How to calculate of Second-order system Frequency and Time domain values?

Question

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

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

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

Answer 1

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. 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!

Answer 2

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.