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 $$ \sigma = \zeta \omega_n $$
Delay time \(t_d\) is the time it takes the response to reach half its final value (the first time)
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 \(\beta = \tan^{-1}(\frac{\omega_d}{\sigma})\)
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
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
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