MRI Physics: Spatial Encoding and Slice Selection

35 flashcardsby Moll
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1
How do gradients give spatial information?
They make the field, and therefore the Larmor frequency, depend on position: f(x) = (γ/2π)·(B0 + Gx·x).
Frequency (or the phase it accumulates) becomes a position label. Without gradients the coil just sums the signal of the whole sample.
2

Worked: frequency offset at x = 4 cm with Gx = 50 mT/m
≈ +85 kHz.
Extra field = 0.05 T/m × 0.04 m = 0.002 T; Δf = 42.58 MHz/T × 0.002 T = 85.2 kHz.
3

How do you tell slice selection, phase encoding and frequency encoding apart on a diagram?
By timing, not by axis:
1) Slice selection: gradient on during the RF pulse.
2) Phase encoding: brief gradient between excitation and readout, ADC off.
3) Frequency encoding (readout): gradient on while the ADC records.
Any axis can do any job.
4

How do you read a pulse-sequence diagram?
All rows share one time axis: read vertically to see what overlaps. Gradient height = strength, width = duration, sign = slope direction, sloped edges = ramps. Signed area = encoding.
A stack of lobes with an arrow means the phase-encode amplitude changes from one TR to the next, not that all are played at once.
5

How does slice selection work?
A gradient is on during a frequency-selective RF pulse, so only spins whose local frequency falls inside the pulse bandwidth are excited: a slab perpendicular to the gradient.
6

What sets slice thickness, and how do you make it thinner?
δz = BW_RF / ((γ/2π)·G). Thinner: stronger gradient or narrower bandwidth (longer pulse).
The denominator is "Hz per metre", so dividing the frequency band by it gives a thickness. The thinnest slice is limited by G_max (and pulse length/power), so the scanner refuses settings that are too thin.
7

Worked: BW = 1000 Hz and a 2 mm slice. What gradient?
G ≈ 11.7 mT/m (= 1.17 G/cm).
G = 1000 Hz / (42.58 × 10^6 Hz/T × 0.002 m) = 0.0117 T/m.
8

How do you move a slice, and worked: same slice at +5 mm?
Change the RF centre frequency: f_RF = f0 + (γ/2π)·G·z_c. Thickness stays the same.
Worked: 1000 Hz per 2 mm = 500 Hz/mm, so z_c = +5 mm needs f_RF = f0 + 2.5 kHz.
9

Why is there a negative lobe after the slice-select gradient of a 90° pulse?
Because the gradient is still on during the second half of the pulse, spins across the slice thickness pick up position-dependent phase and partly cancel. An opposite lobe of about half the slice-gradient area undoes this.
Half, because excitation effectively happens at the pulse centre.
10

Why does a slice-selective 180° pulse need no rephasing lobe?
The arrangement is symmetric: phase picked up from the gradient before the 180° is reversed by the pulse and cancelled by the identical gradient after it.
(Any crusher lobes placed around it are symmetric too.)
11

Which anatomical plane is which, and which gradient selects it?
Axial splits upper/lower, coronal frontack, sagittal leftight; the selecting gradient depends on how the subject lies.
x, y, z are scanner axes (z along the bore). Human supine, head along z: Gz → axial, Gy → coronal, Gx → sagittal. Prone rodent along the bore (brain's front–back axis along z): Gz → coronal, Gy → axial (horizontal), Gx → sagittal.
Trap: Gz does not always mean axial; check the localizer.
12

How do you select an oblique slice?
Switch on two or three gradients together so the combined slope points along the desired slice normal.
The field still points along z; only the direction in which its strength changes is tilted.
13

What does frequency encoding give you?
A gradient during recording makes each position precess at its own frequency; the Fourier transform of the signal separates the frequencies = positions → a 1D profile.
14

Why is a frequency-encoded profile a projection?
Each profile point sums all spins in a thin slab perpendicular to the gradient.
Two objects at the same x but different y land in the same point, so a second encoding direction is needed.
15

How does phase encoding store position?
A brief gradient before readout makes spins at different y precess at different speeds while it is on. When it switches off the speed difference vanishes, but the accumulated phase difference stays as a y-label: φ(y) = γ·Gy·y·τ.
Like runners sprinting for a moment at speeds depending on their lane, then all slowing to the same pace: their head-starts remain.
16

Why can one readout not separate the y positions?
All y positions at one x share the same readout frequency; their different phases simply add into one complex number. One equation, many unknowns.
Repeating with different phase-encode areas (one k-space line each) gives enough equations for the Fourier transform to solve. Frequency carries x, phase carries y.
17

What does a k-space coordinate mean?
k = (γ/2π)·∫G(t)dt (cycles/m): the gradient area accumulated so far. It says which stripe pattern the signal is currently weighted with.
Gradient on → you move through k-space; gradient off → you stand still; bigger gradient → you move faster.
18

Explain the signal equation term by term
S(kx,ky) = ∬ ρ(x,y) · e^(−i2π(kx·x + ky·y)) dx dy
ρ(x,y) = how much signal each location has. e^(−iφ) = an arrow of length 1 at angle φ: complex numbers are just bookkeeping for arrows with direction. The coil adds every location's arrow → one number per k. That is a Fourier transform, so an inverse FT recovers ρ. (The slides write j for i.)
19

What does one k-space sample represent?
The whole object's signal weighted by one stripe pattern (one spatial frequency).
Every k-space point holds information from every pixel, and every pixel depends on every k-space point.
Trap: one k-space line is not one row of tissue.
20

What is spatial frequency, e.g. kx = 100 cycles/m?
How finely a pattern varies across space: 100 cycles/m = one stripe cycle every 10 mm.
Larger |k| → finer stripes. Not the same as Larmor frequency, which is in Hz (cycles per second).
21

What does the centre of k-space (k = 0) contain?
The total signal of the slice: no encoding phase, so all arrows add in step. Usually the brightest point.
It is the "DC" term of the image, not the signal from the centre of the object.
22

Centre vs edge of k-space
Centre (low k): coarse shapes, overall brightness and contrast. Edges (high k): fine detail and sharp edges.
Centre only → blurry but recognisable image; edges only → outlines without the smooth background.
23

Order of events in one GRE repetition
1) RF + Gz: excite the slice.
2) Negative Gz lobe: slice rephase.
3) Gy lobe: choose this TR's k-space row.
4) Negative Gx prephaser: jump to the left edge of the row.
5) Positive Gx with ADC on: sweep across the row; the echo (kx = 0) is at TE.
Repeat after TR with a new Gy.
24

Why is there a readout prephaser?
It moves the trajectory to −kx_max so the readout sweeps through the centre to +kx_max, putting the echo in the middle of the flat part of the readout gradient.
Without it, sampling would start at k = 0 and the first points would be taken during the ramp.
25

Order of events in one spin-echo repetition
1) 90° + Gz: excite the slice.
2) Gz rephaser.
3) Gy phase encode.
4) 180° + Gz at TE/2: refocus the same slice.
5) Gx prephaser, then Gx readout with the ADC centred on TE.
Repeat after TR with the next Gy.
26

Spin echo: what happens to encoding applied before the 180° pulse, and what sign must a prephaser there have?
The 180° flips its sign (k → −k, the mirror point through the centre).
A phase-encode lobe before the 180° therefore acts with opposite sign (reconstruction accounts for it). A readout prephaser placed before the 180° must have the same polarity as the readout; placed after it, opposite polarity as in GRE.
Trap: "prephasers are always negative."
27

Why does a spin echo have 3 slice lobes, 2 read lobes and 1 phase lobe?
Slice axis: with the 90°, the rephaser, and with the 180°. Read axis: prephaser + readout. Phase axis: one phase-encode lobe per TR.
28

Why must the 180° be slice-selective in multislice spin echo?
A non-selective 180° would also invert Mz in every other slice. Those slices would then be excited later from disturbed magnetization, adding unwanted T1 effects.
A selective 180° only refocuses the slice that was excited.
29

Rephaser vs rewinder vs spoiler
Rephaser: cancels unwanted phase spread so the wanted signal adds up (e.g., after slice selection). Rewinder: undoes encoding after readout so every TR starts the same. Spoiler: deliberately scrambles leftover Mxy so it cannot form unwanted echoes.
None of them stops T1 recovery. Identify the job from timing and area, not polarity.
30

2D multislice vs 3D acquisition
2D: excite thin slices one at a time; phase + readout encoding per slice. 3D: excite a thick slab; phase-encode in two directions + readout; 3D Fourier transform.
Why 3D: thin contiguous partitions, good SNR (whole slab gives signal every TR), isotropic voxels for reformatting. Cost: Ny × Nz repetitions.
31

In a 3D sequence, what does the extra Gz lobe family do?
Partition encoding: a second phase encoding that labels z positions inside the excited slab.
Gz also selects the slab during the RF pulse: same axis, different job.
32

Besides gradients, what else can provide spatial information?
The sensitivity profiles of multiple receive coils (parallel imaging, e.g., SENSE/GRAPPA).
Each coil "sees" the object differently, which lets you skip some phase-encode steps.
33

How is a single spectroscopy voxel localized, and why no readout gradient?
Three slice-selective RF pulses on three axes (e.g., 90°–180°–180°): only their intersection produces the final echo. The echo is recorded with gradients off, because frequency must report chemistry, not position.
34

How do G/cm and mT/m relate, and what is the slides' kHz/cm shortcut?
1 G/cm = 10 mT/m, because 1 T = 10,000 G.
For 1H, γ/2π = 4257.6 Hz/G (= 42.58 MHz/T), so 1 G/cm ≈ 4.26 kHz/cm. Example: 1 kHz across 2 mm = 5 kHz/cm ≈ 1.17 G/cm = 11.7 mT/m.
35

What does the ideal k-space signal equation assume?
The rotating (demodulated) frame and no relaxation, so the signal is a pure Fourier transform of ρ(x,y).
Real data are also weighted by flip angle (B1+), coil sensitivity (B1−) and T2/T2* decay.
Trap: decay during a long readout breaks the ideal and blurs the image.
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