Inverse Laplace Transform via Circuit Analysis [HELP]

Question

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.

Answer 1

Let us denote by $a$ the value $1/\sqrt 2$ to have an easy typing. The the partial fraction decomposition over complex numbers of the given fraction is indeed $$ \begin{aligned} \frac{1}{s^2 + \sqrt{2}s + 1} &= \frac{1}{(s^2 + 2as + a^2)+a^2} = \frac1{2ia}\left(\frac 1{s+a-ia}-\frac 1{s+a+ia}\right) \\ &=\frac K{s+a-ia}-\frac {K^*}{s+a+ia} \ . \\[3mm] \text{Here:} & \\ K &=\frac 1{2ia} =-\frac 1{2a}i =\frac 1{2a}\left(\cos\left(-\frac\pi2\right) +i\sin\left(-\frac\pi2\right)\right) \\ &=a\left(\cos\left(-\frac\pi2\right) +i\sin\left(-\frac\pi2\right)\right)\ , \\ |K| &= a\ ,\\ \theta &=-\frac \pi 2\ . \end{aligned} $$ The inverse Laplace transform is thus using the table: $$ 2ae^{-at}\cos(at+\theta) = 2ae^{-at}\sin(at) = \sqrt 2\cdot e^{-t/\sqrt 2}\sin(t/\sqrt 2) \ . $$

Answer 2

Once we complete the square, we can use the sine formula and Frequency Shift Theorem to evaluate the inverse transform:

If we accept that $$\mathcal{L}(\sin(at)) = \frac{a}{s^2+a^2}$$

and $$\mathcal{L}(e^{ct}f(t)) = F(s-c)$$

where $F(s) = \mathcal{L}(f(t))$, we can take our original fraction:

$\begin{align} \mathcal{L}^{-1}(\frac{1}{s^2+\sqrt{2}s+1}) & = \mathcal{L}^{-1}(\frac{1}{(s+\frac{1}{\sqrt{2}})^2+1/2})\\ & = \mathcal{L}^{-1}(\sqrt{2}\frac{\frac{1}{\sqrt{2}}}{(s+\frac{1}{\sqrt{2}})^2+1/2})\\ & = \sqrt{2}*\exp{\frac{-t}{\sqrt{2}}}*\sin(\frac{t}{\sqrt{2}}) \end{align}$

In that last step, we combined the two formulae above, as our fraction was in the form of $\mathcal{L}(\sin(at))$, but shifted by $c = \frac{-1}{\sqrt{2}}$, creating the '$\exp{\frac{-t}{\sqrt{2}}}$' term in the final answer.

Were you to continue the partial fraction decomposition method directly, you would end up with two exponentials terms that you could manipulate into the same answer above using the identity:

$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$

Answer 3

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