Jawaban Contoh Soal Transformasi Laplace 5d1k27

Jawaban contoh soal Transformasi Laplace

1. Hitung: L [10 sin 4t + 4t 2] Jawab: L [10 sin 4t + 4t 2] = L[10 sin 4t ]  L[4t 2 ] = 10L[sin 4t ]  4L[t 2 ] 4 (3) = 10. 2  4. 3 2 s 4 s 40 4.2! = 2  3 s  16 s 40 8 = 2  3 s  16 s 2. Hitung : L[e 5t (sin 2t  sin 4t )] Jawab : L[e 5t (sin 2t  sinh 4t )]  L[e 5t sin 2t ]  L[e 5t sinh 4t ].......(1) L[sin 2t ] 

2 s 4 2

2 ( s  5) 2  4 2 = 2 s  10s  25  4 2 = 2 ……………….(2) s  10s  29 4 L[sinh 4t ]  2 s  16 4 L[e 5t sinh 4t ]  ( s  5) 2  16 4 = 2 s  10s  25  16 4 = 2 ………………(3) s  10s  9 Sehingga persamaan (2) dan (3) disubstitusikan pada persamaan (1), sehingga 2 4 L[e 5t (sin 2t  sin 4t )] = 2 - 2 s  10s  29 s  10s  9 L[e 5t sin 2t ] 

5,0, t  3 3. Hitung : L[ F (t )] , jika F(t) =  0, t  3 Jawab: 

L[ F (t )]   e  st F (t )dt 0 3



0

3

=  e  st .5.dt   e  st .0.dt 3

=  e  st .5.dt 0

3

= 5 e  st dt 0

5 =  e  st ]30 s 5 = (1 e 3s ) s

cos(t  2 / 3), t  2 / 3 4. Hitung : L[ F (t )] , jika : F (t )   0, t  2 / 3 

L[F(t)] =  e  st F (t )dt 0

2 / 3

=

e

 st



.0.dt 

0 

=

e 

e 

 st

cos(t  2 / 3)dt

2 /3  st

cos(t  2 / 3)dt

2 /3

Subs. u  (t  2 / 3) , du = dt Batas integrasi, t    u   t  2 / 3  u  0 

L[F(t)]   e  s (u  2 / 3) cos udu 0

= e

 2 / 3



e

 st

cos udu

0

= e 2 / 3 L[cos u] s = e 2 / 3 . 2 s 1 2 / 3 se = 2 s 1

t 5. Hitung : L[( ) 2 ] ! 4 Jawab: (3) L[t 2 ]  3 s 2! = 3 s 2 = 3 s 1 2 2 L[( ) ] = 4 . s 3 4 ( ) 1/ 4 8 = 3 3 4 .s 1 = 3 8s