I How do I compute the second derivative of a one-dimensional array?

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To compute the second derivative of a one-dimensional array, one must first differentiate the data twice, which is typically done using numerical methods if the data represents discrete samples. For practical applications, such as in Excel, users can create columns for velocity and acceleration by calculating the change in position over time intervals. It's important to clarify that derivatives represent changes relative to another variable, like time, and not just differences in values. The discussion also emphasizes the need for clear definitions of the data context, particularly when dealing with arrays derived from X-ray values sorted into bins. Understanding the underlying data structure and its relationship to time is crucial for accurate calculations.
  • #31
btb4198 said:
Yeah, I am going to remove it. Sorry, I never code a function to take a derivative of a graph before and I could not find any YouTube videos on how to do it by hand.
See my post #27. It is an instruction.
 
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  • #32
btb4198 said:
So Basically to take the first derivative, I am going to go through the array and subtract index + 1 from index.
So:
double value = spectrumData[i + 1] - spectrumData;
You need to get your units right in your derivative calculations (both for the numerator and denominator). The units for the numerator look to be "counts" right? And for the denominator, you have bins separated by some delta-energy in keV? So the units of the first derivative would be delta-counts per delta keV?
 
  • #33
berkeman said:
So the units of the first derivative would be delta-counts per delta keV?
It looks like the horizontal bins are spaced linearly? What is the increment in delta-keV per bin? It looks like a small fraction of a keV delta per bin?
 
  • #34
berkeman said:
You need to get your units right in your derivative calculations (both for the numerator and denominator). The units for the numerator look to be "counts" right? And for the denominator, you have bins separated by some delta-energy in keV? So the units of the first derivative would be delta-counts per delta keV?
Yes, that is all correct.
 
  • #35
berkeman said:
It looks like the horizontal bins are spaced linearly? What is the increment in delta-keV per bin? It looks like a small fraction of a keV delta per bin?
1655149886661.png


From this picture is looks like the bins increase by one .
 
  • #36
Another question, if I were to place this Spectrum into an FFT would my sampling rate(fs) be 1hz ?
 
  • #37
btb4198 said:
From this picture is looks like the bins increase by one .
One what? One eV? Could be -- please take care to always include units with the numbers. Thank you.
 
  • #38
btb4198 said:
From this picture is looks like the bins increase by one .
I don't think so. The increments along the horizontal axis (eV) increase by 2000 eV. Since you're attempting to find an approximation to the derivative, is your math background sufficient for you to understand what you're trying to do? That is, have you taken calculus sometime in the past?
 
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  • #39
btb4198 said:
Another question, if I were to place this Spectrum into an FFT would my sampling rate(fs) be 1hz ?
No. You already have a plot of counts per fraction-of-an-eV. What domain do you want the FT to transform that into? You can't randomly do FFT processing on data and get meaningful information.
 
  • #40
berkeman said:
You can't randomly do FFT processing on data and get meaningful information.
In short, GIGO...
 
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  • #41
berkeman said:
No. You already have a plot of counts per fraction-of-an-eV. What domain do you want the FT to transform that into? You can't randomly do FFT processing on data and get meaningful information.
In the example excel file, could you put that into an FFT? Or no ?
anyhow, The Output of the DPP is Very noise, so I wanted to filter out the noise.
 
  • #42
btb4198 said:
In the example excel file, could you put that into an FFT? Or no ?
anyhow, The Output of the DPP is Very noise, so I wanted to filter out the noise.
Your thread is now temporarily locked. I will send you a Private Message (PM) in a few minutes to try to salvage it.
 
  • #43
After the PM discussion with the OP, there is no reason to re-open this thread. Thanks everybody for trying to help the OP out.
 

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