Derivation of 2nd order damped rise time

Question

Given the transfer function: $$frac{omega_0}{s^2 + 2omega_0zeta s + omega_0^2}$$
How would you derive the rise time of the system from 0% SS to 100% SS.
$$t_r = frac{pi - arccos{frac{1}{zeta}}}{omega_0 sqrt{1 - zeta^2}}$$
I attempted to obtain this by using the solution to the 2nd order ODE: $$y(t) = A exp{(-omega_0zeta t)} cos{(omega_0 sqrt{1-zeta^2}t)}$$ from $dot{y}(0) = 0$, $y(0) = -A$ and, $y(t_r) = 0$, where $A$ is the steady state value, however, the solution is obviously incorrect.

hun · Accepted Answer

I realise now that the solution was wrongly assumed, using the BCs you would get $$y(t) = -A  exp{(-omega_0zeta t)}( cos{(omega_0 sqrt{1-zeta^2}t)} + frac{zeta}{sqrt{ 1-zeta^2 } }sin{(omega_0 sqrt{1-zeta^2}t)}) $$
Leading to
$$frac{sqrt{1-zeta^2}}{zeta} + tan{omega_0 sqrt{1-zeta^2} t_r} = 0$$
where $ arctan{frac{sqrt{1-zeta^2}}{zeta}} equiv arccos{frac{1}{zeta}}$
Then $$omega_0 sqrt{1 - zeta^2} t_r = pi - arctan{ frac{sqrt{1 - zeta^2}}{zeta} }$$ and the rest is trivial.

Derivation of 2nd order damped rise time

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