Talk:Fourier transform infrared spectroscopy

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don't get it[edit]

It states what the difference is between it and standard spectroscopy (lots of colours at once and in various different patterns), but not why you'd want to do that. —Preceding unsigned comment added by 82.24.47.178 (talk) 00:57, 19 August 2010 (UTC)

The answer lies in the advantages described in the section on the Michelson interferometer, below. Petergans (talk) 08:43, 19 August 2010 (UTC)

conceptual introduction[edit]

Reflectance (ATR) is also important. There are pictures of ATR devices in Wiki commons, but in FTIR a cylindrical rather than a flat crystal is used. I didn't think it worth going into so much technical detail. Petergans (talk) 21:05, 1 August 2010 (UTC)

Secondly, I don't think that the illustrated interferogram is real. It looks like a sinc function which would imply a monochromatic source. Petergans (talk) 21:27, 1 August 2010 (UTC)

Good point, it is too specific to say the light passes through the sample. I'm sure it can be reworded somehow. :-)
You're right that the interferogram has more peaks than a typical spectrum. I don't know where it's from. I have a few interferograms on my computer, I guess I could plot and upload one when I get a chance? :-) --Steve (talk) 21:58, 1 August 2010 (UTC)

I've just looked at Frustrated total internal reflection. I believe that this is the same as ATR. Some general clean-up is needed. I don't think that the link as it stands is helpful.Petergans (talk) 08:48, 2 August 2010 (UTC)

Another point. "while a disadvantage is that FTIR cannot use as sensitive a detector as dispersive measurements can". I wonder if this is any longer true. See Mercury cadmium telluride#infrared detection.Petergans (talk) 15:09, 2 August 2010 (UTC) Also mct vs DTGS comparison

Detector elements could be same, but FTIR detection is crippled by restriction on electronics. As a result, sensitivity of dispersive measurements is greatly superior to FTIR. It is not only lock-in detection, which gives orders of magnitude better S/N, even low-pass electronic filter helps a lot for NIR detectors - those filters are restricted in FTIR by the scanning speed. Materialscientist (talk) 23:32, 2 August 2010 (UTC)
For the sensitivity claim, I just copied it out of a book without double-checking. I don't want to say whether I think it's right or wrong without looking into it more.
For "frustrated total internal reflection", I just moved a redirect from one page to another . The acronym is legitimate, i.e. people really do occasionally use the initials "FTIR" to stand for "Frustrated Total Internal Reflection" rather than "Fourier Transform Infrared spectroscopy". If you have a problem with the frustrated total internal reflection article, you should discuss it at that page, which I haven't looked at. (The way I learned it, it's closely related to ATR but not quite the same, or at least the terms are used in slightly different situations.) Anyway, this is a disambiguation-type link, I think it needs to be there even if the frustrated total internal reflection article is bad. :-) --Steve (talk) 01:09, 3 August 2010 (UTC)

Detectors[edit]

There should be a section on detectors. Can anyone help? I created a stub on triglycine sulfate which, amazingly, was (as of 5 Aug 2010) not even mentioned in infrared detector. An important point is that PbS and related detectors are useful in NIR and are often available in UV/Vis spectrometers with a wavelength range that extends into the NIR.Petergans (talk) 16:30, 5 August 2010 (UTC)

Interferogram Plot[edit]

I think there is an error in the plot of the interferogram: Since the axis is labeled Intensity negative values make no sense. I have seen a different version of this plot in one of my lectures with the center of the axis being 0.5, which, in my opinion would make more sense —Preceding unsigned comment added by 78.142.92.52 (talk) 20:18, 5 January 2011 (UTC)

Yes, for sure the limit of large displacement (center of the axis) should be 50% of the maximum intensity. It should not be labeled "0", but it is. I don't have any interferogram data to make a new plot. Do you? --Steve (talk) 23:22, 5 January 2011 (UTC)
For all that, this is a bad illustration as it looks like a sinc function. A sinc function is the FT from a box function, so the plot is not, as the caption states, "An interferogram from an FTIR spectrometer". Petergans (talk) 12:04, 6 January 2011 (UTC)
Many measurements target a tiny peak at a certain wavenumber - such changes would not be seen in the interferogram. Materialscientist (talk) 03:14, 7 April 2011 (UTC)

Weird terminology[edit]

In the section "far-infrared FTIR" there is the phrase "For the relatively long wavelengths of the far infrared (~10 μm), tolerances are adequate, whereas for the rock-salt region tolerances have to be better than 1 μm."

What the heck is the 'rock-salt region'??? This is used again in the mid-IR section. I've read mainly about thermal and near-IR FTIR, but I've never heard this term, and a google search shows up mainly this article and copies of it. Is this a real term? -- snheath@gmail.com (6 April 2011)

Many of these terms are obscure jargon and are to be avoided, but I've uncommented a history section which explains the rock-salt range. Materialscientist (talk) 03:11, 7 April 2011 (UTC)

New interferogram picture[edit]

I just replaced the interferogram image (from File:Ftir-interferogramEn.png to File:FTIR-interferogram.svg). I took this one myself, so there is no question that it is a real interferogram from a real FTIR machine (unlike the previous one). I'm happy to edit the image if there are any suggestions :-) --Steve (talk) 04:54, 11 June 2011 (UTC)

Conceptual introduction[edit]

I made this change, reverting an edit by 7homas.martin to the version I had originally written.

7homas.martin, you said that my version had a "mistake about how the Michelson interferometer works", but I see no mistake. Can you tell me what the mistake is?

The reason I like my version better is that I think it is clearer as an introduction for someone who was previously unfamiliar with anything about FTIRs. For example, 7homas's version says "The light shines into a certain configuration of mirrors, called a Michelson interferometer, that allows to record an interferogram as the mirror is moving (due to wave interference)." The way this sentence is written, the reader is assumed to already know what an interferogram is. Most readers do not. The next sentence says "The interferogram is the mathematical equivalent of the spectrum", which is misleading and confusing. The meaning of "mathematical equivalent" is not obvious, even to a mathematician. I assume that 7homas means to say something like "The interferogram contains the same information as the spectrum and can be transformed into the spectrum." But a reader will understand this sentence as meaning something like "The interferogram is the same as the spectrum", which is not correct. I don't know whether they are "mathematically equivalent" when they are different functions, with different ranges, resolutions, noise properties, etc. I would say they are "fourier transforms of each other", but I would not say they are "mathematically equivalent to each other" without any more specifics about what I mean by that.

I just went through the first two sentences as an example. Like these, I found that all the changes were making things more confusing, especially for people who are not already experts on the subject of FTIRs. Therefore I changed it back.

Now that I've explained what I dislike about your version, 7homas, you can explain what you found confusing or misleading or incorrect about my wording! Hopefully we will wind up with something we can all agree on. :-) --Steve (talk) 13:26, 5 October 2012 (UTC)

(copied from User talk:Sbyrnes321)
I really understand what you mean but the problem is that viewing a Michelson interferometer as a wavelength filter is not right at all. In fact, that is how a spectrometer works. More over the usual mistake about the Michelson interferometer is to compare it to a spectrometer which it is not. You must know that all wavelengths are passing through a Michelson interferometer and that what you see on one of the output ports is the sum of the interference pattern of each wavelength for a certain optical path difference. I know that the way a Michelson interferometer works is not easy to understand but it may be better to avoid false description (I am sorry for my poor english). This is a description taken from Michelson interferometer : "The Michelson interferometer's detector in effect monitors all wavelengths simultaneously throughout the entire measurement, increasing the integration time and the total number of photons monitored" --7homas.martin (talk) 15:06, 5 October 2012 (UTC)
Thanks Sbyrnes321 for copying this. In fact the only thing I don't like is when you say that the Michelson interferometer "allows some wavelengths to pass through but blocks others". I think this is not correct and present the Michelson as a dispersive spectrometer which actually filters the spectrum to measure the intensity for a certain wavelength. --7homas.martin (talk) 14:31, 5 October 2012 (UTC)
I don't understand your statement: "all wavelengths are passing through a Michelson interferometer". Do you agree that if a Michelson interferometer is set to a path difference of 1mm, then light at a wavenumber of 1005cm-1 or 1015cm-1 or 1025cm-1 are 100% blocked by the interferometer? And that light at a wavenumber of 1000cm-1 or 1010cm-1 etc. are 100% transmitted? If you agree with those, why is it not correct to say that a Michelson interferometer is a wavelength filter? --Steve (talk) 13:42, 5 October 2012 (UTC)
Okay, I work on interferometers for astronomy, in the visual band (not in the infrared) so I don't know precisely how the instrument you are using works. But one thing I'm sure is that when you set your interferometer at a certain optical path difference (OPD), it does not act as a grating prism, and does not filter the light. The intensity you are measuring on the output is the sum of the interference patterns of all the interfering wavelengths of the incoming light (that is why we can say that "all wavelengths are passing through a Michelson interferometer"). If it was filtering light you would be recording a spectra on the output as you change the optical path difference but that is not the case. For example if you have a monochromatic beam on the input, as you will change the OPD you will see on the output an infinite sine wave (the interferogram is in this case the exact interference pattern of a monochromatic beam). If you now add another monochromatic light at a different wavelength you will see on the output (as you are recording the interferogram) that the pattern has changed, it is now the sum of two sine waves with different wavelengths. You will see something like the second figure of this article. Another way to see it is to consider that generally you only have access to one output port on a Michelson interferometer but there are two. When you see no light on one output port this is because the sum of the interference patterns for each wavelength is destructive. But the remaining energy is not "absorbed" by the Michelson, nor filtered, you will find that all the light as gone on the second output port (which might be the input port on a simple Michelson) because when the interferences are destructive on one port they are constructive on the other one. I hope that will clarify my point :) --7homas.martin (talk) 14:30, 5 October 2012 (UTC)
A Michelson interferometer is basically the same thing whether you use it for astronomy or for chemistry.
Let's talk about something very specific: "a Michelson interferometer, fixed at OPD = 1mm, where one of the ports goes back towards the input (or otherwise gets thrown out), and the other port is what we use and call the 'output' or 'transmitted' port".
If I shine light of wavenumber 1000cm-1 through this specific thing, a fraction F1 is transmitted. If I shine light of wavenumber 1005cm-1 through this same thing, a fraction F2 is transmitted.
Maybe you have the opinion that F1=F2? If so, it seems to me that you do not understand how a Michelson interferometer works!! I think you should re-read your textbooks or discuss with your colleagues...
Maybe you have the opinion that F1=100% and F2=0%. That's great, we agree!! But maybe you do not think that this thing is a "filter" because you think that "filter" means the same thing as "monochromator" or "narrow-band pass filter". Of course a Michelson interferometer is not a narrow-band pass filter. But I don't think you're using the word "filter" correctly. Usually people use the word "filter" to mean "anything that blocks or transmits or redirects light differently depending on the wavelength of the light". For example, there are high-pass filters, and notch filters, and filters with five different stop-bands, etc. None of these exotic filters will "be recording a spectra on the output", but that does not imply they are not "filters". A monochromator is a very specific type of filter; not all filters are monochromators. :-)
Maybe you have the opinion that F2 is not quite 0%, it is 0.01% because the interferometer is not perfect. Maybe that is what you mean when you say "all wavelengths are passing through a Michelson interferometer"? Well, I suppose you're right ... I can see how my description oversimplifies in this and related ways. I don't view it as terribly misleading -- I think it is kind of a pedantic objection -- but I am willing to work on rewording it to be more technically accurate. :-) --Steve (talk) 20:06, 5 October 2012 (UTC)
Well, I understand what you mean. Then tell me, if it was a filter : what are the filtered frequencies ? (because a filter in optics or electronics filters frequencies, if it just attenuates the signal of all frequencies it is not a filter). The fact is that, as I told you before, when you have nothing at the output all frequencies are participating in giving no signal at the output and that is the big advantage of using Michelson interferometers : you use every single photon to produce each data point of your interferogram (even if you see nothing at the output, strange isn't it ;) ). All my point is that using the word filter is misleading and confusing because : the general mistake about Michelson interferometers is to believe they work as a spectrograph which filter the frequencies.
Above all, I would like someone else (someone who wrote parts of this article) to help us in resolving this obvious conflict. I don't have time to look for someone now, but I will do it as soon as possible. One last thing : please try to avoid insulting sentences, I have read a bunch of books about Michelson interferometers and this instrument is my speciality but I think that has only little to see with what we are talking about. --7homas.martin (talk) 13:32, 6 October 2012 (UTC)
I want to be sure we agree on facts. Above, I was talking about "a Michelson interferometer, fixed at OPD = 1mm, where one of the ports goes back towards the input (or otherwise gets thrown out), and the other port is what we use and call the 'output' or 'transmitted' port". Do you agree with me that light of wavenumber 1000cm-1, 1010cm-1, 1020cm-1, etc. is 100% transmitted through this thing? And do you agree with me that light of wavenumber 1005cm-1, 1015cm-1, 1025cm-1, etc. is 0% transmitted through this thing?
This is not a question about terminology or pedagogy, just a straightforward factual question. I want to be sure that we agree on this very basic thing first. We can discuss terminology and pedagogy afterwards! :-) --Steve (talk) 14:13, 7 October 2012 (UTC)
I agree with you about that. My point (but that may not be so important after all) is that I don't like to describe the Michelson interferometer as a filter. And that is just a problem of terminology. I would prefer to say more generally : "The light shines into a certain configuration of mirrors, called a Michelson interferometer, that creates an interference pattern (the interferogram) which can be recorded by moving one of the mirrors". That avoid the possibility of understanding the Michelson interferometer as a simple kind of filter (narrow-band pass) because it is, to me, a common mistake. Maybe we can agree on that :) I am sorry for having changed a paragraph without talking about it first. I must confess I'm new to this. --7homas.martin (talk) 18:02, 9 October 2012 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── OK, this is helpful. My original wording omitted the word "filter" and I am happy to continue avoiding it (even though I personally think it's not misleading). Your suggestion seems problematic to me because the term "interference pattern" evokes something quite different. Some people do not know what an "interference pattern" is at all, but the rest will almost definitely be imagining an interference pattern in space. For example, I did a google image search for "interference pattern", and all the results were things like this, i.e. two-slit interference. As an FTIR scans, there is never this kind of "interference pattern" in the output: At every moment, the cross section of the beam is spatially uniform in both spectrum and intensity (assuming the FTIR is properly apertured). The oscillatory "pattern" only emerges in an abstract graph, like a graph of the spectrum at a given moment, or the interferogram plot. So "interference pattern" is evoking a misleading image.

Back to my old wording, I said:

The light shines into a certain configuration of mirrors, called a Michelson interferometer, that allows some wavelengths to pass through but blocks others (due to wave interference). The beam is modified for each new data point by moving one of the mirrors; this changes the set of wavelengths that pass through.

I could be wrong, but I think your objection stems from the fact that I did not clearly distinguish "the Michelson interferometer at any given moment, i.e. with a certain mirror configuration" from "the Michelson interferometer as the mirror is scanned". In the first sentence, I meant the former, but you read it as the latter, and I get the impression that you're not accustomed to thinking about the former at all. So let me try to improve...

The light shines into a Michelson interferometer—a certain configuration of mirrors, one of which is moved by a motor. As this mirror moves, each wavelength of light in the beam is periodically blocked, transmitted, blocked, transmitted, by the interferometer, due to wave interference. Different wavelengths are modulated at different rates, so that at each moment, the beam coming out of the interferometer has a different spectrum.

Does that help? :-) --Steve (talk) 19:14, 10 October 2012 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── That's really better, thanks :). The fact that it does not work as a simple filter is more clear this way. --7homas.martin (talk) 19:46, 10 October 2012 (UTC)

The basic concept[edit]

I've not read all through the rambling discussion above. It seems to be missing the point. The Michelson interferometer works by modulating the light from a "white" light source. The modulated light is passed through the sample, which absobs at specific wavelengths by specific amounts. The light is then demodulated to give back the spectrum of the original source, minus some intensity at wavelengths absorbed. The spectrum of the sample is obtained by removing the spectrum of the source (ratio-ing). The principal advantage of this procedure is that light intensity is measured at all wavelengths simultaneously. In the same way, astronomical measurements of emission spectra are collected at all wavelengths by passing the starlight through an interferometer, recording the intensity as a function of retardation, and then FT to recover the spectrum. In fact I believe that the interferometer was first developed for astronomy (Connes) and only later for ir spectroscopy.

More technically, mirror movement produces, by mechanical means, a Fourier transform of the spectrum of the source from the wavelength domain to the distance co-domain. The Fourier transform of the interferogram returns the data back to the wavelength domain. Petergans (talk) 13:30, 2 November 2012 (UTC)

I agree with everything you wrote. If you have a problem with any descriptions in the article don't be shy to say so :-) --Steve (talk) 22:28, 2 November 2012 (UTC)

I have a discussion type explanation of "The Basic Concept"

Imagine an oscilloscope screen shot of an audio recording containing many frequencies. Imagine a jumble of waves and wiggles. The signal can be digitized, and the fourier transform can be applied to create an Intensity versus Frequency graph.

I will take it for granted, that using Fourier Transform calculations, the raw signal from an audio microphone can be converted to a spectrum of frequencies. by digitizing the incoming signal, for a group of frequencies, that are, say 1MegaHertz and under.

The problem with Infrared frequencies, is that, first our detectors do not create an electrical wave corresponding to the very high frequency waves in the terahertz range. Also, in the current state of the art of digitizing a waveform, I think 100 Giga Samples per second is very difficult to pull off.

The incoming IR light signals, coming in with frequencies in the TeraHertz range, would be impossible to directly convert to an electrical signal, and then digitize it, at such fast frequencies.

The Michelson interferometry arrangement with a moveable mirror performs an essential function of converting an incoming Infrared frequency into a much lower corresponding Audio range frequency.

For example: Imagine a source emitting a single IR frequency, say one with a frequency of 3 TeraHertz, with a wavelength of:(Speed of light)/(number of waves) = (3.00x10exp13 mm/sec)/(3x10exp12 waves/sec) = 1mm

Now, in the interferometry setup with the moving mirror, let us choose a mirror moving at 1 cm/sec. This changes the travel distance, so after one second moving away, the new light beam has to travel an extra cm going to the mirror, as well as an extra cm, coming back from the mirror. When this beam is recombined with it's split copy, that did not change path length, there is a total of 2 cm containing 20 waves with length 1mm, which would constructively, and destructively interfere creating a sequence of dark and light cycles arriving at the IR detector at a rate of 20 Hz. So we see that the FTIR mirror setup will convert an originally 3 Thz wave into a 20 Htz wave. Likewise, if we study a 300 Terahertz wave, it will be converted to a 2000 Hz wave. Waves such as these are in the audio frequency range, in fact, if the raw signal from the FTIR detector was hooked to an amplified speaker, we could 'hear' the FTIR spectrum with our ears. But the point I wish to make, is that it is easy to see how the Fourier transform can now be done on this converted set of frequencies, to create the Intensity .vs. frequency graph, typical of what we take for granted, when using a modern digital oscilloscope, to analyze a raw scope probe signal.

In summary, I am taking it as a leap of faith, as to how the Fourier transform actually does its magic, of converting a raw signal into one that is a graph of intensity .vs. frequency. I AM pointing out that the arrangement of the moving and stationary mirrors within the FTIR instrument, can be thought of as converting the high TeraHertz range frequencies of IR waves, and converting them to a much lower 'Audio' range frequency, with each IR frequency having a calculated value in the lower frequency detected by the FTIR detector.

Thank you for considering this and I hope it adds some insight into how the FTIR works.

Bunrabbit (talk) 07:00, 27 May 2014 (UTC)

Resolution[edit]

Our Professor told us, that you have to make for a better resolution either the way of the mirror longer or the part of the spectrum smaller, you are measuring. Can anyone explain me why - and add it after in the article? I couldn't find it in any book... Thx! --Minihaa (talk) 13:52, 24 January 2013 (UTC)

Interferometer diagram[edit]

The interferometer diagram shows a "coherent light source." In the case of the FTIR, the light entering the interferometer is definitely not coherent. Otherwise, you wouldn't need an FT, you'd have a scanning dispersive spec. — Preceding unsigned comment added by Cyclotronics (talkcontribs) 14:29, 13 May 2014 (UTC)

"Coherent" presumably means spatially coherent (i.e. collimated) not temporally coherent (i.e. monochromatic). So I wouldn't quite say that the diagram is "wrong". I would say it's confusing and misleading. I would be very happy if you downloaded the file, deleted the word "coherent", and then re-uploaded it as a new version. --Steve (talk) 16:05, 14 July 2014 (UTC)