Transformasi Laplace 2.1 RUMUS-RUMUS PEMBUKTIAN TRASFORMASI LAPLACE PDF

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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) didefinisik...

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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: ∞

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 ta

⎩

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...