Standard form of 2nd order transfer function (Laplace transform)?

Question

Say I had a 2nd order system such as: \begin{equation} A \dfrac{d^2y(t)}{dt^2} + B \dfrac{dy(t)}{dt} + C y(t) = D \ x(t) \end{equation} Dividing both sides by A: \begin{equation} \dfrac{d^2y(t)}{dt^2} + 2 \zeta \omega_0 \dfrac{dy(t)}{dt} + \omega_0^2 \ y(t) = \dfrac{D}{A} \ x(t) \end{equation} Hence, the transfer function is: \begin{equation} H(s) = \dfrac{\dfrac{D}{A}}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} \end{equation} But I read some texts and they all list the standard form of the transfer function for a second-order system as: \begin{equation} H(s) = \dfrac{\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} \end{equation} Why is this? Thank you.

Answer 1

The "standard" form you believe you have is in fact a low-pass 2nd order filter. Here's a picture that might explain things: -

The standard form listed above applies to all types of 2nd order filter i.e. low-pass, high-pass etc..

Note that the numerator changes dependant on what type of filter it is and in your question, the numerator is D/A.

D/A could be made to be whatever you want and this can turn the filter into a low-pass or a high-pass etc..

Information taken from here and my brain

Answer 2

The differential equation you provided corresponds to a second order low pass system.

The numerator in your expression can be written as,

$$\frac{D}{A} = \frac{D\times C}{A \times C} = \omega_0^2\times \frac{D}{C} = \omega_0^2 A_0$$

And you can write the transfer function as:

$$H(s) = \dfrac{A_0\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2}\tag1$$

This expression, given in (1) is the standard form of transfer function of 2nd order low pass system. What is given in equation (2) is transfer function of 2nd order low pass system with unity gain at DC.

$$H(s) = \dfrac{\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2}\tag2$$