UNIT 4: Identifying and Classifying Image Artifacts

Question 1

What is an artifact?

Answer 1

Any false visual feature on a medical image that simulates tissue or obscures tissue.
Artifacts can be defined as an unwanted signal. Artifacts can be vendor specific and can vary by vendor.

Question 2

In digital radiography (DR), three classifications of artifacts can be described:

Answer 2

-image receptor
-software
-object

Question 3

Image receptor artifacts are caused by

Answer 3

• Dust, dirt, scratches
  -Pixel malfunction
  -Ghost images

Question 4

Software Artifacts are caused by

Answer 4

• Histograms
  -Range/scaling
  -Image compression

Question 5

Object Artifacts are caused by

Answer 5

-Patient positioning
-Collimator/partition
-Backscatter

Question 6

Responsibilities of the radiographer

Answer 6

• Ensure patient is prepared for the exam
• Ensure equipment is functioning properly
• Technical considerations

Question 7

Responsibilities of the Radiologist

Answer 7

• Interpret entire image
• Follow up with ordering physician

Question 8

What are some ways radiographers can reduce image artifacts?

Answer 8

-Ensure image receptors are kept clean
-Optimal kVp and sufficient mAs
-Parallel Collimation
-Proper histogram selection
-Avoid windowing

Question 9

What is a low signal artifact and how does it appear in a radiograph?

Answer 9

Artifacts that cause beam attenuation of low signal to the IR
-Ex:
• Dust, dirt, scratches
• Jewelry, cell phones, snaps
• Sponges, sheets, towels
• Wet hair, hair pins/ holders
• Wheelchair or stretcher rails
• Residual contrast material
• Equipment artifacts
• Software artifacts

Question 10

What is a high signal artifact and how does it appear in a radiograph?

Answer 10

Artifacts that cause an increase signal to the IR
-Ex:
• Backscatter from damage to IP
• Exposure field outside IR
• High exposure to IR
• Lacks collimation
• Background radiation(CR)
• Ghost images

Question 11

Artifacts produced by dust can be corrected easily with

Answer 11

proper cleaning (unless the dust is internal to the optics of a computed radiography (CR) imaging system. Dust on any section of the CR optical path—mirrors and lenses—cannot be corrected by the radiologic technologist and will require professional service)

Question 12

Scratches or a substantial malfunction of pixels likely will require

Answer 12

replacement of the image receptor

Question 13

Answer 13

Residual glue on IR

(Damage to CR plates)

Question 14

Answer 14

Debris on IR mimics foreign bodies

(Damage to CR plates)

Question 15

Answer 15

-Dust inside the Cassette
-Resolution: Clean the Image Plate with approved solution

Question 16

Answer 16

-Dust or damage to Light Guide
-Resolution: Call service provider to clean/repair equipment

Question 17

Answer 17

-Damaged DR Detector
-Resolution: Replacement of the Detector

Question 18

Answer 18

-Liquid Contamination
-Resolution: Replacement of the Detector

Question 19

The appearance of ghost images occurs because of

Answer 19

incomplete erasure of an previous image on a CR image receptor

Question 20

Usually, ghost images can be corrected by

Answer 20

additional signal erasure techniques

Question 21

If a CR image receptor has not been used for 24 hours, it should be

Answer 21

erased again before use. When a completely erased image receptor is processed, the resultant image should be uniform and free of artifacts

Question 22

Answer 22

-Phantom/Ghost Images
-Caused by Incomplete erasure of IR

Question 23

Answer 23

CR Background radiation, IR not used for days

Question 24

Answer 24

CR IR Came Apart

Question 25

Irradiation of a digital radiographic image receptor by the raw x-ray beam may show

Answer 25

variations over the image, producing an irregular pattern that could interfere with diagnosis

Question 26

With irregular patterns, a preprocessing manipulation known as

Answer 26

flatfielding, resulting in a uniform response to a uniform x-ray beam

Question 27

Flatfielding is a preprocessing software correction that is performed to

Answer 27

equalize the response of each pixel to a uniform x-ray beam

Question 28

Flatfielding corrects the anode heel-effect

Answer 28

Question 29

If a lot of scattering material is present behind the image receptor, backscatter radiation can cause a

Answer 29

phantom/ghost image

Question 30

If a phantom artifact is discovered, the back side of the image receptor should be

Answer 30

shielded to reduce backscatter x-rays

Question 31

Answer 31

-Backscatter Radiation -Resolution: Collimate, place lead strip behind Cassette

Question 32

Answer 32

Incorrect histogram selection

Question 33

Answer 33

-The exposure field must be properly collimated, sized, and positioned. Exposure field recognition errors leads to histogram analysis errors because signal outside the exposure field is included in the histogram. The result is very dark or very light or very noisy images. -In DR, proper collimation has the added value of defining the image histogram. If improperly collimated, the histogram can be improperly analyzed, resulting in an artifact

Question 34

Answer 34

-Normal chest x-ray of newborn • Proper cassette size and orientation • Center area of interest to IR • Collimate to area of interest

Question 35

Answer 35

Systems can recognize even-numbered x-ray exposure fields that are centered and cleanly collimated. Multiple image exposures on 1 CR IR is not recommended unless the unexposed portion is shielded. Note the loss of contrast on A when compared to B.

Question 36

Answer 36

For the image histogram to be properly analyzed, each collimated field should consist of 4 distinct collimated margins. 3 collimated margins usually works, but fewer than three can result in artifacts.

Question 37

Answer 37

If multiple fields are projected onto a single image receptor, each must have clear, collimated edges and margins between each field. This process, called partitioning, allows two or more images to be projected on a single image receptor. Partitioning of multiple digital images on a single image receptor results in proper separation and collimation of each image.

Question 38

Answer 38

Alignment of the exposure field on the IR is important. When the image field is not orientated with the size and dimensions of the IR, image artifacts can appear.

Question 39

How can this image be improved?

Answer 39

• Use appropriate sized IR • Center body part to IR • Parallel collimation • Tighter collimation • Proper patient positioning

Question 40

Answer 40

-Cause: Poor grid alignment-grid lines seen. -Resolution: The CR should be centered to the grid to avoid this. -Grid lines can also be seen by scanning the IR in the same direction as the grid lines. Grid frequency of 103 lines per inch or greater recommended for mobile grids to avoid this artifact.

Question 41

Answer 41

-Exposure artifact -Quantum Mottle (Starvation)

Question 42

Answer 42

-Exposure artifact -Data Drop (Detector Saturation)

Question 43

Answer 43

Foreign body: swallowed coin

Question 44

Answer 44

Foreign body: screw

Question 45

Answer 45

Foreign body: Shot in the eye with a pellet gun

Question 46

Answer 46

Foreign body: Swallowed Safety Pin

Question 47

Answer 47

Foreign body: aspirated washer

Question 48

Answer 48

Swallowed Watch Battery

Question 49

Answer 49

Foreign body: Swallowed Magnets

Question 50

Answer 50

Foreign body: Swallowed Marble

Question 51

Answer 51

Foreign body: Aspirated Almond

Question 52

Answer 52

Foreign body: Swallowed Pencil

Question 53

Answer 53

Foreign body: Swallowed Turkey Bone

Question 54

Answer 54

Foreign body: Piece of Wood

Question 55

Answer 55

Foreign body: Piece of Glass

Question 56

Answer 56

Foreign body: Stepped on a Nail

Question 57

Answer 57

Foreign body: Acupuncture Needles

Question 58

Answer 58

Foreign body: Surgical Sponge

Question 59

Answer 59

Foreign body: Bullet

Question 60

Answer 60

Foreign body: Sharpnel

Question 61

Answer 61

Foreign body: Swallowed packets of illicit drugs

Question 62

Answer 62

Foreign body: push up bra

Question 63

Answer 63

Foreign body: extension arm of vacuum

Question 64

Answer 64

Foreign body: Low signal artifact

Question 65

International Electrotechnical Commission (IEC)

Answer 65

-Developed and published an international standard of exposure index and deviation index ( IEC 62494-1) and the DICOM storage and transmission of these numbers in 2008 -The American Association of Physicists in Medicine (AAPM Task Group # 116) and the ImageGently Campaign adopted these standards in 2010

Question 66

Exposure Indicator or Index

Answer 66

*the numeric value assigned to the exposure to the IR -The base exposure indicator number for all systems designates the middle of the detector operating range -Exposure indicator numbers outside the range indicate overexposure and underexposure.

Question 67

High **S** number means

Answer 67

Underexposed

Question 68

Low **S** number means

Answer 68

Overexposed

Question 69

Fuji, Philips, and Konica systems

Answer 69

• the exposure indicator is known as the **S, or sensitivity number** • The S number is the amount of luminescence emitted at 1 mR at 80 kVp, and it has a value of 200 **• The higher the S number with these systems, the lower the exposure.** -For example, an S number of 400 is half the exposure of an S number of 200, and an S number of 100 is twice the exposure of an S number of 200. • The numbers have an **inverse relationship** to the amount of exposure so that each change of 200 results in a change in exposure by a factor of 2. **-Higher S # =Underexposed** **-Lower S # = Overexposed**

Question 70

Kodak Carestream

Answer 70

• uses **exposure index, or EI, as the exposure indicator** • A 1 mR exposure at 80 kVp combined with aluminum/copper filtration yields an EI number of 2000 • An EI number plus 300 (EI + 300) is equal to a doubling of exposure, and an EI number of minus 300 (EI − 300) is equal to a halving of exposure. • The numbers for the Kodak Carestream system have a **direct relationship** to the amount of exposure so that each change of 300 results in change in exposure by a factor of 2.

Question 71

Agfa Systems

Answer 71

• exposure indicator is the **lgM, or logarithm of the median exposure** • An exposure of 20 µGy at 75 kVp with copper filtration yields an lgM number of 2.6 • Each step of 0.3 above or below 2.6 equals an exposure factor of 2. • An lgM of 2.9 equals twice the exposure of 2.6 lgM, and an lgM of 2.3 equals an exposure half that of 2.6 **• The relationship between exposure and lgM is direct**

Question 72

Example 1: An exposure of 80 kVp and 10 mAs results in an S# of 600. What change would be needed to bring the S# into the suggested manufacturer’s range of 100-400?

Answer 72

Double the mAs to get 300 for the S number -Explanation: the S# is Fuji, Konica or Phillip's indicator numbers – An S# of 200 in the middle of the exposure indicator range for these systems. The S# in inversely proportional to the exposure intensity at the IR, therefore an S# of 600 indicates an underexposure. If the radiographer double the mAs to 20 mAs the S# should decrease to approximately half to around 300. Since the median is 200, this is slightly underexposed still, but within the range of 100-400 as suggested by the manufacture.

Question 73

Example 2: On a Agfa system an exposure of 85 kVp at 40mAs of a CTL hip resulted in a lgM of 2.0. What change in mAs is needed to bring the lgM to the median of 2.6?

Answer 73

Double the mAs TWICE (40x2= 80, 80x2= 160) to get the exposure indicator (IgM) to. 2.6 -Explanation: The lgM is the logarithm of the median exposure. The median exposure for this system is 2.6. The exposure indicator is directly related, therefore a lgM of 2.0 is underexposed. For each doubling of the mAs a 0.3 increase of the lgM would result. If the radiographer increase the mAs 4X to 160 mAs, this change would increase the lgM by 0.6 and bring it to about 2.6, which is the median suggested by the manufacturer.

Question 74

Deviation Index

Answer 74

the deviation of the actual EI to the target EI. The target EI (EIT) is the middle of the EI range for a particular exposure system. The actual EI is divided by the EI target. The formula for deviation index is: DI = 10 x log10 (EI/EIT) The AAPM recommends 5 DI ranges (rows) with 4 limits.

Question 75

What can a properly designed Deviation Index chart inform radiographers?

Answer 75

how far from the target EI the actual exposure, assist in making accept/consult/repeat decisions and assess potential changes to future exposures

Question 76

Using the example Cannon RadPro Chart posted below answer the following question: An AP mobile chest exam for central line placement is made with 110 kVp and 3 mAs on a large patient with a 8:1 grid. All critical anatomy is present, positioning is acceptable and the Central line is visible within the SVC. The EI # is 405. Based on the information given and your assessment of the DI chart, what should the radiographer do?

Answer 76

Usually you would divide mAs by 1.25 according to the chart, but because all anatomy is present and the Central line is visible in the image, even though the EI is higher that the EIT, indicating overexposure (b/t 26-58%) it is still within the acceptable range. The image should be accepted.

Question 77

Using the example Cannon RadPro Chart posted below answer the following question: A soft tissue neck is taken on a geriatric patient with a HX of possibly swallowing a small fish bone using 80 kVp at 5 mAs. • The image is well positioned and all essential anatomy is visible. The EI# is 174. Given the information available and your assessment of the DI chart, what should the radiographer do?

Answer 77

The EI indicates underexposure below 37%. This image is likely to suffer from several photographic quality problems: 1. Exposure: Given an exposure this low, the image would demonstrate quantum mottle noise. The noise would make visualizing objects with similar subject contrast difficult because when noise is increased, image contrast resolution decreases. The action that would solve this problem is to increase mAs in order to decrease image noise. Without any other changes mas should be multiplied by 1.5. 2. Contrast: A small fish bone is difficult to visualize due to its similar subject contrast with the structures of the neck. This is a problem with contrast noise due to kVp. As kVp is Increased, noise due to Compton scatter increases, compounding the problem above; loss of contrast due to increased noise. The action that would solve this problem would be to decrease kVp. Without any other changes kVp should be reduced by 15%. To summarize: Problems – loss of contrast due to noise from quantum mottle and contrast noise due to high kVp. Solution- increase mAs and decrease kVp. The new technique should be 70 kVp at 15 mAs. Here is the math: 80 kvp at 5 mAs: Original technique • Step 1 - 80 kVp at 7.5 mAs; 5 x 1.5 = 7.5 • Step 2 - 70 kVp at 15 mAs - Decrease kVp using 15 % rule to maintain the new exposure (decreases Compton scatter, increases subject contrast – decreases contrast noise, increase contrast resolution)