question · question. when slice thickness is changed from 2.5 mm to 5.0 mm, image noise: a....
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Question
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Answer
B. The same for both patients
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A. The patient dose will be higher for the smaller patient
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Answer
B. The radiation dose will increase
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Answer
C. Will decrease with increasing patient size
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Answer
C. 10-50 mGy
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A. Increasing slice thickness increases the number of photons (signal) used to acquire the image and decreases noise therefore increasing image contrast. B. Spatial resolution will worsen because each voxel will be larger. This will cause more averaging of attenuation coefficients due to more tissue types in the slice, worsening spatial resolution.D. PVA will increase due to the larger voxels covering multiple tissue types.
Answer
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Answer
A. decreases As the pixel size decreases, there are fewer photons passing through it. This increases noise thus lowering the SNR.
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Answer
B. increases
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Question
Increasing the kVp in CT while maintaining the mAs constant will result in a decrease in which of the following?
A. Anode heat loadingB. CTDIvol
C. Subject contrastD. Partial volume averaging
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Answer
Increasing the kVp in CT while maintaining the mAs constant will result in a decrease in which of the following?
A. Anode heat loadingB. CTDIvol
C. Subject contrastD. Partial volume averaging
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When slice thickness is changed from 2.5 mm to 5.0 mm, image noise:
A. Decreases by ½ B. Decreases by 40%C. Increases by 40%D. Doubles
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When slice thickness is changed from 2.5 mm to 5.0 mm, image noise:
A. Decreases by ½ B. Decreases by 40%C. Increases by 40%D. Doubles Sqrt(5.0/2.5) = 1.4
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What determines the minimum possible reconstructed slice thickness on a multi-slice CT unit?
A. Detector widthB. Pitch factorC. Detector binningD. Bed translation speed
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Answer
What determines the minimum possible reconstructed slice thickness on a multi-slice CT unit?
A. Detector widthB. Pitch factorC. Detector binningD. Bed translation speed
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The CTDIvol displayed on the dose summary page following an abdomen scan of an adult patient is 40 mGy. The effective diameter of the patient is 38 cm. The actual dose to the patient in this case would be:
A. Less than 40 mGyB. 40 mGyC. More than 40 mGy
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Answer
The CTDIvol displayed on the dose summary page following an abdomen scan of an adult patient is 40 mGy. The effective diameter of the patient is 38 cm. The actual dose to the patient in this case would be:
A. Less than 40 mGyB. 40 mGyC. More than 40 mGy
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Question
Reducing which of the following will minimize partial volume artifacts in CT?
A. Section thicknessB. Scan timeC. Focal spot sizeD. Scan length
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Answer
Reducing which of the following will minimize partial volume artifacts in CT?
A. Section thicknessB. Scan timeC. Focal spot sizeD. Scan length
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Which of the following CT exams has the highest effective dose?
A. Abdomen/PelvisB. HeadC. C-spineD. Chest
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Answer
Which of the following CT exams has the highest effective dose?
A. Abdomen/PelvisB. HeadC. C-spineD. Chest
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Answer
A. Dose length product
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Which one of the following scenarios will result in the highest skin dose to the patient?
A. B. C. D.
Question
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Which one of the following scenarios will result in the highest skin dose to the patient?
A. B. C. D.
B. Short SSD plus large SID results in higher patient dose. Use of the grid also requires higher patient dose.
Answer
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Increasing the usage of electronic magnification mode in most fluoroscopy systems results in:
A. Increased quantum mottleB. Greater spatial distortionC. Increased patient doseD. More scattered radiationE. More minification
Question
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Increasing the usage of electronic magnification mode in most fluoroscopy systems results in:
A. Increased quantum mottleB. Greater spatial distortionC. Increased patient doseD. More scattered radiationE. More minification
C. The cost of using higher mag modes is radiation dose.
Answer
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AAPM PC: If we modify the SID from 72” to 40”, which of the following will occur?
A. Radiation dose to patient will decrease by 4xB. Image spatial resolution will increaseC. Image noise will increaseD. The object of interest will appear larger on the image
Question
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AAPM PC: If we modify the SID from 72” to 40”, which of the following will occur?
A. Radiation dose to patient will decrease by 4xB. Image spatial resolution will increaseC. Image noise will increaseD. The object of interest will appear larger on the image
D. Mag factor is defined as SID/SOD. Decreasing SID also decreases SOD, but the ratio of SID/SOD increases since OID stays the same. Appearance of more magnification
Answer
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AAPM PC: What is the single most important component of a radiographic system for determining patient dose:
A. Focal spot sizeB. X-ray generator power ratingC. X-ray generator type (3-phase, high freq, falling load)D. Parameter settings on the AECE. Table attenuation
Question
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AAPM PC: What is the single most important component of a radiographic system for determining patient dose:
A. Focal spot sizeB. X-ray generator power ratingC. X-ray generator type (3-phase, high freq, falling load)D. Parameter settings on the AECE. Table attenuation
D. The AEC drives how much exposure is getting to the detector, which drives the x-ray tube output.
Answer
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AAPM PC: Detection of a large, low contrast object in a noisy image can be improved by:
A. Applying edge enhancementB. Applying image smoothingC. Increasing window widthD. Digitally magnifying the image
Question
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AAPM PC: Detection of a large, low contrast object in a noisy image can be improved by:
A. Applying edge enhancementB. Applying image smoothingC. Increasing window widthD. Digitally magnifying the image
B – applying smoothing reduces noise without reducing contrast. Edge enhancement (A) increases noise and will decrease conspicuity. Increasing WW (C) will decrease apparent noise, but also decreases display contrast which hurts conspicuity. Digitally magnifying the object forces the eye to concentrate on noise. Often, it helps to minify b/c that increases averaging of pixels in the eye and effectively smooths the image.
Answer
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AAPM PC: Which of the following will improve low contrast resolution in a radiographic image?
A. A change from 10:1 to an 8:1 gridB. Move the patient closer to the image receptorC. Reduce mAsD. Use a smaller field of view
Question
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AAPM PC: Which of the following will improve low contrast resolution in a radiographic image?
A. A change from 10:1 to an 8:1 gridB. Move the patient closer to the image receptorC. Reduce mAsD. Use a smaller field of view
D – Using a smaller FOV results in less scatter contribution, and less scatter reaching the image receptor. As scatter decreases, low contrast resolution increases. A) would allow more scatter through, B) is the opposite of an ‘air gap’ technique, C) worsens stats, increases noise.
Answer
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AAPM PC: For a KUB on an avg patient, what would be a reasonable technique considering dose, contrast, noise, and
motion:
A. 75 kVp, 400 mA, 50 msB. 120 kVp, 800 mA, 15 msC. 50 kVp, 100 mA, 500 msD. 75 kVp, 100 mA, 25 msE. None of the above
Question
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AAPM PC: For a KUB on an avg patient, what would be a reasonable technique considering dose, contrast, noise, and
motion:A. 75 kVp, 400 mA, 50 msB. 120 kVp, 800 mA, 15 msC. 50 kVp, 100 mA, 500 msD. 75 kVp, 100 mA, 25 msE. None of the above
A – rule of thumb on these, ~ 80 kVp and 20mAs for AP work.
Answer
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Answer
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55©UW and DPS
AAPM PC: Assuming AEC, radiograph at level of kidneys. For a pregnant patient, what is best way to minimize fetal dose?
A. Use high kVp, since it will lower mAs and decrease doseB. Wrap abdomen in lead apron to cover fetusC. Collimate to cover smallest regions possibleD. Have patient lay prone instead of supineE. Remove grid
Question
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AAPM PC: Assuming AEC, radiograph at level of kidneys. For a pregnant patient, what is best way to minimize fetal
dose?A. Use high kVp, since it will lower mAs and decrease doseB. Wrap abdomen in lead apron to cover fetusC. Collimate to cover smallest regions possibleD. Have patient lay prone instead of supineE. Remove grid
C
Answer
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AAPM PC: The patient skin dose will be reduced by using _________.
A. More added filtrationB. Higher grid ratioC. Lower kVpD. Smaller focal spot sizeE. None of the above
Question
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AAPM PC: The patient skin dose will be reduced by using _________.
A. More added filtrationB. Higher grid ratioC. Lower kVpD. Smaller focal spot sizeE. None of the above
A is the best answer. Added filtration increases the average photon energy via preferential attenuation of lower energy x-rays. Higher avg photon energies increase the proportion of x-rays passing through the patient without interacting.
Answer
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Question
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Answer
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Question
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Answer
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Question
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Answer
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Answer
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The decision is made to add 1 mm Al permanently to the filtration of an x-ray beam. This is done to reduce ____.
A. Load on the x-ray tubeB. Scatter into the detection systemC. Maximum field sizeD. Overall system latitudeE. Patient skin dose
Question
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The decision is made to add 1 mm Al permanently to the filtration of an x-ray beam. This is done to reduce ____.
A. Load on the x-ray tubeB. Scatter into the detection systemC. Maximum field sizeD. Overall system latitudeE. Patient skin dose
E. Most of the soft (i.e. low energy) radiation in an x-ray beam is absorbed in the patient and does not contribute to the image. Hardening the beam, by adding filtration, reduces the patient’s skin dose for the same radiation reaching the detector.
Answer
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What is the effect of magnification on mammography patient dose?
A. IncreasesB. DecreasesC. Has no effect
Question
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What is the effect of magnification on mammography patient dose?
A. IncreasesB. DecreasesC. Has no effect
A. Entrance skin exposure and breast dose are reduced because no grid is used but are increased due to the shorter focus to breast distance.
Answer
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Compression in mammography results in increased:
A. Breast doseB. Geometric blurringC. Patient motionD. ScatterE. None of the above
Question
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Compression in mammography results in increased:
A. Breast doseB. Geometric blurringC. Patient motionD. ScatterE. None of the above
E. Patient motion, blurring, scatter and dose all reduce due to compression of breast.
Answer
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Answer
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Question
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Answer
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Question
The bulk modulus describes which of the following characteristics of a medium?
A. stiffnessB. densityC. impedanceD. amplitude
Question
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AnswerThe bulk modulus describes which of the following characteristics of a medium?
A. stiffness
The bulk modulus describes the resistance to compression of a medium under pressure. “Compressibility” “Rigidity” or “Stiffness” are terms you may see.
Though they are related, Density and Bulk Modulus are not the same property.
Answer
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Question
Sound travels fastest in which of the following?
A. lungB. boneC. bloodD. fat
Question
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AnswerSound travels fastest in which of the following?
A. lungB. boneC. bloodD. fat
The speed of sound in a medium is dependent on two properties—density and a property known as the bulk modulus. The bulk modulus is a measure of compressibility. The less compressible a given medium is, the faster sound will travel through it given its mechanical wave properties. Sound travels significantly slower through air than bone for this reason.
Answer
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Question
Medical ultrasound typically operates in which of the following frequency ranges?
A. 1–100 HzB. 1–100 kHz C. 1–10 MHz D. 1–10 GHz
Question
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AnswerMedical ultrasound typically operates in which of the following frequency ranges?
A. 1–100 HzB. 1–100 kHz C. 1–10 MHz D. 1–10 GHz
Modern ultrasound transducers operate on the order of 1 to 10 MHz for clinical procedures. Human hearing is on the order of 20 to 20,000 Hz.
Doppler shifts are in the low kHz range.
Answer
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Question
If the frequency used to image a liver is 8 MHz, what is the approximate wavelength of the ultrasound beam in tissue?
A. 0.2 mm B. 2.0 mm C. 2.0 cm D. 20cm
Question
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Question
Answer
If the frequency used to image a liver is 8 MHz, what is the approximate wavelength of the ultrasound beam in tissue?
A. 0.2 mm B. 2.0 mm C. 2.0 cm D. 20cm
Can we solve this? If so, how?
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AnswerAnswer
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AnswerAnswer
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Question
If the relative intensity between the original and received ultrasound is cut in half, this will account for a loss of how many decibels?A. -1 B. -3C. -20 D. -100
Question
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88©UW and DPS
AnswerIf the relative intensity between the original and received ultrasound is cut in half, this will account for a loss of how many decibels?
A. -1 B. -3 C. -20D. -100
Given the equation relative intensity (db) = 10log (I2/I1):(db) = 10log (1/2) = 10(-0.301) = -3
Answer
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AnswerAnswer
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Question
The main disadvantage of using multiple focal zones for ultrasound imaging is decreased:
A. lateral resolutionB. temporal resolution C. axial resolution D. elevational resolution
Question
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Answer
The main disadvantage of using multiple focal zones for ultrasound imaging is decreased:
A. lateral resolutionB. temporal resolution C. axial resolution D. elevational resolution
When using multiple-layer focal zones, the system needs more time to process the ultrasound beam transmitting and receiving from multiple distances before stitching the image together. This will decrease frame rate and thus decrease temporal resolution.
Answer
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Question
In order to resolve an object in the axial plane using ultrasound, the boundaries between objects must be separated by a minimum of:
A. less than one spatial pulse length B. less than half a spatial pulse length C. greater than one spatial pulse length D. greater than half a spatial pulse length
Question
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Answer
In order to resolve an object in the axial plane using ultrasound, the boundaries between objects must be separated by a minimum of:
A. less than one spatial pulse length B. less than half a spatial pulse length C. greater than one spatial pulse length D. greater than half a spatial pulse length
In order to resolve an object in the axial plane using ultrasound, the boundaries between objects must be separated by greater than one half of the spatial pulse length. If the separation between two objects is less than this, the returning signals will overlap and be seen as originating from one object. Using higher frequencies will reduce wavelength (and SPL) thus improve axial resolution.
Answer
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Answer
• Axial resolution (linear, range, longitudinal or depth resolution) is the ability to separate two objects lying along the axis of the beam
• The minimal required separation distance between two boundaries is ½ Spatial Pulse Length (about ½ λ) to avoid overlap of returning echoes
• Higher frequencies reduce Spatial Pulse Length, improving axial resolution however, increases attenuation
• Axial resolution remains constant with depth
Answer
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Question
Lateral resolution is primarily dependent on which of the following in ultrasound?
A. depthB. frame rate C. amplitude D. spatial pulse length
Question
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AnswerLateral resolution is primarily dependent on which of the following in ultrasound?
A. depthB. frame rate C. amplitude D. spatial pulse length
Lateral resolution is dependent on beam width. For both linear and curvilinear transducers, the beam width changes with depth in the near and far fields. Therefore, lateral resolution is directly dependent on depth as well. The frequency should be adjusted accordingly when imaging in order to focus directly on the organ of interest.
Answer
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Question
Reflective properties of sound at a tissue interface are determined by all of the following properties of tissues except ____________.
A. Acoustic impedanceB. DensityC. Speed of soundD. Attenuation coefficient
Question
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Answer
Reflective properties of sound at a tissue interface are determined by all of the following properties of tissues except ____________.
A. Acoustic impedanceB. DensityC. Speed of soundD. Attenuation coefficient
The primary factor determining the reflective properties is the change in acoustic impedance. Acoustic impedance: Z = ρ · c, units are rayl [kg/m2-sec]
Answer
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Question
Question
The reflection coefficient is greatest for which of the following tissue interfaces?
A. liver-fatB. muscle-lungC. muscle-boneD. fat-muscleE. cornea-aqueous humour
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Question
Question
The reflection coefficient is greatest for which of the following tissue interfaces?
A. liver-fatB. muscle-lungC. muscle-boneD. fat-muscleE. cornea-aqueous humour
Material Acoustic Impedance, Z
Air 0.00043Lung 0.18Fat 1.42Castor Oil 1.40Water 1.48Soft Tissue ~1.45Brain 1.56Blood 1.6Kidney 1.61Liver 1.64Muscle 1.63Eye Lens 1.83Bone 6.12
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Question
Answer
The reflection coefficient is greatest for which of the following tissue interfaces? A. liver-fatB. muscle-lungC. muscle-boneD. fat-muscleE. cornea-aqueous humour
Muscle lung has the greatest difference in Z (mainly due to air/lung being very very low) Muscle-bone would be the next greatest.
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• Q: What is the artifact seen in this image of Adenomyomatosis of the gall bladder?
• A. mirror image artifact • B. refraction artifact• C. comet tail artifact • D. aliasing
Ultrasound Review
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• Q: What is the artifact seen in this image of Adenomyomatosis of the gall bladder?
C. comet tail artifact
Ultrasound Review
Adenomyomatosis is a diseased state of the gallbladder in which the gallbladder wall is excessively thick. Cholesterol crystals within the lumina are responsible for the artifact in this case. When there is a serious mismatch in acoustic impedance between surfaces, reverberations from highly reflective interface are repeatedly reflected and arrive at the transducer later than the actual echo. Erroneously received reflections create bands at lower depths are called comet tails.
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• Q: What is the artifact seen in this image of Adenomyomatosis of the gall bladder?
C. comet tail artifact
Ultrasound Review
Comet tails are a type of reverberation artifact in which sequential echoes are closely spaced and display a triangular tapered shape. Other reverberation artifacts can appear as multiple equally spaced signals extending into the deep field, as shown in the image of a palpable mass to the right.
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• Q: What is the name of the artifact that occurs when the Doppler sampling rate is less than twice the Doppler frequency shift?
• A. aliasing artifact• B. mirror image artifact• C. reverberation artifact• D. erroneous sample artifact
Ultrasound Review
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• Q: What is the name of the artifact that occurs when the Doppler sampling rate is less than twice the Doppler frequency shift?
A. aliasing artifact
Ultrasound Review
In the pulsed Doppler imaging, sampling rate or pulse repetition frequency (PRF) is set by the sonographer. Sampling rate (PRF) must be at least twice the maximum frequency shift present to satisfy the Nyquist sampling criterion. (ie- PRF of at least 20 Khz is required for a Doppler shift of 10 Khz) One half of the PRF is called Nyquist frequency limit.
Doppler shifts above the Nyquist frequency limit wrap around and are displayed as a low-frequency shift—an artifact. Some ways to avoid this artifact include increasing the PRF, moving the color baseline up or down, or increasing the velocity scale.
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• Q: Identify the artifact depicted by the arrow in the image below
• A. mirror-image artifact• B. ring artifact• C. banding artifact• D. echo artifact
Ultrasound Review
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• Q: Identify the artifact depicted by the arrow in the image below
• A. mirror-image artifact
Ultrasound Review
Structures located in front of highly reflective surfaces may scatter sound off-axis.The delayed arrival of these signals is interpreted as a mirror image at a deeper location by the transducer.
This image of the right hepatic lobe shows an echogenic lesion in the right hepatic lobe (cursors) and a duplicated echogenic lesion (arrow) equidistant from the diaphragm overlying the expected location of lung parenchyma.