Fourier Transform of sampled reflectance measurements

Question

Goal: Calculate the complex index of refraction ($\hat n = n +jk)$ from reflectance measurements.

Data:

Reflectance measurements for various materials were taken with a FTIR (Nicolet™ iS50 FTIR, from Thermofisher). The spectral range of the reflected measurements are from 400nm - 15um, they are reported in both wavenumber and wavelength (relationship: $\hat\nu = \frac{1}{\lambda} [cm^{-1}]) $. Note, that the full wavelength range was captured with three different detectors.

To simplify the matter (and give access to data) I have been using reflectance measurements form the following NASA database https://speclib.jpl.nasa.gov/library. I have chosen SiO2 from the database (2um-15um) which can be navigated to by selecting:

Select Spectral Type-> Minerals

Filter by class-> Silicate

Wavelength-> VSWIR+TIR

Quartz SiO_2

Brief Theory:

Here's a short summary: https://shimadzu.com.au/sites/default/files/Appl_FTIR_Polymer_specular_reflectance_055_en.pdf

The imaginary part of an analytical signal can be found from the real part alone through the Hilbert transform, e.g. $\tilde {x_c}(t) = x_r(t) +j x_i(t)$ meaning $x_i(t) = HT{x_r(t)} $. https://epdf.pub/hilbert-transforms-volume-1-encyclopedia-of-mathematics-and-its-applications.html

Changing mindset to the optical domain, and the question at hand, noting that most optical literature calls the Hilbert transform the Kramer Kronig. The FTIR measures spectral reflectance (i.e. Intensity).

$$\tilde r = re^{j\phi} = \sqrt{R}e^{j\phi}$$

$$R = |\tilde r|^2$$

• Where r is the reflectivity

• R is the reflectance (This is what is measurable, intensity)

• $\phi$ is the phase change at the surface caused by the absorption of the material.

Through the Fresnel equations the complex refractive index can be calculated from the below equations.

Dispersion: $$n(\nu) = \frac{1-R(\nu)}{1 + R(\nu) - 2\sqrt{R(\nu)}cos(\phi(\nu))}$$

Absorptive index: $$k(\nu) = \frac{-2\sqrt{R(\nu)}sin(\phi(\nu))}{1 + R(\nu) - 2\sqrt{R(\nu)}cos(\phi(\nu))}$$

So, the goal is to take the measured reflectance data and calculate the phase through the Hilbert Transform, as shown below.

$$\phi(\nu_g) = \frac{2\nu_g}{\pi} \int_0^\infty \frac{ln\sqrt{R(\nu)}}{\nu^2 - \nu_g^2}$$

• where $\nu$ is the wavenumber

Since I have discrete data (over a finite range) the Hilbert transform can't be calculated directly. It's usually calculated through Maclaurin's method or the double Fourier Transform (Discrete Fourier Transform (DFT) in this case).

The approximation of the Hilbert transform using the double FT is given shown in the below equation.

$$\phi(\nu_g) = 4 \int_0^\infty cos(2\pi \nu_gt)dt \int_0^\infty ln\sqrt{R(\nu)}sin(2\pi \nu_gt)dv$$

My Question:

My question pertains to the Fourier transform (FT) of the sampled data. All the equations I see on line show the FT of evenly spaced sample data with respect to "time", which I don't know. My data is evenly sampled with respect to wavenumber $\nu$, can I still use the FT but replace the sampling time (1/$\Delta$t) with (1/$\Delta\nu$)? If yes, is there anything extra I need to consider with the respect to the reluctance data.

The last equation showing the double Fourier transform has a intergal with respect to time, but I don't have any sampling time information, is there a way around this, or some conversion?

I found the following relationships online but can't quite see how to apply them.

$$f = \frac{1}{t} ....||||.... \nu = \frac{1}{\lambda}$$

$$w = 2\pi f ....||||.... k = \frac{2\pi}{\lambda}$$

Answer 1

This elaborates on my comment and my linked FTIR example. Once the FTIR spectrum is obtained, the phase spectrum can be obtained via the Hilbert Transform of the spectrum and that can be performed as a convolution. The Hilbert Transform can be performed by various software packages. In Igor Pro (v. 6.3), the Help text on Hilbert Transforms (page 723 of the Manual), is simply:

In the linked example, the IR spectrum of ethyl acrylate was obtained via a Mueller optical calculus simulation. For present illustrative purposes, the spectrum was adjusted in two ways. First, everything below 800 wavenumbers was set to zero. This is typically where the noise is worst because IR detectors are relatively insensitive and noisy compared with detectors in higher energy ranges. Second, the spectrum was zero filled from 4001 to 4095 wavenumbers, in order to achieve a total of 4096 wavenumber values. This spectrum is shown below:

Performing the Hilbert Transform, as per the Igor Pro command in the first figure, then yields the phase spectrum:

Maybe this helps and maybe it doesn't. But it points to a way forward: find a reflectance spectrum in the refereed literature and the dispersion and absorption spectra that were computed from it. This has to be something about which there is no doubt, i.e., it will be the standard. Then try using software to compute the Hilbert transform and see if the result agrees with the standard. Lastly, if you still want to do so, then pursue the pathway you have outlined in your question text. Hope this helps a bit!

Answer 2

When you say Fourier Transform (FT) it usually implies the continuous case. When you have discrete samples, the Discrete Fourier Transform is used. They are very similar, but it is important to keep them distinct as concepts do not directly translate from one to the other.

The DFT is a mathematical transform that does not care about the units of the domain, so yes you should be able to use it. Whether it is appropriate, or what it would mean, I can't tell from what you have described.

I did a quick search for your equations and could not find them. Perhaps you can provide a link. They do not look "proper" to me. It seems to me that the $2R(\nu)cos(\phi(\nu))$ term in the denominators should be $2\sqrt{R(\nu)}cos(\phi(\nu))$.

I am struck with the similarity to equations (31) and (38) here:

• Exponential Smoothing with a Wrinkle

Here they are:

$$ \sum_{k=0}^{\infty} { a^k \cos( \alpha k ) } = \frac{ 1 - a \cos \alpha }{ 1 - 2 a \cos \alpha + a^2 } \tag {31} $$

$$ \sum_{k=0}^{\infty} { a^k \sin( \alpha k ) } = \frac{ a \sin \alpha }{ 1 - 2 a \cos \alpha + a^2 } \tag {38} $$

The denominators take the form of the Law of Cosines:

$$ C^2 = A^2 + B^2 - 2 A B \cos( \theta ) $$

Quite similar, but a bit different. Whether the similarity has any significance will take further study. Coincidences in math tend to be meaningful.

Also, you may find this other article of mine interesting as well:

• Off Topic: Refraction in a Varying Medium