I have measured the spectrum of the LED in my interferometry set-up, and now I want to calculate the coherence length from it. A commonly found formula is $l_{coh} = \frac{c}{\Delta f}$, sometimes with an additional factor for the shape of the spectrum. However, I have only ever found factors for Gaussian and Lorentzian lineshapes, and mine is neither.

I am looking for either a way to determine this factor, or derive the coherence length from the spectrum in a different way.

Thanks in advance.

  • $\begingroup$ To the best of my (ancient) knowledge, any formulae beyond the basic 1/bandwidth are estimates only, rather than strict derivations from the fundamental properties of a given LED material structure. Further, the rules change dramatically if you are operating in the Superluminescent regime. What is your spectral shape, and how did you determine that, by the way? //PS: this page may be of some help: en.wikipedia.org/wiki/Wiener%E2%80%93Khintchine_theorem $\endgroup$ Mar 31, 2014 at 11:39
  • $\begingroup$ I have measured the spectra using a diffraction grating spectrometer. The spectra look a lot like a Lorentzian shape, but slightly 'slanted' - just like the red spectrum in this image. $\endgroup$
    – Oebele
    Mar 31, 2014 at 12:38

2 Answers 2


Presumably you have measured your spectrum as a function of wavelength, so you have $\mathscr{F}(\lambda)$, which is an power per unit wavelength. You must now convert this power per unit frequency spectrum.

So we seek $\mathscr{G}(f)$ where $\mathscr{G}(f)\,|df| = \mathscr{F}(\lambda)\,|d\lambda|$; given $c = f\,\lambda$ we have:

$$d\lambda = -\frac{df}{f}\,\lambda = - \frac{c}{f^2} df$$

so that

$$\mathscr{G}(f) = \frac{c}{f^2} \mathscr{F}\left(\frac{c}{f}\right)$$

So now we have derived our power spectral density $\mathscr{G}(f)$ from your experimental $\mathscr{F}$ spectrum as a function of wavelength. This is then converted to an autocorrelation function of time by the Wiener-Khinchin theorem:

$$\tilde{\Gamma}(t) = \int_{-\infty}^\infty e^{2\pi\,i\,f\,t} \mathscr{G}(f) \,df=2\,\mathrm{Re}\left(\int_{f_{min}}^{f_{max}} e^{2\pi\,i\,f\,t} \frac{c}{f^2} \mathscr{F}\left(\frac{c}{f}\right) \,df\right)$$

where $[f_{min},\,f_{max}]$ is your experimental measurement interval.

So now you convert your autocorrelation as a function of time to autocorrelation as a function of shift displacement $x = c\,t$, so your final autocorrelation function will be:

$$\Gamma(x) = \tilde{\Gamma}\left(\frac{x}{c}\right) = 2\,\mathrm{Re}\left(\int_{f_{min}}^{f_{max}} \exp\left(2\pi\,i\,f\,\frac{x}{c}\right) \frac{c}{f^2} \mathscr{F}\left(\frac{c}{f}\right) \,df\right)$$

So now you have to decide how you will define your coherence length: common definitions include (1) the shift displacement $x$ at which $\Gamma(x)$ is $1/e$ times $\Gamma(0)$ and (2) the rms spread:

$$\sqrt{\frac{\int_0^\infty x^2\,\Gamma(x)^2\,dx}{\int_0^\infty \Gamma(x)^2\,dx}}$$

  • $\begingroup$ There he goes again, stealing the answer from my comment :-) Nicely presented. $\endgroup$ Mar 31, 2014 at 13:38
  • $\begingroup$ Very nice answer. Of course the spectrum I measured is discrete. If I am not mistaken, the integral should therefore be transformed to this sum: $$\Gamma(x) = \tilde{\Gamma}\left(\frac{x}{c}\right) = 2\,\mathrm{Re}\left(\sum_{f_{min}}^{f_{max}} \exp\left(2\pi\,i\,f\,\frac{x}{c}\right) \frac{c}{f^2} \mathscr{F}\left(\frac{c}{f}\right) \,\Delta f\right)$$ with $\Delta f$ being the difference in wavelength between sample points, converted to the frequency domain. Please correct me if I'm wrong. $\endgroup$
    – Oebele
    Apr 1, 2014 at 9:47
  • $\begingroup$ @AleStrooisma This is right. Be aware that the FFT in many implementations runs from 0 to the $f_N$, the Nyquist frequency, then jumps to $-f_N$ and runs back to 0. Mathematica is like this, for example, and you need to rotate your datasets accordingly. Also, the fact that your spectrum is measured with a grating confirms that the spectrum is a power per unit wavelength: the diffraction angle is proportional to spatial frequency. $\endgroup$ Apr 1, 2014 at 12:05
  • $\begingroup$ I finally found the time to implement this, and to do the necessary reading up on the Wiener-Khinchin theorem. What I still wonder about is why exactly you take 2 times the real component of the result of the Fourier transform. This question is raised especially because I can easily take the 1/e point of the absolute values, but not from the real part. $\endgroup$
    – Oebele
    Apr 7, 2014 at 11:46
  • $\begingroup$ The factor of 2 comes from the convention used to define the "analytic signal", a.k.a. the complex representation of the real data provided by your equipment. Some authors use a representation without the factor of two; in my experience the factor of two is more common. You have to be a little careful to make sure that you always use a set of formulae that are all consistent with one convention. $\endgroup$
    – garyp
    Apr 10, 2014 at 14:54

A detector generally provides values that are proportional to the power spectral density (PSD) of the radiation hitting it. Energy per time interval (collection time), per spectral interval (pass band, slit width, pixel width, ...).

The Fourier transform of the PSD is the time-domain autocorrelation function of the radiation. The autocorrelation function gives you a direct window on the correlation time. The FT will be peaked at zero (perfect correlation when the signal is on top of itself) and decline from there as the correlation reduces (imperfect correlation when the signal is offset from itself). Once you have the autocorrelation function you will have to make some judgement about how to get the correlation time from it. The time at which the correlation falls to 1/2 its peak might be a good choice. Multiply that by $c$ and you have your correlation length.

Note that you will get slightly different answers if you choose some other criterion for the correlation time, for example if you choose $1/e$ of peak value. Also, the spectral sensitivity of your detector will change things a bit. Moral: at the end of the day you might do no better than $c/\Delta f$. It depends on what you want to do with the information once you have it.

Update I see that @Wet... has posted a more detailed response. He or she correctly adds the necessity of converting your wavelength data to frequency data.

  • $\begingroup$ Finally I have enough rep to give you the upvote you deserve. This answer adds some theory to the equations of @Wet, making it easier to understand what is happening. $\endgroup$
    – Oebele
    Apr 10, 2014 at 14:12

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