0
\$\begingroup\$

I want to design a second order sallen key low pass filter. I want to attenuate a 100Hz signal to around 10%. I need to use a 2nd order low pass filter. How could I calculate the corner frequency such that I will achieve this 10% attenuation at 100Hz? In this case a Butterworth Filter has to be used.

\$\endgroup\$
3
  • \$\begingroup\$ A lowpass of 2nd order is determined not only by its corner frequency but also the form of the pass-band (maximally flat oder with a certain peaking). Hence, you have to select one of the corresponding approximations (Bessel or Butterworth or Chebyshev I or Chebyshev II ). \$\endgroup\$
    – LvW
    May 31 '18 at 8:33
  • \$\begingroup\$ In this case a Butterworth filter is going to be used \$\endgroup\$
    – suyol854
    May 31 '18 at 8:39
  • \$\begingroup\$ Didn't you ask this same question on dsp.SE? I can't find it, did you delete it? There was a comment in there... \$\endgroup\$ May 31 '18 at 8:48
2
\$\begingroup\$

Attenuating to 10% means the dB reduction is 20 dB because 20 log 10 = 20 dB.

A 2nd order low-pass filter will normally reduce the high frequencies at 40 dB per decade therefore, you need to have the filter's cut-off frequency half a decade below 100 Hz at 31.62 Hz (\$\sqrt{10\times 100}\$).

Here's an approximation: -

enter image description here

This was modelled with an RLC low pass filter interactive tool having a butterworth response (\$\zeta\$ = 0.7071). Make sure that if you are using a sallen key filter the op-amp has a suitable GBWP or it might not meet your expectations.

Note that I didn't go to great lengths to choose LC to make exactly a cut-off 31.62 Hz but it's near enough to demonstrate that the marker at 100.66 Hz is 20 dB down.

\$\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.