Question

I'm having trouble figuring out how to evaluate a particularly ugly inverse laplace form and need help; posted below in a moment.

Answer 1

\[L^{-1}\left\{Y(s)\right\}= L^{-1} \left\{\vphantom{} \frac{s+3}{[(s+3)^{2}+4](s+1)} \right\}\]

Any immediate ideas? Because I can see this getting ugly real quick. My first thought is to try to expand the denominator and then recombine it as a different set of factors possibly?

Answer 2

are u allowed to use convolution ?

Answer 3

it will be very easy if we use it
take two separe terms\[F_1(s)=\frac{ s+3 }{ (s+3)^2 +4}\\and\\ F_2(s)=\frac{ 1 }{ s+1 }\\\]

Answer 4

Not allowed to use convolution, hee

Answer 5

Very last problem in the section one prior XP

Answer 6

now take the inverse laplace of the first function\[f_1(t)=L^{-1}(F_1(s))=L^{-1}\frac{ s+3 }{ (s+3)^2+4 }=e^{-3t}*L^{-1}\frac{ s }{ s^2+4 }\\=e^{-3t}.\cos(2t)\]

Answer 7

\[f_2(t)=L^{-1}\frac{ 1 }{ s+1 }=sint\\now\\finally ~~f(t)=\int\limits_{0}^{t}f_1(u).f_2(t-u)du\]
this will make it a lot easier
and if u're not allowed then laborious partial fraction
\[\frac{ s+3 }{ (s^2+6s+13)(s+1) }=\frac{ A }{ s+1 }+\frac{ Bs+C }{ s^2+6s+13 }\]
this way-----

Answer 8

Alright, so the partial fraction approach is to bust it up, thank you so much. Just was making sure I wasn't missing something.

I also promise I'll do those convolution problems you posted!

Answer 9

yes try that :)
sole purpose of convolution is to reduce the complexity