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Full Version: Why do we not measure FID but the gradient recalled echo?
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Hi,

I always thought FID can not be measured directly because a ringing in the receiver coil for a few milliseconds renders that impossible.

BUT then I see a beautiful FID recorede on this page under FID
http://mri-q.com/free-induction-decay.html.

Who knows the reason, why we record the gradient recalled echo, instead of the FID as a whole?

Thanks,
Harry
Dear Harry1,

The FID shown in the image you cited was one I recorded several years ago from a uniform phantom without the use of any spatial encoding gradients. The oscilloscope was triggered to begin recording at several hundred microseconds after the RF pulse, so any noise/oscillations early in the FID were excluded.

The main reason we generally don't typically measure the FID for routine imaging is that (unless special methods are used) we need to apply gradients to localize the signal in space. A test-tube NMR laboratory spectrometer doesn't require spatial localization and so an FID can be used there.

You may wish to read the "Advanced Discussion" from the following Q&A http://mriquestions.com/free-induction-decay.html

ADE
In the last few months, both GE and Siemens have come out with "silent scan" sequences as commercial products. These sequences have ultra-short TE values and do sample the FID primarily instead of a GRE or SE.
I had a similar question in my MRI physics class and with a friend but neither were really able to help me understand. If the initial 90 pulse in a spin echo sequence produces a FID that, according to the MRI in Practice book, is too fast, how can GRE sequences use flip angles less than 90 but measure those signals?

I am very green and have only been learning this for 5 weeks.
not expert, however I googled and found this is maybe related to k-space:

If the 90-FID imaging sequence presented in the previous chapter was used, we would record only half of k-space. We would like to produce the equivalent of an echo in the center of our acquisition window when the frequency encoding gradient is turned on. This would give us both the left and right halves of k-space.


https://www.sciencedirect.com/topics/neu...tion-decay
https://www.cis.rit.edu/htbooks/mri/chap-8/chap-8.htm
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