Sciencemadness Discussion Board

FTIR - FFT procedure

CodeHD - 15-7-2014 at 12:00

Hi,

I just joined this board, because I need some external comments, opinions and advice. As a Master's student, I currently work with an FT-Spectrometer. For some technical reasons, I recently tried to recreate/reverse engineer the precise algorithm the software supplied by the manufacturer (Bruker) uses to transform the interferogram. This has now been very much successful, with some little difference remaining that hopefully Bruker will help me with. During this, however, I have run into some trouble with understanding some basic principles about FTIR and the FFT that are given in the literature. I hope that someone here can help me with this. Here's my chain of thoughts (I appologize in advance for the long post, but I can't really describe it quicker):

In the literature, it is stated that the resolution only depends on the maximum optical path difference. In other words, there is a position where all spectral components are in phase - the peak in the interferogram (Zero Path difference - ZPD). Then you can either measure an asymmetric 'single sided' interferogram or a symmetric double sided interferogram. The resolution however is determined only by the path difference between ZPD and the end of the long arm, the length of the asymmetric arm only determines the "phase resolution" for the phase correction procedure.

Now, I imagine a theoretical interferogram consisting only of a single sine component. Fourier theory tells me, that the width of the sinc function produced from this depends on the length of my data. In this case, we can not assign any point of ZPD, because we can have arbitrarily many maxima. I can then continue to add more sine components, each adding another sinc function to my spectrum. The width of each sinc is still determined by the total length of the interval, which has not changed.

Now, in FTIR, there just happens to be this location - ZPD - where all my components are (approximately) in phase, which causes a large peak with a rapidly decaying envelope for a continuum source. I fail to see completely how this correlation of phase should suddenly influence the resolution. The total length of the interval has still not changed.

I can make some sense of it since I recreated the algorithm, when I read about one step that (so far) I have only found being described in this article on page 11 and 12:

http://mmrc.caltech.edu/FTIR/Understanding%20FTIR.pdf

The authors are multiplying the symmetric part of the interferogram by a ramp function. The reason given is not counting the symmetric part twice. In light of my description given earlier, I don't think that I am counting anything twice. It is additional information just like any other part of the interferogram.

The way I could rather understand it's necessity is if I want to apply apodisation to a single sided interferogram. Since the signal (the interferogram) is largest at ZPD, and hence my SNR is best here, I want to center my apodisation at this point. For a symmetric apodisation function and a single sided interferogram, this means that one end will be close to zero and the other end won't. If I now use a ramp in addition, this would bring both ends down and improve sidelobe suppression (this I have seen in practice). The this 'not counting twice' would also reduce my resolution.

I have never seen this ramp function being mentioned anywhere except for this one article. I wonder why other sources still state the resolution dependence as they do, and if there is to much physical meaning being interpreted into the existence of the ZPD-peak?

Maybe someone of you has some deeper knowledge of all this and can help me with my questions, I would be very happy.

Thanks,
CodeHD

Marvin - 16-7-2014 at 11:19

I'm not sure how much help I can be but ZPD is intrinsic to interferometry. Zero path difference essentially means you don't have an interferometer, the paths are combined in exactly the same way they were separated, so you learn nothing spectroscopically. It's only when you move away from this point that you get any spectral information as phases start to cancel.

At ZPD the thing is symmetrical so it doesn't really matter which arm you change to get this information, add a path difference to one side gets you the same result as adding it to the other side. The two copies are the same, thinking about it like that I'm amazed it gets you any new information. Prioritising the area around ZPD may simply boost the low resolution information that is already over represented.