Jawaban Contoh Soal Transformasi Laplace

Oleh Lukman Hakim

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Transkrip Jawaban Contoh Soal Transformasi Laplace

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

Judul: Jawaban Contoh Soal Transformasi Laplace

Oleh: Lukman Hakim


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