Appendix-settling and peak time

T_s and T_p

$$G(s) = \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2} \\ = \frac{\omega_n^2}{(s+\zeta\omega_n + j \sqrt{1-\zeta^2}\omega_n)(s+\zeta\omega_n - j \sqrt{1-\zeta^2}\omega_n)}\\ = \frac{A}{s+\zeta\omega_n + j \sqrt{1-\zeta^2}\omega_n} - \frac{A}{s+\zeta\omega_n - j \sqrt{1-\zeta^2}\omega_n}$$

here

$$-2A j \sqrt{1-\zeta^2}\omega_n = \omega_n^2 \\ A = \frac{j\omega_n}{2\sqrt{1-\zeta^2}}$$

So

$$G(s) = \frac{j\omega_n}{2\sqrt{1-\zeta^2}} [\frac{1}{s+\zeta\omega_n + j \sqrt{1-\zeta^2}\omega_n} - \frac{1}{s+\zeta\omega_n - j \sqrt{1-\zeta^2}\omega_n}]\\ g(t) = \frac{j\omega_n}{2\sqrt{1-\zeta^2}} [e^{-(\zeta\omega_n + j \sqrt{1-\zeta^2}\omega_n)t} -e^{-(\zeta\omega_n - j \sqrt{1-\zeta^2}\omega_n)t}]\\ = \frac{j\omega_n}{2\sqrt{1-\zeta^2}}e^{-\zeta\omega_n t} [-2j\sin(\sqrt{1-\zeta^2}\omega_n t)]\\ = \frac{\omega_n}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_n t} \sin(\sqrt{1-\zeta^2}\omega_n t)$$

So the output:

$$Y(s) = \frac{1}{s} G(s)\\ y(t) = u(t) * g(t) = \int^t_{0_-} g(\tau) d\tau\\ \mathbb{L}[y'(t)] = sY(s) - y(0_-) = sY(s) = G(s)$$

So we have

$$y'(t) = g(t) = \frac{\omega_n}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_n t} \sin(\sqrt{1-\zeta^2}\omega_n t)$$

Now we want to calculate y(t), consider

$$\int^t_0 e^{-a\tau}\sin(b\tau) d\tau\\ = \int^t_0 \text{Im}[e^{(-a+jb)\tau}] d\tau\\ = \text{Im}[ \int^t_0 e^{(-a+jb)\tau} d\tau ]\\ = \text{Im}[ \frac{1}{a-jb} [1 - e^{(-a+jb)t}] ]\\ = \text{Im}[ \frac{a+jb}{a^2 + b^2} [1-e^{-at}\cos(bt) -je^{-at}\sin(bt)] ]\\ = \frac{1}{a^2+b^2} [-ae^{-at}\sin(bt) + b - be^{-at}\cos(bt)]\\ = \frac{b}{a^2+b^2} - \frac{e^{-at}}{a^2+b^2} \sqrt{a^2+b^2}\sin(bt+\phi)$$

Here

$$\tan(\phi) = \frac{b}{a}$$

in all

$$y(t) = \int^t_{0_-} g(\tau) d\tau\\ = \frac{\omega_n}{\sqrt{1-\zeta^2}} \int^t_{0_-} e^{-\zeta\omega_n \tau} \sin(\sqrt{1-\zeta^2}\omega_n \tau) d\tau\\ = \frac{\omega_n}{\sqrt{1-\zeta^2}} [ \frac{\sqrt{1-\zeta^2}\omega_n}{\omega_n^2} - \frac{e^{-\zeta\omega_n t}}{\omega_n} \sin(\sqrt{1-\zeta^2}\omega_n t + \phi)]\\ = 1 - \frac{e^{-\zeta\omega_n t}}{\sqrt{1-\zeta^2}} \sin(\sqrt{1-\zeta^2}\omega_n t + \phi)$$

Here

$$\tan(\phi) = \frac{\sqrt{1-\zeta^2}}{\zeta}\\ \sin(\phi) = \sqrt{1-\zeta^2}\\ \cos(\phi) = \zeta$$

Calculate T_s

For T_s

$$T_s = \frac{4}{\zeta \omega_n}\\ e^{-\zeta\omega_n T_s} = e^{-4} \approx 0.02$$

Calculate overshoot and T_p

set y’(T_p) = 0

$$y'(t=T_p) = \frac{\omega_n}{\sqrt{1-\zeta^2}}e^{-\zeta\omega_n T_p} \sin(\sqrt{1-\zeta^2}\omega_n T_p) = 0\\ \sqrt{1-\zeta^2}\omega_n T_p = \pi\\ T_p = \frac{\pi}{\sqrt{1-\zeta^2}\omega_n}$$

here we have

$$1 + \% = y(t=T_p) \\ = 1 - \frac{e^{-\zeta\omega_n T_p}} {\sqrt{1-\zeta^2}} \sin(\sqrt{1-\zeta^2}\omega_n T_p + \phi)\\ = 1 - \frac{e^{-\zeta\omega_n T_p}} {\sqrt{1-\zeta^2}} \sin(\pi + \phi)\\ = 1 + \frac{e^{-\zeta\omega_n T_p}} {\sqrt{1-\zeta^2}} \sqrt{1-\zeta^2}\\ = 1 + e^{-\zeta\omega_n T_p} = 1 + e^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}}$$

So, we have

$$\% = e^{-\pi \frac{\zeta}{\sqrt{1-\zeta^2}}}\\ -\frac{\ln(\%)}{\pi} = \frac{\zeta}{\sqrt{1-\zeta^2}}\\ \frac{-\ln(\%)}{\sqrt{\pi^2 + \ln^2(\%)}} = \zeta$$