MEDPHYS 580

Question 1

Draw an X-ray tube.

Answer 1

Include: anode, cathode, focusing cup, vacuum envelope, tube housing, rotor, bearings, stator, transformer oil, output port

Question 2

Explain the basic principles of operation of an x-ray tube. What component of the x-ray tube is the source of electrons? How are the electrons accelerated to the anode?

Answer 2

Electrons are ‘boiled’ off the cathode through thermionic emission. After electrons are emitted, they are electrostatically repelled from each other and accelerated towards the anode.

A negatively-biased focusing cup (-100 V in respect to the cathode) is used to focus this beam. Shape of the cup controls the width of the beam.

Electrons hit the rotating anode and through bremsstrahlung release x-ray photons.

Question 3

What percent (%) of electron energy absorbed in the anode is converted to x-rays?

Answer 3

Only 1% of the electron energy absorbed in the anode is converted to x-rays.

The remaining ~ 99% is converted to heat in the anode via soft and hard collisions.

Question 4

What modulates the number of electrons available in x-ray tube operation?

Answer 4

Filament current.

Question 5

What are grid-controlled x-ray tubes used for ? What is specific to them?

Give two examples.

Answer 5

Grid-controlled x-ray tubes employ large negative biasing (e.g. -3000 V) to completely turn off the electron beam. This design is used when a very fast turn-on / turn-off of x-ray output is desired.

Examples: pulsed fluoroscopy and cardiac angiography.

Question 6

What are the two geometries of anodes?

Answer 6

Two geometries:

1. Reflective geometry (x-rays emerge on the same side as the incoming electrons)
2. Transmission geometry (x-rays emerge on the opposite side as the incoming electrons)

Question 7

List properties of a good anode material.

What is the best material to use?

Answer 7

1. High Z
2. High heat capacity
3. High melting point

Best material to use is Tungsten (W) (Z=74) covered with a small amount of Rhenium (Re) to improve ductility.

Question 8

What do kV, kVp, mA, ms, and mAs describe?

Answer 8

kV = accelerating voltage of the tube

kVp= peak kV

mA= tube current flowing from cathode to anode

ms= time tube is active for (often pulsed)

mAs= product of mA and ms

Question 9

If the x-ray tube is operated with a particular accelerating potential, in kV, then what is the energy of each electron striking the anode, in units of keV?

Answer 9

Electrons striking the target will have kinetic energy in keV equal to the kV.

10 kV accelerating potential= 10keV electron energy

Question 10

In the spectrum of photon energies produced in an x-ray tube what is the maximum photon energy? Where does the peak occur?

Answer 10

Spectrum of photons will have the maximum energy (in keV) equal to the kV, but peak energy about ⅓ of that amount.

Question 11

When energetic electrons strike the target material in the anode, what are the types of electron-matter interactions that can occur? Which interaction leads to production of Bremsstrahlung radiation? Which interaction leads to characteristic x-rays?

Answer 11

Types of interactions: Bremsstrahlung (Coulomb effects) and Compton scattering

Deceleration of electrons due to Coulomb effects leads to Bremsstrahlung radiation (spectrum with highest photon energy equal to kV and peaks at ⅓ of the maximum).

Compton scattering produces characteristic photons via fluorescence yield form the knocked-out electrons.

Question 12

What is the K-shell binding energy for tungsten? What are the four K-lines of tungsten listed in the notes?

Answer 12

~69.5, ~67, ~59, and ~57 keV

Question 13

How characteristic x-rays are produced when electrons strike the target? How do the characteristic x-ray energy relate to electron binding energies. What energy must the incident electron have for the K-shell to participate in characteristic x-ray production?

Answer 13

When a Compton scatter interaction occurs in the anode, it may knock loose an electron from one of the orbital shells if the incident electron has enough kinetic energy to overcome the binding energy of the orbital electron.

Once knocked free, the vacant space in the orbital may be filled by an electron from a higher shell. When this happens, the dropping electron emits a characteristic photon of ~69.5, ~67, ~59, or ~57 keV (for Tungsten).

To produce these, the kV must exceed their energy.

Question 14

What is the shape of the bremsstrahlung energy spectrum from a thick target, according to the simple Kramers model?

How does the total energy of the spectrum (area under the spectrum) depend on: mAs, kV, and Z?

Answer 14

Kramers model: Intensity vs photon energy; linear relationship with characteristic energy peaks. Curve intersects the x-axis at tube voltage (kV) in keV.

Higher kV shifts the spectrum further right, increasing total energy by a squared factor (double kV makes for quadruple area= quadruple energy).

kV ∝ (Area)²= (total energy)²

Higher mAs will shift spectrum slope/height. Changes total energy linearly (double mAs makes double energy).

mAs ∝ Area= total energy

Question 15

Draw how the shape of the Kramers spectrum changes when the mAs is doubled.

Answer 15

Maximum energy stays the same; curve cuts y-axis higher up; area under the curve doubles

Question 16

Draw how the shape of the Kramers spectrum changes when the kV is doubled.

Answer 16

Maximum energy doubles (x-axis intercept doubles); y-intercept doubles; area under the curve quadruples

Question 17

Why do we need a collimator?

What is the purpose of the oil in the tube?

Answer 17

The objective of the collimator is to produce a beam that is proper “quality” , desired dimensions and positioned correctly on the patient.

Tube housing is filled with oil for better heat dissipation. Radiolucent window allows x-rays to go out.

Question 18

What is beam hardening?

Answer 18

Beam hardening is the removal of lower energy photons from the x-ray beam.

Filtration hardens the beam, removing non penetrating dose-depositing low-energy photons.

Question 19

What is the electron source in the x-ray tube?

Answer 19

Filament is electron source- electrons are ‘boiled off’ through thermionic emission* due to the current run through it via a separate, low (~10) V circuit.

Higher filament current emits more electrons, but raises temperature- too hot breaks it.

Question 20

Line focus principle. How does an angled anode enable a large actual focal spot, but a small projected focal spot (as viewed from the detector).

Why is this geometry beneficial? How is the effective focal spot size related to the actual focal spot size, for a given anode angle?

Answer 20

Larger focal spot handles heating better and can be used for higher energy applications, but smaller focal spots provide higher-resolution images. This reflective geometry allows some of both.

Larger anode angles make for larger focal spots. Note this geometry only affects one axis.

Effective focal spot: F X F

Actual focal spot on the anode surface: F X L where L=F/sin∅ (∅ is the anode angle)

Question 21

How does field-of-view (FOV) depend on anode angle?

Answer 21

Maximum field of view along the cathode-anode axis (left and right) depends on anode angle. Larger angles enable larger FOVs.

Question 22

Heel effect. How does intensity vary along the cathode-anode axis? What is the cause of the heel effect? How does the severity of the heel effect depend on anode angle?

Answer 22

X-Rays are generated within the target, and must first travel out. Heel Effect refers to how some are absorbed before they can escape. Electrons ‘deeper’ in the anode (further from cathode) must travel further to escape. Shallower anode angles result in larger heel effect.

In this way, emitted spectrum is weaker further from the cathode.

Question 23

What is the purpose of the rotor and stator?

Answer 23

Anode rotates to help dissipate heat from the interactions (note: only 1% of electron energy goes to photons, the rest to heat!). Rotor and stator make a motor to rotate the anode. Stator uses magnetic induction to make rotor spin.

Question 24

How long must the cathode filament be to achieve a 1 x 1 mm effective focal spot size with a 10 degree anode angle?

Answer 24

see image

Question 25

Automatic exposure control: What is the basic principle of operation of a radiography phototimer system, and what parameters can the operator control? How does a fluoroscopic automatic exposure control (AEC) compare to the radiography phototimer?

Answer 25

AEC modulates kV, mA, and t (mA\*t =mAs) to maintain constant detector (not necessarily patient!) exposure under changing conditions (like moving table relative to source, imaging thicker body part, etc.) Operator chooses kV. Radiography Phototimer: integrates over detector charge to terminate tube once a set point of exposure is reached. Simply changes t (mAs), nothing else. Often comes with many preset conditions. Fluoroscopic Phototimer: adjusts all 3 parameters in real-time, since fluoro is in real time, and may involve moving patient couch or machine arm.

Question 26

Know the definitions of fluence and energy fluence, and their units. How do fluence and energy fluence vary with distance from a point-like source of x-rays? Be able to scale the fluence specified at a particular distance r_A to some new distance r_B. How do fluence and energy fluence scale with mAs?

Answer 26

Fluence is photons/mm², Energy Fluence is that times energy keV\*photons/mm² Both fall off as 1/R² due to geometric beam divergence. Both are linearly proportional to mAs.

Question 27

Understand exponential attenuation. Calculate the number of primary x-rays transmitted through a slab of material, given: thickness, linear attenuation coefficient, and the number of incident x-rays.

Answer 27

N= x-rays transmitted N₀= incident x-rays t= thickness mu= linear attenuation coefficient

Question 28

What are the three important x-ray-matter interactions, in the diagnostic energy range? Understand the concept of partial attenuation coefficients and partial mass attenuation coefficients.

Answer 28

In the diagnostic range, we see Rayleigh Scattering, Compton Scattering, and Photoelectric Effect. Each of these has its own attenuation coefficient for how each process individually attenuates, summing these gives total attenuation coefficient. To decouple attenuation from medium phase, divide by density to get mass attenuation coefficients.

Question 29

How does dose correlate to collision Kerma and exposure?

Answer 29

Collision Kerma is energy lost from electrons in J/kg, while Exposure is charge over mass C/kg.

Question 30

What is 1 Roentgen equal to, in SI units (Coulombs/kg) ?

Answer 30

1 Roentgen is 0.876 cGy in air kerma

Question 31

What different components of an x-ray tube which filter the x-ray beam

Answer 31

Beam is filtered initially by the anode (Heel Effect), then the oil, then the window. Filters can also be added.

Question 32

What does a “K-edge filter” do to an x-ray spectrum?

Answer 32

A k-edge causes a large spike in the spectrum due to an energy threshold now permitting more electrons to interact.

Question 33

Draw a typical filtered x-ray spectrum.

Answer 33

See picture

Question 34

What is the definition of half value layer?

Answer 34

A half-value layer (HVL) refers to the thickness of a specific material needed to reduce the exposure of a beam to one half of what it’d be without the material.

Question 35

Calculate the half value layer of a monoenergetic x-ray beam, given the attenuation coefficient at the appropriate energy.

Answer 35

see equation -ln(0.5)= 0.693

Question 36

Understand the concept of equivalent photon energy and the “1/3 to ½ kVp” rule of thumb.

Answer 36

Equivalent energy of a diagnostic x-ray beam is generally ⅓ to ½ the maximum of the spectrum (kV).

Question 37

Know a value for mR/mAs @ 100 cm, for some kV and total Al filtration. Apply this value to calculate an exposure value.

Answer 37

100kV and 10mR/mAs at 100cm is 2mm Al. (inherent 1mm Al, plus 1mm added).

Question 38

What is the dependence of beam intensity (or energy fluence) on kV at the following locations: i) at the anode (Kramers theory), ii) after inherent and added filtration but before the patient, and iii) after patient filtration?

Answer 38

see picture Intensity dependance on kV increases (powers i)V², ii) V^3-4, iii)V^4-6)

Question 39

Understand the basic stages found in all x-ray detectors (input x-rays, converter, secondary quanta, sensor, readout)

Answer 39

1. 1. **Input x-rays** that were made in the anode fall to the converter/absorber 2. **Converter** converts x-rays into many secondary quanta 1. Indirect conversion: uses phosphor to make optical photon quanta 2. Direct conversion: uses photoconductors to make electrical quanta 3. **Secondary quanta** fall onto an array of **sensors** or film that form a spatial distribution of the input quanta 4. Spatially distributed secondary quanta from the sensor result in a **readout** image

Question 40

What are the names of the two main types of converters, and what kind of secondary quanta do they produce?

Answer 40

1. **Indirect Conversion** - uses a phosphor layer to generate many optical-range photons from incident x-ray photons. Photosensitive elements react to this light. May cause some blurring in initial conversion. 2. **Direct Conversion** - uses a photoconductor to produce electron-hole pairs from incident x-ray photons. This charge is stored on element capacitors.

Question 41

Understand the basic operating principles of screen/film, image intensifier/TV, imaging plate, flat panel detector, and solid-state CT detectors.

Answer 41

**Screen/Film** - silver halide crystals react to photons, become dark spots on an otherwise transparent film. Produces a permanent non-digital image. Indirect Conversion. **Image Intensifier/TV** - bottle-shaped, takes input photons, converts to optical photons via CsI phosphor, then electrons, then accelerates down tube via bias- amplifies signal, then makes optical photons for viewing/recording via camera. Indirect Conversion. **Imaging Plate** - Like screen, but does not re-emit absorbed light immediately, read off later with laser. Indirect Conversion, makes a digital image from readout. **Flat Panel Detector** - May be direct or indirect, uses many small a-Si elements to store electric charge after secondary quanta for a digital output. **Solid-State CT** - Uses segmented scintillator to produce secondary light photons, read off with 2D array of photodiodes.

Question 42

Compare spreading of secondary quanta versus converter thickness, for i) unstructured phosphors, ii) structured phosphors, and iii) photoconductors.

Answer 42

In indirect conversion, unstructured phosphors allow secondary quanta to spread more than structured ones like CsI, which restricts it to the columnar shapes the phosphor is grown into. Photoconductors, in direct conversion, don’t allow any spreading of the signal.

Question 43

What aspect of detector performance is affected by **spreading of secondary quanta**?

Answer 43

Spreading of secondary quanta like this lessens spatial resolution, blurring signal.

Question 44

What aspect of detector performance is affected by **converter thickness?**

Answer 44

In indirect conversion, Greater thickness allows for more spreading of the secondary quanta (bad), but increases detector efficiency (good), since it allows for more assurance that photons will interact within the medium. (trade off resolution vs signal strength).

Question 45

List desirable properties of a converter material.

Answer 45

1. high Z 2. high density 3. high yield of secondary quanta per unit x-ray energy absorbed 4. ability to fabricate in large area 5. fast response / low afterglow

Question 46

Describe the difference between an energy-integrating detector, a photon-counting detector, and an energy-resolved photon-counting detector.

Answer 46

Energy-Integrating: As above, the signal is the sum total of all the x-ray energy deposited in the medium. Photon-Counting: Instead, counts the individual pulses of signal Energy-Resolved Photon-Counting: can weight each counted pulse by its energy, can ditch low-energy pulses to optimize SNR

Question 47

Definition of radiographic contrast?

Answer 47

Contrast is the fluence difference between one region and another.

Question 48

Be able to calculate the contrast of an object, given the appropriate material thicknesses, densities, and mass attenuation coefficients.

Answer 48

See picture.

Question 49

Know the approximate expression for contrast in the low-contrast limit, for both added and embedded objects.

Answer 49

See picture.

Question 50

Know the tissues/materials in the body that give rise to radiographic contrast. For each material, what physical properties are responsible for the contrast?

Answer 50

Mainly: Soft Tissue, Fat, Bone, Air, and Contrast Agents Mass Attenuation coefficient, density, and thickness determine how each affects contrast.

Question 51

Understand the distinction between radiographic contrast and detected contrast

Answer 51

Detected contrast, rather than relying on idealized fluences, depends on the actual measured signals between two points. C= (S₁-S₂)/S₁ Signal may be a voltage, current, or charge resultant from the x-rays. Most notably, this accounts for noise and off-focus radiation.

Question 52

Understand what radiographic contrast depends on and what it doesn't depend on.

Answer 52

Depends on: kV and thickness Doesn't depend on: mA, exposure time, or mAs Radiographic Contrast depends on kV because, at lower kV, the beam will be lower energy, meaning its photons are more readily attenuated. This increases the radiographic contrast. kV ⇡ C ⇡ Thickening the material causes beam hardening, which reduces radiographic contrast (since it makes the spectrum overall more high energy, and thus, more resistant to attenuation). Thickness ⇡ C⇣

Question 53

What is the relationship between the mean and variance of a Poisson-distributed random variable.

Answer 53

Mean and variance are the same in Poisson distribution = lambda

Question 54

What is the relationship between the noise level and mAs? if the mAs is increased by a factor of 3, by what factor does the noise standard deviation change? By what factor does the background signal-to-noise ratio change?

Answer 54

Fluence is linearly proportional to mAs, so N is too. Increasing mAs by 3x will increase noise by sqrt(3x), and increase N by 3x. This will increase overall SNR by sqrt(3x).

Question 55

What does the autocovariance function quantify?

Answer 55

SDNR combines contrast and SNR (sometimes SDNR is called CNR). See image.

Question 56

Use the Rose model to calculate the required fluence to detect an object with given area and contrast (or given quantities that allow you to determine contrast and area). What is a typical value for the parameter “k”? What does “k” represent?

Answer 56

Rose Model determines detectability threshold where N is photons incident, R is the whole image area, A is the area with differing contrast C (that we want to detect), and DQE is intrinsic detector efficiency. K is whatever SDNR threshold is required to detect the object. Example value of 5 or so.

Question 57

How can you interpret contrast-detail curve?

Answer 57

Objects on or above the curve are detectable in the system at hand. A lower curve is better, as it allows more things to be detected in the system.

Question 58

What is SPR?

Answer 58

SPR stands for scatter to primary ratio. It shows how many scattered photons there are in respect to the primary photons.

Question 59

What is SF?

Answer 59

SF stands for scatter fraction. It shows what fraction of the beam energy is from scatter, and what’s from primary photons

Question 60

How does the presence of scatter reduce contrast? Be able to write expressions for contrast with and without scatter.

Answer 60

Question 61

What is CDF?

Answer 61

CDF stands for Contrast degradation factor.

Question 62

How is grid ratio defined?

Answer 62

Grid ratio is defined as r= h/D where h is the grid height and D is interseptum width.

Question 63

What is grid cutoff?

Answer 63

Cutoff refers to places on the grid where septa and beam geometry do not allow any photons through. This can be adjusted by having the septa placed at some angle at the further edges of the grid.

Question 64

Given the mean x-ray fluence incident on an ideal counting aperture, and the area of the aperture, be able to calculate: mean counts, variance in the counts, standard deviation in the counts, counts signal-to-noise ratio, and the noise in normalized image data.

Answer 64

In a normalized image, a simple scalar (g) is multiplied onto mean counts and standard deviation. SNR is unchanged, but, when normalized, noise becomes proportional to 1/sqrt(N), the inverse of SNR.

Question 65

What is the definition of grid primary transmission?

Answer 65

Primary transmission= (Transmitted primary)/ (Incident primary) Secondary transmission= (Transmitted secondary)/ (Incident secondary)

Question 66

What is grid selectivity?

Answer 66

Question 67

What is Bucky factor?

Answer 67

Grids reduce all incoming radiation. In film detectors, some baseline amount of radiation is needed to form a proper image- this is corrected by increasing the mAs by this factor B.

Question 68

When a grid is used, how does the detected scatter fraction change?

Answer 68

Question 69

Does a grid reduce patient exposure requirements to achieve a given SDNR?

Answer 69

Grids *usually* lower exposure for a given SDNR, but there are some tradeoffs and it is not guaranteed.

Question 70

What is magnification?

Answer 70

m= SID/SOD magnification = source to image distance / source to object distance m= image size/ object size

Question 71

What is the convolution model of blurring, and the concept of a point spread function (PSF), and scaling relationships?

Answer 71

In image space, convolving a Point Spread Function (PSF) with points on the image to account for blurring that occurs due to physical limitations (focal spot not a perfect point, detector elements not perfect points, patient motion). These are further affected by any magnification in the setup.

Question 72

For a rectangular focal spot of width ‘fs’ at the source plane, what is the focal spot PSF width at the image plane? What is the focal spot PSF width referenced back to the object plane?

Answer 72

The focal spot PSF is a rect function with width equal to the focal spot, with some adjustment for magnification. At image plane: fs(m-1) At object plane: fs(m-1)/m

Question 73

For a model of linear object motion in a plane, what is the width of the PSF in the image plane and at the object plane? How does this PSF depend on object velocity and exposure time?

Answer 73

Rect function with width reliant on velocity and exposure time (it’s assumed that the motion is of constant velocity during scan) At image plane: mvt At object plane: vt

Question 74

For a detector aperture of width ‘a’ at the image plane, what is the width of the PSF at the image plane and at the object plane?

Answer 74

At image plane: a At object plane: a/m

Question 75

What are the optical transfer function (OTF) and modulation transfer function (MTF)?

Answer 75

To mathematically account for the PSF blurring, we convolve the lot of them together into one big function. But convolutions suck. So instead we Fourier Transform. This makes Sinc (since each PSF is a rect, and the FT of a rect is a Sinc) functions, called Optical Transfer Functions OTF. We can then simply multiply the ~3 functions together to get a system OTF. The absolute value of the OTF is the MTF, Modulation Transfer Function. This is because negative parts would have a 180 degree phase shift (bad).

Question 76

T/F Convolution in the spatial domain corresponds to multiplication in the frequency domain.

Answer 76

True

Question 77

What is the Fourier transform of the rect function? If the rect function has width ‘a’, what is the spatial frequency of the “first zero” of the sinc function? If the rect is normalized to unit area, what is the value of the sinc function at zero spatial frequency?

Answer 77

The Fourier Transform of a rect function is a Sinc function. If the rect has width a, the sinc has zeros at n/a, so larger width rectangles make the first zeros come quicker and faster. Physically, this is as if larger PSF make MTFs discard higher frequencies ‘sooner’ A unit area rect makes sinc(0) = 1 Sinc(0)=a\*h (where h is the height of rect)

Question 78

How do “narrow” functions in the spatial domain transform to the frequency domain?

Answer 78

It transforms to wide functions in the frequency domain. A narrow function in the spatial domain means the points don’t spread much. This corresponds to a wide frequency function, which means higher frequencies are preserved by the MTF. Remember: the first zero comes at 1/a where a is rect width - wider rects have zeros with lower values!

Question 79

Calculate the system MTF, given the MTFs corresponding to the individual blurring processes in an imaging system. Where is the first zero of the system MTF ?

Answer 79

System MTF is simply all of the individual MTFs multiplied together. Accordingly, the first zero of the system MTF is equal to the lowest zero of the individual MTFs.

Question 80

**Focal spot vs. detector blurring.** 1. How does a) focal spot, b) detector and c) motion **blurring** behave with respect to magnification? 2. Given a *focal spot size (f)* and *detector aperture size (a)*, determine the optimal magnification.

Answer 80

a) Focal spot MTF goes down with magnification b) Detector MTF goes up with magnification c) Motion MTF doesn't depend on magnification 2. Optimal magnification occurs where the first zero of the focal spot MTF (at the object plane) is equal to the first zero of the detector aperture MTF (at the object plane).

Question 81

What should the **sampling interval** be to ensure no loss of information?

Answer 81

Given maximum frequency u_Nyquist, the sampling spacing ∇x should follow:

Question 82

Given a particular sampling interval, what is the Nyquist frequency? What happens when the analog signal to be sampled contains spatial frequencies above the Nyquist frequency?

Answer 82

In general, u_Nyquist ≤ 1/ (2 ∇x) Failing to meet these criteria results in signal aliasing, making replications over our proper image.

Question 83

What is the regulatory limit on fluoroscopic skin entrance exposure rate, in R/min? In mGy/min air kerma?

Answer 83

Limit is 10 R/min or ~88mGy/min for typical fluoroscopic systems

Question 84

Under what imaging geometry is the regulatory limit on skin entrance exposure measured? Are there regulatory limits on skin entrance exposure rate for fluorography or DSA modes?

Answer 84

30cm from image screen. There are not regulatory limits on Cine or DSA modes.

Question 85

Are there regulatory limits on skin entrance exposure rate for fluorography or DSA modes?

Answer 85

There are not regulatory limits on Cine or DSA modes.

Question 86

What is the distinction between **stochastic** and **deterministic** radiation-induced effects? For the patient, what is the main deterministic effect of concern? What is the main stochastic effect of concern?

Answer 86

**Stochastic:** At **lower or high doses**, not necessarily proportional in effect to dose. **Always delayed acting**, can still be death. **Deterministic:** occurs at **high dose (\>1 Sv)**, worsening with higher doses. Effects may be **acute or delayed**, can be death.

Question 87

What is Kerma-Area-Product (KAP) ? Is KAP constant or changing for a particular beam? Why?

Answer 87

KAP is kerma (K) at some distance multiplied by beam area (A) at some distance. KAP = K_c × Area *Remember, you must scale dose and area to the same distance from the source before taking the product.* Is a constant value for a whole beam, since kerma falls off as 1/R^2 while area increases by the same amount.

Question 88

Does KAP depend on distance from the source? Why?

Answer 88

No. ## Footnote KAP does not depend on distance (since kerma falls off by the same rate as area increases), but does depend on beam area.

Question 89

How is KAP correlated to dose?

Answer 89

KAP is proportional to x-rays entering patient, which accordingly means higher KAP has higher dose and scatter (either by increasing beam area, or energy).

Question 90

Modern fluoroscopic systems monitor air kerma as well as KAP. Where is the air kerma defined?

Answer 90

Air Kerma for these systems is measured at a reference point **15cm closer to the source from isocenter**.

Question 91

What practices would minimize radiation dose to the patient and/or staff ?

Answer 91

Reduce patient dose: 1. Image thinner part of the patient 2. Use collimation 3. Keep patient close to the detector 4. Keep patient far from the source 5. Use less magnification 6. Use shallower arm angle 7. Use fluoro instead of acquisition mode 8. Keep settings as low as possible Reduce staff dose: 1. stay behind shielding 2. don't get too close

Question 92

What is DQE?

Answer 92

DQE stands for Detective Quantum Efficiency Also DQE= NEQ/ N

Question 93

What is NQE?

Answer 93

NQE stands for Noise-equivalent quanta Conceptualize as: “An image generated by an imperfect system looks the same as an image that was generated by a perfect system using fewer photons. The NEQ is the mean of that reduced number of photons.”

Question 94

What is QDE?

Answer 94

QDE is the quantum detector efficiency. It is the best case scenario for DQE, refers to how many of the incident photons actually interact within the detector.

Question 95

What are the signal and noise propagation equations in respective detector gain stages?

Answer 95

See image.

Question 96

How does stochastic gain degrade DQE?

Answer 96

Stochastic gain degrades DQE by adding in the variance of its own gain, as well as amplifying existing variances by the gain

Question 97

Derive an expression for the line integral of attenuation coefficient in terms of the number of x-rays transmitted along a ray and the number of x-rays entering the patient.

Answer 97

Question 98

What is the definition of the Hounsfield Unit (CT number).

Answer 98

Hounsfield units are derived from the attenuation coefficient relative to water at some point, normalized.

Question 99

Write an equation for a line parallel to the y_r-axis and intersecting the x_r-axis at a specific value.

Answer 99

Holding x_r constant, this makes a line (equation). This simulates the x-ray beams’ orientation

Question 100

If a measurement is made along a specific line in object space, what does that correspond to in sinogram space?

Answer 100

A line intergal through object space maps to a **point** in the sinogram space. ## Footnote The point in sinogram space will be located at (phi, x_r), as per the line image space.

Question 101

What do all of the lines through a fixed object point map to, in sinogram space?

Answer 101

All the lines through a fixes object point map to a **cosine curve** in the sinogram space. According to: x_r = r×cos(theta- phi) , this maps out a cosine curve with amplitude r, and initial ‘phase shift’ theta. If we were mapping to a y_r-phi system, we’d have a sine curve instead.

Question 102

How is sinogram space filled for 1st generation CT? For 3rd generation CT?

Answer 102

**1st generation CT:** does parallel acquisitions. This ‘fills’ sinogram space up from the bottom, in a raster pattern. Collects all x_r for some phi, increases phi and repeats. ## Footnote **3rd generation CT:** fills it out via curves through space- will have to continue for some additional angle based on fan angle. Fills up like a glass of water, with each detector element making a single vertical line

Question 103

Explain the meaning of Central Slice Theorem.

Answer 103

If you take a single line through image space and Fourier Transform it, it’s equal to a single line (at that same angle) taken through the Fourier Transform of the image.

Question 104

Calculate the required angular interval given the detector sampling pitch and the field-of-view size.

Answer 104

This can also be called ‘the number of views’ needed. For units of degrees:

Question 105

What are the definitions and relationships between CTDI100, CTDIw, and CTDIvol.

Answer 105

CTDI stands for CT Dose Index.

Question 106

What is the pitch of a helical scan?

Answer 106

Where N = number of slices collected, T = slice thickness, and I = table increment between scans.

Question 107

What is the CTDIvol for a helical scan based on CTDIw?

Answer 107

Question 108

What is DLP?

Answer 108

DLP stands for dose length product. Where CTDI_vol is units of mGy, and ScanLength is length (cm), NOT time

Question 109

What is the effective mAs for a scan?

Answer 109

Effective mAs scales each slice by how long it is exposed. Pitch \<1 (lingers/double doses) results in an increased mAs, while pitch \>1 (skips sections) will lower it.

Question 110

How does the noise level in the reconstructed image (standard deviation in HU) depend on the effective mAs?

Answer 110

Increasing effective mAs reduces noise by a factor of 1/root(mAs) Also based on slice thickness and, though unchangeable, DQE

Question 111

What artifacts does beam hardening produce?

Answer 111

Beam hardening artifacts arise from how the nonuniform attenuation of photons can shift a beam’s energy spectrum. They can manifest as **cupping** or **streaking artifacts**. **Cupping artefact:** the center of an image is darker, due to those beams having traveled through more material and thus being more hardened. Streaking aftereffect: dark stretches appear between two high-attenuation inserts (like bones).

Question 112

How does the linearization correction method work?

Answer 112

Linearization correction is performed by **determining the effects of beam hardening** for a material experimentally, i**n advance**, then re-correcting measured values based on the divergence seen experimentally.

Question 113

What is fluoroscopy?

Answer 113

Fluoroscopy is performed in real time during interventional procedures. Often not recorded and relatively low dose per frame (lower SNR).

Question 114

What is Fluorography?

Answer 114

Fluorography works on the same physics as fluoroscopy, but is recorded for diagnostic purposes, and thus comes with higher SNR, thus greater dose per frame. **Often called Cine or DSA, rather than fluorography.**

Question 115

What is angiography?

Answer 115

Angiography is specifically imaging of blood vessels via injected contrast agents. These are also recorded for diagnostic purposes, but can also be done live during procedures.

Question 116

What are typical frame rates for fluoroscopy? What is the difference between continuous and pulsed fluoroscopy?

Answer 116

Typically ~15 frames/second, or 15 pulses/second in pulse mode. In pulse mode, the beam itself is pulsed in time with frame acquisition for less exposure.

Question 117

Why is subtraction of a mask image helpful? What is a potential image artifact in digital subtraction angiography (DSA)?

Answer 117

This is helpful because it removes background anatomy, making the desired blood vessels easier to see. It can, however, cause misregistration artifacts (possibly from patient motion) which would throw off this cancelling effect.