# How to choose order and cut-off frequency for low-pass Butterworth filter?

When using IMU data (accelerations and angular velocities), many papers use a 4th order low-pass butterworth filter with a cutoff frequency of 5 Hz. However, none of them really justify why those filters have been chosen.

Can anyone please explain how the order and cutoff frequency are chosen for a filter of this nature? My data has a sampling frequency of 125 Hz an $$n$$th-order Butterworth filter in the $$s$$-domain has a magnitude function that is two straight lines connected with a soft corner at $$\omega_0$$ and -3 dB. the straight line at the left is a flat 0 dB but the line on the right (assuming you're looking at log frequency and dB) has a slope of $$-20n$$ dB/decade drop. That is the same as $$-6n$$ dB/octave drop, if you like your log frequency in units of octaves. So you have to figure out if the drop is steep enough too whack the frequency components you want whacked.
$$\Big| H_n(j \omega) \Big|^2 = \frac{1}{1 + \left(\frac{\omega}{\omega_0}\right)^{2n}}$$
\begin{align} 20 \log_{10} \big| H_n(j \omega)\big| &=-10\log_{10}\left(1 + \left(\tfrac{\omega}{\omega_0}\right)^{2n}\right) \\ \\ &\approx \begin{cases} 0 \qquad & \omega \ll \omega_0 \\ -10\log_{10}\left( \left(\tfrac{\omega}{\omega_0}\right)^{2n}\right) & \omega_0 \ll \omega \end{cases} \\ \\ &= \begin{cases} 0 \qquad & \omega \ll \omega_0 \\ 20 n \log_{10}( \omega_0) - 20 n \log_{10}( \omega) & \omega_0 \ll \omega \end{cases} \\ \\ \end{align}