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


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.

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

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

  • $\begingroup$ In a typical FTIR experiment, the spectrum is typically in wavenumbers, i.e., reciprocal centimeters, rather than in wavelengths. This is due to the way the interferograms are generated by the moving mirror modulation. You might have a look here to get an idea how FTIR works. And, to answer your question, time is not essential in Fourier transforms: look at the abscissa units at the link: they are in cm (for moving mirror movement). So the inverse FT gets the desired reciprocal cm units. Hope this helps! $\endgroup$
    – Ed V
    Mar 31, 2020 at 21:06
  • $\begingroup$ Give me a second and I'll update the question with more information. Thanks for you response and link. $\endgroup$
    – Jesse T-P
    Apr 1, 2020 at 23:14
  • $\begingroup$ I am upvoting this question because it is very much improved and I am deleting my other (moot) comments. In an FTIR with moving mirror, the modulation frequency, in Hz, equals 2 x the moving mirror's speed, in cm/s, x the wavenumber of the light, in $cm^{-1}$. The interferogram is sampled in time and the sampling frequency has to be fast enough to avoid aliasing. Then the inverse Discrete FT yields the spectrum in wavenumbers. At 2 µm, the wavenumber value is 5000 $cm^{-1}$, so if the mirror scan speed is 1 cm/s, then the sampling frequency would have to be above 20 kHz, preferably faster. $\endgroup$
    – Ed V
    May 2, 2020 at 17:43

2 Answers 2


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:

Igor Pro Hilbert Transform

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:

Ethyl acrylate spectrum

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

Phase spectrum of EA

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!

  • $\begingroup$ Great post ED.I've been delving into the literature understanding how an FTIR works, needles to say my understanding has increased tremendously, From the capturing of the interferogram (voltage vs wavenumber) to the Fourier Transform of the interferogram, ect. The reference to the manual will be of great help, Thanks. As a final note I'm also exploring the maximum entropy method to recover the phase of the reflectivity, which is looking promising, updates to follow. $\endgroup$
    – Jesse T-P
    May 3, 2020 at 19:24
  • $\begingroup$ Happy to be of a little help! Best of success with your efforts! $\endgroup$
    – Ed V
    May 3, 2020 at 22:08

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:

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:

  • $\begingroup$ @EdV Thanks for the kind words. First timers not responding is actually way too common. I suspect they generally don't come back because they found their answer elsewhere, but you can take solace in that answers may be helpful to others besides the OP. Sometimes they come back weeks later, go figure. $\endgroup$ Apr 1, 2020 at 11:40
  • $\begingroup$ Thanks for the response I'll update the question with more meaningful information, sorry for the lazy question. $\endgroup$
    – Jesse T-P
    Apr 1, 2020 at 23:17
  • $\begingroup$ @JesseT-P As I saw it, your question was "Can I use the DFT on things other than time series data?" Easy answer: Affirmative, no further context needed for that. My curiosity is whether my intuition about your equations is correct. It is a common misperception that the DFT can "pluck out any frequencies" in a signal. That is not true in general. Thus, a little more information might be useful in seeing if you are applying it properly. $\endgroup$ Apr 1, 2020 at 23:36
  • $\begingroup$ Well I updated the question with some links and a better layout of the theory. Yes, My original question was answered, mabey I should ask another one showing why my results aren't matching up, this might be better suited for the stackoverflow or Math channel though. $\endgroup$
    – Jesse T-P
    Apr 2, 2020 at 0:30
  • $\begingroup$ @JesseT-P That's pretty thorough. Unfortunately I don't have the time to dive headfirst into this. I'll ponder it though. I think you are most likely to find some subject matter experts in the Physics forum for this one. $\endgroup$ Apr 2, 2020 at 0:42

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