7 views (last 30 days)

Show older comments

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)

or

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

Question:

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)) ?

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!