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I am trying to compute the phase evolution of a proton spin in the presence of a slice selection gradient. I have the following properties for the slice selection gradient and the RF excitation:

[Slice selection gradient]
Peak amplitude: 0.015 T/m
Duration: 0.0042 s.

[RF excitation]: Sinc pulse
Duration: 0.004 s.
flip angle: 30 degrees
center frequency: -2000 Hz

The reason the two durations are slightly different is that the RF pulse begins when the gradient has reached its peak amplitude (after a small ramp time).

Now, I want to track the phase evolution of the signal during the whole process. To do so, I have discretized the signal in time steps of $10^{-6}$ seconds and what makes sense to me is to numerically integrate the frequency over time.

i.e.

$$ \phi(z) = \int \gamma \ G_z z \ dt $$

and I will do this for the whole time when the gradient and the RF pulse is applied (including the ramp up and ramp down time). Does this sound correct?

The reason I ask is that in the book "Handbook for MRI pulse sequence" there is an alternate formula in terms of the RF bandwidth and the isodelay parameter and the phase is given as:

$$ \Delta \phi = 2 \pi \Delta f \Delta t_1 $$

where $\Delta f$ is the bandwidth and $\Delta t_1$ is the isodelay parameter. I am unable to develop any intuition for this formula. So using the method I described before and this one, are they equivalant in terms of the final phase evolution? If yes, I think the numerical integration might be more accurate as it takes the phase dispersion during the gradient ramp time into account?

[EDIT]: I implemented this and what I find is that the phase dispersion that I compute using numerical integration from the beginning of the RF pulse is twice as long as the second formula (which sort of makes sense as the isodelay is approximately half the length of the RF pulse). Why do we not compute the phase dispersion from the beginning of the RF pulse?

Another thing I wanted to clarify was the purpose of doing this. Is this because the MRI slice that will be excited has a certain thickness and the frequency variation along this thin slice would cause the spins to get out of phase with each other and eventually result in signal loss? However, if we assume a true 2D infinitesimally thin slice, this would not be necessary as all the spins in the slice would experience the same frequency and not go out of sync.

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    $\begingroup$ Sounds good to me. However, I do not see a reason why you would have twice the phase dispersion... Maybe some subtle difference in the understanding of what a bandwidth is? If I remember correctly, this book sometimes defines a bandwidth "one-sided" and assume that you would go this value in both directions around the center. Why this is done: Yes, there is signal loss from the slice selection gradient. That's the reason why there is a refocussing gradient after the excitation. $\endgroup$ – M529 May 23 '16 at 21:27
  • $\begingroup$ Thank you for commenting. Another thing I read somewhere now is that the phase dispersion is supposed to start happening from the centre of the RF pulse. Something to do with the traverse mag. can be sen to happen around the RF peak and THEN we start looking at the phase dispersion. This would explain the factor of 2 disparity but I am not sure yet why the dispersion would only be calculated when the magnetisation vector is in the traverse plane. $\endgroup$ – Luca May 24 '16 at 6:09
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    $\begingroup$ I am not an expert on RF pulses, but as far as I can remember, for a sinc-pulse, the magnetization vector just jiggles around a bit before the center of the rf pulse. This is kind of intuitive, when you look at the pulse: there are many positive and negative $B_1$-values that nearly cancel each other out, except at the main lobe. So at this time, there is an significant amount of magnetization in the transverse plane and from then on, the dephasing due to the gradient becomes significant. Hence, the "strange" starting point for the calculations. $\endgroup$ – M529 May 24 '16 at 8:56
  • $\begingroup$ Yeah, so basically the dephasing only starts to happen after the spins are excited. I have one other doubt though...I can adjust my calculation to start at the middle but say that gradient offset is -1000Hz. In that case, the estimated phase offset for the spins will be negative. However, looking at the alternate formula the bandwidth and isodelay parameters will always be positive. So, the computed phase dispersion is always positive. Does this not matter? I must say I have issues getting head around phase. $\endgroup$ – Luca May 24 '16 at 9:32

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