- #1
Ephant
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Can someone verify if the following computations I saw is correct, that a 2nd order Butterworth filter only has 7.5% more noise than a brick wall filter? also how do you convert for example these ##H(f) = 1## for 0 to ##F## to math fonts like the one with equation shown?
"It is specified for the BMA-200 low-pass filter: for every octave (doubling of frequency), filter attenuates by an additional -12dB, which means H(2f) = 0.25 H(f), H(2*2f) = H(4f) = 0.25*0.25 H(f), and so on.
If ##D## is the spectral noise density (some nV/sqrt(Hz)), and we assume it is constant for all frequencies (it isn't, it tends to get bigger at very low frequencies, but it often stays constant at higher frequencies), then for the brick-wall digital filter like g.USBamp the (rms) noise voltage amplitude is
$$V_n = \sqrt{\int_0^\infty D^2 H(f)^2 d f} = D \sqrt{F}$$
where ##F## is the bandwidth in Hz. This is because for a brick-wall filter H(f) = 1 within that band, and immediately zero outside.
For the Butterworth filter on the BMA-200, the same is
$$V_n = \sqrt{\int_0^\infty D^2 H(f)^2 d f} = D \sqrt{ 1.155 F } = 1.075 D \sqrt{F}$$
because ##H(f) = 1## for 0 to ##F##, but for ##f \ge F##, is ##H(f) = 10^{-0.6 \log(f / F) / \log(2) }## (i.e. -12dB per octave). (I had to ask Maxima for the integral, and check it a couple of times. )
Augh! The correction factor 1.075 is much smaller than I thought now that I calculated it out. The less vertical low-pass edge only adds about 7.5% noise compared to the brick-wall filter.
I did not expect that. Thanks for pushing me to check the actual math, and not just assume! Assuming makes an ass out of me :-[."
[Mentor Note: LaTeX fixed up a bit]
"It is specified for the BMA-200 low-pass filter: for every octave (doubling of frequency), filter attenuates by an additional -12dB, which means H(2f) = 0.25 H(f), H(2*2f) = H(4f) = 0.25*0.25 H(f), and so on.
If ##D## is the spectral noise density (some nV/sqrt(Hz)), and we assume it is constant for all frequencies (it isn't, it tends to get bigger at very low frequencies, but it often stays constant at higher frequencies), then for the brick-wall digital filter like g.USBamp the (rms) noise voltage amplitude is
$$V_n = \sqrt{\int_0^\infty D^2 H(f)^2 d f} = D \sqrt{F}$$
where ##F## is the bandwidth in Hz. This is because for a brick-wall filter H(f) = 1 within that band, and immediately zero outside.
For the Butterworth filter on the BMA-200, the same is
$$V_n = \sqrt{\int_0^\infty D^2 H(f)^2 d f} = D \sqrt{ 1.155 F } = 1.075 D \sqrt{F}$$
because ##H(f) = 1## for 0 to ##F##, but for ##f \ge F##, is ##H(f) = 10^{-0.6 \log(f / F) / \log(2) }## (i.e. -12dB per octave). (I had to ask Maxima for the integral, and check it a couple of times. )
Augh! The correction factor 1.075 is much smaller than I thought now that I calculated it out. The less vertical low-pass edge only adds about 7.5% noise compared to the brick-wall filter.
I did not expect that. Thanks for pushing me to check the actual math, and not just assume! Assuming makes an ass out of me :-[."
[Mentor Note: LaTeX fixed up a bit]
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