Question

For the closed-loop system with \(G_p(s) = \frac{14.4}{s(1 + 0.1s)}\) and \(G_c(s) = 1\), the unit-step response shows damped oscillations. The damped natural frequency is \(\underline{\hspace{2cm}}\) rad/s. (Round off to 2 decimal places.)

Hint:

The damped natural frequency is always lower than the undamped natural frequency and depends on the damping ratio \(\zeta\).

Answer 1

Correct Answer: 10.8

Open-loop transfer function: \[ G(s) = \frac{14.4}{s(1 + 0.1s)} \] Closed-loop characteristic equation: \[ 1 + G(s) = 0 \] \[ 1 + \frac{14.4}{s(1 + 0.1s)} = 0 \] \[ s(1 + 0.1s) + 14.4 = 0 \] \[ 0.1s^2 + s + 14.4 = 0 \] Divide by 0.1: \[ s^2 + 10s + 144 = 0 \] Standard second-order form: \[ s^2 + 2\zeta\omega_n s + \omega_n^2 = 0 \] Thus: \[ \omega_n^2 = 144 $\Rightarrow$ \omega_n = 12 \] \[ 2\zeta\omega_n = 10 $\Rightarrow$ \zeta = \frac{10}{24} = 0.4167 \] Damped natural frequency: \[ \omega_d = \omega_n \sqrt{1 - \zeta^2} \] \[ \omega_d = 12 \sqrt{1 - 0.4167^2} \] \[ = 12 \sqrt{0.8264} = 12 \times 0.908 = 10.90\ \text{rad/s} \] This lies in the expected range 10.80 to 11.00 rad/s.