Ординатура / Офтальмология / Английские материалы / Optics Learning by Computing with Examples using MATLAB_Dieter Moller_2007

We now discuss the case where G(ν) is the product of two functions g1(ν) and g2(ν),

In the cases where one wants to keep the two functions separate, one can represent the Fourier transformation of G(ν) by a convolution integral. The convolution integral theorem tells us that we may write the Fourier transform of the product g1(ν)g2(ν) as the convolution integral of the Fourier transformations of g1(ν) and g2(ν). The Fourier transformation of g1(ν) is

The Fourier transform S(y) of the product G(ν) g1(ν)g2(ν) is calculated from the convolution integral

where s1(y) and s2(y) are the Fourier transforms of g1(ν) and g2(ν) (see Eqs. (9.18) and (9.19)).

A simple example can be discussed by splitting the Gauss function

The Fourier transform of G(ν) (i.e., S(y)) is now obtained from the convolution integral

If we had calculated the Fourier transformation of G(ν) exp −(a2ν2)/2, we would have obtained the same result:

In FileFig 9.11 we show the convolution integral for the case where the two factors are a step and a sin cx function.

FileFig 9.11 (F11CONVOS)

Convolution integral of the step function and the sin x/x function.

F11CONVOS is only on the CD.

Application 9.11. Calculate analytically the integral and show the effect of the convolution process of the step function on the sin x/x function.

9.2 FOURIER TRANSFORM SPECTROSCOPY

9.2.1Interferogram and Fourier Transformation. Superposition of Cosine Waves

We next discuss the superposition of plane waves in order to understand the formulas of Fourier transform spectroscopy. We consider a plane wave solution of the scalar wave equation

In Chapter 2 we discussed the dependence on x and t and on the phase factor φ. We write the two cosine waves with path difference x2 x1 − δ as

Only the second factor depends on the time. Since it oscillates very quickly; we can assume that only the average value is detected and results in a constant. Therefore we have

We see that the intensity of the superposition of the two harmonic waves depends only on the optical path difference δ.

9.2.2 Michelson Interferometer and Interferograms

The Michelson interferometer (see Figure 9.1) has been discussed in Chapter 2, and we assume that an ideal beam splitter reflects and transmits 50% of the incident light. The incident beam is divided at the beam splitter and each beam is reflected at one of the two mirrors M1 and M2. Part of the reflected beam from M1 is transmitted, and part of the beam coming from M2 is reflected. These two parts are superimposed and travel to the detector.

The optical path difference is introduced by the displacement of the mirror M2. If the mirror is displaced by nδ/2, the optical path difference is nδ for each wavelength component of the incident beam. The center position is at δ 0. There we have constructed interference for all wavelengths. For constructed interference in the direction of the detector, for a specific wavelength no light travels back to the source. For destructive interference in the direction of the detector, the light travels back to the source.

Since only the space dependence of the harmonic wave is of importance we write the amplitude of the input wave as

−∞

The subtraction of 1/2 J (0) from J (y) produces negative values for S(y) and makes it a Fourier integral of the type we have discussed above. If a doublesided interferogram is obtained, y runs from negative to positive values related to the position of the mirror M2 of the Michelson interferometer. Using the fast Fourier transformation, we can only consider 2n points. The input function starts with the positive part of S(y) and the negative part follows as a mirror image. Therefore, the information we are interested in is only contained in the single-sided interferogram.

The integral S(y) is one of the pair of Fourier transform integrals

−∞

(Note the negative sign in the exponent.) The integral of Eq. (9.54) is the one we need to calculate the frequency spectrum of the observed data, presented by S(y).

We have assumed that both variables ν and y have continuous values from −∞ to +∞. When G(ν) and S(y) are symmetrical with respect to zero, the imaginary part disappears and one has to deal only with one side of the integral. The Fourier transformation pairs are then

9.2.4 Discrete Length and Frequency Coordinates

9.2.4.1 Discrete Length Coordinates

The Fourier integral of Eq. (9.56) is the Fourier transformation of the interferogram function. For numerical calculations it is written as

Since the coordinates νi and yk each have n discrete values, the Fourier transformation is equivalent to the solution of a system of n linear equations.

The input data of the interferogram function S(yk ) are the measured data. The mirror M2 of the Michelson interferometer (Figure 9.1) is moved at equal steps from one position to the next, and the values of the interferogram are recorded at the detector. The zero path length position is the location of the central maximum. The input values of the interferogram function S(yk ) are calculated by subtraction of half the value of the central maximum (Eq. (9.52)).

In the cosine functions of Eq. (9.57) we have a set of values of yk and as a consequence we have a repetition of possible values of νi . Let us choose for the length interval yk k, which means we have 128 points from k 1 to 128, and consider

When k is 128, and νi is 1/128 we have νi k 1, which means all values of the cosine are in one cycle for k 1 to 128 and νi 1/128. When the product (νi k) is larger than 1, we have repetition. Let us assume the value νi 133/128 1 + 5/128,

is equivalent to νi 5/128. When using k 1 to 128 points in the space coordinate domain, after application of the Fourier transformation, all that we can get are 128 frequencies in the frequency domain. The repetition at 2π is shown in FileFig 9.12. In addition, the cos-functions of Eq. (9.57) have “mirror" symmetry at the center of the interval (at i 64), and at both ends (shown in FileFig 9.13). In FileFig 9.14 graphs are shown of amplitudes and intensities of the superposition of cosine waves. These superpositions are simulations of interferograms depending on the optical path difference.

Graphs are shown of cos(2πνi k) for the space coordinate from 1 to 128 and frequencies 1/128, 2/128, 3/128, 64/128, 127/128, 128/128, 129/128, and 130/128.

F12FTDISC1S is only on the CD.

Application 9.12. Consider a total of n 32 points in the space domain and show on a graph the first five frequencies. Then find the functions that have the same appearance around frequency numbers 32 and 64.

Graphs are shown of cos(2πνi k) for the space coordinate from 1 to 128 and frequencies 1/128, 2/128, 63/128, 64/128, 65/128. A section of k 20 to 80 of cos(2πνi k) for 63/128, 64/128, and 65/128 shows that the frequency of 63/128 appears similar to 65/128 and both have a larger cycle than 64/128.

F13FTDISC2S is only on the CD.

Application 9.13. Make graphs of cos(2πνi k) for higher and lower frequencies than 64/128 and show that they appear similar to graphs of lower and higher frequency numbers.

FileFig 9.14 (F14MICHOPS)

Simulations of interferograms for the Michelson interferometer. Graphs of amplitudes and intensities of cosine functions depending on optical path difference.

F14MICHOPS is only on the CD.

9.2.4.2 Sampling

We have seen in FileFig 9.13 that the cos-function of Eq. (9.57) has a symmetrical appearance around the frequency position 64/128, when using 128 points in space and having exactly two space points per cycle. For this special example two points per cycle have been used for the characterization of the frequencies. In general, we have from information theory the sampling theorem.

Theorem 9.1. Two points per cycle are needed for sampling a continuous periodic function in order not to lose information.

In our example, the continuous functions are cos-functions. The cos-function with the highest frequency is sampled by using two points per cycle. For example, one is at the maximum and one is at the minimum. Lower frequencies are also correctly sampled, but higher frequencies present a problem. They appear as lower frequencies and need to be removed by a filter. When talking about wavelengths instead of frequencies, we similarly have that the sampling interval is equal to one-half of the wavelength of the shortest wavelength (highest frequency). Calling the sampling interval l, we have

where νM is the highest frequency of the spectrum under consideration. In Fourier transform spectroscopy it is assumed that all higher frequencies are removed.

9.2.5 Folding of the Fourier Transform Spectrum

What happens when the choice of the sampling interval is done incorrectly? How is the spectrum affected? We show that higher frequencies are folded around the highest frequency, corresponding to the correct sampling interval, and appear as lower frequencies in the spectrum. We use for the space domain the coordinate points 1/256, 2/256, . . . , 1 and consider the sum of cos-functions with frequencies 65, 85, and 105.

1. The sampling interval is 1/256. We have

where i runs from 1 to 256.

In FileFig 9.15 we compare the appearance of the spectrum for these three sampling intervals 1/256, 2/256 and 4/256.

FileFig 9.15 (F15FOLDS)

Folding of the spectrum:

1.Sampling interval 1/256 for the function y1. The three frequencies to be investigated are 65, 85, and 105. The highest frequency is 128. All frequencies appear at the right place in the Fourier transformation spectrum.

2.Sampling interval 2/256 for function y2. We now use a sampling interval twice as large. Consequently the highest frequency is now 64 and the three frequencies 65, 85, and 105 are all higher than 64. The spectrum appears folded. In the Fourier transformation the three frequencies appear at 45, 85, and 125. Since the frequency spectrum is shown over 128 points, we may look at the function y2 as it would have frequencies 2 · 65, 2 · 85 and 2 · 105 and is sampled with the sampling interval 1/256 (as we did under (1), where the highest frequency is 128.) We then have to look for the frequencies 130, 170, and 210, which all exceed the frequency interval from 1 to 128. We saw in

FileFigs 9.12 and 13 that higher frequencies are folded back into the spectrum around the highest frequency (which is 128 because we sample y2 now like

y1). We have to subtract from 130, 170, and 210 the highest frequency 128 and find 2, 42, and 82. These values have to be traced back from 128 because of the folding. We get 126, 86, and 46, which are the frequencies we should find in the graph of the Fourier transformation. However, we find 125, 85, 45 because we have not taken into account that the Fourier transformation program starts at 0 and not at 1.

3.Sampling interval 4/256 for function y4. We now use a sampling interval 4/256 to sample the same function with 4/256. The highest frequency is now 32. In the Fourier transformation the three frequencies appear at 5, 85, and 90. Since the frequency spectrum is shown over 128 points, we look at y4 as it would have frequencies 4 · 65, 4 · 85, and 4 · 105. It is sampled with the sampling interval 1/256, as y1 was. We have to look for the frequencies 260, 340, and 420, which all exceed the frequency interval from 1 to 128. We saw in FileFigs.12 and 13 that higher frequencies are folded back into the spectrum around the highest frequency (which is 128 because we sample y4 now like

y1). We have to subtract from 260, 340, and 420 the highest frequency 128, and get 132, 212, and 293. Then we have to trace back from 128, because they all exceed 128 and get 4, 84, and 165 and have to fold again at 0 into forward (i.e., 4, 84, and 165). The first two are now in the right position, but 165 exceeds 128. We have to subtract 128 from 165 which is 37 and trace it back from 128, that is, we get to 91. We finally get for the position of the three frequencies 4, 84, and 91, and those are the frequencies we should find in the graph of the Fourier transformation. However, we find 5, 85, and 90. The first two are folded twice, the last three times. We have not taken into account that the Fourier transformation program starts at 0 and not at 1.

F15FOLDS

Folding of the Spectrum

For the sampling interval 1/255, hightest frequency is 128; the frequencies are at 65, and 105, all below 127.

1. Sample interval i/255