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Hi all,

I have a single raman peak and i tried to fit a lorentzian and a voigt profile....I am attaching both fitting with their residuals. In the left part of the fitting there is a small trend in the residuals showing a slight inconsistency (or at least it looks like that). I am not expecting another peak so its a single fitting. Both fittings are quite similar with the voigt profile yielding a slightly better fitting. Does anybody have any ideas of this small inconsistency on the left part of the fitting or do you think its just nothing? Yhanks very much.

Konstantinos Lorentzian fitting.jpg (79.09 KB) Voigt fitting.jpg (83.14 KB)
Hi Konstantinos,

I have been out of Raman for number of years, but did loads way back. There is not much to go on, but here are some ideas:

1. Do you have data from a reference sample that you *know* yields an ideal symmetric peak that fits to your line shape within the noise level? - this would provide evidence to rule out instrumentation effects.

2. Is you sample in the solid state? A slight asymmetry could be caused by internal stresses/defects within the sample. I have seen this type of thing with silicon (where one is limited to sampling near the surface due to the opacity of the sample - especially for short excitation wavelengths) and diamond (with deliberate asymmetric stresses, or with known defects.

I think I would need more information about your sample to speculate any further.

Hope this helps,
Kurt

May 15, 2017 at 05:49 am - Permalink

Hi Kurt,

Thanks very much for your response......Actually this peak is related to the sulphate-oxygen stretching of free sulphate ions in aqueous sulphate. So essentially the specimen is just liquid (aqueous sodium sulphate).....It looks a bit assymetric but only slightly according to my eyes at least....I also considered a number of reasons but as far as i know from literature it is expected to be symmetric. I have a series of these spectra as a function of time and at the moment i fitted a lorentzian for all of them in a consistent way to see if there are any changes.

May 15, 2017 at 06:04 am - Permalink

OK. So that rules out my solid-state ideas. What about the first point? Do you get symmetric Raman bands from reference samples? For example you could use organic liquids?
One other idea that springs to mind is what is your x-axis? Is this in units of energy or wavelength? If you were to transform your axis would the peak then become (more) symmetric? In your literature, what axis units were used to report symmetrical peaks?

[edit]: This article may be relevant:
https://link.springer.com/article/10.1007%2FBF00972713?LI=true

Regards,
Kurt

May 15, 2017 at 06:33 am - Permalink

Hi Kurt,

I do not think that it is instrumental because i have symmetric profiles from other liquid solutions....The axis is in wavenumbers and in general references are in the wavenumber scale. I hardly see any Raman spectra recorded in nanometers but i get your point....maybe the conversion induces this.....I guess there must be some kind of physical reason giving me this which i am missing at the moment.....What do you think about the amount of assymetry? I mean to me it looks quite subtle and this is why i am using Lorentzian (i also check the Voigt) to quantify intensities and areas. However, i do the same thing consistently for all my spectra to be able to compare if needed.

May 15, 2017 at 06:35 am - Permalink

A few points of interest ...

* The fundamentals should drive the choice of peak shape, not the need to remove an empirically discovered residual by "fudging" the peak shape. Which peak shape best for the fundamentals?

* From the fundamentals of the probe -> event -> signal, Raman responses should/will give symmetrical response (peak) on the energy scale. This is the wavenumber not wavelength scale.

* The baseline below the peak is a straight line. The extension through the entire spectrum is not. What happens when you fit the baseline first to remove it to zero identically and then fit the peak? What effect does this have?

* When you fit a symmetrical peak of any shape (Gauss, Lorentz, Voigt) to an asymmetrical peak, you will always get a residual.

When all else is properly done (proper peak shape chosen and proper baseline removed), what is left must itself have a fundamental basis. The use of a PCA/PCR method may provide another option to discover whether this peak requires other components.

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

May 15, 2017 at 07:06 am - Permalink

Thanks very much for these points.....I have isolated that peak of interest and i fitted a linear baseline. As you can see in the graphs attached, i fitted the baseline and the peak together as generally suggested from literature tutorials. The voigt profile yields a shape factor of ~ 3 indicating the prominent Lorentzian character of my peak. I tried both voigt and lorentzian just to see what i get. This small residual is always there 9using both profiles) indicating a slight assymetry. Now, whether this assymetry is due to any physical reason or due to an additional component its a matter of debate. However, i am not expecting any additional component according to the specific material system and the relevant literature on this. This is why i have the feeling that something causes this subtle assymetry on the left side.

May 15, 2017 at 07:29 am - Permalink

What's the pH / concentration of your solution?
Try to fit a second peak on the 'left' side of your peak (x-axis is missing in the images). It's intensity could correlate with species like NaOSO3- or HSO4- (Check against the respective equilibrium constants...)

HJ

May 15, 2017 at 08:26 am - Permalink

Konstantinos Chatzipanagis wrote:
Thanks very much for these points.....I have isolated that peak of interest and i fitted a linear baseline. As you can see in the graphs attached, i fitted the baseline and the peak together as generally suggested from literature tutorials.


The baseline is NOT linear across the entire spectrum. In this case, the sequence to fit a baseline + peak together is EXACTLY same as removing the baseline first and then fitting the peak because a linear baseline function has no dependence on the shape of the peak and peak fitting is co-addition (not convolution).

I suspect a major part if not all of the residuals arise because the wrong baseline is used.



What happens when you remove the curvature of the baseline to make it linear and then fit the peak?

--
J. J. Weimer
Chemistry / Chemical & Materials Engineering, UAH
wrongbaseline.jpg (133.31 KB)

May 15, 2017 at 08:42 am - Permalink

Thanks for this John.....I had this suspicion as well but because i restricted the fitting range i thought it would not really affect.....If i guess right you suggest that if i remove the background using another baseline function (not linear since its not linear in the entire spectrum) and then perform the fitting i may see that the fitting becomes better, is that right?

May 15, 2017 at 10:25 am - Permalink

Konstantinos Chatzipanagis wrote:
Thanks for this John.....I had this suspicion as well but because i restricted the fitting range i thought it would not really affect.....If i guess right you suggest that if i
remove the background using another baseline function (not linear since its not linear in the entire spectrum) and then perform the fitting i may see that the fitting becomes better, is that right?


You have two options. Method 1 is "brute force". Method 2 is the elegant (and possibly more rigorous) method.

Method 1
--
Fit just the background to a parabolic (quadratic) function over the range shown by the display. Use the cursors to exclude the peak region. Subtract the result from the raw data (and of course save this in a separate wave). Then, fit the resultant wave (with a perfectly zero baseline) to a peak of your choice.

Method 2
--
Modify the fit function to use a parabolic baseline + a Lorentz (or Voigt) peak simultaneously. Fit over a region that is as broad as your display.

I suspect the improper baseline function is the real problem. Your residual is NEGATIVE going, not positive going. This is a LOSS not a gain (of a new peak).

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

May 15, 2017 at 01:10 pm - Permalink

He's using Multipeak Fit 2 even though he has only one peak, a perfectly fine thing to do. Jeff, you are saying to fit a parabolic baseline, which doesn't exist in Multipeak Fit, but he could use a cubic baseline. If it is truly parabolic presumably the extra order will be fit with a small coefficient.

John Weeks
WaveMetrics, Inc.
support@wavemetrics.com

May 15, 2017 at 01:59 pm - Permalink

Yes, absent a quadratic a cubic baseline would do. I only suggested a second order because the baseline has no inflection point. The third order term will as you say likely go to zero.

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

May 15, 2017 at 08:23 pm - Permalink

Thanks both of you very much....It looks like a cubic function works better....However, i noticed that that the area values do not really change significantly....I have a difference of about 3% which to me makes sense since even with a linear baseline the fitting is still adequate. I guess sometimes getting rid of the background and then perform fitting is maybe better, right?

May 16, 2017 at 07:21 am - Permalink

I have always felt that it was better to fit the baseline along with the peaks because that way you don't have to worry about selecting background areas that have negligible "contamination" from the tails of the peaks. That would be particularly true for a Lorentzian peak shape because it has heavy tails. Since you have just one peak and lots of background, perhaps that isn't a significant problem. But I still think it's better to fit them together- using just the outer areas to fit a baseline separately means that you aren't getting the full benefit of the entire range of data; you're fitting the baseline to a data set with a big hole in the middle.

The small difference in the area isn't surprising- the misfit using the linear baseline is pretty small. Consider the area under the residual curve compared to the area of the peak.

John Weeks
WaveMetrics, Inc.
support@wavemetrics.com

May 16, 2017 at 09:00 am - Permalink

I agree with John. When possible, always fit the baseline and the peak(s) at the same time.

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

May 16, 2017 at 02:40 pm - Permalink