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Encontrar la transformada inversa de F(s)

F (S)= 10s (s 2 +2 s+18 )

10

s (s2 +2 s +18 )= A

s + Bs+C

s2 +2 s+18

A= 10

s2 +2 s +18 | s= 0 = 5910 =( 59 s + Bs+C s 2 +2 s+18 )(s)(s 2 +2 s+18 )10 =

5

9(s 2 +2 s+18 )+B s2 +Cs

10 =(59 +B)s 2 +(109 +C )s+105

9+B= 0

B= 5

9

10

9+C = 0

C = 10

9

F (s)= 59 s

59 ( s+2s 2 +2 s+18 )

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F ( s)=5

9 s

5

9 ( s+ 2(s + 1 )2 + 17 ) F (s)=

5

9 s

5

9 [ s+1

(s+1 )2

+17 ]

5

9 [ 1

(s+1 )2

+17 ] F (s)= 5

9 s

5

9 [ s+1(s+1 )2 +( 17 )2 ] 59 17 [ 17(s +1 )2 +( 17 )2 ]f (t )= 5

9

5

9e t cos 17 t

5

9 17e t Sen 17 t

f (t )=5

9 5

9 et

(cos 17 t 1

17 Sen 17 t )

Dibujar la funcin f (t )

f (t )= 59

5

9e t ( 11.03 cos 17 t 1 17 1.03 Sen 17 t )1.03

f (t )= 59

5

9e t (cos (13.63 )cos 17 t Sen (13.63 )Sen 17 t )1.03

f (t )= 59

0.57 e t cos (13.63 )(cos 17 t 13.63 )

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