0
$\begingroup$

As I understand it, in MRI, a line in k-space is usually acquired while the readout gradient is applied. So during the application of this gradient the protons spin at different frequencies along the gradient.

Now, I sample the MRI signal during the application of this gradient. My confusion is that I can sample the signal at arbitrary resolution and the inverse FFT will produce the image with as many pixels as the sampling resolution.

To make the question a bit concrete. let's say I have simulated 128 spins at a distance of 1 mm apart. This will give me an object with the FOV of 0.128 meters.

Now say during the readout I sample 10000 equidistant points. So, effectively I have collected 10000 complex k-space values. An inverse FFT would also give me an image with 10000 pixels. Now, how do I resolve this 10000 pixel image into a 128 pixel image where each pixel corresponds to the corresponding spin.

$\endgroup$
2
$\begingroup$

I would expect that you simply cut the outer parts of the image away. Certainly, there is always a hassle to figure out the proper pixel in the center, but that is a detail of the FT-algorithm employed.

Remeber: Sampling denser than you need increases the FOV. Hence, the image you simulate should simply be larger with the original image (that you get by simulating 128 sampling points in the setting you described above) sitting at its center. Cutting off the outer void parts is therefore the technique you need, and not an interpolation technique.

$\endgroup$
  • $\begingroup$ Ahhhh...Thanks for that explanation. I was not sure whether I need to interpolate or not. Yes, these MRI relations are very tricky :) Thank you very much! I owe you a few beers over the course of last few weeks! $\endgroup$ – Luca Jun 28 '16 at 18:02
  • 1
    $\begingroup$ They are not that tricky. Or maybe they are and you just get used to them...? $\endgroup$ – M529 Jun 28 '16 at 18:40
  • $\begingroup$ Perhaps. I have only started to learn about MRI at my university. Hopefully I will understand it at some point! $\endgroup$ – Luca Jun 28 '16 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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