Limitations to X-ray Image Quality

Question 1

What are the three main limitations to x-ray image quality?

Answer 1

• Unsharpness.
• Scatter.
• Noise.

Question 2

What does low unsharpness mean? What factors affect the unsharpness in an image?

Answer 2

Low unsharpness corresponds to the ability to be able to distinguish smaller objects. Objects should also appear sharper.

The following factors affect unsharpness:
- Focal spot size.
- Imaging geometry (e.g. focus-object and object-detector distance).
- Imaging receptor design (e.g. pixel size).
- Patient movement.

Question 3

How does focal spot size affect unsharpness?

Answer 3

• For a larger focal spot, the amount of penumbra increases. This increases the unsharpness in the image.
• The focal spot size also varies across the image. It will be smaller towards the anode side of the image due to the projection of the actual focal spot from the angled anode target resulting in less unsharpness/blur at this side.

Question 4

How does source-image distance (SID) distance affect unsharpness?

Answer 4

A shorter SID will result in a more magnified image and increased penumbra and unsharpness.

Question 5

How does focus-object distance affect unsharpness?

Answer 5

A smaller focus-object distance will produce a magnified image with increased penumbra and unsharpness.

Question 6

How are the unsharpness effects associated with reduced focus-object distance avoided in magnification mammography acquisitions? What are the issues with this?

Answer 6

• The fine focus is selected on the system.
• Tube loading effects mean a longer exposure time is required increasing the likelihood of movement blur.

Question 7

How does compression in mammography affect unsharpness?

Answer 7

Compression causes the focus-breast distance to increase and the breast moves closer to the detector. This will decrease unsharpness, thus improving resolution.

Question 8

State the equation for blurring b for an object at a distance x from a focal spot of size f and a distance y from a detector. How is this corrected to determine the geometric unsharpness (U_g) in object plane? What does this equation suggest is required to minimise U_g.

Answer 8

By considering geometry of the similar triangles created by this setup, we know that:
y/x = b/f
=> b = yf/x
M = SID/SOD = (x + y)/x
=> b = f(M-1)

U_g = f(M-1)/M
=> U_g = yf/(x + y)
=> U_g is minimised with a small focal spot (f), a large focus-object distance (x) and a small object-detector distance (y).

Question 9

What factors determine receptor unsharpness (U_r) for DR and CR detectors? What is the equation?

Answer 9

• Unsharpness in DR detectors will be limited pixel size.
• Unsharpness in CR detectors will be limited by laser spot size.
• This can be calculated as U_r = F/M where F is the intrinsic receptor unsharpness (typically the pixel size for a DR detector).

Question 10

How are geometric and receptor unsharpness combined?

Answer 10

Defining the unsharpness as the FWHM of each contribution, the total unsharpness can be determined by adding the geometric and receptor unsharpness in quadrature.

Question 11

How can motion unsharpness be minimised?

Answer 11

• Use an exposure time that is as short as possible. There are limits on this, however, due to the size of the focal spot and anode heating effects.
• Increasing focus-detector distance and reducing object-detector distance will reduce motion effects.
• Compression can immobilise the breast in mammography.

Question 12

How does scatter affect image quality?

Answer 12

Scattered photons contain no positional information regarding the distribution of x-ray attenuation coefficients in the material. This degrades image quality by reducing contrast.

Question 13

What are the equations for contrast in the absence of scatter and with scatter included?

Answer 13

In the absence of scatter, contrast is defined as:
C = 1 - e^(x(mu_1 - mu_2))
where x is the thickness of the object and mu_1 and mu_2 are the attenuation coefficients of the object and the background, respectively.
With scatter included, this becomes:
C = (1 - e^(x(mu_1 - mu_2))/(1+R)
where R is the scatter-to-primary ratio (typically around 1 or above) and 1/(1+R) is the contrast degradation factor.

For a signal trace, this can be calculated as the difference in signal between an object and background as a proportion of the background signal.

Question 14

How does an anti-scatter grid reduce scatter?

Answer 14

The grid consists of strips of high-attenuation material (e.g. Pb) separated by low attenuation material. Primary photons will pass through the low attenuation regions of the grid (assuming they are parallel). Scattered photons will be incident at a different angle and, therefore, will be preferentially attenuated by the high-attenuation regions. It should be noted that some primary photons will be attenuated and some scattered photons will still pass through the grid unattenuated.

Question 15

Explain and give equations for the grid ratio and lines per mm for an anti-scatter grid. How to these factors affect scatter reduction?

Answer 15

• Grid ratio r = h/D where h is the length of the Pb bars and D is the width of the low-attenuation material. An increased grid ratio will make the likelihood of a scattered photon reaching the detector low as the acceptance angle will be small.
• Line-pairs per mm N = 1/(D+d) where d is the Pb bar thickness. Increased line-pairs will reduce the amount of scattered photons reaching the detector.

Question 16

How does the divergence of an x-ray beam cause issues for a parallel anti-scatter grid? What is a solution to this?

Answer 16

• An increased amount of primary photons will be attenuated by the grid at the periphery of the image due to beam divergence.
• A focussed grid can be used get around this issue. The Pb strips are angled to match the divergence of the beam. This means they are only useable over a small range of focus-to-grid distances.

Question 17

What issues are associated with the use of an anti-scatter grid?

Answer 17

• Total air kerma at the detector is reduced (primary transmission ~ 0.6). This means an increase in patient dose is required to maintain SNR, despite the effects of reduced scatter. This is known as the bucky factor.
• Grid artefacts associated with the grid attenuating primary photons. Normally, a moving grid design is employed to blur grid artefacts.

Question 18

How does an air gap minimise scatter in magnification mammography? What are the issues associated with this?

Answer 18

• As the object is some distance from the detector (i.e. there is an air gap), scattered photons are more likely to be attenuated before reaching the detector or they are more likely to scatter outside of the detector area and not be detected.
• Geometric unsharpness will increase due to the increased magnification unless a smaller focal spot is used.

Question 19

List and explain some techniques other than an anti-scatter grid or air gap that can be used to reduce scatter.

Answer 19

• Collimation will reduce the volume of exposed tissue and, therefore, the amount of scatter.
• Using a lower kV will increase the probability of photoelectric interaction (where this is approximately proportional to Z^3/E^3). This will mean photons are absorbed rather than scattered. However, this will correspond to an increase in patient dose.
• Compression in mammography will reduce the volume of tissue exposed and, therefore, reduce the amount of scatter.

Question 20

What are the typical noise components apparent in an x-ray image?

Answer 20

• Quantum noise.
• Fixed pattern noise.
• Electronic noise.
• Anatomical noise.
• Film noise (granularity) in screen-film.

Question 21

What is quantum noise? Provide an equation.

Answer 21

Quantum noise arises due to the random nature of photon interaction processes. They are described by Poisson statistics. The standard deviation gives a measure of quantum noise:
rho = sqrt(N) where N is the number of photons detected.

Question 22

Provide an equation for signal-to-noise ratio (SNR) and explain. What is the effect of increasing the number of photons at the detector?

Answer 22

• SNR = sqrt(N) as signal is proportional to N and noise is proportional to sqrt(N) => N/sqrt(N) = sqrt(N).
• SNR will increase. Both signal and noise will increase but as the signal will increase more rapidly, the appearance of noise in the image will be reduced. This will mean an increase in contrast.

Question 23

Why should x-ray systems be quantum noise limited?

Answer 23

Quantum noise limited systems can be optimised by varying dose to alter image quality to a level that is adequate for diagnostic purposes.

Question 24

What causes fixed pattern noise? How does it vary with dose? How can it be reduced in DR systems?

Answer 24

• Fixed pattern noise results from variations in pixel sensitivity, filter thickness, tabletop attenuation etc.
• Its magnitude is dose dependent where it can be thought of as a signal (i.e. increased signal corresponds to increased fixed pattern noise).
• It can be reduced in DR systems with flat field corrections.

Question 25

What causes electronic noise? How does it vary with dose?

Answer 25

• Arises from detector electronics, thermal effects, contribution from other sources in the room.
• It is generally assumed to be constant with dose and will only be significant at low detector doses (assuming a properly functioning system).

Question 26

How is the total noise determined from the quantum, fixed pattern and electronic noise? Why is this useful?

Answer 26

• Assuming each of these noise sources are independent, the total noise can be determined by adding each contribution in quadrature.
• As the relationship between air kerma and each noise source is known (i.e. quantum noise = a(AK)^0.5, fixed pattern noise = b(AK) and electronic noise = c), the total measured noise in the image can be used to determine the contribution from each type of noise. Detector air kerma is determined by linearising the image and taking the standard deviation in an ROI.

Question 27

What causes anatomical noise? How are these effects reduced in mammography?

Answer 27

• Anatomical noise is any extraneous anatomy that masks clinical detail.
• Spot compression can move overlying tissue. Tomosynthesis can reduce masking effects associated with overlying tissues.

Question 28

Explain the concept of spatial frequency in a radiological image. What are Fourier series useful for in this sense?

Answer 28

• A radiological image can be broken down into a series of sine waves of different spatial frequencies and amplitude.
• High frequencies are associated with small objects and areas of change in signal (e.g. edges).
• Low frequencies are associated with large objects and more gradual signal changes.
• Fourier series can be used to breakdown an image into its component spatial frequencies and their amplitudes.

Question 29

What is the modulation transfer function (MTF)? Give an equation. Typically, how effective are imaging systems at transmitting low and high spatial frequencies?

Answer 29

• The MTF explains how the actual spatial frequencies of an object are transmitted by the imaging system. It can be calculated as:
  MTF = Image modulation/Input object modulation
• Low spatial frequencies are transmitted better than high spatial frequencies.

Question 30

What is the Nyquist frequency and what limits this in a digital detector? Provide an equation. In terms of pixels, what is the highest frequency that could be produced in a digital detector?

Answer 30

• The Nyquist frequency is the highest frequency that a system can accurately re-produce.
• This is limited by pixel size in digital detectors.
• It is determined as 1/(2 x Sampling distance).
• The maximum frequency that could be produced by a digital detector is one pixel on/one pixel off.

Question 31

Is the MTF better for DR or CR? Is the Nyquist frequency better for DR or CR? What is the outcome of this?

Answer 31

• The MTF is higher for DR when compared to standard CR - there is a better transmission of frequencies.
• The Nyquist frequency is higher for standard CR when compared to DR - there higher theoretical limiting spatial resolution.
• Despite the increased Nyquist frequency, the level of signal produced at these frequencies for standard CR is limited.

Question 32

How can the modulation transfer function (MTF) be derived mathematically and practically? What would it be in an ideal system?

Answer 32

• By taking the Fourier transform of the point or line spread function.
• Practically, a line spread function can be determined by differentiating the edge spread function from an image of a precisely machined edge.
• In an ideal system, the MTF would be 1 for all spatial frequencies. In practice, it drops with increasing spatial frequency.

Question 33

What is limiting spatial resolution? How does unsharpness affect limiting spatial resolution?

Answer 33

• Limiting spatial resolution is the ability of a system to distinguish fine details.
• Increased geometric unsharpness will degrade limiting spatial resolution.

Question 34

What does a limiting spatial resolution bar pattern test determine? Why is this test beneficial? What are the associated issues?

Answer 34

• Limiting spatial resolution bar pattern tests determine what the highest discernible spatial frequency is.
• This test is beneficial as it gives an quick and simple practical assessment of a system’s spatial resolution performance.
• However, this test provides no information on how the system performs at other spatial frequencies. Results also vary between operator.

Question 35

What is the signal-to-noise ratio (SNR) equation in terms of pixel values? What does a high SNR mean?

Answer 35

• SNR = pv/sigma_pv where pv is the linearised pixel value and sigma_pv is the standard deviation of linearised pixel values in a region of interest.
• A high SNR is a good thing but a sudden increase could indicate a change in the dose or dose rate at the detector or blurring.

Question 36

What is contrast-to-noise ratio (CNR), how is it measured and what is its equation? What additional information does CNR consider that SNR does not?

Answer 36

• CNR is also known as the signal difference-to-noise ratio (SdNR).
• It can be measured using a test object which contains a high and low contrast region.
• CNR can then be calculated as:
  (M1 - M2)/sqrt((sigma_1^2 + sigma_2^2)/2) i.e. the difference in pixel values between the objects (M1 and M2) divided by the geometric mean of the standard deviations (sigma_1 and sigma_2) of the same two regions of interest.
• The signal difference (M1 - M2) will depend on beam quality which is not considered in SNR.

Question 37

What are the issues associated with using the standard deviation to characterise noise? What technique can be used instead?

Answer 37

• It gives an indication of noise but gives no information on the type (e.g. spatial frequency) of the noise present.
• Noise power spectrum (NPS) can be used instead.

Question 38

What is the noise power spectrum (NPS)? How is it determined? What do we expect from a typical NPS?

Answer 38

• NPS gives information on the amount of noise present at each spatial frequency.
• To determine NPS the Fourier transform of a uniform raw linearised image is taken.
• A typical NPS will have high levels of noise at low spatial frequency which flattens off at increased frequencies. There is a cut off at the Nyquist frequency. A spike in an NPS may indicate anti-scatter grid artefacts.

Question 39

What is a threshold contrast detail diameter test?

Answer 39

A thin, flat test object containing discs of different thicknesses and diameters is imaged using specific exposure parameters (this ensures specific range of contrast levels). The discernibility of these contrast details is then graded to give a practical assessment of image quality - it is essentially a measure of SNR.

Question 40

Describe the shape of a contrast detail curve.

Answer 40

• A contrast detail curve is the percentage threshold contrast vs detail diameter (typically plotted on log-log axes).
• Below the curve will correspond to objects that are visible and above the curve objects that are not visible.
• This means that a contrast detail curve closer to the axes will correspond to better performance.

Question 41

Relate the detection of small and large objects in threshold contrast detail tests to the modulation transfer function (MTF).

Answer 41

• For small contrast details (i.e. increased spatial frequency), MTF and, therefore, signal will be lower meaning they will be more difficult to detect.
• For large contrast details (i.e. reduced spatial frequency), MTF and, therefore, signal will be increased meaning detection is easier.

Question 42

What are the advantages and disadvantages of threshold contrast detail tests?

Answer 42

Advantages:
- Practical measure of SNR.
- Semi-quantitative.
Disadvantages:
- Variation between observers (although computer automated reading now available).
- Tests are not clinically realistic in that a flat, low dynamic range test object is used.