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Pertemuan 18 &19 SIFAT-SIFAT TRANSFORMASI LAPLACE 1. Sifat Kelinearan Jika C1 dan C2 konstan,
L[F1 (t )] = f1 (s ) & L[F2 (t )] = f 2 (s )
maka: a) L[C1 F1 (t )] = C1 f1 (s ) dan L[C2 F2 (t)] = C2 f 2 (s ) b) L[C1 F1 (t ) ± C2 F2 (t )] = L[C1 F1 (t )] ± L[C2 F2 (t )] = C1 f1 (s ) ± C2 f 2 (s )
contoh :
[
]
L 4t 2 − 3 cos 2t + 5e − t = 4 L[t 2 ] − 3L[cos 2t ] + 5 L[e − t ] = 4.
2! S 1 −3 2 +5 2 +1 2 S − (−1) S S +2
= 4.
2 1 S −3 2 +5 3 S +1 S S +4
=
8 3S 5 − 2 + 3 S S + 4 S +1
2. Sifat Translasi Pertama atau Pergeseran Jika
[
Contoh :
[
]
L[F (t )] = f (s ) → L e at F (t ) = f (s − a )
]
L e -t sin 2t = f [ s − (−1)] = f ( s + 1)
F(t) = sin2t
f [ s ] = L[sin2t] =
2 2 = 2 2 s +2 s +4 2
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f [ s + 1] =
2 (s + 1) 2 + 4
[
]
Jadi L e - t sin 2t = f ( s + 1) =
2 2 2 = 2 = 2 2 (s + 1) + 4 s + 2s + 1 + 4 s + 2s + 5
3. Sifat Translasi Pertama atau Pergeseran Jika
L[F (t )] = f (s ) dan G (t ) =
Maka
L[G (t )] = e − as . f (s )
{(
F t −a ) ; t >a o ; t
Contoh :
F(t ) =
{
(t -2)3 ; t > 2 o ; t <2
F (t − 2) = (t − 2) 3 F (t ) = t 3 f (s ) = L[ F (t )] = L[t 3 ] =
3!
s
L[F (t )] = e − as . f (s ) = e − 2 s .
3+1
=
3.2.1 6 = 4 s4 s
6 6e −2 s = 4 s4 s
4. Sifat Pergantian Skala Jika L[F (t )] = f (s ) → L[F (at )] =
1 a
s f a
Contoh : L[sin 3t ] =
F (t ) = sin t f ( s ) = L[t ] = L[sin t ] =
1 1 = 2 2 s +1 s +1 2
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1 1 9 9 s = 2 . = 2 f( )= s 3 9 S +9 ( )2 + 1 s + 1 3 9
Jadi L[sin 3t ] =
1 s 1 9 3 = 2 f = . 2 3 3 3 S + 9 S + 9
5. 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
µ
dF = lim [e - st F (t )] oµ - lim L µ→∞ dt µ → ∞ F (t ) dF = lim [ st ] oµ - lim L →∞ µ µ→∞ e dt
∫ F (t ) d[e
- st
]
o
µ
∫ F (t ).e
- st
(-s) dt
o
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
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Jadi :
dF = s . L[F(t)] − F(0) L dt
Sehingga
d 2F dF dF L = s . L[ ]− (0) 2 dt dt dt
d 2F dF L = s . [ s . L[F(t)] − F(0)] − (0) 2 dt dt
d 2F dF L = s 2 .L[F(t)] − s .F(0) − (0) 2 dt dt
juga
d 3F dF d 2F 2 L = s − s − . L[F(t)] . (0) (0) 3 dt dt 2 dt d 3F dF d 2F L = s 3 .L[F(t)] − s 2 .F(0) − s . (0) − (0) 3 dt dt 2 dt
Kesimpulan : dF L = s . L[F(t)] − F(0) dt d 2F dF L = s 2 .L[F(t)] − s .F(0) − (0) 2 dt dt d 3F dF d 2F L = s 3 .L[F(t)] − s 2 .F(0) − s . (0) − (0) 3 dt dt 2 dt
b) Analog: d nF dF d (n −1) F (0) − ... − (n −1) (0) L n = s n .L[ F (t )] − s n −1.F (o ) − s n − 2 . dt dt dt
contoh : 1. Selesaikan dengan Transformasi Laplace
x(0) = 0 ! Jawab :
dx + x = et dt
dx + x = e t dengan syarat awal dt
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L[
dx + x] = L[e t ] dt
L[
1 dx ] + L[ x] = dt s −1 1 s −1 1 s. L[x] − 0 + L[ x] = s −1 1 ( s + 1) L[x] = s −1
s. L[x] − x(0) + L[ x] =
L[x] =
1 A B = + ( s − 1)( s + 1) ( s − 1) ( s + 1)
1 1 1 = = ( s − 1)( s + 1) (1 + 1) 2 di mana s-1 = 0 , maka s=1
A = ( s − 1).
1 1 1 1 = = =− ( s − 1)( s + 1) (−1 − 1) − 2 2 di mana s+1 = 0 , maka s= -1
A = ( s + 1).
L[x] =
A B + ( s − 1) ( s + 1)
1 1 − 2 L[x] = 2 + ( s − 1) ( s + 1) 1 1 − 2 ] X = L−1 [ 2 + ( s − 1) ( s + 1) 1 1 − X = L−1 [ 2 ] + L−1 [ 2 ] ( s − 1) ( s + 1)
X=
1 −1 1 1 1 ] − L−1 [ ] L [ ( s − (−1)) 2 ( s − 1) 2
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X=
1 t 1 −t e − e 2 2
2. Selesaikan dengan Transformasi Laplace
d 2x − x = e 2t dengan syarat 2 dt
dx (0) = 0 ! dt
awal x(0) = 0 dan Jawab:
d 2x − x = e 2t 2 dt L[
d 2x − x] = L[e 2t ] 2 dt
L[
1 d 2x ] − L[ x] = 2 s−2 dt
s 2 .L[x] − s.x(0) −
1 dx (0) − L[ x] = dt s−2
s 2 .L[x] − s.0 − 0 − L[ x] =
s 2 .L[x] − L[ x] =
( s 2 − 1).L[ x] =
1 s−2
1 s−2
1 s−2
L[ x] =
1 ( s − 2)( s 2 − 1)
L[ x] =
1 A B C = + + ( s − 2)( s + 1)( s − 1) ( s − 2) ( s + 1) ( s − 1) 1 1 1 1 = = = ( s − 2)( s + 1)( s − 1) (2 + 1)(2 − 1) (3)(1) 3 di mana s-2 = 0 , maka s=2
A = ( s − 2).
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1 1 1 1 = = = ( s − 2)( s + 1)( s − 1) (−1 − 2)(−1 − 1) (−3)(−2) 6 di mana s+1 = 0 , maka s=-1
B = ( s + 1).
1 1 1 1 1 = = = =− ( s − 2)( s + 1)( s − 1) (1 − 2)(1 + 1) (−1)(2) − 2 2 di mana s-1 = 0 , maka s=1
C = ( s − 1).
L[ x] =
A B C + + ( s − 2) ( s + 1) ( s − 1)
1 1 1 − 2 L[ x] = 3 + 6 + ( s − 2) ( s + 1) ( s − 1) 1 1 1 − 2 ] x = L−1 [ 3 + 6 + ( s − 2) ( s + 1) ( s − 1) 1 1 1 − x = L−1 [ 3 ] + L−1 [ 6 ] + L−1 [ 2 ] ( s − 1) ( s − 2) ( s + 1)
x=
1 −1 1 1 1 1 1 ] + L−1 [ ] − L−1 [ ] L [ ( s − (−1)) 2 ( s − 1) 3 ( s − 2) 6
1 1 1 x = e 2 t + e −t − e t 3 6 2
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6. Transformasi Laplace dari Integral * Jika
L[F (t )] = f (s )
t 1 maka L ∫ F (µ ) dµ = f (s ) o s contoh :
t L ∫ sin 2µ . dµ = o Jawab: F ( µ ) = sin 2µ , maka F (t ) = sin 2t f ( S ) = L[ F (t )] = L[sin 2t ] =
2 2 = 2 2 s +2 s +4 2
t 1 1 2 2 Jadi L ∫ sin 2 µ . dµ = f (s ) = [ 2 ]= 2 s s +4 s ( s + 4) o s * Perkalian dengan t n Jika
L[F (t )] = f (s )
[
]
dn maka L t F(t ) = (− 1) [ f (s )] = (− 1)n . f ( n) (s ) n ds n
n
Contoh :
[ ]
L t.e 2t = Jawab :
F (t ) = e 2t , maka f (s ) = L[ F (t )] = L[e 2t ] = f (1) (s ) = (−1)( s − 2) −1−1 .( s − 2) ' f (1) (s ) = (−1)( s − 2) − 2 .(1) =
−1 ( s − 2) 2
1 = ( s − 2) −1 s−2
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[ ]
Jadi L t.e 2t = (− 1) . f (1) (s ) = (−1).
*
1
1 −1 = 2 ( s − 2) ( s − 2) 2
Pembagian dengan t Jika
L[F (t )] = f (s )
maka
F (t ) F (t ) = ∫ f (u ) du ; dengan syarat lim ADA L t →o t t s ~
contoh:
sin t L = t Jawab :
F (t ) = sin t , maka f (s ) = L[F (t )] = L[sin t ] = sehingga f (u ) =
sin t L = t
1 1 = 2 , 2 s +1 s +1 2
1 u +1 2
~
~
s
s
∫ f (u ) d u = ∫
1 du = 2 u +1
~
2
=
s
= [arc tg x]∞s = arc tg ∞ - arc tg s =
SOAL-SOAL LATIHAN:
du
∫1+ u π 2
- arc tg s
