Looking at some examples in the "handbook of MRI Pulse Sequences" by Bernstein, I see the following:

So if I can generate an RF pulse with a bandwidth of 2 kHz and I want a 3 mm slice thickness, the gradient amplitude can be chosen with:

$$ G_z = \frac{2\pi \Delta f}{\gamma \Delta_z} $$

Plugging the values, this gives a gradient strength of 15.6 mT/m.

Now, my question is that will this gradient select a 3 mm thick slice at the magnet isocenter?

Now say in my MR experiment I am acquiring 128 such slices. Now, if I want to excite the "bottom most slice" than I need to adjust the RF bandwidth and the change in RF BW is given by:

$$ \delta f = \frac{\gamma G_z \delta_z}{2 \pi} $$

Here now, my $\delta_z = -64 * 0.003 m$. Hence, the new RF BW will need to be

$$ f = \Delta f - 64 \Delta f = -126 khz $$

I am not sure if this calculation is correct. The RF value seems quite high to be although I am not sure what is the range of modern MR scanner RF generators.


1 Answer 1


The bandwidth of the RF pulse does not change at all - it is the center-frequency of the RF-pulse that needs to be adjusted. The slice thickness is related to the frequency spread caused by your slice-selection gradient (as you see in the first formula). Hence, to achieve the same slice thickness for each slice, the RF bandwith may not change, but the RF center frequency.

  • $\begingroup$ Thanks for the answer and clarifying this issue. So, if I consider the rotating frame of reference than the center frequency of the RF pulse to excite the slice at the magnet isocenter would be 0 Hz? So, I need to make the center freq. -126 Knz to excite the lower most slice? $\endgroup$
    – Luca
    May 20, 2016 at 10:03
  • $\begingroup$ Yes, that sounds right. $\endgroup$
    – M529
    May 20, 2016 at 10:06

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