# area under the phase encoding gradient in MRI

I have a question about how I could compute the area (duration and amplitude) for the phase encoding gradient in a typical MRI gradient echo sequence. So, let us assume that my field of view in the phase-encode direction is 20 cms and I have a k-space matrix of size 256. From the tutorial here, I can see that the k-space spacing is $1/20e-3$ per meter. Let us also assume that the maximum gradient strength I can achieve is 10mT/m.

I am now stuck and not sure how I can compute the area that I need under the phase encode area. I am aware that the area will change at each phase encoding step and at least from the pulse sequence diagrams that I see for a GE sequence, it seems that the amplitude is adjusted (going for negative max to positive max) and the gradient time is kept constant during each phase encode step.

However, I am at a loss as to how to compute this area?

• By the way: If you are interested in MRI, there is the "Handbook of MRI Pulse Sequences" from Bernstein et al., which is a very good resource of practical information. I suppose every MRI sequence programmer has a copy of this book in her or his desk. – M529 May 14 '16 at 9:32
• Yes, I have ordered it today! I am actually trying to teach things myself, so money is a scarce resource but it seems this was a worthwhile investment. – Luca May 14 '16 at 9:38

Actually, it is as simple as the frequency encoding gradient. Both encoding gradients look terribly different, but in the end they work the very same way.

1. The outer $k$-space position $k_\textrm{max}$ is related to your resolution, e.g. $1\,\textrm{mm}$, or in terms of the FOV: $FOV/N$ where $N$ is the number of image pixels.

2. For the steps between two $k$-space lines, $\Delta k = 1/FOV$

With $k(t) = \frac{\gamma}{2\pi}\int_0^t G(t^\prime)\,\textrm{d}t^\prime$ you can find that for the largest gradient moment $M_\textrm{max} = \frac{\pi(N-1)}{\gamma FOV}$. The steps in between scale this down linearly.

Pay attention to the line numbering, because the central line in $k$-space should be acquired with zero gradient moment.

• Thank you so much for the reply. Could you clarify on how you got the largest gradient area $M_max$ expression? – Luca May 14 '16 at 9:37
• 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? – M529 May 14 '16 at 13:40
• 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 :) – Luca May 14 '16 at 13:45
• 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. – Luca May 17 '16 at 8:42
• @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. – M529 May 17 '16 at 9:14