Try to calculate the following inverse Laplace transform. ccL^(-1){(1−1e^(−2s))/(s(s+2))}

Kayden Mills

Kayden Mills

Answered question

2022-11-07

Try to calculate the following inverse Laplace transform. L 1 { 1 1 e 2 s s ( s + 2 ) }

Answer & Explanation

Faith Wise

Faith Wise

Beginner2022-11-08Added 17 answers

L 1 { 1 e 2 s s ( s + 2 ) } = L 1 { 1 s ( s + 2 ) e 2 s s ( s + 2 ) } = L 1 { 1 / 2 s 1 / 2 s + 2 e 2 s s ( s + 2 ) } = 1 2 L 1 { 1 / s } 1 2 L 1 { 1 / ( s + 2 ) } L 1 { e 2 s s ( s + 2 ) }
and
L 1 { 1 s ( s + 2 ) } = 1 2 L 1 { 1 / s } 1 2 L 1 { 1 / ( s + 2 ) } = 1 2 ( 1 ) 1 2 ( e 2 t ) = f ( t )
and
L 1 { e a s F ( s ) } = f ( t a ) U ( t a ) ,     a > 0
where, here, a = 2
Jared Lowe

Jared Lowe

Beginner2022-11-09Added 5 answers

You can write
1 1 e 2 s s ( s + 2 ) = 1 2 [ A ( s ) A ( s ) e 2 s ]
with
A ( s ) = 1 s 1 s + 2
So you can find
a ( t ) = L 1 ( A ( s ) ) = 1 + e 2 t
and
L 1 ( A ( s ) e 2 s ) = u ( t 2 ) a ( t 2 )

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