding the Laplace tranform of a tep function the name of which pay homage to the pioneering electrical engineer Oliver Heaviide ( ). The formal definition run a follow. Definition 2.1 Heaviide unit tep function, or imply the unit tep function, i defined a { t < H(t) = 1 t. Since H(t) i preciely the ame a 1 for t >, the Laplace tranform of H(t) mut be the ame a the Laplace tranform of 1, i.e. 1/. The witching on of an arbitrary function i achieved imply by multiplying it by the tandard function H(t), o if F(t) i given by the function hown in Fig. 2.1 and we multiply thi function by the Heaviide unit tep function H(t) to obtain H(t)F(t),Fig.2.2 reult. Sometime it i neceary to define what i called the two ided Laplace tranform e t F(t)dt which make a great deal of mathematical ene. However the additional problem that arie by allowing negative value of t are evere and limit the ue of the two ided Laplace tranform. For thi reaon, the two ided tranform will not be purued here. P. Dyke, An Introduction to Laplace Tranform and Fourier Serie, 13 Springer Undergraduate Mathematic Serie, DOI: 1.17/ _2, Springer-Verlag London 214

2 14 2 Further Propertie of the Laplace Tranform Fig. 2.1 F(t), a function with no well defined tarting value Fig. 2.2 H(t)F(t), the function i now zero before t = 2.2 Derivative Property of the Laplace Tranform Suppoe a differentiable function F(t) ha Laplace tranform f (), we can find the Laplace tranform L{F (t)} = e t F (t)dt of it derivative F (t) through the following theorem. Theorem 2.1 L{F (t)} = e t F (t)dt = F() + f().

3 2.2 Derivative Property of the Laplace Tranform 15 Proof Integrating by part once give L{F (t)} = [ F(t)e t] + e t F(t)dt = F() + f() where F() i the value of F(t) at t =. Thi i an important reult and lie behind future application that involve olving linear differential equation. The key property i that the tranform of a derivative F (t) doe not itelf involve a derivative, only F() + f() which i an algebraic expreion involving f (). The downide i that the value F() i required. Effectively, an integration ha taken place and the contant of integration i F(). Later, thi i exploited further through olving differential equation. Later till in thi text, partial differential equation are olved, and wavelet are introduced. Let u proceed here by finding the Laplace tranform of the econd derivative of F(t). We alotate thi in the form of a theorem. Theorem 2.2 If F(t) i a twice differentiable function of t then L{F (t)} = 2 f () F() F (). Proof The proof i unremarkable and involve integrating by part twice. Here are the detail. L{F (t)} = e t F (t)dt = [ F (t)e t] + e t F (t)dt = F () + [ F(t)e t] + 2 e t F(t)dt = F () F() + 2 f () = 2 f () F() F (). The general reult, proved by induction, i L{F (n) (t)} = n f () n 1 F() n 2 F () F (n 1) () where n i a poitive integer. Note the appearance of n contant on the right hand ide. Thi of coure i the reult of integrating thi number of time. Thi reult, a we have aid, ha wide application o it i worth getting to know. Conider the reult

4 16 2 Further Propertie of the Laplace Tranform L(in(ωt)) = ω 2 + ω 2. Now, o uing the formula d (in(ωt)) = ω co(ωt) dt L(F (t)) = f() F() with F(t) = in(ωt) we have o ω L{ω co(ωt)} = 2 + ω 2 L{co(ωt)} = 2 + ω 2 another tandard reult. Another appropriate quantity to find at thi point i the determination of the value of the Laplace tranform of t F(u)du. Firt of all, the function F(t) mut be integrable in uch a way that g(t) = t F(u)du i of exponential order. From thi definition of g(t) it i immediately apparent that g() = and that g (t) = F(t). Thi latter reult i called the fundamental theorem of the calculu. We can now ue the reult L{g (t)} =g() g() to obtain L{F(t)} = f () = g() where we have written L{g(t)} =g(). Hence which finally give the reult g() = f () ( t ) L F(u)du = f ().

5 2.2 Derivative Property of the Laplace Tranform 17 The following reult i alo ueful and can be tated in the form of a theorem. { } F(t) Theorem 2.3 If L(F(t)) = f () then L = f (u)du, auming that t { } F(t) L a. t Proof Let G(t) be the function F(t)/t, o that F(t) = tg(t). Uing the property L{tG(t)} = d d L{G(t)} we deduce that f () = L{F(t)} = d { } F(t) d L. t Integrating both ide of thi with repect to from to give [ { }] F(t) { } F(t) { } F(t) f (u)du = L = L t t = L t ince { } F(t) L a t which complete the proof. The function Si(t) = t in u u du define the Sine Integral function which occur in the tudy of optic. The formula for it Laplace tranform can now be eaily derived a follow. Let F(t) = in t in the reult ( ) F(t) L = f (u)du t to give ( ) in t L = t = du 1 + u 2 [ tan 1 (u) ] = π 2 tan 1 () ( ) 1 = tan 1.

6 18 2 Further Propertie of the Laplace Tranform We now ue the reult ( t ) L F(u)du = f () to deduce that ( t L ) in u u du = L{Si(t)} = 1 ( ) 1 tan Heaviide Unit Step Function A promied earlier, we devote thi ection to exploring ome propertie of Heaviide unit tep function H(t). The Laplace tranform of H(t) ha already been hown to be the ame a the Laplace tranform of 1, i.e. 1/. The Laplace tranform of H(t t ), t >, i a little more enlightening: L{H(t t )}= H(t t )e t dt. Now, ince H(t t ) = fort < t thi Laplace tranform i L{H(t t )}= t e t dt = ] [ e t Thi reult i generalied through the following theorem. t = e t. Theorem 2.4 (Second Shift Theorem) If F(t) i a function of exponential order in t then L{H(t t )F(t t )}=e t f () where f () i the Laplace tranform of F(t). Proof Thi reult i proved by direct integration. L{H(t t )F(t t )}= = = H(t t )F(t t )e t dt t F(t t )e t dt (by definition of H) = e t f (). F(u)e (u+t ) du (writing u = t t ) Thi etablihe the theorem.

7 2.3 Heaviide Unit Step Function 19 The only condition on F(t) i that it i a function that i of exponential order which mean of coure that it i free from ingularitie for t > t. The principal ue of thi theorem i that it enable u to determine the Laplace tranform of a function that i witched on at time t = t. Here i a traightforward example. Example 2.1 Determine the Laplace tranform of the ine function witched on at time t = 3. Solution The ine function required that tart at t = 3iS(t) where S(t) = { in tt 3 t < 3. We can ue the Heaviide tep function to write S(t) = H(t 3) in t. The econd hift theorem can then be ued by utiliing the ummation formula o in t = in(t 3 + 3) = in(t 3) co(3) + co(t 3) in(3) L{S(t)} =L{H(t 3) in(t 3)} co(3) + L{H(t 3) co(t 3)} in(3). Thi may eem a trange tep to take, but in order to ue the econd hift theorem it i eential to get the argument of both the Heaviide function and the target function in the quetion the ame; in thi cae (t 3). We can now ue the econd hift theorem directly to give or L{S(t)} =e 3 1 co(3) e 3 in(3) L{S(t)} =(co 3 + in 3)e 3 /( 2 + 1). 2.4 Invere Laplace Tranform Virtually all operation have invere. Addition ha ubtraction, multiplication ha diviion, differentiation ha integration. The Laplace tranform i no exception, and we can define the Invere Laplace tranform a follow. Definition 2.2 If F(t) ha the Laplace tranform f (), that i L{F(t)} = f ()

8 2 2 Further Propertie of the Laplace Tranform then the invere Laplace tranform i defined by and i unique apart from null function. L 1 { f ()} =F(t) Perhap the mot important property of the invere tranform to etablih i it linearity. We tate thi a a theorem. Theorem 2.5 The invere Laplace tranform i linear, i.e. L 1 {af 1 () + bf 2 ()} =al 1 { f 1 ()}+bl 1 { f 2 ()}. Proof Linearity i eaily etablihed a follow. Since the Laplace tranform i linear, we have for uitably well behaved function F 1 (t) and F 2 (t): L{aF 1 (t) + bf 2 (t)} =al{f 1 (t)}+bl{f 2 (t)} =af 1 () + bf 2 (). Taking the invere Laplace tranform of thi expreion give which i the ame a af 1 (t) + bf 2 (t) = L 1 {af 1 () + bf 2 ()} al 1 { f 1 ()}+bl 1 { f 2 ()} =L 1 {af 1 () + bf 2 ()} and thi ha etablihed linearity of L 1 { f ()}. Another important property i uniquene. It ha been mentioned that the Laplace tranform wa indeed unique apart from null function (function whoe Laplace tranform i zero). It follow immediately that the invere Laplace tranform i alo unique apart from the poible addition of null function. Thee take the form of iolated value and can be dicounted for all practical purpoe. A i quite common with invere operation there i no ytematic method of determining invere Laplace tranform. The calculu provide a good example where there are plenty of ytematic rule for differentiation: the product rule, the quotient rule, the chain rule. However by contrat there are no ytematic rule for the invere operation, integration. If we have an integral to find, we may try ubtitution or integration by part, but there i no guarantee of ucce. Indeed, the integral may not be poible to expre in term of elementary function. Derivative that exit can alway be found by uing the rule; thi i not o for integral. The ituation regarding the Laplace tranform i not quite the ame in that it may not be poible to find L{F(t)} explicitly becaue it i an integral. There i certainly no guarantee of being able to find L 1 { f ()} and we have to devie variou method of trying o to

9 2.4 Invere Laplace Tranform 21 do. For example, given an arbitrary function of there i no guarantee whatoever that a function of t can be found that i it invere Laplace tranform. One neceary condition for example i that the function of mut tend to zero a. When we are certain that a function of ha arien from a Laplace tranform, there are technique and theorem that can help u invert it. Partial fraction implify rational function and can help identify tandard form (the exponential and trigonometric function for example), then there are the hift theorem which we have jut met which extend further the repertoire of tandard form. Engineering text pend a coniderable amount of pace building up a library of pecific invere Laplace tranform and to way of extending thee via the calculu. To a certain extent we need to do thi too. Therefore we next do ome reaonably elementary example. Note that in Appendix B there i a lit of ome invere Laplace tranform. Example 2.2 Ue partial fraction to determine { } L 1 a 2 a 2 Solution Noting that a 2 a 2 = 1 [ 1 2 a 1 ] + a give traight away that { } L 1 a 2 a 2 = 1 2 (eat e at ) = inh(at). The firt hift theorem ha been ued on each of the function 1/( a) and 1/( +a) together with the tandard reult L 1 {1/} =1. Here i another example. Example 2.3 Determine the value of L 1 { 2 ( + 3) 3 }. Solution Noting the tandard partial fraction decompoition 2 ( + 3) 3 = ( + 3) ( + 3) 3 we ue the firt hift theorem on each of the three term in turn to give L 1 { 2 ( + 3) 3 } = L L 1 6 ( + 3) 2 + L 1 9 ( + 3) 3 = e 3t 6te 3t t2 e 3t

10 22 2 Further Propertie of the Laplace Tranform where we have ued the linearity property of the L 1 operator. Finally, we do the following four-in-one example to hone our kill. Example 2.4 Determine the following invere Laplace tranform (a) L 1 ( + 3) ( 1)( + 2) ; (b) ( 1) L ; (c) e 7 L 1 2 ; (d) L ( + 3) 3. Solution All of thee problem are tackled in a imilar way, by decompoing the expreion into partial fraction, uing hift theorem, then identifying the implified expreion with variou tandard form. (a) Uing partial fraction decompoition and not dwelling on the detail we get + 3 ( 1)( + 2) = ( 1) + 1 6( + 2). Hence, operating on both ide with the invere Laplace tranform operator give L ( 1)( + 2) = L L 1 3( 1) + 1 L 1 6( + 2) = L L L uing the linearity property of L 1 once more. Finally, uing the tandard form, we get { } L = 3 ( 1)( + 2) et e 2t. (b) The expreion i factoried to ( + 4)( 2) which, uing partial fraction i 1 6( 2) + 5 6( + 4). Therefore, taking invere Laplace tranform give L = 1 6 e2t e 4t.

11 2.4 Invere Laplace Tranform 23 (c) The denominator of the rational function doe not factorie. In thi cae we ue completing the quare and tandard trigonometric form a follow: So = ( 1) + 1 ( 1) 2 = + 4 ( 1) L = 3L 1 ( 1) ( 1) L 1 2 ( 1) = 3e t co(2t) + 5e t in(2t). Again, the firt hift theorem ha been ued. (d) The final invere Laplace tranform i lightly different. The expreion e 7 ( 3) 3 contain an exponential in the numerator, therefore it i expected that the econd hift theorem will have to be ued. There i a little fiddling that need to take place here. Firt of all, note that L 1 1 ( 3) 3 = 1 2 t2 e 3t uing the firt hift theorem. So L 1 e 7 ( 3) 3 = { 12 (t 7) 2 e 3(t 7) t > 7 t 7. Of coure, thi can uccinctly be expreed uing the Heaviide unit tep function a 1 2 H(t 7)(t 7)2 e 3(t 7). We hall get more practice at thi kind of inverion exercie, but you hould try your hand at a few of the exercie at the end.

12 24 2 Further Propertie of the Laplace Tranform 2.5 Limiting Theorem In many branche of mathematic there i a neceity to olve differential equation. Later chapter give detail of how ome of thee equation can be olved by uing Laplace tranform technique. Unfortunately, it i ometime the cae that it i not poible to invert f () to retrieve the deired olution to the original problem. Numerical inverion technique are poible and thee can be found in ome oftware package, epecially thoe ued by control engineer. Inight into the behaviour of the olution can be deduced without actually olving the differential equation by examining the aymptotic character of f () for mall or large. In fact, it i often very ueful to determine thi aymptotic behaviour without olving the equation, even when exact olution are available a thee olution are often complex and difficult to obtain let alone interpret. In thi ection two theorem that help u to find thi aymptotic behaviour are invetigated. Theorem 2.6 (Initial Value) If the indicated limit exit then lim F(t) = lim f(). t (The left hand ide i F() of coure, or F(+) if lim t F(t) i not unique.) Proof We have already etablihed that L{F (t)} =f() F(). (2.1) However, if F (t) obey the uual criteria for the exitence of the Laplace tranform, that i F (t) i of exponential order and i piecewie continuou, then e t F (t)dt e t F (t) dt = 1 M e t e Mt dt a. Thu letting in Eq. (2.1) yield the reult. Theorem 2.7 (Final Value) If the limit indicated exit, then lim F(t) = lim f(). t Proof Again we tart with the formula for the Laplace tranform of the derivative of F(t)

13 2.5 Limiting Theorem 25 L{F (t)} = e t F (t)dt = f() F() (2.2) thi time writing the integral out explicitly. The limit of the integral a i lim Thu we have, uing Eq. (2.2), e t F (t)dt = lim lim T T e t F (t)dt = lim lim T {e T F(T ) F()} = lim F(T ) F() T = lim F(t) F(). t lim F(t) F() = lim f() F() t from which, on cancellation of F(), the theorem follow. Since the improper integral converge independently of the value of and all limit exit (a priori aumption), it i therefore correct to have aumed that the order of the two procee (taking the limit and performing the integral) can be exchanged. (Thi ha in fact been demontrated explicitly in thi proof.) Suppoe that the function F(t) can be expreed a a power erie a follow F(t) = a + a 1 t + a 2 t 2 + +a n t n +. If we aume that the Laplace tranform of F(t) exit, F(t) i of exponential order and i piecewie continuou. If, further, we aume that the power erie for F(t) i abolutely and uniformly convergent the Laplace tranform can be applied term by term L{F(t)} = f () = L{a + a 1 t + a 2 t 2 + +a n t n + } = a L{1}+a 1 L{t}+a 2 L{t 2 }+ +a n L{t n }+ provided the tranformed erie i convergent. Uing the tandard form the right hand ide become a + a a 2 3 L{t n }= n! n n!a n +. n+1

14 26 2 Further Propertie of the Laplace Tranform Hence f () = a + a a n!a n +. n+1 Example 2.5 Demontrate the initial and final value theorem uing the function F(t) = e t. Expand e t a a power erie, evaluate term by term and confirm the legitimacy of term by term evaluation. Solution L{e t }= lim F(t) = F() = t e = 1 lim f() = lim + 1 = 1. Thi confirm the initial value theorem. The final value theorem i alo confirmed a follow:- lim F(t) = lim t t e t = The power erie expanion for e t i lim f() = lim + 1 =. e t = 1 t + t2 2! t3 tn + +( 1)n 3! n! L{e t }= ( 1)n + 3 n+1 = 1 ( ) 1 = Hence the term by term evaluation of the power erie expanion for e t give the right anwer. Thi i not a proof of the erie expanion method of coure, merely a verification that the method give the right anwer in thi intance. 2.6 The Impule Function There i a whole cla of function that, trictly, are not function at all. In order to be a function, an expreion ha to be defined for all value of the variable in the pecified range. When thi i not o, then the expreion i not a function becaue it i

15 2.6 The Impule Function 27 not well defined. It may not eem at all enible for u to bother with uch creature, in that if a function i not defined at a certain point then what ue i it? However, if a function intead of being well defined poee ome global property, then it indeed doe turn out to be worth conidering uch pathological object. Of coure, having taken the deciion to conider uch object, trictly there need to be a whole new mathematical language contructed to deal with them. Notion uch a adding them together, multiplying them, performing operation uch a integration cannot be done without preliminary mathematic. The general conideration of thi kind of object form the tudy of generalied function (ee Jone 1966 or Lighthill 197) which i outide the cope of thi text. For our purpoe we introduce the firt uch function which occurred naturally in the field of electrical engineering and i the o called impule function. It i ometime called Dirac δ function after the pioneering theoretical phyicit P.A.M. Dirac ( ). It ha the following definition which involve it integral. Thi ha not been defined properly, but if we write the definition firt we can then comment on the integral. Definition 2.3 The Dirac-δ function δ(t) i defined a having the following propertie δ(t) = t, t = (2.3) h(t)δ(t)dt = h() (2.4) for any function h(t) continuou in (, ). We hall ee in the next paragraph that the Dirac-δ function can be thought of a the limiting cae of a top hat function of unit area a it become infiniteimally thin but infinitely tall, i.e. the following limit δ(t) = lim T T p(t) where t 1/T 1 T p (t) = 2 T 1/T < t < 1/T t 1/T. The integral in the definition can then be written a follow: h(t) lim T p(t)dt = lim h(t)t p (t)dt T T provided the limit can be exchanged which of coure depend on the behaviour of the function h(t) but thi can be o choen to fulfil our need. The integral inide the limit exit, being the product of continuou function, and it value i the area under the curve h(t)t p (t). Thi area will approach the value h() a T by the following argument. For ufficiently large value of T, the interval [ 1/T, 1/T ]

16 28 2 Further Propertie of the Laplace Tranform will be mall enough for the value of h(t) not to differ very much from it value at the origin. In thi cae we can write h(t) = h() + ɛ(t) where ɛ(t) i in ome ene mall and tend to zero a T. The integral thu can be een to tend to h() a T and the property i etablihed. Returning to the definition of δ(t) trictly, the firt condition i redundant; only the econd i neceary, but it i very convenient to retain it. Now a we have aid, δ(t) i not a true function becaue it ha not been defined for t =. δ() ha no value. Equivalent condition to Eq. (2.4) are:- h(t)δ(t)dt = h() and + h(t)δ(t)dt = h(). Thee follow from a imilar argument a before uing a limiting definition of δ(t) in term of the top hat function. In thi ection, wherever the integral of a δ function (or later related derivative ) occur it will be aumed to involve thi kind of limiting proce. The detail of taking the limit will however be omitted. Let u now look at a more viual approach. A we have een algebraically in the lat paragraph δ(t) i ometime called the impule function becaue it can be thought of a the hape of Fig. 2.3, the top hat function if we let T. Of coure there are many hape that will behave like δ(t) in ome limit. The top hat function i one of the implet to tate and viualie. The crucial property i that the area under thi top hat function i unity for all value of T, o letting T preerve thi property. Diagrammatically, the Dirac-δ or impule function i repreented by an arrow a in Fig. 2.4 where the length of the arrow i unity. Uing Eq. (2.4) with h 1 we ee that δ(t)dt = 1 which i conitent with the area under δ(t) being unity. We now ak ourelve what i the Laplace tranform of δ(t)? Doe it exit? We upect that it might be 1 for Eq. (2.4) with h(t) = e t, a perfectly valid choice of h(t) give δ(t)e t dt = δ(t)e t dt = 1. However, we progre with care. Thi i good advice when dealing with generalied function. Let u take the Laplace tranform of the top hat function T p (t) defined mathematically by t 1/T 1 T p (t) = 2 T 1/T < t < 1/T t 1/T.

17 2.6 The Impule Function 29 Fig. 2.3 The top hat function Fig. 2.4 The Dirac-δ function The calculation proceed a follow:- L{T p (t)} = 1/T T p (t)e t dt 1 = 2 Te t dt = [ T2 ] 1/T e t = [ T 2 T ] 2 e /T.

18 3 2 Further Propertie of the Laplace Tranform A T, hence e /T 1 ( ) 1 T + O T 2 T 2 T 2 e /T O ( ) 1 T which 2 1 a T. In Laplace tranform theory it i uual to define the impule function δ(t) uch that L{δ(t)} =1. Thi mean reducing the width of the top hat function o that it lie between and 1/T (not 1/T and 1/T ) and increaing the height from 2 1 T to T in order to preerve unit area. Clearly the difficulty arie becaue the impule function i centred on t = which i preciely the lower limit of the integral in the definition of the Laplace tranform. Uing - a the lower limit of the integral overcome many of the difficultie. The function δ(t t ) repreent an impule that i centred on the time t = t.it can be conidered to be the limit of the function K (t) where K (t) i the diplaced top hat function defined by t t 1/2T 1 K (t) = 2 T t 1/2T < t < t + 1/2T t t + 1/2T a T. The definition of the delta function can be ued to deduce that h(t)δ(t t )dt = h(t ) and that, provided t > Letting t lead to L{δ(t t )}=e t. L{δ(t)} =1 a correct reult. Another intereting reult can be deduced almot at once and expree mathematically the property of δ(t) to pick out a particular function value, known to engineer a the filtering property. Since h(t)δ(t t )dt = h(t )

19 2.6 The Impule Function 31 with h(t) = e t f (t) and t = a we deduce that L{δ(t a) f (t)} =e a f (a). Mathematically, the impule function ha additional interet in that it enable inight to be gained into the propertie of dicontinuou function. From a practical point of view too there are a number of real phenomena that are cloely approximated by the delta function. The harp blow from a hammer, the dicharge of a capacitor or even the ound of the bark of a dog are all in ome ene impule. All of thi provide motivation for the tudy of the delta function. One property that i particularly ueful in the context of Laplace tranform i the value of the integral t δ(u u )du. Thi ha the value if u > t and the value 1 if u < t. Thu we can write or t t { < u δ(u u )du = 1 t > u δ(u u )du = H(t u ) where H i Heaviide unit tep function. If we were allowed to differentiate thi reult, or to put it more formally to ue the fundamental theorem of the calculu (on function one of which i not really a function, a econd which i not even continuou let alone differentiable) then one could write that δ(u u ) = H (u u ) or tate that the impule function i the derivative of the Heaviide unit tep function. Before the pure mathematician end out lynching partie, let u examine thee looe notion. Everywhere except where u = u the tatement i equivalent to tating that the derivative of unity i zero, which i obviouly true. The additional information in the albeit looe tatement in quotation mark i a quantification of the nature of the unit jump in H(u u ). We know the gradient there i infinite, but the nature of it i embodied in the econd integral condition in the definition of the delta function, Eq. (2.4). The ubject of generalied function i introduced through thi concept and the intereted reader i directed toward the text by Jone and Lighthill. All that will be noted here i that it i poible to define a whole tring of derivative δ (t), δ (t), etc. where all thee derivative are zero everywhere except at t =. The key to keeping rigorou here i the property h(t)δ(t)dt = h(). The derivative have analogou propertie, viz.

20 32 2 Further Propertie of the Laplace Tranform h(t)δ (t)dt = h () and in general h(t)δ (n) (t)dt = ( 1) n h (n) (). Of coure, the function h(t) will have to be appropriately differentiable. Now the Laplace tranform of thi nth derivative of the Dirac delta function i required. It can be eaily deduced that e t δ (n) (t)dt = e t δ (n) (t)dt = n. Notice that for all thee generalied function, the condition for the validity of the initial value theorem i violated, and the final value theorem although perfectly valid i entirely uele. It i time to do a few example. Example 2.6 Determine the invere Laplace tranform and interpret the F(t) obtained. { L 1 2 } Solution Writing = and uing the linearity property of the invere Laplace tranform give { L 1 2 } { } 1 2 = L 1 {1} L = δ(t) in t. Thi function i inuoidal with a unit impule at t =. Note the direct ue of the invere L 1 {1} =δ(t). Thi arie traight away from our definition of L. It i quite poible for other definition of Laplace tranform to give the value 2 1 for L{δ(t)} (for example). Thi may worry thoe reader of a pure mathematical bent. However, a long a there i conitency in the definition of the delta function and the Laplace tranform and hence it invere, then no inconitencie { arie. The example given above will alway yield the ame anwer L 1 2 } 2 = δ(t) in t. The mall variation poible in the definition of the + 1 Laplace tranform around t = do not change thi. Our definition, viz.

21 2.6 The Impule Function 33 L{F(t)} = e t F(t)dt remain the mot uual. { Example 2.7 Find the value of L 1 3 } Solution Uing a imilar technique to the previou example we firt ee that = o taking invere Laplace tranform uing the linearity property once more yield { L 1 3 } { } 2 = L 1 {} L = δ (t) co t where δ (t) i the firt derivative of the Dirac-δ function which wa defined earlier. Notice that the firt derivative formula: with F (t) = δ (t) co t give L{F (t)} =f() F() L{δ (t) co t} = F() which i indeed the above reult apart from the troubleome F(). F() i of coure not defined. Care indeed i required if tandard Laplace tranform reult are to be applied to problem containing generalied function. When in doubt, the bet advice i to ue limit definition of δ(t) and the like, and follow the mathematic through carefully, epecially the wapping of integral and limit. The little book by Lighthill i full of excellent practical advice. 2.7 Periodic Function We begin with a very traightforward definition that hould be familiar to everyone: Definition 2.4 If F(t) i a function that obey the rule F(t) = F(t + τ) for ome real τ for all value of t then F(t) i called a periodic function with period τ.

22 34 2 Further Propertie of the Laplace Tranform Periodic function play a very important role in many branche of engineering and applied cience, particularly phyic. One only ha to think of pring or alternating current preent in houehold electricity to realie their prevalence. Here, a theorem on the Laplace tranform of periodic function i introduced, proved and ued in ome illutrative example. Theorem 2.8 Let F(t) have period T > o that F(t) = F(t + T ). Then L{F(t)} = T e t F(t)dt 1 e T. Proof Like many proof of propertie of Laplace tranform, thi one begin with it definition then evaluate the integral by uing the periodicity of F(t) L{F(t)} = = T 3T + e t F(t)dt 2T e t F(t)dt + 2T T e t F(t)dt + + e t F(t)dt nt (n 1)T e t F(t)dt + provided the erie on the right hand ide i convergent. Thi i aured ince the function F(t) atifie the condition for the exitence of it Laplace tranform by contruction. Conider the integral nt (n 1)T e t F(t)dt and ubtitute u = t (n 1)T. Since F ha period T thi lead to nt which give (n 1)T T e t F(t)dt = e (n 1)T e u F(u)du n = 1, 2,... T e t F(t)dt = (1 + e T + e 2T + ) = T e t F(t)dt 1 e T e t F(t)dt on umming the geometric progreion. Thi prove the reult.

23 2.7 Periodic Function 35 Here i an example of uing thi theorem. Example 2.8 A rectified ine wave i defined by the expreion determine L {F(t)}. { in t < t < π F(t) = in t π < t < 2π F(t) = F(t + 2π) Solution The graph of F(t) ihowninfig.2.5. The function F(t) actually ha period π, but it i eaier to carry out the calculation a if the period wa 2π. Additionally we can check the anwer by uing the theorem with T = π. With T = 2π we have from Theorem 2.8, 2π e t F(t)dt L{F(t)} = 1 e T where the integral in the numerator i evaluated by plitting into two a follow:- 2π e t F(t)dt = π e t in tdt + 2π π e t ( in t)dt. Now, writing I{} to denote the imaginary part of the function in the brace we have π { π e t in tdt =I } e t+it dt ] π [ 1 =I i e t+it { } 1 =I i (e π+iπ 1) { } 1 =I i (1 + e π ). So Similarly, 2π π π e t in tdt = 1 + e π e t in tdt = e 2π + e π Hence we deduce that

24 36 2 Further Propertie of the Laplace Tranform Fig. 2.5 The graph of F(t) (1 + e π ) 2 L{F(t)} = (1 + 2 )(1 e 2π ) 1 + e π = (1 + 2 )(1 e π ). Thi i preciely the anwer that would have been obtained if Theorem 2.8 had been applied to the function F(t) = in t < t < π F(t) = F(t + π). We can therefore have ome confidence in our anwer. 2.8 Exercie 1. If F(t) = co(at), ue the derivative formula to re-etablih the Laplace tranform of in(at). 2. Ue Theorem 2.1 with t in u F(t) = u du to etablih the reult.

25 2.8 Exercie 37 { } in(at) L t { = tan 1 a }. 3. Prove that 4. Find 5. Determine { t v } L F(u)dudv = f () 2. { t L } co(au) co(bu) du. u { 2int inh t L t }. 6. Prove that if f () indicate the Laplace tranform of a piecewie continuou function f (t) then lim f () =. 7. Determine the following invere Laplace tranform by uing partial fraction (a) (c) (e) 2(2 + 7) + 9, > 2 (b) ( + 4)( + 2) 2 9, 2 + 2k 2 ( 2 + 4k 2 ), (d) 1 ( + 3) 2, 1 ( 2) 2 ( + 3) Verify the initial value theorem, for the two function (a) 2 + co t and (b) (4 + t) Verify the final value theorem, for the two function (a) 3 + e t and (b) t 3 e t. 1. Given that L{in( t)} = k e 1/4 3/2 ue in x x near x = to determine the value of the contant k. (You will need the table of tandard tranform Appendix B.) 11. By uing a power erie expanion, determine (in erie form) the Laplace tranform of in (t 2 ) and co (t 2 ). 12. P() and Q() are polynomial, the degree of P() i le than that of Q() which i n. Ue partial fraction to prove the reult

26 38 2 Further Propertie of the Laplace Tranform { } P() L 1 = Q() n k=1 where α k are the n ditinct zero of Q(). 13. Find the following Laplace tranform: (a) H(t a) P(α k ) Q (α k ) eα kt (b) (c) f 1 = f 2 = { t + 1 t 2 3 t > 2 { t + 1 t 2 6 t > 2 (d) the derivative of f 1 (t). 14. Find the Laplace tranform of the triangular wave function: { t t < c F(t) = 2c t c t < 2c F(t + 2c) = F(t). 15. Find the Laplace tranform of the generally placed top hat function: F(t) = { 1h a t < a + h otherwie. Hence deduce the Laplace tranform of the Dirac-δ function δ(t a) where a > i a real contant.

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