Inactive Tutor answered 07/20/20
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L-1{e-ss2/(s2 + 5)} = L-1{e-s(1 - 5/(s2 + 5)} = δ(t - 1) - √5u(t -1)sin(√5(t - 1)).
Answer is D.)
Fizaa A.
asked 07/19/20The Inverse Laplace Transform of
is:
A.)
B.)
C.)
D.)
Inactive Tutor answered 07/20/20
L-1{e-ss2/(s2 + 5)} = L-1{e-s(1 - 5/(s2 + 5)} = δ(t - 1) - √5u(t -1)sin(√5(t - 1)).
Answer is D.)
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