enter image description hereInverse 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 (s+$\frac{\sqrt{2}}{2})^2$ + $\frac{1}{2}$ , then I can solve for s, it will make it as ((s+ $\frac{\sqrt{2}}{2}$) + $\frac{\sqrt{2}}{2}j$)((s+ $\frac{\sqrt{2}}{2}$) - $\frac{\sqrt{2}}{2}j$)

So now, according to my professor and the sheet of paper is to do Partial Fraction Decomposition of this which is absurd to me because of complex roots it has:

$\frac{1}{s^2 + \sqrt{2}s + 1}$

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

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

Partial Fraction of Complex root will be

= $\frac{K}{(s+ \frac{\sqrt{2}}{2}) + \frac{\sqrt{2}}{2}j} $ + $\frac{K^*}{(s+ \frac{\sqrt{2}}{2}) - \frac{\sqrt{2}}{2}j} $

right here I am stuck and don't know what to do.


1 Answer 1


To solve this issue, you need to be familiar with Euler's formula: $$\sin(wt) = \frac{e^{jwt}-e^{-jwt}}{2j}$$ You definitely started on the right track with this issue, only finding the right value for $K$ and $K^*$ Might be hard. So lets continue from there: $$K\left(s+\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}j\right)+K^*\left(s+\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}j\right) = 1$$ $$Ks+K^*s = 0 \rightarrow K = -K^*$$ $$K\left(\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}j\right)+K^*\left(\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}j\right) = 1$$ $$-K\sqrt{2}j = 1$$ $$K = \frac{1}{\sqrt{2}}j = \frac{1}{2}\sqrt{2}j \rightarrow K^*=-\frac{1}{2}\sqrt{2}j$$ Right, I suppose you know this inverse Laplace transform rule: $$\frac{1}{s-a} \rightarrow e^{at}$$ Let substitute the found expression in it: $$\frac{1}{2}\sqrt{2}je^{(-\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}j)t}-\frac{1}{2}\sqrt{2}je^{(-\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}j)t}$$ $$\frac{1}{2}\sqrt{2}je^{-\frac{1}{2}\sqrt{2}t}\left(e^{-\frac{1}{2}\sqrt{2}jt}-e^{\frac{1}{2}\sqrt{2}jt}\right)$$ Next, it might be useful to know that: $$j = \frac{1}{-j}$$ So substituting that in the equation leads to: $$\sqrt{2}e^{-\frac{1}{2}\sqrt{2}t}\left(\frac{e^{\frac{1}{2}\sqrt{2}jt}-e^{-\frac{1}{2}\sqrt{2}jt}}{2j}\right)$$ As you can see, Euler's formula neatly fits in here. resulting in the final answer: $$\sqrt{2}e^{-\frac{1}{2}\sqrt{2}t}\sin(\frac{1}{2}\sqrt{2}t)$$

This might be considered the hard way, explaining every step only using one of the most basic inverse Laplace transforms. However, for the future, I suggest you add these inverse transforms to your list as well: $$Ke^{at}\sin(bt) = \frac{Kb}{(s-a)^2+b^2}$$ $$Ke^{at}\cos(bt) = \frac{K(s-a)}{(s-a)^2+b^2}$$ As the damped periodic response is a very normal physical response, so expect to see a lot of those.

EDIT: to append to my earlier answer, I see you already had the correct formula at hand (albeit slightly more elaborate). The goal is to find $K$ and its complex conjugate $K^*$. As shown in your table. the magnitude of $K$ and its angle $\angle K$ are used to describe the magnitude of the (undamped) oscillation and the phase angle. These can be easily found as follows: $$|K| = \sqrt{K\cdot K^*}$$ $$\theta = \arctan\left(\frac{\mathfrak{Im}(K)}{\mathfrak{Re}(K)}\right)$$ Since in this case $K$ is pure imaginary, its easy to see that $|K| = 0.5\sqrt{2}$ and $\theta = -0.5\pi$.

  • $\begingroup$ I solved this way, but let me post what they want in my post, let me re-edit. I solved it this way at first until after the instruction. $\endgroup$
    – EM4
    Commented Aug 5, 2020 at 15:10
  • $\begingroup$ I re-edit my post with the formulas they want me to use, sorry about the vague.. $\endgroup$
    – EM4
    Commented Aug 5, 2020 at 15:14
  • $\begingroup$ I edited my post containing the probably lacking detail. If anything remains unanswered, feel free to ask $\endgroup$
    – Petrus1904
    Commented Aug 5, 2020 at 15:35
  • $\begingroup$ how you got |K| and $\theta$ ? $\endgroup$
    – EM4
    Commented Aug 5, 2020 at 16:15
  • 1
    $\begingroup$ you forgot the square root for $|K|$ and my mistake for $\theta$ you need to use the atan2 function. you cannot divide by the real part of $K$ as it is 0. as for the atan2, its a more elaborate arctangent that completely describes the entire unit circle: en.wikipedia.org/wiki/Atan2 $\endgroup$
    – Petrus1904
    Commented Aug 6, 2020 at 17:59

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