Ultrasound, like any wave, carries energy that can be absorbed by the medium carrying it, producing effects that vary with intensity. When focused to intensities of 103103 to 105105 W/m2W/m2, ultrasound can be used to shatter gallstones or pulverize cancerous tissue in surgical procedures. (See Figure 2.) Intensities this great can damage individual cells, variously causing their protoplasm to stream inside them, altering their permeability, or rupturing their walls through cavitation. Cavitation is the creation of vapor cavities in a fluid—the longitudinal vibrations in ultrasound alternatively compress and expand the medium, and at sufficient amplitudes the expansion separates molecules. Most cavitation damage is done when the cavities collapse, producing even greater shock pressures.

Figure 2: The tip of this small probe oscillates at 23 kHz with such a large amplitude that it pulverizes tissue on contact. The debris is then aspirated. The speed of the tip may exceed the speed of sound in tissue, thus creating shock waves and cavitation, rather than a smooth simple harmonic oscillator–type wave.

Most of the energy carried by high-intensity ultrasound in tissue is converted to thermal energy. In fact, intensities of 103103 size 12{"10" rSup { size 8{3} } } {} to 104W/m2104W/m2 size 12{"10" rSup { size 8{4} } "W/m" rSup { size 8{2} } } {} are commonly used for deep-heat treatments called ultrasound diathermy. Frequencies of 0.8 to 1 MHz are typical. In both athletics and physical therapy, ultrasound diathermy is most often applied to injured or overworked muscles to relieve pain and improve flexibility. Skill is needed by the therapist to avoid “bone burns” and other tissue damage caused by overheating and cavitation, sometimes made worse by reflection and focusing of the ultrasound by joint and bone tissue.

In some instances, you may encounter a different decibel scale, called the sound pressure level, when ultrasound travels in water or in human and other biological tissues. We shall not use the scale here, but it is notable that numbers for sound pressure levels range 60 to 70 dB higher than you would quote for ββ size 12{β} {}, the sound intensity level used in this text. Should you encounter a sound pressure level of 220 decibels, then, it is not an astronomically high intensity, but equivalent to about 155 dB—high enough to destroy tissue, but not as unreasonably high as it might seem at first.

When used for imaging, ultrasonic waves are emitted from a transducer, a crystal exhibiting the piezoelectric effect (the expansion and contraction of a substance when a voltage is applied across it, causing a vibration of the crystal). These high-frequency vibrations are transmitted into any tissue in contact with the transducer. Similarly, if a pressure is applied to the crystal (in the form of a wave reflected off tissue layers), a voltage is produced which can be recorded. The crystal therefore acts as both a transmitter and a receiver of sound. Ultrasound is also partially absorbed by tissue on its path, both on its journey away from the transducer and on its return journey. From the time between when the original signal is sent and when the reflections from various boundaries between media are received, (as well as a measure of the intensity loss of the signal), the nature and position of each boundary between tissues and organs may be deduced.

Reflections at boundaries between two different media occur because of differences in a characteristic known as the acoustic impedance ZZ size 12{Z} {} of each substance. Impedance is defined as

where ρρ is the density of the medium (in kg/m3kg/m3) and vv size 12{v} {} is the speed of sound through the medium (in m/s). The units for ZZ are therefore kg/(m2·s)kg/(m2·s).

Table 1 shows the density and speed of sound through various media (including various soft tissues) and the associated acoustic impedances. Note that the acoustic impedances for soft tissue do not vary much but that there is a big difference between the acoustic impedance of soft tissue and air and also between soft tissue and bone.

At the boundary between media of different acoustic impedances, some of the wave energy is reflected and some is transmitted. The greater the difference in acoustic impedance between the two media, the greater the reflection and the smaller the transmission.

The intensity reflection coefficient aa size 12{a} {} is defined as the ratio of the intensity of the reflected wave relative to the incident (transmitted) wave. This statement can be written mathematically as

where Z1Z1 size 12{Z rSub { size 8{1} } } {} and Z2Z2 size 12{Z rSub { size 8{2} } } {} are the acoustic impedances of the two media making up the boundary. A reflection coefficient of zero (corresponding to total transmission and no reflection) occurs when the acoustic impedances of the two media are the same. An impedance “match” (no reflection) provides an efficient coupling of sound energy from one medium to another. The image formed in an ultrasound is made by tracking reflections (as shown in Figure 3) and mapping the intensity of the reflected sound waves in a two-dimensional plane.

The applications of ultrasound in medical diagnostics have produced untold benefits with no known risks. Diagnostic intensities are too low (about 10−2W/m210−2W/m2 size 12{"10" rSup { size 8{ - 2} } "W/m" rSup { size 8{2} } } {}) to cause thermal damage. More significantly, ultrasound has been in use for several decades and detailed follow-up studies do not show evidence of ill effects, quite unlike the case for x-rays.

Figure 3: (a) An ultrasound speaker doubles as a microphone. Brief bleeps are broadcast, and echoes are recorded from various depths. (b) Graph of echo intensity versus time. The time for echoes to return is directly proportional to the distance of the reflector, yielding this information noninvasively.

The most common ultrasound applications produce an image like that shown in Figure 4. The speaker-microphone broadcasts a directional beam, sweeping the beam across the area of interest. This is accomplished by having multiple ultrasound sources in the probe’s head, which are phased to interfere constructively in a given, adjustable direction. Echoes are measured as a function of position as well as depth. A computer constructs an image that reveals the shape and density of internal structures.

Figure 4: (a) An ultrasonic image is produced by sweeping the ultrasonic beam across the area of interest, in this case the woman’s abdomen. Data are recorded and analyzed in a computer, providing a two-dimensional image. (b) Ultrasound image of 12-week-old fetus. (credit: Margaret W. Carruthers, Flickr)

How much detail can ultrasound reveal? The image in Figure 4 is typical of low-cost systems, but that in Figure 5 shows the remarkable detail possible with more advanced systems, including 3D imaging. Ultrasound today is commonly used in prenatal care. Such imaging can be used to see if the fetus is developing at a normal rate, and help in the determination of serious problems early in the pregnancy. Ultrasound is also in wide use to image the chambers of the heart and the flow of blood within the beating heart, using the Doppler effect (echocardiology).

Whenever a wave is used as a probe, it is very difficult to detect details smaller than its wavelength λλ size 12{λ} {}. Indeed, current technology cannot do quite this well. Abdominal scans may use a 7-MHz frequency, and the speed of sound in tissue is about 1540 m/s—so the wavelength limit to detail would be λ=vwf=1540 m/s7×106Hz=0.22 mmλ=vwf=1540 m/s7×106Hz=0.22 mm size 12{λ= { {v rSub { size 8{w} } } over {f} } = { {"1540""m/s"} over {7 times "10" rSup { size 8{6} } "Hz"} } =0 "." "22""mm"} {}. In practice, 1-mm detail is attainable, which is sufficient for many purposes. Higher-frequency ultrasound would allow greater detail, but it does not penetrate as well as lower frequencies do. The accepted rule of thumb is that you can effectively scan to a depth of about 500λ500λ size 12{λ} {} into tissue. For 7 MHz, this penetration limit is 500×0.22 mm500×0.22 mm, which is 0.11 m. Higher frequencies may be employed in smaller organs, such as the eye, but are not practical for looking deep into the body.

Figure 5: A 3D ultrasound image of a fetus. As well as for the detection of any abnormalities, such scans have also been shown to be useful for strengthening the emotional bonding between parents and their unborn child. (credit: Jennie Cu, Wikimedia Commons)

In addition to shape information, ultrasonic scans can produce density information superior to that found in X-rays, because the intensity of a reflected sound is related to changes in density. Sound is most strongly reflected at places where density changes are greatest.

Another major use of ultrasound in medical diagnostics is to detect motion and determine velocity through the Doppler shift of an echo, known as Doppler-shifted ultrasound. This technique is used to monitor fetal heartbeat, measure blood velocity, and detect occlusions in blood vessels, for example. (See Figure 6.) The magnitude of the Doppler shift in an echo is directly proportional to the velocity of whatever reflects the sound. Because an echo is involved, there is actually a double shift. The first occurs because the reflector (say a fetal heart) is a moving observer and receives a Doppler-shifted frequency. The reflector then acts as a moving source, producing a second Doppler shift.

Figure 6: This Doppler-shifted ultrasonic image of a partially occluded artery uses color to indicate velocity. The highest velocities are in red, while the lowest are blue. The blood must move faster through the constriction to carry the same flow. (credit: Arning C, Grzyska U, Wikimedia Commons)

A clever technique is used to measure the Doppler shift in an echo. The frequency of the echoed sound is superimposed on the broadcast frequency, producing beats. The beat frequency is FB=∣f1−f2∣FB=∣f1−f2∣ size 12{F rSub { size 8{b} } = \lline f rSub { size 8{1} } - f rSub { size 8{2} } \rline } {}, and so it is directly proportional to the Doppler shift (f1−f2f1−f2 size 12{f rSub { size 8{1} } - f rSub { size 8{2} } } {}) and hence, the reflector’s velocity. The advantage in this technique is that the Doppler shift is small (because the reflector’s velocity is small), so that great accuracy would be needed to measure the shift directly. But measuring the beat frequency is easy, and it is not affected if the broadcast frequency varies somewhat. Furthermore, the beat frequency is in the audible range and can be amplified for audio feedback to the medical observer.

Example 2: Calculate Velocity of Blood: Doppler-Shifted Ultrasound

Ultrasound that has a frequency of 2.50 MHz is sent toward blood in an artery that is moving toward the source at 20.0 cm/s, as illustrated in Figure 7. Use the speed of sound in human tissue as 1540 m/s. (Assume that the frequency of 2.50 MHz is accurate to seven significant figures.)

1. What frequency does the blood receive?
2. What frequency returns to the source?
3. What beat frequency is produced if the source and returning frequencies are mixed?

Figure 7: Ultrasound is partly reflected by blood cells and plasma back toward the speaker-microphone. Because the cells are moving, two Doppler shifts are produced—one for blood as a moving observer, and the other for the reflected sound coming from a moving source. The magnitude of the shift is directly proportional to blood velocity.

Strategy

The first two questions can be answered using fobs=fsvwvw±vsfobs=fsvwvw±vs size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{s} } } } right )} {} and fobs=fsvw±vobsvwfobs=fsvw±vobsvw size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } +- v rSub { size 8{"obs"} } } over {v rSub { size 8{w} } } } right )} {} for the Doppler shift. The last question asks for beat frequency, which is the difference between the original and returning frequencies.

Solution for (a)

(1) Identify knowns:

• The blood is a moving observer, and so the frequency it receives is given by
  fobs=fsvw±vobsvw.fobs=fsvw±vobsvw. size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } +- v rSub { size 8{"obs"} } } over {v rSub { size 8{w} } } } right )} {}

  (6)
• vbvb size 12{v rSub { size 8{b} } } {} is the blood velocity (vobsvobs size 12{v rSub { size 8{"obs"} } } {} here) and the plus sign is chosen because the motion is toward the source.

(2) Enter the given values into the equation.

f obs = 2, 500 , 000 Hz 1540 m/s + 0 . 2 m/s 1540 m/s f obs = 2, 500 , 000 Hz 1540 m/s + 0 . 2 m/s 1540 m/s size 12{f rSub { size 8{"obs"} } = left (2,"500","000"" Hz" right ) left ( { {"1540"" m/s"+0 "." "2 m/s"} over {"1540 m/s"} } right )} {}

(7)

(3) Calculate to find the frequency: 20,500,325 Hz.

Solution for (b)

(1) Identify knowns:

• The blood acts as a moving source.
• The microphone acts as a stationary observer.
• The frequency leaving the blood is 2,500,325 Hz, but it is shifted upward as given by
  fobs=fsvwvw–vb.fobs=fsvwvw–vb. size 12{f rSub { size 8{"obs"} } =f rSub { size 8{s} } left ( { {v rSub { size 8{w} } } over {v rSub { size 8{w} } +- v rSub { size 8{b} } } } right )} {}

  (8)

  fobsfobs is the frequency received by the speaker-microphone.

• The source velocity is vbvb size 12{v rSub { size 8{b} } } {}.
• The minus sign is used because the motion is toward the observer.

The minus sign is used because the motion is toward the observer.

(2) Enter the given values into the equation:

f obs = 2, 500 , 325 Hz 1540 m/s 1540 m/s − 0 . 200 m/s f obs = 2, 500 , 325 Hz 1540 m/s 1540 m/s − 0 . 200 m/s size 12{f rSub { size 8{"obs"} } = left (2,"500","325"" Hz" right ) left ( { {"1540"" m/s"} over {"1540 m/s " - 0 "." "200"" m/s"} } right )} {}

(9)

(3) Calculate to find the frequency returning to the source: 2,500,649 Hz.

Solution for (c)

(1) Identify knowns:

• The beat frequency is simply the absolute value of the difference between fsfs size 12{f rSub { size 8{s} } } {} and fobsfobs size 12{f rSub { size 8{"obs"} } } {}, as stated in:
  fB=∣fobs−fs∣.fB=∣fobs−fs∣.

  (10)

(2) Substitute known values:

∣ 2, 500 , 649 Hz − 2, 500 , 000 Hz ∣ ∣ 2, 500 , 649 Hz − 2, 500 , 000 Hz ∣ size 12{ lline 2,"500","649"`"Hz" - 2,"500","000"`"Hz" rline } {}

(11)

(3) Calculate to find the beat frequency: 649 Hz.

Discussion

The Doppler shifts are quite small compared with the original frequency of 2.50 MHz. It is far easier to measure the beat frequency than it is to measure the echo frequency with an accuracy great enough to see shifts of a few hundred hertz out of a couple of megahertz. Furthermore, variations in the source frequency do not greatly affect the beat frequency, because both fsfs size 12{f rSub { size 8{s} } } {} and fobsfobs size 12{f rSub { size 8{"obs"} } } {}would increase or decrease. Those changes subtract out in fB=∣fobs−fs∣.fB=∣fobs−fs∣.

Unless otherwise indicated, for problems in this section, assume that the speed of sound through human tissues is 1540 m/s.