how to get a spectrum profile with points/bars
dilatoryyang
Tue, 05/29/2012 - 01:04 pm
Hi,I'm a new learner of the igor, and now I have to make a spectrum with it. I have the position and intensity for each transition from calculation, such that I can plot many points or bars(by setting the mode to "sticks to zero"), but how to get a spectrum which looks real and could be compared with an experimental spectrum. Is that possible if I give those peaks an equal width, but actually I don't know how to do it, would anyone can help me with this? That would be much appreciated.
What type of spectrum are you wanting to create? It sounds to be a plot of intensity versus energy, where "peaks" have some characteristic shape (Gauss, Lorentz ...). What method creates the experimental spectrum, for example FTIR, Raman ... or something else? Alternatively, is this a distribution plot of some type?
Could you provide a few more details please.
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J. J. Weimer
Chemistry / Chemical & Materials Engineering, UAHuntsville
May 29, 2012 at 01:17 pm - Permalink
May 29, 2012 at 01:47 pm - Permalink
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com
May 29, 2012 at 02:46 pm - Permalink
May 29, 2012 at 03:15 pm - Permalink
The width isn't either a standard deviation width, or FWHM. It's sqrt(2)*std dev, which is what Igor's built-in curve fitting Gauss function uses (it's historical). You might want to change that.
The function is in the procedure window GaussFromLocations.ipf. I have added a test output from the function to your original Graph2.
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com
May 29, 2012 at 04:31 pm - Permalink
May 29, 2012 at 06:37 pm - Permalink
May 30, 2012 at 04:37 pm - Permalink
Convolving two distributions is different from overlapping them, both mathematically and with regard to the inherent physical meaning behind the operation. There is some ambiguity in your request that suggests, you may need to understand the differences a bit better.
In case you really mean convolution, look at the Convolve operation.
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J. J. Weimer
Chemistry / Chemical & Materials Engineering, UAHuntsville
May 31, 2012 at 05:45 am - Permalink
John Weeks
WaveMetrics, Inc.
support@wavemetrics.com
May 31, 2012 at 09:22 am - Permalink
John is right ... convolving a peak shape function with a delta just gives the peak centered at the position of the delta. Overlapping, symmetrical peaks are resolved according to the equation ...
R ~ 2*\Delta{}x / (1.7*(FWHM_1 + FWHM_2))
where R is the resolution, \Delta{}x the separation, and FWHM_i the full-width at half-maximum of each peak.
--
J. J. Weimer
Chemistry / Chemical & Materials Engineering, UAHuntsville
June 1, 2012 at 07:44 am - Permalink