The transfer function relating a step input to the output of a control system is:

$$16/(s^3+0.8s^2+16s)$$

The natural frequency $$omega_n$$ of the system and the damping ratio $$zeta$$ are most nearly:

A.   $$omega_n=2" rad/s; "zeta=0.1$$
B.   $$omega_n=2" rad/s; "zeta=0.2$$
C.   $$omega_n=4" rad/s; "zeta=0.1$$
D.   $$omega_n=4" rad/s; "zeta=0.2$$

Page 227 of our FE Reference Handbook provides information on where our natural frequency and damping ratio is contained within our second order control system model:

First, we can factor out the 1/s in our expression since this represents the Laplace of the unit step function. Once we factor this out, we get an expression that resembles the one above.

$$\begin{gathered} F\left(s\right)=\frac{16}{s^3+0.8s^2+16s} \\ F\left(s\right)=\frac{16}{s(s^2+0.8s+16)} \\ F\left(s\right)=\frac{1}{s}·\frac{16}{s^2+0.8s+16} \end{gathered}$$

Our second part of the expression after the multiplication sign now represents the expression which we can derive both our natural frequency and damping ratio from:

$$\begin{gathered} \frac{K{{\omega }_n}^2}{s^2+2\zeta {\omega }_{n}s+{{\omega }_n}^2}=\frac{16}{s^2+0.8s+16} \\ {{\omega }_n}^2 = 16 \\ {\omega }_n=4 \end{gathered}$$

And now solving for damping ratio:

$$\begin{gathered} 2\zeta {\omega }_{n}s = .8s \\ 2\zeta {\omega }_n = .8 \\ \zeta {\omega }_n = .4 \\ \zeta \left(4\right) = .4 \\ \zeta = .1 \end{gathered}$$

Therefore, our natural frequency, ${\omega }_n$, is 4 rad/s while our damping ratio, $\zeta$, is equal to .1. This corresponds to answer C.