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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:
{ } dttFesfL st == )()(F(t) 0 Transformasi Laplace untuk beberapa fungsi dasar:
No.
1. 1
s1
s > 0
2. nt ; (m = 1, 2, 3, K 1ns
!n+
s > 0
3. pt ; (p > -1) 1p
)1p(
s ++ s > 0
4. ate as
1
s > a
5. t cos 22 +s
s
s > 0
No.
6. t sin 22
+s
s > 0
7. Cosh at 22 as
s
s > a
8. Sinh at 22 as
a
s > a
Buktikan: { } !s
11 L =
Bukti:
-
= 0 st dt)t(Fe)}t(F{L [ ] dt e lim dt )1( 1
o
st-
o
==
steL
s1- lim )ed( e
s1- lim st-
o
st-
o
ste
==
( ) ( )s11-0
s1 -e e lim
s1 - os- ===
Buktikan: 2s
1}t{L =
Bukti :
( )[ ]
dt )( o = tFetFL st
[ ] dt t.e lim dt )( o
st-
o
==
tetLst
.d(t)]e -e[t.lim
s1 - )t.d(e
lims1 -
oo
st-
o
st-st-
===
=+=
o
st-
o
st (1)dtelims1]e[t/.lim
s1 -
=+=
o
st-
o
st (1)dtelims1]e[1/s.lim
s1 -
s1
s1.
s1]1[.
s10dt 1)(
s1 )0(
s1 - 2
o
==+=+= Le st
Buktikan: 3
2
s
2)t{L =
Bukti:
dttetL
dttFetFL
st
st
2
0
2
0
}(
)()}({
==
-
=
=
p st
p
stp
p
stdets
dtet
0
2
0
2
)(1lim
lim
=
=
p stpst
p
p st
p
tdeets
edts
0
20
2
0
2
)( lim1
)( lim1
=
=
00.
22
00
2
20limlim1
)2(limlim1
dtteee
ps
dtteet
s
stspspp
p st
p
pstp
{ }[ ]
=
=
2
121
2001
ss
tLs
32 2}(
stL =
Buktikan: 1n
n
s
!n}{t L +=
Bukti:
1nn
121232
112
10n
s
!n}(t L
s
!2
s
12
s
2}{t L
s
!1
s
1 {t} L
s
!031
}{t L {1} L
+
++
+
+
=
===
==
===
M
Buktikan [ ] ( )11 ++= nn sntL Bukti:
Fungsi Gamma: dxx e 1nx0)n(
= ( ) +=+
o
1-1n dx 1 xen x
-
( ) =+o
n dx 1 xen x misalkan : x = st
dx = s.dt
( )
=+o
n s.dt st)( 1 sten
( ) =+o
nn s.dt ts 1 sten
( )
+=+o
n1n dt ts1 sten ( ) ][.s1 1n ntLn +=+
[ ] ( )11 ++= nn sntL
maka [ ] ( )11+n 1Sn! ++== nn sntL [terbukti]
Buktikan : as
L =1}{e at
Bukti:
dt)t(Fe{F(t)} L st0
= [ ] dt lim dt )(
oo +
==
atstatstat eeeeL
a)]-d[-s(t lima)-(s-
1 dt lim o
)(
o
)( ==
tastas ee
]e
1[lima)-(s-
1 ]e[lima)-(s-
1
o)(o
)(
tastas
==
]1lim1lim[a)-(s-
10).()( asas ee
=
]10.[a)-(s-
1]1lim0[a)-(s-
10 == e
a-s1}{e at =L [terbukti]
-
RUMUS FULER
biabia eee =+ ) sin (cos bibee abia +=+
+=+=+=
beibeibe
bi
bi
bi
cos2ebsin cos
bsin cos
bi
2ee
bcosbibi +=
2ee
t costiti +=
==+=
bieibeibe
bi
bi
bi
sin 2ebsin cos
bsin cos
bi
ieeb
bibi
2sin
= ie ti
2e t sin
ti
=
Buktikan 22s
st) (cos L +=
Bukti:
( )
[ ][ ]
++=
+=+=
+=+=
+=
==
is1
is1
21
}e{L}e{L21
}e{L}e{L21
dteedtee21
dt eee21
dt2
eee
dttcoset} {cosL
dt)t(Fe)}t(F{L
itit
titi
tist
0
tist
0
ititst
0
ititst
0
st
0
st
0
-
[terbukti] }{cos
)1(
221
))((21}{cos
22
22
222
+==
=
+++=
sstL
ss
iss
isisisistL
Buktikan: 22s
}t{sinL +=
( )
[ ]
[terbukti] s
)t{sinL
)1(s
i2i2
1is
)isisi2
1
)is)(is()is(is
i21
is1
is1
i21
}e{L}e{Li2
1
dteedteei2
1
dt eeei2
1
dt2i
eee
dttsinet} {sinL
dt)t(Fe)}t(F{L
22
22
222
itit
itst
0
itst
0
ititst
0
ititst
0
st
0
st
0
+=
=
++=++=
+=
=
=
=
=
==
Buktikan: 22 as
s}at{coshL =
Bukti:
-
{ }{ } { }[ ]
[terbukti] as
s}at{coshL
as
s221
)as)(as(as as
21
e Le L21
ee L 21
2ee
L}at{coshL
22
22
atat
atat
atat
==
+++=
+=
+=
+=
Buktikan: 22 as
a at}{sinh L =
Bukti:
{ }{ } { }[ ]
[terbukti] as
a at}(sinh L
as
a221
as
asas21
)as)(as()a(s as
21
as1
as1
21
e Le L21
ee L 21
2ee
L at}{sinh L
22
22
22
at-at
atat
atat
==
++=++=
+=
==
=
2.2 BEBERAPA SIFAT PENTING DARI TRANSFORMASI LAPLACE 1. Kelinearan
{ } { } { })()(
)(F )(F )(c)(c
2211
22112211
sfcsfctLctLctFtFL
+=+=+
Contoh:
-
{ }
15
438
)1(15
23!24
}{ 5}2{cos 3}{t L 452cos34t
23
2212
2
2
+++=++=
+=+
+
sss
s
sss
s
eLtLetL
t
t
2. Translasi Pertama atau pergeseran
Jika f(s) {F(t)} L = Maka { } )as(f)t(Fe L at = Contoh:
5s2s
2
41s2s
2
4)1s(
2
1)f(s2t}sin {e L
4s
2
2s
2 2t}{sin L f(s)
2t}sin {e L
222
t
222
t
++=+++=++=+=
+=+===
K
3. Translasi Kedua atau Pergeseran
Jika f(s) {F(t)} L = dan
=
at0
at)at(F)t(G
Maka )s(fe {G(t)} L -as =
Contoh:
44133 6123!3}{)(
ssstLsf ==== +
dan
=atatt
tG0
)2()(
3
4
s2
4s2
-as
s
e6
s
6e
)s(fe{G(t)} L maka ==
=
4. Perubahan Skala
Jika f(s){F(t)} L =
Maka
=as
fa1
{F(a.t) L
-
Contoh:
93
99
31
31
31 3t}{sin L
99
99
1
19
1
13
13
11
1s1 L{sin t}f(s)
3t}{sin
22
2
222
222
+=+=
=+=
+=+=
+
=
+=+==
=
ss
f
s
ssssf
s
L K
Soal:
1. K= }3{5t L 2. K= }et2{ L t2 3. K 5t} cos {3 L = 4. K 2t} cos 5 2tsin {6 L = 5. { } K=+ 22 )1(t L 6. K= }t2cos{e L t 7. K= }t2cos{t L 3 8. { } K=+ t2 e2)(t L 9. { } K=2t coshe L -4t 10. { }
)52(21
)1(21 sine L 2
2
++++=
ss
ss
tt K
11. Jika
>
=
5t,1
5tt,2t)t(F
16. Tentukan L {F(t)} jika =1t0,0
1t,1)-(t)t(F
2
-
2.3 TRANSFORMASI LAPLACE TURUNAN DAN INTEGRAL Transformasi Laplace dari Turunan
)0(F{F(t)} LsdtdF
L =
dst
)0(dt
Fd)0(
dt
Fds)0(
dtdf
s)0(Fs{F(t)} Lsdt
Fd L
)0(dt
Fd)0(
dtdf
s)0(Fs{F(t)} Lsdt
Fd L
)0(dtdf
)0(Fs{F(t)} Lsdt
Fd L
3
3
2
2234
4
4
2
223
3
3
22
2
M
=
=
=
Bukti :
Transformasi Laplace dari Derivatif
Jika ( )[ ] ( )sftFL = maka a) ( )0)]([. FtFLS
dtdFL =
Bukti:
( )[ ]
dt )( o = tFetFL st
dt )]([ o = dt
tFdedtdFL st
( )
]td[F e lim o
st-=
dtdFL
( ) ( )
]d[e tF lim-]tF [elim o
st-o
st- =
dtdFL
( ) ( )
dt (-s).etF lim-]e
tF [lim o
st-ost =
dtdFL
-
( ) ( )
dt.etF lim.]e
F [lim o
st-s +=
s
dtdFL
( ) ( )
dttFe .]e
F(0)[lim]e
F [lim o
st-s.0s += sdt
dFL
L[F(t)] .F(0)0 sdtdFL +=
Jadi :
F(0)L[F(t)] . =
sdtdFL
Sehingga (0)]L[ . 22
dtdF
dtdFs
dtFdL =
(0)F(0)]L[F(t)] .[ . 22
dtdFss
dtFdL =
(0)F(0).L[F(t)]. 222
dtdFss
dtFdL =
juga (0)(0).L[F(t)]. 22
23
3
dtFd
dtdFss
dtFdL =
(0)(0)..F(0)L[F(t)]. 22
233
3
dtFd
dtdFsss
dtFdL =
Kesimpulan :
F(0)L[F(t)] . =
sdtdFL
(0)F(0).L[F(t)]. 222
dtdFss
dtFdL =
(0)(0)..F(0)L[F(t)]. 22
233
3
dtFd
dtdFsss
dtFdL =
b) Analog:
( ) ( )( ) (0)...(0)..)]([. 11
21
=
n
nnnn
n
n
dtFd
dtdFsoFstFLs
dtFdL
Transformasi Laplace di atas dipakai untuk menyelesaikan PD yang mempunyai syarat-syarat awal (initial condition)
-
Contoh:
Selesaikanlah transformasi Laplace texdtdx =+ dengan syarat awal 0 (0) =x !
0 x(0) ==+ texdtdx
( )1)1(1 {x} L
11 {x} L )1(
1s1 {x} L 0 {x} Ls
1s1 {x} L (0) x {x} Ls
}{e {x} dtdx
}{e dtdx
t
t
+==+=+=+
=+
=
+
ss
ss
LLL
LxL
Misal: L {x} = y
)1()1(1
11)1)(1(1
++=
++=+=
sBsA
sB
sA
ssy
= 1suntuk
21
A A21
0A21
)11( B)11( A 1
==+=
++=
= 1suntuk
21
B B21
B201
)11( B)11( A 1
==+=
++=
1s21
1s21
)1s)(1s(1
++
=+
-
121
121
)1)(1(1
++
=
+=
ssy
ssy
++
=
==
121
121
}{}{
1
1
ssLx
yLxyxL
tt eex
sL
sLx
21
21
11
21
11
21 11
+=
++
+=
Contoh 2:
Selesaikan dengan transformasi Laplace t22
2
exdt
xd = dengan syarat awal
==
0)0(dtdx
0)0(x
{ }{ }2-s
1 {x} )0()0( {x}
e }{
e
2
2t2
2
2t2
2
22
2
=
=
=
=
LdtdxxsLs
LxLdtxdL
LxdtxdL
exdtxd t
Misal: L {x} = y
-
)1)(1)(2()1)(2()1)(2()1)(1(1
112)1)(1)(2(1
)1)(2(1
21)1(
21
2100.
2
2
2
2
+++++=+++=+=
====
sssssCssBssA
sC
sB
sA
sssy
ssy
sys
syys
sysys
= 1suntuk
21
B B21
0B201
)11)(21(A )11)(21( B)11)(11( A 1
==+=
+++++=
= 1suntuk
61
C C61
c6001
)11)(21( C)11)(21( B)11)(11( A 1
==++=
+++++=
= 2suntuk
31 A 31
0031)12)(22(A )12)(22( )12)(12(A 1
==++=
+++++=
A
AB
1s61
1s21
2s31
y
1sC
1sB
1sA
y
)1s)(1s)(2s(1
y
++
+=
+++=+=
++
+
=
++
+=
==
1s71
L1s
21
L1s
31
Lx
1s61
1s21
2s31
Lx
}y{Lx
y}x{L
1
1
1
-
ttt
1
e61
e21
e31
x
1s1
L61
1s1
L21
2s1
L31
x
+=
++
+
=
Tranformasi Laplace dari Integral
Jika L {F(t)} = f(s), maka:
s)s(f
du )u(F Lt
0=
Contoh: { }
{ }
)4(2
42
)(du2u sin
42)(
2s22t}{sin
du2u sin
2
2
t
0
2
22
t
0
+=
=
=+=+=
=
ss
ss
ssfL
ssf
L
L K
Perkalian dengan nt
Jika L {F(t)} = f(s),
maka ( )[ ] ( ) ( )[ ] ( ) ( )sfsfdsdtL nnn
nnn )(.11tF ==
-
Contoh :
{ }
2
2
'11
1
2
2
2
)2(1
)1()2(1)2()2(1)(
)2(2
1)(
21}{)}({
)(
t
==
=====
=
=
s
ssssf
ss
sf
seLtFL
etF
eL
t
t
t K
2
2
12t
)2s(
1)2s(
11
)s(f)1(}e{t L
===
Pembagian dengan t
Jika L{F(t)} = f(s), maka:
du )(t
)F( ftLs=
-
Contoh:
s tg2
s tgarc tg
tgarc
1
11
d )(t
sin t
11)(
11
11f(s)
{sint t} )}({t
sin t
2
2
2
222
arc
arc
d
d
fL
f
ss
LtFL
L
s
s
s
s
==
=+=
+
=
+=+=+=
==
K
=
=
=
=
tgarc 2
2
01
2
90 cos90sin 90 tg 0
00
tg
tg
Soal:
1. K= at}sin {t L 2. K= at} cos{t L 3. K 3t}cosh {t L = 4. K dt tsin et 3t-
0=
5. K=
tee
Lbt-at
6. K=
ttsinh
L
7. K dt t
ee t6t3
0=
8. Jika
>=
1t,t
1t0,t2)t(F maka K {F(t)} L =
-
10. K= dtt tsin22
0