A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. CHAPTER 2 A REVIEW ON LAPLACE TRANSFORMATION 2.1LAPLACE TRANSFORMATION Laplace Transformation is one of the most efficient methods in solving ordinary and partial differential equations.The effectiveness of the Laplace Transform is due to its ability to reduce the problem of solving a differential equation to an algebraic problem.If the Laplace Transformation is appliedtonon-homogeneouslineardifferentialequation,thesolutioncanbeobtaineddirectly without first solving the corresponding homogeneous equation.Another advantage is that it takes careoftheinitialconditionssothatininitialvalueproblems,thedeterminationofageneral solution is avoided. L{ }= = 0stdt ) t ( f e ) s ( F ) t ( f 2.2LAPLACE TRANSFORMATION OF COMMON FUNCTIONS A.L{ }a - s1eat=Provided s > a B.L{ } e0= L{ }s10 - s11 = = C.L{ } tn= 1 ns! n+Provided n > 0 D.L{ } bt sin =2 2b sb+ E.L{ }2 2b ssbt cos+= 2.3LINEARITY PROPERTY OF LAPLACE TRANSFORMATION A.L{ } C f(t) C =L{ } (t) f B.L{ }1 2 1C g(t) C f(t) C = +L{ }2C f(t) +L{ } g(t) 2.4LAPLACE TRANSFORMS OFbt sinhANDbt cosh A.L{ }2 2b sbbt sinh=Provided s > b B.L{ }2 2b sshbt os c=Provided s > b 2.5FIRST SHIFTING PROPERTY OF LAPLACE TRANSFORMATION L{ }( )) a s ( F dt ) t ( f e ) t ( f e 0t a s at = = A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. Summary of Formula (t) fL{ } ) s ( F (t) f =(t) fL{ } ) s ( F (t) f =1 s1 t aea s1 nt1 ns! n+ n att e( )1 na s! n+ bt sin2 2b sb+bt sin eat ( )22b a sb+ bt cos2 2b ss+bt cos eat ( )22b a sa s+ bt sinh2 2b sbbt sinh eat ( )22b a sb bt cosh2 2b ssbt cosh eat ( )22b a sa s Examples: Find the Laplace Transforms of the following (t) f : A.5 t 3 sin 4 3t (t) f2+ =L{ } 3 ) s ( F (t) f = =L{ } 4 t2L{ } 5 t 3 sin +L{ } 1L{ }s5 9 s12 s6 s153 s34s2!3 ) s ( F (t) f2 3 2 2 3++ =|.|

\|+|.|

\|+|.|

\|= = B.( ) t 2 sin sin t 2 cos cos t 2 sin 2t sin (t) f = = = L{ } = = ) s ( F (t) fL{ }4 s2t 2 sin2+ = C.( ) t 2 cos e21e21t 2 cos 121e t cos e (t) ft 3 t 3 t 3 2 t 3 + =((

+ = =L{ }21) s ( F (t) f = =L{ }21e3t -+L{ } cos2t e-3t L{ }( ) ( )((

+ + +++=((

+ + ++|.|

\|+= =4 3 s3 s3 s121

2 3 s3 s21 3 s121) s ( F (t) f2 2 2 2.6PROBLEM SET 1.3 t 2 t (t) f3+ = 2.( )t 4 2e 1 t (t) f + =3.t 3 sinh (t) f = 4.t cos t sin 6 (t) f = A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. 5. 2t 3sin (t) f2= 6. -5t 3te 3 e (t) f = 7.t 4 sin 2 t 2 cos 4 (t) f = 8.( ) = t cos (t) f 9.t 2 cos (t) f2= 10. ( ) 1 - 2te (t) f= 11.( )33 t (t) f = 12.( )2te t (t) f + = 13.t sinh e (t) ft 2= 14.t 3 cos e (t) f2 t= 15. ( ) t 1 5e t (t) f= 16. |.|

\|+ =2t 3 sin e (t) ft 4 17.t cos t sin e (t) f2 2 t = 18.( ) t 4 sin t e (t) ft 3 = 19.t 3 cosh t (t) f3= 20.t sinh t sin (t) f = 2.7INVERSE LAPLACE TRANSFORMATION L -1 { } ) t ( f (s) F = TheInverseTransformcanbefoundonlyif(s) F isinitsstandardformasgiveninthe Table of Summaryof Formula.If not, theMethod of Partial Fraction Expansion is used to reduce the(s) Fto the standard forms with known inverses. Examples:Find the Inverse Laplace Transforms of the given(s) F : A. s44 s6(s) F2=L -1{ }26) t ( f (s) F = =L -142 s2 2 2)` L -14 t 2 sinh 3s1 =)` B. ( ) ( ))`+ =+ + =+ =9 1 s1 s9 1 s 2 s1 s10 s 2 s1 s(s) F2 2 2 L -1{ } = = ) t ( f (s) FL -1 ( )t 3 cos e9 1 s1 st2=)`+ C. ( )( )( )( )( ) ( ) ( ) ( )5 4 5 5 5 53 s163 s33 s163 s3 s 33 s16 3 s 33 s7 s 3(s) F++=+++=+ +=+=L -1{ }! 33) t ( f (s) F = =L -1 ( ) ! 4163 s! 34)`+ L -1 ( ) )`+53 s! 4 L -1{ } ( )t 3 - 3 t 3 - 4 t 3 - 3e t t 4 361e t32e t21) t ( f (s) F = = =A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. D. ( ) ( ) 2 sD1 sCsBsA2 s 1 - s s4 s 2 s 3s 2 s s4 s 2 s 3(s) F2 22 32 3 42 3+++ + =+ = + = Partial Fraction Expansion, Method of Undetermined Coefficients: ( ) ( ) ( ) ( ) ( ) ( ) 1 - s Ds 2 s Cs 2 s 1 - s B 2 s 1 - s As 4 s 2 s 32 2 2 3+ + + + + + = when 0 s = ,( )( ) B 2 2 1 - B 4 - = = , 2 B =1 s = ,( )( ) C 3 3 1 C 3 4 2 3 = = = -1 C =-2 s = , ( ) ( ) ( )( ) D 12 3 4 D 36 4 4 2 8 3 = = = 3 D =coeff. of 3s ,3 1 A D C A 3 + = + + = 1 A = 2 s31 s1s2s1(s) F2++ + =L -1{ } = = ) t ( f (s) FL -1 ! 12s1+)` L -1)`2s! 1 L -131 s1+)` L -1 )`+ 2 s1 L -1{ }t 2 - te 3 e t 2 1 ) t ( f (s) F + + = = 2.8PROBLEM SET 1. s3s4(s) F3 = 2. 4 s62 s5(s) F++= 3. ( )52 s20(s) F+= 4. ( )16 s4 s 4(s) F2+= 5. 8 s 4 s6(s) F2+ = 6. 64 ss 4(s) F2= 7. 5 s 4 ss 2 5(s) F2 = 8. 15 s 2 s1 s(s) F2 + += 9. 13 s 6 s1 s 3(s) F2+ + += 10. ( )33 s9 s 4(s) F= 11. ( ) ( ) 4 s 1 s2 s(s) F2+ ++= 12. ( ) 1 s s1 s(s) F2 2++= 13. 2 s 3 s1 s 2(s) F32+ += 14. s 5 s 2 s10 s 4 s(s) F2 32+ ++ += 15. 1 - s4 s 2 s(s) F42+ = A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. 2.9LAPLACE TRANSFORM OF DERIVATIVES Given Function:f(t) y = Laplace of the nth Derivative : L nnnsdt) t ( y d=)` L{ } ) 0 ( y ) 0 ( ' ' y s ) 0 ( ' y s ) 0 ( y s ) t ( y' ) 1 n ( 3 n 2 n 1 n Example: L 555sdt) t ( y d=)` L{ } ) 0 ( y ) 0 ( ' ' ' sy ) 0 ( ' ' y s ) 0 ( ' y s ) 0 ( y s ) t ( yiv 2 3 4 2.10SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICENTS USING LAPLACE TRANSFORMATION Three Essential Steps in Solving Differential Equations Using Laplace Transformation: 1.Thegivendifferentialequationistransformedintoanalgebraicequationcalledthe subsidiary equation) t ( F y ) t ( adtdy) t ( a ...dty d) t ( adty d) t ( adty d) t ( a0 12 n2 n2 n1 n1 n1 nnnn= + + + + +. 2.The subsidiary equation is solved by purely algebraic manipulations. 3.The solution of the subsidiary equation is transformed back to the original function so that it becomes the required solution of the given differential equation. Consider the Linear Differential Equation with Constant Coefficients: ) t ( F y a Dy a ... y D a y D a y D a0 12 n2 n1 n1 nnn= + + + + + ( ) ) t ( F y a D a ... D a D a D a0 12 n2 n1 n1 nnn= + + + + + With initial conditions: 0C ) 0 ( y =2C ) 0 ( ' ' y = 1C ) 0 ( ' y =3C ) 0 ( ' ' ' y = Steps in Solving the Differential Equation: 1.Take the Laplace Transform of both sides of the given linear differential equation. 2.Apply the initial conditions and solve the resulting algebraic equation to obtainL{ } ) s ( F ) t ( y = . 3.Find the inverse transform. L -1 L{ } = = ) t ( y ) t ( yL -1{ } ) s ( F ) t ( f ) t ( y = Example: Solve0 y 2 y' 5 y'' 4 y''' = + + + 0 ) 0 ( ' y ) 0 ( y = = 3 ) 0 ( ' ' y = Solution: Take the Laplace Transform of both sides of the given differential equation. L{ }= + + + y 2 y' 5 y'' 4 y'''L{ } 0 A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. |3sL{ } | |2 2s 4 ) 0 ( ' ' y ) 0 ( ' sy ) 0 ( y s ) t ( y + L{ } | ) 0 ( ' y ) 0 ( sy ) t ( y | s 5 +L{ } | 2 ) 0 ( y ) t ( y + L{ } 0 ) t ( y = Apply the initial conditions: 0 ) 0 ( ' y ) 0 ( y = = 3 ) 0 ( ' ' y =3sL{ }2s 4 3 ) t ( y + L{ } s 5 ) t ( y +L{ } 2 ) t ( y + L{ } 0 ) t ( y =( ) 2 s 5 s 4 s2 3+ + +L{ } 3 ) t ( y =L{ }( ) ( )2 2 31 s 2 s32 s 5 s 4 s3) s ( F ) t ( y+ +=+ + += = Breaking the) s ( Finto its Partial Fraction Expansion, ( ) ( ) ( )2 21 sC1 sB2 sA1 s 2 s3+++++=+ + Multiply both sides of the equation by its LCD, ( ) ( ) ( ) ( ) 2 s C 2 s 1 s B 1 s A 32+ + + + + + = When 1 s =3 C =2 s =3 A =0 s = C 2 B 2 A 3 + + = ( ) 3 2 B 2 3 3 + + = 3 B = Substitute the arbitrary constants on the Partial Fraction Expansion, L{ }( )21 s31 s32 s3) s ( F ) t ( y++++= = Take the Inverse Laplace Transform, L -1 L{ } 3 ) t ( y ) t ( y = =L -132 s1)`+ L -131 s1+)`+ L -1 ( ) )`+21 s1 t t t 2te 3 e 3 e 3 ) t ( y + = Example: Solvet cos 4 y y'' = 0 ) 0 ( y = 1 ) 0 ( ' y = Solution: Take the Laplace Transform of both sides of the given differential equation. L{ } 4 y y'' = L{ } t cos|2sL{ } | ) 0 ( ' y ) 0 ( sy ) t ( yL{ })`+=1 ss4 ) t ( y2 Apply the initial conditions: 0 ) 0 ( y = 1 ) 0 ( ' y =2sL{ } 1 ) t ( yL{ }1 ss 4) t ( y2+=( ) 1 s2L{ }1 s1 s 4 s1 s1 s s 411 ss 4) t ( y22222+ + +=+ + += ++=A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. L{ }( )( ) ( )( ) ( ) 1 s 1 s 1 s1 s 4 s1 s 1 s1 s 4 s) s ( F ) t ( y222 22 + ++ += ++ += = Breaking the) s ( Finto its Partial Fraction Expansion, ( )( ) ( )( )1 ss 2 D1 sC1 sB1 sA1 s 1 s 1 s1 s 4 s2 2 22++++++= + ++ + Multiply both sides of the equation by its LCD, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 s 1 s s 2 D 1 s 1 s C 1 s 1 s B 1 s 1 s A 1 s 4 s2 2 2 + + + + + + + + = + + When 1 s =( ) 4 A 2 = 21A =1 s = ( ) ( ) 2 2 B 1 4 1 = + +23B =0 s = C B A 1 + =C23211 + = 0 C =Eq. coeff. of 3s :D 2 B A 0 + + = D 223210 + + = 1 D = Substitute the arbitrary constants on the Partial Fraction Expansion, L{ }1 ss 21 s231 s21) s ( F ) t ( y2+++= = Take the Inverse Laplace Transform, L -1 L{ }21) t ( y ) t ( y = =L -1 231 s1+)`+ L -121 s1)` L -1 )`+1 ss2 t cos 2 e23e21) t ( yt t + = 2.11PROBLEM SET 1.0 y 13 ' y 6 y'' = + + 3 ) 0 ( y = 1 ) 0 ( ' y = 2.0 y 12 ' y y'' = 3 ) 0 ( y = 5 ) 0 ( ' y = 3.0 y 2 y' 3 y''' = 1 ) 0 ( y = 11 ) 0 ( ' y = 13 ) 0 ( ' ' y = 4.0 y 4 y' 4 ' ' y y''' = + + + 0 ) 0 ( ' y ) 0 ( y = = 5 ) 0 ( ' ' y = 5. t 2e 10 y y'' = + 0 ) 0 ( ' y ) 0 ( y = = 6. te 6 y 2 ' y 3 y''= + 3 ) 0 ( ' y ) 0 ( y = = 7. t 3e 10 y 5 ' y 4 y''= + + 4 ) 0 ( y = 0 ) 0 ( ' y =A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. 8.t cos e y y''t 2 = + 0 ) 0 ( ' y ) 0 ( y = = 9.4 t y 4 y'' + = + 1 ) 0 ( y = 0 ) 0 ( ' y = 10.t 18 ' y 3 y'' = + 0 ) 0 ( y = 5 ) 0 ( ' y = 11.t 8 2 y 4 y'' = 0 ) 0 ( y = 5 ) 0 ( ' y = 12.t 2 sin 6 y y'' = + 3 ) 0 ( y = 1 ) 0 ( ' y = 13.t 2 cos 4 y 4 ' y 4 y'' = + 2 ) 0 ( y = 5 ) 0 ( ' y = 14.t cos 4 t sin 8 y 5 ' y 2 y'' = + 1 ) 0 ( y = 3 ) 0 ( ' y = 2.12LAPLACE TRANSFORM OF INTEGRALS L{ }s1dt ) t ( f t= L{ } ) s ( Fs1) t ( f = Examples: Find the Laplace Transforms of the following integrals: A. L{ }s1dt sin4t e t2t -= L{ }( )((

+ +=16 2 s4 s1sin4t e22t - B.L{ }s1dt e t t4t - 5= L{ }( ) ( ) ((

+=((

+=6 64t - 54 s120 s14 s! 5 s1e t C.L{ }s4dt t 2 cosh 4 t= L{ }((

=4 ss s4t 2 cosh2 2.13DIFFERENTIATION OF LAPLACE TRANSFORMATION L{ } ( )nnn ndsd1 ) t ( f t =L{ } ( ) ) s ( Fdsd1 ) t ( fnnn = Example: Find the Laplace Transform of the following function, (t) f : L{ } ( )222 2dsd1 t 3 cos t =L{ }( ) ( )( ) (((

+ +=((

+=2222 229 ss 2 s 9 sdsd9 ssdsdt 3 cosL{ }( )( ) ( ) ( )( ) ( )( )( )422 22222229 ss 2 9 s 2 s 9 s 2 9 s9 ss 9dsdt 3 cos t++ +=(((

+=A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. L{ }( ) ( )( )( )( ) ( ) ( ) | |( )422 2 2422 22229 ss 9 2 9 s 9 s s 29 s9 s s 9 s 4 9 s s 2t 3 cos t+ + + + =++ + =L{ }( )( )( )( )( )( )322322322 229 s27 s s 29 ss - 27 s 29 s2s - 18 9 s s 2t 3 cos t+=+=++ + = 2.14COROLLARY OF THE DIFFERENTIATION OF LAPLACE TRANSFORMATION t1) t ( f =L -1 )`) s ( Fdsd Example: Find the Inverse Laplace Transform of the following function,(s) F : ( ) ( ) ( ) 4 s ln s ln 2 4 s ln s ln4 ssln ) s ( F2 2 222+ = + =||.|

\|+=t1) t ( f =L -1 t1) s ( Fdsd =)` L -1( ) | |t14 s ln s ln 2dsd 2 =)`+ L -1 )`+ 4 s2s-s2 2 ( ) ( )tt sin 4t 2 cos 1t2t 2 cos 2 2t1) t ( f2= = = 2.15INTEGRATION OF LAPLACE TRANSFORMATION L =)` s t) t ( f L{ }= sds ) s ( F ds (t) f Example: Find the Laplace Transform of the following function,(t) f : L =)`st tt 3 sin e L{ }( )+=((

+ += s s2-t31 sarctan33ds9 1 s3ds t 3 sin eL |.|

\| + =|.|

\| + =)`31 sarctan2 31 sarctan arctantt 3 sin e t 2.16COROLLARY OF THE INTEGRATION OF LAPLACE TRANSFORMATION t ) t ( f =L -1{ } sds ) s ( F Example: Find the Inverse Laplace Transform of the following function: ( )225 s 4 s2 s(s) F+ + += A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. t ) t ( f =L -1 ( )t ds5 4s s2 s s22=)`(((

+ + + L -1( )( ) { }+ + + s2 -2ds 2 s 5 4s st ) t ( f =L -1 ( )2t 15 s 4 s 21 s12=)`(((

+ + L -1 ( ))`+ +s25 s 4 s1 2t) t ( f=L -1 2t5 s 4 s1 1 2=)`+ + L -1 2t5 s 4 s1 2=)`+ + L -1 ( ))`+ + + 1 4 s 4 s1 2 2t) t ( f = L -1 ( ) 2t sin te1 2 s1 t 22=)`+ + 2.17PROBLEM SETS 2.17.1DIFFERENTIATION AND INTEGRATION OF LAPLACE TRANSFORMATION 1.( ) = t sin t ) t ( f 2.t cos t sin t 2 ) t ( f2= 3.t cos e t ) t ( ft 2= 4.t cos t ) t ( f2 2= 5.( ) t 4 sin e t ) t ( ft 3 = 6. t4 t 3 cos) t ( f=7. t2 e 3) t ( ft 2= 8. tt sin 4) t ( f2= 9. tt 4 sin e) t ( ft 3= 10. tt 2 cos e) t ( ft 4= 2.17.2COROLLARY OF THE DIFFERENTIATION AND INTEGRATION OF LAPLACE TRANSFORMATION 1.( ) 3 s arctan (s) F + = 2. ||.|

\|=4 ssln (s) F24 3. ||.|

\|+ + =1 s 2 s2 s 2 sln (s) F22 4. ||.|

\|++=3 s9 sln (s) F2 5. ( )((

+=2 s s2 sln (s) F2 6. ( ) ((

+=1 s1 sln (s) F24 7. ( )2216 ss 4(s) F+= 8. ( )229 ss 3(s) F=A COMPILATION OF LECTURE NOTES IN CONTROL SYSTEMCHAPTER 2 CATADMAN, LIZETTE IVY G. 9. ( )223 s 2 s1 s(s) F + += 10. ( )( )2210 s 6 s3 s 5(s) F+ = 2.18LAPLACETRANSFORMOFUNITSTEPFUNCTIONS-SECONDSHIFTINGPROPERTY OF LAPLACE TRANSFORMATION L{ }ase ) a t ( u ) t ( f= L{ } ) a t ( f + Examples: Find theL{ } ) a t ( u ) t ( f . A.L{ }se ) t ( u int s= L( ) { } t in s +t sin ) t ( f = = a( ) t sin sin t cos cos t sin t sin ) a t ( f = + = + = + L{ }se ) t ( u int s= L{ }1 se1 s1e t in s -2s2s+=|.|

\|+= B.L( ) { }s 2 2e ) 2 t ( u 4 - t= L( ) | | { }24 - 2 t +( )24 t ) t ( f = 2 a =( ) | | ( ) 4 t 4 t 2 t 4 - 2 t ) a t ( f2 2 2+ = = + = +L( ) { }s 2 2e ) 2 t ( u 4 - t= L{ }|.|

\|+ = + s4s4s2e 4 t 4 t2 3s 2 2 C. 3 t3 t 11 t 0

1t2) t ( f