The difference between the value at Peak Time and the steady state is known as the Maximum Overshoot .
M_p = x(t_p) - x(\infty)$$ Evaluating at both $t=t_p, t=\infty$ We can get that:\begin{align} M_p &= \frac{Ka}{\omega_n^2}{1-e^{-\xi\omega_nt_p}[cos\omega_dt_p+\frac{\xi}{\sqrt{1-\xi^2}}\sin{\omega_dt_p}]}-\frac{Ka}{\omega_n^2} \ &= \frac{Ka}{\omega_n^2}e^{-\xi\omega_nt_p}[cos\omega_dt_p+\frac{\xi}{\sqrt{1-\xi^2}}\sin{\omega_dt_p}] \end{align}
From [[Peak Time#final-eq|eq: t_p]] we can state $$\omega_dt_p = \pi \Rightarrow \cos\omega_dt_p = -1, \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\sin\omega_dt_p =0$$ Subbing all of this in to the equation before gives us:M_p = \frac{Ka}{\omega_n^2}e^{-\xi\omega_nt_p}
Alternatively, given that $x(\infty) = \frac{Ka}{\omega_n^2}$ and that $t_p = \frac{\pi}{\omega_d}$:M_p = x(\infty)e^{-\frac{\pi\xi}{\sqrt{1-\xi^2}}}
This being said however, we typically define the maximum overshoot in reference to the $x(\infty)$ meaning that the definition is: #### Final Eq:M_p = e^{-\frac{\pi\xi}{\sqrt{1-\xi^2}}}, \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }M_p = 100e^{-\frac{\pi\xi}{\sqrt{1-\xi^2}}}%