I recognize that many articles attempting to describe the basics of MRI under simplify a lot of technical details. I would like to confirm the following understanding about the free induction decay following an RF pulse...as I think these detail simplifications are interfering with the 'correct', foundational understanding.

Specifically, various sources (even research papers) describe the FID signal as a single cosine (or sine) wave at one frequency (the larmor frequency $\omega_0$) multiplied with a scaled (based on longitudinal magnetization and flip angle) exponentially decaying function. For example, the form: $\cos(\omega_ot)\cdot M_{\alpha}\cdot e^{\frac{t}{T_2^*}}$. This has led many papers I have read to say something like, "The FID signal oscilates at a frequency of $\omega_0$".

Here is my confusion: the whole premise of the transverse magnetization decay (even in the absence of magnetic inhomogeneities) is that it is caused by fluctuations in the magnetic field which lead to subpopulations of protons experiencing magnetic fields slightly deviating away from the $B_0$ field. This results in these subpopulations of protons having a different precession frequency: let's say $\omega_i'=\omega_0 + b_i$, where $i$ is a given proton subpopulation.

Necessarily, then, there are many different protons precessing at different frequencies. Because the measured signal at our receive coils is capturing the sum magnetic moment from all of these different subpopulations of protons, our FID signal is clearly comprised of many different frequencies.

I have never taken a raw FID signal and put it through a 1D Fourier transform. I suspect, though, that if I did, and then applied an inverse Fourier Transform, I would not recover the formula: $\cos(\omega_ot)\cdot M_{\alpha}\cdot e^{\frac{t}{T_2^*}}$. Instead I would recover something like:

$$\gamma_1\cos((\omega_o+b_1)t)\cdot M_{\alpha}\cdot e^{\frac{t}{T_2^*}}+\gamma_2\cos((\omega_o+b_2)t)\cdot M_{\alpha}\cdot e^{\frac{t}{T_2^*}}+\gamma_3\cos((\omega_o+b_3)t)\cdot M_{\alpha}\cdot e^{\frac{t}{T_2^*}}+\cdots$$

where $\gamma_i$ are different weights reflecting the fractional contribution that a given proton subpopulation $i$ precessing at a frequency $\omega_0+b_i$ gives to the observed FID.

Is this the correct interpretation?

If so, what is the benefit of consistently describing the FID as a signal oscillating at a single frequency? Is it simply for instructional purposes?


1 Answer 1


A pure single frequency signal must have a constant amplitude.

A signal with non constant amplitude, like $\cos(\omega_0 t)\ M_{\alpha}\ e^{\frac{t}{T_2^*}}$ already contains multiple frequencies as is. There is no need to have the weighted sum you mention.

Furthermore, the shorter the $T_2^*$ the broader the range of frequencies in $\cos(\omega_0 t)\ M_{\alpha}\ e^{\frac{t}{T_2^*}}$, which corresponds to the situation with a broader range of dephasing sources.

Now, it is true that $\cos(\omega_0 t)\ M_{\alpha}\ e^{\frac{t}{T_2^*}}$ is an approximation that does not generally exactly match the data. But it is a good first approximation and it already contains the feature that you are intending to capture with your more complicated formula.

  • $\begingroup$ Thank you for your comments. I assume you are saying that $\cos(\omega_0 t)\ M_{\alpha}\ e^{\frac{t}{T_2^*}}$'s fourier transform contains a 'broad spectrum' (and therefore many frequencies centered at $w_0$) due to the Lorentzian behavior of the decaying exponential. Is that correct? $\endgroup$
    – S.C.
    Nov 11, 2022 at 4:45
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    $\begingroup$ @S.C. yes, that is correct. And the shorter the T2* the broader the spectrum, which makes physical sense. $\endgroup$
    – Dale
    Nov 11, 2022 at 4:54
  • $\begingroup$ If you're willing, I have another question. Maybe this is not answerable unless you are familiar with the history of MRI. I am having a little philosophical chicken or the egg dilemma here (because some of your comments make me think there is a bit of circularity in the rationale). Did we (physicists) choose to use $\cos(\omega_0 t)\ M_{\alpha}\ e^{\frac{t}{T_2^*}}$ as a representative signal strictly because it resembled the raw data 'to a first approximation'? $\endgroup$
    – S.C.
    Nov 11, 2022 at 5:09
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    $\begingroup$ @S.C. I don’t think the historical happenstance is particularly important. Science is full of many false starts, retrospective understandings, and correct predictions. I don’t know which this is and it doesn’t matter. Regardless of the history the observations match the current theory well, so our current understanding is supported by evidence regardless of the path we took to get this understanding $\endgroup$
    – Dale
    Nov 11, 2022 at 5:32
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    $\begingroup$ Thank you for the input. Cheers~ $\endgroup$
    – S.C.
    Nov 11, 2022 at 5:37

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