6.2 Transforms of Derivatives and Integrals.

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1 SEC. 6.2 Transforms of Derivaives and Inegrals. ODEs Change of scale. If l( f ()) F(s) and c is any APPLICATION OF s-shifting posiive consan, show ha l( f (c)) F(s>c)>c (Hin: In Probs find he ransform. In Probs find Use ().) Use his o obain l(cos v) from l(cos ). he inverse ransform. Show he deails of your work. 24. Inverse ransform. Prove ha l is linear. Hin: e ke a cos v Use he fac ha l is linear e 4.5 sin 2p 36. sinh cos p INVERSE LAPLACE TRANSFORMS (s p) 2 (s ) 3 Given F(s) l( f ), find f (). a, b, L, n are consans. Show 2 4 he deails of your work (s 22) 4 s 2 2s 3.2s.8 5s p s s s 2 ps 24p 2 s a L 2 s 2 n 2 p (s 2)(s 3) s a (s ) 2 a 2 (s ) 3 2 4s s s 6 s 2 6 s s 2 s 2 (s a)(s b) 43. 2s 44. a (s k) bp s 2 6s 8 (s k) 2 p k (s a) k (s a) Transforms of Derivaives and Inegrals. ODEs The Laplace ransform is a mehod of solving ODEs and iniial value problems. The crucial idea is ha operaions of calculus on funcions are replaced by operaions of algebra on ransforms. Roughly, differeniaion of f () will correspond o muliplicaion of l( f ) by s (see Theorems and 2) and inegraion of f () o division of l( f ) by s. To solve ODEs, we mus firs consider he Laplace ransform of derivaives. You have encounered such an idea in your sudy of logarihms. Under he applicaion of he naural logarihm, a produc of numbers becomes a sum of heir logarihms, a division of numbers becomes heir difference of logarihms (see Appendix 3, formulas (2), (3)). To simplify calculaions was one of he main reasons ha logarihms were invened in pre-compuer imes. THEOREM Laplace Transform of Derivaives The ransforms of he firs and second derivaives of f () saisfy () (2) l( f r) sl( f ) f () l( f s) s 2 l( f ) sf () f r(). Formula () holds if f () is coninuous for all and saisfies he growh resricion (2) in Sec. 6. and f r() is piecewise coninuous on every finie inerval on he semi-axis. Similarly, (2) holds if f and f r are coninuous for all and saisfy he growh resricion and f s is piecewise coninuous on every finie inerval on he semi-axis.

2 26 CHAP. 6 Laplace Transforms EXAMPLE 6 Shifed Daa Problems This means iniial value problems wih iniial condiions given a some insead of For such a problem se., so ha gives and he Laplace ransform can be applied. For insance, solve ys y 2, y( 4 p) 2 p, yr( 4 p) 2 2. Soluion. We have and we se p 4 4 p. Then he problem is y s y 2( 4 p), where y Using (2) and Table 6. and denoing he ransform of by Y ( ) y(). y, we see ha he subsidiary equaion of he shifed iniial value problem is s2 Y s # 2 p (2 2) Y 2 s 2 s, Solving his algebraically for Y, we obain Y hus 2 (s 2 )s 2 p 2 (s 2 )s The inverse of he firs wo erms can be seen from Example 3 (wih v ), and he las wo erms give cos and sin, y l ( Y ) 2( sin ) 2 p( cos ) 2 p cos (2 2) sin 2 2 p 2 sin. (s 2 )Y 2 s 2 2 p 2 ps s s 2. Now 4 p, sin (sin cos ), so ha he answer (he soluion) is 2 2 p y () 2 p, yr() 2 2 s ps y 2 sin cos INITIAL VALUE PROBLEMS (IVPS) Solve he IVPs by he Laplace ransform. If necessary, use parial fracion expansion as in Example 4 of he ex. Show all deails. yr 5.2y 9.4 sin 2, y() PROBLEM SET 6.2 yr 2y, y().5 ys yr 6y, y(), yr() 28 ys 9y e, y(), yr() ys 4 y, y() 2, yr() ys 6yr 5y 29 cos 2, y() 3.2, yr() 6.2 ys 7yr 2y 2e 3, y() 3.5, yr() ys 4yr 4y, y() 8., yr() 3.9 ys 4yr 3y 6 8, y(), yr() ys.4y.2 2, y() 25, yr() ys 3yr 2.25y , y(), yr() SHIFTED DATA PROBLEMS Solve he shifed daa IVPs by he Laplace ransform. Show he deails. 2. ys 2yr 3y, y(4) 3, yr(4) 7 3. yr 6y, y() 4 4. ys 2yr 5y 5, y(2) 4, yr(2) 4 5. ys 3yr 4y 6e 23, y(.5) 4, yr(.5) OBTAINING TRANSFORMS BY DIFFERENTIATION Using () or (2), find l( f ) if f () equals: 6. cos 4 7. e a 8. cos sin 2 v 2. sin 4. Use Prob cosh 2

3 SEC. 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) PROJECT. Furher Resuls by Differeniaion. Proceeding as in Example, obain (a) and from his and Example : (b) formula 2, (c) 22, (d) 23 in Sec. 6.9, (e) (f ) INVERSE TRANSFORMS BY INTEGRATION Using Theorem 3, find f() if l(f ) equals: s 2 s>4 s 3 2ps s(s 2 v 2 ) s 4 s s 28. 3s 4 s 4 9s 2 s 4 k 2 s s 3 as 2 l( cos v) s2 v 2 l( cosh a) s2 a 2 (s 2 a 2 ) 2, l( sinh a) (s 2 v 2 ) 2 2as (s 2 a 2 ) PROJECT. Commens on Sec (a) Give reasons why Theorems and 2 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. 2. 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. 2. 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 ).

4 SEC. 6.3 Uni Sep Funcion (Heaviside Funcion). Second Shifing Theorem (-Shifing) 223 We se s and and hen equae he sums of he and erms o zero, obaining (all values rounded) (s ) 4,, 9( )A, A.2776 (s ) 4,, 9( )B, B (s 3 -erms) A B D, D (s 2 -erms) A B D K, K Since K # 4, we hus obain for he firs erm I in I I I I s s s s # 4 s From Table 6. in Sec. 6. we see ha is inverse is i ().2776e 2.644e cos sin 4. This is he curren i() when 2p. I agrees for 2p wih ha in Example of Sec. 2.9 (excep for noaion), which concerned he same RLC-circui. Is graph in Fig. 63 in Sec. 2.9 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 2ps. Since cos 4( 2p) cos 4 and sin 4( 2p) sin 4, he second shifing heorem (Theorem ) gives he inverse i 2 () if 2p, and for 2p i gives i 2 ().2776e (2p) 2.644e (2p) cos sin 4. Hence in i() he cosine and sine erms cancel, and he curren for 2p is i().2776(e e (2p) ) 2.644(e e (2p) ). s 3 s 2 I goes o zero very rapidly, pracically wihin.5 sec. C = 2 F R = Ω L =. H E() Fig. 25. RLC-circui in Example 4 PROBLEM SET 6.3. 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. 2 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. 2. ( 2) 3. 2 ( 2) 4. cos 4 ( p) 5. e ( p>2) 6. sin p (2 4) 7. e p (2 4) 8. 2 ( 2) 9. 2 ( 3 2). sinh ( 2). sin (p>2 p) 2 7 INVERSE TRANSFORMS BY THE 2ND SHIFTING THEOREM l( f ) Find and skech or graph f () if equals 2. e 3s >(s ) ( e ps )>(s 2 9) 4. 4(e 2s 2e 5s )>s 5. e 3s >s (e s e 3s )>(s 2 4) 7. ( e 2p(s) )(s )>((s ) 2 )

5 224 CHAP. 6 Laplace Transforms IVPs, SOME WITH DISCONTINUOUS INPUT Using he Laplace ransform and showing he deails, solve 8. 9ys 6yr y, y() 3, yr() 9. ys 6yr 8y e 3 e 5, y(), yr() 2. ys yr 24y 44 2, y() 9>2, yr() 5 2. ys 9y 8 sin if p and if p; y(), yr() ys 3yr 2y 4 if and 8 if ; y(), yr() 23. ys yr 2y 3 sin cos if 2p and 3 sin 2 cos 2 if 2p; y(), yr() 24. ys 3yr 2y if and if ; y(), yr() 25. ys y if and if ; y(), yr() 26. Shifed daa. ys 2yr 5y sin if 2p and if 2p; y(p), yr(p) 2e p Shifed daa. ys 4y 8 2 if 5 and if 5; y() cos 2, yr() 4 2 sin MODELS OF ELECTRIC CIRCUITS 28 3 RL-CIRCUIT Using he Laplace ransform and showing he deails, find he curren i() in he circui in Fig. 26, assuming i() and: 28. R k ( ), L H, v if p, and 4 sin V if p 29. R 25, L. H, v 49 e 5 V if and if 3. R, L.5 H, v 2 V if 2 and if 2 3. Discharge in RC-circui. Using he Laplace ransform, find he charge q() on he capacior of capaciance C in Fig. 27 if he capacior is charged so ha is poenial is and he swich is closed a. V RC-CIRCUIT Using he Laplace ransform and showing he deails, find he curren i() in he circui in Fig. 28 wih R and C 2 F, where he curren a is assumed o be zero, and: 32. v if 4 and 4 # 6 e 3 V if v if 2 and ( 2) V if v() V if.5.6 and oherwise. Why does i() have jumps? C LC-CIRCUIT v() R Fig. 28. Problems Using he Laplace ransform and showing he deails, find he curren i() in he circui in Fig. 29, assuming zero iniial curren and charge on he capacior and: 35. L H, C 2 F, v 99 cos V if p 3p and oherwise 36. L H, C.25 F, v 2 ( 3 3 ) V if and if 37. L.5 H, C.5 F, v 78 sin V if p and if p R L C L v() Fig. 26. Problems 28 3 v() Fig. 29. Problems C Fig. 27. Problem 3 R 38 4 RLC-CIRCUIT Using he Laplace ransform and showing he deails, find he curren i() in he circui in Fig. 3, assuming zero iniial curren and charge and: 38. R 4, L H, C.5 F, v 34e V if 4 and if 4

6 SEC. 6.4 Shor Impulses. Dirac s Dela Funcion. Parial Fracions ys 6y 4d( 3p), y() 2, yr() 5. ys y d( p) d( 2p), y(), yr() Se (n )p in he nh inegral. Take ou e (n)p from under he inegral sign. Use he sum formula for he geomeric series. 6. ys 4yr 5y d( ), y(), yr() 3 (b) Half-wave recifier. Using (), show ha he 7. 4ys 24yr 37y 7e d( half-wave recificaion of sin v in Fig. 37 has he 2), y(), yr() Laplace ransform 8. ys 3yr 2y (sin d( )), y(), v( e ps>v ) yr() l( f ) (s 2 v 2 )( e 2ps>v ) 9. ys 4yr 5y 3 u( )4e e d( ), v y(), yr().. ys 5yr 6y d( (s 2p) u( p) cos, v 2 )( e ps>v ) y(), yr() (A half-wave recifier clips he negaive porions of he. ys 5yr 6y u( ) d( 2), y(), yr() curve. A full-wave recifier convers hem o posiive; see Fig. 38.) 2. ys 2yr 5y 25 d( p), y() 2, (c) Full-wave recifier. Show ha he Laplace ransform yr() 5 of he full-wave recificaion of sin v is 3. PROJECT. Heaviside Formulas. (a) Show ha for v ps a simple roo a and fracion A>(s a) in F(s)>G(s) we coh have he Heaviside formula s 2 2 v 2v. A lim s:a (s a)f(s). G(s) f () (b) Similarly, show ha for a roo a of order m and fracions in Fig. 37. πω / 2 πω / 3 πω / Half-wave recificaion we have he Heaviside formulas for he firs coefficien and for he oher coefficiens 4. TEAM PROJECT. Laplace Transform of Periodic Funcions (a) Theorem. The Laplace ransform of a piecewise coninuous funcion f () wih period p is () F(s) G(s) A k A m (s a) m A m (s a) A s a furher fracions A m lim s:a (s a) m F(s) G(s) (m k)! lim s:a l( f ) ps p e s f () d e Prove his heorem. Hin: Wrie d mk (s a) m F(s) ds c d, mk G(s) m Á k, Á, m. p (s ). 2p p Á. f () Fig. 38. Full-wave recificaion (d) Saw-ooh wave. Find he Laplace ransform of he saw-ooh wave in Fig. 39. f () Fig. 39. Saw-ooh wave 5. Saircase funcion. Find he Laplace ransform of he saircase funcion in Fig. 4 by noing ha i is he difference of k>p and he funcion in 4(d). f () k k πω / 2 πω / 3 πω / p 2p 3p p 2p 3p Fig. 4. Saircase funcion

7 SEC. 6.5 Convoluion. Inegral Equaions 237 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 2b d s s 2 Y(s) # s 2 s s 2 s2 s s(s 2 ). (s 2 s )>s cancels on boh sides, so ha solving for Y simply gives Y(s) s s 2 and he soluion is y() cosh PROBLEM SET CONVOLUTIONS BY INTEGRATION Find:. * 2. * 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 2 9. y() y() d. y() y() sin 2( ) d sin 2. y() ( )y() d 2. y() y() cosh ( ) d e 3. y() 2e y()e d e 4. y() y()( ) d CAS EXPERIMENT. Variaion of a Parameer. (a) Replace 2 in Prob. 3 by a parameer k and invesigae graphically how he soluion curve changes if you vary k, in paricular near k 2. (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 2 ) f * g f * g 2 (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 26 INVERSE TRANSFORMS BY CONVOLUTION Showing deails, find f () if l( f ) equals: (s.5)(s 4) (s a) 2 2ps (s 2 p 2 ) 2 s(s 3) v e as s 2 (s 2 v 2 ) s(s 2) s(s 2 9) (s 2 )(s 2 25) 8s 25. (s 2 36) 2 ys v 2 y r(), y() K, yr() K 2 y v sin v * r() K cos v K 2 v sin v. 26. Parial Fracions. Solve Probs. 7, 2, and 23 by parial fracion reducion.