CHAPTER 8 Forced Equaion and Syem 8 The aplace Tranform and I Invere Tranform from he Definiion 5 5 = b b {} 5 = 5e d = lim5 e = ( ) b {} = e d = lim e + e d b = (inegraion by par) = = = = b b ( ) ( ) 3 { e } e e d e d lim e + + + + b = = = lim = b ( ) ( ) 4 { e } e e d e d e co+ in = = = (inegraion by par wice) b + 4 + 4 b 5 { in} e ind lime = = (inegraion by par wice) + 9 6 { co3} e co3d 7 { } e e e e = d = lim b 3 = (inegraion by par wice) 3 4 4 4 8 f () = e d = e = e = e 4 { } ( ) ( ) 9 { f () } = ( ) e d+ e d e d e d e d ( e ) = + = { f () } = ( ) e d = 3 ( e ) + e b 767
768 CHAPTER 8 aplace Tranform Tranform wih Tool a b c + + = + + = + + 3 { a b c } a{} b{} c { } + = + = + = + + + { e } {} { e } 3 { e e } { e } { e } ( ) + = + = + = + 4 3 + + = + + = + + + + 4 4 { 3 e in} 3{} {} { e in} { e 3 } { e } 3 { e } 6 + = + = + 5 ( ) ( + ) ( + ) 3 3 3 3 6 { e 4e co3} { e } 4 { e co3} 3 ( ) ( ) 4( ) ( ) 6 + + = + = + 4 + 3 + +9 a a a a + 3a e e = e e = = a + a a 7 { } { } { } + = + = + + 3 3 8 { e in} { e } { in} ( + 3) ineariy of he aplace Tranform 9 Uing baic properie of he inegral yield { + } = () + () = () + () = c{ f} + c { g} cf cg e cf cg d c e f d c e g d I There a Produc Rule? Clearly, no produc rule exi, a here i no produc rule for inegral For example, bu {} = { } = 3, o {} {} = { 4 }
aplace Tranform of Damped Sine and Coine Funcion SECTION 8 The aplace Tranform and i Invere 769 (a) (b) ( ) ( ) { } a+ ik aik a+ ik e = e d = = aik a + k ( ) Breaking hi ino real and complex par yield he deired reul ( + ) { } ( ) { co in } { ( co )} { in } a ik a a a e = e k+ i k = e k +i e k Maching real and complex par of he oluion yield he aplace ranform formula a a for e co k and e in k aplace Tranform of Hyperbolic Funcion e e b = = = = b b b b { inhb} { e } { e } ( b) ( + b) b b b e + e b b { coh b} = = { e } + { e } = + = Uing Hyperbolic Funcion 3 (a) { coh b inh b} ( b) ( + b) b b b b b e + e e e = b b b b = ( e + + e ) ( e + e ) = {} = 4 4 e + e = + b b b b b b (b) { coh b} = = e + e = { e } + { e } ( b) ( + b) 3 3 Power Rule e = n 4 (a) Given { } n d, we inegrae by par, leing u n = and dv = e d, o ge e n = + b n n b { } lim e n On he righ ide, he lef-hand erm become in he limi (for > ); n he inegral erm become { n } The reul follow immediaely d
77 CHAPTER 8 aplace Tranform (b) Performing inegraion by par n ime yield n n! { } = e d n Inegraing give he final anwer a n! n { } = n + Muliplier Rule d d d 5 f () = e f d = e f d f e d F = = d d d { } () ( ) () () ( ) Muliplier Applicaion { } The muliplier rule (Problem 5) ay o evaluae f( ) ranform of f ( ), geing F ( ) Then o ge he ranform we differeniae ( ) Tha i, d { f () } = { f () } d, we can fir ignore he and ake he F and change he ign a a 6 { e } { e } d d = = d d a = a 7 { in 3} { in3} ( ) d d 3 6 = = d d 9 = + + 9 ( ) e + e 8 coh b b b b b b { b} = = { e + e } = { e } + { e } d d = d + b d + b = + ( b) ( + b) 9 { a} { a} d d a 3 co = 3 co = 3 3 d d = a + + a ( )
3 { inh } { e } { e } ( + ) ( ) SECTION 8 The aplace Tranform and i Invere 77 e e d d = = = d + d + = Exponenial Shif { } () () ( ) 3 a a ( a) e f () = e e f d = e f d = F a Uing he Shif { } To find e a f ( ) f ( ), and hen replace in F() { f() } a, Problem 3 ay we can ignore he exponenial funcion e, ake he ranform of = by ( a) Tha i, a { e f() } = F( a) n! 3 { n a n e } We fir compue F( ) = { } = n n a Then { e } F( a) + n! = = n a + ( ) 33 { e in } + 4 We fir compue F( ) = { in } = = = + 4 Then { e in } F( ) ( ) 34 { e co3 } + 9 We fir compue F( ) = { co3} = Then { e co3} F( ) ( ) = + = + + 9 +
77 CHAPTER 8 aplace Tranform 35 { e coh 3 } 9 We fir compue F( ) = { coh3} = = = 9 Then { e coh 3} F( ) ( ) 36 { e 3 inh } We fir compue F( ) = { inh } = = + = + 3 3 Then { e inh } F( 3) ( ) 37 { e in3} a We ue boh he muliplier rule (Problem 5) and he e law (Problem 3) for he aplace ranform I make no difference which one i ued fir Here he ranform i compued fir: Uing he exponenial law, we hen find Finally, from he muliplier rule, ineariy of he Invere 38 Uing he lineariy of, we wrie 3 { in 3} = + 9 3 { e in 3} = + 9 ( ) 6( ) ( ) d 3 { e in 3} = = d ( ) + 9 + 9 { { ( )} { ( )}} ( ) { { }} { ( )} a F b G a F b G af bg + = + { } = ( ) + ( ) Taking he invere ranform of each ide yield which prove he lineariy of Ou of Order { af( ) + bg( ) } = a F( ) + b G( ), 39 For any conan α, we can pick can be large enough o ha > α, which implie ha > α, which, in urn, implie ha e > e α Hence evenually e will be greaer han e α
SECTION 8 The aplace Tranform and i Invere 773 Invere Tranform The key for Problem 4-54 i o rewrie each funcion in erm of funcion lied in he hor aplace ranform able, on page 47 of exbook Thee ranform are alo included in he longer able inide he back cover of he ex 4 3 = 3 = 4 3 7 + + 3 7 = + 3 + 7 = + 3e + 3 4 43 5 = + 3 3 4 + = 3e + 4e 3 + 3 5 3 5 in 3 3 + ( 3) 3 = 3 3 44 The parial fracion decompoiion ( ), by he lineariy of he invere ranform A B = + + 3 + 3 Equaing coefficien of like erm and olving for A and B yield + 3 3 3( + 3) 3 Hence = 3 = ( e ) i equivalen o ( 3) A = and 3 = + A+ B B = 3 45 + + + Compleing he quare in he denominaor yield + + = + + + + 9 ( ) Hence, by he exponenial law, + ( + ) e = = + + ( + ) + 9 co3 46 47 = + 4+ 4 = e ( + ) ( ) 3 5 + = 3 3 + 4 6+ 5 ( 3) + 6 3 7 4 3 7 = 3 + e 3 = co4 + in4 ( 3) + 6 ( 3) + 6
774 CHAPTER 8 aplace Tranform 48 49 + + Facoring he denominaor and wriing a a parial fracion give + + A B = = + + + + ( )( ) Solving for A and B yield A = and B = Hence 3 3 + e e = + = + 3 + 3 + 3 3 5 + 6 Facoring he denominaor and wriing a a parial fracion give 5 5 A B = = + + + + ( )( ) 6 3 3 Solving for A and B yield A = and B =, o he invere ranform i 5 + 4 5 e e = + = + + 6 + 3 Facoring he denominaor and wriing a a parial fracion give + 4 + 4 A B = = + + + ( )( ) 3 Solving for A and B yield A = 3 and B =, o he invere ranform i + 4 3 3e e = = + 5 7 = + 4+ 7 7 3 7 e in 3 3 = ( + ) + ( 3) 3 5 6 + = + 4+ 3 ( ) ( ) + + + + 9 + 3 = + 4 e = ( co3+ in3) ( + ) + 9 ( + ) + 9
SECTION 8 The aplace Tranform and i Invere 775 53 4 ( + 4) Here we ue parial fracion o wrie 4 ( + 4) A B C+ D = + + + 4 Muliplying he equaion by he denominaor on he lef-hand ide yield 3 ( ) ( ) ( ) 4= A+ C + B+ D + 4A +4 B Comparing coefficien and olving he reuling equaion for A, B, C, and D yield A =, B =, C =, and D = Hence 54 4 = = = in ( + 4) + 4 + 4 3 ( + )( + 4) We ue parial fracion o wrie 3 ( + )( + 4) A + B C + D = + + + 4 Muliplying he equaion by he denominaor on he lef-hand ide, we ge ( A C) 3 ( B D) ( A C) ( B D) 3= + + + + 4 + + 4 + Comparing coefficien and olving he reuling equaion for A, B, C, and D yield A =, B =, C =, and D = Hence 3 in in = = = ( + )( + 4) + + 4 + + 4 Compuer Exploraion 55 Suden Projec Suggeed Journal Enry 56 Suden Projec
776 CHAPTER 8 aplace Tranform 8 Solving DE and IVP wih aplace Tranform Fir-Order Problem, y = ( ) y = y y = The aplace ranform yield he equaion { } = + Subiuing in he iniial condiion and olving for he aplace ranform yield { y} Taking he invere ranform yield y ( ) = ( ) y y, y = = + The aplace ranform yield he equaion { y} y { y} Subiuing in he iniial condiion give { y} Taking he invere ranform yield ( ) = ( ) 3 y y e, y = y = = = e = The aplace ranform yield he equaion { y} { y} Subiuing he iniial condiion and olving for he aplace ranform give { y} = + ( ) Taking he invere ranform yield y ( ) = e+ e + = ( ) 4 y y e, y = + + = + The aplace ranform yield he equaion { y} { y} Subiuing he iniial condiion and olving for he aplace ranform give { y} = ( ) y = e e Taking he invere ranform yield ( )
SECTION 8 Solving DE and IVP wih aplace Tranform 777 Tranformaion a Work 5 y 3y y, + = y ( ) =, ( ) y = Taking he aplace ranform yield he equaion Solving for he aplace ranform yield Hence, he oluion i ( ) 6 y y 4, + = y ( ) =, ( ) { } { } ( ) { } y 3 y + y = 3 3 { y} = = = + 3+ y = e + e y = 4 Taking he ranform yield he equaion Solving for he aplace ranform yield ( )( ) { } { } y + 4+ y = 4 4 = + + { y} ( + ) + ( +) Uing parial fracion we wrie he fir erm of { y} a 4 A B C = + + = + + + + + ( ) Similarly, we wrie he la fracion of { y} a D E = + = + + + + ( ) Puing hee expreion ogeher yield he ranform 3 { y} = + + + Taking he invere ranform yield he oluion of he iniial-value problem + = y ( ) =, ( ) 7 y 9y e, y = ( ) 3 y e = + + Taking he ranform yield he equaion { y} ( ) + 9 { y} = +
778 CHAPTER 8 aplace Tranform Solving for he aplace ranform yield { y} = + = + 9 9 + 9 + + + + + + +9 ( )( 9) Combining erm and aking he invere yield 8 y + 9y = co3, ( ) y =, ( ) y = Taking he ranform yield he equaion Solving for he aplace ranform yield ( ) co3 in y = e + 3 { } ( ) + + 9 { } = + 9 y y { y} = + + 9 ( + 9) Noice ha he fir erm i half of he negaive derivaive of Thu, he invere ranform + 9 of he fir erm i imply in 3 Breaking he econd erm ino wo par yield 6 co3 in 3 Hence, he anwer i 3 9 y 3y y 6, + + = y ( ) =, ( ) y() = in 3+ co3 in 3 6 3 y = Taking he ranform yield he equaion Solving for he aplace ranform yield { } ( ) 3 { } { } y + y + y = 6 { y} Hence, he final reul i ( ) { } 6 = + ( + )( + ) ( + )( + ) 3 6 3 = + + + + + + 3 4 = + + + { } y = y = 3 4 e + e
SECTION 8 Solving DE and IVP wih aplace Tranform 779 y + y + y =, y() =, y () = Taking he aplace ranform of he equaion Y() + Y() + Y() = + Y( ) = = = (by parial fracion) ( + + ) 3 + + + + 4 + = (by compleing he quare) 3 + + 4 3 + = 3 3 3 + + + + 4 4 We obain (/ ) 3 (/ ) 3 y () = e in e co 3 y + y + y= in, y() =, y () = Taking he aplace ranform of he equaion () + () + () = + Y Y Y + Y( ) = = + (by parial fracion) ( + )( + + ) + + + + = + (by compleing he quare) + 3 + + 4 3 + = + + + + + + + 4 4 3 3 3 3 y () = co+ e in + e 3 co (/ ) (/) 3
78 CHAPTER 8 aplace Tranform y + y + y= e, y() =, y () = Taking he aplace ranform of he equaion () + () + () = + + Y( ) = = (by parial fracion) ( + )( + + ) + + + Y Y Y = (by compleing he quare) + 3 + + 4 3 + = + 3 + + + + + 4 4 3 3 (/ ) () 3 in y = e + e 3 (/ ) 3 e co General Soluion 3 y y = Taking he aplace ranform of each ide and calling y( ) Solving for { y} yield { y} Hence, { } { } = y A B y + A + B = = A + B + = A and y ( ) = B, we ge ( ) = A + B + y() = Acoh+ Binh + inh = Acoh + ( B+ ) inh (uing parial fracion)
4 y + 3y + y= Taking he aplace ranform of each ide and calling y( ) { } { } Solving for { y} yield { } SECTION 8 Solving DE and IVP wih aplace Tranform 78 ( ) { } y A B+ 3 y A + y = + 3 y = A + B + 3+ + 3+ + 3 = A + B ( + )( + ) ( + )( + ) = A and y ( ) = B, we ge = A + + B + (uing parial fracion) + + + + y = Ae + Ae Be + Be = C e + C e Hence, ( ) where C = A+ B and C = A B Raiing he Sake 5 y y y y 6e, + = y( ) = y ( ) y ( ) = = + = The ranform yield { y} { y} { y} { y} Solving for he { y} yield 3 6 6 = { y} ( )( 3 + ) The denominaor facor furher, and we can hen ue parial fracion, o 6 A B C D = + + + + ( ) ( + ) ( ) ( ) 3 3 3 3 3 3 = + 3 4 4 + ( ) ( ) 3 3 3 3 Thu he oluion i y() = e e + e e 4 4 (4) 6 y y, = y ( ) =, y ( ) =, y ( ) =, ( ) The ranform yield 4 { y} 3 + { y} = y = Solving for he aplace ranform yield ( )( ) + { y} = = + Thu he oluion o he IVP i y ( ) co ( )( + )( + ) =
78 CHAPTER 8 aplace Tranform Which Grow Faer? 7 For k > and y() =, conider he wo DE y = ky, y = k y() d, k > For he fir equaion i hould be familiar ha y () = e bu we olve he equaion uing aplace ranform Y() = ky() Y() = k k y () = e The econd equaion can alo be olved uing aplace ranform Y() Y () = k Y() = k k k e + e y () = coh ( k) = k For value of k uch ha < k <, he oluion o he econd equaion will evenually oupace he fir, while if k, he oluion o he fir equaion will oupace he econd To ee hi, compue he limi of he raio of he wo oluion k k e e ( k k) lim = lim = lim e k k k e + e e When k k <, hi limi i indicaing ha he denominaor i much larger han he numeraor When he revere i rue, he limi i infinie A he value of k =, he limi i exacly, indicaing ha he oluion o he fir equaion i eenially wice he oluion o he econd aplace Tranform Uing Power Serie 8 The power erie for e i f e x x x n ( ) = = = + + + n= n! n! F () = = = + + + n! n+ n+ 3 n= n=
SECTION 8 Solving DE and IVP wih aplace Tranform 783 Thi i a geomeric erie wih he fir erm and a common raio of The cloed form for hi erie i F () = = Operaor Mehod 9 y + 3y + y= The differenial equaion can be rewrien a ( D + )( D+ ) y =, or a he yem of equaion ( D+ ) v= ( D + ) y= v (Simply ubiuing he econd equaion ino he fir will yield he original equaion) Solving he fir equaion for v uing aplace ranform: V () v() + V () = v() v() V() = + = + + ( + ) + + v () = ( v() ) e + Subiuing hi ino he econd equaion yield ( D+ ) y = ( v() ) e v() Y () y() + Y () = + + ( + ) Noe here ha here wa really no need o find he expreion for v() unle you are inereed in v() y() v() Y() = + + + ( + )( + ) ( + )( + ) = y() + v() + + (by parial fracion) + + + + ( + ) y () = y() e + v()( e e ) + e + e +
784 CHAPTER 8 aplace Tranform 6 y + y= ( D + ) y = 6 y () D + = Beel Funcion wih IDE Suden ab Projec Compuer Exploraion Suden ab Projec aplace Vibraion 3 Suden ab Projec Suggeed Journal Enry 4 Suden Projec 6 = D + D D + D 4 6 6 ( ) (by dividing + ino ) 6 4 4 = + + + + + = + 3 36 7 3 36 7
SECTION 83 The Sep Funcion and he Dela Funcion 785 83 The Sep Funcion and he Dela Funcion Sepping Ou The funcion f() fir ha he value of for <, hen a from o Hence we wrie hi a ( ) ep( ) f = a However, a = he funcion hif from a o b We, herefore, ubrac a and add b, yielding ( ) ep( ) ( ) ep( ) f = a + ba Finally, when =, he funcion hif from b o c, o we ubrac b and add c Thu for all we have ( ) ep( ) ( ) ep( ) ( ) ep( ) f = a + ba + cb Following he procedure in Problem, we wrie he funcion a ( ) ( ) ep( ) ( ) ep( 3) f = + e + e 3 Following he procedure in Problem, we wrie he funcion a ( ) ( 4 ) ep( ) ( 4 ) ep( 4) f = + + + 4 Following he procedure in Problem, we wrie he funcion a ( ) inπ ep( ) inπ ep( 4) f = Geomeric Serie 5 f ( ) ep( ) ep( ) ep( 3) = + + +, { } e e e e 3 e = + + + = + + + + = e 3 { f () } ( e ) ( e ) ( e ) 6 f ( ) ep( ) ep( ) ep( 3) = + + + +, 3 { } e e e = + + + + = + + + + = e { ()} 3 f ( e ) ( e ) ( e ) =, 4 8 7 f () ep( ) ep( ) ep( 3) { f () } 3 e e e ( e ) = = 4 8 = e ( e ) 8 f ( ) ep( ) ep( ) = +, 3 e e e { f () } = + = + e
786 CHAPTER 8 aplace Tranform 9 f ( ) ep( ) ep( ) = +, ( ) e e e e e = + = + = + e 3 { f () } e ( e ) Piecewie-Coninuou Funcion In he following problem we ue he alernae delay rule c { f ( ) ep( c) } = e f ( + c) { } The funcion can be repreened by ( ) = ep( ) ep( ) + ( ) ep( ) ep( ) f Uing he Alernae form of he Delay Theorem we ge { f () } = e { + } + e { } e { } = e e + e e + e ( = e + e = e + e ) () ( ) 3 f = + ( ep ) ( ep( ) ep( 3) ) e 3 3 { f () } = + e e = e + e f ( ) = ( ) ep( ) ep( 3) + ep( 3), 3 3 3 3 3 3 e e e e e e e { f () } = e {} e { + } + = + = π a ( ) 3 f () = bin ep( a) π ( a) π ( ), ( + a) π a π { f () } = b in be in a a 4 () ( ), π ( ) b π b = be in = + a + a a + a π f = in ( ep ) + ep( ) ep( ), e e { f () } π = e π + ( ) + a ( ) ( a e ) π
5 The wo par of he ine funcion can be wrien a SECTION 83 The Sep Funcion and he Dela Funcion 787 ( ) = in ( π )( ep( ) ) in ( π )( ep( ) ep( ) ), { f () } = { inπ} e in( π( + ) ) e { inπ + } + e in ( π( + ) ) f { } ( ) { } π π π π π = + e + e + e = ( + 3e + e ) + π + π + π + π + π Tranforming Dela 3 { δ + δ + 3δ 3 } = e + e + 3e π π { δ δ π + δ π } = e + e { δ δ + δ } = e + e π π { δ δ π δ π } e e 6 ( ) ( ) ( ) 7 ( ) ( ) ( ) 8 ( ) ( ) ( ) 9 ( ) ( ) ( ) + + + = + + + aplace Sep by Sep f ( ) = ep( ) The aplace ranform of i ranform i e { f } = f ( ) = ep( ) + ep( ) f ( ) = ( ep ) ( ) Noe ha hi i he funcion f ( ) ranform, of (which i anwer i 3 f ( ) = in( π ) ep( π ), wherea he ranform of ep ( ) { f } e e = + i e Hence, he = hifed o he righ one uni Thu, we ake he aplace ) and muliply i by he hifing facor (which i e { f } = Thi i he ine funcion hifed o he righ π Hence, he ranform i π e { f } = + e ) Hence he final
788 CHAPTER 8 aplace Tranform 4 f ( ) = e ep( 3) The funcion here ha no been hifed, o we mu perform he hif and wrie he funcion a ( ) e 3 ep 3 e 3 Hence, he ranform i 5 f ( ) = ep( e ) 3 33 3 e e { f} = e = Someime i i ueful o pu funcion a argumen of he ep funcion, o i wiche off and on a pecial poin depending on he funcion In hi cae, he funcion in he argumen e i alway poiive (for > ), o he funcion ep ( e ) = for > Hence, 6 f ( ) = ep ( ) { f } = We wrie he expreion in he form f ( c) ep( c) We do hi by wriing ep( ) = ( ) ep( ) + 4( ) ep( ) + 4ep( ) Hence, he ranform i Invere Tranform 4 4 { f} = e + + 3 7 e Facoring ou he e, he funcion i imply ; he invere ranform i The facor e here i a ep( ) muliplied in hi erm Hence, he invere ranform i ep( ) mean Once again, he graph of hee funcion are all graph of familiar funcion delayed, and example can be een in he ex 8 e The invere ranform of i Hence, he invere ranform i e = ( ) ep( )
SECTION 83 The Sep Funcion and he Dela Funcion 789 9 e 3 The invere ranform of 3 i 3 e Hence, he invere ranform i e 3( ) = e ep( ) 3 3 4 e + 4 The invere ranform of + 4 i 4 e Hence, he invere ranform i 4 e 4( 4) = e ep( 4) + 4 3 e ( + ) Delayed Exponenial f() Wriing he parial fracion decompoiion, yield = + + ( ) 5 Hence, hi expreion ha he invere ranform e The given funcion ha he invere ranform (ee figure): 3 4 e ( + ) ( ) ( ) ( e ) = ep 3 3 4 e e + e e We break hi ino four par, yielding an invere aplace ranform 3 4 e e + e e ( ) ( ) ( ) ( = ep ep + ep 3 ep 4 )
79 CHAPTER 8 aplace Tranform Tranforming Soluion 33 x = ep( ), ( ) x = The aplace ranform of he equaion i ( ) X e = e = Taking he invere ranform yield he oluion of he iniial-value problem Solving for X ( ) yield X( ) ( ) ( ep ) ( ) x = x ( ) = ( ) ep( ) f() = ep ( ) 3 4 34 x ep( ) ep( ) ( ) = +, x = The aplace ranform of he equaion i ( ) X Solving for X ( ) yield ( ) X e e = + e e = + Taking he invere ranform yield he oluion of he iniial-value problem ( ) ( ) ep( ) ( ) ep( ) x = + 35 x + x= ep( 3), x ( ) =, ( ) x = The aplace ranform of he equaion i X 3 e + = ( ) X( ) Solving for X ( ) yield ( ) X 3 e = + + + ( ) Taking he invere ranform yield he oluion of he iniial-value problem ( ) = in + ep( 3) co( 3) x x ()= ( ) ep( ) + ( ) ep ( ) 3 4 f ()= ep ( ) + ep ( ) Forcing erm and repone 3 x ()= in () ep ( 3) ( co( 3) ) f ()= ep ( 3) 3 6 9 Forcing erm and repone
36 x + x= ep( π ) ep( π ), x ( ) =, ( ) The aplace ranform of he equaion i X π π e e + = ( ) X( ) Solving for X ( ) yield ( ) X π π e e = + + + + ( ) ( ) Uing a parial fracion decompoiion, we wrie = ( + ) + SECTION 83 The Sep Funcion and he Dela Funcion 79 x = 3 3 f () 3 6 x () 9 Repone o one quare pule Taking he invere ranform yield he oluion of he iniial-value problem, ( ) ( ) ( ) ( π ) ( ( )) ( π ) x = in + + co ep co ep Periodic Formula 37 In Problem, we oberved ha he aplace ranform of he ingle riangular wave on he inerval [, ] wa e + e { ingle wave} = f ( ) e d = Uing equaion (4) in he ex, he aplace ranform of he periodic riangular wave on [, ) i e + e { f () } = e y f() 3 4 Triangular wave 5 6 38 From Problem, { ingle wave} = ( ) f e e + e = d Hence, he aplace ranform of he periodic riangular wave of period 3 i e + e { f () } = e 3 y 3 4 5 6 Modified riangular wave
79 CHAPTER 8 aplace Tranform 39 Fir, we find he aplace ranform of he ingle wave-form which i ( ) = ep( ) f, <, { ingle wave} ( ) f e d = We deermine he aplace ranform by applying he alernae form of he Delay Theorem inead of inegraing 3 4 { ingle wave} = e { ( + ) } = e + = e + The aplace ranform of he periodic wave-form of period on [, ) i = e = e e + { f () } { ingle wave} 4 From Problem 3 (wih a= π, b= ), we have π { ingle wave} = ( + e ) + Hence, he aplace ranform of he full-recified periodic wave wih period π on [, ) i y Sawooh Wave 5 6 π + e { f () } = π e + 4 From Problem 3 (wih a= π, b= ), we have π { ingle wave} = ( + e ) + Hence, he aplace ranform of he half-recified periodic wave wih period π on [, ) i π + e { f () } = π e + Full wave recificaion Half-wave recificaion
SECTION 83 The Sep Funcion and he Dela Funcion 793 4 From Problem 4, we have { ingle wave} π e = e + + π ( ) ( ) ( 4 + π ) π + π e e 4 + π = e Hence, he aplace ranform of he periodic funcion of period 4 on [, ) i y 3 4 5 6 7 { f () } π e e = e 4 + e π + ( ) 43 From Problem, we have y { ingle wave} e e = + 3 e + e = 3 3 4 5 6 7 Hence, he aplace ranform of he periodic funcion of period 4 on [, ) i 3 e + e { f () } = 4 e 44 From Problem 5, we have π { ingle wave} = ( + 3e + e ) + π π ( + 3e + e ) = + π Hence, he aplace ranform of he periodic funcion of period on [, ) i π + 3e + e { f } = e + π y 3 4 5 6
794 CHAPTER 8 aplace Tranform 45 From Problem 8, we have y { ingle wave} = e f ( ) d = e d = ( e ) Hence, he aplace ranform of he periodic funcion of period on [, ) i e { f () } = e 3 3 4 Square wave 5 6 Sawblade 46 Inpu x + x= f (), ( ) x = where ( ) ep( ) ep( ) f = y f () Taking he aplace ranform of he DE yield X( ) { f } =, + 5 x () e e where { f () } = Uing wo parial fracion decompoiion, awooh wave inpu 4 6 = + and =, + + + + ( ) ( ) we obain e e X( ) = e e = + + + + + Therefore, he oluion of he iniial-value problem i () ( ) ( ) ( ) ( ) ( ) ( ) x = + e e ep e ep Square Wave Inpu 47 x + x= f ( ), x( ) x ( ) = = where ( ) ep( ) ep( ) f = + Taking he aplace ranform of he DE yield X( ) where e e { f () } = + = { f }, + Repone o a quare wave inpu
Hence X( ) = e + e ( + ) SECTION 83 The Sep Funcion and he Dela Funcion 795 We wrie he parial fracion decompoiion a = ( + ) + ( ) X = e e e e + + + Therefore, he oluion of he iniial-value problem i () = ep( ) + ep( ) ep( 3) + co+ co( ) ep( ) co( ) ep( ) + co( 3) ep( 3) = ( ) ( ) + ( ) ( ) ( ) ( ) x co ep co ep co ep 3 co 3 Solve on Impule 48 x = δ ( ), ( ) x = Taking he aplace ranform of boh ide of he DE yield X ( ) = or X( ) Hence, we have x ( ) = = 49 x = δ( ) δ( ), ( ) x = Taking he aplace ranform of boh ide of he DE yield The invere i X( ) e =, or X( ) ( ) ep( ) x = e = 5 x + x= δ ( π ), x( ) x ( ) = = Taking he aplace ranform of boh ide of he DE yield The invere i ( ) ( ) X e π + =, or X( ) ( ) in ( π ) ep( π ) x = ; i graph i he ine funcion aring a = π π e = +
796 CHAPTER 8 aplace Tranform 5 x + x=δ ( π) + δ( π), x ( ) =, ( ) x = Taking he aplace ranform of boh ide of he DE yield ( ) ( ) π π + X = e + e Solving for X ( ) yield The invere i π π X( ) = e e + + + + ( ) in in( π ) ep( π) in( π) ep( ) x = + π 5 x + x= δ ( π ), x ( ) =, ( ) x = Taking he aplace ranform of boh ide of he DE yield ( ) ( ) X e π + = Solving for X ( ) yield ( ) e X The invere i π = + Repone o wo impule + + Repone o an impule a π ( ) co in ep( ) x = + π aplace wih Forcing Funcion 53 Suden ab Projec Suggeed Journal Enry 54 Suden Projec
SECTION 84 The Convoluion Inegral and he Tranfer Funcion 797 84 The Convoluion Inegral and he Tranfer Funcion Convoluion Properie Verify f ( g h) = ( f g) h Mehod Compuaion of hee properie from he inegral definiion of convoluion require lenghy ubiuion and exchanging of he order of inegraion I can be done from he definiion, bu we will rely on he convoluion heorem in hi mehod o how ha i i rue ( ( )) = () ( ) = ()[ () ()] = [ () ()] () = ( ) () (( ) ) f g h F g h F G H F G H f g H f g h Mehod Uing he inegral definiion of convoluion, ( ) f ( g h) = f g( w ) h( w ) dw w = f ( w ) g( w w) h( w) dw dw eing u = w w in he inner inegral (reaing w a a conan) yield = f ( w) g( u ) h( w u)( du) dw w w = f ( w ) g( u ) h( w u) du dw w = f ( w ) g( u ) h( w u ) du dw Exchanging he order of he inegraion (a very nonrivial ep) yield = f ( w) g( u) h( w u) dw u du eing w = u + u (reaing u a a conan) in he inner inegral u = + f ( ( u u)) g( u) h( u) du du Again, wih ha nonrivial exchanging of he order of inegraion, u = f ( u u ) g( u ) du h( u ) du = f( u ) g( u ) du h = ( f g) h
798 CHAPTER 8 aplace Tranform Prove f ( g + h) = f g + f h 3 Prove f = [ ] f ( g + h) = f( w) g( w) + h( w) dw = f( w) g( w) f( w) h( w) dw = f ( w) g( w) dw f( w) h( w) dw = f g + f h Calculaing Convoluion f = f ( w)() dw = dw = 4 5 6 = ()() d = = = ( w) dw= = ( w) dw= 7 3 w w = ( w) wdw= = 3 6 8 To find, for any k, where k repreen he number of ime appear in he produc, (Noe ha Problem 7 i he iniial k = cae) Then for k = 3, : 4 5 3 3 w w = = ( w) w dw 6 = 6 6 4 5 5 = 5! Uing hi reul and Problem 7, we make he following conjecure for any k: = (k )! For k 3, we confirm hi conjecure a follow: k k Proof by inducion: Aume ha * = for ome k (k )! 3
For he k + cae: ( ) = (k )! which prove i rue for k =, 3, 4, SECTION 84 The Convoluion Inegral and he Tranfer Funcion 799 k = ( ) ( k + )! k w w dw k k = w w dw (k )! k k+ w w = (k )! k k + k+ k+ = (k )! k k + = = (k )! k(k + ) (k + )! k+ k+ 9 ( w) w w w e e = e e dw= e dw= e e e inh = + = a a a( w) aw a aw aaw a a e e = e e dw = e dw e e e inh a = a = + = a a a Fir-Order Convoluion Equaion I = b a a oluion o he equaion a = b? e find ou Thi reul mean ha = b a b b a = ( a) dw a a b = [ bw] = b = b a in a oluion o a = b The main problem lie in he fac ha i no longer a muliplicaive ideniy if you are conidering o be muliplicaion, nor i a an invere for he operaion anymore
8 CHAPTER 8 aplace Tranform Convolued Soluion x = f ( ), ( ) x = Taking he aplace ranform of boh ide of he DE yield X( ) { f} f Solving for X ( ) yield X ( ) = { f} = { } { } The invere yield he oluion () () ( τ ) 3 x = f ( ), ( ) x = y = f = f dτ = Taking he aplace ranform of boh ide of he DE yield X( ) { f} Solving for X ( ) yield X ( ) = + { f} = + {} { f} The invere yield he oluion 4 x + x= f ( ), ( ) x = () () ( τ ) y = + f = + f dτ Taking he aplace ranform of boh ide of he DE yield Solving for X ( ) yield X( ) + X( ) = { f} = The invere yield he oluion 5 x + x= f ( ), ( ) x = X ( ) = { f} = { e } { } + f () () ( τ ) ( τ ) x = e f = e f dτ Taking he aplace ranform of boh ide of he DE yield Solving for X ( ) yield ( ) ( ) { } X + X = f The invere yield he oluion + + + e f ( ) = + { f} = + { } { } X ( τ ) () = + () = + ( τ ) τ x e e f e e f d
SECTION 84 The Convoluion Inegral and he Tranfer Funcion 8 6 x + x= f ( ), x ( ) =, ( ) x = Taking he aplace ranform of boh ide of he DE yield ( ) ( ) { } X X f + = } + + + f Solving for X ( ) yield X ( ) = + { f} = + { in } { The invere yield he oluion () co in () co in ( ) ( ) x = + f = + τ f τ dτ 7 x + 3x + x= f ( ), ( ) ( ) x = x = Taking he aplace ranform of boh ide of he DE yield Solving for X ( ) yield The invere yield he oluion ( 3 ) X( ) { f} + + = { f} { f} { f} X ( ) = = = { e } { f} { e } { f } + 3+ + + () () () ( ) () ( ) ( τ) ( τ) τ τ x = e f e f = e e f = e e f d Tranfer and Impule Repone Funcion 8 x = f ( ), ( ) x = Taking he ranform of he DE yield X = ( ) = { f} or X ( ) { f} The ranfer funcion i he coefficien of { f } or in hi cae, Tranfer Funcion = The impule repone funcion I( ) i he invere ranform of hi funcion Hence, I( ) = The oluion in erm of he ranfer funcion i () = () () = ( τ ) ( τ) τ = ( τ) τ x I f I f d f d
8 CHAPTER 8 aplace Tranform 9 x + ax = f ( ), ( ) x = Taking he ranform of he differenial equaion yield The ranfer funcion i he coefficien of { f }, or = + a X( ) + ax( ) = { f} or X ( ) { f} Tranfer Funcion = + a The impule repone funcion I( ) i he invere ranform of hi funcion Hence, a ( ) e I = The oluion in erm of he ranfer funcion i x + x= f ( ), x( ) x ( ) a ( τ ) () = () () = ( τ ) ( τ) τ = ( τ) τ x I f I f d e f d = = Taking he ranform of boh ide of he equaion, yield ( ) + ( ) = { } or X ( ) = { f} X X f The ranfer funcion i he coefficien of { f }, or + Tranfer Funcion = + The impule repone funcion I( ) i he invere ranform of hi funcion Hence, ( ) in I = The oluion in erm of he ranfer funcion i () = () () = ( τ ) ( τ) τ = ( τ) ( τ) τ x I f I f d in f d
x + 4x + 5x= f ( ), ( ) ( ) SECTION 84 The Convoluion Inegral and he Tranfer Funcion 83 x = x = Taking he ranform of he differenial equaion yield ( ) + 4 ( ) + 5 ( ) = { } or X( ) X X X f The ranfer funcion i he coefficien of { f }, or The impule repone funcion i Tranfer Funcion = () + + ( ) in ( + ) + I = = e { f} { f} ( ) = = + 4+ 5 + + ( τ ) The oluion i x() = I() f () = I( τ ) f ( τ) dτ = e in ( τ) f ( τ) dτ Invere of Convoluion Theorem Find 3 Find Then Then 3 F ( ) = o ha f ( ) = = (Problem 4) F= f = = ( ) o ha ( ) (Problem 6) 4 Find w w Then F( ) = o ha f( ) = e = () e dw= [ e ] = e ( + ) + 4 5 Find ( ) 4 F () = w w w ( ) = 4 = 4 ( ) = 4 ( ) (Inegraion by par) + 4 f e w e dw w e e = e
84 CHAPTER 8 aplace Tranform 6 Find ( + ) F () = + f() = in = ( w)inwdw = ( w)( co w) ( co w)( ) dw (Inegraion by par) = in+ 7 Find 8 Find ( + ) F () = + + f() = in in = in( w)inwdw ( + k ) = ( co+ co( w) ) dw (by Trigonomeric Ideniy) = wco in( w) 4 = in co F () = + k + k f ( ) = ink ink = in k( w)inkwdw = ( co k + co( k kw) ) dw (by Trigonomeric Ideniy) = wco k in( k kw) k 4k = in k cok k
SECTION 84 The Convoluion Inegral and he Tranfer Funcion 85 Nonzero Iniial Sae 9 Find he aplace ranform for he oluion o ax + bx + cx = f ( ), x() = x, x () = x in erm of h() where ah + bh + ch = δ (), h() = h () = Fir noe ha H() = a + b + c Now olving he differenial equaion, Solving for X(), a X ax ax + bx bx + cx = F () () () () X () = [ F() + ax + ax +bx ] a + b + c Noe ha olving for x() i no required ax + ax + bx = H() F() + a + b + c Nonzero Pracice 3 x + x + x= δ ( ), x() =, x () = X() H() e = + + + 3 + = H() e + 3 3 3 + + + + 4 4 3 3 x() = h() δ ( ) + e co 3e in 3 x + 4x= 4co, x() =, x () = 4 X() = H() + + + 4 4 = H() + + + 4 + 4 x() = h() 4co+ co in
86 CHAPTER 8 aplace Tranform Fracional Calculu 3 / / I/() = ( ) = ( w) dw π π = [ w] = π π / / I/() = ( ) = ( w)( w) dw π π / 3/ 3/ 4 = w w π 3 = 3 π / / I/( a + b+ c) = ( ( a + b+ c) ) = ( a( w) + b( w) + c) ( w) dw π π = ( a b c) w ( a b) w ( a ) w π + + + + 3 5 / 3/ 5/ = ( a b c) ( a b) ( a ) π + + + + 3 5 / 3/ 5/ 6 4 = a b c π + + 5 3 33 / / I ( I ( f))( ) = f( w) dw / I/( I/( f))( ) = ( I/( f)) π Applying he convoluion heorem o each ide yield: ( indicae he aplace ranform) π [ I ( I ( f))( ) ] = [ I ( f) ] / / / π = π F () = π π F () Since aplace ranform have unique coninuou invere, ( ) I/ I/( f) ( ) = f( w) dw
/ d d / I / 34 (a) () = ( () ) d d d = d π = π SECTION 84 The Convoluion Inegral and he Tranfer Funcion 87 (ee problem 3) d d / (b) () = ( I () ) / d d / d 4 d 3 π 3/ = = π (ee problem 3) d d / (c) ( a + b + c) = / ( I/( a + b + c) ) d d d 6 4 = a b c (ee problem 3) d π + + 5 3 8 3/ / / = a b c π + + 3 Trendy Saving 35 f e 6 () = (a) (b) A = A+ e A = 6 8, () 8 6 () A = e e Invemen and Saving 36 A = A+ e A = 4, () A ( ) = e e 4 4( w) w 5 () = ( )( ) 3347 or $334,7 A e e dw Coniency Check 37 Suden Projec
88 CHAPTER 8 aplace Tranform ake Polluan 38 P = P+ e P = 5, () 4 4 P () = e e = e e dw= e e 3 3 5 ( w) 5w 5 Radioacive Decay Chain 39 A = A+ e A =, () A () = e e = ( w) w e e dw = 99e 99e Volerra Inegral Equaion 4 y () ywdw ( ) = Y() = Y() Y() = + y () = e 4 4 y () = ( wywdw ) ( ) 3 () = + in( ) ( ) y wywdw Y() = Y() Y() = + y () = in 6 Y() = + Y() 4 + + Y() = = + 6 3 5 y () = + 6( ) 6 5! 6 4
SECTION 84 The Convoluion Inegral and he Tranfer Funcion 89 43 44 () w w y = e + e ywdw ( ) = e+ e ywdw ( ) Y() = + Y() Y() = y () = y () = co+ in( wywdw ) ( ) Y() = + Y() + + Y() = y () = General Soluion of Volerra Equaion 45 y () = g () + k ( wywdw ) ( ) Y() = G() + K() Y() G () Y() = K ( ) ooking for he Curren 46 (a) The inegrodifferenial equaion i I () + I () + 5 Id () = V () where V() = 4 ep ( ) (b) Uing he iniial condiion I() = I () = and aking he aplace Tranform give I () + I () + 5() I= { V ()} = { V() } V() 4 = e = 4e where I() i he aplace Tranform of I() Solving for I() give I () = (4 e ), ( + 5)
8 CHAPTER 8 aplace Tranform which i he produc of he ranfer funcion H() = ( + 5) of V () and he aplace Tranform e 5 5 Hence I ( ) =4 4 ep ( ) ( e 5 e ) =, ( + 5) + 5 5 + + + becaue = ( 5) ( 5) ( 5) (c) 5 5 5( ) I () = e e + ( 96+ ) ep ( ) e + 5 5 5 5 Tranfer Funcion for Circui 47 Q () + Q() = V () and Q() = I() = C Applying aplace ranform Q () + Q() = { V ()} C Q() + Q() = { V ()} C C C () C H = = = C C + + C + C C C h () = in = in C C C Q () = H () { V ()} Taking he invere aplace Tranform Q () = H() { V()} { } = h ()* V() C = in * V ( ) C
SECTION 84 The Convoluion Inegral and he Tranfer Funcion 8 48 I () + RI() = V () Applying aplace ranform yield Then { } I() + RI() = { V ()}, I () = { V ()}, + R o ha H( ) = i he ranfer funcion + r I () = H () { V ()} = e * V () R Inereing Convoluion 49 y + y = in Y () + Y= + Y() = + + y () = in in Duhamel Principle 5 ay + by + cy = f (), y() = y () = az + bz + cz =, z() = z () = () + () + () = () a Y by cy F Y() = F(), and a + b + c () + () + () = Z() =, o az bz cz a + b + c Y() = Z() F() y () = z () F () Uing Duhamel Principle 5 y y = f(), y() = y () = z z =, z() = z () = ha oluion o z () = e + e y () = ( e + e ) f()
8 CHAPTER 8 aplace Tranform Inereing Inegral Equaion 5 ywdw ( ) = y () y () Y () = Y () Y () Y() =, or Y() = y () = or y () = Noe y() = i eaily verified Alo, y() = yield Suggeed Journal Enry 53 Suden Projec dw = and * = ()() dw =
SECTION 85 aplace Tranform Soluion of inear Syem 83 85 aplace Tranform Soluion of inear Syem aplace for Syem x = y, x() = y = x, y() = x = x X() (), o () = X X = + X() = = = + + in x() = co x = x y, x() = y = x+ 4 y, y() = x = 4 x X() (), o () = 4 X X 4 = 4 X() = = 4 5 6 + 4 5+ 6 5+ 6 = = + 5+ 6 5+ 6 e x () = e
84 CHAPTER 8 aplace Tranform 3 x = y, x() = y = x+ y+ e y = 4 3, () x = 4 3 x+ e X() (), o () = 3 X + 3 X = + 4 4 3 X() = = 3 3 + + + 4 4 3 + + 3+ 3+ 4 = + + 3+ 3+ 4 6+ 5 6 + + = = ( 3 )( 4) 4 6+ 8+ 5 8 + ( 3+ )( 4) 4 4 5e 6e + e () = 5e 8 4 e + e 4 x = y, x() = y = x+ co, y() = x = x+ co X() = (), o () + = X X + + ( ) + X() = = = + + + ( + ) in x() = in+ co
SECTION 85 aplace Tranform Soluion of inear Syem 85 5 3 x = y+ e, x() = y = x+ 3 y, y() = 3 e x = 3 x+ X() = () 3, o () 3 3 X + 3 X = 3 X() = 3 3 3 3 + 3+ + = = + ( 3+ )( 3) 3 e + e x () = 3 e + e e 6 x = y+, x() = y = 3x+ 4y 4, y() = x = 3 4 x+ 4 X() = (), o () 3 4 X + 4 3 4 X = 4 4 X() = = 3 4 4 4 3 3 + 4 3 3 + ( 4+ 3) ( ) ( 3) = = 3+ + 4 3 3 + ( 4 3) ( ) ( 3) + 3 3 e + e () x = 3 3 3 + e e
86 CHAPTER 8 aplace Tranform 7 4 x = +, () = x x x() ( I A) ( F( ) + x ) = { } 4 = + ( 3) = = + 4 + 4 3 6( 3) ( ) 4 = 3 e + 3 8 3 3 4 x = +, () = x x x() ( I A) ( F( ) + x ) = { } 3 3 4 + = + ( ) = = + 3 4+ 3 + 6+ + 3 7 = 6e
SECTION 85 aplace Tranform Soluion of inear Syem 87 9 5 e x = +, () = 5 3 6 x x e x() = {( I A) ( F( ) + x ) } 3 6 5 e e 5 = + 3 3 = + + x = +, () = 3 4 x 4 x x() ( I A) ( F( ) + x ) = { } 3 4 4 = + ( 4)( + ) + ( 4) = ( 4+ 3) 3( + ) + ( 4) e e 3 3 = + + General Soluion of inear Syem x = x+ y, x() = x y = 4 x+ y, y() = y x = 4 x x x X() = (), o () = y 4 X 4 X y
88 CHAPTER 8 aplace Tranform x =x 4 y, x() = x y = x y, y() = y x x X () = 4 = y 3 4 y 3 3 x = 4 y 3 3 + + ( + ) ( 3) 4( + ) 4( 3) x = + + y ( + ) ( 3) ( + ) ( 3) 3 3 e + e e + e 4 4 x x () = 3 3 y e + e e + e 4 x = x x 4 + 4 x X() = (), o () = y X + X y + 4 x + 4 x X() = = y 5 + + + + y + 4 + 4 5 5 x ( ) 4 ( ) 4 + + + + + + + + x = = + y + y + + 5 + + 5 ( + ) + 4 ( + ) + 4 e co e in x() = x e in e co y
SECTION 85 aplace Tranform Soluion of inear Syem 89 More Complicaed inear Syem 3 x+ 4x+ y =, x() = x x+ y=, y() = 4 x + x= 4 + 4 () (), o () X + = = = X X + 4 X() = = 34 + 3 4 5( ) 5( 4) + = = 3 + 3 4 5( + ) 5( 4) 4 e e 5 5 x () = 3 4 e + e 5 5 4 Higher-Order Syem x = x, x () = 4 x = x, x () = 3 x = x, x () = 3 4 3 x = x, x () = 4 4 x = x X X() = X(), o () =
8 CHAPTER 8 aplace Tranform 3 3 X () = = 4 3 3 3 + + 3 + + = 4 3 + + 3 + + + + + ( ) ( + ) + + + + + ( ) ( + ) + = + = + ( ) ( ) + + + ( ) ( + ) + + e + e + in coh+ in e e + co x inh + co () = = coh in e + e in inh co e e co Finding General Soluion x = a x+ a y+ f (), x() = c 5 y = ax+ a y+ f(), y() = c x = Ax+ f c c X = AX + F I A X = + F () () (), o ( ) () () c c c X = I A + IA F c x() = x () + x (), ( ) ( ) ( ) ( ) h p where each erm i he invere aplace ranform of he erm above Therefore only on A, c, and c, while x ( ) depend only on A, f() and f () p x ( h ) depend
SECTION 85 aplace Tranform Soluion of inear Syem 8 Drug Meabolim =, () = 6 x kx x x = kx kx, x () = k x= k k x k k X() = (), o () = k k X k + k X k + k X() = = k + k k k k k k = = k ( k)( + k) ( k) ( + k) k e x() = k k e e Ma-Spring Syem 7 We apply m = m = and k = k = k 3 = o he yem of DE in Example 4 o obain x = x+ ( y x) = x+ y y =( yx) y= xy o ha [ X( ) ] = X( ) + Y( ) [ Y()] = X() Y(), o + X() = Y() + X() + + Y() = = 4( ) + + + + 4 4 4 8 3 + + 3 + = + + = + 4 4 + 8 + 3 + 3 + 3 in + in x () 3 = y () 3 in in 3 +
8 CHAPTER 8 aplace Tranform 8 We apply m = m = and k = k = k 3 = o he yem of DE in Example 4 o obain x = y x y = ( yx) o ha = [ X () ] Y () X () Y =Y X () [ () ()], o + X() = Y() + X() + + 4 Y() = = 4( ) + + + + + ( + ) ( + ) = = ( ) ( + ) + + in x () y () = in Noe ha hi oluion end he cener of ma moving o he righ wih a conan velociy of in accordance wih he law of claical mechanic A Three-Comparmen Model dx 9 = R ux kx+ kx (given) d dx = + k x k x d k x + k x dx3 = k3x k3x3 d 3 3 3
SECTION 85 aplace Tranform Soluion of inear Syem 83 Vibraion wih a Free End (a) x =4x ( x x), x() = x () = x = ( x x ), x () =, x () = 6 x+ x= + () (), o () + 6 6 X + X = = X + 6 + X() = 4 = + + 7 + + 6 4 + 7 + = ( + 6) 4 + 7 + Comparing aplace x = +, () = x x + X() = (), o () X + X = + + X() = = = ( + ) = + ( )( ) + + e + x() = x = +, () = 3 x 6 x X() = () 6, o () 6 3 X + 3 X = 3 X() = 6 = 6 = 6 = 3 5 6 + ( 3) 3 x() = e 3
84 CHAPTER 8 aplace Tranform 3 x = +, () = x x X() = (), o () X + X = + ( ) () X = = = = + ( ) e x () = e Suggeed Journal Enry I 4 Suden Projec Suggeed Journal Enry II 5 Suden Projec