In the following link (at bottom) we can read:
A single Butterworth low pass filter of order N has a frequency response H(f) = 1/sqrt(1 + (f/f0)^(2N))
At f=f0, the response is 1/sqrt(2). We consider this the definition of cutoff frequency of a filter: the frequency where it transfers 1/sqrt(2) times the original signal.
If you apply the filter twice, the response is squared: H(f) = 1/(1 + (f/f0)^(2N))
Now we need to look for the frequency at which the response is 1/sqrt(2). It is no longer f0. Rather, it is the f for which:
H(f) = 1/sqrt(2)
1/(1 + (f/f0)^(2N)) = 1/sqrt(2)
The solution is:
f = f0 * (sqrt(2)-1)^(1/(2N))
For N=2, we get f = f0 * 0.8022 which was already mentioned in this thread.
In the context the correction factor (Cf) will be calculated as:
Cf = (sqrt(2)-1)^(1/(2*order))
Fc = Fc/ Cf
For the high pass filter the correction is the same? Or we should use a different equation of this Cf = (sqrt(2)-1)^(1/(2*Order)) ?