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

c : ff t(y1)

N : last(c) N 128 j : 0 . . . 128

Frequency peaks are at 65, 85, 105.

2.Sample interval 2i/255

For the sampling interval 2/255, highest frequency is 64: the original frequencies are at 65, 85, 105: they are all larger 64 and appear folded.

c : ff t(y2)

N : last(cc) N 128

j : 0 . . . 128

Frequency peaks appear at 45, 85, 125.

3.Sample interval 4i/255

For the sampling interval 4/255, highest frequency 32, the frequencies are higher than 1 times 32 and 2 times 32.

ccc : ff t(y4)

N : last(ccc) N 128 j : 1 . . . 128

Freguency peaks appear

65 at → 125

85 at → 185

105 at → 45.

Application 9.15. Use the three frequencies 15, 34, and 97. Determine the appearance in the Fourier transformation as discussed for the original set (65, 85, 105) of frequencies.

9.2.6 High Resolution Spectroscopy

In high resolution spectroscopy one is often interested in investigating a narrow bandwidth of frequencies for high resolution. The sampling interval l in the length domain is related to the highest frequency νM in the 1/length domain. When increasing the number of points N in the length domain, one also increases the number N of points in the frequency domain. Since the sampling interval determines the frequency interval from 0 to νM , a higher number of points in the length domain results in a higher number of points in the frequency domain. One gets higher resolution in the interval from 0 to νM . The resolution is inverse proportional to the total length L of the interferogram. The length may be expressed as L N l, and one has

A high resolution interferometer uses a large optical path difference L in order to make ν small. If we take, for example, L 50 cm, we have for ν (1/100) cm−1 and for the resolving power R νM / ν at 100 microns (equal to 100 cm−1)

The number of points in length space is equal to the number of frequency intervals in frequency space, which is

and is also equal to R. To record 10,000 points, assuming about 3 sec for each point, would take 9 hours. The information we obtain in this way is a spectrum from 1 to 100 cm−1 with resolution of 1/100 cm−1.

In the case where one is only interested in the study of a section of the spectrum one may use folding of the spectrum. Let us assume that we are interested in the section from 2/3 of 100 cm−1 to 100 cm−1. In that case we use a bandpass

FIGURE 9.2 We assume that a bandpass filter eliminates the spectrum from 1 to 66 cm−1. The spectrum is folded by using a sampling interval of 3 times l, where l is the sampling interval, corresponding to the highest frequency 100 cm−1; that is, 1 1/2 · 100 cm−1 0.005 cm 50 microns. The frequency scale for the sections changes the direction to the highest value for even or odd factors of folding.

filter and eliminate all other frequencies, and fold the spectrum three times by choosing a sample interval three times larger (see Figure 9.2). This is similar to what we discussed above in FileFig 9.15, where we studied the folding of a sum of cos-functions sampled with intervals 1/256, 2/256, and 4/256.

We have νM /3 for the width of the spectral sections to be studied with the length L of the interferogram. Consequently, the sampling interval has the value 3 l L/3333, instead of l L/(10, 000). The total number of points is reduced from 10,000 to 3333, but the length of the interferogram remains the same. Therefore we obtain the same high resolution in a smaller spectral region with one-third of the points. A very important fact is that we have to use the bandpass filter we assumed to apply. The spectra of the two sections of 1 to 33 cm−1 and 33 cm−1 to 66 cm−1 are folded onto the spectrum of 66 cm−1 to 100 cm−1. If they contain spectral information, the spectrum will be “messed up," like the Fourier transforms in FileFig 9.15. We obtain the same high resolution for a smaller part of the spectrum, using a larger sampling interval and fewer points. In our case the spectrum is obtained in one-third of the time.

In Figure 9.3 we show the background spectrum, taken with a Michelson interferometer and a bandpass filter. The bandpass has to have a width of νM /integer.

FIGURE 9.3 The background spectrum taken with a Michelson interferometer and a bandpass filter. (From J. Kachmarsky et al., Far-infrared high-resolution Fourier transform spectrometer: applications to H2O, NH3, and NO2 lines. Applied Optics, 15, 1976, p 708–713.)

FIGURE 9.4 High-resolution spectrum of water in the 104 to 107 cm−1 region obtained by folding the spectrum six times. (From J. Kachmarsky et al., Far-infrared high-resolution Fourtier transform spectrometer: applications to H2O, NH3, and NO2 lines. Applied Optics, 15, 1976, p 708–713.)

In Figure 9.4 we show a high-resolution spectrum of water in the 104 to 107 cm−1 region, obtained with a sampling interval six times larger and in a time span six times shorter. We state the procedure in more general terms as follows. A filter is used to eliminate all parts of the spectrum with frequencies higher than νM and lower than νm, where m stands for minimum. The value of νm has to be chosen in such a way that q νM /(νM − νm) is an integer. We then obtain with N sample intervals, each calculated by using

366 9. FOURIER TRANSFORMATION AND FT-SPECTROSCOPY

a spectrum of width νM /q and a resolution of ν 1/(2L), where L Nl. The scale has its highest value νM /q and runs forward for q odd and backward for q even (Figure 9.2).

9.2.7 Apodization

The Fourier integrals, as shown in Eq. (9.55) and (9.56),

have a range of integration from 0 to ∞. The discrete Fourier transform, as shown in Eq. (9.57),

has a finite summation range, instead of the infinite large integration range. In the section on Fourier transformation, we reduced the integration range for the Fourier transformation of the slit from −∞ to ∞ to a range from −a to a. We did this because the width of the slit is 2a, and outside the interval from −a to a there is no information. We may multiply the interferogram function S(y) in the integral of Eq. (9.70) with a step function p(y), which is 1 in the interval from −a to a and otherwise zero. We introduce into Eq. (9.70) the step function p(y),

−∞

and reduce the infinite range of integration to a finite range −a to a, which we may write as

From the point of view of calculating a Fourier transformation over a limited range of space, Eq. (9.73) is similar to the discrete Fourier transformation of Eq. (9.71). In FileFig 9.17 we discuss an example of the Fourier transformation of a cos-function integrated over a finite range. We use the function yk cos(2πf k/255), with frequency f 31, plotted over a range of k 0 to 400 and do the multiplication with a step function pk of width 256 points. The result of the real Fourier transformation is that we obtain a peak at f 31, not infinitely narrow and with a loop extending to negative values. Since negative intensities do not appear in spectroscopy, it is desirable to find a procedure to avoid this artifact. The situation can be corrected by using in place of pk , which

3. Fourier transformation of y × p: we have to use 255 points

4. Triangle function

5. Fourier transformation of y × p; we have to use 255 points

N 128

j : 0 . . . N

c : ff t(ay)

N : last (c).

Application 9.16.

1.Use the apodization function qqk (1 − k/255)2 and calculate the apodized function and the Fourier transform. Compare with the use of pk and qk .

2.Study the impact of the reduction of resolution. Repeat the procedure for a sum of five cosine functions of five different frequencies, such as 30, 35, 40, 45, and 50. Study the resolution using the apodization functions pk , qk , and qqk .

APPENDIX 9.1

A9.1.1 Asymmetric Fourier Transform Spectroscopy

We have discussed the use of the Michelson interferometer for spectroscopy. The sample was assumed to be positioned either before the light was divided at the beam splitter or after the light was recombined. The reflection and transmission properties of the two arms of the interferometer were assumed to be equal and the only asymmetry was the difference in the path. To calculate the spectrum we could use the “real" Fourier transformation.

We now assume that the sample is at the surface of one of the mirrors of the Michelson interferometer. The reflection properties of the two arms of the interferometer are then different and the complex Fourier transformation must be applied. As a result of using the complex Fourier transformation one has more information available (complex numbers instead of real numbers). We may calculate not only the intensity (square of amplitude) but also obtain phase information. If in the case of reflection for a certain frequency range, the reflected amplitude and the phase change at reflection are known, the optical constants n and K may be calculated. We start with Eqs. (9.34) and (9.35), describing the two beams reflected from the two mirrors in the Michelson interferometer. We assume that the light is reflected in the fixed arm by a mirror producing the same magnitude of the amplitude and a phase shift of π. Using complex notation we