# 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