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I am having some troubles deriving the formula for the roots for different types systems.. I am not quite sure if they are correct (pretty sure they aren't).

$y(s) = \frac{s+2\zeta\omega_n}{s^2 + 2\zeta\omega_ns +\omega_n^2}$

I don't see how i should derive them for different scenarios

In my book it is stated that for an underdamped system $\zeta < 1$ is the roots S1,2 = $-\zeta\omega_n \pm i\omega_n \sqrt{1-\zeta^2}$ which i don't get.

I've derived it to be $\frac{-2\zeta\omega_n \pm \sqrt{4\zeta^2\omega_n^2- 4\omega_n^2}}{2}$

I don't see how the hell they got the other one, and how they are for the other cases!!??

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Well, just look at the root!

$$\frac{-2\zeta\omega_n \pm \sqrt{4\zeta^2\omega_n^2- 4\omega_n^2}}{2} =\frac{-2\zeta\omega_n \pm 2\omega_n\sqrt{\zeta^2- 1}}{2}$$ And that is: $$-\zeta\omega_n \pm i\omega_n\sqrt{1-\zeta^2}$$ Elaboration: $$\sqrt{4\zeta^2\omega_n^2- 4\omega_n^2}=\sqrt{4\omega_n^2(\zeta^2-1)}=$$ $$\sqrt{4\omega_n^2}\sqrt{\zeta^2-1}=2\omega_n\sqrt{\zeta^2-1}$$ Different scenarios:

  • $\zeta=0$ no dampening, formula $S_{12}=i\omega_n$
  • $\zeta<1$ dampening, solution is complex, oscillatory
  • $\zeta=1$ critical dampening, the formula will be $S_{12}=-\zeta\omega_n$
  • $\zeta>1$ overdamped, two different real roots, will take longer to converge than criticaly dampened, formula $S_{12}=-\zeta\omega_n \pm \omega_n\sqrt{\zeta^2-1}$
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  • $\begingroup$ Well.. that makes sense, but how come for the other cases, they do not exist in my text book.. $\endgroup$
    – newbiemath
    Apr 18, 2014 at 13:09
  • $\begingroup$ The other cases arise when you put $\zeta$=1 for crit dampened, and $\zeta$>1 for the overdamping. And well I forgot to put the imaginary unity in front of it. $\endgroup$
    – WalyKu
    Apr 18, 2014 at 13:14
  • $\begingroup$ They are just cases, $\zeta$ is the dampening factor. For different $\zeta$ you will get different behaviour. The easiest way to see this is by putting the values of $\zeta$ for the different scenarios in the formula, and looking how it changes. $\endgroup$
    – WalyKu
    Apr 18, 2014 at 13:22
  • $\begingroup$ Yes.. I know that the other cases arrives for different values of the damping ratio, but do they have a formula like the underdamped here? and $i$ is stil missing from the $4\omega_n^2$ $\endgroup$
    – newbiemath
    Apr 18, 2014 at 13:22
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    $\begingroup$ Ahh... ofCourse, $\sqrt{\zeta^2 - 1} =>\sqrt{-1(1 -\zeta^2)} => \sqrt{-1} \sqrt{(1 -\zeta^2)} $ $\endgroup$
    – newbiemath
    Apr 18, 2014 at 13:40

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