I am a bachelor physics student and as an assignment we had to perform measurements on an FT spectroscopy setup.


Our setup consisted of a Michelson interferometer through which the light of two sources namely a HeNe-laser and a sodium lamp, passed at the same time. The light of the interference pattern was split into two. One of the beams was passed through a polarizer to filter out the laser light (partially). Both beams were sent to their respective detectors which measured the intensity in a single spot of the interference pattern.

One of the mirrors of the interferometer was attached to a moving stage which was moving at a constant velocity during the measurement.

Here is a drawing of the setup:

Fourier transform spectroscopy setup

The detectors measured the intensity during the time the stage was moving. We therefore received intensity vs. time as data. Detector 1 measured the intensity of the sodium light + the weakened laser light. Detector 0 measured the intensity of both the laser light and the sodium light (the latter was negligible since the laser light was so much stronger than the sodium light).

The data from detector 0 was used for calibration. We assumed the laser light consists of a single wavelength of light. We measured however that there were varying frequencies in the intensity. These variations must therefore be caused by flaws in the setup (like faulty calibration or inconsistency in the speed of the stage).

The variations were corrected by moving the data points along the time axis such that a nice sinusoid remained. The frequency of that sinusoid was chosen to be the average frequency of the original data (I am not sure if this is true. We were not told the exact details of the data correction procedure.) To prepare the data for Fast Fourier Transform the data points were moved along the sinusoid such that they were spaced equally horizontally.

Here is an illustration of the procedure:

Data correction procedure

Because the variations caused by flaws in the setup were the same in the signals from both detectors (since they were measuring at the same time) the corrections of the data from detector 0 were applied to the data from detector 1 as well.

The data from detector 1 was Fourier transformed (using the FFT algorithm). The frequencies found from this were scaled down by a factor of $\frac{2v}{c}$, where $v$ was the speed of the stage. (I am leaving out the derivation.) We scaled the frequencies back up such that the frequency of the laser peak matched its optical frequency and turned those into wavelengths.

This gave the following spectrum:

Sodium spectrum

Now onto the main problem. How do I determine the spectral resolution of this setup? Spectral resolution is the smallest difference in wavelength the spectrometer can detect at a certain wavelength.

The time for measurements has long passed so it's no longer possible to make new measurements.


We were told that because the laser peak is sharpest in our spectrum its width gives us the resolution. But I do not understand how these are related. Can someone explain that? If the width of the laser peak does not give the resolution, what would and why?

What I am also interested in are the factors that influence the resolution.


I was thinking about a hypothetical case where the detectors could measure infinitely fast producing a continuous signal. The setup would also have a special laser with a variable wavelength instead of a sodium lamp.

By collecting the data from detector 0 which is measuring the intensity of both lasers and applying a continuous Fourier transform (we are skipping the data correction procedure) a spectrum is obtained.

Because of the variations in the frequencies each laser will produce a normal distribution (a Gaussian) in the spectrum. The spectrum then consists of a sum of two Gaussians.

One method of finding out how many peaks there are in a spectrum is counting the maxima. If the Gaussians are far apart we can see two maxima. But when two Gaussians get closer and closer together (the distance between their maxima decreases), at some point their sum will only have a single maximum before the tops of the two Gaussians have overlapped. At this point the two peaks can no longer be distinguished and our method will tell us there is only one peak.

For the following plot we took the sum of two identical Gaussians $e^{-(x-a)^2}+e^{-(x+a)^2}$. The two Gaussians are a distance $2a$ apart. By taking the derivative with respect to $x$ and setting it equal to $0$, the extrema can be found. In the plot the $x$-coordinate of the extremas are plotted against $a$. WolframAlpha plot

The minimum distance that the Gaussians have to be apart such that two maxima can be seen could be called the resolution. This distance is a function of the amplitudes and standard deviations of the two Gaussians which are themselves determined by the quality of the setup.

To find the resolution of this hypothetical setup one could decrease the difference in wavelength between the two lasers till only one maximum is visible and repeat this for all wavelengths.

However, this is just a hypothetical setup and not the real one. In the real one, the detectors measure a finite amount of points per second, there is data correction, FFT and a finite amount of point per nanometer in the spectrum and it's not clear how the resolution is determined.

  • $\begingroup$ Always always always always always plot spectra like this on a semi-log scale (vertical axis logarithmic). Always. $\endgroup$
    – DanielSank
    Commented Jun 3, 2014 at 1:01

2 Answers 2


This question has a lot of extraneous information etc., but I think that's because you aren't really sure what to ask. There is one very simple basic fact you may not know:

Suppose you have a series of data points with separation on the x-axis of $\delta x$. If you compute the discrete Fourier transform of your data you get a complex number for each point on a new axis which has dimensions of $1/x$. For example, if your data represents a time series, so the x-axis is time, then the discrete Fourier transform data is on a $1/\text{time} = \text{frequency}$ axis.

Now, if the total extent of your measurement is $X$, then the resolution in the Fourier transform data is $1/X$. For example, if you take data for 1 second, then your frequency resolution is 1 Hz. If you take data for 1000 seconds, your frequency resolution is 1 mHz.

I'm pretty sure this is the part of your problem you didn't already understand, but if it's not, leave a comment.

  • $\begingroup$ You were right. I didn't understand that. But the resolution of the Fourier transform just gives the upper-bound of the resolution of the setup. If there are many problems with the setup the peaks could get wider and the resolution of the setup is decreased while the Fourier transform resolution remains the same. $\endgroup$
    – m009
    Commented Jun 4, 2014 at 11:48
  • $\begingroup$ @m009: Correct. You seem to already understand those effects so I didn't comment on them. $\endgroup$
    – DanielSank
    Commented Jun 4, 2014 at 14:27

Let's build up a model by considering three stages:

  1. Suppose the resolution of your measurement system would be "perfect", $\delta f_{meas}=0Hz$, that the He:Ne laser runs at a single (perfectly fixed) frequency, $\delta f_{laser}=0Hz$, and that you are measuring for an infinity long time interval, $\delta f_{FFT}=0Hz$. Assuming that you have no systematic error in your setup, you would be able to measure the laser frequency perfectly: The frequency spectrum consists of a single Dirac-$\delta$-peak.
  2. Next, let's assume that the laser is not running with a perfectly fixed frequency, but has a finite frequency width, $\delta f_{laser}\ne 0Hz$. Although, the frequency spectrum still consists of a single peak, this peak is no longer "infinity" thin. Instead, the peak has a finite width -- the functional form is often taken to be Lorentzian.
  3. Finally, let's add the finite frequency resolution of your measurement system. Each point of the spectrum measured in (2), becomes a "distribution" with finite resolution $\delta f_{meas}$. Mathematically, we measure the function $$ [spectrum_{laser} *resolution_{meas}](f) $$ where "$*$" is the convolution operator -- think of a convolution as a moving average. Usually, the resolution function of the measurement system is considered to be a Gaussian (=normal distribution) with zero mean value, and standard deviation $\delta f_{meas}$.

Using the general result from step (3) we see that the frequency width of the laser and the finite resolution of the measurement system are "coupled". Even if we knew the exact functional form of the laser spectrum, we must use a (rather complicated) regression method to obtain the resolution of the measurement system. However, what if the frequency spectrum of the laser consists of a multiple $\delta$-peaks with frequency $\{f_1, f_2, \cdots, f_k\}$?

The answer to this question is given by the so called shift theorem of convolutions: Assuming a Gaussian resolution function, the result is $\sum_{i=1}^k N(f_{i}, \delta f_{meas})$ -- a sum of Gaussian distributions with mean value $f_{i}$, and standard deviation $\delta f_{meas}$. Hence, if the laser spectrum consists of peaks, which are "much narrower" than the resolution of our measurement system, we are able to obtain a "good estimate" of the measurement resolution without using complicated regression methods: We simply fit a Gaussian distribution function to each of the peaks and evaluate their "width" (=standard deviation). This is what your supervisor was referring to.


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