Is the cut-off frequency for each high order butterworth filter the same or different?

Question

From reading I see that I can calculate my cut-off frequency for a given Butterworth filter using the equations below. Let's say then that I have a 8th order filter, and I'm going to make it out of four stages. Do I calculate the cut-off frequency once using n = 8? Or do I calculate a different cut-off frequency for each stage i.e., n=2, n=4, n=6, n=8?

This is what I can't wrap my head around yet, are all my stages the same or does each one need slightly different gain and frequency settings to achieve my final response?

Answer 1

The natural resonant frequency of a 2nd order low pass filter will have a pole zero diagram like this: -

If you are unfamiliar with pole-zero diagrams see if this helps: -

If you follow that, then go back to the first diagram and you should realize that the undamped natural resonant frequency (\$\omega_0\$) has a value anywhere on the semi circle and it is the Q of the filter how far round from the jw axis the two poles are. Given that any order of butterworth filter has all its poles on the same semi circle, the answer to your question is: -

Is the cut-off frequency for each high order butterworth filter the same or different?

If you mean the natural resonant frequency then YES!

If you mean the 3dB point of each filter's response on the jw axis (the axis that pertains to "real-life" measurements on a spectrum analyser) then NO!

Butterworth pole-zero diagram: -

The above shows the poles of a 10th order butterworth filter - note that all the poles lie on the same circle and therefore all the individual filters (5 x 2nd order) have the same natural resonant frequency.

Taken from here.

Answer 2

@confused - you have posted another question (asking about the same) - and there I gave you already some answers. Look into the tables I have recommended - and you will see that all 4 stages must have the same pole frequency (which for the Butterworth case are identical to the cut-off frequencies). However, the pole-Q values are different. If the gain values of each stage are the same or not - depends on you or the selected topology, respectively (unity gain, gain-of-two, equal component design).

EDIT: Answer to your question regarding pole-Q:

The classical normalized denominator D(s) of a 2nd order lowpass transfer function is

D(s)=wp^2 + s*wp/Qp + s^2

with pole (angular) frequency wp=wc (cut-off) and pole-Q=Qp which gives you the amount of peaking of the transfer function at w=wp=wc.

This denominator form has to be compared with the actual denominator which is derived from the circuit. From this you can derive the formulas for the various componenets R and C. Note that these formulas, of course, can be found also in the variuous publications on Sallen-Key filters.