Inverse Laplace Transform via Circuit Analysis [HELP] Inverse Laplace Transform $\frac{1}{s^2 + \sqrt{2}s + 1}$
so what I did it changed the denominator to complete the square format which is $\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}$, then I can solve for $s$, it will make it as
$$
\left(\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i\right)
\left(\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i\right)
$$
So now, to the sheet of paper is to do Partial Fraction Decomposition of this which is absurd to me because of complex roots it has:
$$
\frac{1}{s^2 + s\sqrt{2} + 1}
= \frac{1}{\left(s+\frac{\sqrt{2}}{2}\right)^2 + \frac{1}{2}}
$$
Partial Fraction of Complex root will be
$$
\frac{K}{\left(s+ \frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2}i}
 + \frac{K^*}{\left(s+ \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2}i}
$$
to follow the formula sheet.
which I got my K = -$i\frac{\sqrt{2}}{2}$ and $K^*$ = $i\frac{\sqrt{2}}{2}$
the problem I get is magnitude and $\theta$ is undefined it makes no sense at all.


You may also use Contour Integration and Residue Theorem so as to find the Inverse of any given function $F(s)$. First, we must know the Inverse Laplace Transform:
$$\mathcal{L}^{-1} \{F(s)\} (t) = f(t) = \frac{1}{2 \pi i} \cdot \int_{\gamma-i \infty}^{\gamma + i \infty} e^{st}F(s) ds $$
for any real $\gamma$ on the right of all poles of $F(s)$. So as to calculate this, we may define two semi-circle closed paths: $C_l$) The left one which must possess all singularities of $F(s)$ and converges for $t>0$; $C_r$) the right converges for $t<0$ and is zero since $e^{st}F(s)$ is analytic (But we do not want this since we assume that the Laplace Transform is well-behaved for $t>0$).
From Contour Integration and Residue Theorem, we have:
$$\int_{C_l} e^{st}F(s) ds = \int_{semi-circle} e^{st}F(s)ds + \int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F(s)ds = 2\pi i \cdot \sum_{i=1}^k \{ \text{Res} (e^{st}F(s), s_i)\}$$
In the second path integral we perform this substitution:
$$s = Re^{i \theta}, \,\, ds = iRe^{i \theta}d \theta$$
Since the third path is a straight line from $-i \infty$ to $i \infty$ at $\Re(z)=\gamma$, the semi-circle's radius will be infinite and we'll integrate in respect to $\theta \in [\frac{\pi}{2}, \frac{3\pi}{2}]$. Thus, we take the limit
$$\int_{semi-circle} e^{st}F(s)ds = \lim_{R \to \infty} \left[ \int_{\pi/2}^{3\pi/2} e^{tRe^{i \theta}}F(Re^{i \theta})i Re^{i \theta} d\theta\right]$$
Now, use Euler's Formula $e^{ix} = \cos(x) + i \sin(x)$ to transform $e^{i \theta}$ and then $e^{itR\sin(\theta)}$. After that, we then have
$$\int_{semi-circle} e^{st}F(s)ds = \lim_{R \to \infty} \left[ i \int_{\pi/2}^{3\pi/2} e^{i \theta} \cdot \underbrace{ \frac{(\cos(tR\sin(\theta)+ i \sin(tR\sin(\theta))}{ e^{-tR cos(\theta)}  } }_{0} \cdot F(Re^{i \theta}) R \, d\theta\right]$$
The underbraced expression goes to zero because: $t>0$, $\cos( \theta) <0 $ and the denominator oscillates between two finite values. In this case, $F(Re^{i \theta}) R$ goes to zero aswell, therefore the limit does exist and the integral is zero:
$$\int_{semi-circle} e^{st}F(s)ds = \lim_{R \to \infty} \left[ i \int_{\pi/2}^{3\pi/2} e^{i \theta} \cdot \underbrace{ \frac{(\cos(tR\sin(\theta)+ i \sin(tR\sin(\theta))}{ e^{-tR cos(\theta)}  } }_{0} \cdot \underbrace{F(Re^{i \theta}) R}_{0} \, d\theta\right] = 0$$
As a result, we have:
$$\int_{C_l} e^{st}F(s) ds = 0 + \int_{\gamma-i\infty}^{\gamma+i\infty}e^{st}F(s)ds = 2\pi i \cdot \sum_{i=1}^k \{ \text{Res} (e^{st}F(s), s_i)\}$$
We will sum the residues of $e^{st}F(s)$. In fact just $F(s)$, because the exponential is analytic on $\mathbb{C}$, therefore no poles. Since the poles are not repeated, we may calculate the residue as follows
$$\text{Res}(f(z), z_0) = \lim_{z \to z_0 \text{(pole)}} [ (z-z_0) f(z)]$$
Here the poles are the values that make $s^2 + \sqrt2 s +1 = 0$. They are:
$$s_{1,2} = - \frac{1}{\sqrt2} \pm \frac{1}{\sqrt2}$$
Thus, the residues are:
$$\text{Res}(e^{st}F(s), s_1) = \frac{e^{-\frac{t}{\sqrt2}+\frac{it}{\sqrt2}}}{i \sqrt2}\\
\text{Res}(e^{st}F(s), s_1) = \frac{e^{-\frac{t}{\sqrt2}-\frac{it}{\sqrt2}}}{-i \sqrt2} $$
Then,
$$ \mathcal{L}^{-1} \{F(s)\} (t) = f(t) = \frac{1}{2 \pi i} \cdot 2\pi i  (e^{-\frac{t}{\sqrt2}} \sqrt2 \sin(t/ \sqrt2) )\\
\mathcal{L}^{-1} \{F(s)\} (t) = f(t) = e^{-\frac{t}{\sqrt2}} \sqrt2 \sin(t/ \sqrt2) 
 $$
You may find other results on Wikipedia and this series will be helpful: https://www.youtube.com/watch?v=iUhwCfz18os&list=PLdgVBOaXkb9CNMqbsL9GTWwU542DiRrPB
