# Second Order System Transient Analysis

The response of a second order system to a unit step input is specified by:

- Delay time \(t_d\)
- Rise time \(t_r\)
- Peak time \(t_p\)
- Maximum overshoot \(M_p\)
- Settling time \(t_s\)

Two useful formulas we'll be using in this topic are:
$$ \omega_d = \omega_n \sqrt{1 - \zeta^2} $$
and
$$ \sigma = \zeta \omega_n $$

fact

Delay time \(t_d\) is the time it takes the response to reach half its final value (the first time)

fact

Rise time \(t_r\) is the time it takes the response to go from 10% to 90% of its final value (sometimes 5% to 95% or 0% to 100%)
\(t_r = \frac{\pi - \beta}{\omega_d}\)
Where \(\beta = \tan^{-1}(\frac{\omega_d}{\sigma})\)

fact

Peak time \(t_p\) is the time the response reaches its first local maximum turning point (the first peak)
\(t_p = \frac{\pi}{\omega_d}\)

The peak time for a critically damped or overdamped case is usually given either as infinity or undefined, depending on the situation

fact

Maximum overshoot \(M_p\) is the percentage of the highest peak compared to the steady state value:
\(M_p = \frac{c(t_p) - c(\infty)}{c(\infty)} \times 100%\)

Sometimes the maximum overshoot is given as \(c(t_p) - c(\infty)\), how much larger the peak is than the steady state value, rather than a percentage, the context will usually make it clear which form we're using

fact

Settling time \(t_s\) is the time required for the response to settle to within 5% (or sometimes 2%) of its final value.
\(t_s = 4T = \frac{4}{\sigma}\) (2% settling value)
\(t_s = 3T = \frac{3}{\sigma}\) (5% settling value)

The settling time is not the first time the curve comes within 5% of its final value, it's the time it takes for the response to be within 5% of the final value and never be further away than that.

practice problems