Jawaban contoh soal Transformasi Laplace

1. F ( t )=1

L {F ( t )}=∫

0

∞

e−st1=f (s )

= Limp→∞

∫0

p

e−st dt

= limp→∞(−1

se−st)

0

p

= limp→∞(− 1

se−∞+1

se0 )

=0+1s

=1s

=f ( s )

2. F ( t )=t

L {F ( t )}=∫0

∞

e−st t dt

= limp→∞

∫0

p

t .−1sd (e−st )

=−1

slimp→∞

te−st−∫0

p

e−st dt

=−1

slimp→∞(te−st+1

se−st)

o

p

=−1

slimp→∞( pe−sp+1

se−sp)−(0e0+ 1

se0)

0

p

=−

1s ((0+0 )−(0+

1s ))

=−1

s (0−1s )

= 1

s2

3. F ( t )=eat

L {F ( t )}=∫0

∞

e−st t eat dt

= limp→∞

∫0

p

e−( s−a) tdt

= 1s−a

limp→∞

[e−( s−a) t ]0p

=

1−(s−a )

limp→∞( 1

e( s−a)∞−

1

s(s−a )0 )

= 1s−a

4. F ( t )=sinat

L {F ( t )}=∫0

∞

e−st sin at dt

=Limp→∞

∫0

p

e−st−1ad (cos at )

= Limp→∞ (−1

acos at .e−st+∫

0

∞1a

cos atd( e− st ))0

p

=Limp→∞ (−1

acos at .e−st+∫

p

∞

− sa

cos at .e−st dt)0

p

=Limp→∞ (−1

acos at .e−st− s

a∫0

∞

e−st .1ad( sinat ))0

p

=Limp→∞ (−1

acos at .e−st− s

a2( e−st sin at−∫

0

p

sin at .d (e−st ))0

p

=Limp→∞(−1

acos at .e−st− s

a2( e−stsinat−∫

0

p

sin at .−se−st ))0

p

=Limp→∞ (−1

acos at .e−st− s

a2e−st sin at− s

2

a2∫0

p

sin at . se−st ))0

p

=Limp→∞

a2

a2+s2 (−1a

cosat .e−st− sa2

sin at .e−st)0

p

= a2

a2+s2 (−cosata .est

− s .sinata2 .est )

=

a2

a2+s2 ((0−0 )−(−1a−0))

= a2

a2+s2 ( 1a )

= a

a2+s2

5. F ( t )=cos at

L {F ( t )}=∫0

∞

e−st cos at dt

=Limp→∞

∫0

p

e−st1ad (sin at )

=Limp→∞ ( 1

asinat .e−st−∫

0

∞1a

sinatd (e−st ))0

p

=Limp→∞ ( 1

asinat .e−st+∫

p

∞sa

sinat .e−st dt)0

p

=Limp→∞( 1

asinat .e−st+ s

a∫0

∞

e−st .1ad (−cos at ))0

p

=Limp→∞ ( 1

asinat .e−st+ s

a2(e−st (−cosat )−∫

0

p

−cosat .d (e−st ))0

p

=Limp→∞ ( 1

asinat .e−st− s

a2(e−st cosat )+∫

0

p

cos at .−se−st dt ))0

p

=Limp→∞( 1

asin at .e−st− s

a2(e−st cosat )− s

2

a2∫0

p

cosat .e−st ))0

p

=Limp→∞

a2

s2+a2 ( 1a

sin at .e−st− sa2

cosat .e− st)0

p

= a2

s2+a2 (sin ata .est

− s . cosata2 .est )

= a2

s2+a2 ((0−0 )−(0− sa2 ))

= a2

s2+a2 ( sa2 )

= a

s2+a2

6.

L {F ( t )}=∫0

∞

e−stF ( t )dt

= limp→∞

∫0

p

e−st tdt

= limp→∞

∫0

p

t .−1sd ( e− st )

=−1

slimp→∞

te−st−∫0

p

e−st dt

=−

1s

limp→∞[ te−st+ 1

se−st ]

0

p

=−1

s [0−1s ]

= 1

s2

=f ( s )

7. L {5 t−3}=L {5 t−3a}=L {5 t }−L{3}

=5 L {t }−3 L{1}

=5

1

s2−3

1s

= 5

s2−3s

8. L {6 sin 2 t−5 cos2 t}=L {6 sin2 t }−L{5 cos2 t}

=6 L {sin 2t }−5 L{cos2 t}

=6

2

s2+4−5

s

s2+4

=12−5 s

s2+4

9. L {( t2+1)2}=L {t4+2 t2+1}

=L {t4}+L{2 t2}+L {1}

=L {t4}+2L {t2}+L {1}

=

4 !

s4 +1+2( 2 !

s2+1 )+1s

=24

s5+ 4

s3+ 1s

10. L {4e5t+6 t2−3sin 4 t+2cos 2t }

=L {4e5 t}+L {6 t2}−L {3sin 4 t}+L {2cos2 t}

=4 L {e5t }+6 L {t2}−3L {sin 4 t }+2 L {cos2t }

=4

1s−5

+62

s3−3

4

s2+4+2

s

s2+4

= 4s−5

+12

s3−12

s2+16+ 2 s

s2+4

11. Jika L {F ( t )}= 6

(s+2)3=f ( s )

maka L {F (3 t )}=1

3f ( s

3)

= 6

3( s3 +2)3

= 6 . 9

( s+6)3

12.L {sin at }= a

s2+a2=f (s )

Misal F ( t )=sinat diperoleh F ' ( t )=acos at , F ''( t )=−a2sin at

sehingga L {sinat }=− 1

a2L¿¿

Dengan menggunakan sifat transformasi Laplace dari turunan-turunan diperoleh

L {sinat }=(− 1

a2 ) ( sf ( s )−sF (0)−F '(0 )) f

=−

1

a2 (s2 a

s2+a2−s(0 )−a)

=− 1

a2 ( as2s2+a2−a)

=− 1

a2 ( as2−as2−a3

s2+a2 )

= a

s2+a2

13. Tentukan L {t sinat}

Jawab

L {sinat }= a

s2+a2, maka menurut sifat perkalian dari pangkat t

ndiperoleh

L {tF ( t )}=(−1 )n dn f ( s )dsn , sehingga

L {t sinat}=(−1 )dds ( a

s2+a2 )

= 2as

( s2+a2)2

14. Tentukan L {t2cosat }

Menurut sifat di atas, L {t2 cosat }=(−1 )2 d

2

ds2 ( ss2+a2 )

= dds ( a

2−s2

(s2+a2 )2)

=2 s3−6a2 s(s2+a2 )3

15.L−1{3 s−12

s2+9 }=L−1{ 3 s

s2+9 }−L−1{12

s2+9 }

=3 L−1{ s

s2+9 }−12 L−1{ 1

s2+9 }

=3 cos 3t−12sin 3 t

3

16. L−1{ 1

s2−9 }=sinh 3 tt

maka

L−1{ 1

(s2−2 s+13 }=L−1{ 1

(s−2 )2+9 }=e2t sinh 3 t3

17. Tentukan L−1{ 3 s+16

s2−s−6 }

Jawab

L−1{ 3 s+16

s2−s−6 }=L−1{ 3 s+16( s+2 )(s−3 )}

3 s+16( s+2 )(s−3 )

= As+2

+ Bs−3

=A( s−3)+B( s+2 )

s2−s−6

=

( A+B )s+(2B−3 A )s2−s−6

atau A+B = 3 dan 2B-3A = 16 atau 2(3-A)–3A=16 sehingga didapat

A = -2 dan B = 5

L−1{ 3 s+16(s+2)( s−3) }=L−1{−2

s+2+ 5s−3 }

=L−1{ −2

s+2 }+L−1{ 5s−3 }

=−2e−2 t+5e3 t

18. Tentukan L−1{ s−1

(s+3)( s2+2 s+2)}

Jawab

L−1{ s−1

(s+3)( s2+2 s+2)}=L−1{ As+3+

Bs+C(s2+2 s+2 )}

As+3

+ Bs+Cs2+2 s+2

=A( s2+2 s+2)+(Bs+C )( s+3 )

( s+3 )(s2+2 s+2 )

=As2+2 As+2 A+Bs2+(3 B+C )s+3C

( s+3 )(s2+2 s+2 )

Sehingga

{ s−1

(s+3)( s2+2 s+2) }={( A+B )s2+(2 A+3 B+C )s+(2 A+3C )(s+3 )(s2+2 s+2) }

Diperoleh A+B = 0, 2A+3B+C=1, 2A+3C=-1

Atau A = −4

5 , B =

45 , dan C =

15

Akhirnya diperoleh

L−1{ s−1

(s+3)( s2+2 s+2)}=L−1{−45

s+3+

45s+

15

(s2+2 s+2 )}L−1{−4

5s+3

+

45s+

15

( s2+2 s+2)}=−45L−1{ 1

s+3 }+ 45 { (s+1)

(s+1)2+1 }=− 4

5e−3 t+ 4

5e−t cos t