How can I find ccL^(-1){(1)/(1+2e^(−s)+e^(−2s))}?

taumulurtulkyoy 2022-10-14 Answered
Inverse Laplace transform of 1 1 + 2 e s + e 2 s
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Answers (1)

honejata1
Answered 2022-10-15 Author has 10 answers
I guess that you would like to find a function g(t) such that h ( s ) g ( s ) = δ ( t ) (where ∗ means convolution), where h ( t ) = δ ( t ) + 2 δ ( t 1 ) + δ ( t 2 )
After taking the Laplace Transform of both parts we get H G = L { δ } = 1 (where ⋅ means multiplication) . Hence, we need to compute the inverse Laplace transform of
G ( s ) = 1 H ( s ) = 1 1 + 2 e s + e 2 s = 1 ( 1 + e s ) 2 = n = 0 ( n + 1 ) ( 1 ) n e n s .
By recalling that L 1 ( e n s ) = δ ( t n ), we obtain
g ( t ) = n = 0 ( n + 1 ) ( 1 ) n δ ( t n ) .
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