- #1

- 284

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It seems to me that both methods should be mathematically equivalent as basically a sum over integrals or an integral over sums, but I can't help but feel like I'm missing something.

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- Thread starter JaWiB
- Start date

- #1

- 284

- 0

It seems to me that both methods should be mathematically equivalent as basically a sum over integrals or an integral over sums, but I can't help but feel like I'm missing something.

- #2

- 807

- 23

How you approach this will depend on what kind of analysis you're doing, and what you know about the noise.

- #3

- 284

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Perhaps it helps if I describe the calculation. The first half of the signal represents a measurement taken under one experimental condition while the second is taken under another condition. We integrate over part of the first half, and (ideally) the same part of the second half, and the final data point is the difference between those integrals. I can either do those integrals on a single waveform, or let the digitizer average a bunch of times and then do the integral. As far as I know, the other students in the lab have always done the latter--I just don't know if it matters.

- #4

- 487

- 3

Perhaps it helps if I describe the calculation. The first half of the signal represents a measurement taken under one experimental condition while the second is taken under another condition. We integrate over part of the first half, and (ideally) the same part of the second half, and the final data point is the difference between those integrals. I can either do those integrals on a single waveform, or let the digitizer average a bunch of times and then do the integral. As far as I know, the other students in the lab have always done the latter--I just don't know if it matters.

Integration is a linear operator, so the sum of integrals is always the same as the integral of sums.

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