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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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