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Pocket-Sized 6 GHz 1 TS/s ET Scope
joeqsmith:
Looking at the previous data I provided, you only plotted channels 1&2. Also note that you only plot 119 of the 150 sweeps. I wrote a simple viewer for your CSV files to show all of the data and have included the data for channel 3. I assume these micro glitches toward the peak are considered normal based on your previous statements.
SJL-Instruments:
--- Quote from: joeqsmith on January 29, 2024, 09:24:53 pm ---Looking at the previous data I provided, you only plotted channels 1&2. Also note that you only plot 119 of the 150 sweeps. I wrote a simple viewer for your CSV files to show all of the data and have included the data for channel 3. I assume these micro glitches toward the peak are considered normal based on your previous statements.
--- End quote ---
The zip file you provided contains only 119 files - perhaps some of them were excluded.
Based on your plot, the largest spike is ~5 mV from the mean over 150 sweeps. We consider this within specification (but on the worse side).
If there is a timing issue, then this would also explain why all the spikes are downwards. (But doesn't explain why they're at the peak only.)
--- Quote from: joeqsmith on January 29, 2024, 09:15:28 pm ---From the 4.3.3 example, there are only three data points that meet that criteria. Showing the raw plus two methods for these three points. Using least squares for linear fit of the inverse Gaussian Error.
Totally lost me but I suspect based on your comment about them being the same, there is something wrong with what I have shown.
--- End quote ---
The inverse Gaussian error function (as described in 4.3.4) should send 0.5 to 0. As a sanity check, verify that:
g(0.5) = 0
g(0.8 ) = 0.5951
g(0.2) = -0.5951
where g(x) = erf^-1(2x-1) as defined in 4.3.4 (implementation note 1).
If you perform a linear fit, it must also be weighted according to 4.3.4 implementation note 1 - otherwise the outer points will blow up the fit.
joeqsmith:
1 Start by selecting only CDFs that meet: "The fit is performed only on the subset of entries satisfying 0.1 < F < 0.9 for numerical stability."
2) g(x) = erf^-1(2x-1) is then calculated for the CDF subset.
2) run a linear fit on this data using some sort of weighted numbers.
3) For the weights, we take the derivative of CDF raised to ^-2
4) multiply the weights by step 2
--- Quote ---The inverse Gaussian error function (as described in 4.3.4) should send 0.5 to 0. As a sanity check, verify that:
g(0.5) = 0
g(0.8 ) = 0.5951
g(0.2) = -0.5951
--- End quote ---
That much I can verify. See attached.
SJL-Instruments:
--- Quote from: joeqsmith on January 29, 2024, 11:48:35 pm ---1 Start by selecting only CDFs that meet: "The fit is performed only on the subset of entries satisfying 0.1 < F < 0.9 for numerical stability."
2) g(x) = erf^-1(2x-1) is then calculated for the CDF subset.
2) run a linear fit on this data using some sort of weighted numbers.
3) For the weights, we take the derivative of CDF raised to ^-2
--- End quote ---
Yes, this should work (if in step 3 you mean g'(F)^-2). The extracted voltage is then just the x-intercept of the fitted line. This is how it's done in the software.
The derivative of g should always be positive - if the blue line is g'(F), then that part looks incorrect. As a check, you should have g'(0) = 1.772.
You should not take the derivative of the CDF - you want the derivative of g evaluated at the CDF.
joeqsmith:
--- Quote from: SJL-Instruments on January 30, 2024, 12:34:11 am ---
--- Quote from: joeqsmith on January 29, 2024, 11:48:35 pm ---1 Start by selecting only CDFs that meet: "The fit is performed only on the subset of entries satisfying 0.1 < F < 0.9 for numerical stability."
2) g(x) = erf^-1(2x-1) is then calculated for the CDF subset.
2) run a linear fit on this data using some sort of weighted numbers.
3) For the weights, we take the derivative of CDF raised to ^-2
--- End quote ---
Yes, this should work. The extracted voltage is then just the x-intercept of the fitted line. This is how it's done in the software.
The derivative of g should always be positive - if the blue line is g'(F), then that part looks incorrect.
--- End quote ---
Well you say that but don't underestimate my ability to screw things up!
I suspect I have an error in (g'(F))^-2.
If I multiply the weights after the least squares, shown in fitter 5.
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