So, I understand how the slice selection gradients work in MRI. So, the frequency offset introduced by the slice selection gradient at a location $z$ relative to the MRI isocenter is given by:

$$ \Delta f = \gamma z G_z $$

Putting some numbers to it, for a gradient of 5 mT/m and a slice thickness of 3 mm, we introduce an offset of $639.7$ Hz per slice.

Now, say I have an RF pulse which has a bandwidth of 2000 Hz and a center frequency of 10 kHz. Given the location of the spin along the slice direction, I want to know whether this spin will be excited by the RF pulse or not.

  • Is it possible to figure that out from this information?

I have a feeling it should be possible from the relation above but I am not sure how the bandwidth parameter will come into play here.

  • Is it a simple matter of checking that if the resonant frequency experienced by the spin falls in the RF center frequency $\pm$ bandwidth / 2.0?

1 Answer 1


Yes, it basically is that simple. If the resonance frequency of the spin falls in the frequency range that the pulse excites, the spin will be manipulated by the RF pulse. Reality usually kicks in when you have imperfect slice profiles (i.e. to achieve a perfect rectangular slice profile, you would need an infinitely long RF sinc-pulse).

Is it possible that you accidentally mixed up angular frequency and normal frequency? I am asking, because your frequency spread and RF bandwidth would excite $\frac{2000\,\textrm{Hz}}{639.7\,\frac{\textrm{Hz}}{\textrm{slice}}} \approx 3.12\,\textrm{slices}$, if I am not mistaken, and... when it comes to frequencies and a ratio is so close to $\pi$, and there is a "/2.0" in the original question, there could be something going on with wrong frequency units.

  • $\begingroup$ Thank you for confirming. I was trying the calculation yesterday with different RF bandwidth, so I might have mixed it up. I will check when I am back at the university and will update the thread! $\endgroup$
    – Luca
    May 21, 2016 at 9:16
  • $\begingroup$ Ya, the issue is that the 5mT/m gradient strength was really picked put of thin air. With those parameters, it should be more like: $ (2000)/(42.57 * 3 * 1000)$ which is more like 15 mT/m. Thank you for pointing that out to me. $\endgroup$
    – Luca
    May 23, 2016 at 11:41

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