Laplace Transforms Chapter 2 Dyke

7/30/2019 Laplace Transforms Chapter 2 Dyke

1/12

The Laplace Transform

11 Introduction

As disciine mthemtics encomsses st rnge of subjects In uremthemtics n imortnt concet is the ide of n xiomtic system wherebyxioms re roosed nd theorems re roed by inoking these xioms ogi-cy These ctiities re oen of itte interest to the ied mthemticin towhom the ure mthemtics of gebric structures wi seem ike tinkering withxioms for hours in order to roe the obious To the engineer, this kind of uremthemtics is een more of n nthem The ue of knowing bout suchstructures ies in the biity to generise the "obious to other res These

generistions re notoriousy unredictbe nd re oen ery surrising In-deed mny sy tht there is no such thing s nonicbe mthemtics ustmthemtics whose iction hs yet to be found

The Lce nsform exresses the conict between ure d iedmthemtics sendidy There is temttion to begin book such s thison iner gebr outining the theorems nd roerties of normed sces Thiswoud indeed roide sound bsis for ture resuts Howeer most iedmthemticins nd engineers woud robby turn o On the other hd,

engineering texts resent the Lce nsform s tookit of resuts with ittettention being id to the underying mthemtic structure, regions of idity or restrictions Wht hs been decided here is to gie brief introduction tothe underying ure mthemtic structures enough it s hoed for the uremthemticin to recite wht kind of creture the Lce nsform is,whist emhsising ictions d giing enty of exes The oint ofiew from which this book is written is therefore denitey tht of the iedmthemticin Howeer ure mthemtic ides some of which cn be quite

1


2/12

2 An ntodction to Lapace Tansfoms and Foie Seies

{()} f(s)

space s space

Figure 1 The Lce nsform ming

extensie wi occur It remns the iew of this uthor tht Lce nsformsony come e when they re used to soe re robems Those who strongydisgree with this wi nd ure mthemtics textbooks on integr trnsformsmuch more to their iking

The min re of ure mthemtics needed to understnd the ndment

roerties of Lce nsforms is nysis nd, to esser extent the normedector sce Anysis in rticur integrtion is needed from the strt it goerns the existence conditions for the Lce nsform itsef howeer is soon rent, ccutions inoing Lce nsforms cn tke cewithout exicit knowedge of nysis Normed ecto sces nd ssocitedner gebr ut the Lce nsform on rm theoretc footing butcn be eft unti itte ter in book imed t second yer undergrdutemthemtics students

1. 2 The Laplace ansform

The denton of the Lce nsform coud hrdy be more strghtforwrdGien suitbe function Ft the Lce nsform written ( s denedby

( = 10 Fte-dtThis bd sttement my stisfy most engineers but not mthemticis Thequestion of wht constitutes "suitbe functon wi now be ddressed Theintegr on the right h innite rnge nd hence is wht is ced n imroerintegr This too needs crefu hndng The nottion {Ft} is used todenote the Lce nsform of the function Ft

Another wy of ooking t the Lce nsform is s ming from ointsn the t domin to oints in the domin Pictoriy Fgure 11 indictesthis ming rocess The time domn wi contin those functions Ft

whose Lce nsform exists wheres the frequency domin contins the


3/12

The Lapace Tansfom 3

mges { F t} Another ect of Lce nsforms tht needs mentonngt ths stge s tht the rbe s often hs to tke comex ues hs menstht s s functon of comex rbe whch n turn ces restrctons onthe (re functon Ft gen tht the mroer ntegr must conerge Muchof the nyss noed n deng wth the mge of the functon Ft n the sne s therefore comex nyss whch my be qute new to some reders

As h been sd erer engneers re qute hy to use Lce nsforms to he soe rety of robems wthout questonng the conergence ofthe mroer ntegrs hs goes for some ed mthemtcns too he rgument seems to be on the nes tht f t ges wht ooks resonbe nswerthen ne! In our ew ths tkes the engneer's mxm "f t n't broke, don't xt too fr hs s rmry mthemtcs textbook therefore n ths oenng

chter we sh be more mthemtcy exct thn s customry n books onLce nsforms In Chter 4 there s some more ure mthemtcs whenFourer seres re ntroduced ht s there for smr reons One mthemtc queston tht ought to be ked concerns unqueness Gen functonFt ts Lce nsform s surey unque from the we dened nture of themroer ntegr Howeer s t ossbe for two derent functons to he thesme Lce nsform? o ut the queston derent but equent wys there functon Nt, not dentcy zero whose Lce nsform s zero?

For ths functon ced u functon, coud be dded to ny sutbe functonnd the Lce nsform woud remn unchnged Nu functons do exstbut s ong we restrct oursees to ecewse contnuous functons ths ceesto be robem Here s the denton of ecewse contnuous

Denition 11 I a teva [0 t say ca be patto e to a te umbe o sub tevas [0 t [t t, [t t3, . [t t th 0 t t t t a ceasg seque ce o tmes a such that a gve u ctot s cotuous

each o these sub tevas but ot ecessay at the e po ts themseves thet s pecese co t uous the teva [0 t

Ony functons tht der t nte number of onts he the sme Lcensform I F t = Ft excet t nte number of onts where they derby nte ues then {F t} = {Ft} We menton ths gn n the nextchter when the nerse Lce nsform s dened

In ths secton, we sh exmne the condtons for the exstence of theLce nsform n more det thn s usu In engneerng texts the sme

denton foowed by n exnton of exonent order s tht s requredhose tht re stsed wth ths cn rtuy sk the next few rgrhs ndgo on study the eementry roertes Secton 3 Howeer some my needto know enough bckground n terms of the ntegrs nd so we deote ttesce to some fundments We w need to ntroduce mroer ntegrs butet us rst dene the Remnn ntegr It s the ntegr we know nd oe nds dened n terms of mts of sums he strct denton runs foows

Let Fx be functon whch s dened nd s bounded n the nter

a x b nd suose tht m nd M re resectey the ower nd uer


4/12

4 An ntodction to Lapace Tansfoms and Foie Sees

bounds of F(x in this interl (written b] see Aendx C Tke set ofoints

X = x X X X X = b

nd wrte 5 = XX Let M be the bounds of F(x in the subinterl(X- , X nd form the sums

=L5

These re clled resectiely the uer nd lower Riemnn sums corresondingto the mode of subdiision It is certinly cler tht There re rietyof wys tht cn be used to rtition the interl ( b nd ech wy will he(in generl dierent M nd leding to dierent nd Let M be theminimum of ll ossible M nd be the mimum of ll ossible A lowerbound or suremum for the set is therefore M(b nd n uer bound orinmum for the set is (b These bounds re of course rough There rexc bounds for nd cl them Jnd I resectely I I = J F(x is sidto be Riemnn integrble in ( b nd the lue of the integrl is I or Jnd isdenoted by

I = J = 1bF(xdxFor the urist it turns out tht the Riemnn integrl is not quite gener

enough nd the Stieltes integrl is ctully required Howeer we will not useths concet which belongs securely in secilist nl stge or grdute texts

The imroer integrl is dened in the obious wy by tking the limit

lim RF(xdx = {o

F(xdxR- a Joroided F(x is continuous in the interl x for eery , nd the limiton the le exists

This is enough of generl theory we now ly it to the Llce nsformThe rmeter is dened to tke the ncresing lues from 0 to o. The con-dition F(xj M is termed "F(x is of exonenti order nd is sekingoosely quite wek condition All olynomil functions nd (of course exo-

nentil functions of the tye k (k constnt re ncluded well boundedfunctions Excluded functions re those tht he singulrities such ln( or/ (x - nd functions tht he growth rte more rid thn exonentilfor exmle Functions tht he nite number of nite discontinuitiesre lso included These he secil role n the theory of Llce nsforms(see Chter 3 so we will not dwell on them here: suce to sy tht functionsuch s

{ 2 < x < 2 + 1

F(x = 0 2 + 1 < x < 2 + 2 where = 0 ,


5/12

The Laplace Tansfom

is one exme However the function

F(x =

{ x rtionx irrtion

is excuded becuse though the discontinuities re nite there e inniteymny of them

We sh now foow stndrd rctice nd use time) insted of x thedummy vribe

1 . 3 Elementary Properties

The Lce nsform h mny interesting nd usefu roerties, the mostfundment of which is inerity It is inerity tht enbes us to dd resutstogether to deduce other more comicted ones d is so bic tht we stte it theorem nd rove it rst

Theoem 12 (Lineaity) I F ( n F ( r o ncon ho LpcTnor x hn

hr n b r rbry connPoof

{F ( + bF(} = 10 (F + bF)ed= 10 Fe + bFe d

= 10 Fed + b10Fe-d= {F ()} + b{F ()}

where we hve ssumed tht

so tht

jF + bF : !F + bl F : (j!M + bM)e

where 3 = x{O, } This roves the theorem

0

In this section we sh concentrte on those roerties of the ce nsformtht do not invove the ccuus The rst of these tes the form of nother

theorem becuse of its generity


6/12

6 An ntodction to Lapace Tansfoms and Foie Seies

Theoem 1.3 (Fist Shift Theoem) I is possib o choos consans Man a such ha F( Met, ha is F( is o xponnia orr hn

.{e-btF(} = f(s + b

proi b a. In pracic i F( is o xponnia orr hn h consan acan b chosn such ha his inquai hos

Poof The roof is strightforwrd nd runs follows

.{e-btF(} lim {T estebtF(dT Jo

= 10e-ste-btF(t)dt (athelimitexists)10 e(s+b)tF(df(s + b

This estblishes the theorem

0

We shl mke considerble use of this once we hve estblished few elementryLlce nsforms This we shl now roceed to do

Exampe 1.4 in h Lapac Tansorm o h uncion F( =

Solution Using the denition of Llce nsform

( lim {T estdT- JoNow we hve tht

this lt exression tends to12 T o.s

Hence we hve the result

( = s


7/12

. The Laace Tansfom

We cn generlise this result to deduce the following resultCoroary "( n . . .}. = s , n oste tegerProof The roof is strghtforwrd

( = 10 ed this time tkng the limit stright wy[ ] 0 1o -e + _ne-ds s

= (

sI we ut n = 2 in ths recurrence retion we obtn

I we sume

then

This estbishes tht

by induction

"( _ n + 1 _ (n + 1 .s s s

7

0

Exampe 1.5 Fn he Lapace Tansorm o {e} an euce he aue o{e} where a s a rea consan an n a pose neer

Soution Using the rst shift theorem with b = a gies

so with

we get

Using F( = the formul

{F(e} = f(s a

1F( = nd f = 2s


8/12

8 An Intodction to Laplace Tansfoms and Fouie Seies

followsLter, we shll generlise this formul further extending to the ce where

n is not n integerWe move on to consider the Llce nsform of trigonometric functions

Seciclly we shll clculte sin(} nd cos( } It is unfortunte, butthe Llce nsform of the other common trigonometric functions tn, cotsc nd sec do not exist they ll hve singulrities for nite he conditiontht the nction ( h to be of exonentil order is not obeyed by ny ofthese singulr trigonometric functions cn be seen, for exmle by notingtht

tn( o /2

nd cot( o 0

for ll vlues of the constnt a. Similrly neither sc nor sec re of exonentilorder

n order to nd the Llce nsform of sin( nd cos( it is best todetermine ( where J he function is comlex vlued butit is both continuous nd bounded for ll so its Llce nsform certnlyexists Tking the Llce nsform

Now,

( = fo = fo [ei-s)t ]0

z - 1

- 1

+z + 1 + 1

( (cos( + sin(= (cos( + (sin(

Equting rel nd imginry rts gives the two results

nd

(cos( =1s +

(sin( =1 . +The linerity roerty h been used here nd will be used in ture without

futher comment


9/12

. he apace ransfom

Gien tht the restriction on the tye of function one cn Lce nsformis wek ie it hs to be of exonenti order nd he t most nite numberof nite jums one cn nd the Lce nsform of y oynomi nycombintion of oynomi with sinusoid functions nd combintions of thesewith exonentis (roided the exonenti functions grow t rte eatwherea is constnt We cn therefore roch the robem of ccuting theLce nsform of ower series It is ossibe to tke the Lce nsformof ower series term by term ong the series uniformy conerges to iecewise continuous function We sh inestigte this further lter in the textmenwhie et us ook t the Lce nsform of functions tht re not eencontinuous

nctions tht re not continuous occur ntury in brnches of eectric

nd contro engineering nd in the softwre industry One ony h to thinkof switches to reise how widesred discontinuous functions re throughouteectronics n comuting

Expe 16 n h Lapac Tansorm o h ncon rprsn b (whr

( = 2 2

{ O < <

0 > 2Soution This function is of the "swtooth riety tht is quite cmmon ineectric engineering There is no question tht it is of exonenti order ndtht

10 estF(exists nd is we dened F( is continuous but not dierentibe This is not

troubesome Crrying out the ccution is itte messy nd the detis cnbe checked using comuter gebr

(F( = 10 est(=1toe-std+12to(2tot)e-tdt

to= [-!est]to

1to est [-2 -est]2to-12to etdS s S to toS

to stto to t2to=--e s 82 8 82 to= :2[esto _ 1+ 8 [e-2sto_ e-sto]=

s

12

[1-eto+e2sto]

=

12[1- esto]2


10/12

0 An ntrodction to aplace Transforms and Forier Series

I he ext chter we shl estige n more de he roeres of dscon-tinuous nctios such the Heisde un se funcio As nroducionto his, et us do the foowing exme

Exampe 1.7 Deene he Lapace Tanso o he sep ncon F) dened by

F) = { 0 0 < ? Soution F) isef s bounded, so here s no queson h s so of exo-neni order The Lce nsform of F) s herefore

CF)) estF)d=

==Here s noher usefu resu

lo ae-stdto

[e-st]S to stoesd

Theorem 18 I C(F()) = s) hen C(F)) =-ds(s)and n ene C(

F)) )

d

(s) .sProof Le us sr wh he denon of Lce nsform

CF)) estF)dnd dierene his wh resec o s o ge

! = e-stF()d e-stF)dsuming bsoute conergence o jusfy nerchnging dierenon d (m-roer iegrion Hence

dCF()) -ds(s) .


11/12

The Lapace Transform 1

One cn now see how to rogress by induction. Assume the resut hods for n,so tht

C(F()) = () d

f(s)s

nd dierentite both sides with resect to s (suming rorite coner-gence roerties to gie

or

So

C(tn+F(t))=(-1 )n+dsn+

f(s)

which estblishes the resut by inuction.

Exampe 19 Deermne he Lapace Tanform of he ncon sin(

0

Soution To eute this Lce nsform we use Theorem 1.8 wth f() =sin(. This gies

d { } 2sC{t m(t)}=-ds 1+s2 = ( 1 +s2)2which is the required resut

1.4 Exercises

1 For ech of the foowing functions determine which h Llce ns-form I it exists nd it; if it does not sy briey why

( n( (b e3 (c e, (d e , (e 1/

(f f() = { 1 f s een0 1f s odd2. Determine from rst rincies the Lce nsform of the folowng

functions

( ek , (b (c cosh(.

3 Find the Lce nsforms of the foowing functions:


12/12

12 An ntrodction to apace Transforms and Forier Series

4. Fin h Lapac ansfom of h funcion () , wh () is gin by

{ < < () = 2 1 < 20 ohwis.

5 Us h popy of Thom 18 o min h foowing Lapac ans-foms

(a) , (b) cos(), (c) cos()

6. Fin h Lapac ansfoms of h foowing funcions:(a) sin( + ) (b) cosh(6)

7 If ( + b) = () min h Lapac asfom of in ms of{} = () a a ni inga

8 Po h foowing chang of sca su:

{()} = ().Hnc ua h Lapac ansfoms of h wo funcions

(a) cos(6), (b) cos(7)

Documents

Laplace Transforms Chapter 2 Dyke