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I'm trying to design a filter with a bandwidth of 150Hz, center frequency of approx. 60Hz, with a roll-off of 40dB/decade. I am following the design of a 2nd order Multiple Feedback Band-Pass filter as discussed in page 315 and 316 of this reference manual.

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For my application, the Q factor should be 60/150, or 0.4. Looking at one of the resistor calculations,

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R3 requires 2(Q^2)+Am > 0 to function properly, otherwise the value will be negative. This suggests the Q-factor should be adequately large for this design of MFB Band-pass to function. In my case, Am needs to be between 0 and -0.32 (gain is inverted in this circuit), which seems like a very low number and would defeat the purpose of an op-amp based filter.

I am confused, however, since I recall reading(cannot remember where) that MFB Band-pass filters are used for their good response to low Q-factors. In this case, was what I read incorrect? Or is a Q-factor of 0.4 simply too low for the MFB band-pass to handle?

Please let me know if more information is required.

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I'm trying to design a filter with a bandwidth of 150Hz, center frequency of approx. 60Hz

If the bandwidth is 150 Hz and the centre frequency is 60 Hz then we can only be talking about a high-pass filter followed by a low-pass filter that have combined 3 dB points at 21 and 171 Hz.

Put another way (and thinking logarithmically), 60 Hz is 2.85 times lower than 171 Hz and, 21 Hz is 2.85 times lower than 60 Hz. And, 171 Hz minus 21 Hz is a bandwidth of 150 Hz.

That's the only realistic way you can define this filter.

And, because of this (the massive asymmetry), you cannot use the Q-factor approximation of centre frequency divided by bandwidth. Reason: the pass-spectrum is so close to DC that the asymmetry of the higher and lower cut-offs are just too different.

Normally we might talk about a selective filter having a centre frequency of (say) 1 MHz with a BW of (say) 10 kHz and then, we can use the Q-factor formula approximation you have used but, it falls-over when we are this close to DC.

So, you have to regard the two cut-off frequencies (low and high) as being formed with a high-pass filter followed by a low-pass filter.

with a roll-off of 40dB/decade

If you want 40 dB/decade on both slopes then, you'll need two 2nd order filters cascaded to achieve this. Design each independently and cascade them. For non-peaking in the passband, use a Butterworth Q-factor of 0.7071 for both high-pass and low-pass filters.

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  • \$\begingroup\$ Thank you, I did not know the Q-factor bandwidth thing was an approximation. Are you also saying that the 2nd order in this element can be separated and seen as "1 order for low-pass, 1 order for high-pass"? \$\endgroup\$ Dec 18, 2023 at 16:57
  • \$\begingroup\$ @HFOrangefish yes, it becomes a 1st order for low-pass and, a 1st order for high-pass. \$\endgroup\$
    – Andy aka
    Dec 18, 2023 at 17:00
  • \$\begingroup\$ @Andy aka Yes, I want to give an additional explanation. When the Q-factor is lower than 0.4 we have two real poles and not a complex-conjugate pole pair. In general, the Q-factor is related to the poles of the transfer function . For a bandpass function, this factor equals the center frequency-to-bandwidth ratio for Q>0.5 only. \$\endgroup\$
    – LvW
    Dec 18, 2023 at 17:06
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    \$\begingroup\$ Yes - I have revised it. \$\endgroup\$
    – LvW
    Dec 18, 2023 at 17:09
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    \$\begingroup\$ @Andy aka yes, of course. We have two real poles for Q<0.5. I really do not know how such a typing error could happen. \$\endgroup\$
    – LvW
    Dec 18, 2023 at 22:25

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