I know what the solution is to this inverse Laplace transform, I just have NO idea how to get there.
L−1(16s(s2+4)2)
Basically, my question is what modification do I have to do to the equation above?
Answer
Using a table, note the form:
f(t)=tsin(at)
F(s)=2as(s2+a2)2
Using this:
f(t)=L−1(16s(s2+4)2)=L−1(2∗2∗s(s2+22)2∗4)=4tsin(2t)
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