1. 1. The speed of sound in tissues is:
1. A. Roughly 1540 m/s
2. B. Roughly 1540 km/s
3. C. Roughly 1540 cm/s
4. D. Roughly 1540 m/min
2. 2. The relationship between propagation speed, frequency, and wavelength is given by the formula:
1. A. Propagation speed = frequency × wavelength
2. B. Propagation speed = wavelength/frequency
3. C. Propagation speed = frequency/wavelength
4. D. Propagation speed = wavelength × period
3. 3. The frame rate increases with:
1. A. Increasing the depth
2. B. Reducing sector angle
3. C. Increasing line density
4. D. Adding color Doppler to B-mode imaging
4. 4. Period is a measure of:
1. A. Duration of one wavelength
2. B. Duration of half a wavelength
3. C. Amplitude of the wave
5. 5. Determination of regurgitant orifice area by the proximal isovelocity surface area (PISA) method is based on:
1. A. Law of conservation of mass
2. B. Law of conservation of energy
3. C. Law of conservation of momentum
4. D. Jet momentum analysis
6. 6. In which situation can you not use the simplified Bernoulli equation to derive the pressure gradient?
1. A. Peak instantaneous gradient across a nonobstructed mitral valve
2. B. Peak gradient across a severely stenotic aortic valve
3. C. Mean gradient across a severely stenotic aortic valve
4. D. Mean gradient across a stenotic tricuspid valve
7. 7. Which of the following resolutions change with increasing field depth?
1. A. Axial resolution
2. B. Lateral resolution
8. 8. With a fixed-focus transducer with crystal diameter 20 mm and wavelength 2.5 mm, what is the depth of the focus?
1. A. 40 m
2. B. 30 mm
3. C. 40 mm
4. D. 4 m
9. 9. A sonographer adjusts the ultrasound machine to double the depth of view from 5 to 10 cm. If sector angle is reduced to keep the frame rate constant, which of the following has changed?
1. A. Axial resolution
2. B. Temporal resolution
3. C. Lateral resolution
4. D. The wavelength
10. 10. Which of the following properties of a reflected wave is most important in the genesis of a two-dimensional image?
1. A. Amplitude
2. B. Period
3. C. Pulse repetition period
4. D. Pulse duration
11. 11. Increasing depth will change all of the following except:
1. A. Pulse duration
2. B. Pulse repetition period
3. C. Pulse repetition frequency
4. D. Duty factor
12. 12. The two-dimensional images are produced because of this phenomenon when the ultrasound reaches the tissue:
1. A. Refraction
2. B. Backscatter
3. C. Specular reflection
4. D. Transmission
13. 13. Attenuation of ultrasound as it travels through tissue is higher at:
1. A. Greater depth
2. B. Lower transducer frequency
3. C. Blood rather than soft tissue like muscle
4. D. Bone more than air
14. 14. The half-intensity depth is a measure of:
1. A. Ultrasound attenuation in tissue
2. B. Half the wall thickness in mm
3. C. Coating on the surface of the transducer
4. D. Half the ultrasound beam width
15. 15. What is the highest pulse repetition frequency (PRF) of a 3 MHz pulsed wave transducer imaging at a depth of 7 cm?
1. A. 21 000 Hz
2. B. 2 333 Hz
3. C. 11 000 Hz
4. D. 2.1 million Hz
16. 16. Examples of continuous wave imaging include:
1. A. Two-dimensional image
2. B. Volumetric scanner-acquired LV image
3. C. Color flow imaging
4. D. Nonimaging Doppler probe (Pedoff)
17. 17. Which of the following manipulations will increase the frame rate?
1. A. Increase depth
2. B. Increase transmit frequency
3. C. Decrease sector angle
4. D. Increase transmit power
18. 18. The lateral resolution increases with:
1. A. Decreasing transducer diameter
2. B. Reducing power
3. C. Beam focusing
4. D. Reducing transmit frequency
19. 19. Axial resolution can be improved by which of the following manipulations?
1. A. Reduce beam diameter
2. B. Beam focusing
3. C. Reduce gain
4. D. Increase transmit frequency
20. 20. Type of sound used in medical imaging is:
1. A. Ultrasound
2. B. Infrasound
3. C. Audible sound
Answers for chapter 1
1. 1. Answer: A.
Speed of sound in tissue is 1540 m/s. Hence, travel time to a depth of 15 cm is roughly 0.1 ms one way (1540 m/s = 154 000 cm/s or 154 cm/ms or 15 cm per 0.1 ms) or 0.2 ms for to and fro travel. This is independent of transducer frequency and depends only on the medium of transmission.
2. 2. Answer: A
Wavelength depends on frequency and propagation speed. It is given by the following relationship: wavelength (mm) = propagation speed (mm/µs)/frequency (MHZ). Hence, propagation speed = frequency × wavelength.
3. 3. Answer: B.
Reducing the sector angle will reduce the time required to complete a frame by reducing the number of scan lines. This increases the temporal resolution. Decreasing the depth will increase the frame rate as well by reducing the transit time for ultrasound. Adding color Doppler will reduce the frame rate as more data need to be processed.
4. 4. Answer: A
Period is the time taken for one cycle or one wavelength to occur. The common unit for period is µs. Period decreases as frequency increases. The relationship is given by the equation: period = 1/frequency. For a 5-MHZ ultrasound the period is 0.2 µs (1/5 million cycles) = 0.2 µs.
5. 5. Answer: A.
The law of conservation of mass is the basis of the continuity equation. As the flow rate at the PISA surface and the regurgitant orifice is the same, dividing the flow rate (cm3/s) by the velocity (cm/s) at the regurgitant orifice obtained by continuous wave Doppler gives the effective regurgitant area in cm2 (regurgitant flow rate in cm3/s divided by flow velocity in cm/s equals effective regurgitant area in cm2).
6. 6. Answer: A.
In a non-obstructed mitral valve flow velocities are low. Significant energy is expended in accelerating the flow (flow acceleration). As the flow velocity is low, energy associated with convective acceleration is low. As viscous losses in this situation are minimal, the other two components (flow acceleration and convective acceleration) of the Bernoulli equation have to be taken into account. In the simplified Bernoulli equation, the flow acceleration component is ignored. Put simply, when you deal with low-velocity signals in pulsatile system, the simplified Bernoulli equation does not describe the pressure flow relationship accurately.
7. 7. Answer: B.
Lateral resolution depends on beam width, which increases at increasing depths. Axial resolution depends on spatial pulse length, which is a function of transducer frequency, pulse duration, and propagation velocity in the medium.
8. 8. Answer: C.
Depth of focus equals squared crystal diameter divided by wavelength multiplied by 4. In this situation, (20 mm)2/(2.5 mm × 4) = 400/10 = 40 mm.
9. 9. Answer: C.
Lateral resolution diminishes at increasing depths owing to beam divergence. Frame rate determines the temporal resolution as temporal resolution is the reciprocal of frame rate. For example, frame rate of 50 fps gives a temporal resolution of 1/50 = 0.02 s or 20 m. Wavelength is a function of the transducer frequency and is independent of depth and frame rate adjustments.
10. 10. Answer: A.
Amplitude or strength of the reflected beam, and its temporal registration, which determines depth registration.
11. 11. Answer: A.
Pulse duration is the characteristic of the pulse and does not change with depth. Increase in depth will increase the pulse repetition period, and hence reduce frequency and the duty factor.
12. 12. Answer: B.
Backscatter or diffuse reflection produces most of the clinical images. Specular reflection reaches the transducer only when the incident angle is 90° to the surface, which is not the case in most of the images produced. Refracted and transmitted ultrasounds do not come back to the transducer.
13. 13. Answer: A.
Attenuation is the loss of ultrasound energy as it travels through the tissue and is caused by absorption and random scatter. It is greater with longer travel path length as it has to go through more tissue. Attenuation is greater at higher frequencies due to shorter wavelength. Attenuation is greatest for air followed by bone, soft tissue, and water or blood.
14. 14. Answer: A.
It is a measure of attenuation and reflects the depth at which the ultrasound energy is reduced by half. It is given by the formula: 6 cm/frequency in MHz For example, for an ultrasound frequency of 3 MHz the half-intensity depth is 2 cm, and for 6 MHz it is 1 cm.
15. 15. Answer: C.
The PRF is independent of transducer frequency and only determined by time of flight, which is the total time taken by ultrasound in the body in both directions. Ultrasound can travel 154 000 cm in a second at a travel speed of 1540 m/s. In other words, at 1 cm depth (2 cm travel distance) the technical limit to the number of pulses that can be sent is 154 000 cm/2 cm = 77 000 s−1 (Hz). Hence, the PRF equals 77 000/depth in cm. For 7 cm depth, the total distance is 14 cm. PRF = 154 000 (cm/s)/14 cm = 11 000 s−1.
16. 16. Answer: D.
Pedoff is a continuous wave Doppler modality for velocity recording. All other modalities utilize the pulsed wave technique, in which each of the crystals performs both transmit and receive functions.
17. 17. Answer: C.
Increase in the frame rate occurs by reducing the sector angle and reducing the depth, the former by reducing scan lines and the latter by reducing the ultrasound transit time. It is independent of transmit frequency and power.
18. 18. Answer: C.
Focusing increases lateral resolution. Increase in transducer diameter and frequency also increases lateral resolution.
19. 19. Answer: D.
Increasing the transmit frequency will reduce the wavelength and hence the spatial pulse length. This will increase the PRF and the axial resolution. Beam diameter and focusing have no effect on axial resolution.
20. 20. Answer: A.
Ultrasound is used in medical imaging. Typical frequency is 2–30 MHz: 2–7 MHz for cardiac imaging, 10 MHz for intracardiac echocardiography, and 20–30 MHz for intravascular imaging. Ultrasound in the 100–400 MHz range is used for acoustic microscopy. Frequency > 20 000 Hz is ultrasound. Audible range is 20–20 000 Hz and frequency < 20 Hz is called infrasound.