PENGEMBANGAN DAN STUDI TABUNG SOLAR KOLE
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
L[e 5t sin 2t ]
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
………………(3)
= 2
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
5,0, t 3
3. Hitung : L[ F (t )] , jika F(t) =
0, t 3
Jawab:
L[ F (t )] e st F (t )dt
= e st .5.dt e st .0.dt
0
3
= e st .5.dt
0
3
3
= 5 e st dt
0
3
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
e
2 / 3
=
e
st
0
=
.0.dt
e
st
cos(t 2 / 3)dt
2 /3
st
cos(t 2 / 3)dt
Subs. u (t 2 / 3) , du = dt
Batas integrasi, t u
t 2 / 3 u 0
2 /3
L[F(t)] e s (u 2 / 3) cos udu
0
= e
2 / 3
e
0
st
cos udu
= 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
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
L[e 5t sin 2t ]
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
………………(3)
= 2
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
5,0, t 3
3. Hitung : L[ F (t )] , jika F(t) =
0, t 3
Jawab:
L[ F (t )] e st F (t )dt
= e st .5.dt e st .0.dt
0
3
= e st .5.dt
0
3
3
= 5 e st dt
0
3
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
e
2 / 3
=
e
st
0
=
.0.dt
e
st
cos(t 2 / 3)dt
2 /3
st
cos(t 2 / 3)dt
Subs. u (t 2 / 3) , du = dt
Batas integrasi, t u
t 2 / 3 u 0
2 /3
L[F(t)] e s (u 2 / 3) cos udu
0
= e
2 / 3
e
0
st
cos udu
= 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