Quantitative MRI mOOCB1 Mapping · Exercise 4-4
How does an off-resonance pulse add phase?
- Warm-up Lab, Larmor or RF frame?
- (a) What does a pulse far off resonance do to M along z?
- The images In which frame is the measured phase?
- (b) Where does the phase come from?
- (c) How does it depend on B1, the offset and the pulse shape?
- Summary Two offsets, ±4 kHz, give B1.
Exercise 4-4 · warm-up
Which frame is it? A quiz on frames of reference
One spin, no relaxation, rectangular pulses. Counter-clockwise seen from +z, as in the T2 lab. Anything at f0 (127.7 MHz at 3 T) is drawn far slower than real. B1 orange, Beff red, M blue with its path.
Exercise 4-4 · (a)
A strong pulse, off resonance, on M along z
Turns at f0, with the spins on resonance: what the receiver sees
Drag to turn the view
Peak B1 = 10 µT. On resonance, this pulse would turn M by . What does it do 4 kHz off resonance? Predict, then play.
Exercise 4-4 · (a) · What we saw
Far off resonance, the pulse barely tips M
- In the frame turning with the RF, the spins feel a static effective field: B1 along x′, plus −Δω/γ along z, what is left of B0 in that frame.
- 4 kHz off, Δω/γ = 94 µT while B1 peaks at 10 µT: Beff leans only 6° from −z. M along z circles it in a narrow cone, and since the pulse rises and falls slowly, M follows Beff back to z: it ends tipped by 0.03°.
- On resonance the same pulse turns M by about 870°; at 250 Hz Beff leans 60° and M ends far from z. A Bloch–Siegert pulse must be far enough off resonance to excite nothing, including spins at their own B0 offset.
Exercise 4-4 · the images
The Bloch–Siegert images: in which frame?
magnitude, pulse at +4 kHz
magnitude, pulse at −4 kHz
Measured at 3 T: one slice, an 8 ms Fermi pulse (500° if it were on resonance) at ±4 kHz before the readout, TE/TR 15/100 ms. Masked to the brain.
If this off-resonance pulse tips the longitudinal magnetization of the on-resonance spins, that tip is measured relative to the z axis of which frame?
The Larmor and RF frames turn about z, at f0 and at fRF: all three frames share one z axis (along B0). So the angle from z, and Mz, are the same in every frame, as in the three frame tabs of part (a).
The frames differ only in where x and y point. That matters for the phase: next question.
Each pixel is the angle of Mxy. Measured from the x axis of which frame?
The receiver demodulates at f0: the phase is Mxy’s angle from x′ of the frame turning at f0, the Larmor frame of the on-resonance spins.
- Not the lab: Mxy spins at 128 MHz there.
- Not the transmit frame: the scans transmit at f0 + 4 kHz and f0 − 4 kHz, two different frames, but share one receive frame: so their phases can be subtracted.
- Not the same in all three: z is shared, but each frame’s x turns at its own rate.
Exercise 4-4 · (b)
M in the transverse plane: where the phase comes from
Turns at f0, with the spins on resonance: what the receiver sees
Drag to turn the view · dashed: where M started
A 90° pulse has put M along +y. Then the same pulse (peak B1 = 10 µT). What does it do to M 4 kHz off resonance: stay put, tip, or turn? Predict, then play.
Exercise 4-4 · (b) · What did you see?
What did M do, and what will it do next?
Questions 1–4: what the last slide showed (Fermi pulse, 10 µT, 4 kHz, M starting along +y). Questions 5–8: predictions for the next slide.
- 4 kHz off resonance, in the Larmor (receive) frame: what did M do during the pulse? It tipped up towards zIt stayed in (or very near) the transverse plane and turned slowly about zIt stayed exactly where it was Far off resonance the pulse can’t tip M: it wobbles at most 6° out of the plane and ends with |Mxy| = 1. But it does turn it, by 0.74 rad (42°) over the 8 ms.
- Which way did it turn? Clockwise seen from +z: its phase went downCounter-clockwise: its phase went upBack and forth, with no net turn From π/2 down to 0.83 rad. With the RF 4 kHz above resonance, the spin is pushed away from the RF frequency: it precesses a little slower, and falls behind the Larmor frame.
- And in the RF (transmit) frame? M stood stillM tipped down to −zM circled Beff about 4000 times a second, staying near the transverse plane In the RF frame the field is B1 plus ΔB0 = −94 µT along z: Beff is only 6° from −z. M, at 90° from z, circles it a little faster than 4 kHz. Subtract the frame’s own 4 kHz and what’s left is the slow turn of question 1.
- So what does the pulse change in the image? Its phase, not its magnitudeIts magnitude, not its phaseBoth, by about the same amount |Mxy| stays at 1: the magnitude images at ±4 kHz differ by under 2 % (slide 5). B1 is written into the phase.
- Predict: if B1 doubles (20 µT), the phase gained over the pulse… stays the samedoublesis 4 × larger The extra precession grows as B1²/offset: quadratic in B1. Twice the B1, four times the phase. Check it on the next slide.
- Predict: if the offset doubles (8 kHz), the phase… doubleshalvesis 4 × smaller As B1²/offset: inversely proportional to the offset. Further off resonance, the pulse disturbs M even less.
- Predict: with the RF at −4 kHz instead of +4 kHz, the phase… is the samehas the same size, but the opposite signis zero The spin is always pushed away from the RF frequency: from below, it speeds up instead of slowing down. That sign flip is how the B1 map is made: see the summary.
- Predict: much closer to resonance (1 kHz), does the phase still grow as B1²/offset? Yes, exactly: 4 × the 4 kHz phaseRoughly, but M starts to tip and the law bendsNo phase at all That law needs B1 ≪ the offset, both as fields. At 1 kHz the offset is only 23.5 µT against B1’s 10 µT: M swings up to 24° out of the plane, and the phase falls below the law (−164° instead of −170°).
Exercise 4-4 · (c)
How the phase depends on B1 and on the offset
Turns at f0, with the spins on resonance: what the receiver sees
Drag to turn the view · dashed: where M started
Run the pulse at a few B1 values and offsets; each run adds a point. Predict the shape first: what happens to the phase when B1 doubles?
Exercise 4-4 · Summary
What the spins showed
- Far off resonance, a strong pulse barely tips M: it sees an effective field almost along z (6° from it at 4 kHz, 10 µT).
- But it adds a phase: while the pulse is on, transverse M precesses a little away from the RF frequency, and gains φBS = KBS·B1² in the receive frame. The magnitude is untouched.
- φBS grows as B1²/offset: twice the B1, four times the phase; twice the offset, half (as long as B1 ≪ the offset). KBS = ∫(γB1,norm(t))²/2ωRF dt is set by the pulse shape and offset: known in advance.
- A real image phase also holds the excitation and receive-coil phase, and each spin’s B0 offset. How do you get B1 out?
Repeat with the pulse on the other side of resonance: the Bloch–Siegert phase flips sign, and nothing else changes. So |φ+ − φ−| = 2KBSB1², and B1 = √(|φ+ − φ−| / 2KBS). Unlike DAM or AFI, T1 and the excitation flip angle don’t enter.