Question

Find the inverse LaPlace Transform of...

Answer 1

\[\frac{ 2e^{-6s} }{ s^2+4 }\]

Answer 2

\[\frac{ 2 }{ s^2+2^2 }e^{-6s}\]Looks like \[F(s)e^{-cs}\]Need it in the form\[f(t-6)u(t-6)\]

Answer 3

\[f(z)=\sin2z\]\[z=t-6\]\[\sin2(t-6)\]Final Answer\[\sin (2t-12)u(t-6)\]Can somebody confirm?

Answer 4

\[\mathcal{L}^{-1} \left\{ e^{- \sigma s}F(s)\right\}=f(t-\sigma)u(t- \sigma)\]
With:
\[F(s)=\frac{2}{s^2+2^2}; \sigma = 6 \implies f(t)=\sin(2t) \implies f(t-\sigma) = \sin(2(t-6))\]
\[\implies \mathcal{L}^{-1} \left\{ \frac{2 e^{-6t}}{s^2+2^2}\right\}=\sin(2(t-6))u(t-6)\]

Answer 5

Thanks

Answer 6

@malevolence19 I just saw this\[g(t)u(t-a)=e^{-as}\mathcal{L} \left\{ g(t+a)\right\}(s)\]so would the answer be\[\sin(2(t+6))u(t-6)\]

Answer 7

on my test I had this as my answer and it was marked correct.\[\sin(2(t+6))u(t-6)\]In class the teacher got your answer. Can you elaborate on why it is your answer?

Answer 8

Honestly, its just because what you had was in the form of the equation I used. The equation you have written down doesn't quite work because we don't have the laplace transform of sin(2(t+a)), we have the laplace transform of sin(2t). See what I mean?

Answer 9

Ya I see what you mean. Thanks for getting back to me

Answer 10

No problem, run any laplace problems you got by me. We might need to make a new page at some point though because all the latex will cause it to lag up. :)

Answer 11

Thanks. If I have any other questions I will start a new post