Second Order Overdamped

This page demonstrates the steps in finding the step response of an overdamped second order system. Here the auxiliary equation has two real roots. It is also assumed that initial conditions are zero.

Assume $a \frac{d^2O}{dt^2} + b \frac{dO}{dt} + c O = c I$ where $I$ is a unit step.

The steady state response, where $O$ is not changing, is when $cO = c$ that is $O = 1$.

The auxiliary equation is $ar^2 + br + c = 0$

If its roots are $r_{1}$ and $r_{2}$, the transient response is $k_{1}e^{r_{1}t} + k_{2}e^{r_{2}t}$

The complete response is $O = 1 + k_{1}e^{r_{1}t} + k_{2}e^{r_{2}t}$

The constants are determined by finding $\frac{dO}{dt}$, evaluating that and $O$ when $t = 0$, forming two simultaneous equations in $k_{1}$ and $k_{2}$ and solving them, as is shown here.

For underdamped, critically damped, or oscillators, please see under damped page or 'other' damped page

Finding general response

Here we assume the roots are $r_{1}$ and $r_{2}$, so : \begin{align} O = 1 + k_{1}e^{r_{1}t} + k_{2}e^{r_{2}t} \\ \cssId{Step1}{{}\frac{dO}{dt} = r_{1}k_{1}e^{r_{1}t} + r_{1}k_{2}e^{r_{2}t}} \\[3px] \cssId{Step2}{{}At \ t = 0, O = 0, so \ 0 = 1 + k_{1} + k_{2}} \\[3px] \cssId{Step3}{{}At \ t = 0, \frac{dO}{dt} = 0, so \ 0 = r_{1}k_{1} + r_{2}k_{2}} \\[3px] \cssId{Step4}{{}So \ k_{1} = -\frac{r_{2}}{r_{1}}k_{2} } \\[3px] \cssId{Step5}{{}Using \ this \ 0 = 1 - \frac{r_{2}}{r_{1}}k_{2} + k_{2} = 1 - \frac{r_{2}-r_{1}}{r_{1}}k_{2} } \\[3px] \cssId{Step6}{{}Thus \ k_{2} = \frac{r_{1}}{r_{2}-r_{1}}} \\[3px] \cssId{Step7}{{}And \ so \ k_{1} = \frac{r_{2}}{r_{1}-r_{2}}} \\[3px] \cssId{Step8}{{}Hence \ O = 1 + \frac{r_{2}}{r_{1}-r_{2}}e^{r_{1}t} + \frac{r_{1}}{r_{2}-r_{1}}e^{r_{2}t} } \end{align}