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Ultrasound Physics and Instrumentation, 5e Final Exam: Version A Answers
(Questions 1-8: T/F)
The Doppler Effect is an apparent change in frequency, because of a change in wavelength, due to relative movement between a source and an observer. Answer: True
Sound waves cannot exist in a vacuum.
Answer: True
- If you reduce the amplitude from 6 Volts to 2 Volts, the power decreases by a factor of 4.
Answer: False (a factor of 3 (6/2= 3) ).
- Rayleigh scattering is highly angle dependent.
Answer: False
- A shorter wavelength represents a higher frequency and occurs when relative motion is toward an observer.
Answer: True (pg. 519)
- A linear array transducer always has some portion of the image that was created with parallel beams.
Answer: True
- If the transmit frequency is 5.000 MHz, and the returning frequency is 4.999 MHz, the Doppler shift would be considered positive, or toward the transducer.
Answer: False (Returning frequency is below transmitted frequency implying a negative shift, or, away from the transducer.)
- The PRF equals 1 divided by the imaging depth in cm multiplied by 13 msec.
Answer: False (13 microseconds)
- The fact that a higher pitch whistle would result in a greater Doppler shift (for the same relative motion) indicates that:
a) the Doppler shift is directly related to the velocity. b) the Doppler shift is directly related to the wavelength. c) the Doppler shift is inversely related to distance. d) the Doppler shift is inversely related to the amplitude. e) the Doppler shift is proportional to the transmit frequency.
Answer: E
- If the transmitted frequency is 3.0 MHz and an observer hears a signal at 2.993 MHz, the Doppler shift is:
a) – 3 kHz b) + 2.993 MHz c) + 7 kHz d) – 7 kHz e) – 3.0 MHz
Answer D (2.993 MHz – 3.000 MHz = – 0.007 MHz = – 7 kHz) (page 524)
- Write the wavelength equation.
Answer:
c
f
- For a given interrogating frequency, if the velocity doubles, what happens to the Doppler shift frequency?
Answer: The Doppler shift doubles since the Doppler frequency is proportional to the operating frequency.
- In the Doppler equation:
a) the cosine is in the numerator. b) the sine is in the numerator. c) the cosine is in the denominator. d) the sine is in the denominator.
Answer: A (page 533)
- Given that the major tick marks for the sweep speed represent 1 second, what is the best estimate for the heart rate represented in this Doppler spectrum?
a) 60 bpm b) Slightly less than 60 bpm c) 75 bpm d) Slightly greater than 75 bpm e) cannot be determined
Answer: B (notice that the cardiac cycle lasts just slightly longer than 1 second. A period of 1 second represents 60 beats in 60 seconds, or 60 beats per minute. Since the time is slightly greater than 60 seconds the heart rate is slightly less than 60 bpm.)
- Using a 7.5 MHz Doppler, a Doppler shift of –6.3 kHz was measured. If the frequency is decreased to 2.5 MHz (with no other changes), what would the Doppler shift be?
Answer: –2.1 kHz
so decreasing the transmit frequency by a factor of 3,
reduces the Doppler shift by a factor of 3.
f Dop fo
- Describe the advantages and disadvantage of PW relative to CW Doppler.
Answer: PW has good range specificity, whereas there is no inherent range specificity with CW. However, CW can detect virtually unlimited velocities whereas PW is limited by Nyquist and aliasing.
- Which statement best describes the flow at the specific time indicated by the white arrow?
a) Accelerating flow toward the transducer b) Accelerating flow away from the transducer c) Decelerating flow toward the transducer d) Decelerating flow away from the transducer e) Constant flow away from the transducer
Answer: C : Note that the spectrum is inverted so flow below the baseline represents flow toward the transducer. At the specific time indicated by the arrow, the velocity is decreasing, indicating deceleration.
- If at a stenosis the blood velocity increased from 0.3 m/sec to 1.5 m/sec, what would happen to the Doppler shift?
a) Would not change b) Increased by a of 1. c) Increase by factor of 5 d) Decrease by a factor of 1. e) Decrease by a factor of 5
Answer: C
f Dop v so increasing the velocity by a factor of 5 (1.5/0.3),
increases the Doppler shift by a factor of 5.
- For adult echo Doppler, at what setting would you expect to set the spectral Doppler wall filters? a) About 25 Hz b) About 50 Hz c) About 200 Hz d) About 800 Hz e) About 1600 Hz
Answer: C
- Which statement is best regarding the dark blue flow indicated by the white arrow?
a) The mean velocity is below 23 cm/sec b) The peak velocity is below 23 cm/sec c) The mean velocity is close to 0 cm/sec d) The mean velocity is close to 40 cm/sec e) The peak velocity is 40 cm/sec f) The peak velocity is 0 cm/sec
Answer: D (Aliasing – the mean velocity is somewhere between 35 and 46 – making 40 the best answer.)
- The fact that the flow is predominantly above the baseline in this spectrum is the result of
a) aliasing. b) the Doppler angle was less than 90 degrees. c) spectral invert. d) the Doppler angle was less than 90 degrees and spectral invert. e) the Doppler angle was less greater than 90 degrees and aliasing.
Answer: C: Notice that the flow above the baseline is labeled as -47 cm/sec implying that spectral invert was active (Note that the angle was greater than 90 degrees since the flow is predominantly above the inverted baseline i.e. the cosine was negative since the angle was greater than 90 degrees).
- At a clinical rotation, you are told by the clinical coordinator that your color PRF is too high while performing a venous study. Which two characteristics of the color image would most likely have indicated this to the clinical coordinator? a) Color aliasing b) Color speckle c) Color blossoming d) Regions of color dropout in the image
Answer: D If the color scales are too high, the color wall filters will be too high, potentially eliminating detection of low velocity flow.
- In which direction is the blood flowing?
a)
b)
c)
d)
e) Cannot be determined
Answer: C
- Write the Nyquist relationship in terms of the maximum Doppler shift and the PRF.
Answer:
Dop^ (max)
PRF
f
- Which of the following would increase the likelihood of aliasing (choose all that apply) a) Decrease the PRF b) Decrease the Doppler gate depth c) Measuring higher velocity flow d) Decrease the operating frequency e) Going from a 50 degree to a 30 degree Doppler angle
Answer: A, C, and E Choice A: Decreasing the PRF makes aliasing more likely Choice C: A higher velocity increases the Doppler shift and hence makes aliasing more likely. Choice E: Using a smaller Doppler angle increases the Doppler shift, increasing the likelihood of aliasing.
- Assume that using an 9 MHz transmit frequency, with a PRF of 6 kHz, the Doppler shift is measured as 9 kHz. At what transmit frequency would aliasing be eliminated? a) 3 MHz b) 4 MHz c) 4.5 MHz d) 6 MHz e) 8 MHz
Answer: D
Dop 2 2
Dop o
PRF kHz
f kHz
f f
so decreasing the operating frequency by a factor of 3 (from 9 to 3 MHz)
results in a Doppler shift of (6 kHz)/3 = 3 kHz which is the limit where alias will end.
(max)
- Referring to the associated color bar, assume that the wall filters are approximately 5% of the scales. Below what velocity would flow signals not present? a) 8.4 cm/sec. b) + 2.4 cm/sec c) + 4.8 cm/sec d) + 1.2 cm/sec e) 0.24 cm/sec
Answer: D (0.05) *(+ 24 cm/sec) = + 1.2 cm/sec
- What is the best estimate of the Doppler angle in the assocaited diagram? a) 30 b) 60 c) 20 d) 150 e) 120
Answer: D
- Describe what happens if wall filters are set too high.
Answer: If the wall filters are set too high, lower velocity signal that produce low frequency shifts are eliminated. This means that low velocities cannot be appreciated.
- Which of the following changes should be made to the associated Doppler spectrum? a) Decrease transmit power b) Increase contrast c) Shift baseline down d) Increase the PRF e) Invert the spectrum
Answer: D
- For the associated picture
a) What modality is presented? Answer: HPRF b) What do the diamonds represent? Answer: Ambiguous gate locations c) What is the advantage of this technique relative to standard PW? Answer: Ability to detect higher velocities without aliasing (Higher PRF) d) What is the disadvantage of this technique relative to standard PW? Answer: There is ambiguity above the desired gate location (not just below)
- Write Snell’s Law
Answer: Snell’s Law: (^) ci sin θt ct sin θi
- Which statement best summarizes this textbook’s description of artifacts? a) Always useful b) Never useful c) Often useful d) Rarely useful e) Almost always never useful except on those rare occasions in which they sometimes are (although normally not) useful.
Answer: C
- Which artifact(s) can create a lateral displacement in an image? (more than one may apply) a) Reverberation b) Grating lobes c) Shadowing d) Mirror artifact e) Refraction artifact
Answer: B and E
- Given the incident beam as shown, which beam would be created through the medium labeled “2”? a) A b) B c) C d) D e) E
Answer: A
- The bright echoes (as indicated by the arrow) exemplify what artifact? a) Reverberation b) Grating lobes c) Ring down d) (Multiple) mirror artifacts e) Enhancement
Answer: A
Medium 1 1500 m/sec
Medium 2 1200 m/sec
A B
C
D
E
- The following schematic (labeled Figure A) indicates a uniform block of tissue with two structures as indicated. In the Figure B, draw what the ultrasound image would look like. (Dotted lines (indicating depth) have been added to make your drawing easier and more accurate.)
Answer:
- Given that the Doppler scales cannot be increased, what change should be made to eliminate the aliasing artifact?
Answer: Use a lower transmit frequency
1540 m/sec
770 m/sec
1540 m/sec
1540 m/sec
Figure A Figure B
- Artifactual spectral broadening increases with (more than one may apply)?
a) Deeper sample volumes b) Larger Doppler angles c) Higher PRFs d) Lower wall filters e) Lower receive gain
Answer: B
- Which of the following statements is not true regarding the gold standard test? (more than one may apply)
a) The gold standard is always perfect. b) The gold standard is the test recognized as the most consistent and accurate test for a particular disease. c) The gold standard is used as the reference against which other test results are compared. d) Almost all gold standard tests are perfect or near perfect. e) The gold standard is “assumed” to be perfect in terms of the statistical comparisons made.
Answer: A and D
When the results of a comparison test match the results of the gold standard test, the results are said to be TRUE.
Which statement is implied by a result which is called a true positive?
a) The comparison test is negative for disease and matches the gold standard. b) The comparison test is positive for disease but does not match the gold standard. c) The comparison test is positive for disease and matched the gold standard. d) The comparison test is negative for disease but does not match the gold standard.
Answer: C
- What statistical parameter is determined by the following equation?
TP
TP FN
Answer: Sensitivity
- What is the term for the parameter specified by the outlined box?
Answer: False Negative (FN)
- Which statement is correct regarding the accuracy?
a) It must be greater than the PPV. b) It must be less than the NPV. c) It must be greater than both the NPV and the PPV. d) It must be lower than both the NPV and the PPV. e) It must be between the NPV and the PPV.
Answer: E
Stable cavitation occurs when the oscillation of the microbubbles does not lead to collapse.
Which of the following statements is true regarding transient cavitation?
a) The best science to date indicates that it seems to be a threshold effect. b) Theory indicates that it is much more likely to happen with lower frequency operation. c) The best science to date indicates that it rarely, if ever, occurs with sound waves. d) Theory indicates that it is much less likely to occur when the tissue temperature exceeds between 40 and 50 degrees Celsius. e) Acoustic measurements made with hydrophones indicate that it occurs more when the transducer surface itself is warm.
Answer: A
TEST
GOLD
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- Which of the following is NOT considered a drawback to conventional tissue harmonic imaging?
a) Degradation in axial resolution. b) Less penetration in the far field c) Increased phase aberration d) increased clutter artifacts e) More than one choice are not drawbacks
Answer: e (both choices C and D)
- How does pulse inversion harmonics work?
a) Transmit one fundamental and one harmonic beam with same phase, invert harmonic beam, and add. b) Transmit two beams with inverted phases and subtract. c) Transmit two beams with inverted phases and add. d) Transmit one fundamental beam, receive the harmonic signal, and invert the phase.
Answer: C)
- In the following statistical grid, fill in the appropriate values according to the following information.
Talamungo Lab tests a series of patients for a specific disease with the following results: 500 total patients tested 87 positive for disease of which 7 do not match the gold standard 50 of the negative tests do not match the gold standard
TEST
GOLD
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- Given the follow statistical table, how many patients in the patient population do not have disease?
Answer: 4 + 72 = 76
- Given the follow statistical table, what is the prevalence of disease for the patients tested?
Answer: 48/100 = 48%
- If the sensitivity of a test if 74% and the specificity is 91%, which cannot be said about the accuracy?
a) The accuracy could be 84% b) The accuracy cannot be 93% c) The accuracy could not be 73% d) All of the above are true e) None of the above are true
Answer: D (accuracy must be somewhere between sensitivity and specificity)
TEST
GOLD
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TEST
GOLD
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