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.
$$f = \frac{1}{t} ....||||.... \nu = \frac{1}{\lambda}$$
$$w = 2\pi f ....||||.... k = \frac{2\pi}{\lambda}$$