COAS_P2_Ch17_it

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solid materials. A typical value for body tissue is 1500 m s–1. Using the wave equation v = f λ, we can calculate the wavelength of 2.0 MHz ultrasound waves in tissue:

λ = νf = 1500

2.0 × 106

λ = 7.5 × 10–4 m ≈ 1 mm

This means that 2.0 MHz ultrasound waves will be able to distinguish detailed features whose dimensions are of the order of 1 mm. Higher-frequency waves have shorter wavelengths and these are used to detect smaller features inside the body.

Using ultrasound in medicine

Chapter 17

Working with ultrasoundUltrasound is any sound wave that has a frequency above the upper limit of human hearing. This is usually taken to imply frequencies above 20 kHz (20 000 Hz), although the limit of hearing decreases with age to well below this figure. In medical applications, the typical frequencies used are in the megahertz range.

Sound waves are longitudinal waves. They can only pass through a material medium; they cannot pass through a vacuum. The speed of sound (and hence of ultrasound) depends on the material. In air, it is approximately 330 m s–1; it is higher in

Ultrasound scanning is routinely used to check the condition of a baby in the womb (Figure 17.1). There do not seem to be any harmful side-effects associated with this procedure, and it can provide useful information on the baby’s development. Indeed, for many children, their first appearance in the family photo album is in the form of an ante-natal (before birth) scan!

This technique has many other uses in medicine. It can be used to detect gallstones or kidney stones (two very painful complaints), so men as well as women may experience this type of scan.

The technique of ultrasound scanning is rather similar to the way in which sailors use echosounding and echolocation to detect the seabed and shoals of fish. Ultrasound waves are directed into the patient’s body. These waves are partially reflected at the boundaries between different tissues and the reflected waves are detected and used to construct the image.

In this chapter, we will look at the principles of ultrasound scanning and we will also look at another technique in which ultrasound is used to measure the rate of blood flow in the body.

Family photos

Figure 17.1 An expectant mother undergoes an ultrasound scan. The image of her baby is built up by computer and appears on the monitor.

hyperlink destination

http://advancedsciences.cambridge.org/ocr/physics_2/17/EL/

Chapter 17: Using ultrasound in medicine

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piezoelectric substances, the maximum value of strain is about 0.1%; in other words, the crystal’s width changes by about 0.1%.In a piezoelectric transducer, an alternating voltage

is applied across the crystal, which then acts as the vibrating source of ultrasound waves. A brief pulse of ultrasound waves is sent into the patient’s body; the transducer then receives an extended pulse of reflected ultrasound waves.

Detecting ultrasoundThe transducer also acts as the detector of reflected ultrasound waves. It can do this because the piezoelectric effect works in reverse: a varying stress applied to the crystal produces a varying e.m.f. across the crystal – see Figure 17.2b. To maximise the effect, the frequency of the waves must match the resonant frequency of the crystal. The optimum size of the crystal is half the wavelength (λ

2) of the

ultrasound waves.

Figure 17.3 shows the construction of a piezoelectric ultrasound transducer. Note the following features:

The crystal is now usually made of polyvinylidene • difluoride. Previously, quartz and lead zirconate titanate were used.The outer case supports and protects the crystal.•

Producing ultrasoundLike audible sound, ultrasound is produced by a vibrating source. The frequency of the source is the same as the frequency of the waves it produces. In ultrasound scanning, ultrasonic waves are produced by a device in which a varying electrical voltage is used to generate ultrasound. The same device also acts as a detector. This device is known as a transducer; this is a general term used to describe any device that changes one form of energy into another.

At the heart of the transducer is a piezoelectric crystal. This is a crystal that has a useful property: when a voltage is applied across it in one direction, it shrinks slightly – see Figure 17.2a. When the voltage is reversed, it expands slightly. So an alternating voltage with frequency f causes the crystal to contract and expand at the same frequency f. We say that the voltage induces a strain in the crystal. In the best

applied stress induced e.m.f.

applied voltage induced strain

a

b ++++++

––––––

+ – connector

outer case

acoustic window

crystal

dampingmaterial

Figure 17.2 The piezoelectric effect. a An applied voltage causes a piezoelectric crystal to contract or expand. b An applied stress causes an induced e.m.f. across the crystal.

Figure 17.3 A section through an ultrasound transducer.





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boundary where the wave speed changes. The reflected waves are then detected and used to construct an internal image of the body.

Figure 17.4 shows what happens when a beam of ultrasound reaches a boundary between two different media. The beam is partially refracted (so that the transmitted beam has changed direction) and partially reflected. This diagram should remind you of the way in which a ray of light is refracted and reflected when it strikes the boundary between two media. It is the change in speed which causes the refraction of a wave.

For ultrasound, we are interested in the fraction of the incident intensity of ultrasound that is reflected at the boundary. This depends on the acoustic impedance Z of each material. This quantity depends on the density ρ and the speed of sound c in the material. Acoustic impedance is defined as follows:

acoustic impedance = density × speed of sound

Z = ρc

The unit of acoustic impedance Z is kg m–2 s−1.

At the base is the acoustic window made from a • material that is a good transmitter of ultrasound.Behind the crystal is a large block of damping • material (usually epoxy resin). This helps to stop the crystal vibrating when a pulse of ultrasound has been generated. This is necessary so that the crystal is not vibrating when the incoming, reflected ultrasound waves reach the transducer.

SAQ 1 Quartz is an example of a piezoelectric material.

The speed of sound in quartz is 5700 m s–1. a Calculate the wavelength of

ultrasound waves of frequency 2.1 MHz in a quartz crystal.

b If the crystal is to be used in an ultrasound transducer, its thickness must be half a wavelength. Calculate the thickness of the transducer.

2 Piezoelectric crystals have many applications other than in ultrasound scanning. For example, they are used: a in gas lighters (to produce a spark) b in inkjet printers (to break up the stream of ink

into droplets) c in guitar pickups (to connect the guitar to an amplifier)

d in the auto-focus mechanism of some cameras (to move the lens back and forth).

For each of these examples, state whether the piezoelectric effect is being used to convert mechanical energy to electrical energy or the other way round.

EchosoundingThe principle of an ultrasound scan is to direct ultrasound waves into the body. These pass through various tissues and are partially reflected at each

Zincidentwave

angle of incidence

reflectedwave

angle of refraction

refracted wave(transmitted)

1 Z 2

Figure 17.4 An ultrasound wave is both refracted and reflected when it strikes the boundary between two different materials.



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Table 17.1 shows values of ρ, c and Z for some materials that are important in medical ultrasonography.

Calculating reflected intensitiesWhen an ultrasound beam reaches the boundary between two materials, the greater the difference in acoustic impedances, the greater the fraction of the ultrasound waves that is reflected. For normal incidence (i.e. angle of incidence = 0°) the ratio of the reflected intensity Ir to the incident intensity I0 is given by:

Ir

I0 = (Z2 – Z1)2

(Z2 + Z1)2

or

Ir

I0 = Z2 – Z1

Z2 + Z1 2

where Z1 and Z2 are the acoustic impedances of the two materials (see Figure 17.4). The ratio Ir

I0indicates the fraction of the intensity of the beam that is reflected.

Now look at Worked example 1.

Material Density ρ/kg m–3 Speed of sound c/m s–1 Acoustic impedance Z/106 kg m–2 s–1

air 1.3 330 0.0004

water 1000 1500 1.50

Biologicalbloodfatsoft tissue (average)musclebone (average; adult)

1060925

106010751600

15701450154015904000

1.661.341.631.716.40

Transducersbarium titanatelead zirconate titanatequartz polyvinylidene difluoride

5600765026501780

5500379057002360

30.829.015.14.20

Table 17.1 Ultrasound properties of some materials important in medical scanning.

A beam of ultrasound is normally incident on the boundary between muscle and bone. Use Table 17.1 to determine the fraction of its intensity reflected.

Step 1 Write down the values of Z1 (for muscle) and Z2 (for bone).

Z1 = 1.71 × 106 kg m–2 s–1 Z2 = 6.40 × 106 kg m–2 s–1

Step 2 Substitute these values in the equation forIr

I0; note that we can use this equation because we

know that the angle of incidence = 0°.

Ir

I0 = (Z2 – Z1)2

(Z2 + Z1)2

= (6.40 – 1.71)2

(6.40 + 1.71)2

= 0.33

Note also that we can ignore the factor of 106 in the Z values because this is a factor common to all the values, so they cancel out.

So 33% of the intensity of ultrasound will be reflected at the muscle–bone boundary.

Worked example 1




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The poor match of impedance between air and tissue means that ultrasound cannot penetrate the lungs. The operator must take care to avoid any bubbles of gas in the intestines. Bones are also difficult to see through. For an ultrasound scan of the heart, the probe must be directed through the gap between two ribs.

SAQ 3 Calculate the acoustic impedance of brain

tissue. (Density = 1025 kg m–3, speed of sound = 1540 m s–1.)

4 Determine the fraction of the intensity of an ultrasound beam that is reflected when a beam is incident normally on a boundary between water and fat. (Use values from Table 17.1.)

5 The ultrasound image shown in Figure 17.5 clearly shows the baby’s skin and some bones. Explain why these show up clearly while softer organs inside its body do not.

6 Explain why ultrasound cannot readily be used to examine the brain. Suggest an alternative scanning technique(s) that can be used for this.

Ultrasound scanningThere are several different types of ultrasound scan which are used in practice. To illustrate the basic principles, we will concentrate on the A-scan and the B-scan.

A-scanThis is the simplest type of scan. A pulse of ultrasound is sent into the body and the reflected ‘echoes’ are detected and displayed on an oscilloscope or computer screen as a voltage against time graph.

Comparing acoustic impedancesA big change in acoustic impedance gives a large fraction of reflected intensity. Inspection of Table 17.1 shows that:

a very large fraction (• Ir

I0 ≈ 99.95%) of the

incident ultrasound will be reflected at an air–tissue boundary

a large fraction will be reflected at a tissue–bone • boundary (as shown in Worked example 1)very little will be reflected at a boundary between • soft tissues including fat and muscle.

This means that bone shows up well in an ultrasound scan, but it is difficult to see different soft tissues (Figure 17.5). Another problem is that the patient’s skin is in contact with air, and 99.95% of the ultrasound will be reflected before it has entered the body. To overcome this, the transducer must be ‘coupled’ to the skin using a gel whose impedance matches that of the skin. This process of impedance matching explains why the patient’s skin is smeared with gel before a scan.

The acoustic impedance of the gel is typically 1.65 × 106 kg m–2 s–1 and that of skin is 1.71 × 106 kg m–2 s–1. With gel between the skin and the transducer, the percentage of the intensity reflected is 0.03%.

Figure 17.5 Ultrasound scan of a fetus at 20 weeks; the baby’s skin is clearly visible, as are its bony skull and ribs.



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where c is the speed of the ultrasound in the bone (see Worked example 2).

Because ultrasound waves are gradually attenuated as they pass through the body (their energy is absorbed so that their amplitude and intensity decrease), the echoes from tissues deeper in the body are weaker and must be amplified.

A-scans are used for some straightforward procedures such as measuring the thickness of the eye lens.

B-scanIn a B-scan, a detailed image of a cross-section through the patient is built up from many A-scans. The ultrasound transducer is moved across the patient’s body in the area of interest. Its position and orientation are determined by small sensors attached to it.Each reflected pulse is analysed to determine the

depth of the reflecting surface (from the time of echo) and the nature of the surface (from the amplitude of the reflected wave). A two-dimensional image is then built up on a screen by positioning dots to represent the position of the reflecting surfaces and

A pulse generator controls the ultrasound transducer. It is also connected to the time base of the oscilloscope. Simultaneously, the pulse generator triggers a pulse of ultrasound which travels into the patient and starts a trace on the screen. Each partial reflection of the ultrasound is detected and appears as a spike on the screen (Figure 17.6).In Figure 17.6, the pulses 1, 2 and 3 are reflected at

the various boundaries. Pulse 1 is the reflection at the muscle–bone boundary at B. Pulse 2 is the reflection at the bone–muscle boundary at C. The time Δt is the time taken for the ultrasound to travel twice the thickness of the bone. Finally, pulse 3 is the reflection at the muscle–air boundary at D. The thickness of the bone can be determined from this A-scan.

time interval between pulses 1 and 2 = Δt

thickness of bone = distance travelled by ultrasound2

thickness of bone = cΔt2

Time

0

A B

1

Δt

voltagepulse sentfromtransducer

2

reflected pulses

bonemuscleultrasoundtransducer

gel

muscle

3

C D

0

Volta

ge

Figure 17.6 An A-scan. Information about the depth of reflecting tissues can be obtained from the positions of the spikes along the time axis; their relative amplitudes can indicate the nature of the reflecting surfaces.

In a particular A-scan, similar to Figure 17.6, the time interval between pulses 1 and 2 is 12 µs. The speed of ultrasound in bone is about 4000 m s−1. Determine the thickness of the bone.

Step 1 Determine the distance travelled by the ultrasound in the time interval of 12 µs.

distance = speed × time

distance = 4000 × 12 × 10−6 = 4.8 × 10−2 m

Step 2 Calculate the thickness of the bone.

The ultrasound wave has to travel twice the thickness of the bone. Hence:

thickness of bone = 4.8 × 10–2

2

= 2.4 × 10−2 m (2.4 cm)

Worked example 2




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SAQ 7 Two consecutive peaks in an ultrasound A-scan

are separated by a time interval of 0.034 ms. Calculate the distance between the two reflecting surfaces. (Assume that the speed of sound in the tissue between the two surfaces is 1540 m s–1.)

8 Explain why an ultrasound B-scan is used to examine a fetus rather than X-rays.

Doppler ultrasoundThe patient in Figure 17.9 is undergoing a different medical application of ultrasound scanning. The doctor is holding an ultrasound probe against her neck to examine the blood flow in a major artery. This relies on the Doppler effect.

The Doppler effect occurs when waves are emitted, reflected or detected by a moving object. In this case, we will consider a moving reflector – the iron-rich blood cells moving along an artery. When a pulse of ultrasound waves is sent along the artery, they are partially reflected back by the cells. The reflected waves have a slightly different wavelength and frequency, an example of the Doppler effect.

with brightness determined by the intensity of the reflection, brighter dots indicating more reflected ultrasound (see Figure 17.7).

Figure 17.8 shows the result of a typical B-scan. Because it takes several seconds for the scanner to move across the body, problems can arise if the organs of interest are moving – this gives a blurred image.

movement ofultrasoundtransducer

B-scandisplay

organ

skin

Figure 17.7 In a B-scan, dots are produced on the screen rather than the pulses as in the A-scan. By moving the transducer, a series of dots on the screen trace out the shape of the organ being examined.

Figure 17.8 An ultrasonic B-scan of an abnormal thyroid gland.

Figure 17.9 Ultrasound examination of the carotid artery in the neck, a form of Doppler ultrasonography.





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10 Figure 17.10 shows an ultrasound transducer pointing towards a blood vessel. The frequency of the ultrasound waves emitted by the transducer is f. The detected ultrasound waves have a higher frequency because of the Doppler effect. The change in the frequency ∆f of the waves is given by the equation:

∆f = 2fν cos θ

c

where v is the speed of the blood, θ is the angle between the incident ultrasound and the direction of the blood flow and c is the speed of the ultrasound in blood. A transducer emitting ultrasound at frequency 6.0 MHz detects a change in frequency of 0.85 kHz when the angle between the emitted waves and the blood vessel is 58°. The speed of ultrasound in blood is 1500 m s–1. The diameter of the blood vessel is 1.2 mm. Determine the speed v of the blood flow in the vessel and the rate of flow of blood (in m3 s–1) through the vessel.

Why does this happen? If the reflecting surface is stationary, nothing unusual happens – the reflected waves are a mirror image of the incident waves, so that their wavelength is unaltered. However, if the reflector is moving away from the source, things are different. Each reflected wave is reflected from a slightly different position, slightly further away from the source. The result is that the string of reflected waves is slightly longer than the incident waves. In other words, the reflected waves are longer – they have an increased wavelength. Because their speed is unaffected, their frequency must decrease.The greater the speed of the receding reflector, the

greater the decrease in the frequency of the ultrasound waves. It follows that the speed of the receding reflector can be deduced from the change in frequency of the waves. (Note that an approaching reflector will increase the frequency of the ultrasound waves.)

Doppler ultrasonography is used to study the movement of a patient’s blood along blood vessels (veins and arteries). It can show up the pulsing movement of the blood as the heart beats. More importantly, it can show whether the blood is moving smoothly (the blood moves at a uniform speed throughout the blood vessel) or whether there is turbulence (blood flows at different speeds in different regions of the vessel). This can be a sign of blockages developing, or of weakening of the walls of the vessel.

Kidney transplant patients are monitored using Doppler ultrasound to determine whether blood is flowing normally through their new organ. The technique is also used in the diagnosis of faulty heart valves.

SAQ 9 Doppler ultrasound can be used to monitor the

action of the heart as it beats. At any instant, some areas of the heart will be moving towards the ultrasound probe while others will be moving away from the probe. How will the wavelength and frequency of ultrasound waves be affected when they reflect from a surface that is moving towards the probe? Explain your answer.

skin

θ

gel

ultrasound transducer

blood vessel

red blood cells

Figure 17.10 For SAQ 10.





258

SummaryUltrasound is a longitudinal mechanical wave with a frequency greater than 20 kHz.• Ultrasound transducers use the piezoelectric effect to generate and detect ultrasound waves.• The acoustic impedance • Z of a material depends on its density ρ and the speed c of sound: Z = ρc.

The fraction of the intensity of an ultrasound wave reflected at a boundary is given by• Ir

I0 = (Z2 – Z1)2

(Z2 + Z1)2 or Ir

I0 = Z2 – Z1

Z2 + Z1 2

To transfer a high proportion of the intensity of an ultrasound pulse into the patient’s body, an • impedance-matching gel must be used with acoustic impedance almost the same as that of the skin.

In ultrasound scanning, an A-scan uses a single pulse to determine the depth and nature of reflecting • surfaces. A B-scan builds up a two-dimensional image from multiple A-scans.

In the Doppler effect, the wavelength and frequency of a wave are altered on reflection by a • moving surface. The frequency of a wave decreases for a receding source and increases for an approaching source.

The Doppler effect can be used to determine the speed of blood in arteries.•

Questions a1 Define acoustic impedance of a material and show it has unit kg m–2 s–1. [2]

The acoustic impedance of air, gel and skin are 430 kg mb –2 s–1, 1.6 × 106 kg m–2 s–1

and 1.7 × 106 kg m–2 s–1 respectively. Calculate the fraction of intensity of the ultrasound that penetrates the i

air–skin boundary and the gel–skin boundary. [6] Explain why gel is smeared on the skin and the transducer before an ii

ultrasound scan. [2] [Total 10]

a2 Describe the principles of the production of a short pulse of ultrasound using a piezoelectric transducer. [5]

continued


259

20 μs

The CRO timebase is set to 20 µs cm–1. The speed of ultrasound in the fetal head is 1.5 × 103 m s–1.

Calculate the size of the fetal head. [4]i State and explain what would be seen on the CRO screen if gel had ii not been

applied between the ultrasound transducer and the skin of the mother. [3]OCR Physics A2 (2825/02) June 2007 [Total 12]

3 The ratio of reflected intensity to incident intensity for ultrasound reflected at a boundary is related to the acoustic impedance Z1 of the medium on one side of the boundary and the acoustic impedance Z2 of the medium on the other side of the boundary by the following equation:

reflected intensityincident intensity

= (Z2 – Z1)2

(Z2 + Z1)2

State the a two factors that determine the value of the acoustic impedance. [2]

continued

The diagram below shows a trace on a cathode-ray oscilloscope (CRO) of an bultrasound reflection from the front edge and rear edge of a fetal head.


260

An ultrasound investigation was used to identify a small volume of substance in ba patient. It is suspected that this substance is either blood or muscle.

During the ultrasound investigation, an ultrasound pulse of frequency 3.5 × 106 Hz passed through soft tissue and then into the small volume of unidentified substance. A pulse of ultrasound reflected from the front surface of the volume was detected 26.5 µs later. The ratio of the reflected intensity to incident intensity for the ultrasound pulse reflected at this boundary was found to be 4.42 × 10–4. The table below shows data for the acoustic impedances of various materials found in a human body.

Medium Acoustic impedance Z/kg m−2 s−1

air 4.29 × 102

blood 1.59 × 106

water 1.50 × 106

brain tissue 1.58 × 106

soft tissue 1.63 × 106

bone 7.78 × 106

muscle 1.70 × 106

Use appropriate data from the table to identify the unknown medium. You imust show your reasoning. [4]

Calculate the depth at which the ultrasound pulse was reflected if the speed iiof ultrasound in soft tissue is 1.54 km s–1. [2]

Calculate the wavelength of the ultrasound in the soft tissue. [2]iiiOCR Physics A2 (2825/02) June 2005 [Total 10]

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