Quantitative MRI mOOCB1 Mapping · Exercise 4-3
Building intuition: why do the two AFI signals depend on B1?
Exercise 4-3 · (a)
With TR1 = TR2, AFI is just SPGR
Drag to turn the view
One magnetization vector in steady state: each pulse tips it by 60° × B1, the readout locks its transverse part (S), the spoiler removes it, and Mz recovers until the next pulse, where it is locked too. Move B1 and T1: do S1 and S2 ever differ?
| Mz just before pulse 1 | |
| Mz just before pulse 2 | |
| S1 (readout after pulse 1) | |
| S2 (readout after pulse 2) | |
| r = S2 / S1 | |
Exercise 4-3 · (b)
Higher or lower? Predict the AFI signals
TR2 = 5 × TR1, each pulse tips by 60° × B1, T1 = 900 ms unless the question says otherwise. S1 is read during TR1, S2 during TR2.
- For any TR, T1 or B1: is the signal read during TR2 (S2) higher or lower than the one read during TR1 (S1)? Pulse 2 tips magnetization that had only TR1 to recover; pulse 1 tips magnetization that had the longer TR2. They’re equal only if TR1 = TR2 (the previous slide). Check it with the sliders on the next slide, then come back: your progress here is kept.
- Is S1 higher or lower than the signal of an SPGR scan with TR = TR1 (20 ms)? About 3 × higher here (0.127 vs 0.037 M0): in AFI, pulse 1 comes after the long TR2, not after 20 ms. Check it: the previous slide (TR1 = TR2 = 20 ms) is an SPGR scan with TR = 20 ms; compare its S with S1 on the next slide. Your progress here is kept.
- And S2, compared with an SPGR scan with TR = TR2 (100 ms)? Lower (0.081 vs 0.165 M0): in AFI, pulse 2 comes after only TR1 of recovery, not after 100 ms.
- If B1 > 1, is S2 higher or lower than at B1 = 1? Lower (0.056 at B1 = 1.2 vs 0.081): with TRs this short, 60° is already far above the Ernst angle (12° for 20 ms, 27° for 100 ms), so a larger flip saturates more. Check it with the sliders on the next slide, then come back: your progress here is kept.
- For a short T1 (500 ms), is S1 higher or lower than for brain T1s at 3 T (about 800–1400 ms)? Higher (0.21 vs 0.13 at 900 ms and 0.085 at 1400 ms): a shorter T1 recovers more during each TR. Check it with the sliders on the next slide, then come back: your progress here is kept.
- If B1 > 1, is the ratio r = S2/S1 higher or lower than at B1 = 1? Lower: r falls as the flip angle grows (1 at 0°, 0.64 at 60°, 0.2 at 90° for TR2 = 5 TR1). That’s what the AFI formula reads B1 from. Check it with the sliders on the next slide, then come back: your progress here is kept.
- If T1 doubles (900 → 1800 ms), is r = S2/S1 higher or lower? About the same (0.638 vs 0.637): S1 and S2 both drop, but their ratio doesn’t depend on T1 as long as both TRs are much shorter than T1. That’s why AFI measures B1 without knowing T1. Check it with the sliders on the next slide, then come back: your progress here is kept.
Exercise 4-3 · (c)
Now with TR2 = 5 × TR1
Drag to turn the view
Same vector, but TR1 = 20 ms and TR2 = 100 ms. Move B1 from 0.5 to 1.5, then T1: how do S1, S2 and the B1 from the AFI formula change?
| Mz just before pulse 1 | |
| Mz just before pulse 2 | |
| S1 (readout after pulse 1) | |
| S2 (readout after pulse 2) | |
| r = S2 / S1 | |
| B1 from the AFI formula | |
Exercise 4-3 · (d)
B1 guessing game: is AFI as easy as DAM?
TR1 = 20 ms
TR2 = 100 ms
Click a pixel to read its value in both images.
The brain of Exercise 4-1, through the same receive coils, now imaged with AFI: α = 60°, TR1 = 20 ms, TR2 = 100 ms (T1 = 900 ms), both on one grey scale.
Press Start: a point appears for 5 seconds. Is the transmit B1+ there above or below 1?
Be quick: a right answer scores up to 1000 points, fewer the longer you take.
Exercise 4-3 · (d) · the answer
B1 guessing game: is AFI as easy as DAM?
No.
- DAM: at B1 = 1 the two images are equal (ratio 1), so “which one is brighter?” tells you if B1 is above or below 1.
- AFI: at B1 = 1 the ratio is 0.64, not 1. You have to judge by eye whether S2 is more or less than 64 % of S1.
- AFI with TR2 = TR1: the ratio is 1, but for every B1: the two images are the same (AFI is just SPGR), so they tell you nothing about B1.
- With TR2 = TR1 it is 1, but for every B1: AFI is then just SPGR.
AFI: S2/S1 = (1 + n cos θ)/(n + cos θ), n = TR2/TR1, θ = 60° × B1, for TRs much shorter than T1. DAM (long TR): S(120°)/S(60°) = 2 cos θ.
Exercise 4-3 · Summary
Reading B1 from AFI images
- With TR1 = TR2, AFI is just SPGR: the two images are identical, so they tell you nothing about B1.
- With TR2 > TR1, the ratio S2/S1 gives B1: the receive field cancels in it, and so does T1, as long as both TRs are much shorter than T1.
- Guessing B1 by eye is much harder than with DAM: at B1 = 1 the two AFI images aren’t equal (S2/S1 = 0.64 here, and it depends on TR2/TR1), so there’s no “which one is brighter?” shortcut like with 60° and 120°.
- High B1 → darker, low B1 → brighter: with TRs this short, a larger flip angle saturates the magnetization more and gives less signal. So why do we need two images?
The receive coils also make whole regions bright or dark: in one image, a dark spot can be high B1 or just far from the coils. Only the ratio S2/S1 cancels the receive field.