0
$\begingroup$

I understand the idea of spatial encoding afforded by the application of a readout gradient along one of the planes (frequency direction) in MRI. In typical 2D image generation, there is also a phase selection gradient, but this is irrelevant to the question.

In 1D I found this image at this point on an online presentation by FundBioImag:

enter image description here

The precessing spins along the y direction are progressively more dephased with time after the application of the negative lobe of the gradient. When the gradient changes polarity the phases of the spins along the axis come back together to emit a RF pulse or signal that peaks at a maximum.

The question is perhaps at a much more basic or electronics level: I get what we are doing to the spins with the gradients, but what prompts these spins to emit a signal?

After all there is no signal immediately emitted after applying the slice and phase gradients in the 2D image acquisition setting:

enter image description here

$\endgroup$
1
$\begingroup$

The gradients do not create the signal, they merely spatially encode the signal.

The signal itself is produced by the RF excitation pulse which tips some of the magnetization into the transverse plane. As long as there is magnetization in the transverse plane there is signal. The signal produced immediately after the peak of the RF pulse is actually the largest signal. It is called the free induction decay.

The reason the pulse sequence diagram does not show the free induction decay is simply because the RF receivers are not turned on at that point. The signal is physically present, but it is not being received so it is not drawn in this diagram.

$\endgroup$
  • $\begingroup$ OK... So the coincidence in the diagram in the OP between the mid-point of the readout gradient (in green) and the maximal signal is simply created by turning on the receiver antenna at the appropriate time while the signal present since the initial RF 90 degree excitation pulse is being further "massaged" by the spatial-encoding readout gradient? $\endgroup$ – Antoni Parellada Nov 25 '18 at 1:17
  • $\begingroup$ No, that is caused because to a first approximation most objects that you image have a boxcar shape in space so they have a sinc shape in k-space. The peak of a sinc is in the center of k-space, so the center of k-space has higher signal due to the spatial encoding, but due to T2 and T2* the signal is lower than it would be if the center of k-space were acquired earlier $\endgroup$ – Dale Nov 25 '18 at 1:44
  • 1
    $\begingroup$ I don't see a contradiction between my comment and your answer - I understand what you mean by the sinc function, and I think it does makes sense. I chalk the confusion to the drawn out nature of my comment, but if there is a clear mistake in what I said, please let me know. $\endgroup$ – Antoni Parellada Nov 25 '18 at 1:50
  • 1
    $\begingroup$ Oh, I misread the “at the appropriate time” part of your comment. Looking back I see there indeed was no conflict $\endgroup$ – Dale Nov 25 '18 at 1:54

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.