SEC. 6.3 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-shifting) 217. Second Shifting Theorem (t-shifting)

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1 SEC. 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) 7. PROJECT. Furher Resuls by Differeniaion. Proceeding as in Example, obain (a) and from his and Example : (b) formula, (c), (d) 3 in Sec. 6.9, (e) (f ) 3 9 INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3, find f() if l(f ) equals: s s>4 s 3 ps s(s v ) s 4 s 7. s 8. 3s 4 s 4 9s s 4 k s 9. s 3 as l( cos v) s v l( cosh a) s a (s a ), l( sinh a) (s v ) as (s a ). 3. PROJECT. Commens on Sec. 6.. (a) Give reasons why Theorems and are more imporan han Theorem 3. (b) Exend Theorem by showing ha if f () is coninuous, excep for an ordinary disconinuiy (finie jump) a some a (), he oher condiions remaining as in Theorem, hen (see Fig. 7) (*) l( f r) sl( f ) f () 3 f (a ) f (a )4e as. (c) Verify (*) for f () e if and if. (d) Compare he Laplace ransform of solving ODEs wih he mehod in Chap.. Give examples of your own o illusrae he advanages of he presen mehod (o he exen we have seen hem so far). f () f (a ) f (a + ) a Fig. 7. Formula (*) 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) This secion and he nex one are exremely imporan because we shall now reach he poin where he Laplace ransform mehod shows is real power in applicaions and is superioriy over he classical approach of Chap.. The reason is ha we shall inroduce wo auxiliary funcions, he uni sep funcion or Heaviside funcion u( a) (below) and Dirac s dela d( a) (in Sec. 6.4). These funcions are suiable for solving ODEs wih complicaed righ sides of considerable engineering ineres, such as single waves, inpus (driving forces) ha are disconinuous or ac for some ime only, periodic inpus more general han jus cosine and sine, or impulsive forces acing for an insan (hammerblows, for example). Uni Sep Funcion (Heaviside Funcion) u( a) The uni sep funcion or Heaviside funcion u( a) is for a, has a jump of size a a (where we can leave i undefined), and is for a, in a formula: () if a u( a) b if a (a ).

2 8 CHAP. 6 Laplace Transforms u() u( a) a Fig. 8. Uni sep funcion u() Fig. 9. Uni sep funcion u( a) Figure 8 shows he special case u(), which has is jump a zero, and Fig. 9 he general case u( a) for an arbirary posiive a. (For Heaviside, see Sec. 6..) The ransform of u( a) follows direcly from he defining inegral in Sec. 6., l{u( a)} e s u( a) d e s # e s d s ` a ; here he inegraion begins a a () because u( a) is for a. Hence () l{u( a)} eas s (s ). The uni sep funcion is a ypical engineering funcion made o measure for engineering applicaions, which ofen involve funcions (mechanical or elecrical driving forces) ha are eiher off or on. Muliplying funcions f () wih u( a), we can produce all sors of effecs. The simple basic idea is illusraed in Figs. and. In Fig. he given funcion is shown in (A). In (B) i is swiched off beween and (because u( ) when ) and is swiched on beginning a. In (C) i is shifed o he righ by unis, say, for insance, by sec, so ha i begins sec laer in he same fashion as before. More generally we have he following. Le f () for all negaive. Then f ( a)u( a) wih a is f () shifed (ranslaed) o he righ by he amoun a. Figure shows he effec of many uni sep funcions, hree of hem in (A) and infiniely many in (B) when coninued periodically o he righ; his is he effec of a recifier ha clips off he negaive half-waves of a sinuosidal volage. CAUTION! Make sure ha you fully undersand hese figures, in paricular he difference beween pars (B) and (C) of Fig.. Figure (C) will be applied nex. f () π π π π π + π (A) f () = 5 sin (B) f ()u( ) (C) f ( )u( ) Fig.. Effecs of he uni sep funcion: (A) Given funcion. (B) Swiching off and on. (C) Shif.

3 SEC. 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) 9 k k (A) k[u( ) u( 4) + u( 6)] (B) 4 sin ( π )[u() u( ) + u( 4) + ] Fig.. Use of many uni sep funcions. _ Time Shifing (-Shifing): Replacing by a in f () The firs shifing heorem ( s-shifing ) in Sec. 6. concerned ransforms F(s) l{f ()} and F(s a) l{e a f ()}. The second shifing heorem will concern funcions f () and f ( a). Uni sep funcions are jus ools, and he heorem will be needed o apply hem in connecion wih any oher funcions. T H E O R E M Second Shifing Theorem; Time Shifing If f () has he ransform F(s), hen he shifed funcion (3) ~ f () f ( a)u( a) b if a f ( a) if a has he ransform e as F(s). Tha is, if l{f ()} F(s), hen (4) l{f ( a)u( a)} e as F(s). Or, if we ake he inverse on boh sides, we can wrie (4*) f ( a)u( a) l {e as F(s)}. Pracically speaking, if we know F(s), we can obain he ransform of (3) by muliplying F(s) by e as. In Fig., he ransform of 5 sin is F(s) 5>(s ), hence he shifed funcion 5 sin ( )u( ) shown in Fig. (C) has he ransform e s F(s) 5e s >(s ). P R O O F We prove Theorem. In (4), on he righ, we use he definiion of he Laplace ransform, wriing for (o have available laer). Then, aking inside he inegral, we have e as e as F(s) e as e s f () d e s(a) f () d. Subsiuing a, hus a, d d in he inegral (CAUTION, he lower limi changes!), we obain e as F(s) e s f ( a) d. a

4 CHAP. 6 Laplace Transforms To make he righ side ino a Laplace ransform, we mus have an inegral from o, no from a o. Bu his is easy. We muliply he inegrand by u( a). Then for from ~ o a he inegrand is, and we can wrie, wih f as in (3), e as F(s) e s f ( a)u( a) d (Do you now see why u( a) appears?) This inegral is he lef side of (4), he Laplace ~ ransform of f () in (3). This complees he proof. e s ~ f () d. E X A M P L E Applicaion of Theorem. Use of Uni Sep Funcions Wrie he following funcion using uni sep funcions and find is ransform. if f () d if p cos if p. (Fig. ) Soluion. Sep. In erms of uni sep funcions, f () ( u( )) (u( ) u( p)) (cos )u( p). Indeed, ( u( )) gives f () for, and so on. Sep. To apply Theorem, we mus wrie each erm in f () in he form f ( a)u( a). Thus, ( u( )) remains as i is and gives he ransform ( e s )>s. Then Togeher, l e u( ) f l a ( ) ( ) b u( ) f a s 3 s s b es l e u a pb f l e a pb a s 3 p p s 8s b e ps> l e (cos ) u a pb f l e asin a pbb u a pb f s eps>. l( f ) s s es a s 3 s s b e s a s 3 p s If he conversion of f () o f ( a) is inconvenien, replace i by p a p pb 8 b u a pb f p 8s b e ps> s eps>. (4**) l{f ()u( a)} e as l{f ( a)}. (4**) follows from (4) by wriing f ( a) g(), hence f () g( a) and hen again wriing f for g. Thus, l e u( ) f e s l e ( ) f e s l e f es a s 3 s s b as before. Similarly for l{ u( p)}. Finally, by (4**), l e cos u a pb f eps> l e cos a pb f eps> l{sin } e ps> s.

5 SEC. 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) f() 4 Fig.. ƒ() in Example E X A M P L E Applicaion of Boh Shifing Theorems. Inverse Transform Find he inverse ransform f () of F(s) es s p e s s p e 3s (s ). Soluion. Wihou he exponenial funcions in he numeraor he hree erms of F(s) would have he inverses (sin p)>p, (sin p)>p, and e because >s has he inverse, so ha >(s ) has he inverse e by he firs shifing heorem in Sec. 6.. Hence by he second shifing heorem (-shifing), f () p sin (p( )) u( ) p sin (p( )) u( ) ( 3)e(3) u( 3). Now sin (p p) sin p and sin (p p) sin p, so ha he firs and second erms cancel each oher when. Hence we obain f () if, (sin p)>p if, if 3, and ( 3)e (3) if 3. See Fig Fig. 3. ƒ() in Example E X A M P L E 3 Response of an RC-Circui o a Single Recangular Wave Find he curren i() in he RC-circui in Fig. 4 if a single recangular wave wih volage is applied. The circui is assumed o be quiescen before he wave is applied. Soluion. The inpu is V 3u( a) u( b)4. Hence he circui is modeled by he inegro-differenial equaion (see Sec..9 and Fig. 4) V Ri() q() C Ri() C i() d v() V 3u( a) u( b)4. C v() i() v() V V /R R a b a b Fig. 4. RC-circui, elecromoive force v(), and curren in Example 3

6 CHAP. 6 Laplace Transforms Using Theorem 3 in Sec. 6. and formula () in his secion, we obain he subsidiary equaion Solving his equaion algebraically for I(s), we ge RI(s) I(s) sc V s 3eas e bs 4. I(s) F(s)(e as e bs ) where V IR F(s) s >(RC) and l (F) V R e>(rc), he las expression being obained from Table 6. in Sec. 6.. Hence Theorem yields he soluion (Fig. 4) i() l (I) l {e as F(s)} l {e bs F(s)} V ha is, i() if a, and K e >(RC) if a b i() c (K K )e >(RC) if a b R 3e(a)>(RC) u( a) e (b)>(rc) u( b)4; where K and K V e b>(rc) V e a>(rc) >R >R. E X A M P L E 4 Response of an RLC-Circui o a Sinusoidal Inpu Acing Over a Time Inerval Find he response (he curren) of he RLC-circui in Fig. 5, where E() is sinusoidal, acing for a shor ime inerval only, say, E() sin 4 if p and E() if p and curren and charge are iniially zero. Soluion. The elecromoive force E() can be represened by ( sin 4)( u( p)). Hence he model for he curren i() in he circui is he inegro-differenial equaion (see Sec..9).ir i i() d ( sin 4)( u( p)). i(), ir(). From Theorems and 3 in Sec. 6. we obain he subsidiary equaion for I(s) l(i).si I I s # 4s s 4 a s e ps b. s Solving i algebraically and noing ha s s (s )(s ), we obain # 4 l(s) (s )(s ) a s s 4 se ps s 4 b. For he firs erm in he parenheses ( Á ) imes he facor in fron of hem we use he parial fracion expansion 4,s (s )(s )(s 4 ) A s B s Ds K s 4. Now deermine A, B, D, K by your favorie mehod or by a CAS or as follows. Muliplicaion by he common denominaor gives 4,s A(s )(s 4 ) B(s )(s 4 ) (Ds K)(s )(s ).

7 SEC. 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) 3 We se s and and hen equae he sums of he and erms o zero, obaining (all values rounded) (s ) 4,, 9( 4 )A, A.776 (s ) 4,, 9( 4 )B, B.644 (s 3 -erms) A B D, D.3368 (s -erms) A B D K, K Since K # 4, we hus obain for he firs erm I in I I I I s s.3368s s # 4 s 4. From Table 6. in Sec. 6. we see ha is inverse is i ().776e.644e.3368 cos sin 4. This is he curren i() when p. I agrees for p wih ha in Example of Sec..9 (excep for noaion), which concerned he same RLC-circui. Is graph in Fig. 63 in Sec..9 shows ha he exponenial erms decrease very rapidly. Noe ha he presen amoun of work was subsanially less. The second erm I of I differs from he firs erm by he facor e ps. Since cos 4( p) cos 4 and sin 4( p) sin 4, he second shifing heorem (Theorem ) gives he inverse i () if p, and for p i gives i ().776e (p).644e (p).3368 cos sin 4. Hence in i() he cosine and sine erms cancel, and he curren for p is i().776(e e (p) ).644(e e (p) ). s 3 s I goes o zero very rapidly, pracically wihin.5 sec. C = F R = Ω L =. H E() Fig. 5. RLC-circui in Example 4 P R O B L E M S E T Repor on Shifing Theorems. Explain and compare he differen roles of he wo shifing heorems, using your own formulaions and simple examples. Give no proofs. SECOND SHIFTING THEOREM, UNIT STEP FUNCTION Skech or graph he given funcion, which is assumed o be zero ouside he given inerval. Represen i, using uni sep funcions. Find is ransform. Show he deails of your work.. ( ) 3. ( ) 4. cos 4 ( p) 5. e ( p>) 6. sin p ( 4) 7. e p ( 4) 8. ( ) 9. ( 3 ). sinh ( ). sin (p> p) 7 INVERSE TRANSFORMS BY THE ND SHIFTING THEOREM Find and skech or graph f () if l( f ) equals. e 3s >(s ) ( e ps )>(s 9) 4. 4(e s e 5s )>s 5. e 3s >s 4 6. (e s e 3s )>(s 4) 7. ( e p(s) )(s )>((s ) )

8 SEC. 6.4 Shor Impulses. Dirac s Dela Funcion. Parial Fracions R, L H, C.5 F, v() kv if 4. and if R, L H, C. F, v 55 sin V if p and if p C R L v() Fig. 3. Problems Fig. 3. Curren in Problem Shor Impulses. Dirac s Dela Funcion. Parial Fracions An airplane making a hard landing, a mechanical sysem being hi by a hammerblow, a ship being hi by a single high wave, a ennis ball being hi by a racke, and many oher similar examples appear in everyday life. They are phenomena of an impulsive naure where acions of forces mechanical, elecrical, ec. are applied over shor inervals of ime. We can model such phenomena and problems by Dirac s dela funcion, and solve hem very effecively by he Laplace ransform. To model siuaions of ha ype, we consider he funcion () >k f k ( a) b if a a k (Fig. 3) oherwise (and laer is limi as k : ). This funcion represens, for insance, a force of magniude >k acing from a o a k, where k is posiive and small. In mechanics, he inegral of a force acing over a ime inerval a a k is called he impulse of he force; similarly for elecromoive forces E() acing on circuis. Since he blue recangle in Fig. 3 has area, he impulse of in () is ak () I k f k ( a) d d. k f k a /k Area = a a + k Fig. 3. The funcion ƒ k ( a) in ()

9 6 CHAP. 6 Laplace Transforms To find ou wha will happen if k becomes smaller and smaller, we ake he limi of as k : (k ). This limi is denoed by d( a), ha is, d( a) lim k: f k ( a). d( a) is called he Dirac dela funcion or he uni impulse funcion. d( a) is no a funcion in he ordinary sense as used in calculus, bu a so-called generalized funcion. To see his, we noe ha he impulse I k of f k is, so ha from () and () by aking he limi as k : we obain f k (3) if a d( a) b oherwise and d( a) d, bu from calculus we know ha a funcion which is everywhere excep a a single poin mus have he inegral equal o. Neverheless, in impulse problems, i is convenien o operae on d( a) as hough i were an ordinary funcion. In paricular, for a coninuous funcion g() one uses he propery [ofen called he sifing propery of d( a), no o be confused wih shifing] (4) g()d( a) d g(a) which is plausible by (). To obain he Laplace ransform of d( a), we wrie f k ( a) 3u( a) u( (a k))4 k and ake he ransform [see ()] l{f k ( a)} ks 3eas e (ak)s 4 e as e ks. ks We now ake he limi as k :. By l Hôpial s rule he quoien on he righ has he limi (differeniae he numeraor and he denominaor separaely wih respec o k, obaining se ks and s, respecively, and use se ks >s : as k : ). Hence he righ side has he limi e as. This suggess defining he ransform of d( a) by his limi, ha is, (5) l{d( a)} e as. The uni sep and uni impulse funcions can now be used on he righ side of ODEs modeling mechanical or elecrical sysems, as we illusrae nex. PAUL DIRAC (9 984), English physicis, was awarded he Nobel Prize [joinly wih he Ausrian ERWIN SCHRÖDINGER (887 96)] in 933 for his work in quanum mechanics. Generalized funcions are also called disribuions. Their heory was creaed in 936 by he Russian mahemaician SERGEI L VOVICH SOBOLEV (98 989), and in 945, under wider aspecs, by he French mahemaician LAURENT SCHWARTZ (95 ).

10 SEC. 6.4 Shor Impulses. Dirac s Dela Funcion. Parial Fracions 7 E X A M P L E Mass Spring Sysem Under a Square Wave Deermine he response of he damped mass spring sysem (see Sec..8) under a square wave, modeled by (see Fig. 33) ys 3yr y r() u( ) u( ), y(), yr(). Soluion. From () and () in Sec. 6. and () and (4) in his secion we obain he subsidiary equaion s Y 3sY Y s (es e s ). Soluion Y(s) Using he noaion F(s) and parial fracions, we obain F(s) From Table 6. in Sec. 6., we see ha he inverse is s(s 3s ) s(s )(s ) s s s. Therefore, by Theorem in Sec. 6.3 (-shifing) we obain he square-wave response shown in Fig. 33, y l (F(s)e s F(s)e s ) f () l (F) e e. f ( )u( ) f ( )u( ) s(s 3s ) (es e s ). ( ) d e () e () ( ) e () e () e () e () ( ). y() Fig. 33. Square wave and response in Example E X A M P L E Hammerblow Response of a Mass Spring Sysem Find he response of he sysem in Example wih he square wave replaced by a uni impulse a ime. Soluion. We now have he ODE and he subsidiary equaion Solving algebraically gives By Theorem he inverse is ys 3yr y d( ), and (s 3s )Y e s. Y(s) e s (s )(s ) a s s b es. if y() l (Y) c e () e () if.

11 8 CHAP. 6 Laplace Transforms y() is shown in Fig. 34. Can you imagine how Fig. 33 approaches Fig. 34 as he wave becomes shorer and shorer, he area of he recangle remaining? y() Fig. 34. Response o a hammerblow in Example E X A M P L E 3 Four-Terminal RLC-Nework Find he oupu volage response in Fig. 35 if R, L H, C 4 F, he inpu is d() (a uni impulse a ime ), and curren and charge are zero a ime. Soluion. To undersand wha is going on, noe ha he nework is an RLC-circui o which wo wires a A and B are aached for recording he volage v() on he capacior. Recalling from Sec..9 ha curren i() and charge q() are relaed by i qr dq>d, we obain he model Lir Ri q C Lqs Rqr q qs qr,q d(). C From () and () in Sec. 6. and (5) in his secion we obain he subsidiary equaion for Q(s) l(q) (s s,)q. Soluion Q (s ) 99. By he firs shifing heorem in Sec. 6. we obain from Q damped oscillaions for q and v; rounding we ge (Fig. 35) , q l (Q) 99.5 e sin 99.5 and v q C.5e sin () v 8 R L 4 A C B v() =? 8 Nework Volage on he capacior Fig. 35. Nework and oupu volage in Example 3 More on Parial Fracions We have seen ha he soluion Y of a subsidiary equaion usually appears as a quoien of polynomials Y(s) F(s)>G(s), so ha a parial fracion represenaion leads o a sum of expressions whose inverses we can obain from a able, aided by he firs shifing heorem (Sec. 6.). These represenaions are someimes called Heaviside expansions.

12 SEC. 6.4 Shor Impulses. Dirac s Dela Funcion. Parial Fracions 9 An unrepeaed facor s a in G(s) requires a single parial fracion A>(s a). See Examples and. Repeaed real facors (s a), (s a) 3, ec., require parial fracions A (s a) A s a, A 3 (s a) 3 A (s a) A s a, ec., The inverses are (A ( A 3 A A )e a A )e a,, ec. Unrepeaed complex facors (s a)(s a), a a ib, a a ib, require a parial fracion (As B)>3(s a) b 4. For an applicaion, see Example 4 in Sec A furher one is he following. E X A M P L E 4 Unrepeaed Complex Facors. Damped Forced Vibraions Solve he iniial value problem for a damped mass spring sysem aced upon by a sinusoidal force for some ime inerval (Fig. 36), ys yr y r(), r() sin if p and if p; y(), yr() 5. Soluion. From Table 6., (), () in Sec. 6., and he second shifing heorem in Sec. 6.3, we obain he subsidiary equaion (s Y s 5) (sy ) Y ( e ps ). s 4 We collec he Y-erms, (s s )Y, ake s 5 s 3 o he righ, and solve, (6) s 3 Y. (s 4)(s s ) (s 4)(s s ) s s e ps For he las fracion we ge from Table 6. and he firs shifing heorem (7) l b s 4 (s ) r e (cos 4 sin ). In he firs fracion in (6) we have unrepeaed complex roos, hence a parial fracion represenaion Muliplicaion by he common denominaor gives As B Ms N. (s 4)(s s ) s 4 s s (As B)(s s ) (Ms N)(s 4). We deermine A, B, M, N. Equaing he coefficiens of each power of s on boh sides gives he four equaions (a) 3s 3 4 : A M (b) 3s 4 : A B N (c) 3s4 : A B 4M (d) 3s 4 : B 4N. We can solve his, for insance, obaining M A from (a), hen A B from (c), hen N 3A from (b), and finally A from (d). Hence A, B, M, N 6, and he firs fracion in (6) has he represenaion (8) s s 4 (s ) 6. Inverse ransform: cos sin e ( cos 4 sin ). (s )

13 3 CHAP. 6 Laplace Transforms The sum of his inverse and (7) is he soluion of he problem for p, namely (he sines cancel), (9) y() 3e cos cos sin if p. In he second fracion in (6), aken wih he minus sign, we have he facor e ps, so ha from (8) and he second shifing heorem (Sec. 6.3) we ge he inverse ransform of his fracion for in he form cos ( p) sin ( p) e (p) 3 cos ( p) 4 sin ( p)4 cos sin e (p) ( cos 4 sin ). The sum of his and (9) is he soluion for p, () y() e 3(3 e p ) cos 4e p sin 4 if p. Figure 36 shows (9) (for p) and () (for p), a beginning vibraion, which goes o zero rapidly because of he damping and he absence of a driving force afer p. y() Driving force y = (Equilibrium posiion) y Dashpo (damping) π π 3π 4π Mechanical sysem Fig. 36. Example 4 Oupu (soluion) The case of repeaed complex facors 3(s a)(s a )4, which is imporan in connecion wih resonance, will be handled by convoluion in he nex secion. P R O B L E M S E T CAS PROJECT. Effec of Damping. Consider a vibraing sysem of your choice modeled by ys cyr ky d(). (a) Using graphs of he soluion, describe he effec of coninuously decreasing he damping o, keeping k consan. (b) Wha happens if c is kep consan and k is coninuously increased, saring from? (c) Exend your resuls o a sysem wih wo d-funcions on he righ, acing a differen imes.. CAS EXPERIMENT. Limi of a Recangular Wave. Effecs of Impulse. (a) In Example in he ex, ake a recangular wave of area from o k. Graph he responses for a sequence of values of k approaching zero, illusraing ha for smaller and smaller k hose curves approach he curve shown in Fig. 34. Hin: If your CAS gives no soluion for he differenial equaion, involving k, ake specific k s from he beginning. (b) Experimen on he response of he ODE in Example (or of anoher ODE of your choice) o an impulse d( a) for various sysemaically chosen a ( ) ; choose iniial condiions y(), yr(). Also consider he soluion if no impulse is applied. Is here a dependence of he response on a? On b if you choose bd( a)? Would d( a ) wih a a annihilae he effec of d( a)? Can you hink of oher quesions ha one could consider experimenally by inspecing graphs? 3 EFFECT OF DELTA (IMPULSE) ON VIBRATING SYSTEMS Find and graph or skech he soluion of he IVP. Show he deails. 3. ys 4y d( p), y() 8, yr()

14 3 CHAP. 6 Laplace Transforms 6.5 Convoluion. Inegral Equaions Convoluion has o do wih he muliplicaion of ransforms. The siuaion is as follows. Addiion of ransforms provides no problem; we know ha l( f g) l( f ) l(g). Now muliplicaion of ransforms occurs frequenly in connecion wih ODEs, inegral equaions, and elsewhere. Then we usually know l( f ) and l(g) and would like o know he funcion whose ransform is he produc l( f )l(g). We migh perhaps guess ha i is fg, bu his is false. The ransform of a produc is generally differen from he produc of he ransforms of he facors, l( fg) l( f )l(g) in general. To see his ake f e and g. Then fg e, l( fg) >(s ), bu l( f ) >(s ) and l() >s give l( f )l(g) >(s s). According o he nex heorem, he correc answer is ha l( f )l(g) is he ransform of he convoluion of f and g, denoed by he sandard noaion f * g and defined by he inegral () h() ( f * g)() f ()g( ) d. T H E O R E M Convoluion Theorem If wo funcions f and g saisfy he assumpion in he exisence heorem in Sec. 6., so ha heir ransforms F and G exis, he produc H FG is he ransform of h given by (). (Proof afer Example.) E X A M P L E Convoluion Le H(s) >[(s a)s]. Find h(). Soluion. >(s a) has he inverse f () e a, and >s has he inverse g(). Wih g( ) we hus obain from () he answer To check, calculae h() e a * e a # d. a (ea ) f () e a and H(s) l(h)(s) a a s a s b # a a s as s a #. s l(ea )l() E X A M P L E Convoluion Le H(s) >(s v ). Find h(). Soluion. The inverse of >(s v ) is (sin v)>v. Hence from () and he firs formula in () in App. 3. we obain sin v sin v h() * v v v sin v sin v( ) d [cos v cos (v v)] d v

15 SEC. 6.5 Convoluion. Inegral Equaions 33 in agreemen wih formula in he able in Sec v c cos v v c cos v sin v v d sin v v d P R O O F We prove he Convoluion Theorem. CAUTION! Noe which ones are he variables of inegraion! We can denoe hem as we wan, for insance, by and p, and wrie F(s) e s f () d and G(s) e sp g( p) dp. We now se p, where is a firs consan. Then p, and varies from o. Thus G(s) e s() g( ) d e s e s g( ) d. in F and in G vary independenly. Hence we can inser he G-inegral ino he F-inegral. Cancellaion of and hen gives e s e s F(s)G(s) e s f ()e s e s g( ) d d f () e s g( ) d d. Here we inegrae for fixed over from o and hen over from o. This is he blue region in Fig. 4. Under he assumpion on f and g he order of inegraion can be reversed (see Ref. [A5] for a proof using uniform convergence). We hen inegrae firs over from o and hen over from o, ha is, F(s)G(s) e s f ()g( ) d d e s h() d l(h) H(s). This complees he proof. τ Fig. 4. Region of inegraion in he -plane in he proof of Theorem

16 34 CHAP. 6 Laplace Transforms From he definiion i follows almos immediaely ha convoluion has he properies f * g g * f f * (g g ) f * g f * g ( f * g) * v f * (g * v) (commuaive law) (disribuive law) (associaive law) f * * f similar o hose of he muliplicaion of numbers. However, here are differences of which you should be aware. E X A M P L E 3 Unusual Properies of Convoluion f * f in general. For insance, * # d. ( f * f )() may no hold. For insance, Example wih v gives sin * sin cos sin (Fig. 4) Fig. 4. Example 3 We shall now ake up he case of a complex double roo (lef aside in he las secion in connecion wih parial fracions) and find he soluion (he inverse ransform) direcly by convoluion. E X A M P L E 4 Repeaed Complex Facors. Resonance In an undamped mass spring sysem, resonance occurs if he frequency of he driving force equals he naural frequency of he sysem. Then he model is (see Sec..8) ys v y K sin v where v k>m, k is he spring consan, and m is he mass of he body aached o he spring. We assume y() and yr(), for simpliciy. Then he subsidiary equaion is s Y v Y Kv. Is soluion is Y Kv. s v (s v )

17 SEC. 6.5 Convoluion. Inegral Equaions 35 This is a ransform as in Example wih v v and muliplied by Kv. Hence from Example we can see direcly ha he soluion of our problem is y() Kv. v a cos v sin v b K v v (v cos v sin v ) We see ha he firs erm grows wihou bound. Clearly, in he case of resonance such a erm mus occur. (See also a similar kind of soluion in Fig. 55 in Sec..8.) Applicaion o Nonhomogeneous Linear ODEs Nonhomogeneous linear ODEs can now be solved by a general mehod based on convoluion by which he soluion is obained in he form of an inegral. To see his, recall from Sec. 6. ha he subsidiary equaion of he ODE () ys ayr by r() (a, b consan) has he soluion [(7) in Sec. 6.] Y(s) [(s a)y() yr()]q(s) R(s)Q(s) wih R(s) and he ransfer funcion. Inversion of he firs erm 3 Á l(r) Q(s) >(s as b) 4 provides no difficuly; depending on wheher 4a b is posiive, zero, or negaive, is inverse will be a linear combinaion of wo exponenial funcions, or of he form (c c )e a>, or a damped oscillaion, respecively. The ineresing erm is R(s)Q(s) because r() can have various forms of pracical imporance, as we shall see. If y() and yr(), hen Y RQ, and he convoluion heorem gives he soluion (3) y() q( )r() d. E X A M P L E 5 Response of a Damped Vibraing Sysem o a Single Square Wave Using convoluion, deermine he response of he damped mass spring sysem modeled by ys 3yr y r(), r() if and oherwise, y() yr(). This sysem wih an inpu (a driving force) ha acs for some ime only (Fig. 43) has been solved by parial fracion reducion in Sec. 6.4 (Example ). Soluion by Convoluion. The ransfer funcion and is inverse are Q(s), hence q() e e. s 3s (s )(s ) s s Hence he convoluion inegral (3) is (excep for he limis of inegraion) y() q( ) # d 3e () e () 4 d e () e(). Now comes an imporan poin in handling convoluion. r() if only. Hence if, he inegral is zero. If, we have o inegrae from (no ) o. This gives (wih he firs wo erms from he upper limi) y() e e (e () e () ) e () e ().

18 36 CHAP. 6 Laplace Transforms If, we have o inegrae from o (no o ). This gives Figure 43 shows he inpu (he square wave) and he ineresing oupu, which is zero from o, hen increases, reaches a maximum (near.6) afer he inpu has become zero (why?), and finally decreases o zero in a monoone fashion. y() y() e () e () (e () e () )..5 Oupu (response) 3 4 Fig. 43. Square wave and response in Example 5 Inegral Equaions Convoluion also helps in solving cerain inegral equaions, ha is, equaions in which he unknown funcion y() appears in an inegral (and perhaps also ouside of i). This concerns equaions wih an inegral of he form of a convoluion. Hence hese are special and i suffices o explain he idea in erms of wo examples and add a few problems in he problem se. E X A M P L E 6 A Volerra Inegral Equaion of he Second Kind Solve he Volerra inegral equaion of he second kind 3 y() y() sin ( ) d. Soluion. From () we see ha he given equaion can be wrien as a convoluion, y y * sin. Wriing Y l(y) and applying he convoluion heorem, we obain Y(s) Y(s) Y(s) s s s. The soluion is s Y(s) s s 4 s s 4 and gives he answer y() 3 6. Check he resul by a CAS or by subsiuion and repeaed inegraion by pars (which will need paience). E X A M P L E 7 Anoher Volerra Inegral Equaion of he Second Kind Solve he Volerra inegral equaion y() ( ) y( ) d sinh. 3 If he upper limi of inegraion is variable, he equaion is named afer he Ialian mahemaician VITO VOLTERRA (86 94), and if ha limi is consan, he equaion is named afer he Swedish mahemaician ERIK IVAR FREDHOLM (866 97). Of he second kind (firs kind) indicaes ha y occurs (does no occur) ouside of he inegral.

19 SEC. 6.5 Convoluion. Inegral Equaions 37 Soluion. By () we can wrie y ( ) * y sinh. Wriing Y l(y), we obain by using he convoluion heorem and hen aking common denominaors Y(s) c a, hence s s b d s s Y(s) # s s s s s s(s ). (s s )>s cancels on boh sides, so ha solving for Y simply gives Y(s) s s and he soluion is y() cosh. P R O B L E M S E T CONVOLUTIONS BY INTEGRATION Find:. *. * sin v 3. e * e 4. (cos v) * (cos v) 5. (sin v) * (cos v) 6. e a * e b (a b) 7. * e 8 4 INTEGRAL EQUATIONS Solve by he Laplace ransform, showing he deails: 8. y() 4 y()( ) d 9. y() y() d. y() y() sin ( ) d sin. y() ( )y() d. y() y() cosh ( ) d e 3. y() e y()e d e 4. y() y()( ) d 5. CAS EXPERIMENT. Variaion of a Parameer. (a) Replace in Prob. 3 by a parameer k and invesigae graphically how he soluion curve changes if you vary k, in paricular near k. (b) Make similar experimens wih an inegral equaion of your choice whose soluion is oscillaing. 6. TEAM PROJECT. Properies of Convoluion. Prove: (a) Commuaiviy, f * g g * f (b) Associaiviy, ( f * g) * v f * (g * v) (c) Disribuiviy, f * (g g ) f * g f * g (d) Dirac s dela. Derive he sifing formula (4) in Sec. 6.4 by using f k wih a [(), Sec. 6.4] and applying he mean value heorem for inegrals. (e) Unspecified driving force. Show ha forced vibraions governed by wih v and an unspecified driving force r() can be wrien in convoluion form, 7 6 INVERSE TRANSFORMS BY CONVOLUTION Showing deails, find f () if l( f ) equals: (s.5)(s 4) (s a) ps (s p ) s(s 3) v e as.. s (s v ) s(s ) s(s 9) (s )(s 5) 8s 5. (s 36) ys v y r(), y() K, yr() K y v sin v * r() K cos v K v sin v. 6. Parial Fracions. Solve Probs. 7,, and 3 by parial fracion reducion.