Peak Time

As mentioned in the Rise Time, an underdamped input subject to a step input begins at rest before overshooting the $x(\infty )$ value and reaching a peak. Due to the damping, the first peak will be the greatest in amplitude assuming a step input.

The time we reach that peak is defined as the peak time $t_p$. As the peak time corresponds to the maximum in the signal we know that the derivate of the signal must be zero at the peak time and therefore:

$$\dot{x}(t_p)=0$$

Differentiating the equation for an underdamped second order system, and simplifying we get:

$$\dot{x}(t)=\frac{Ka}{{\omega }_n}{e}^{-\xi {\omega }_n{t}_p}(\frac{{\xi }^2{\omega }_n}{\sqrt{1-{\xi }^2}}+{\omega }_d)\sin {\omega }_d{t}_p$$

However at peak time we know that $\dot{x}(t)=0$, giving us the potential solutions of:

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

Given that $Ka\ne 0$ we know the first solution isnt true, in the same way, with any $t_p$ we can also surmise that solution 2 isnt true, in addition to this, ${\omega }_n\ne 0$ meaning that solution 3 isnt true, leaving us only with:

$$\sin {\omega }_d{t}_p=0⟹{\omega }_d{t}_p=n\pi ,n=1,2,3,...$$

Again, as we are only looking for the first instance we can rearrange to get:

Final Eq:

$$t_p=\frac{\pi }{{\omega }_d}$$