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$\frac{0.25}{s^2+0.5s}$

Can I use formulas for 2nd order systems in this pdf?

If not, how can I understand if this sytem is underdamped, damped, etc?

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    $\begingroup$ Welcome to Engineering.SE. Could you add some information about why you think that you can or can't do what you are asking? Also, providing more background information will be helpful to people who aren't currently studying exactly what you are studying. $\endgroup$
    – hazzey
    Commented Jun 11, 2015 at 17:58

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No, you cannot.

As you already guessed, it's very difficult to fit $$\dfrac{0.25}{s^2+0.5s}$$ into the described format $$\dfrac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$ There is simply no value for $\omega_n$ and $\zeta$ where this equation holds.

You probably know that transfer functions in the Laplace (s) domain have a direct relation to differential equations. I come from a mechanical engineering background, so I'll take a mechanics example. The 'second order system' you describe is a model for a basic forced mass-spring-damper: i.e., $m\ddot{x}+d\dot{x}+kx=F$. Here, a dot indicates a derivative.

Image courtesy Wikipedia

Image courtesy Wikipedia. I use $d=B$, since $d$ stands for damper (not sure why anyone would use $B$).

We can take the Laplace transform easily, knowing that a derivative corresponds to a multiplication with $s$. $$mX(s)s^2+dX(s)s+kX(s)=F(s)$$ or, rearranging to get the transfer function $$\dfrac{X(s)}{F(s)}=\dfrac{1}{ms^2+ds+k}$$ Hey, look at that! That's our second order system, but now a physical interpretation.

Now, how is your transfer function different? It has $k=0$! In other words, it does not have a spring, just a damper and a force! You can now easily see that this thing will never oscillate when given a step response - so a natural frequency $w_n$ is silly to describe.

We can translate this mass/spring/damper idea to an inductor (mass), resistor (damper) capacitor (spring) -problem (assuming a circuit in series). Just imagine this circuit without a capacitor - it will not oscillate!

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  • $\begingroup$ To add to what was already said, looking at the numerator you would have $\omega_n^2 = 0.25$ and at the denominator $\omega_n^2 = 0$ so you can see what the problem is... $\endgroup$
    – am304
    Commented Jun 12, 2015 at 8:45
  • $\begingroup$ @am304 I thought that was so obvious that I just said 'There is simply no value...' $\endgroup$
    – Sanchises
    Commented Jun 12, 2015 at 10:25
  • $\begingroup$ yes my comment was more for the OP's benefit $\endgroup$
    – am304
    Commented Jun 12, 2015 at 11:18
  • $\begingroup$ The systems can also be seen as a first order system multiplied with an integrator. For example, the velocity of a DC motor can be written as a first order transfer function and by multiplying it with an integrator you have a transfer function which describes the position. $\endgroup$ Commented Nov 3, 2016 at 21:59
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You can initially consider the system to be $$\frac{\frac{0.25}{\epsilon}\epsilon}{s^2+0.5s+\epsilon}$$

This gives $\omega =\sqrt{\epsilon }$ and $\zeta =\frac{0.25}{\sqrt{\epsilon }}$.

In the limit $\epsilon \to 0$, $\omega \to 0$ and $\zeta \to \infty$. From the latter we conclude that it is damped to the hilt.

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