Echo Physics MCQ- Answers List

The correct answer is E: 2 cm.

While current day transducers have multiple crystals, it is important to know the basics of a simple single crystal transducer, as the principles are similar.

The figure shows the beam emanating from a crystal of length D.

The beam actually has three dimensions - lateral (side to side), axial (also called radial, longitudinal or depth) and elevation (slice thickness). The beam is typically hourglass shaped with a minimal width at the natural focus equal to D/2.

The area before the natural focus is called the Fresnel zone; the area after is the Fraunhofer zone, where the beam diverges and becomes wider. At twice the natural focus the beam width is the same as the diameter of the crystal.

The distance to the natural focus is called the 'near zone length" and is calculated with the following equation:

Near zone length = D^2/4*wavelength

where D is the diameter of the crystal.
In tissue since c = 1540m/s this is also equivalent to
D^2*frequency/6
where diameter is mm and frequency is in MHz.

In this question if the near zone length is at 6 cm, then at 12 cm, it is at twice the near zone length and the beam width will be the same as the crystal diameter again, or 2 cm, so this is the correct answer. At the natural focus the beam width would be 1 cm.

The correct answer is A: 0.5 cm.

Look at the figure shown.

The focus of the beam is where the highest lateral resolution will be as this is where the beam is the narrowest. The narrower the beam, the better the transducer can distinguish two objects that are close together, hence the higher the resolution. We know that at the focus the beam width is approximately half the diameter of the crystal of half of 1 cm which is 0.5 cm. None of the other answers are correct. We do not need to know the frequency of the transducer to answer this question.

The correct answer is C: 3 cm.

Recall that the pulse repetition period (PRP) is the time that passes between pulses sent out by the transducer. In between pulses the transducer is listening back for signal. The longer the depth, the more time will be needed to listen and hence the PRP will be longer.

As a general useful rule of thumb, in human tissue (assuming a propagation velocity of 1540 meters/second), the depth is 1 cm for every 13 microseconds of a pulse repetition period. In this case since we are told the PRP is 39 microseconds, the depth is 39/13 or 3 cm.

In fact, any time you are given the depth, you also should be able to calculate the PRP (just multiply the depth in cm by 13 and you will have your answer in microseconds). This is one variation of how this type of question could be asked.

The correct answer is B: 2 cm

One needs to know the equation shown in the figure.
D here is in mm as is NZL
Here we have NZL = 5*5*5/6 = 20.8mm. This is closest to 2 cm.

The correct answer is E: 4 KHz. This question requires knowledge of the Doppler shift equation which is shown in the figure.

The correction answer is C: 75 Hz

To generate a 2D cine video image, the ultrasound machine strings together a number of still images or frames. For each image it collects one line of data at a time as scanning occurs across a plane so it can generate the frame as shown in the figure .

To generate a 2D image you need a large number of scanned lines. The more lines you have the longer it will take to scan that plane and create an the frame before the machine can start all over on the next frame. The machine then puts these frames together and we perceive a moving 2D video image. For a decent video, the human eye needs about 30 frames/sec or 30 Hz as minimum the frame rate.

To calculate the frame rate we first calculate the time needed to get one line in a frame. This is from the pulse repetition period (PRP), or the time between ultrasound pulses. Recall that the PRP is just the viewing depth in cm x 13--giving us the answer in microseconds. So here 4cm x 13 = 52 microseconds for each line.

Now we have 256 lines for one frame in this question. So if each line takes 52 microseconds, then 256 lines will take 52 x 256 or a total of 13,312 microseconds. This is the same as saying 13.3 milliseconds per frame.

Thus the frame rate (number of frames per second) would be 1/13.3 milliseconds or 0.075 Kilohertz which is 75 Hz.

Note that for answers D and E, the units are megahertz (MHz) which are very high for a 2D frame rate and nonsensical. They can be eliminated easily.

The correct answer is C: 1.25 m/s To answer this question you need to know the Doppler equation and how to use it. It is shown in the figure.

The correct answer is B: 3 MHz. To answer this question you need to know that frequency = 1/period. In this case 1/33.3 x 10^-8 seconds. It helps to convert this to microseconds since we want to deal with megahertz (MHz) ultimately. Remember, the range for diagnostic ultrasound is 2 to 20 MHz). So we can use 0.333 x 10^-6 seconds as the period. The frequency is therefore 1/.333*10^-6 which is 3 MHz. None of the other answers are correct and you have enough information from what is given to answer the question.

The correct answer is E: 385 micrometers. Recall the range of diagnostic ultrasound is 77 to 770 micrometers. Knowing this alone one can answer the question without having to do any calculations. To calculate this directly we use the formula that wavelength = 1540m/s (the average speed of ultrasound in tissue)/ frequency which is 1540m/second/4MHz = 385 micrometers. None of the other answers A-D make sense and are orders of magnitude off.

The correct answer is B. This question is testing your knowledge of the Huygen's Principle and how phased arrays work.

The correct answer is D: One cycle per 0.001 seconds.

The normal range of frequency of typical diagnostic ultrasound is 2 to 12 Megahertz (MHz or 10^6 cycles per second) so choice A is out since it falls within this range. For the others we must convert them to frequencies on the MHz scale.

Gigahertz (GHz) is 10^9 cycles per second, and 0.01 GHz is equal to 10 MHz so choice E is eliminated.

Frequency is the inverse of period. Period gives the time per cycle. Frequency is the number of cycles per second. So 1 divided by 0.2 microseconds per cycle (or 1/0.2x 10^-6) is the same as 5 x 10^6 or 5 million cycles per second which is 5 MHz. So choice B is out.

3,000 Kilohertz (KHz or 10^3 cycles per second) is the same as 3 MHz so choice C is out.

This leaves choice D. One cycle per second 0.001 seconds is the same as saying there are 1000 cycles in 1 second. This is only 1 KHz and is not in the range of diagnostic ultrasound.

The correct answer is B: Absorption decreases with increasing depth.
Absorption actually increases exponentially with increasing depth so this statement is not correct.

Absorption is defined as the conversion of ultrasound energy into heat as it passes through a medium (Choice D) and this is the main cause of thermal bioeffects (Choice E). It occurs the most in dense tissues like bone and the least in tissues with water (Choice C). It also increases linearly with increasing frequency (Choice A)

The correct answer is E: intensity. Intensity is a better indicator of bioeffects than power (Choice B) or amplitude (Choice A)., Remember than power is proportional to amplitude^2 and is a better measure of bioeffects than amplitude alone.

Intensity is defined as power/beam area and measured in Watts/cm^2. A higher intensity ultrasound beam is more likely to increase the risk of bioeffects. Beam area is therefore not correct either (Choice C) and is inversely proportional to intensity and therefore the risk of bioeffects. Frequency is not directly related to bioeffects so choice E is correct.

The correct answer is C: Voltage is not an appropriate measure of amplitude for an ultrasound transducer. This statement is false. Voltage can be used as a measure of amplitude for a transducer. The higher the voltage the greater the amplitude of the electrical signal from a transducer. In addition amplitude can be used to describe any of the 4 major acoustic variables of sound which are 1) pressure 2) density 3) temperature and 4) particle location so Choice A is incorrect. Power, usually measured in Watts, is proportional to amplitude squared so Choice B is incorrect. A larger amplitude sound wave can improve penetration in some cases so Choice D is incorrect. Because amplitude is related to power and thereby also related to intensity, it relates to the risk of bioeffects, so Choice E is in correct.

The correct answer is A: As the wavelength increases the frequency decreases in a particular medium. Frequency and wavelength are inversely proportional to each other as seen in the equation: propagation velocity = frequency * wavelength.

Propagation velocity is determined by the medium only. For soft tissues we use a value of 1540 meters/second. So as the frequency increases for a given medium, the velocity will not change but rather the wavelength will decrease. Hence choice B is incorrect. For a given frequency bone will have a longer wavelength than soft tissue since the propagation velocity in bone is higher at around 4000 meters/sec. Soft tissue has a lower velocity of 1540 m/s and wavelength is directly proportional to propagation velocity. Hence choice C is incorrect. Finally as we seen in the bone vs soft tissue example, as the stiffness increases for an object the propagation velocity increases causing the wavelength to increase. Hence choice D is incorrect.

The correct answer is C: Pulse repetition period. Having good working knowledge of all of these variables is important for understanding basic principles of ultrasound physics. Doing many questions on these topics should help your understanding.

Let's go through them one by one. Axial resolution is defined as spatial pulse length or SPL divided by 2. Spatial pulse length is just the distance of the all of the cycles in a pulse and is calculated as cycle wavelength x number of cycles in a pulse. Since changing the frequency changes the wavelength (they are inversely related) then SPL is also affected. The bottom line is that axial resolution is affected when we change frequency. Increasing frequency, decreases SPL and decreases axial resolution. Choice A is therefore not correct.

Pulse duration (PD) is the period of one cycle x the number of cycles. Since frequency and period are inversely related, changing frequency definitely changes PD. Choice D is therefore not correct.

Pulse repetition period (PRP) is based only on the depth/and velocity of propagation. Neither of these depend on the frequency. The depth can be set by the user and the velocity of propagation is a property of the medium. Hence choice C is correct.

Duty Factor is the PD/PRP or the fraction of time spent transmitting a pulse. For PW Doppler this number is usually very low. For CW Doppler this number if 1 (always transmitting). Since Duty Factor depends on PD and PD depends on frequency, this Duty Factor also depends on frequency and so choice B is not correct.

The correct answer is A: Bone. Bone has a much higher absorption rate than other tissues as seen in the other answer choices, and therefore bone will have the greatest temperature rise. Ultrasound attenuation is so significant with bone, this is one of the reasons we try to avoid it when imaging the heart.

The correct answer is D: Changes in polarization. This question involves knowledge of the so-called acoustic variables of sound: properties which we can measure as a result of sound interacting with a medium. There are 4 acoustic variables: pressure, temperature, density and particle location. Polarization is a property of an electromagnetic wave, and is not an acoustic variable. It refers to the orientation of the electrical field an electromagnetic wave.

The correct answer is D. You need to know the Doppler shift equation to answer this question as shown.
You also need to have a rough sense of the difference in ultrasound propagation velocity for different media with blood being the fastest and lung tissue (mostly air) being the slowest and blood being in the middle.

Changing the angle by only 20 degrees will only change the shift by about 6 percent (Cosine of 0 is 1 and cosine of 20 is 0.94) Changing the velocity is a 2 fold difference and changing the transducer frequency also exactly 2 fold. However since the speed of ultrasound in lungs is only 500 m/s and in blood it is roughly 1500 m/s this is a 3 fold change. It will impact the Doppler shift the most and hence is the correct answer.

The correct answer is B: M mode. This form of ultrasound has the highest temporal resolution since there is a just one scan line along which information is gathered. An easy way to conceptualize temporal resolution is as frame rate. How many frames per second are being generated. For M mode this can be up to 1800 frames/second .For 2 D Doppler (Choice C) the ultrasound machine has to scan over a plan, one scan line at a time to generate an image frame; this takes time so the temporal resolution is not as good and may be on the order of 30-75 frames/second. By the same principals, 3D Doppler (Choice D) is worse since the machine has to scan in two different plane orthogonal to each other. Finally color Doppler (Choice A) also has poor temporal resolution since along each line of scanning, the ultrasound machine has to send a whole packet of scan lines to get information needed for color Doppler. Thus temporal resolution is worse than 2D Doppler.

The correct answer is A: Color Doppler > 2 D > PW Doppler > M mode

Of these modalities, color Doppler has the lowest temporal resolution. The ultrasound machine sends out a whole packet of acoustic lines across each scan line to generate the color information as well the as the 2D image. 2D requires multiple scan lines and will take longer than a modality with just one scan line. PW Doppler requires time to listen for a signals before sending out another pulse but is along only one scan line. M mode has the highest temporal resolution as it is displaying all data along one scan line over time

The correct answer is C: CW has a maximal detectable velocity of about 1.5 m/s.

PW has a maximal detectable velocity where it aliases, but CW does not have maximal velocity detection per se. With CW there is constant transmission and receiving of signals (Choice B). The one downside is range ambiguity; you cannot pinpoint the location of the highest velocity along the line of interrogation as you can with PW (Choice A). The Pedoff probe uses only CW (Choice D).

The correct answer is D: Bone. Bone has the highest propagation velocity as shown in the table below.

Propagation velocity is defined as how fast a sound wave can travel through a medium. It's calculation is not straightforward but depends on both the stiffness and the density of the medium. The higher the stiffness the faster the velocity. The more dense the medium the lower the velocity. However the catch is that many materials which are quite dense are also very stiff and usually the stiffness factor (called bulk modulus) wins out.

The actual equation is

velocity = square root (bulk modulus/density)

Hence bone (which is denser but also much stiffer than air) has the highest propagation velocity. Air has the lowest propagation velocity since it is not stiff at all. Hence lung tissue which has a lot of air has a low propagation velocity as well. Fat has a slightly lower propagation velocity than blood.

In general, we use a soft tissue average of 1540m/sec for ultrasound. This is a handy number of keep in mind for other ultrasound physics questions.

The correct answer is C: Increases by a factor of 9. Power is proportional to the amplitude squared. Voltage is a measure of amplitude for transducers. Therefore if the voltage is increase from 3 to 9 or by a multiple of 3, then the increase in power is by a multiple of 3^2 or 9. None of the other answer choices are accurate and there is enough information given to answer the question.

The correct answer is D. Most techniques for focusing a beam make the focal point shallower. These include use of a lens, curving the surface of the transducer and using phasing, which is an electronic technique where electrical impulses are sent out at different times to create a beam.

See the figure. Recall the natural focal length (also called the near zone length) is D^2*frequency/6. Therefore at a given frequency, the only thing which will change the focus past it's natural focus point is to change the crystal diameter - in effect creating a new, longer natural focus point. This is not trivial and would require a new transducer set up.

The correct answer is A: Increases by a factor of 2.

This question requires you to know that for pulsed wave Doppler the frequency depends on both the thickness of the crystal and the velocity of propagation of sound within the crystal.

This is given by the equation: frequency = (propagation velocity in the crystal)/(2x crystal thickness).

Thus if we cut the thickness of the crystal in half then the frequency will increase 2 fold. None of the other answers are correct.

Note that for continuous wave (CW) Doppler the frequency depends only on the transmit voltage since the crystal is constantly being stimulated and is always "on" emitting ultrasound. Thus it is not dependent on crystal thickness. Hence choice E would be correct if this were CW Doppler.

The correct answer is D: There are three main types: specular reflection, Rayleigh scattering and refraction. Refraction is not a type of reflection, it is an entirely different process where by differences in the velocity of travel within a medium changes the direction of propagation at an interface. The three main types are specular, backscattering and Rayleigh scattering. Rayleigh scattering occurs when the wavelength of the object is similar to that of the ultrasound. (Choice D) Specular reflection occurs when the object has a much larger wavelength than the ultrasound. The angle of incidence equals the angle of reflection and a mirror image is created. Good specular reflectors in the body include bones and tendons or large calcifications. Backscattering occurs when the a boundary has irregularities that are about the same size and the ultrasound wavelength.

The correct answer is D: Destructive interference is ideal at the surface of the matching layer so that the reflection of ultrasound that occurs at the interface of the match layer does not become part of image generated.

For this to occur the matching layer should be 1/4 a wavelength since the trip through the layer would be 1/4 a wavelength or 90 degrees and the trip back would be another 90 degrees for a total of 180 degrees. Thus the wave back to the crystal will be 180 out of phase with wave being sent out from the crystal, causing destructive interference and cancelling out these reflected waves from the matching layer as shown in the figure below.

The matching layer should have an impedance that is intermediate between that of the crystal and tissue to allow for more transmission and less reflection. Remember the higher the acoustic impedance the more the reflection. The whole point of the matching layer is to increase transmission to the tissue.

The correct answer is A: It estimates how much damping is occurring. Quality factor is defined as the resonant frequency/bandwidth and is unitless. Imaging transducers typically have relatively low quality factors (compared to other types of transducers such as therapeutic transducers) with a large bandwidth. This is a result of the backing material which causes damping. Quality factor is unrelated to duty factor which is a percent of how much time the transducer is sending a pulse (transmission time/listening time)

The correct answer is C: The wavelength is determined only by the properties of the medium through which the ultrasound travels. Wavelength is actually determined by both the speed at which the ultrasound travels within the medium (c) as well as the frequency of transmission from the source. In fact c = f * wavelength.

All of the other choices are correct. The type of scattering at a surface depends on the wavelength (specular, Rayleigh scattering for example) so E is correct. The typical range for diagnostic ultrasound is 2 MHz to 20 MHz. This corresponds to wavelengths of 77-770 micrometers (assuming a speeds of 1540m/s in the medium) so B is correct. Axial resolution is in part defined by wavelength so choice A is correct. And choice D defines wavelength; it is a distance of one cycle which is from a peak to peak or from valley to valley.

The correct answer is B: Particles vibrate perpendicular to the direction of the wave propagation. Sound is a mechanical wave, meaning by definition it has to have a physical interaction with a medium in order to propagate (choice D). This is as opposed to an electromagnetic wave (choice C) where a medium is not necessary an propagation can even occur in a vacuum.

Mechanical waves can be either longitudinal or transverse. Sound is a LONGITUDINAL wave, meaning that particles vibrate back and forth PARALLEL to the direction of propagation. This is shown in the figure below [echo.physics.longitudinal wave.jpg]. The areas of higher densities of particles are called compressions and the areas of lower densities are called rarefactions (choice A).

In contrast a TRANSVERSE, wave the particles move perpendicular to the direction of the propagation (choice B), as shown in figure.

The correct answer is C: The effect involves transformation of electrical energy to mechanical energy.

By definition a substance that exhibits the piezoelectric effect is one that can convert electrical energy to mechanical energy (ultrasound) and can also convert mechanical energy back into electric energy. Therefore by definition it is bidirectional. Naturally occurring substances have poor efficiency and can be modified to increase sensitivity and bandwidth, using high temperatures and magnetic fields to align the crystal lattice in a process called "poling". Quartz exhibits the piezoelectric effect, but not very efficiently.

The correct answer is A: It is the best determinant of the likelihood of cavitation. The mechanical index (MI) is in fact the best indicator of cavitation. It is related to the peak rarefactional pressure divided by the transducer frequency, so it is inversely proportional to the frequency, making Choice B incorrect.

Cavitation is more likely to occur with a higher MI so Choice C is not correct. Typically the MI has to be reduced during contrast studies, since with contrast agents there is significant impedance mismatch between blood and the contrast agent, a lower MI transducer can produce significant signal. A high MI can cause bubble destruction and cavitation.

The correct answer is D: Commercial contrast bubbles of the size needed to pass through the pulmonary capillary bed do not naturally resonate in the range of frequency used for diagnostic ultrasound. In fact, fortunately the natural resonant frequency of the bubbles at this size is exactly in the range picked up by diagnostic ultrasound (2 to 20 MHz). The typical contrast bubble size is 1-5 microns.

All of the other answers are correct.

The correct answer is C: the backing material. This is designed to cause dampening which decreases the number of cycles in the pulse that occur naturally from the crystal. This in effect decreases the pulse duration (time of the pulse which is time per cycle x number of cycles in a pulse) and leads to an increase in bandwidth. It also improves axial resolution by decreasing spatial pulse length (length of a pulse which is length of a cycle x number of cycles in a pulse).

The purposes of the other parts, shown in figure below [insert US transducer components figure}are as follows
case insulation - insulates from outside electrical interference
lens - focuses the beam
piezoelectric crystal - converts the electrical energy applied into sound energy to transmit to the patient
matching layer - minimizes the acoustic impedance mismatch between the crystal and the human body to allow more ultrasound waves to be transmitted

The correct answer is B. Phase delay is a technique where by changing in timing of the electrical signal to each crystal in the transducer can make the ultrasound beam change direction and sweep from side to side or change the focus [insert transducer phase delay figure]. It was a big technological advance achieved with small crystals that removed the need for moving parts in a transducer. The sweeping from side to side is what creates a 2D echo image. Now one limitation is that the focus can only be made lower than the natural focus but not deeper than the natural focus on the probe. Given all this information only B is is not true and therefore the correct response. All of the other answers are true.

The correct answer is A. Lateral resolution is equal to the lateral beam width. The smaller the diameter of the beam width, the higher the resolution. Having a high resolution means being able to resolve objects that are close to each other. [insert lateral resolution illustration figure]. If two objects are in the same beam, they cannot be resolved as two objects; they will appear as one object. Thus a narrower beam will result in better lateral resolution.

All of the other choices increase the beam width or are unrelated. Spatial pulse length is related to axial resolution, not lateral resolution. Decreasing the frequency will increase the beam width and result in worse lateral resolution. Increasing the crystal diameter will increase the beam width as well. Frame rate relates to temporal resolution, not lateral resolution.

The correct answer is C. To answer this question you need to know the Doppler shift equation (see figure). Because this question asks about the magnitude only and not direction (positive or negative change in frequency) we do not care about whether the object is moving toward (increase in frequency, Doppler shift is positive) or away (decrease in frequency, Doppler shift is negative) from the observer. We only care about the absolute change.

From the equation we can see that the Doppler shift is directly proportional to the object velocity, and emitting frequency and the cosine of the angle in line from the observer. Hence the combination of these that is the greatest is the answer. B has a higher velocity than A, so A can be eliminated. D has the same velocity and frequency as B but is at a 10 degree angle. We know that the cosine of anything other than 0 degrees will be less than 1 so D can be eliminated. Similarly E can be eliminated since it's shift will similar to D as the velocity is only 2.5 m/s but the frequency is doubled at 10 MHz. Between B and C then we must ask ourselves is a doubling of the velocity from 5m/s in B to 10m/s in C enough to compensate being off by 10 degrees in angle. Suffice to say that we are commonly off by 10 degrees and it doesn't effect the result by 2 fold. In point of fact the cosine of 10 degrees is 0.985 so it is only 1.5% less. Hence answer C will have the greatest magnitude of Doppler Shift.

The correct answer is C: Amplitude. All of the other parameters are determined by the ultrasound source: period, frequency and wavelength or by the medium: propagation velocity. Period and wavelength are inverses of each other, hence these could have easily been eliminated without much other knowledge since if one was adjustable then naturally the other would have to be also. Propagation velocity is dependent on the medium and can also be eliminated. Finally wavelength is related to propagation velocity and frequency only by the equation c (propagation velocity) = frequency x wavelength, and therefore also can be eliminated. This leaves only amplitude, which is expressed in decibels (dB) and is commonly modified by the ultrasonographer.

The correct answer is Choice E: None of the above. This is because all of the above parameters apply to PW Doppler. As shown below in the figure [insert Advanced wave parameters], Pulse Duration (Choice A) is the time for the whole pulse (number of cycles x the period, or time for each cycle). Spatial Pulse Length (Choice B) is the distance of the pulse (number of cycles x the wavelength, or distance for each cycle. Pulse Repetition Period (Choice D) is the time between pulses. Duty Factor is the amount of time the transducer spends sending a pulse and can be calculated by the Pulse Duration over the Pulse Repetition Period. This is typically a very low number for PW Doppler whereas for CW Doppler it is 1 by definition since the transducer is constantly sending out ultrasound waves.

The correct answer is E: B mode Doppler. This is a good question to review the different ultrasound modalities and know which ones are scanned and non scanned.

A scanned modality means the ultrasound transducer scans across a 2 dimensional plane to collect data and create a 2 dimension image. Older transducers used to have mechanically moving parts to move a set of crystals in this fashion. Modern transducers use electrical scanning techniques and do not have moving parts.

PW and CW Doppler are non-scanned since they only interrogate velocities along a line that is chosen by the user. Information is given only as a set of velocities seen along that line (Y axis) across time (X axis).

M mode is similar in that it gives information only along a line of interrogation, however it given us distances (Y axis) across time (X axis)

A mode ultrasound is not really used clinically. It uses only one line of interrogation like M mode but plots ultrasound intensity (Y axis) across time (X axis)

B mode ultrasound is a scanned modality that gives us 2 D images as the ultrasound beam scans across a 2 D plane. It was the major breakthrough that allowed for more practical application of ultrasound for diagnostic purposes.

The correct answer is D: Backing material. The backing material is used to decrease ringing and decreases spatial pulse length improving axial resolution. It does not change image amplitude.

Transmit voltage determines how much power the ultrasound has as it exits the transducer and goes to the patient. TGC gains,or transducer receiver gains are settings on the machine the user can alter to change the amplification of the signal the machine receives, with a goal to modify the brightnes of the image displayed on the screen.

The correct answer is B: penetration is worse compared to fundamental imaging.

Harmonic imaging means transmitting at a fundamental frequency and receiving signals at some multiple of that fundamental frequency, usually at the 2nd harmonic which corresponds to twice the fundamental frequency. The harmonic response is non linear with increasing intensity and varies significantly with imaging depth. The main advantage of harmonic imaging is to decrease clutter signals and also imaging artifacts. Ideally there should be little overlap between the fundamental (trasmitting) and harmonic (receiving) bandwidths so that the quality of the harmonic image is not degraded.
Because the harmonic frequency is by definition higher than the fundamental frequency and because at greater imaging depths, there is increased rate of attenuation for hamornic signals, penetration is worse with harmonic imaging.

The correct answer is A. The beam will be refracted away from a perpendicular line to the interface between the media. Since medium 2 has a faster propagation velocity for ultrasound it will be refracted away from the midline as shown in the figure below [insert refraction low to high velocity medium]. This is based on Snells' law that relates the sine of the angle of incidence and refraction to the velocity of travel through the two mediums. The higher the velocity the greater the angle of refraction. The other choices are therefore not correct.

The correct answer is A. In fact the Doppler shift is DIRECTLY proportional to the cosine of the angle of insonification and not inversely proportional. This question requires knowledge of the Doppler equation shown below [insert Doppler shift picture and equation]. As we can see the amount of shift is directly proportional to velocity of the object (Choice D is therefore incorrect), the frequency of the ultrasound transducer (Choice B is incorrect) as well as the cosine of the angle between the beam and moving object. The cosine of 0 degrees is 1 so at an angle of 0 degrees, we have the highest Doppler shift. Speed of ultrasound in a medium is in the denominator and so faster travel of the beam through a medium will mean less Doppler shift (Choice C is incorrect)

The correct answer is D: The axial resolution using harmonic imaging is not as good. This is because the transmit pulse uses a narrow bandwidth during harmonic imaging so as to decrease the overlap between fundamental (transmitting) and harmonic (receiving) bandwidths. This is done by putting more cycles within each pulse and thereby increasing the spatial pulse length (number of pulses x length of each pulse). Since axial resolution is spatial pulse length/2 it increases also and hence axial resolution is worse (larger number is worse, smaller is better). Harmonic imaging is used to decrease artifacts so C is not correct. It does not change the frame rate appreciably and it does not change the amount of reflection, neither of which would necessarily cause a valve leaflet to appear thicker in either case. Harmonic imaging does not cause more reverberations.

The correct answer is C: The percent reflection between two surfaces can be calculated using only their acoustic impedances.

Acoustic impedance is a property of a substance that is related to proportional to BOTH the density and velocity of sound propagation through it (B is therefore incorrect). The higher the density of the object the higher the acoustic impedance. Thus we would expect that blood would have a lower acoustic impedance than tissue (Choice D is therefore incorrect)

Acoustic impedance is commonly abbreviated by the letter Z and is measured in the units of Rayls. It is a key factor that helps us understand how much of an ultrasound source will be reflected back vs transmitted through an object. The larger the impedance mismatch between two surfaces at an interface, the LESS ultrasound is transmitted and the MORE is reflected back (choice A is therefore incorrect). See the figure below. [Insert percent reflection and acoustic impedance]
The equation to calculate the percent reflection at an interface is based solely on the acoustic impedances of the two surfaces is shown in the figure as well ( from surface Z1 to surface Z2) The concept is important because piezoelectric crystals have a high acoustic impedance compared to the human body. Hence matching layers are used in transducers to make the final interface of the probe to decrease this difference. This is also why we use ultrasound gel. It helps increase the amount of transmission by decreasing the difference in acoustic impedance at the surface. This is also how contrast bubbles work. They have a higher acoustic impedance than blood cells and act as many small tiny reflectors of ultrasound to help in visualization.

Hence choice C is correct.

The correct answer is B: Excessive energy can cause formation of gas bubbles in the medium. There are two major forms of bioeffects: thermal and mechanical (Choice A is not correct). Thermal effects relate to increased temperature in a medium from the interaction of a sound wave with the medium. This temperature increase is usually on the order of several degrees Celsius. High temperatures can be bad for the local cells and tissues.

Mechanical bioeffects include cavitation, where excess energy can cause small gas bubble formation in the medium (Choice B). Cavitation occurs usually during the sound wave's peak rarefactional pressure and not during compression (Choice C is not correct). The bubbles can be either stable or transient (Choice E is not correct). In transient cavitation there is formation and then collapse and implosion of the bubble with localized temperatures that can be very high (up to 10,000 Kelvin!) (Choice D is not correct). Stable cavitation causes bubbles which can be long lived.

The correction answer is C: Refraction is not a form of attenuation. Both refraction, reflection (Choice B) and also absorption are all forms of attenuation.

Because reflection is fundamental to diagnostic ultrasound, and it is form of attenuation, modern diagnostic ultrasound does indeed rely in part of attenuation, making Choice A incorrect. Choice D basically defines attenuation and is true, so it is not the correct answer.

The correct answer is C, turning on multi-focus. The ability to track the motion of an object over time is governed by temporal resolution or frame rate. Frame rate is 1/(time to scan one frame). Any measure that increases the time to scan one frame will reduce the frame rate and thus the temporal resolution. With multi-focus, to scan each line requires two pulses to different depths, so it will take longer to scan one frame. Similarly, scanning a wider and deeper sector takes more time and lowers frame rate. Lastly, the more dense the sector with lines to be scanned, the longer scanning one frame takes and thus the lower the frame rate.

The correct answer is B, 5KHz.

This question tests your conceptual knowledge about the limits of Pulsed wave Doppler. Since the machine sends out pulses and listens inbetween, it can misjudge events in time and space if listening time is too long. There is therefore a maximum upper limit at which PW Doppler can detect a Doppler shift, called the Nyquist limit.

It is equal to half of the pulse repetition frequency (PRF), which is the frequency of transmitting pulses (how often the machine sends out a pulse). PRF is the reciprocal of the time period between pulses, known as the pulse repetition period (PRP). Recall that period and frequency are reciprocals.

Since we are given a time between pulses of 0.1ms (PRP) the PRF is 1/0.1 or 10KHz. Since the Nyquist limit is PRF/2, the answer is 5KHz, the maximum Doppler shift we can detect. Anything above this will cause aliasing.

The correct answer is B, 7 KHz. To answer this question you need to simply know the equation for the Nyquist limit.

The Doppler frequency above which there is aliasing is called the Nyquist limit. We are used to thinking about this in term of velocited on the ultrasound machine, but in reality the limit is a frequency shift, which is then translated by the machine into a velocity. The Nyquist limit is equal to half the pulse repetition frequency (PRF) which is how often the machine sends out a pulse. We are given a PRF of 14 KHz, hence 14/2 is 7 KHz

The correct answer is D:PZT crystal>matching layer> ultrasound gel>skin.

The PZT crystal is very dense and has the highest impedance followed by the matching layer which is an intermediary layer that has an impedance inbetween that of gel and the crystal so as increase the amount of transmission that occurs. Gel has the same function and has an impedance in between that of the matching layer and the skin in order to increase ultrasound transmission/penetrance to the body. This is because the larger the difference in impedance between two surfaces, the more signal that will reflect and the less that will penetrate. Gel and the matching layer aim to decrease these large inherent differences in impedances.

The correct answer is D: Shift the baseline. Aliasing occurs when the Pulsed Wave Doppler cannot resolve the Doppler frequencies it receives, because they are above the Nyquist limit. This is equal to the pulse repetition frequency (PRF), which is how often the transducer sends out a pulse, divided by two. Events occurs much faster in time that this are not resolvable and cause aliasing, as shown in the figure below [figure.PW.aliasing]. Note that signal is both above and below the baseline in a haphazard fashion.

All of the techniques listed above can minimize aliasing except for shifting the baseline which will not remove true aliasing. CW Doppler (Choice A) is not really subject to aliasing and can resolving high velocity shifts. A lower frequency transducer (Choice B) will shrink the available spectrum and cause less aliasing as higher frequency transducers will cause more Doppler shift (recall that doppler shift is proportional to transducer frequency) and more aliasing. Chaning the avaliable scale on the machine (Choice C) to increase the Nyquist limit should decrease aliasing as will changing the angle of insonification. If you angle away from a higher velocity jet (Choice E) and are off axis, there will naturally be less aliasing.

The correct answer is C: There is increased temporal resolution.

Color power Doppler is a modality that measures ampltiude of Doppler shift only and not direction or velocity. There is thus only one color for flow and you cannot tell which way it is going using this modality alone. As such it has increased ability to pick up low flows that might have multiple directions (like ASD flow). It also is not affect much by large changes in angle as it does not care about velocity information.

For the same reason, Color Power Doppler does not alias since it is not using velocity information. It is more succeptible to motion artifact however and actually has lower frames rates than conventional Color Doppler, meaning decreased temporal resolution.

The correct answer is B: backscatter.
Backscatter occurs when the wavelength of the ultrasound is similar to irregularities in the boundary. It results in scattering in many different directions but all backward and usually therefore toward the transducer. This form therefore is the most useful for generating an image.

Rayleigh scattering occurs when the wavelength of the reflecting object is much smaller than the wavelength of ultrasound sending waves in all different directions, only some of which get back to the transducer. Specular reflection occurs when the object has a much larger wavelength than the ultrasound. The angle of incidence equals the angle of reflection and a mirror image is created, but this doesn't make up a large portion of ultrasound image creation since the more of the beam that reflects, the less that transmits. Refraction is a different phenomenon altogether where the beam is bent by different in speed of ultrasound through different media.

The correct answer is B: The intensity of the artifact increases with an increasing angle.

Side lobe artifacts occur because not all of the energy from the transducer remains in a single central beam (Choice D) and are 3 dimensional just as the beam is 3 dimensional (Choice A). In fact the intensity DECREASES with increasing angle. The farther out from the central beam, the weaker the side lobe. Artifacts are created since the machine interprets signal returning from the side lobes as coming from the central beam. These signals are usually much weaker than those from coming back from the main beam unless there is a strong reflector (Choice C), in which case they are called dominant side lobes.

The correct answer is D: Better at getting on axis dimension measurement of the LV. M mode has higher temporal resolution (1000-3000 Hz compared to up to 60 Hz for 2D echocardiography) and higher or equalent spatial resolution. As such it is better able to tell you about fine movments in flutter valves, such as the diastolic mitral valve fluttering seen in aortic insufficiency. One pitfall of M mode measurement is that they may not be on axis, depending on how the probe is aimed at the LV in the parasternal long axis. This is why 2 D measurements sometimes may be more reliable.

The correct answer is A: Filters exclude high velocities and low intensity reflectors

Tissue Doppler imaging is commonly performed when assessing diastolic function. In this setting, the goal to assess the movment of the mitral annulus in diastole.

The ultrasound machine filters needed for tissue Doppler imaging are opposite that of standard spectral Doppler imaging. Standard spectral Doppler (Pulsed Wave and Continuous Wave Doppler) imaging focuses on blow flow which has relatively high velocities ( 1-5 meters/second) compared to tissue (5-20 cm/second). Thus filters are needed to exclude high velocities (large Doppler shift) signals returning to the probe. Red blood cells are also very weak reflectors as opposed to tissue which is a very strong reflector. Thus during spectral Doppler, filters are set to exclude high intensity reflectors. With Tissue Doppler you want to instead exclude these low intensity reflectors to focus on the tissue. This is why one some machines because of the filters, the gain needs to be turned down significantly to get good images for tissue Doppler. Hence only answer A is correct as with Tissue Doppler the filters exclude high velocity signals and low intensity reflectors.