Pertemuan-15 & 16
2.4 INVERS DARI TRANSFORMASI LAPLACE
Jika L {F(t)} = f(s), maka F(t) disebut Invers Transformasi Laplace dari F(s), ditulis L−1{s(s)} = F( t ) .
Sifat Invers Transformasi Laplace:
1.
L−1 {c1 ⋅ f1 ( s ) + c2 ⋅ f 2 ( s )} = c1 ⋅ F1 (t ) + c2 ⋅ F2 (t )
2.
L−1{f (s − a)} = e at ⋅ F( t )
3.
L−1{f ( n ) (s)} = ( −1) n ⋅ t n F( t )
4.
5.
6.
7.
t
⎧ f (s) ⎫
L−1 ⎨
⎬ = ∫0 F(μ) ⋅ dμ
⎩ s ⎭
⎧ ⎛ s ⎞⎫
L−1 ⎨f ⎜ ⎟⎬ = a ⋅ F(at )
⎩ ⎝ a ⎠⎭
{
}
L−1 e − as ⋅ f (s) = μ( t − a) F( t − a)
−1
L
t
{f (s) ⋅ g (s)} = ∫0
F(μ) G(t − μ) dμ
Contoh:
1.
⎧ 4 − 5s ⎫
L−1 ⎨ 3 / 2 ⎬ = K
⎭
⎩s
4
5
s
−
⎧ 1 ⎫
⎧ 1 ⎫
⎫
⎧
L−1 ⎨ 3 / 2 ⎬ = 4 L-1 ⎨ 3 / 2 ⎬ − 5L-1 ⎨ 3 / 2 ⎬
⎭
⎩s
⎭
⎩s
⎭
⎩s
n!
γ (n + 1)
n
L {t } =
= n +1
s n +1
s
⎧ γ (n + 1) ⎫
t n = L−1 ⎨ n +1 ⎬
⎩ s
⎭
⎧ 1 ⎫
t n = γ (n + 1) L-1 ⎨ n+1 ⎬
⎩s ⎭
tn
⎧ 1 ⎫
= L−1 ⎨ n+1 ⎬
γ (n + 1)
⎩s ⎭
⎧ 1 ⎫
⎧ 1 ⎫
L−1 ⎨ 3 / 2 ⎬ = L−1 ⎨ 1 +1 ⎬
⎩s ⎭
⎩s2 ⎭
1
1
1
t2
t2
2t 2
=
=
=
γ ( 12 + 1) 12 γ ( 12 )
π
⎧ 1 ⎫
⎧ 1 ⎫
L−1 ⎨ 1 / 2 ⎬ = L−1 ⎨ − 1 +1 ⎬
⎩s ⎭
⎩s 2 ⎭
−1
−1
=1
t 2
t 2
t 2
=
=
=
γ (- 12 + 1) γ ( 12 )
π
⎧ 4 − 5s ⎫
⎧ 1 ⎫
⎧ 1 ⎫
L−1 ⎨ 3 / 2 ⎬ = 4 L−1 ⎨ 3 / 2 ⎬ − 5 L−1 ⎨ 1 / 2 ⎬
⎩ s
⎭
⎩s ⎭
⎩s ⎭
1
2t 2
= 4⋅
=
π
1
2
− 5⋅
8t − 5 ⋅ t
π
− 12
t
− 12
π
2.
⎧ 1 ⎫
L−1 ⎨ 2
⎬ =K
⎩ s + 2s ⎭
1
1
A
B
=
= +
s + 2s s ( s + 2) s s + 2
2
Kiri dan kanan dikali dengan KPK : s(s+2)
untuk s = 0 →
1 = A ( s + 2) + B.s
1 = A(0 + 2) + B.0
1 = 2A + 0
1 = 2A →
untuk s = −2 →
A=
1
2
1 = A (−2 + 2) + B ⋅ (−2)
1 = 0 − 2B
1
1 = 2B → B = 2
1
1
−
1
= 2 + 2
s 2 + 2s s s + 2
1
− 12 ⎫
⎧ 1 ⎫
−1 ⎧ 2
=
+
L−1 ⎨ 2
L
⎬
⎨
⎬
⎩ s + 2s ⎭
⎩s s + 2⎭
1 ⎧1 ⎫ 1 ⎧ 1 ⎫
= L−1 ⎨ ⎬ − L−1 ⎨
⎬
2 ⎩s ⎭ 2 ⎩s + 2⎭
1
1
= ⋅ 1 − e −2t
2
2
1 1
= − e −2t
2 2
1
= 1 − e −2t
2
(
3.
)
⎧ 2s + 3 ⎫
L−1 ⎨ 2
⎬ =K
⎩ s − 2s + 5 ⎭
2s + 3
2s + 3
2s + 3
= 2
=
s − 2 s + 5 ( s − 2 s + 1) + 4 ( s − 1) 2 + 4
s ( s − 1)
2 s − 2 + 5 2( s − 1) + 5
5
=
−
=
+
2
2
2
( s − 1) + 4 ( s − 1) + 4 ( s − 1) + 4 ( s − 1) 2 + 4
2
92
⎫
5
⎧ 2s + 3 ⎫
−1 ⎧ 2( s − 1)
L−1 ⎨ 2
+
⎬=L ⎨
⎬
2
2
⎩ s − 2s + 5 ⎭
⎩ ( s − 1) + 4 ( s − 1) + 4 ⎭
⎧ ( s − 1) ⎫
⎫
1
−1 ⎧
= 2 L−1 ⎨
⎬ + 5L ⎨
⎬
2
2
⎩ ( s − 1) + 4 ⎭
⎩ ( s − 1) + 4 ⎭
⎧ s ⎫
⎧ 1 ⎫
= 2 ⋅ e t ⋅ L−1 ⎨ 2
+ 5e t L1 ⎨ 2
2⎬
2⎬
⎩s + 2 ⎭
⎩s + 2 ⎭
5
⎧ 2 ⎫
= 2 e t cos 2t + e t L−1 ⎨ 2
2⎬
2
⎩s + 2 ⎭
5
= 2 e t cos 2t + e t sin 2t
2
1 t
= e (4 cos 2t + 5 sin 2t )
2
4.
⎧ 1 ⎫
L−1 ⎨
⎬ =K
⎩ 5+3⎭
⎧ 1 ⎫⎪
⎧ 1 ⎫
−1 ⎪
L−1 ⎨
=
L
⎬
⎨
1 ⎬
⎪⎩ ( s + 3) 2 ⎪⎭
⎩ s + 3⎭
⎧1⎫
= e −3t ⋅ L−1 ⎨ 1 ⎬
⎩s2 ⎭
⎧ 1 ⎫
= e −3t ⋅ L−1 ⎨ − 1 +1 ⎬
⎩s 2 ⎭
−1
=e
− 3t
=e
− 3t
=
5.
t 2
⋅
γ (− 12 + 1)
⋅
e 3t ⋅ t
t
− 12
γ ( 12 )
− 12
π
⎧ 3s − 4 ⎫
L−1 ⎨
=K
5⎬
⎩ (25 − 3) ⎭
3s − 4
3s − 4
3s − 4
=
=
5
5
5
(2 s − 3)
3⎞
⎧ ⎛
3 ⎞⎫
5 ⎛
2
s
s
−
2
⎜
⎟
⎟⎬
⎨ ⎜
2 ⎠⎭
⎝ 2⎠
⎩ ⎝
1⎞ 1
⎛
⎜ 3s − 4 ⎟ +
2⎠ 2
=⎝
5
3⎞
⎛
25 ⎜ s − ⎟
2⎠
⎝
3⎞ 1
⎛
3⎜s − ⎟ +
2⎠ 2
= ⎝
5
3⎞
5⎛
2 ⎜s − ⎟
2⎠
⎝
Transformasi Laplace
93
⎧ ⎛
3⎞ 1⎫
⎪ 3⎜ s − ⎟ + ⎪
⎧⎪ 3s − 4 ⎫⎪
2⎠ 2⎪
⎪
L−1 ⎨
= L−1 ⎨ ⎝
5⎬
5 ⎬
⎪⎩ (2s − 3) ⎪⎭
3⎞ ⎪
⎪ 5⎛
⎪ 2 ⎜⎝ s − 2 ⎟⎠ ⎪
⎭
⎩
⎡ ⎧ ⎛
⎧
⎫⎤
1 ⎞⎫
1
⎢ ⎪3 ⎜ s − ⎟ ⎪
⎪
⎪⎥
1 ⎢ -1 ⎪ ⎝
2 ⎠ ⎪ −1 ⎪
⎪⎥
2
= 5 L ⎨
+L ⎨
⎬
⎬
5
5
2 ⎢ ⎪⎛
1 ⎞ ⎪⎥
1⎞ ⎪
⎛
⎪
s− ⎟ ⎥
⎢
⎜s − ⎟ ⎪
⎪⎩ ⎜⎝
2⎠ ⎭
2 ⎠ ⎪⎭⎥⎦
⎢⎣ ⎪⎩ ⎝
⎡
⎧
⎫
⎧
⎫⎤
⎢
⎪
⎪
⎪
⎪⎥
1 ⎢ -1 ⎪
1
1
⎪ 1 −1 ⎪
⎪⎥
=
+ L ⎨
3L ⎨
⎬
⎬
4
5
32 ⎢
3 ⎞ ⎪ 2 ⎪⎛
3 ⎞ ⎪⎥
⎛
⎪
s− ⎟
s− ⎟ ⎥
⎢
⎪⎩ ⎜⎝
⎪⎩ ⎜⎝
2 ⎠ ⎪⎭⎥⎦
2 ⎠ ⎪⎭
⎢⎣
3 32 t −1 ⎧ 1 ⎫ 1 32 t −1 ⎧ 1 ⎫
=
e L ⎨ 4⎬+
e L ⎨ 5⎬
32
⎩ s ⎭ 64
⎩s ⎭
3 32 t −1 ⎧ 1 ⎫ 1 32 t −1 ⎧ 1 ⎫
e L ⎨ 3+1 ⎬ +
e L ⎨ 4+1 ⎬
32
⎩ s ⎭ 64
⎩s ⎭
3
4
3 32 t
t
1 32 t
t
=
e
+
e
32
γ (3 + 1) 64 γ (4 + 1)
=
3 32 t
e
32
3 32 t
=
e
32
=
=
t 3 1 32 t t 4
+
e ⋅
3! 64
4!
3
t
1 32 t
t4
⋅
+
e ⋅
3.2.1 64
4.3.2.1
⋅
1 32 t 3
1 32 t 4
e t +
e t
64
1536
3
t
3
t
3
24 e 2 t 3 e 2 ⋅ t 4
=
+
1536
1536
t
3
t
3
t
24 e 2 + 3 + e 2 t 4
=
1536
t 3 (24 + t ) + e 2
=
1536
Soal:
1.
2.
3.
4.
5.
94
⎧ 2s − 8 ⎫
L−1 ⎨ 2
⎬ =K
⎩ s + 36 ⎭
⎧2 − s ⎫
L−1 ⎨ 3 / 2 ⎬ = K
⎩s
⎭
⎧⎪ s 3 − s 2 + s − 1 ⎫⎪
L−1 ⎨
⎬ =K
⎪⎩
⎪⎭
s5
⎧ 3s − 16 ⎫
L−1 ⎨ 2
⎬ =K
⎩ s − 64 ⎭
⎫
⎧
s2 + 2
L−1 ⎨ 2
⎬ =K
2
⎩ ( s + 10)( s + 20) ⎭
6.
7.
8.
⎧ 4s ⎫
L−1 ⎨ 2
⎬ =K
⎩ s + 16 ⎭
⎧ 1 ⎫
L−1 ⎨ 2
⎬ =K
⎩ 4s + 9 ⎭
⎧⎪ (2s + 1) 2 ⎫⎪
L−1 ⎨
⎬ =K
⎪⎩ s 5
⎪⎭
⎧
⎫
s
L−1 ⎨
⎬ =K
+
+
(
s
3
)(
s
5
)
⎩
⎭
2
⎧ ⎫
10. L−1 ⎨ 3 ⎬ = K
⎩s ⎭
9.
⎧⎪ e −5s ⎫⎪
⎬ =K
⎪⎩ s ⎪⎭
11. L−1 ⎨
⎧⎪ 2e − s − e −2 s ⎫⎪
⎬ =K
s
⎪⎩
⎪⎭
12. L−1 ⎨
⎧
8 − 10s ⎫
=K
2⎬
⎩ (s + 1)(s − 2) ⎭
13. L−1 ⎨
⎧
⎛ s + 6 ⎞⎫
⎟⎬ = K
⎝ s + 2 ⎠⎭
14. L−1 ⎨ln ⎜
⎩
⎧
⎫
1
15. L−1 ⎨ 2
=K
3⎬
⎩ (s + 1) ⎭
-oo0oo2.5
Fungsi-fungsi Periodik
T
Misalkan F(t) mempunyai perioda T > 0, sehingga F(t+T) = F(t), maka L{F (t )} =
∞
L{F (t )} = ∫ e − st F (t ) dt
0
T
2T
3T
0
T
2T
L{F (t )} = ∫ e − st F (t ) dt + ∫ e − st F (t ) dt + ∫ e − st F (t ) dt + ...
Dalam integral pertama : t = u → du = dt
Transformasi Laplace
95
∫e
− st
F (t ) dt
0
1 - e -sT
Dalam integral kedua : t = u + T → du = dt , F(u + T ) = F (u )
Dalam integral ketiga : t = u + 2T → du = dt , F(u + 2T ) = F (u )
.
.
.
Dan seterusnya.
Maka
T
2T
3T
0
T
2T
L{F (t )} = ∫ e − st F (u ) du + ∫ e − s (u +T ) F (u + T ) du + ∫ e − s (u + 2T ) F (u + 2T ) du + ...
T
T
T
0
0
0
L{F (t )} = ∫ e − su F (u ) du + ∫ e − su e − sT F (u ) du + ∫ e − su e − 2 sT F (u ) du + ...
T
T
T
0
0
0
L{F (t )} = ∫ e − su F (u ) du + e − sT ∫ e − su F (u ) du + e − 2 sT ∫ e − su F (u ) du + ...
(
)
T
L{F (t )} = 1 + e − sT + e − 2 sT + ... ∫ e − su F (u ) du
0
T
L{F (t )} =
a
e − su F (u ) du , di mana a = 1 dan r = e − sT
1 − r ∫0
L{F (t )} =
1
e − su F (u ) du
1 − e − sT ∫0
T
T
L{F (t )} =
∫e
F (u ) du
0
1 − e − sT
T
L{F (t )} =
− su
∫e
− st
, di mana u = t
F (t ) dt
0
1 − e − sT
Contoh :
⎧sin t , 0 < t < π
Grafik fungsi F(t ) = ⎨
⎩ 0 , π < t < 2π
Meluas secara periodic dengan perioda 2π .
Carilah L{F (t )}
Solusi :
96
!
Karena T = 2π , maka
T
L{F (t )} =
∫e
F (t ) dt
,
0
1 − e − sT
2π
L{F (t )} =
− st
∫e
− st
F (t ) dt
0
1 − e − s ( 2π )
π
L{F (t )} =
π
0
1− e
π
L{F (t )} =
2π
− st
− st
∫ e sin t dt + ∫ e (0) dt
∫e
0
− st
− 2πs
sin t dt + 0
( )
1 − e −πs
2
⎡ e − st
π⎤
(− s ⋅ sin t − 1⋅ cos t ) ⎥
⎢
2
2
0⎦
(− s ) + 1
L{F (t )} = ⎣
−πs
−πs
1+ e 1− e
(
)(
)
e − sπ
e − s ( 0)
(
)
(− s ⋅ sin 0 − 1⋅ cos 0)
π
π
−
s
⋅
sin
−
1
⋅
cos
−
2
s2 +1
L{F (t )} = s + 1
(1 + e−πs )(1 − e−πs )
e − sπ
e − s ( 0)
(
(
)
)
(− s ⋅ 0 − 1⋅1)
−
s
⋅
0
−
1
⋅
−
1
−
2
2
s
+
1
s
+
1
L{F (t )} =
(1 + e −πs )(1 − e−πs )
e − sπ
1
+ 2
2
L{F (t )} = s +−1πs s +−π1s
1+ e 1− e
(
)(
1 + e −πs
s2 +1
L{F (t )} =
−πs
1 + e 1 − e −πs
(
Transformasi Laplace
)(
)
)
97
L{F (t )} =
1 + e −πs
1
⋅
2
−πs
s + 1 1 + e 1 − e −πs
L{F (t )} =
(1 − e )(s
98
(
)(
1
−πs
2
)
+1
)