Echo Physics Questions

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 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 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: 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.