Pertemuan-13 & 14
Transformasi Laplace 2.1 RUMUS-RUMUS PEMBUKTIAN TRASFORMASI LAPLACE PENGERTIAN TRANSFORMASI LAPLACE Transformasi Laplace adalah proses mengubah fungsi F(t) dari fungsi waktu ke fungsi kompleks f(s) dari operasi kompleks S.
Transformasi Laplace dari suatu fungsi F(t) didefinisikan sebagai: ∞
L {F(t)} = f ( s ) = ∫ e − st ⋅ F (t ) ⋅ dt 0
Transformasi Laplace untuk beberapa fungsi dasar: No. 1.
1
2.
t n ; (m = 1, 2, 3, K
3.
s>0
1 s
t p ; (p > -1)
n! s
γ ( p +1) s
4.
e
5.
cos ωt
at
s>0
n +1
s>0
p +1
1 s−a
s>a
s 2 s + ω2
s>0
ω s +ω2
s>0
No.
sin ωt
6.
2
7.
Cosh at
s s2 − a2
s> a
8.
Sinh at
a s2 − a2
s> a
1 s
Buktikan: L {1} = ! Bukti:
∞
L{F( t )} = ∫ e − st ⋅ F( t ) ⋅ dt 0
∞
L [1 ] =
∫
μ
e − st (1 ) dt
= lim
μ→∞
o
= lim μ→∞
1 s
μ
∫e
- st
dt
o
⎛ 1 ⎞ e - st d( e - st ) = lim ⎜ - e − st ⎟ μ→∞ s ⎝ ⎠
∫ o
(
μ
o
)
1 1 1 = - lim e - μs − e o = - (0 - 1) = s μ →∞ s s 1 s2
Buktikan: L{t} =
Bukti : L [F (t )] =
∞
∫e
− st
F ( t ) dt
o
∞
L [t ] =
∫
μ
e − st ( t ) dt
= lim
μ→∞
o
1 = - lim s μ→∞
μ
∫
t.d(e
- st
) = -
∫ t.e
- st
dt
o
1 lim [t. e - st s μ→∞
μ o
μ
-
∫e
- st
.d(t)] =
o
o
μ
μ 1 1 = - lim[t/.e st ] + lim ∫ e -st (1)dt = o s μ →∞ s μ →∞ o μ
μ 1 1 = - lim[1/s.e st ] + lim ∫ e -st (1)dt = o s μ →∞ s μ →∞ o
= -
1 1 (0) + s s
Buktikan: L{t 2 ) =
∞
∫e
− st
(1) dt = 0 +
o
2 s3
Bukti: ∞
L{F (t )} = ∫ e − st ⋅ F (t ) ⋅ dt 0
∞
L(t 2 } = ∫ e − st ⋅ t 2 dt 0
1 1 1 1 . L [1 ] = . = 2 s s s s
p
= lim ∫ t 2 e − st dt p →∞ 0
= lim − p →∞
1 p 2 − st t ⋅ e d (− st ) s ∫0
p 1 = − lim ∫ t 2 d (e − st ) s p →∞ 0 p 1 = − lim ⎡t 2 e − st 0p − ∫ e − st d (t 2 )⎤ ⎢ ⎥⎦ 0 s p →∞ ⎣
⎤ p 1 ⎡ ⎧ t2 ⎫ = − ⎢ lim ⎨ st 0p ⎬ − lim ∫ e − st (2t )dt ⎥ s ⎣ p →∞ ⎩ e ⎭ p →∞ 0 ⎦ 2 2 ∞ ⎤ 1⎡ p 0 = − ⎢ lim sp − lim s.0 − 2∫ e − st ⋅ t dt ⎥ 0 p →∞ e s ⎣ p →∞ e ⎦ 1 [0 − 0 − 2 L{t}] s 1⎡ 1⎤ = − ⎢− 2 ⋅ 2 ⎥ s⎣ s ⎦
=−
L(t 2 } =
2 s3
Buktikan: L {t n } =
n! s
n +1
Bukti: L {1} = L {t n } =
1 0! = 0 +1 3 s
1 1! = s 2 s1+1 2 2 ⋅1 2! L {t 2 } = 3 = 2 +1 = 2 +1 s s s M n! L (t n } = n +1 s L {t} =
[ ]
Buktikan L t n = Bukti:
γ (n + 1) s n +1 ∞
Fungsi Gamma: γ ( n ) = ∫ e − x x n −1dx 0
∞
γ (n + 1) = ∫ e − x x n +1-1 dx o
∞
γ (n + 1) = ∫ e − x x n dx
misalkan : x = st
o
dx = s.dt ∞
γ (n + 1) = ∫ e − st (st) n s.dt o
∞
γ (n + 1) = ∫ e − st s n t n s.dt o
∞
γ (n + 1) = s n +1 ∫ e − st t n dt
γ (n + 1) = s
[ ]
L tn =
o
n +1
.L[t n ]
γ (n + 1) s n +1
[ ]
maka L t n =
n! γ (n + 1) = → [terbukti] S n +1 s n +1
Buktikan : L {e at } = Bukti: L {F(t)} = ∫
∞
0
1 s−a
e − st ⋅ F( t ) ⋅ dt ∞
[ ] ∫e
L e at =
μ
− st
( e at ) dt
o
μ
= lim
μ→∞
=
∫
e − ( s − a ) t dt =
o
= lim
μ→∞
∫e
− st + at
dt
o
1 lim - (s - a) μ → ∞
μ
∫e
− ( s − a )t
d[-s(t - a)]
o
μ 1 1 1 lim[e −( s − a )t ] = lim[ −( s − a ) t o - (s - a) μ →∞ - (s - a) μ →∞ e
=
1 1 1 [ lim ( s − a ) μ − lim ( s − a ).0 ] → ∞ → ∞ μ μ - (s - a) e e
=
1 1 1 [0 − lim 0 ] = .[0 − 1] μ →∞ e - (s - a) - (s - a)
L {e at } =
1 → [terbukti] s-a
μ
] o
RUMUS FULER
e a +bi = e a ⋅ e bi e a +bi = e a (cos b + i sin b) e bi = cos b + i sin b e −bi = cos b − i sin b + e bi + e −bi = 2 cos b cos b =
e bi + e − bi e ωti + e − ωti → cos ωt = 2 2
e bi = cos b + i sin b e −bi = cos b − i sin b − e bi − e −bi = 2i sin b
sin b =
e bi − e − bi eωti − e −ωti → sin ωt = 2i 2i
Buktikan L (cos ωt) =
s s + ω2 2
Bukti: L{F( t )} = ∫
∞
0 ∞
L{cos ωt} = ∫
0
=∫
∞
0
= = = = =
e − st ⋅ F( t ) ⋅ dt e − st ⋅ cos ωt ⋅ dt ⎛ e ωti + e − ωti e − st ⋅ ⎜⎜ 2 ⎝
(
⎞ ⎟ ⋅ dt ⎟ ⎠
)
1 ∞ − st ωti e e + e − ωti dt 2 ∫0 ∞ 1 ⎡ ∞ − st ωti e ⋅ e dt + ∫ e − st ⋅ e − ωti dt ⎤ ∫ ⎢ ⎥⎦ 0 0 2⎣ 1 L{e ωti } + L{e − ωti } 2 1 L{e ωit } + L{e − ωit } 2 1⎡ 1 1 ⎤ + 2 ⎢⎣ s − ωi s + ωi ⎥⎦
[ [
] ]
L{cos ωt} =
1 ⎡ s + ωi + s − ωi ⎤ 2 ⎢⎣ ( s − ωi )( s + ωi ) ⎥⎦
1 2s ⋅ 2 2 s − ω 2i 2 s = 2 s − ω 2 (−1) s L{cos ωt} = 2 → [terbukti] s + ω2 =
Buktikan: L{sin ωt} = L{F( t )} = ∫
∞
e − st ⋅ F( t ) ⋅ dt
0 ∞
e − st ⋅ sin ωt ⋅ dt
L{sin ωt} = ∫
0
=∫
⎛ e ωti − e − ωti e − st ⋅ ⎜⎜ 2i ⎝
∞
0
= = = = = = = L{sin ωt ) =
ω s + ω2
(
⎞ ⎟ ⋅ dt ⎟ ⎠
)
1 ∞ − st ωti e e − e − ωti dt 2i ∫0 ∞ 1 ⎡ ∞ − st ωit e ⋅ e dt − ∫ e − st ⋅ e − ωit dt ⎤ ⎥⎦ 0 2i ⎢⎣ ∫0 1 L{e ωit } − L{e − ωit } 2i 1 ⎡ 1 1 ⎤ − 2i ⎢⎣ s − ωi s + ωi ⎥⎦ 1 s + ω i − ( s − ωi ) ⋅ 2i (s − ωi )(s + ωi ) 1 s + ω i − s + ωi ) ⋅ 2i s 2 − ω2 i 2 1 2 ωi ⋅ 2i s 2 − ω2 ( −1) ω → [terbukti] 2 s + ω2
[
Buktikan: L{cosh at} = Bukti:
2
]
s s − a2 2
⎧⎪ e at + e − at ⎫⎪ L{cosh at} = L ⎨ ⎬ 2 ⎪⎩ ⎪⎭ 1 = L e at + e − at 2 1 = L e at + L e − at 2 1 s+a + s−a = 2 (s − a)(s + a) 1 2s = ⋅ 2 2 s − a2 s L{cosh at} = 2 → [terbukti] s − a2
{ } [ { } { }]
Buktikan: L {sinh at} =
a s − a2 2
Bukti: ⎧⎪ e at − e − at ⎫⎪ L {sinh at} = L ⎨ ⎬ 2 ⎪⎩ ⎪⎭ 1 = L e at − e − at 2 1 = L e at − L e - at 2 1⎡ 1 1 ⎤ = ⎢ − 2 ⎣ s − a s + a ⎥⎦ 1 s + a − (s − a) = ⋅ 2 (s − a)(s + a) 1 s+a−s+a = ⋅ 2 s2 − a2 1 2a = ⋅ 2 2 s − a2 a L (sinh at} = 2 → [terbukti] s − a2
{ } [ { } { }]
2.2 BEBERAPA SIFAT PENTING DARI TRANSFORMASI LAPLACE 1. Kelinearan
L{c1 ⋅ F1 (t ) + c 2 ⋅ F2 (t )} = c1 ⋅ L {F1 (t )} + c2 ⋅ L {F2 (t )} = c1 ⋅ f1 ( s ) + c2 ⋅ f 2 ( s )
Contoh:
{
L 4t 2 − 3 cos 2t + 5e − t
}
= 4 L {t } − 3L {cos 2t} + 5 L {e −t } 2
2! s 1 − 3⋅ 2 + 5⋅ 2 +1 2 s s +2 s − (−1) 8 3s 5 = 3− 2 + s s + 4 s +1 = 4⋅
2. Translasi Pertama atau pergeseran Jika L {F(t)} = f(s)
{
}
Maka L e at ⋅ F( t ) = f (s − a) Contoh: L {e − t sin 2t} = K f(s) = L {sin 2t} =
2 2 = 2 2 s +2 s +4 2
L {e − t sin 2t} = f(s + 1) 2 2 2 = = = (s + 1) 2 + 4 s 2 + 2s + 1 + 4 s 2 + 2s + 5
3. Translasi Kedua atau Pergeseran ⎧F ( t − a ) t > a t
Jika L {F(t)} = f(s) dan G( t ) = ⎨ Maka L {G(t)} = e -as ⋅ f (s) Contoh:
f ( s ) = L{t 3 } =
3! 3 ⋅ 2 ⋅1 6 = = 4 3+1 s s4 s
⎧(t − 2) 3
t>a
⎩
t
dan G (t ) = ⎨
0
maka L {G(t)} = e -as ⋅ f (s) 6 6e − 2 s = 4 s4 s
= e − 2s ⋅
4. Perubahan Skala Jika L {F(t)} = f(s) 1 a
⎛s⎞ ⎝a⎠
Maka L {F(a.t) = ⋅ f ⎜ ⎟
Contoh:
L {sin 3t} = K 1 1 = 2 2 s +1 s +1 1 1 = 2 = 2 s s +9 +1 +1 9 9
f(s) = L{sin t} =
2
1 ⎛s⎞ f⎜ ⎟= 2 ⎝3⎠ ⎛ s ⎞ ⎜ ⎟ ⎝ 3⎠ 9 = 2 s +9 1 ⎛1⎞ L {sin 3t} = ⋅ f ⎜ ⎟ 3 ⎝ 3⎠ 1 9 3 = ⋅ 2 = 2 3 s +9 s +9 Soal: 1. L {5t − 3} = K 2. 3. 4.
L {2 t 2 − e − t } = K
5.
L (t 2 + 1) 2 = K
L {3 cos 5t} = K L {6 sin 2t − 5 cos 2t} = K
{
}
−t
6.
L {e
7.
L {t 3 ⋅ cos 2 t} = K
8. 9. 10.
{ L {e L {e
⋅ cos 2 t} = K
} cosh 2t } = K ⋅ sin t} = K
L (t + 2) 2 ⋅ e t = K -4t
−t
2
→
1 s +1 − 2 2( s + 1) 2( s + 2s + 5)
⎧5 0 < t < 3 maka L {F(t)} = … ⎩0 t > 3
11. Jika F(t ) = ⎨
{ L {e
}
12. L e 4t ⋅ cosh 5t = K 13.
−2 t
}
(3 cos 6t − 5 sin 6t) = K
2π 2π ⎧ ⎪cos (t − 3 ) , t > 3 14. Jika F (t ) = ⎨ maka L {F(t)}= … 2π ⎪0 , t< 3 ⎩ ⎧2t , t ≤ t ≤ 5 ⎩1 , t > 5
15. Tentukan L {F(t)} jika F(t ) = ⎨
⎧⎪(t - 1) 2 , t > 1 ⎪⎩0 , 0 < t <1
16. Tentukan L {F(t)} jika F(t ) = ⎨
2.3 TRANSFORMASI LAPLACE TURUNAN DAN INTEGRAL Transformasi Laplace dari Turunan ⎧ dF ⎫ L ⎨ ⎬ = s ⋅ L {F(t)} − F(0) ⎩ dt ⎭
⎧⎪ d 2 F ⎫⎪ df L ⎨ 2 ⎬ = s 2 ⋅ L {F(t)} − s ⋅ F(0) − (0 ) dt ⎪⎩ dt ⎪⎭ ⎧⎪ d 3 F ⎫⎪ df d2F L ⎨ 3 ⎬ = s 3 ⋅ L {F(t)} − s 2 ⋅ F(0) − s ⋅ (0) − 2 (0) dt ⎪⎩ dt ⎪⎭ dt df d2F d3F ⎪⎧ d 4 F ⎪⎫ L ⎨ 4 ⎬ = s 4 ⋅ L {F(t)} − s 3 ⋅ F(0) − s 2 ⋅ (0) − s ⋅ 2 (0) − 3 (0) dt ⎪⎩ dt ⎪⎭ dt dt M dst
Bukti : Transformasi Laplace dari Derivatif
Jika L[F (t )] = f (s ) maka
⎡ dF ⎤ a) L ⎢ ⎥ = S .L[ F (t )] − F (0) ⎣ dt ⎦ Bukti: L [F (t )] =
∞
∫e
− st
F ( t ) dt
o
⎡ dF ⎤ L⎢ ⎥ = ⎣ dt ⎦
∞
∫e
− st
o
⎡ dF ⎤ = lim L⎢ ⎣ dt ⎥⎦ μ → ∞
d [ F ( t )] dt dt
μ
∫e
- st
d[F (t )]
o
⎡ dF ⎤ = lim [e - st F (t )] oμ - lim L⎢ μ→∞ ⎣ dt ⎥⎦ μ → ∞ F (t ) ⎡ dF ⎤ = lim [ st ] oμ - lim L⎢ μ→∞ ⎣ dt ⎥⎦ μ → ∞ e
μ
∫ F (t ) d[e
- st
]
o
μ
∫ F (t ).e o
- st
(-s) dt
F (μ ) ⎡ dF ⎤ L⎢ = lim [ s μ ] + s . lim ⎥ μ μ→∞ → ∞ e ⎣ dt ⎦
μ
∫ F (t ).e
- st
dt
o
F (μ ) F(0) ⎡ dF ⎤ L⎢ = lim [ s μ ] − lim [ s.0 ] + s . ∫ e - st F (t )dt ⎥ → ∞ → ∞ μ μ e e ⎣ dt ⎦ o ∞
⎡ dF ⎤ L⎢ = 0 − F(0) + s . L[F(t)] ⎣ dt ⎥⎦
Jadi :
⎡ dF ⎤ L⎢ = s . L[F(t)] − F(0) ⎣ dt ⎥⎦
Sehingga
⎡d 2F ⎤ dF dF ]− (0) L⎢ = s . L[ 2 ⎥ dt dt dt ⎣ ⎦
⎡ d 2F ⎤ dF (0) L⎢ = s . [ s . L[F(t)] − F(0)] − 2 ⎥ dt dt ⎣ ⎦ ⎡ d 2F ⎤ dF (0) L⎢ = s 2 .L[F(t)] − s .F(0) − 2 ⎥ dt dt ⎣ ⎦
juga
⎡ d 3F ⎤ dF d 2F (0) − (0) L⎢ = s 2 .L[F(t)] − s . 3 ⎥ dt dt 2 ⎣ dt ⎦ ⎡ d 3F ⎤ dF d 2F (0) − (0) L⎢ = s 3 . L[F(t)] − s 2 .F(0) − s . 3 ⎥ dt dt 2 ⎣ dt ⎦
Kesimpulan : ⎡ dF ⎤ L⎢ ⎥ = s . L[F(t)] − F(0) ⎣ dt ⎦ ⎡ d 2F ⎤ dF (0) L⎢ = s 2 .L[F(t)] − s .F(0) − 2 ⎥ dt dt ⎣ ⎦ ⎡ d 3F ⎤ dF d 2F (0) − (0) L⎢ = s 3 . L[F(t)] − s 2 .F(0) − s . 3 ⎥ dt dt 2 ⎣ dt ⎦
b) Analog: ⎡ d nF ⎤ d (n −1) F n n −1 n − 2 dF L ⎢ n ⎥ = s .L[ F (t )] − s .F (o ) − s . (0) − ... − (n −1) (0) dt dt ⎣ dt ⎦ Transformasi Laplace di atas dipakai untuk menyelesaikan PD yang mempunyai syaratsyarat awal (initial condition)
Contoh: Selesaikanlah transformasi Laplace
dx + x = e t dengan syarat awal x (0) = 0 ! dt
dx + x = e t → x(0) = 0 dt ⎧⎛ dx ⎞⎫ L ⎨⎜ + x ⎟⎬ = L {e t } ⎠⎭ ⎩⎝ dt ⎧ dx ⎫ L ⎨ ⎬ + L {x} = L {e t } ⎩ dt ⎭ 1 s ⋅ L {x} − x (0) + L {x} = s −1 1 s ⋅ L {x} − 0 + L {x} = s −1 1 ( s + 1) L {x} = s −1 1 L {x} = ( s + 1)(s − 1) Misal: L {x} = y
y=
1 A B = + ( s − 1)( s + 1) s + 1 s − 1
1 = A( s − 1) + B( s + 1) untuk s = −1 → 1 = A ( −1 − 1) + B ( −1 + 1) 1 = −2 A + 0 1 1 = −2 A → A = − 2 untuk s = 1 → 1 = A (1 − 1) + B (1 + 1) 1 = 0 + 2B 1 = 2B →
1 1 − 1 2 = + 2 (s − 1)(s + 1) s + 1 s − 1
B=
1 2
1 ( s − 1)( s + 1) 1 1 − y= 2 + 2 s +1 s −1 y=
L{x} = y x = L−1{ y} 1 ⎫ ⎧ 1 − ⎪ ⎪ x = L−1 ⎨ 2 + 2 ⎬ ⎪ s + 1 s − 1⎪ ⎩ ⎭
1 −1 ⎧ 1 ⎫ 1 −1 ⎧ 1 ⎫ L ⎨ ⎬ ⎬+ L ⎨ 2 ⎩ s + 1⎭ 2 ⎩ s + 1⎭ 1 1 x = − e −t + e t 2 2 x=−
Contoh 2: Selesaikan dengan transformasi Laplace
d2x − x = e 2 t dengan syarat awal 2 dt
d 2x − x = e 2t 2 dt ⎫ ⎧d 2x L ⎨ 2 − x ⎬ = L e 2t ⎭ ⎩ dt
{ }
⎧d 2x ⎫ L ⎨ 2 ⎬ − L{x} = L e 2t ⎩ dt ⎭ 1 dx s 2 ⋅ L {x} − s ⋅ x(0) − (0) − L {x} = s-2 dt
{ }
Misal: L {x} = y
⎧x ( 0 ) = 0 ⎪ ⎨ dx (0) = 0 ⎪⎩ dt
1 s−2 1 s2 y − y = s−2 1 ( s 2 − 1) ⋅ y = s−2
s 2 ⋅ y − s.0 − 0 − y =
1 ( s − 2)( s 2 − 1) A B C 1 y= = + + ( s − 2)( s − 1)( s + 1) s − 2 s − 1 s + 1 ( s − 2)( s − 1)( s + 1) 1 = A( s − 1)( s + 1) + B( s − 2)( s + 1) + C ( s − 2)( s − 1) y=
untuk s = 1 → 1 = A (1 − 1)(1 + 1) + B (1 − 2)(1 + 1) + A (1 − 2)(1 + 1) 1 = 0 − 2B + 0 1 = −2 B →
B=−
1 2
untuk s = −1 → 1 = A ( −1 − 1)( −1 + 1) + B ( −1 − 2)( −1 + 1) + C ( −1 − 2)( −1 + 1) 1 = 0 + 0 + 6c 1 = 6C → untuk s = 2 →
C=
1 6
1 = A (2 − 1)(2 + 1) + B (2 − 2)(2 + 1) + A (2 − 2)(2 + 1) 1 = 3A + 0 + 0 1 1 = 3A → A = 3
1 (s − 2)(s − 1)(s + 1) A B C y= + + s −1 s −1 s +1 1 1 1 − 3 2 y= + + 6 s − 2 s −1 s +1
y=
L{x} = y x = L−1{y} 1 ⎫ 1 ⎧ 1 − ⎪ ⎪⎪ ⎪ x = L−1 ⎨ 3 + 2 + 6 ⎬ ⎪ s − 2 s − 1 s + 1⎪ ⎪⎭ ⎩⎪ ⎧ 1 ⎫ ⎧ 1⎫ ⎪ ⎪ ⎪⎪ − 2 ⎪⎪ −1 ⎪ 3 ⎪ + x=L ⎨ L ⎬ ⎨ ⎬+L ⎪ s − 1⎪ ⎪ s − 1⎪ ⎪⎩ ⎪⎭ ⎪⎩ ⎪⎭
⎧ 1 ⎫ ⎪⎪ 7 ⎪⎪ ⎨ ⎬ ⎪ s + 1⎪ ⎪⎩ ⎪⎭
1 −1 ⎧ 1 ⎫ 1 ⎧ 1 ⎫ 1 ⎧ 1 ⎫ L ⎨ ⎬− L⎨ ⎬+ L ⎨ ⎬ 3 ⎩ s − 2 ⎭ 2 ⎩ s + 1⎭ 6 ⎩ s + 1⎭ 1 1 1 x = e−t − et + e−t 3 2 6
x=
Tranformasi Laplace dari Integral Jika L {F(t)} = f(s), maka: t f ( s) L ⎧⎨∫ F( u) du ⎫⎬ = 0 s ⎩ ⎭
Contoh:
L
{∫
t
}
sin 2u du = K
0
L {sin 2t} = f (s) = L
{∫ sin 2u du}= t
0
=
=
2 s + 22 2 2 s +4 f (s) s 2 s2 4 s 2 2 s ( s + 4) 2
Perkalian dengan t n Jika L {F(t)} = f(s),
[
]
maka L t n F(t ) = (− 1)
n
dn [ f (s )] = (− 1)n . f ( n) (s ) n ds
Contoh :
{
}
L t ⋅ e 2t = K F (t ) = e 2t L{F (t )} = L{e 2t } =
1 s−2
1 = ( s − 2) −1 s−2 f ′( s ) = −1( s − 2) −1−1 ( s − 2) ' f ( s) =
= −1( s − 2) −2 (1) −1 = ( s − 2) 2 L {t ⋅ e 2t } = ( −1)1 f ′(s) −1 = −1 ⋅ (s − 2) 2 1 = (s − 2) 2
Pembagian dengan t Jika L{F(t)} = f(s), maka:
⎧ F(t ) ⎫ ∞ L⎨ ⎬ = ∫ f ( μ ) du ⎩ t ⎭ s
Contoh:
⎧ sin t ⎫ L⎨ ⎬ =K ⎩ t ⎭ L{F (t )} = L {sint t} 1 1 = 2 2 s +1 s +1 1 f (μ ) = 2 μ +1 f(s) =
2
∞ ⎧ sin t ⎫ L⎨ ⎬ = ∫s f ( μ ) dμ ⎩ t ⎭ ∞ 1 ∫s μ 2 + 1 ⋅ dμ ∞ dμ =∫ s 1+ μ2
= arc tg μ
∞ s
= arc tg ∞ − arc tg s =
π 2
− arc tg s
sin 90 0 cos 90 0 π 1 tg = 2 0
tg 90 0 =
tg
π 2
π
2
=∞ = arc tg ∞
Soal: 1. 2. 3.
L {t ⋅ sin at} = K
4.
∫0
5. 6. 7. 8. 9.
L {t ⋅ cos at} = K L {t ⋅ cosh 3t} = K ∞
t ⋅ e - 3t ⋅ sin t dt = K
⎧⎪ e -at − e − bt ⎫⎪ L⎨ ⎬ =K t ⎪⎩ ⎪⎭ ⎧ sinh t ⎫ L⎨ ⎬ =K ⎩ t ⎭ e −3 t − e −6 t dt = K t ⎪⎧t 2 , 0 < t ≤ 1 Jika F(t ) = ⎨ maka L {F(t)} = K ⎪⎩0 , t > 1 ⎧2 t , 0 ≤ t ≤ 1 maka L {F(t)} = K Jika F(t ) = ⎨ ⎩t , t > 1 ∞
∫0
10.
∞
∫0
sin 2 t dt = K t2