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

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? • Are you making a single filter with 8 poles, or are you making 4 different filters with 2, 4, 6, and 8 poles? The formulas for n=8 will tell you how to make a filter with 8 poles. The filters for n=4 will tell you how to make a butterworth filter with 4 poles. – The Photon Nov 21 '14 at 17:55
• I just want to make one filter, I guess that means one filter with 8 poles? – confused Nov 21 '14 at 19:04

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.

• Where did that 3d plot come from? I like it! – Scott Seidman May 12 '16 at 11:13
• @ScottSeidman I drew it myself but you are very welcome to use or abuse it. – Andy aka May 12 '16 at 12:02
• @confused - is there anything in this answer you need clarification on? – Andy aka Jun 23 '17 at 15:32

@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.

• Ok so each stage has the same cut off frequency which I calculate by plugging my passband into the formula in my original question. I don't understand what pole-Q values are yet, I'll read some more on that. – confused Nov 21 '14 at 18:55