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

laplace transformtransfer function

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.

Best Answer

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

enter image description here

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