0
\$\begingroup\$

Ok I'm sure people are getting tired of seeing my filter questions but here's another one. With my new understanding of 2nd order Butterworth filter design I set out to try to design an 8th order. I made a big excel sheet that helped calculate all the values I'd need for each stage and verified that they were all produced the same frequency and the appropriate Q for the stage. The Q of the last stage is pretty high according to the table at about 2.5.

Anyway I got a high output bump in my filter response after that last stage. Is that what's expected or have I done something wrong? If I turn down the Q of the last stage I can dial the bump down.

Here's what it looks like. I tried searching for pictures but most of what I found looks like the ideal response.

The plot below shows the response for each stage on top of one another.

enter image description here

Improve this question
\$\endgroup\$
4
  • 1
    \$\begingroup\$ That's how a high order filter works. That bump compensates for the early rolloff in the other stages to keep the overall response flat right up to cutoff. You do need to consider the signal levels in all stages : if you put that stage ahead of the others, a high level signal at its frequency peak would be clipped. \$\endgroup\$ – Brian Drummond Nov 24 '14 at 23:39
  • \$\begingroup\$ (+1 Brain D.) The one gotcha you need to watch out for is that gain in the final stage. Don't let the signal level clip. (dynamic range etc.) \$\endgroup\$ – George Herold Nov 25 '14 at 0:05
  • \$\begingroup\$ I need to add that I once made a 6 pole Butterworth, but hated the step response and switched to Bessel. \$\endgroup\$ – George Herold Nov 25 '14 at 0:51
  • \$\begingroup\$ Yes - each 2nd order stage with a pole-Q value higher than Qp=0.7071 shows such a gain peaking in the vicinity of the pole frequency. \$\endgroup\$ – LvW Nov 25 '14 at 10:11
1
\$\begingroup\$

Confused - something must be wrong in your calculation - at least for the 3rd stage. According to the Qp values as given in the TI document (I gave you the reference earlier in another thread) the 3rd stage (second order) of an 8th order Butterworth filter must have a value Qp=0.9. As I have mentioned earlier, each stage with Qp>0.7071 shows a gain peaking at the pole frequency (for Butterworth identical to the 3dB cutoff frequency). However, in your figure the 3rd stage has no peaking and, thus, a pole Q<0.7071.

EDIT: More than that, the peaking of the last stage seems to be too large. How many dB? For a Q-factor Qp=0.9 the peaking is only app. 1 dB.

Improve this answer
\$\endgroup\$
0
\$\begingroup\$

The final stage of a butterworth filter (order greater than 2) should always peak up. This peaking will of course be countered by the gradual roll-off from previous stages to produce the maximally flat response demanded by the butterworth design. The final stage is made to peak up for the simple reason that if it were the first stage of the filter, the op-amp outputs would clip on a large signal input so, lower Q circuits come first then higher Q circuits come last in the chain.

If you have got a peak in the overall response then you have miscalculated or misinterpreted something incorrectly. Butterworth is maximally flat and it means just that (flat).

PS, how did you get on with my answer in this: Is the cut-off frequency for each high order butterworth filter the same or different?

Improve this answer
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.