I am trying to implement a fourth order Butterworth high pass filter under a Sallen-Key topology. Two frequencies, one at the passband and another for the stopband, are set to be at some specific gain/attenuation.

Consulting the Butterworth tables, I was able to design a functioning circuit with two cascaded Sallen-Key networks. The higher frequency (passband side) has no problems and is within my expectation. The problem is with the lower frequency. The filter fails to assign the lower frequency to the set attenuation level.

This is because the Butterworth filter has a linear 80 db/dec since it is a fourth order filter. I am under the impression that the slope cannot be changed. With that, I am confused as to what possible steps I should take to solve the problem.

  • \$\begingroup\$ Can you adjust the Butterworth corner frequency? If so, then try to keep the higher frequency of the pair within the passband.... a variable corner frequency will change the attenuation of the lower frequency of the pair, even with constant 80 dB/dec slope. \$\endgroup\$
    – glen_geek
    Mar 20, 2019 at 17:18
  • \$\begingroup\$ Numbers are important here. \$\endgroup\$
    – Andy aka
    Mar 20, 2019 at 17:28
  • \$\begingroup\$ Possible duplicate of Understanding the exact meaning of dB/decade in a Bode plot \$\endgroup\$ Mar 20, 2019 at 17:33
  • \$\begingroup\$ All filter asymptotes are N*20dB/dec . The only variable is N defined by the number of independent reactance parts (L or C) in any active or passive filter. ( note 2 caps in series are not independent and are lumped as 1) Using software tools it is now trivial to design any order filter to any shape \$\endgroup\$ Mar 20, 2019 at 17:35
  • \$\begingroup\$ user262213, if you cannot meet the stopband attenuation (and when you have not made any error) there is no way-out - you must use a higher order filter (n=5 0r n=6) \$\endgroup\$
    – LvW
    Mar 20, 2019 at 18:17

2 Answers 2


Yes. Rolloff slope is part of the definition of a Butterworth filter.

If you have passband/stopband specifications and filter order specification, then it's likely that your problem is over-constrained. You have to loosen at least one of those constraints of passband/stopband/order so that the filter can be designed.

Or, you can use a different filter type.


It's 20dB/dec per order. A fourth order butterworth is actually 2 second order filters (if your using the sallen key implementation) with the poles at the same (or similar) location.

With a butterworth low pass, if more attenuation is needed the corner frequency can't be moved, increasing the order is the only way to increase attenuation if the stop band is not sufficient. The order increases by even numbers.

enter image description here Source: https://en.wikipedia.org/wiki/Butterworth_filter

  • \$\begingroup\$ No - it is not correct that a Butterworth filter can be realized using a cascade of 4 first-order filters. Each of this filter will have a Q-value smaller (or equal) 0.5. This does not allow a Butterworth response.The common way is to use two second order filters with different Q values (each above Q=0.5) \$\endgroup\$
    – LvW
    Mar 20, 2019 at 18:13
  • \$\begingroup\$ OK - I have realized that you now have modified your answer correspondingly. Fine. \$\endgroup\$
    – LvW
    Mar 20, 2019 at 18:34
  • \$\begingroup\$ Due to the OP's high stop bands, I was thinking more along the lines of low pass filtering. \$\endgroup\$
    – Voltage Spike
    Mar 20, 2019 at 18:38

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