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May 17, 2016 at 9:39 comment added Luca Thanks for clearing that up. I thought so as well but quite unsure of things at the moment.
May 17, 2016 at 9:14 comment added M529 @Luca Always the total area with the ramps: If we were not bound by the laws of physics, we could switch gradients instantaneously. However, in reality this is impossible and we have to ramp up the gradients. The gradient ramps do also contribute to the magnetic field (and therefore to the gradient moment) each spin experiences and must therefore be taken into account in the encoding process.
May 17, 2016 at 9:03 comment added Luca May I ask if this should be the total area under the gradient or just the area under the flat part of the gradient i.e. during the time when the gradient is at the peak?
May 17, 2016 at 8:42 comment added Luca I think you are right. So, $k_{max} = \frac{1}{2} (N-1) \Delta k_y$ and then the area is related to it by a factor of $\frac{\gamma}{2\pi}$ to give the relation you gave.
May 14, 2016 at 13:45 comment added Luca Yeah, I was basically trying to understand a simple gradient echo sequence. I think the $\delta k$ is correct. Maybe the $k_{max}$ term. However, as you can tell, my understanding it quite limited :)
May 14, 2016 at 13:40 comment added M529 I am sorry, I assume I have to correct my answer. I tried to come up with an easy to understand explanation, but I assume that there is a problem in the definition of either $\Delta k$ or $k_\texrrm{max}$ (basically a factor of 2 is missing somewhere). This may arise from the exact definition of the Fourier-transform in textbooks, but I have not found a stringent, clear solution to the problem. I assume this also is where your problem comes from, right?
May 14, 2016 at 9:37 comment added Luca Thank you so much for the reply. Could you clarify on how you got the largest gradient area $M_max$ expression?
May 13, 2016 at 21:02 vote accept Luca
May 13, 2016 at 19:50 history answered M529 CC BY-SA 3.0