How does the author get from one step to another?

Question

I have to apply convolution theorem to find the inverse Laplace transform of a given function. I know that convolution is applied when the given function is multiplication of two functions.

The question is as follows:
$$ \frac{s^2+2s+3}{(s^2 + 2s+2)(s^2+2s+5)} $$
Now, the author goes on to factorize the top and bottom by using completing the square method and ends up with the following:

$$ \frac{(s+1)^2 + 2}{[(s+1)^2 + 1)][(s+1)^2 + 4]} $$
I do not understand what he does after that because he skips the intermediate steps and states no reasons what so ever. So given the factorized form of the function above, I need to calculate the inverse Laplace using convolution theorem.

I can see that there is a shift. There will be a $e^{-t}$

Answer 1

In addition to the convolution formula and the shift formula, you will also want to know what the Laplace transform of the derivative of a function is (in terms of the Laplace transform of the original function), as well as this: $$ \mathcal{L}(\sin(at)\cdot u(t)) = \frac{a}{s^2 + a^2} $$

Where $u(t)$ is the step function

Answer 2

The essence of solving Laplace transform problem is just a pattern matching. The following is an overkill-point-of-view solution to apply the available concepts as much as possible.

LaTeX code is given as follows for the sake of completeness.

\documentclass[preview,border=12pt]{standalone}
\usepackage{amsmath}
\usepackage{mathrsfs}

\def\Laplace#1{\mathscr{L}\left\{#1\right\}}
\def\InverseLaplace#1{\mathscr{L}^{-1}\left\{#1\right\}}

\begin{document}
Consider the following formulae as we will use it to solve your problem.
\begin{gather}
\Laplace{f(t)}=F(s)\\
\Laplace{e^{at}f(t)}=F(s-a)\\
\Laplace{\int_0^t f(t-\tau)g(\tau)\,\textrm{d}\tau}=F(s)G(s)\\
\int_{-\infty}^{+\infty} f(x)\delta(x-a)\,\textrm{d}x=f(a)\\
\Laplace{1} =\delta(t)\\
\Laplace{\sin \omega t} =\frac{\omega}{s^2+\omega ^2}\\
\Laplace{\cos \omega t} =\frac{s}{s^2+\omega ^2}
\end{gather}
The last form of your problem
\begin{align*}
    \frac{(s+1)^2+2}{\left[(s+1)^2+1\right]\left[(s+1)^2+4\right]} 
    &=
    \frac{1}{(s+1)^2+1}
    \left[
        \frac{(s+1)^2+4}{(s+1)^2+4}-\frac{2}{(s+1)^2+4}
    \right]\\
    &= 
    \frac{1}{(s+1)^2+1}
    \left[
        1-\frac{2}{(s+1)^2+4}
    \right]
\end{align*}
is useful as we can infer the partial inverse.
\begin{align*}
\InverseLaplace{\frac{1}{(s+1)^2+1^2}} &= e^{-t} \sin t\\
\InverseLaplace{1} &= \delta (t)\\
\InverseLaplace{\frac{2}{(s+1)^2+2^2}} &= e^{-t} \sin 2 t
\end{align*}
Thus
\begin{align*}
\InverseLaplace{\frac{(s+1)^2+2}{\left[(s+1)^2+1\right]\left[(s+1)^2+4\right]}}
&=
\int_0^t e^{-(t-\tau)} \sin (t-\tau) \left[\delta(\tau) - e^{-\tau} \sin 2 \tau\right]\,\textrm{d}\tau\\
&=
\int_0^t e^{\tau-t} \sin (t-\tau) \delta(\tau) \, \textrm{d}\tau - e^{-t} \int_0^t  \sin (t-\tau)  \sin 2 \tau \, \textrm{d}\tau\\
&=
e^{-t} \sin t  - \frac{e^{-t}}{3} \left(2\sin t - \sin 2t\right)\\
&=
\frac{e^{-t}}{3} \left(\sin t + \sin 2t\right)
\end{align*}

\end{document}

Answer 3

Consider

$$F(s) = \frac{s^2+2}{(s^2+1)(s^2+4)}$$

We can decompose this fraction as follows:

$$F(s) = \frac{A}{s^2+1} + \frac{B}{s^2+4} = \frac{(A+B) s^2+(4 A+B)}{(s^2+1)(s^2+4)}$$

Thus, $A+B=1$ and $4 A+B=2$ which means that $A=1/3$ and $B=2/3$. Thus, using the fact that the ILT of $1/(s^2+\omega^2)$ is $\sin{(\omega t)}/{\omega}$, we find that the ILT of $F(s)$ is

$$f(t) = \frac13 (\sin{t}+\sin{2 t})$$

To complete the ILT, use the fact that $F(s+1)$ transforms back to $f(t) e^{-t}$, so we simply multply $f(t)$ by $e^{-t}$ and we are done.