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In nuclear magnetic resonance relaxation theory, in order to determine relaxation rates from transition probabilities, one needs the spectral density of random magnetic field fluctuations around the spin of interest. To get the spectral density, one needs to take the autocorrelation function of the fluctuations first, before Fourier transforming it to get the spectral density.

My question is - Why not Fourier transform the fluctuations as is? Why do you need to go through the autocorrelation function first?

Many thanks.

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  • $\begingroup$ see Wiener-Khinchin theorem $\endgroup$ Commented Oct 6, 2022 at 10:56
  • $\begingroup$ By definition of Spectral Density, as the Fourier transform of the autocorrelation. If you perform only Fourier transform, you just get the Fourier transform of the fluctuations. For relaxation, you need the autocorrelation (or its transform, the PSD, power spectral density). The actual question could be: how to efficiently and accurately evaluate PSD from time measurements of the fluctuations? Two main ways: 1. Compute autocorrelation first, and then transform. 2. Transform and then multiply transormed signals, as shown in it.wikipedia.org/wiki/Teorema_di_Wiener-Chin%C4%8Din $\endgroup$
    – basics
    Commented Oct 6, 2022 at 11:17
  • $\begingroup$ I linked wiki page in Italian, because it looks richer than the English one. Use translator of your browser, if needed. They first talk about deterministic signals, and then about stochastic signals, with integrals in mean $\endgroup$
    – basics
    Commented Oct 6, 2022 at 11:20

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Using the so-called Wiener-Khintchin theorem the usual definition of spectral density is that it is the Fourier transform of the autocorrelation function, so it is natural for calculating the spectrum. Of course, your question is not really answered by this. The reason for not using the periodogram to estimate the spectrum is because it is not a consistent estimator even for a the normal (Gaussian) process, details below.

Let $X(t)$ be the stochastic process and take the Fourier transform of a finite segment of length $T$: $$\hat X_T(u)=\int_{-\frac{T}{2}}^{\frac{T}{2}}\mathrm{d}t\, e^{-\mathfrak j 2\pi ut} X(t) \tag{1}\label{1}$$ Look at the resulting spectral estimator, also called the periodogram: $$S(u,T))=\frac{|\hat X_T(u)|^2}{T} \tag{2}\label{2}$$ If now you let $T\to \infty$ you would like that $S(u,T)$ converge at frequency $u$ to some constant non-fluctuating value, i.e., $$\lim_{T\to \infty} \mathbf{Var}{S(u,T)}=0 \tag{3}\label{3}.$$ Unfortunately, this fails even for a normal process. In fact, with some easy manipulation of the moments of a normal rv you can show that $$\mathbf{E}\{|S(u,T)|^2\} \ge |\mathbf{E}\{S(u,T)|\}^2 \tag{4}\label{4}$$ and unless the spectral content is null this does not converge. See for details [1].

To remove that fluctuation you must also include an ensemble averaging operation on $\eqref{2}$, formally written as $$ \tilde S(u)=\lim_{T\to \infty} \mathbf {E}\{\frac{|\hat X_T(u)|^2}{T}\} \tag{5}\label{5} $$ This now is a consistent estimate, and with the assumption of ergodicity you replace the ensemble average with a time integral and you end up with the Wiener-Khintchin formula.

[1] Davenport- and Root: An Introduction to the Theory of Random Signals and Noise, pp105-108

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  • $\begingroup$ Nice answer! "...to remove that fluctuation you must also include an ensemble averaging operation on (3)." (I guess you mean (2), or am I missing something?). $\endgroup$
    – Quillo
    Commented Oct 6, 2022 at 12:51
  • $\begingroup$ @Quillo yes of course, thank you for catching it. $\endgroup$
    – hyportnex
    Commented Oct 6, 2022 at 12:57

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