Saturday 22 December 2018

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


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.




No comments:

Post a Comment

arduino - Can I use TI's cc2541 BLE as micro controller to perform operations/ processing instead of ATmega328P AU to save cost?

I am using arduino pro mini (which contains Atmega328p AU ) along with cc2541(HM-10) to process and transfer data over BLE to smartphone. I...