clearly not, since ℒ(1 - cos(6t)) = 36/(s(s^2+6^2))
and the s^2+9 should have been a clue to look for sin/cos (3t), not 6t.
I get a result of 1/6 t sin(3t)
see whether you can find where you went astray.
Question:
How do we find the inverse laplace transformation of
s/[ ((s^2) + 9)^2 ]
My attempt at the question:
I noted that s/[ ((s^2) + 9)^2 ] can be written as follows:
s/[ ((s^2) + 9)^2 ] ={ s/[(s^2) + 9] }* { 1/[(s^2) + 9] }
By further simplying, I got,
s/[ ((s^2) + 9)^2 ] =(1/3) * { s/[(s^2) + 9] }* { 3/[(s^2) + 9] }
The RHS now seems to be a multiplication of two functions F(S) and G(S) such that
F(S) = s/[ ((s^2) + 9)^2 ]
and
G(S) = 1/[(s^2) + 9], which gives,
s/[ ((s^2) + 9)^2 ] = (1/3)* F(S) * G(S)
Now ILT(Inverse Laplace Transform) of s/[ ((s^2) + 9)^2 ] = ILT[ (1/3)* F(S) * G(S) ] = (1/3) * ILT [ (F(S)*G(S) ]
From the Convolution theorem, we know that,
ILT( F(S) * G(S) ) = integration from 0-t { f(u)*g(t-u) du } , where f(t) = ILT(F(S)) and g(t) = ILT(G(S))
Hence,
ILT of s/[ ((s^2) + 9)^2 ] = -cos(6t) | limits from 0 - t,
which gives the answer as, 1 - cos(6t)
Is this approach correct?
1 answer