HELP)) Signal deconvolution using JMP

[#1]

Hello, I'm writing because I have a problem to solve.
The company I work for wants to deconvolution the following Original

The Y-axis is the measured Original value.
The X-axis is a kind of energy intensity expressed in measuring equipment.

What I want to do is to separate(deconvolution) the waveform by finding the peak value according to the X value of the Original Signal expressed for the given Y values.

for example, Black line is Original signal, 

red line is created signal which has peak at 531ev

blue line is created signal which has peak at 532ev...

I would appreciate your help as we are unable to find a suitable Add-In or method for Discussion.

if you could create an Add-In so that we can do this repeatedly.

[#2]

Are there equation(s) that model the red and blue lines? Do you already know their peak values? 

[#3]

First of all, I wish you a happy New Year.

I don't know the formula. The problem is found in our lab using the other software, and we think we can use JMP to isolate the signal.

So, the ultimate goal is to create an add-in to get each peak value by each x-value

[#4]

Take a look at the distribution platform and see if it does what you need.

I don't know if it is the right answer for the right reason. 

I tried it with Y:Binding and Freq:Original Counts and then selected 2nd red triangle->Continuous Fit->Fit Normal 3 mixture. 

Two of means look promising

[#5]

Thank you for your help.
I uploaded a new file.
Deconvolution for 531, 532 x-axis values with existing software
Added value.
Maybe it's using the Lorenz or Gaussian equation, but I don't know the exact formula.

Based on the range of x values, peak values, and Curve, we think we should find a formula to place peak values at x values 531, 532 on the Original Line.

I'd appreciate any further comments.

[#6]

I'm well beyond my comfort zone. Maybe someone else. They probably need to know what the answer will look like:

1: two numbers, like 531, 532

2: a graph, like yours

3: a set of coefficients and exponents for a formula for the red curve, and another set for the blue curve.

The problem, I think, is we don't know what the general form of (3) looks like. It might help to know a little more about how you got the existing results.

Maybe someone with JMP PRO can play with the Functional Data Explorer and see what that can do.

[#7]

Thank you for your interest in this matter.

I thought it was easy, but it wasn't as easy as I thought.

Considering that there is a separate software that specializes in deconvolution,

it probably won't be easy to solve.

Thank you for your help again, and if find a way, I think it will help those who do similar analysis.

I will continue to find a way and share solutions.

[#8]

I think that @Craige_Hales solution is good. I was going to suggest it myself, but he beat me to it!

You are estimating the location of the peak height. It seems that you expect whole integer values. Why? Energy in eV can take on any rational value. If you need integers, then round the unbiased estimate.

[#9]

Thanks @Mark_Bailey,  for your advice.

The link below is a page that I found,

http://www.cma4ch.org/chemo/pubbl/peakdeconv.html

but it's similar to the process I've been looking for,

So the file I attached( 'GVisco_deconvoltution1.12_2Gauss.xls') looks similar to the solution I was looking for.
I think someone implemented Excel Deconvolution, but I'm not sure how to apply it to JMP.
I'm leaving a link and an attachment in case who wondering about similar problems

Thanks

[#10 · Accepted solution]

As your reference states, PLS is also a classic chemometrics models for spectra. JMP does PLS modeling.