FR35 · MRI

MRI k-Space: Skip Lines, Fold the Head, Fill the Gaps

Each line of k-space costs one repetition. Which lines can you skip, and what does the image lose? Skipping k-space rows saves scan time, and which rows you skip decides what the image loses. Skipping every other row halves the field of view, so two copies of the head overlap. Keeping only central rows blurs fine detail. Random skipping leaves faint ripples; a fit that favours few wavelet details removes part of them, roughly halving the error at double speed in a 64-row model.

A head phantom is scanned one k-space line at a time. Skip lines regularly and the head folds; skip them at random and a sparse fit recovers part of the detail.

Scan time $2.13\,\mathrm{min}$

Acceleration (all lines / recorded lines) $1$

Image error against the full noise-free scan $0.895\,\%$

Small-dot contrast kept, relative to the full scan $99.9\,\%$

Choose which lines to record

Which rows to record

Keep one line in

1

R

Field of view

256

\mathrm{FOV}\;(\mathrm{mm})

Receiver noise relative to mean image brightness

0.01

\sigma/\bar{m}

How to fill the rows not recorded

Drag the head, or click a k-space row to drop or add it.

Fit weight, centre rows and seed

Sparsity weight of the wavelet fit

0.008

Rows always kept at the centre (random pattern)

8

Seed of the line draw and the receiver noise

3521

Model, units and sources

$d_{rc}=N\,\Delta k^2\sum_j\rho_ja_jb_j\,\frac{J_1(2\pi\kappa_j)}{\kappa_j}\,e^{-i2\pi\mathbf{k}\cdot\mathbf{r}_j}+\varepsilon_{rc},\qquad \kappa_j=\sqrt{(a_jk_u)^2+(b_jk_v)^2}$

Each ellipse has an exact Fourier transform, so the k-space grid is sampled from the continuous phantom, not from a pixel image. The image is the unitary inverse discrete Fourier transform of the grid. One eighth of the recorded rows outside the centre are held out: the weight curve is fitted without them and scored on them, and the final image then uses every recorded row.

Lustig, Donoho and Pauly choose the weight so the residual matches the noise level and solve with nonlinear conjugate gradients; this Lab uses held-out rows and an accelerated soft-threshold iteration (50 steps) on the same objective.

MIT 6.3000 MRI handout (2025), pp. 15–21 · Lustig, Donoho and Pauly (2007)

Why random skipping can be repaired and regular skipping cannot

Model assumptions

Two-dimensional Cartesian model on a 64 by 64 grid in the rotating frame. The phantom is a sum of uniform ellipses defined for this course, so every k-space sample is its exact analytic transform; seeded complex Gaussian noise is added. Each phase-encode line costs one repetition time of 2 s, the value used in the MIT 6.3000 handout. There is no relaxation contrast, no coil sensitivity, no motion and no off-resonance. The sparse fit is an iterative soft threshold of Haar wavelet details that keeps every recorded sample; its weight is checked on lines held out from the fit. Numbers are a teaching model, not scanner performance.

k-space record (log scale; pink ticks: held-out rows)

Reconstruction: missing lines set to zero

Ideal reference: phantom averaged per pixel

Sparsity weight chosen from held-out rows

More numbers and how they are checked

Rows recorded: $0$
Rows held out to check the fit weight: $0$
Row spacing in spatial frequency: $0$
Largest spatial frequency on the grid: $0$
Pixel width: $0$
Receiver noise relative to mean image brightness: $0$
Held-out row error at the chosen weight: $0$
Weight with the smallest held-out error: $0$
Weight with the smallest error against the full scan: $0$
Full noise-free scan against the ideal phantom: $0$

Errors are root-mean-square differences of complex images divided by the norm of the full noise-free scan on the same grid. Dot contrast is the image value at the three small dots minus the value between them, as a share of the same quantity in the full scan. Leaving rows at zero can never predict a held-out row, so its held-out error is 100 %.

Physics tutorial

An image assembled one k-space line at a time

BackgroundIn a two-dimensional scan, a short gradient pulse along y winds a phase pattern across the head before each readout. The readout gradient along x then sweeps one row of spatial frequencies. One row costs one repetition, so the number of rows sets the scan time.

Why it mattersThe MIT 6.3000 handout works out that 256 rows at 2 s each take about 8.5 minutes. Recording fewer rows saves time, but the image is a Fourier sum over the grid, so the missing rows have to be paid for somewhere.

Start with the essentials

Focus question: Each line of k-space costs one repetition. Which lines can you skip, and what does the image lose?
One-sentence intuition: The row spacing $\Delta k_y$ fixes the field of view $1/\Delta k_y$ . Keep every second row and the spacing doubles, so the field of view halves and two copies of the head overlap. Skip rows at random instead and the missing energy spreads into faint noise-like ripples, which a fit that prefers sparse wavelet coefficients can partly remove.

Core mathematical model

Every sample sees the whole head

d(k_x,k_y)=\iint\rho(x,y)\,e^{-i2\pi(k_xx+k_yy)}\,\mathrm{d}x\,\mathrm{d}y,\qquad k_y=\bar\gamma\int G_y\,\mathrm{d}t

One sample is the overlap of the whole head with one stripe pattern, which is the colour pattern drawn on the stage. The sign follows the MIT 6.3000 handout.

Regular skipping folds the image

\hat m(y)=\frac{1}{R}\sum_{j=0}^{R-1}m\!\left(y+j\,\frac{\mathrm{FOV}}{R}\right)

Zero-filling every $R$ -th row gives the average of $R$ copies shifted by a fraction of the field of view (wrapped around the grid).

A sparse fit for random skipping

\hat m=\arg\min_m\ \tfrac12\left\|M\mathcal{F}m-d\right\|_2^2+\lambda\left\|\Psi m\right\|_1

This is the Lagrangian form in Lustig, Donoho and Pauly (2007). Here $\Psi$ is a three-level Haar transform and the weight $\lambda$ is checked on rows held out from the fit.

Common difficulties

A k-space sample is not a pixel

Typical misconceptionEach point of k-space belongs to one point of the image, so a missing row leaves a missing strip in the head.

Better mental modelEach sample is a stripe pattern across the whole head. Dropping rows near the centre removes the broad brightness; dropping outer rows removes edges and small dots. No single strip of the image goes missing.

Run the experiment

01
Watch the grid fill
Start with Every line: the full scan and watch the stage and the k-space record together.
What to observe: The first rows are near the centre and their stripes are wide; later rows sit further out and their stripes are fine. The brightest samples sit at the centre of the grid.
02
Skip every other row
Choose Skip every other line: the head folds.
What to observe: The scan time halves, and a second copy of the head appears shifted by half the field of view, as in the MIT handout.
03
Skip at random, then fit
Compare Random lines, gaps left empty with Random lines, sparse fit fills gaps, then click rows of the k-space record to drop or add them.
What to observe: No copy appears. The sparse fit roughly halves the error and brings the small dots back, but not fully: at this matrix size line skipping buys about a factor of two.

MRI k-Space: Skip Lines, Fold the Head, Fill the Gaps

An image assembled one k-space line at a time

Core mathematical model

Every sample sees the whole head

Regular skipping folds the image

A sparse fit for random skipping

Common difficulties

A k-space sample is not a pixel

Run the experiment

Watch the grid fill

Skip every other row

Skip at random, then fit

Related experiments and articles

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