Second-order transducer frequency and step response

Given: $\omega_n = 1000$ rad/s, $\zeta = 0.7$.

(a) Frequency response at 400 rad/s

Frequency ratio $r = \omega/\omega_n = 0.4$.

$$M = \frac{1}{\sqrt{(1 - r^2)^2 + (2\zeta r)^2}} = \frac{1}{\sqrt{(0.84)^2 + (0.56)^2}} = \frac{1}{\sqrt{0.7056 + 0.3136}} = 0.9905$$ $$\phi = -\tan^{-1}\left(\frac{2\zeta r}{1 - r^2}\right) = -\tan^{-1}\left(\frac{0.56}{0.84}\right) = -33.7^\circ$$

The amplitude error is only 0.95%.

(b) Overshoot

$$M_p = \exp\left(\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}\right) = \exp\left(\frac{-\pi(0.7)}{0.7141}\right) = \exp(-3.079) = 0.0460 = 4.60\%$$

(c) Time-domain parameters

Damped frequency:

$$\omega_d = \omega_n\sqrt{1-\zeta^2} = 1000(0.7141) = 714.1\ \text{rad/s}$$

Peak time: $t_p = \pi/\omega_d = 4.40$ ms.

Rise time (0 to 100%): $t_r = \dfrac{\pi - \cos^{-1}\zeta}{\omega_d} = \dfrac{3.1416 - 0.7954}{714.1} = 3.29$ ms.

2% settling time: $t_s \approx \dfrac{4}{\zeta\omega_n} = \dfrac{4}{700} = 5.71$ ms.

Answer: $M = 0.9905$, lag $= 33.7^\circ$, overshoot $= 4.6\%$, $\omega_d = 714$ rad/s, $t_p = 4.40$ ms, $t_r = 3.29$ ms, $t_s = 5.71$ ms.