Page 86 - LAPLACE TRANSFORM
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20 50
I(s) [100 + ] =
s s
100s 20 50
I(s) [ + ] =
s s s
100s + 20 50
I(s) [ ] =
s s
50 s
I(s) = x [ ]
s 100s+20
50 50 1
I(s) = = 1 = 1
100 +20 100( + ) 2 ( + )
5 5
1
2
I(s) = 1
( + )
5
By applying the inverse Laplace transformation method;
1 1
i(t) = ℒ −1 [ ]
2 1
(s + )
5
−
=
When t = 0.3 seconds;
1 1 − (0.3)
1
1 − t
( ) = e 5 = e 5
2 2
1
= ( )(0.941) = .
2
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