Question

Find the inverse Laplace transform of

$G\left(s\right)=\frac{s}{s^2+2s+2};\sigma >-1$

Answer 1

Given $L^{-1}\left\{\frac{s}{s^2+2s+1+1}\right\}$

$=L^{-1}\left\{\frac{s}{(s+1{)}^2+1}\right\}$

Adding and subtracting $1$ from the numerator, we get

$=L^{-1}\left\{\frac{s+1-1}{(s+1{)}^2+1}\right\}$
Using the property of $L^{-1}\left\{f\right(s)+g(s\left)\right\}=L^{-1}\left\{f\right(s\left)\right\}+L^{-1}\left\{g\right(s\left)\right\}$

$\therefore L^{-1}\left\{\frac{s}{s^2+2s+2}\right\}=L^{-1}\left\{\frac{s+1}{(s+1{)}^2+1}\right\}-L^{-1}\left\{\frac{1}{(s+1{)}^2+1}\right\}$......(i)
Since, we have $L^{-1}\left\{\frac{b}{(s-a{)}^2+b}\right\}=e^{at}\sin bt$
Substitute $a=-1$ and $b=1$, we get
$L^{-1}\left\{\frac{1}{(s+1{)}^2+1}\right\}=e^{-t}\sin t$.......(ii)
Similarly, we have $L^{-1}\left\{\frac{s-a}{(s-a{)}^2+b^2}\right\}=e^{at}cosbt$
Substitute $a=-1$ and $b=1$, we get
$L^{-1}\left\{\frac{s+1}{(s+1{)}^2+1}\right\}=e^{-t}\cos t$.......(iii)

Substitute equation (ii) and (iii) in equation(i),

$\therefore L^{-1}\left\{\frac{s}{s^2+2s+2}\right\}=e^{-t}\cos t-e^{-t}\sin t$