Anti-aliasing advice for de-interleaved signals
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From: PhilipOrr on 27 Jul 2010 06:47 >On 7/26/2010 12:45 AM, Michael Plante wrote: >> Jerry wrote: > >>> ...that precedes ADC. (You wrote DAC, but I don't think you meant >>> that.) ... >> >> Perhaps a nitpick, but he wrote DAQ, not DAC, and I've seen that in the >> context of Labview/NI to mean "Data AcQuisition unit". I.e., a packaged >> ADC. That term threw me off a bit too when I first heard it, because it is >> pronounced like "DAC", but really means "ADC". Fun stuff... > >Thanks for the heads-up. I apologize to P.O. for my poor reading. > >Jerry >-- >Engineering is the art of making what you want from things you can get. > No problem, you have all been immensely helpful :-)
From: PhilipOrr on 28 Jul 2010 09:56 If anyone is still interested: What I have done to improve the noise level is sample at 16 times the switching frequency, to give me 8 samples per switch state (where previously I had only a single sample). Then, I use the mean of these 8 to get the differential measurement. The noise is vastly improved. Is this known as 'oversampling'? And is it, in effect, providing anti-aliasing before producing the low-rate differential signal? Philip
From: Jerry Avins on 28 Jul 2010 12:24 On 7/28/2010 9:56 AM, PhilipOrr wrote: > If anyone is still interested: > > What I have done to improve the noise level is sample at 16 times the > switching frequency, to give me 8 samples per switch state (where > previously I had only a single sample). > > Then, I use the mean of these 8 to get the differential measurement. The > noise is vastly improved. > > Is this known as 'oversampling'? And is it, in effect, providing > anti-aliasing before producing the low-rate differential signal? The result shows that (at least much of) the noise is there before the switch. If most of the noise power is at higher frequencies than your signal of interest, there are more effective ways to reduce its effect than simple averaging. If the noise has a Gaussian distribution, averaging provides an 8:1, or 18 dB SNR improvement. Aside from compensating for offsets, why go differential? Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
From: Fred Marshall on 28 Jul 2010 14:00 PhilipOrr wrote: > If anyone is still interested: > > What I have done to improve the noise level is sample at 16 times the > switching frequency, to give me 8 samples per switch state (where > previously I had only a single sample). > > Then, I use the mean of these 8 to get the differential measurement. The > noise is vastly improved. > > Is this known as 'oversampling'? And is it, in effect, providing > anti-aliasing before producing the low-rate differential signal? > > Philip Calculating the mean is a form of lowpass filtering. So, you're getting a lowpassed version of what's coming out of the switch. That it improves the result suggests that there's indeed higher frequency components and you've been undersampling. One might ask, "What's the issue with undersampling noise anyway?" Well, as you've experienced here, it removes the ability to filter out some of the noise. In this case, you've increased the bandwidth by a factor of 8 and then lowpassed - which may get rid of 8X the noise bandwidth. Many noises add (subtract) as the sqrt of the bandwidth. So, it appears, you'd get 9dB I believe, if the noise is truly spread across the higher bandwidth. And, by not sampling at the higher rate, you were aliasing the noise down into a narrower bandwidth - thus increasing the noise power. And, particularly in this case, the noise is likely quite random. So, differencing accentuates the higher frequency components by 3dB. .... well, this is pretty much arm-waving and I may have made a mistake or two but the idea is there. Fred
From: Jerry Avins on 28 Jul 2010 14:20
On 7/28/2010 2:00 PM, Fred Marshall wrote: > ... which may get rid of 8X the noise > bandwidth. Many noises add (subtract) as the sqrt of the bandwidth. So, > it appears, you'd get 9dB I believe, if the noise is truly spread across > the higher bandwidth. I was thinking 8x voltage ratio when I wrote 18 dB. A silly goof. Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� |