Question

I'm trying to use s-axis translation properties to evaluate an inverse laplace transform, and I got a solution, but I'm missing part of the book's soln and can't figure out why; could somebody take a look at this?

Answer 1

\[L^{-1} \left\{\vphantom{} \frac {s}{s^2+4s+5} \right\}\]

Answer 2

\[L^{-1} \left\{\vphantom{} \frac{s}{s^2+4s+5} \right\} = L^{-1} \left\{\vphantom{} \frac {s}{(s+2)^2+1} \right\}\]

Answer 3

Fitting the general form of an inverse laplace transform of cosine(kt), \[L^{-1} \left\{\vphantom{} \frac {s}{s^2 + k^2} \right\}\]...whoop, wait, it's right there, I see it, nevermind. The numerator has to be subtracted something to fit that same form of \[s \pm c\] in order to be dealt with in that way? Wait, maybe?

Answer 4

\[L^{-1} \ Y(s) = L^{-1} \left\{\vphantom{} \frac{s}{(s+2)^2+1} \right\};\]
\[L^{-1} Y(s) +L^{-1} \left\{\vphantom{} \frac{2}{s^2+4s+5} \right\} = L^{-1} \left\{\vphantom{} \frac{s}{s^2+4s+5} \right\} +L^{-1} \left\{\vphantom{} \frac{2}{s^2+4s+5} \right\}\]
\[L^{-1} Y(s) +L^{-1} \left\{\vphantom{} \frac{2}{s^2+4s+5} \right\} =L^{-1} \left\{\vphantom{} \frac{s+2}{s^2+4s+5} \right\} \]
\[L^{-1} Y(s)=L^{-1} \left\{\vphantom{} \frac{s+2}{s^2+4s+5} \right\} -L^{-1}\left\{\vphantom{} \frac{2}{s^2+4s+5} \right\} \]

Answer 5

Okay so far. Next step would be to use the completed-square form of each denominator.

Answer 6

\[F(s+2)=\frac{s+2}{(s+2)^2+1}~~\implies~~F(s)=\frac{s}{s^2+1}\]
Use the fact that
\[\mathcal{L}^{-1}\{F(s-c)\}=e^{-cs}f(t)\]

Answer 7

Yep, taking a shot now:

Answer 8

http://i.imgur.com/XmHIdo1.png This look right?

Answer 9

Yes that's correct.