0
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.