Page 94 - LAPLACE TRANSFORM

I(s) = A + (Bs+C)
2
(s + 3) (s + 5)

− ( s+ )

I(s) = +
2
(s+3) (s + 5)

− { 3s+5 }
I(s) = +
2
(s+3) (s + 5)

− 1 (3s+5)
I(s) = { } + {
2
2
(s+3) (s + (√5 ) )
− 1 3s 5
I(s) = { } + { + }
2
2
2
2
(s+3) (s + (√5 ) ) (s + (√5 ) )
− 1 s 5
I(s) = { } + { (3). + }
2
2
2
2
(s+3) (s + (√5 ) ) (s + (√5 ) )

By applying the inverse Laplace transform method;

−
-3t
i(t) = e + { (3). cos √5t + sin √5t }

− -3t
i(t) = e + cos √5t + sin √5t }

-3t
( ) = [ − e + √ + √ ]

Calculate the value of VL (t) when t = 0.1sec

−9 9 3
-3t
i(t) = e + cos √5t + sin √5t
14 14 14

d i(t) −9 -3t 9 3
= (−3) e + ( ) (− √5 ) sin √5t + ( ) √5) cos √5t
dt 14 14 14

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