L3 · Chapter 5

Sampling, wraparound and resolution: what one missing line does

Sandbox Physics · 2026-10-29

Chapter 4 showed that scan time is set almost entirely by the number of phase-encode lines. The obvious way to save time is to record fewer lines. But every position in k-space matters: the spacing between samples sets the field of view and the outermost sample sets the pixel. This chapter works out what the two simplest ways of recording fewer lines do to the image. Skipping every other line shrinks the image to half the field of view with two copies on top of each other; keeping only the centre lines blurs it. Finally it looks at noise and the price recording fewer lines pays in signal-to-noise ratio.

The step sets the field of view, the range sets the pixel

Discrete k-space samples are $\Delta k$ apart. The matching image repeats in space with period $1/\Delta k$ , so the field of view is $\mathrm{FOV}=1/\Delta k$ . Water outside the field of view does not vanish; it folds in from the opposite side. This is the same folding as along the readout in Chapter 3, now along the phase-encode direction.

In the other direction, the spatial frequency of the outermost line, $k_{\max}$ , sets the finest stripe that can be resolved. $N$ lines cover a range of $N\Delta k$ , so the pixel width is $\Delta y=1/(N\Delta k)=\mathrm{FOV}/N$ . The full reconstruction in the MIT handout uses every sample with $-128<k_x,k_y<128$ .

The “Field of view too small: wraparound” preset of the k-space experiment shrinks the field of view from 256 mm to 180 mm. The head phantom is 230 mm tall, so 25 mm at the top and 25 mm at the bottom fall outside and fold back from the opposite edge. It is 184 mm wide, so 2 mm also folds back at each side.

\mathrm{FOV}=\frac{1}{\Delta k},\qquad \Delta y=\frac{1}{N\Delta k}=\frac{\mathrm{FOV}}{N}

The step and the field of view are reciprocals, and so are the sampled range and the pixel.

y_{\mathrm{image}}=y-\mathrm{FOV}\cdot\operatorname{round}\!\left(\frac{y}{\mathrm{FOV}}\right)

Water outside the field of view appears after subtracting a whole number of fields of view from its position.

Each excitation records one line of k-space: the phase-encoding gradient picks

k_y

and the readout sweeps

k_x

. Solid lines are recorded and dashed lines skipped, as when every other line is kept. The spacing

\Delta k=1/\mathrm{FOV}

sets the field of view; the extent

N\Delta k

sets the pixel. Original schematic.

Source: MIT 6.3000 Signal Processing · M. Lustig

Skipping every other line: two half-strength copies

The MIT handout asks what the image becomes if only the even columns are measured. Write “keep the even lines, fill the odd ones with zeros” as a multiplying factor. Line $n$ is multiplied by $\tfrac12(1+e^{i\pi n})$ , which is 1 for even lines and 0 for odd ones.

The factor $e^{i\pi n}=e^{i2\pi k_y(\mathrm{FOV}/2)}$ is exactly the shift-theorem phase that moves the object by half the field of view. So the zero-filled image is the average of the original and a copy shifted by half the field of view: each copy at half strength, the two copies offset by half the field of view. With a 256 mm field of view the copies are 128 mm apart; half of the top of the head lies over the bottom and half of the bottom lies over the top.

Lustig and co-workers call this coherent aliasing: equispaced undersampling with zero filling gives a superposition of shifted replicas, and the original and its replicas are equally likely, so they cannot be told apart. The benchmark program checks the identity above on the full noise-free data; every pixel agrees to $9.7\times10^{-17}$ or better.

m_{R=2}(y)=\tfrac12\left[m(y)+m\!\left(y+\tfrac{\mathrm{FOV}}{2}\right)\right]

Keeping one line in

R

gives the average of

R

copies offset by

\mathrm{FOV}/R

Source: MIT 6.3000 Signal Processing · M. Lustig

Centre lines only: the image blurs

Another way to save time is to record only the lines near the centre. The line spacing is unchanged, so the field of view is unchanged and nothing folds; but $k_{\max}$ shrinks and the pixel along the phase-encode direction grows. With only the centre 16 lines (a 4-fold acceleration), the resolution along $y$ goes from 4 mm to $256/16=16\,\mathrm{mm}$ , while the readout direction stays at 4 mm.

Near the bottom of the phantom sit three small dots 8 mm across, their centres 16 mm apart along $x$ . A vertical resolution of 16 mm is twice the dot diameter, so each dot is smeared up and down and its peak drops. The readout shows only 48.2% of the full-scan dot contrast left, with a 31.6% image error.

Lustig and co-workers note that most of an image’s energy lies close to the k-space origin. That is why the large-scale brightness survives when only the centre is kept; the high-frequency samples that were dropped carry the edges and the detail.

\Delta y_{\mathrm{centre}}=\frac{1}{N_{\mathrm{lines}}\,\Delta k}=\frac{\mathrm{FOV}}{N_{\mathrm{lines}}}

With centre lines only, the pixel width is set by the number of lines kept.

Source: MIT 6.3000 Signal Processing · M. Lustig

The noise cost of fewer lines

What follows is this course’s own derivation, using the experiment’s normalization: the discrete Fourier transform is unitary, meaning the forward and inverse transforms are each divided by $\sqrt N$ . A unitary transform preserves total energy, so if every k-space sample carries independent noise of standard deviation $\sigma$ , every pixel of the full-scan image also carries noise of standard deviation $\sigma$ .

Record a fraction $f$ of the lines and fill the rest with zeros, and only $f$ of the noise energy remains: the per-pixel noise variance becomes $f\sigma^2$ . What happens to the signal depends on which lines are kept. Skipping every other line leaves each copy at half strength while the noise standard deviation falls only to $1/\sqrt2$ , so the signal-to-noise ratio drops to $1/\sqrt2$ of the full scan. Keeping the centre lines leaves the brightness of large regions unchanged and halves the noise standard deviation ( $f=1/4$ ), so large regions actually gain a factor of two in signal-to-noise ratio, at the cost of pixels four times taller.

The benchmark program checks the $f\sigma^2$ prediction with 8 random seeds: the measured noise variance of the zero-filled images divided by the prediction is 1.0013.

\operatorname{Var}\!\left[\tilde m(\mathbf r)\right]=f\,\sigma^2

\tilde m

is the zero-filled image. For a unitary transform,

f

is the fraction of lines recorded.

Source: MIT 6.3000 Signal Processing · M. Lustig

Worked example: a 180 mm field of view

Problem: in the k-space experiment, set the field of view to 180 mm with the matrix still $64\times64$ . Find the k-space step, the pixel width, and where the top 25 mm of the head lands.

The step is $\Delta k=1/0.18\,\mathrm m=5.56\,\mathrm{m^{-1}}$ and the pixel $180/64=2.81\,\mathrm{mm}$ . The pixel is finer than with the 256 mm field of view, yet the scan time is unchanged at 64 lines and 2.13 minutes.

The field of view covers $-90$ to $+90\,\mathrm{mm}$ . The top of the head is at $+115\,\mathrm{mm}$ and appears one field of view lower, at $115-180=-65\,\mathrm{mm}$ . So the strip from 90 to 115 mm lies over the bottom of the image from $-90$ to $-65\,\mathrm{mm}$ , right on top of the three small dots at $y=-85\,\mathrm{mm}$ . To avoid wraparound the field of view must be at least 230 mm, which gives a 3.59 mm pixel.

Source: MIT 6.3000 Signal Processing · M. Lustig

Try it in the Lab

Compare two ways of recording fewer lines

Choose “Skip every other line: the head folds”. Predict two heads 128 mm apart, each half as bright. Check against the image and note the 49.8% error and the 54.9% dot contrast.
Then choose “Centre lines only: a blur”. It records half as many lines as skipping, yet nothing folds. Watch the small dots smear vertically and read the 48.2% contrast.
Choose “Field of view too small: wraparound”, find the strip that folds from the top of the head to the bottom and confirm that it covers the small dots. Then drag the field-of-view slider to 232 mm; the wraparound should disappear.

Exercises with explained answers

With a 256 mm field of view, one line in four is recorded and the rest are zero-filled. What does the image look like?
Show the answer
The average of four copies offset by $256/4=64\,\mathrm{mm}$ , each at a quarter of the strength. Dragging the “Keep one line in” slider to 4 in the experiment shows it.
With a 256 mm field of view and only the centre 32 lines, what is the resolution along the phase-encode direction, and what happens to the three 8 mm dots?
Show the answer
The resolution is $256/32=8\,\mathrm{mm}$ , about the dot diameter, so the dots are still smeared somewhat but are much clearer than with 16 lines.
Skipping every other line halves the time. How many repeats of that scan must be averaged to bring the signal-to-noise ratio back to the full-scan level?
Show the answer
The ratio falls to $1/\sqrt2$ and averaging $n$ repeats raises it by $\sqrt n$ , so two repeats, which takes the time back to that of the full scan. The folding is still there; removing it needs the methods of Chapter 6.

Sources and what we recomputed

MIT 6.3000 Signal Processing, Magnetic Resonance Imaging lecture handout (6 May 2025), data and examples from E. Adalsteinsson
Read in full. Page 9: the proton gyromagnetic ratio rounded to 42.58 MHz/T. Pages 11–13: gradient fields, the phase integral and the definition of k as the time integral of the gradient. Pages 15–18: k-space data, the complex image and its magnitude. Page 19: 256 lines at 2 s each, about 8.5 minutes. Pages 20–21: keeping only even columns of k-space. Pages 24–36: two-coil unfolding, including the coil pair whose sensitivities cannot separate the folded pixels. The handout uses the sign convention of the Fourier kernel that this course follows.
M. Lustig, D. Donoho and J. M. Pauly, Sparse MRI: the application of compressed sensing for rapid MR imaging, Magnetic Resonance in Medicine 58, 1182–1195 (2007), doi:10.1002/mrm.21391
Read in full. Fig. 1 and the three requirements of compressed sensing; Fig. 2 (a 256-sample sparse signal, 8-fold undersampled); Eq. 1 (spread of the interference) and its derivation in Appendix B, which the text calls Appendix II; the remark that single-slice 2DFT interference is D^{1/4} larger; Eq. 2 (transform point spread); Eqs. 3–4 (constrained reconstruction); Appendix A, Eqs. A1–A3 (Lagrangian form and nonlinear conjugate gradients, 80 to 200 iterations); Results and Discussion (2.4-fold brain, 5-fold angiography from 22 s to 4.4 s, loss of low-contrast features). The in-vivo images are described, not reproduced.