## Control System Model

Consider a control system's transfer function, which relates a step input to output, is 25/(s^3+0.6s^2+25s). Estimate the natural frequency, steady-state gain, and damping ratio.

Hint
$$\frac{Y(s)}{R(s)}=\frac{K\omega _{n}^{2}}{s^2+2\zeta\omega_ns+\omega _{n}^{2} }$$$where $$K$$ is the steady state gain, $$\zeta$$ is the damping ratio, and $$\omega_n$$ is the undamped natural frequency ( $$\zeta=0$$ ). Hint 2 In the second order form, our transfer function can be written as: $$\frac{Y(s)}{R(s)}=\frac{25}{s^3+0.6s^2+25s}$$$
One standard second-order control system model is
$$\frac{Y(s)}{R(s)}=\frac{K\omega _{n}^{2}}{s^2+2\zeta\omega_ns+\omega _{n}^{2} }$$$where $$K$$ is the steady state gain, $$\zeta$$ is the damping ratio, and $$\omega_n$$ is the undamped natural frequency ( $$\zeta=0$$ ), In the second order form, our transfer function can be written as: $$\frac{Y(s)}{R(s)}=\frac{25}{s^3+0.6s^2+25s}=\frac{25}{s(s^2+0.6s+25)}$$$
To find the natural frequency, $$\omega_n$$ , we know $$\omega_n^{2}=25$$ , based on the transfer function format. Thus,
$$\omega_n=\sqrt{25}=5$$$To find the steady-state gain, $$K$$ , we know $$K\omega _{n}^{2}=25$$ , based on the transfer function format. Thus, $$K=\frac{25}{\omega _{n}^{2}}=\frac{25}{5^2}=\frac{25}{25}=1$$$
To find the damping ratio, $$\zeta$$ , we know $$2\zeta\omega_n=0.6$$ , based on the transfer function format. Thus,
$$\zeta=\frac{0.6}{2\omega_n}=\frac{0.6}{2(5)}=\frac{0.6}{10}=0.06$$\$
$$\omega_n=5$$ , $$K=1$$ , $$\zeta=0.06$$