Second Order Other damped

This page demonstrates the steps in finding the step response of a critically damped or oscillatory second order system. This is achieved by adapting the underdamped case for when the imaginary part or real part of the roots of the auxiliary equation are zero.

As shown on my underdamped page, with zero initial conditions, for roots $r_{1} = a+ib$ and $r_{2}= a-ib$, the step response is

$O = 1 - e^{at} (cos(bt) - \frac{a}{b} sin(bt)) $.

If $b$ is zero, there are two repeated roots of $a$, and the system is critically damped. We can find the response by seeing what happens to $O$ as $b$ tends to zero.

If instead $a$ is zero, the two roots are imaginary, $ib$ and $-ib$, and the system oscillates. We can find the response by seeing what happens to $O$ when $a$ is zero.

We can use the same formula, and the facts that

$e^{0t}=1$
When $b$ approaches zero $cos(bt)$ approaches $1$
When $b$ approaches zero $sin(bt)$ approaches $bt$

Finding general response

We separately adapt the underdamped response when $b$ is zero and instead when $a$ is zero : \begin{align} O = 1 - e^{at} (cos(bt) - \frac{a}{b} sin(bt)) \\ \cssId{Step1}{{}When \ b \ approaches \ 0, O = 1 - e^{at} (1 - \frac{a}{b} bt) } \\[3px] \cssId{Step2}{{}Hence \ O = 1 - e^{at} (1 - at) } \\[3px] \cssId{Step3}{{}So \ O = 1 - e^{at} + at e^{at}} \\[3px] \cssId{Step4}{{}Instead \ if \ a \ is \ 0, O = 1 - e^{0} (cos(bt) - \frac{0}{b} sin(bt)) } \\[3px] \cssId{Step5}{{}So \ O = 1 - cos(bt) } \end{align}