MRI Safety and Screening Quiz

K-space and Rapid Imaging Quiz

K-Space

  1. A point in k-space corresponds to
    1. The amount of a particular spatial frequency in the entire imaged object
    2. The amount of a particular spatial frequency at a particular point in the imaged object
    3. The magnitude and phase of the signal arising from a particular point in the imaged object
    4. The spin density of the signal arising from the entire imaged object.

    K-space is an array of numbers representing spatial frequencies in the MR image. The individual points (kx,ky) in k-space do not correspond one-to-one with individual pixels (x,y) in the image. Each k-space point contains spatial frequency and phase information about every pixel in the final image. Link to Q&A discussion

  2. Which of the following statements about k-space is false?
    1. The kx and ky axes of k-space generally correspond to the horizontal (x-) and vertical (y-) axes of the image.
    2. Each point in k-space contains spatial frequency and phase information about every pixel in the final image.
    3. Each pixel in the image maps to every point in k-space.
    4. Points near the center of k-space correspond to points in the center of the image.

    Each point in k-space contains spatial frequency and phase information about every pixel in the final image. So choice (d) is false. Link to Q&A discussion

  3. Which of the following statements about k-space is true?
    1. The center of k-space provides information about the edges and details of the imaged object.
    2. The periphery of k-space provides information about the general shape and contour of the object.
    3. The raw data at the exact center of k-space always has a larger magnitude than any other point.
    4. k-space must be filled with data in a specific order.

    Only (c) is true. There are two reasons the central area of k-space has the largest magnitude. First, the central row (ky = 0) is acquired with no phase-encoding gradient (and hence no destructive wave interference caused by phase-encoding steps). Secondly, the central column of k-space (kx = 0) coincides with the peak of the MR echo. Link to Q&A discussion

  4. In routine anatomic MR images, what points in k-space have the lowest amplitudes?
    1. The four corners
    2. The center
    3. The middle of the top and bottom edges
    4. The middle of the left and right edges

    Signals from the four corners have the highest spatial frequencies in both directions and correspond to data locations where the MR signals are weakest. For time-savings in spiral imaging, the four corners are often excluded from data sampling with only minimal loss of resolution noted. Link to Q&A discussion

  5. The “k” of k-space refers to “wavenumber”, an index for spatial frequency, which is the number of waves or cycles per unit distance. An acceptable unit for “k” could therefore be
    1. Cycles per second
    2. mm−1
    3. cm/sec
    4. Line pairs

    The wavenumber (k) is simply the reciprocal of the wavelength. Since the wavelength is measured in units of distance, the units for wavenumber are (1/distance), such as 1/mm = mm−1. Other acceptable units would include line pairs per mm and cycles per mm. Link to Q&A discussion

  6. Which statement about the location of spatial frequencies in k-space is true?
    1. A point along the kx axis corresponds to horizontal stripes in the image
    2. A point along the ky axis corresponds to vertical stripes in the image
    3. A point at a 45º angle from the the kx axis corresponds to stripes at 45º in the image
    4. If the imaged object is small relative to the field-of-view, its k-space frequency representation will be mostly confined to the center.

    Option (c) is true. Options (a) and (b) are reversed. Concerning incorrect choice (d), the physical size of the object imaged and its frequency expanse in k-space are inversely related. In other words, small objects have ripples far out into the periphery of k-space, while larger objects have their spectral energies more concentrated at the center. Link to Q&A discussion

  7. Consider a conventional 2DFT spin-echo pulse sequence with frequency encoding along the x-axis and phase-encoding along the y-axis. Which statement is false?
    1. For a given phase-encode step (ky), digitized data from the early part of the echo is placed at the far left side of k-space (negative kx values).
    2. For a given phase-encode step (ky), digitized data from the early part of the echo is placed at the far right side of k-space (positive kx values).
    3. For a given phase-encode step (ky), digitized data from the peak of the echo is placed at the the point (kx=0, ky=0).
    4. For a given phase-encode step (ky), digitized data from the peak of the echo is placed at the the point (kx=0, ky).

    Option (c) is false, while option (d) is true. Link to Q&A discussion

  8. Concerning the filling of k-space by a conventional 2DFT spin-echo pulse sequence with frequency encoding along the x-axis and phase-encoding along the y-axis, which statement is false?
    1. The central point of k-space (kx=0, ky=0) corresponds to the constant term in the Fourier representation of the image.
    2. Except for the central row (ky=0), each row of k-space is acquired with the same constant value of the phase-encoding gradient.
    3. The central row (ky=0) of k-space is acquired with no phase-encoding gradient.
    4. The central column of k-space (kx=0) coincides with the peak of the MR echo for each phase-encoding step.

    Option (b) is false. In 2DFT spin-echo imaging, the phase encoding gradient is stepped in magnitude from cycle to cycle, not held constant. The other statements are correct. Link to Q&A discussion

  9. The row-by-row filling of k-space by a conventional 2DFT spin-echo pulse sequence is often referred to as what trajectory?
    1. Cartesian
    2. Radial
    3. Zig-Zag
    4. Blipped

    Traditionally the Cartesian (row-by-row) method was used nearly exclusively for all pulse sequences, but today many different patterns are widely encountered. Link to Q&A discussion

  10. Disadvantages of radial sampling methods include all of the following except
    1. Sensitivity to motion artifacts.
    2. Exclusion of the corners of k-space.
    3. Need to run a calibration scan to estimate gradient delays and distortion.
    4. Acquired data points not in format suitable for Fast Fourier Transformation.

    In radial acquisition the center of k-space is oversampled and continuously updated due to the overlapping "spokes" that repeatedly pass through this region. This redundancy can be exploited to detect and correct for movement if the signal from the k-space center changes between views. Additionally, all radial spokes make equal contributions to the image (unlike Cartesian sampling where just a few lines through the center of k-space set overall image contrast and noise levels). So motion on just one or a few radial views is not likely to severely degrade image quality. Link to Q&A discussion

  11. The non-uniform raw data from a radial or spiral acquisition must be placed into a rectangular (Cartesian) matrix format for efficient computations. This process is referred to as
    1. Morphing
    2. Gridding
    3. Pigeon-holing
    4. Compressing

    “Gridding“ is the commonly used term for this process. Although various methods exist, the typical gridding algorithm first involves multiplication of original data with a set of density compensation weights and convolution with a gridding kernel. The resultant data is then interpolated and placed into the uniformly spaced matrix (the "grid") where a discrete Fourier transform is performed. Finally, the field-of-view is trimmed and the transformed data multiplied by an apodization correction function. Link to Q&A discussion

  12. How many phase cycles are needed to uniquely differentiate each pixel along a certain direction in an image?
    1. 0.5
    2. 1.0
    3. 1
    4. 2.0

    One full phase cycle is needed to uniquely differentiate each pixel. If there are N pixels of width (Δw) spanning the field-of-view (FOV), then N phase cycles are needed. Link to Q&A discussion

  13. If the data sampling rate is halved in the frequency-encode direction (x), what happens to the spacing (Δkx) between points in k-space?
    1. It remains unchanged.
    2. It is halved.
    3. It is doubled.
    4. It is quadrupled.

    This is equivalent to sampling every other kx value, so compared to the original pattern, the spacing Δkx) will double Link to Q&A discussion

  14. If the data sampling rate is halved in the frequency encode direction (x), what happens to the field of view in that direction?
    1. It remains unchanged.
    2. It is halved.
    3. It is doubled.
    4. It is quadrupled.

    We use the formula Δk = 1/FOV. So doubling Δk means the FOV is halved. Link to Q&A discussion

  15. If the data sampling rate is halved in the frequency encode direction (x), what happens to the pixel size in that direction?
    1. It remains unchanged.
    2. It is halved.
    3. It is doubled.
    4. It is quadrupled.

    The pixel width Δw = 1/ kFOV. As long as the kFOV = 2kmax is unchanged, the pixel width will remain unchanged also. Link to Q&A discussion

  16. Supposed the data sampling rate in the frequency encode direction (x) is unchanged, but the sampling time is reduced so that only the central ⅓ of k-space is sampled. What happens to the pixel width and FOV in that direction?
    1. The pixel size is tripled and the FOV is unchanged.
    2. The pixel size is unchanged and the FOV is tripled.
    3. Both the pixel size and FOV are reduced by ⅓.
    4. Both the pixel size and FOV are tripled.

    Because the data sampling rate is unchanged, Δkx is unchanged and the points are spaced out as before. So since Δk = 1/FOV, the FOVx is unchanged. By sampling only the center ⅓ of k-space, the kFOV is now only ⅓ as large. And since pixel width Δw = 1/ kFOV, this means the pixel width has tripled. Hence answer (a) is correct. Link to Q&A discussion

  17. Suppose that the frequency encoding portion of the 2D spin warp imaging was left alone, but the number of phase-encoding steps was cut in half (while leaving the maximum and minimum amplitudes of the phase-encoding gradient unchanged). What happens to the spacing (Δky) between points in the phase-encode direction (y)?
    1. It remains unchanged.
    2. It is halved.
    3. It is doubled.
    4. It is quadrupled.

    Cutting the number of phase encode steps (Ny) in half in this scenario doubles the increment between successive steps, so Δky is doubled. Link to Q&A discussion

  18. In the question above, what does the phrase “while leaving the maximum and minimum amplitudes of the phase-encoding gradient unchanged” mean in terms of kFOV in the y-direction?
    1. It remains unchanged.
    2. It is halved.
    3. It is doubled.
    4. It is quadrupled.

    The phrase basically stipulates that kFOV is unchanged. Link to Q&A discussion

  19. Continuing on with the same scenario as in questions 17 and 18, what happens to the pixel width and the phase FOV?
    1. The pixel size is doubled and the FOV is unchanged.
    2. The pixel size is unchanged and the FOV is doubled.
    3. Both the pixel size and FOV are halved.
    4. The pixel size is unchanged and the FOV is halved.

    The scenario in these last three questions is to create a rectangular field of view with a 1:2 frequency:phase ratio. Because Δky is doubled, the FOVy is halved. Furthermore, because both the FOVy and Ny have been reduced by one-half, pixel size in the phase-encode direction (FOVy/ Ny) and overall spatial resolution of the image are also unchanged. Link to Q&A discussion

  20. Concerning obtaining a rectangular field of view by reducing the number of phase encoding steps as described above, which of the following statements is false?
    1. This option is most beneficial when imaging the spine and extremities.
    2. Spatial resolution is unchanged.
    3. Signal to noise ratio is unchanged.
    4. Wrap-around artifacts in the phase-encode direction are more likely to occur.

    Option (c) is false. Because fewer phase-encoding steps and hence fewer echo signals are acquired, the signal-to-noise ratio (SNR) is proportionately lower compared to full FOV imaging. Link to Q&A discussion

  21. Provided no phase errors occur during data collection, what type of symmetry does k-space exhibit?
    1. x-axis symmetry
    2. y-axis symmetry
    3. z-axis symmetry
    4. conjugate symmetry

    Provided no phase errors occur during data collection, k-space possesses a peculiar mirrored property known as conjugate (or Hermitian) symmetry. Conjugate symmetry applies to pairs of points (like P and Q) that are located diagonally from each other across the origin of k-space. If the data at P is the complex number [a+bi], the data at Q is immediately known to be P's complex conjugate, [a−bi]. Link to Q&A discussion

  22. Which statement concerning phase-conjugate symmetry (partial Fourier) imaging is false?
    1. Over 50% savings in imaging time can be achieved by the use of this technique.
    2. More than half of the phase-encoding lines must be sampled to correct for phase errors.
    3. Spatial resolution is preserved.
    4. For half-Fourier imaging, signal-to-noise is reduced by about 30%.

    Phase-conjugate symmetry techniques use data from the top half of k-space to estimate data in the lower half of k-space. In theory, phase-conjugate symmetry allows one to acquire data using only half the normal number of phase-encoding steps, thus potentially reducing imaging time by up to one-half. However, to correct for phase errors, more than half of the phase-encoding lines must be sampled, so the actual time savings is less than 50%. Link to Q&A discussion

  23. Which statement concerning read-conjugate symmetry imaging is false?
    1. There is no direct time savings.
    2. Flow and motion artifacts along the frequency-encode direction are reduced.
    3. Only the first part of the echo is sampled.
    4. The technique allows for shorter TE values.

    The direction of read-conjugate symmetry is along the readout (frequency-encode) axis. Only the last part of the echo is sampled, so choice (c) is false. The other statements are true. Link to Q&A discussion

  24. Which of the following statements about zero filling is false?
    1. It may be used either for 2D or 3D imaging
    2. It is used to expand the imaging matrix in the frequency-encoded direction
    3. It improves the apparent spatial resolution of the image.
    4. After the first doubling of matrix size with zeros, no added benefit is realized.

    Zero-filling is routinely used to expand the image matrix size in the phase-encoded direction, which may be either within-plane (2D) or through-plane/slab select (3D). Zero filling acts as method to interpolate the signals from neighboring voxels, giving the image a smoother and less "pixel-ly" appearance. Link to Q&A discussion

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Rapid Imaging (FSE/EPI)

  1. Fast/Turbo Spin-Echo Imaging methods are commercial applications of which original technique?
    1. RARE
    2. MEDIUM
    3. ROAST
    4. SMASH

    Fast spin echo (FSE) imaging, also known as Turbo spin echo (TSE) imaging, are commercial implementations of the RARE (Rapid Acquisition with Relaxation Enhancement) technique originally described by Hennig et al in 1986. Link to Q&A discussion

  2. Which notation best describes the FSE/TSE technique?
    1. 90º−180º−echo−90º−180º−echo−90º−180º−echo−…..
    2. 90º−180º−echo−180º−echo−180º−echo−180º−echo−…..
    3. 90º−180º−echo−echo−echo−echo−…..
    4. 90º−180º−echo−90º−echo−180º−echo−90º−echo−180º−echo−…..

    The FSE/TSE pulse sequence uses a series of 180º-refocusing pulses after a single 90º-pulse to generate a train of echoes. Link to Q&A discussion

  3. Advantages of FSE/TSE over conventional spin-echo imaging include all except
    1. Decreased tissue energy deposition
    2. Reduced susceptibility artifacts
    3. Reduced imaging time
    4. Increased resolution or signal-to-noise in same imaging time

    FSE/TSE offers significant reduction of imaging time by scanning multiple lines of k-space nearly simultaneously. This savings may be used to lengthen TR, allowing more time for recovery of longitudinal magnetization and hence improved signal-to-noise. A higher number of phase-encoding steps may be used, allowing improvement in spatial resolution. Susceptibility-induced signal losses are reduced. However, the multiple 180º deposit considerable energy in tissue, leading to high specific absorption rates (SARs). Link to Q&A discussion

  4. What is the main conceptual structural difference between FSE/TSE and Multi-echo Conventional Spin Echo (multi-CSE) sequences?
    1. The number of 90º-pulses within a TR interval
    2. The pattern of 90º- and 180º-RF pulses within a TR interval
    3. The pattern of phase-encoding gradients within a TR interval
    4. The pattern of frequency-encoding gradients within a TR interval

    The FSE/TSE technique changes the phase-encoding gradient for each of these echoes (a multi-CSE sequence collects all echoes in a train with the same phase encoding). Link to Q&A discussion

  5. What is the effective echo time (TEeff)?
    1. The time between the 90º-pulse and first echo
    2. The time between the 90º-pulse and last echo
    3. The time between successive echoes in the train
    4. The time between the 90º-pulse and echo obtained at the center of k-space

    The effective echo time (TEeff) is the time at which the central line of k-space (ky = 0) is being filled. Link to Q&A discussion

  6. What is the echo spacing (ESP)?
    1. The time between the 90º-pulse and first 180º-pulse
    2. The time between the 90º-pulse and last echo
    3. The time between successive echoes in the train
    4. The time between the 90º-pulse and echo obtained at the center of k-space

    The echo spacing is the time between any two successive echoes. Link to Q&A discussion

  7. What is the echo train length (ETL) or turbo factor (TF)?
    1. The number of echoes in a TR interval
    2. The time between echoes in a TR interval
    3. The number of RF pulses in a TR interval
    4. The number of RF pulses required to reach the center of k-space

    ETL or TF is simply the number of echoes in a FSE train between successive 90º-pulses. Imaging time is reduced proportionally by this factor. Link to Q&A discussion

  8. If a conventional spin-echo sequence with a certain TR/TE/spatial resolution takes 8 minutes to perform, an otherwise identical FSE sequence with an echo train length (turbo factor) of 8 would take how long?
    1. 1 minute
    2. 2 minutes
    3. 4 minutes
    4. 8 minutes

    With an ETL/TF = 8, 8 lines of k-space are transversed for each TR interval, reducing imaging time by a factor of 1/8. Link to Q&A discussion

  9. What image changes would not occur with increasing echo train length (ETL) or turbo factor (TF)?
    1. Greater T2-weighting
    2. Increased signal-to-noise
    3. Spatial blurring
    4. Higher signal from fat

    Longer ETL's are associated with a decrease in overall signal-to-noise ratio (SNR) and contrast-to-noise ratio (CNR) because the later echoes in the train are smaller in amplitude. Link to Q&A discussion

  10. What image changes would not be seen with increasing echo spacing (ESP)?
    1. Increased motion artifacts
    2. Increased susceptibility artifacts
    3. Increased edge-related artifacts
    4. Increased signal-to-noise

    Increasing echo spacing (ESP) permits the use of longer TE's but adversely impacts SNR and CNR. Motion, susceptibility, and edge-related artifacts increase. In general, increased ESP has predominantly deleterious consequences on image quality; the shortest permitted ESP should therefore be chosen in most applications Link to Q&A discussion

  11. What is the reason fat looks brighter on FSE/TSE than on Conventional SE images?
    1. FSE/TSE have more intrinsic T1-weighting
    2. Multiple FSE/TSE 180º-pulses disrupt J-coupling in fat molecules
    3. Rapid phase-encode gradient switching inf FSE/TSE disrupts diffusion of fatty acids
    4. It’s just a visual illusion; the absolute fat signal is unchanged.

    Although several factors may contribute to the FSE "bright fat" phenomenon, the dominant mechanism is thought to be T2 prolongation secondary to disruption of J-coupling interactions that take normally place between adjacent fat protons. Link to Q&A discussion

  12. Comparing FSE/TSE and CSE, which statement is false?
    1. FSE/TSE shows increased sensitivity to susceptibility changes.
    2. FSE/TSE demonstrates increased magnetization transfer effects.
    3. FSE/TSE demonstrates spatial blurring
    4. FSE/TSE demonstrates pseudo-edge enhancement

    FSE/TSE sequences are less sensitive to susceptibility changes, as the 180°-refocusing pulses occurring at very short intervals allow relatively little time for susceptibility-induced dephasing. Link to Q&A discussion

  13. Which pulse would be considered a driven-equilibrium pulse if played at the end of an MR sequence?
    1. +90º
    2. +180º
    3. −90º
    4. −180º

    Driven equilibrium (DE), also known as fast recovery, is technique in which a −90º "flip-back" pulse is used to help restore longitudinal magnetization at the end of an MR sequence. Link to Q&A discussion

  14. Comparing the tissue energy deposition of RF-pulses, which is true?
    1. Both 90º- and 180º-pulses deposit approximately the same energy
    2. A 180º-pulse deposits twice the energy as a 90º-pulse.
    3. A 180º-pulse deposits half the energy as a 90º-pulse.
    4. A 180º-pulse deposits four times the energy as a 90º-pulse.

    Energy deposition from an RF-pulse is proportional to the square of the flip angle (α²). Thus, a 180°-pulse deposits 4x the energy of a 90°-pulse. Link to Q&A discussion

  15. Which pulse sequence feature is not commonly implemented as described in 3D-FSE sequences like CUBE/SPACE/VISTA?
    1. Very short TE’s (2-4 ms)
    2. Medium length echo trains (Turbo factors 8-16)
    3. Reduced flip angles
    4. Partial Fourier imaging with zero-interpolation filling.

    3D-FSE sequences use extremely long echo trains, with perhaps 100-200 echoes. Other features include non-selective RF pulses, parallel imaging, and optimized k-space trajectories. Link to Q&A discussion

  16. Why is a zig-zag k-space trajectory commonly used for echo planar imaging instead of a consistent Cartesian (left to right) trajectory as in spin-echo imaging?
    1. It offers better spatial resolution.
    2. It has fewer artifacts.
    3. It is faster to perform.
    4. It is done in the phase-encode direction rather than frequency-encode direction.

    The zig-zag trajectory is popular because it is easy to program and faster, as it doesn’t require rewinding the readout gradient back to −kmax after each echo. The spatial resolution is the same, and the zig-zag is prone to more artifacts, like Nyquist N/2 ghosts. Link to Q&A discussion

  17. All of the following statements about single-shot FSE (HASTE) are true except
    1. Gradient echo signal generation is used for speed.
    2. The total number of echoes may exceed 200.
    3. Phase-conjugate symmetry is exploited to limit phase-encode steps.
    4. Common applications include MR cholangiography and MR myelography.

    Implicit in its name SS-FSE/HASTE is a fast-spin echo technique, meaning that spin-echoes, not gradient echoes are recorded. Link to Q&A discussion

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Parallel Imaging

  1. Which of the following statements about parallel imaging is false?
    1. It can be used with virtually all pulse sequences.
    2. It requires multiple receiver coils.
    3. It can be performed in any direction.
    4. Its primary purpose is to reduce the number of required phase-encoding steps.

    The allowed direction(s) in which PI can be performed depends on the arrangement of coils in the array. Specifically, the coils must should have different sensitivity profiles along the direction chosen for PI. PI cannot not be used with just a single surface, head, knee, or large body coil having only one pair of quadrature output channels. Link to Q&A discussion

  2. Which of the following is not a disadvantage of parallel imaging?
    1. Increased susceptibility artifacts.
    2. Increased aliasing artifacts.
    3. Coil calibration artifacts.
    4. Increased noise in a non-uniform fashion with reduced SNR.

    Phase-related distortions in the MR signal due to susceptibility are reduced (not increased) by the PI acquisition and reconstruction process. This is especially advantageous in echo-planar sequences. Link to Q&A discussion

  3. Comparing image domain PI techniques (like SENSE/ASSET) and k-space PI techniques like (GRAPPA/ARC), which statement is false?
    1. For high R(acceleration)-values, image domain PI techniques provide slightly higher SNR.
    2. Image domain PI performs better in heterogeneous body regions (like the chest)
    3. K-space PI techniques display less aliasing and are more useful when small FOVs are needed.
    4. K-space PI techniques are more effective at minimizing susceptibility distortions.

    Image domain PI techniques like SENSE/ASSET perform poorly in heterogenous body regions. This is because accurate coil sensitivity maps may difficult to obtain. The lungs are a prime example. Link to Q&A discussion

  4. What is the proper order of steps in acquiring an image-domain PI image (like SENSE/ASSET)? Abbreviations: A = Acquire partial k-space data; G = Generate coil sensitivity maps; R = Reconstruct individual coil images; U = Unfold/combine partial FOV images.
    1. A — G — R — U
    2. G — A — U — R
    3. G — A — R — U
    4. A — G — U — R

    The proper order is Generate/Acquire/Reconstruct/Unfold (c). Link to Q&A discussion

  5. What is the proper order of steps in acquiring an k-space PI image (like GRAPPA/ARC)? Abbreviations: A = Acquire partial k-space data; C = Combine data into magnitude image; E = Estimate missing k-space lines; G = generate individual coil images
    1. A — C — E — G
    2. A — E — G — C
    3. E — A — G — C
    4. E — A — C — G

    The proper order is Acquire/Estimate/Generate/Combine (b). Link to Q&A discussion

  6. Which of the following statements about CAIPIRINHA sequence is false?
    1. It can be classified as an image-domain PI technique.
    2. Acceleration factors (R) of 4 are typical.
    3. It is most commonly used for 3D dynamic liver imaging.
    4. It uses shifting k-space sampling patterns to reduce pixel aliasing and overlap.

    CAIPIRINHA is clearly a k-space (not image domain) PI technique, as it depends on k-space data sampling and manipulation prior to reconstruction. Link to Q&A discussion

  7. What happens to the imaging time in a PI study as the acceleration factor (R) is increased from 2 to 8?
    1. It decreases by a factor of 4
    2. It decreases by a factor of 2
    3. It decreases by a factor of √2
    4. It decreases by a factor of 1/√2

    Imaging time is inversely related to R, so if R quadruples (from 2 to 8), the imaging time is reduced by a factor of 4 Link to Q&A discussion

  8. What happens to the signal-to-noise ratio (SNR) in a PI study as the acceleration factor (R) is increased from 2 to 8?
    1. It decreases by a factor of 4
    2. It decreases by a factor of 2
    3. It decreases by a factor of √2
    4. It decreases by a factor of 1/√2

    SNR is inversely related to √R, so if R quadruples (from 2 to 8), the SNR is reduced by a factor of √4 = 2. Link to Q&A discussion

  9. Which statement about the g-factor is false?
    1. It depends on the number and location of coils.
    2. It depends on the direction of phase-encoding
    3. It varies by location across the image.
    4. Typical values range from about 10 to 100.

    The g-factor depends on 1) number and location of surface coils, 2) coil loading, 3) plane of imaging, 4) direction of phase-encoding within the scan plane, and 5) voxel location within the imaged region. With typical coil designs in common use, g-factors ranging from 1.0 to 2.0 across the imaging volume are commonly measured. Link to Q&A discussion

  10. The g-factor in the middle of a homogenous phantom imaged with PI is 2.0 while that near its periphery is 1.0. How do the SNR of the center and periphery compare?
    1. SNR (center) is one-half as large as SNR (periphery).
    2. SNR (center) is equal to SNR (periphery).
    3. SNR (center) is twice as large as SNR (periphery).
    4. SNR (center) is four times as large as SNR (periphery).

    SNR is inversely related to the g-factor, so if the g-factor doubles, the SNR is only ½ as large. Link to Q&A discussion

  11. Artifacts specific to parallel imaging include all of the following except
    1. Coil sensitivity artifacts
    2. Fold-over (SENSE) ghosts
    3. Chemical shift
    4. Spatially dependent noise

    Chemical shift artifacts are unaffected by the PI reconstruction process and are thus not unique. Link to Q&A discussion

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