Quantitative MRI mOOCB1 Mapping · Exercise 4-5
B1 map application example: VFA T1 mapping
Exercise 4-5 · refresher
Variable flip angle (VFA): SPGR at 2 or more flip angles
Steady-state signal Eq. 2.5
$$S_n = A\,\frac{(1-E_1)\sin\theta_n}{1-E_1\cos\theta_n},\qquad E_1 = e^{-TR/T_1}$$A = M0e−TE/T2* × receive gain: whatever is the same at every angle.
Linearized Eqs. 2.7–2.9
$$\frac{S_n}{\sin\theta_n} = E_1\,\frac{S_n}{\tan\theta_n} + A(1-E_1)\quad\Rightarrow\quad T_1 = -\frac{TR}{\ln(\text{slope})}$$Each angle is one point on a line: two angles are the minimum.
Example: T1 = 1000 ms, TR = 15 ms. ○ = Ernst angle; ● = the two angles at 0.71 × its signal. 📖 More in the T1 Mapping chapter · Variable flip angle
Exercise 4-5 · refresher
VFA: where B1 comes in
θn is the nominal flip angle, the one you set. What the voxel actually experiences is θ = B1 θn.
Steady-state signal, with B1 Eq. 2.5
$$S_n = A\,\frac{(1-E_1)\sin(\textcolor{#d2691e}{B_1}\theta_n)}{1-E_1\cos(\textcolor{#d2691e}{B_1}\theta_n)},\qquad E_1 = e^{-TR/T_1}$$A = M0e−TE/T2* × receive gain: whatever is the same at every angle.
Linearized, with B1 Eqs. 2.7–2.9
$$\frac{S_n}{\sin(\textcolor{#d2691e}{B_1}\theta_n)} = E_1\,\frac{S_n}{\tan(\textcolor{#d2691e}{B_1}\theta_n)} + A(1-E_1)\quad\Rightarrow\quad T_1 = -\frac{TR}{\ln(\text{slope})}$$Each angle is one point on a line: two angles are the minimum.
Exercise 4-5
Why measure B1 at all?
Can we fit for B1 instead?
Exercise 4-5 · predict first
Can we fit for B1?
Linearized
$$\frac{S_n}{\sin(\textcolor{#d2691e}{B_1}\theta_n)} = E_1\,\frac{S_n}{\tan(\textcolor{#d2691e}{B_1}\theta_n)} + A(1-E_1)$$General (non-linear)
$$S_n = A\,\frac{(1-E_1)\sin(\textcolor{#d2691e}{B_1}\theta_n)}{1-E_1\cos(\textcolor{#d2691e}{B_1}\theta_n)},\qquad E_1 = e^{-TR/T_1}$$- Can we use this linearized equation to fit for B1? Yes: B1 comes out of the slopeYes: B1 comes out of the interceptNo: you need B1 to place the points in the first place Both axes, Sn/sin(B1θn) and Sn/tan(B1θn), need B1 before a single point can be plotted. The line only gives E1 (slope) and A (intercept), so B1 has to come from somewhere else.
- What about the general, non-linear equation: under ideal circumstances (no noise), is it mathematically possible to fit for B1 with it? Yes: B1 is one more unknown, alongside T1 and ANo: B1 cancels outOnly if T1 is already known Mathematically, yes: B1 sits inside the sin and cos and doesn’t cancel, so with enough noise-free angles a non-linear fit can solve for it together with T1 and A. Whether that holds up with noise is what the next slides test.
- Under ideal circumstances (no noise), what is the minimum number of flip angles needed to fit for B1 with the general equation? 1234 or more Three unknowns, A, T1 and B1, so at least three measurements. If A were known, two would do: that is the setup on the next slide.
Exercise 4-5 · your protocol
You are scanning a brain at 3 T: where would you put your three flip angles?
No B1 map: the fit solves for A, T1 and B1 from your three measurements. You expect B1 to be within the normal range in the brain, 0.7–1.3. The signal curves are hidden until you scan, so choose your TR and angles yourself. Click on the plot to place an angle, drag to move it.
| true T1 | fitted T1 | T1 error | fitted B1 |
True B1 = 1.10 in this voxel. No noise yet. Solid: true curves; dots: your measurements; dashed: the fit.
Exercise 4-5 · predict first
Real images have noise
1. How many flip angles do you think are sufficient to fit for B1 when the SNR is 100 for WM at the Ernst angle?
2. Is it still more efficient to acquire VFA with your angles and no B1 map, than to add an AFI B1 map?
Assume AFI with TR1 = TRVFA and TR2 = 5 TR1: one AFI scan takes as long as 6 VFA angles. With a B1 map, VFA needs only 2 angles.
Exercise 4-5 · the check
With noise: VFA alone vs an AFI B1 map
Simulating…
The T1 fit is much better conditioned when B1 is measured, not fitted: only A and T1 are left to find, so T1 is far more precise for the same scan time.
One WM voxel: T1 = 900 ms, true B1 = 1.1, TR = 15 ms, SNR 100 at the Ernst angle (the same noise in every image), 300 noisy repeats per point.
VFA alone: N angles evenly spaced from 3° to 90°, fitting A, T1 and B1 (B1 within 0.7–1.3).
AFI + 2 angles: AFI at 60°, TR2 = 5 TR1 → B1; then VFA at , fitting A and T1 with that B1.
Exercise 4-5 · try it yourself
VFA fit simulator
Starts on the previous slide's setup: your guessed number of angles, evenly spaced 3°–90°, a WM voxel (T1 900 ms, B1 1.1), TR 15 ms, SNR 100. Change anything, or let A be known.
| B1 | T1 | |
| true | – | – |
| fitted, this acquisition | – | – |
| 300 draws: median ± std | – | – |
| precision (std ÷ true) | – | – |
| bias (median) | – | – |
| fits at a range limit | – | |
Highlighted: precision (how much repeated scans disagree) and bias (how far their median is from the truth).
Exercise 4-5 · Summary
VFA T1 mapping and B1
- VFA T1 fits need accurate B1 values for accurate T1 estimates.
- B1 maps are used by scaling the nominal flip angle voxelwise: θ = B1 θn.
- VFA can fit for B1, but the fit isn’t well conditioned.
- A measured B1 map is more time-efficient and more precise than fitting for B1 (at SNR 100: AFI + 2 angles, 3.8 % T1 std in 8 TR, vs 14 % with 10 angles fitting B1).