Saturday, 25 January 2014

Tough inverse Laplace transform



I know what the solution is to this inverse Laplace transform, I just have NO idea how to get there.



L1(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)=L1(16s(s2+4)2)=L1(22s(s2+22)24)=4tsin(2t)


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...