Rise Time

An underdamped response in a system passes through the steady state value twice per oscillation. The time at which this first happens is known as the rise time $t_r$. This provides an indication on how quickly the system initially responds to an input, if the output is equal to the steady state value at rise time we can state:

$$x(t_r)=x(\infty )$$ $$\begin{array}{rl}\frac{Ka}{{\omega }_n^2}& =\frac{Ka}{{\omega }_n^2}\left\{1-e^{-\zeta {\omega }_n{t}_r}\left[\cos {\omega }_d{t}_r+\frac{\zeta }{\sqrt{1-{\zeta }^2}}\sin {\omega }_d{t}_r\right]\right\}\\ \Rightarrow 1& =1-e^{-\zeta {\omega }_n{t}_r}\left[\cos {\omega }_d{t}_r+\frac{\zeta }{\sqrt{1-{\zeta }^2}}\sin {\omega }_d{t}_r\right]\\ \Rightarrow 0& =e^{-\zeta {\omega }_n{t}_r}\left[\cos {\omega }_d{t}_r+\frac{\zeta }{\sqrt{1-{\zeta }^2}}\sin {\omega }_d{t}_r\right]\end{array}$$

Which has the potential solutions:

$$\begin{array}{rlr}{e}^{-\xi {\omega }_n{t}_r}& =0,& \cos {\omega }_d{t}_r+\frac{\xi }{\sqrt{1-{\xi }^2}}\sin {\omega }_d{t}_r=0\end{array}$$

However since we know that for any $t_r$ , $e^{-\xi {\omega }_n{t}_r}\ne 0$, we get:

$$\cos {\omega }_d{t}_r+\frac{\xi }{\sqrt{1-{\xi }^2}}\sin {\omega }_d{t}_r=0$$

Then solving for the rise time we get:

$${\omega }_d{t}_r=-{\tan }^{-1}(\frac{\sqrt{1-{\xi }^2}}{\xi })+n\pi ,n=1,2,3,..$$

However as we are only looking for the first instance this can be simplified further to:

Final Eq:

$$t_r=\frac{1}{{\omega }_d}[\pi -{\tan }^{-1}(\frac{\sqrt{1-{\xi }^2}}{\xi })]$$