how to get a spectrum profile with points/bars

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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.
dilatoryyang wrote:
... I have to make a spectrum ... but actually I don't know how to do it ....


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.

--
J. J. Weimer
Chemistry / Chemical & Materials Engineering, UAHuntsville

May 29, 2012 at 01:17 pm - Permalink

Yeah, it is many bars of intensity vs frequency. It's a frequency spectrum for resonance enhanced multiphoton ionization(REMPI), the Gaussian shape should be OK. I just want to compare this calculated spectrum to experimental REMPI spectrum to get some information of experimental condition, such as rotational temperature, population on each rotational quantum state of molecules.

May 29, 2012 at 01:47 pm - Permalink

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REMPI_ONOFFv3.pxp

May 29, 2012 at 03:15 pm - Permalink

I didn't take the time to figure out the code that computed your calculated spectrum. Instead I wrote a quick Igor function that takes as input the X and Y waves with the computed spectrum and computes a new wave with evenly-spaced points and specified gaussian width, having a gaussian peak for each of the elements of the computed spectrum waves. That may serve as the basis for new ideas for you.

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
REMPI_ONOFFv3JW.pxp

May 29, 2012 at 04:31 pm - Permalink

There is one more thing I forgot, the convolution process would happen when 2 Gaussian peaks overlap to some extend, is there such function in igor?

May 30, 2012 at 04:37 pm - Permalink

dilatoryyang wrote:
There is one more thing I forgot, the convolution process would happen when 2 Gaussian peaks overlap to some extend, is there such function in igor?


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.

--
J. J. Weimer
Chemistry / Chemical & Materials Engineering, UAHuntsville

May 31, 2012 at 05:45 am - Permalink

The function I wrote just computes a Gaussian peak for each point in the original waves and adds them together. Convolving a Gaussian with a series of delta functions does exactly that, so I don't think there's a difference in this case. Because my function just sums a bunch of Gaussian peaks, they will overlap if they are close together, where "close" means within some number of standard deviations of the center locations.

John Weeks
WaveMetrics, Inc.
support@wavemetrics.com

May 31, 2012 at 09:22 am - Permalink

johnweeks wrote:
The function I wrote just computes a Gaussian peak for each point in the original waves and adds them together. Convolving a Gaussian with a series of delta functions does exactly that, so I don't think there's a difference in this case. Because my function just sums a bunch of Gaussian peaks, they will overlap if they are close together, where "close" means within some number of standard deviations of the center locations.


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