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I am trying to understand the workings of an MRI scanner and wanted to figure out how to simulate it by solving Bloch equations in the rotating frame of reference.

I have a small query regarding the apparent Larmor frequency in the rotating frame of reference. So, if $\omega_0$ is the Larmore frequency in the laboratory frame of reference and in the rotating frame the coordinate system is rotating about the $z$ axes with a frequency $-\omega_r$. In this case, now the apparent Larmor frequency would be: $w_{eff} = \omega_0 - \omega_r$.

Now, I have an RF pulse which is entered on the Larmor frequency and has some bandwidth which means that it will excite spins with resonant frequencies +-100 Hz around the Larmor frequency and say I have a slice selection gradient which achieves this frequency difference.

Now, it is clear that for the spins which are in the centre, they will have an apparent resonant frequency of 0 Hz and the other spins would have an apparent resonant frequency +- 100 Hz depending on their position along the z-axes. So, now the time dependent effective field experienced by a spin due to the RF pulse, would this be something like:

$$ B_x(t) = rf(t) \cos(2 \pi \omega_{eff} t + \phi_{rf}) \\ B_y(t) = rf(t) \sin(2 \pi \omega_{eff} t + \phi_{rf}) $$

Here $rf(t)$ is the amplitude of the rf waveform at time $t$. $w_{eff}$ is the apparent resonant frequency in the rotating frame of reference for a given spin and $\phi_{rf}$ is the phase of the applied RF field. I am not sure if this is correct and wanted to verify.

I am not sure if this should belong to the physics forum but I never get any answers there regarding MRI. But please feel free to migrate if this is not appropriate here.

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  • $\begingroup$ It looks good. Isn't there a reference that you could also check? I suppose this is part of most/every textbook. $\endgroup$ – M529 Jul 3 '16 at 7:37
  • $\begingroup$ There are references but I get myself confused. They do not really talk about the RF waveform really. At least not ones that I have seen. Thanks for confirming it. $\endgroup$ – Luca Jul 3 '16 at 9:25

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