A Brief Introduction To The Fourier Transform in FTIR

Transcription

1 Abstract A Brief Introduction To The in FTIR The Fourier transform is a method of converting a periodic signal from the time domain (x-axis of a graph is time) to the frequency domain (x-axis is frequency). In the case of FTIR we are converting from the distance domain (distance is a function of time) to the frequency domain (in wavenumbers). This document is not intended to be a rigorous examination of the subject (that would require a whole book) but to provide an introduction to the basics from a humble organic chemist s point of view. The mathematics and computer science behind the discrete Fourier transform methods used in modern instruments are awesome and powerful. I encourage students who are interested to examine the subject further. You will find yourself exploring subjects in physics, mathematics, computer science and analytical chemistry that may launch you into a whole new direction of study and career choices. I chose organic chemistry and so I lack most of the tools required to understand this topic but I will do my best. Please correct me where I go wrong. The Fourier transform converts a signal function to a spectrum and a spectrum to a signal function. Consider what the gentleman in the cartoon above looks like and you may get an idea of what it can do. 1 1 Cartoon from It's Hot in Here - Wacky Works by Virgil Partch of Collier's, Virgil Partch, Robert M. McBride & Company, Image obtained from on March 6, Version 0.1 Page 1 of 15

2 The Imagine a signal, perhaps from a laser, with a given frequency. After passing it through an interferometer we can observe the wavelength by the voltage changes in the photodetector as we move the mobile mirror (recall my attempt at introducing you to the interferometer, a simple yet amazing instrument used in many areas of scientific endeavor). The voltage fluctuates with the same wavelength as the laser light. And so we have a voltage signal similar to the graph in figure 1. We could simply measure the wavelength from the graph and get the wavenumber easily. But can we can also do this using the. Here is a signal from the detector. We can easily determine the frequency by looking at the signal. y = cos(5x) Figure 1: An example signal. x-axis is distance, y-axis is voltage. We could easily determine the frequency by inspecting the graph. In this case we imagine a frequency of 5 cm. The function plotted here is y = cos(5x) 1 Here is a simplified version of the Fourier transform. y = n 0 Signal(x) Test(x) dx (1.1) where n is the span of the calculation. For best results n = but no one has that much time on their hands. Observe that we multiply the function that represents our signal by a function that represents a test signal. If we integrate the product we will get a zero value if the frequencies are not equal, and a maximum value when the frequencies are equal. 2 By repeating the operation at every frequency in the range of interest we can determine which frequencies are contributing to a signal. In our example we expect to see only one single frequency, let us see if that bears out. 2 Please don t call me a liar when you learn that it is not quite that simple after a few more pages. Version 0.1 Page 2 of 15

3 In our system above we have a periodicity of 5, representing a signal with a frequency of 5 cm. We see this signal in figure 1. The signal function is cos(5x). If we test it 1 3 against a test function with the same periodicity we should get a maximum value for the area under the curve (the integration). First visually examine the graph of the product of the signal function and the test function as shown in figure 2. The area of the product function will add up to a large value. Multiply the signal by a test frequency. If the frequencies are identical, the integration of the result will be a non-zero value. y = cos(5x) cos(5x) cos(5x) cos(5x) dx = 1 Figure 2: The signal multiplied by a test function with the same frequency. Observe how every part of the curve is above zero. The function plotted above is y = cos(5x) cos(5x). The integration will be a maximum value. Now let us try test functions with periodicities that are not identical to the signal. In figures 3 and 4 we see that the product of the signal and test functions has equal area above and below the zero and the total area under the curve will add up to zero within a given period. If the frequencies are different the integration will be zero. y = cos(5x) cos(6x) cos(5x) cos(6x) dx = 0 Figure 3: The signal multiplied by a test function with a different frequency. The function plotted in this figure is y = cos(5x) cos(6x). The integration will be zero. 3 I know that IR frequencies are in the 500 to 4500 cm 1 range but this is just a simplified example. Version 0.1 Page 3 of 15

4 The area of the product function includes equal positive and negative regions and totals up to zero. y = cos(5x) cos(3x) cos(5x) cos(3x) dx = 0 Figure 4: The signal multiplied by a test function with yet another frequency. The function above is y = cos(5x) cos(3x).. The integration will again be zero. So we just need to repeat this operation with every test signal that we want. Let us test the observed signal against every frequency from 1 to 10 cm 1. We will use the following algorithm for v from 1 to 10 y = end do 2π 0 cos(5x) cos(v x) dx plot(x, y) on a graph (1.2) The region integrated is 0 to 2π radians, which represents 1 cm in the scheme we are using. (recall that cos(x) completes a single cycle of 2π radians, so cos(5x) represents a signal with 5 oscillations in the same period). 4 After running the algorithm (or performing it manualy it is only 10 operations after all) we have the following data set plotted on a graph in figure 5. 4 See the appendix for the calculations that generated the figures in this document. All the plots were created using a mathematics package. I used Maple but there are many other choices. Version 0.1 Page 4 of 15

5 cos(5x) cos(5x) dx = 1 Only the existing frequency gives a value in the Fourier Transform. cos(5x) cos(3x) dx = 0 cos(5x) cos(6x) dx = 0 Figure 5: After performing a simplified Fourier transform. Observe how the frequency of the signal is identified by the peak in the graph. We have converted the signal from a sinusoidal function of distance (cm) to a function of frequency (cm 1 ). 5 The Solves A Mystery What frequencies combine to create the signal presented in figure 6? Can you find the answer by inspecting the graph or do you need a little help? Perhaps a Fourier transform, will help. Can you tell me what frequencies are in this signal? Figure 6: A mystery interferogram. Distance (cm) We can use the Fourier transform method to convert the signal from a changing intensity as a function of the distance the mirror moved to a changing intensity as a function of the frequency. The result will show us the component frequencies as a series of discrete peaks. So perform the algorithm from equation 1.2 and test the signal for frequencies from 1 to 10 cm 1. The result is plotted in figure 7. 5 I know that I have ranted against connecting dots in graphs that s for business, not science I have said. The graph in figure 5 should be represented as a bar graph with a bar at 100% for 5 cm 1 and the rest of the bars at zero. But I must admit that it looks simpler as a scatter plot with connecting lines. Call me a hypocrite, I can take it. Version 0.1 Page 5 of 15

6 We see that the signal contained frequencies of 5 and 8 cm 1. Figure 7: A Fourier transform of the signal from figure 7. Its Not So Simple Real and Imaginary Spectra The Fourier transform presented in the above is very simple. It will not detect only detect cosine signals (sinusoidal signals with a maximum value at x=0) and will not detect a sine wave in the same signal. A sine wave can be considered to be orthogonal to a cosine wave (you cannot use one to create the other). Another way to look at it is that sine waves are out-of-phase with cosine waves. You can concert one to the other with a phase difference of π / 2 radians (90º). If a frequency is out-of-phase and we cannot see it with the cosine Fourier transform we someone might say it doesn t exist. But it does it is imaginary. We can see this imaginary signal using the sine Fourier transform. So, in doing a Fourier transform, your computer will calculate both the real (cosine) and imaginary (sine) Fourier transforms and combine the results to give you your final spectrum. The calculations involved are outside of my ability to explain but we will explore this issue again when we discuss Fourier transform in NMR signals. y = y = n 0 n 0 Signal cos(v x) dx Signal sin(v x) dx "real" "imaginary" For the rest of this introduction we will stay with the cosine Fourier transform only to keep things simple. Integers Are Whole Numbers But They re Not Real Numbers Our spectra have been generated by computing the Fourier transform using test signals of integer wavenumbers. But the electromagnetic spectrum is a continuum, not a set of discrete frequencies. What if we use test functions with values in between the integers? How about 1, 1.1, 1.2, cm 1 as test frequencies? We will tweak our Fourier transform algorithm and run it using the signal from figure 6. (1.3) Version 0.1 Page 6 of 15

7 for v from 1 to 10 by 0.1 y = end do 2π 0 Signal cos(v x) dx plot(x, y) on a graph This algorithm tests the signal function against frequencies from 1 to 10 cm in steps of cm. The Signal function is now known to be Signal = cos(5x) + cos(8x) so we 1 expect to see peaks for frequencies of 5 and 8 cm. The resulting data points are plotted 1 in figure 8. (1.4) y = cos(5x) + cos(8x) Distance (cm) Figure 8: A Fourier transform (right) of the signal from figure 6 (left) using smaller steps. Although the value of the transform is zero on the integer frequencies, we see strange side-lobe signals. The result has the same points as the transform from figure 7 and 10 additional points inbetween each of those. Observe the side-lobes around each peak. This is a result of the fact that the signal cuts off suddenly in a square edge because I can t run the transform out to infinity. The Fourier transform of a square wave has the exact side-lobe pattern we see in the frequency domain. So we see the frequencies in the signal and we see the frequencies that are used to describe a square wave. In a spectrum with many frequencies and many weak signals, thes side-lobes will interfere with interpretation. Apodization: Quieting that ringing in your ears y = 2 0 [ cos(5x) + cos(8x) ] cos(v x) dx How do we improve the result of such a Fourier transform as the one in figure 8? The problem is that we stopped the integration at a square edge. We can t integrate to infinity, but we can bring the signal down to zero in the integration span. Once it s zero, going to infinity won t matter. The shape of the function that is used to bring the signal down to zero will also result in artifacts like the side-lobes of the square wave (in a way, the square wave is the most simple apodization function). There are many, many apodization functions and this area is still an active field of research. One of the simplest is a simple triangle. In figure 9 we see the signal from figure 6 with a triangular apodization function applied and the corresponding Fourier transform. Version 0.1 Page 7 of 15

8 To minimize ringing artifacts the signal must be zero at the extremes. y = n x n We can make this happen using an apodization function. Distance (cm) The artifacts are smaller now. There are better apodization functions than a simple triangle. Figure 9: Apodization of the signal reduces line shape artifacts. Here we see the signal after a simple triangular apodization function is applied (left) and the Fourier transform of the modified signal (right). I will stick to the triangular apodization function (also known as a Bartlett apodization function) in this simplified introduction. There are many choices for these functions. 6 Which does your instrument use? Resolution Depends on Information We have so far only integrated our signal over a distance of 1 cm 1. This is a very small distance compared to the wavelengths involved in our make-belief signal. Observe that the peaks in the Fourier transform have a line width. They are not simply just sticks at the given frequency (Dirac delta functions) but resemble distribution functions. The line width is the resolution of the spectrum. In our simple signal we would expect infinitely thin lines at frequency values of 5 and 8 cm 1 after the Fourier transform. This is because we only have those two exact frequencies in the signal. But to have a perfect Fourier transform we need an infinitely long signal. The longer we extend the integration, the more information we are including in the signal and the higher the resolution. However, in real life the signal is decaying away to near zero as the wavelengths fall out of phase in the inteferogram. At some point instrumental noise will dominate so we don t want to extend our integration too far. A longer mirror movement will give us a higher resolution and a noisier spectrum. There is an optimum that you will choose depending on the quality of the signal and your resolution needs. In most cases we don t need extremely high resolution in IR spectroscopy. Figure 10 shows an example of increasing the span of the signal in the Fourier transform. 6 For a very brief introduction to other apodization methods see the link below to the Shimadzu corporation s website. They make and sell FTIR instruments and have a useful website with tips and helpful information. Version 0.1 Page 8 of 15

9 A span of 0 to 2π A span of 0 to 6π A span of 0 to 20π Figure 10: Increasing resolution by increasing the span of the signal. The more information in the signal, the higher the resolution of the Fourier transform. Observe how the peaks get sharper as the span of the signal is increased. The triangular apodization was applied over each span to minimize side lobes. Also, notice how the span of the side-lobes is related to the line width. The pattern of lobes is part of the line shape for each signal and a narrower line also means a narrower span for the side-lobes. In the final spectrum the line width is narrower than the spacing of our steps in the algorithm (less than 0.1 cm 1 ). So we don t see the side-lobes because there are not enough points in the graph to resolve them. In this case we might want to increase our digital resolution so we are testing the signal at more points. Digital Resolution The number of points that we collect in an interferogram will determine the spacing between data points. The more points the narrower the spacing. This is the digital resolution. If we don t collect enough points then narrow features will not be resolved. The digital resolution must be greater than the expected resolution of the spectrum. Your instrument is not producing a continuous algebraic function like the ones that I used to create the above plots. It is collecting voltage data at regular points along the mirror movement of the interferometer. In my youth there was such a thing as too many points as computer memory was expensive, but your phone has 100 times more memory in it than the workstation that I used in graduate school. Digital resolution used to be something to fret over with cost-benefit relationships that mattered. Not anymore collect all the points you want. In FTIR the digital resolution is usually set by the frequency of the reference laser and not a decision that you control. But always consider this concept if you are working with different spectrometers where it can be changed easily. Figure 11 shows the interferogram of our signal at various digital resolutions. Notice how a low digital resolution results in a signal that makes no sense. Too low of a resolution will cause the discrete Fourier transform to produce strange results. Version 0.1 Page 9 of 15

10 Figure 11: A demonstration of the effect of digital resolution. As we capture fewer points in our data file, the signal starts to loose definition. Too low of a digital resolution will give incorrect results. The computer will use the data points in the file and perform a discrete Fourier transformation, which is a computational method that tests data points generated by your test functions against the data points in the file. The sum of the products will be the Fourier transform result for each tested frequency. This technique is called the fast Fourier transform. It is a discrete method and it requires a number of data points equal to a power of 2 (e.g. 1024, 2048, 4096, etc ). If you don t have enough data points you can pad the file with zeros after apodization (you took the signal down to zero at the end of the file anyway, what s a few more?). This is called zero filling. Looking Ahead So remember that the spectrum that you see on the screen of your FTIR starts as a voltage signal created by a photodetector and translated by an analog-digital converter chip. This creates a digital data file that is interpreted by your computer after some mathematical processing. Different apodization functions and other mathematical techniques can improve resolution, enhance certain regions, reduce noise and perform other useful purposes. Or they can just completely fool you and present incorrect results if not used properly. Always remember that what you see is not real, it is an interpretation of a signal that is measured by an instrument. And the most important instrument is you. Below, in figure 12, we see a simulated interferogram that combines 3000 frequencies from 500 to 3500 wavenumbers. I then subtracted part (but not all) of the signal in three different sections of the spectrum. The Fourier transform of that signal is present as well. The Fourier transform was performed using the algorithm presented here but that was a mistake. It took ALL night! And I think my computer overheated and was damaged somehow. This is because Maple was computing integrations for a continuous trigonometric function with over 3000 terms and was doing it 3000 times to complete the algorithm. Algebraic methods are exact but numeric methods are fast and cheap (computer resource cheap) but approximate. A numeric method like the fast Fourier transform algorithm uses matrices and linear algebra two things computers excel at to quickly provide an almost accurate list of frequencies and intensities based on the signal. Version 0.1 Page 10 of 15

11 The FFT algorithm is the key to our modern lifestyle. If you make a better one, you will make serious money. Signal from photodetector Optical Path Difference (cm) 3000 frequencies are present in this signal. Some are subtracted in full or in part. Can you tell me which frequencies have been attenuated? This is the kind of complex signal that Fourier transform can analyze to give us a human-readable spectrogram Figure 12: A simulated interferogram from an FTIR experiment (left) and the Fourier transform of the signal (right). Observe how the Fourier transform not only gives a list of frequencies that are present, it gives us the relative intensities as well. How does this compare to a typical FTIR spectrum? Signal 1000 to 1080 cm 1, 25% transmittance Does this resemble a FTIR spectrum? 2850 to 2950 cm 1, 40% transmittance 1750 to 1800 cm 1, 10% transmittance MICRONS NICOLET 20SX FT-IR % T R A N S M I T T A N C E A B S O R B A N C E WAVENUMBERS Figure 13: FTIR spectrum of ethyl butyrate. This is the Fourier transform an interferogram recorded by the instrument. We will be revisiting most of these topics in more detail when we explore NMR spectroscopy. These computational methods and their consequences are more important to the NMR user than they are to the everyday FTIR user. Version 0.1 Page 11 of 15

12 Appendix: Maple Commands How did I generate the plots in this introduction and how can you do it yourself? Its easy! I encourage you to explore the world of combining frequencies in signals and performing your own simplified Fourier transforms to observe the effects of changing important parameters. Perhaps you might choose to explore different apodization functions. Below are some maple commands for generating the plots in this document. I am not going to define every command but the help pages of Maple will reveal any answers that you need. Creating a signal function. S := (w) -> cos(w*x*2*pi); This is the same as saying S w where w is the wavenumber that you will input into the function. x is the distance covered by the wave (x-axis of the signal plot). Using the signal function You can now generate a signal plot just by plotting the function directly Signal := S(7); plot(signal, x = 0.. 1); This sequence of commands creates a Signal that is a cosine wave with 7 oscillations in a period (2π). The function is created so that when x=1 we are at 2π radians. So we will see 7 waves in a unit length. If we imagine that the unit is 1 cm then the wave has a frequency of 7 cm 1. Combining signals We can combine frequencies just be adding up signal functions. Signal := S(7)+S(4)+S(9); plot(signal, x = 0.. 1); of a Signal ( ) = cos( 2π x w) The algorithm for the simple cosine fourier transform that I used is outlined below. I am sure many of you will be able to write a much better one or figure out how to use the fourier command in Maple. (Yes there is a fourier transform procedure built into Maple but I can t get it to make a plot. Can you?). Version 0.1 Page 12 of 15

13 X := []; Y := []; n := 10; inc :=.1; for b from 0 by inc to 10 do Test := S(b); A := Signal*Test; F := int(a, x = 0.. n); X := [op(x), b]; Y := [op(y), F]; end do: plot(x,y); Explore these commands. Type?plot,options for ways to make your plots look the way you want. Have fun! Version 0.1 Page 13 of 15

14 Bibliography I m an organic chemist. I know very little math but I do love it. As I was preparing this lecture I encountered the following resources that you might want to explore if the subject of Fourier transform has excited you. A useful textbook that I found is Brian C. Smith, Fundamentals of Infrared Spectroscopy, 2011, CRC Press, Boca Raton A Series of useful articles on FTIR including the nature of the instrument, the interferometer, Fourier transform and data processing. Gronholtz, J. and Herres, W., Understanding FT-IR data processing. Part 1: Data acquisition and Fourier transformation, Comp. App. Lab. 2, 216 (1984). Herres, W. and Gronholtz, J., Understanding FT-IR data processing. Part 2, Instruments and Computers 3, 10 (1985). Herres, W. and Gronholtz, J., Understanding FT-IR data processing. Part 3, Instruments and Computers 3, 45 (1985). I can t find either of the journals online in any form; they may be extinct. I accessed the documents at Some articles on the importance of Fourier transform in our lives. A full free textbook on Fourier transform from a course at Stanford University. I don t know if it is free but it s on the web. A lab exercise in FTIR Fourier transform A brief overview of Fourier transform for a Stanford neuroscience course. The Schimadzu corporation has a great series of articles about FTIR. There are several great short articles on preparing samples that you should read. You can find the all at For this review I read the article on Fourier transform and apodization from the Schimadzu FTIR Talk magazine. Version 0.1 Page 14 of 15

15 And always remember Wikipedia and Google. I found so many resources it was impossible to read them all. About KeynoteChemistry.com This document was obtained at KeynoteChemistry.com. I created this website to share the documents that accompany the videos on the KeynoteChemistry YouTube Channel. Visit the website to find more videos and documents related to spectroscopy, biochemistry and organic chemistry. This document is made available under the Creative Commons Attribution- NonCommercial-NoDerivatives 4.0 International License. Version 0.1 Page 15 of 15