Questions on ultrasound imaging

Question 1

Describe how a piezoelectric transducer both generates and detects ultrasound.

Accepted Answer

To generate: an alternating pd applied across the crystal makes it deform and vibrate at the driving frequency, launching an ultrasound pulse (1). To detect: a returning echo deforms the crystal, which produces a measurable pd across it (1).

Question 2

An echo returns 90 μs after a pulse enters soft tissue, where ultrasound travels at 1540 m s⁻¹. Calculate the depth of the reflecting boundary.

Accepted Answer

d = ct/2 = (1540 × 90 × 10⁻⁶)/2, the factor of two accounting for the there-and-back journey (1) d = 6.9 × 10⁻² m ≈ 6.9 cm (1)

Question 3

State what is meant by the acoustic impedance of a medium, and give its defining equation.

Accepted Answer

Acoustic impedance is the product of the density of a medium and the speed of sound in it: Z = ρc (1). Its unit is kg m⁻² s⁻¹ (1).

Question 4

A scanner uses ultrasound of frequency 3.0 MHz. Calculate the wavelength of this ultrasound in soft tissue, where its speed is 1540 m s⁻¹.

Accepted Answer

Use of λ = c/f (1). λ = 1540/(3.0 × 10⁶) = 5.1 × 10⁻⁴ m = 0.51 mm (1).

Question 5

The acoustic impedance of bone is 7.8 × 10⁶ kg m⁻² s⁻¹ and its density is 1900 kg m⁻³. Calculate the speed of sound in bone.

Accepted Answer

Rearranging Z = ρc gives c = Z/ρ = 7.8 × 10⁶/1900 (1) = 4105 m s⁻¹ ≈ 4.1 × 10³ m s⁻¹ (1).

Question 6

State which nuclei in the body give an MR scanner its signal, and explain why the word nuclear in nuclear magnetic resonance does not mean the patient is given a dose of ionising radiation.

Accepted Answer

The hydrogen nuclei, single protons, of the water the body is largely made of (1). No nucleus decays and none is broken up: each is used only as a small magnet, tipped by radio waves and re-emitting radio waves, and radio photons are far too low in energy to ionise anything (1).

Question 7

The protons of the body's water precess at 42.6 MHz for each tesla of magnetic flux density. Calculate their precession frequency in the 1.5 T field of a clinical MR scanner, and name the region of the electromagnetic spectrum this frequency lies in.

Accepted Answer

f = 42.6 × 1.5 (1) f = 63.9 MHz, which is a radio frequency (1)

Question 8

Calculate the acoustic impedance of liver tissue, of density 1075 kg m⁻³, in which sound travels at 1590 m s⁻¹.

Accepted Answer

Z = ρc = 1075 × 1590 (1) Z = 1.71 × 10⁶ kg m⁻² s⁻¹, to the three significant figures the speed of sound carries (1)

Question 9

A boundary lies 3.5 cm below the skin. Calculate the time after the pulse enters the tissue at which its echo returns (speed of sound in tissue 1540 m s⁻¹), and state why the scanner must finish listening before it sends the next pulse.

Accepted Answer

t = 2d/c = 2 × 0.035/1540 (1)
t = 45 μs (1)
If a new pulse left before the deepest echoes of the last one arrived, the scanner could not tell which pulse an echo belonged to, and depths would be assigned wrongly (1)

Question 10

Distinguish between an A-scan and a B-scan.

Accepted Answer

An A-scan plots echo strength against time along a single line through the body, giving depths along that one direction (1). A B-scan sweeps the beam and turns each echo into a brightness point, building a two-dimensional image of the boundaries (1).

Question 11

Ultrasound meets a boundary between muscle (Z = 1.70 × 10⁶ kg m⁻² s⁻¹) and liver (Z = 1.63 × 10⁶ kg m⁻² s⁻¹). Calculate the fraction of the incident intensity that is reflected.

Accepted Answer

I_r/I_i = (Z₂ − Z₁)²/(Z₂ + Z₁)² (1) = (0.07/3.33)² (1) = 4.4 × 10⁻⁴, about 0.04% (1)

Question 12

Explain, in terms of acoustic impedance, why coupling gel is needed between the transducer and the skin.

Accepted Answer

Air's acoustic impedance is thousands of times smaller than tissue's (1), so at an air-skin boundary the impedance mismatch is enormous and nearly all the intensity reflects before entering the body (1). The gel displaces the air with a medium whose impedance is close to that of tissue, so the pulse is transmitted rather than reflected at the surface (1).

Question 13

A scanner sends 5000 pulses each second and must receive all the echoes from one pulse before it sends the next. Show that the greatest depth this scanner can image is about 15 cm. The speed of ultrasound in tissue is 1540 m s⁻¹.

Accepted Answer

Listening time per pulse = 1/5000 = 200 μs (1). Evidence of use of d = ct/2 for the deepest echo (1). d = (1540 × 200 × 10⁻⁶)/2 = 15.4 cm ≈ 15 cm (1).

Question 14

An A-scan, rather than a B-scan, is used to measure the length of a patient's eye before surgery. Suggest why an A-scan is sufficient for this task.

Accepted Answer

Only depths along a single line through the eye are needed, not a two-dimensional image (1). An A-scan gives exactly this: the echo times along one direction convert directly to the distances of the eye's surfaces (1).

Question 15

Explain why an MR scan is usually the instrument of choice for imaging soft tissue such as the brain, and state what property of the tissue the returning radio signal depends on.

Accepted Answer

Protons tipped out of line by the radio-frequency pulse relax back into line with the field, re-emitting a radio signal (1). The relaxation time depends on the tissue the protons sit in, so soft tissues that attenuate X-rays almost identically still return distinguishable signals, giving fine soft-tissue contrast (1). MR uses no ionising radiation, so this detail costs the patient no dose (1). (Board note: the relaxation-time step is extension for AQA, whose specification states that de-excitation relaxation times will not be examined.)

Question 16

Explain why the radio-frequency pulse an MR scanner sends into the patient has to be at the precession frequency of the protons, and state what the protons do once the pulse has ended.

Accepted Answer

The protons precess about the scanner's magnetic field at a frequency fixed by the flux density there (1). Energy is absorbed from the pulse only when it arrives at that same frequency, which is the resonance the technique is named after, and the energy absorbed tips the protons away from the field direction (1). When the pulse ends they relax back into line with the field and give that energy up again as a radio signal, which receiving coils around the patient detect (1).

Question 17

A hospital owns one MR scanner and several CT scanners, and most of its urgent imaging is done with CT. Give three reasons, other than the quality of the image, why CT is used for most of that work.

Accepted Answer

A CT scan takes seconds where an MR scan keeps the patient still for tens of minutes, so an unwell or restless patient can be imaged and more patients can be seen in a day (1). MR needs a superconducting magnet, which makes the scanner far more expensive to install and to run, so a hospital has fewer of them (1). MR is barred to a patient whose implant is labelled MR Unsafe, because the strong field could disturb or move it, while CT can image that patient (1).

Question 18

The acoustic impedance of air is about 430 kg m⁻² s⁻¹ and of soft tissue about 1.63 × 10⁶ kg m⁻² s⁻¹. Calculate the fraction of intensity reflected at an air-tissue boundary, and comment on the result.

Accepted Answer

I_r/I_i = ((1.63 × 10⁶ − 430)/(1.63 × 10⁶ + 430))² (1)
= 0.9989 ≈ 0.999 (1)
Nearly everything reflects, which is exactly why an air gap makes imaging impossible and why the gel must exclude it (1)

Question 19

In an A-scan of an organ, echoes return 52 μs and 78 μs after the pulse enters the tissue (c = 1540 m s⁻¹). Calculate the depth of each surface of the organ and hence its thickness along the beam.

Accepted Answer

Front surface: d = ct/2 = (1540 × 52 × 10⁻⁶)/2 (1) = 4.0 cm (1) Back surface: d = (1540 × 78 × 10⁻⁶)/2 = 6.0 cm (1) Thickness = 2.0 cm ≈ 2.0 cm (1)

Question 20

A student says higher-frequency ultrasound always gives a better scan. Discuss whether this is true.

Accepted Answer

Higher frequency means shorter wavelength, so finer detail can be resolved, which argues for high frequency (1). But attenuation in tissue increases with frequency, so a high-frequency pulse fades before reaching deep structures (1). The choice is a trade: high frequency for shallow, detailed work such as eye measurements, lower frequency for deep organs (1).

Question 21

Give two reasons why ultrasound, rather than X-rays, is used for antenatal scanning, and one limitation of ultrasound imaging.

Accepted Answer

Ultrasound is non-ionising, so it carries no radiation dose to the foetus (1), and it images in real time, showing movement (1). A limitation: it cannot image through bone or air-filled spaces, since the impedance mismatch reflects the pulse, so structures behind the skull or lungs are hidden (1).

Question 22

A sonographer can use one of two coupling gels. Gel P has acoustic impedance 1.50 × 10⁶ kg m⁻² s⁻¹ and gel Q has acoustic impedance 1.10 × 10⁶ kg m⁻² s⁻¹. The acoustic impedance of skin is 1.63 × 10⁶ kg m⁻² s⁻¹. Deduce which gel transmits the greater fraction of the ultrasound intensity into the patient, supporting your answer with calculations.

Accepted Answer

Evidence of use of I_r/I_i = (Z₂ − Z₁)²/(Z₂ + Z₁)² at the gel-skin boundary (1). Gel P: (0.13/3.13)² = 0.17% reflected (1). Gel Q: (0.53/2.73)² = 3.8% reflected (1). Gel P should be chosen: its impedance is closer to skin's, so about 99.8% of the intensity enters the patient against about 96% for gel Q (1).

Question 23

A scanner converts echo times to depths using a speed of 1540 m s⁻¹ everywhere. A reflecting boundary lies at the bottom of a layer of fat 3.0 cm thick, in which ultrasound actually travels at 1450 m s⁻¹. Determine the depth the scanner displays for this boundary, and the error in the displayed depth.

Accepted Answer

Echo time t = 2d/c = 2 × 0.030/1450 = 41.4 μs (1). The scanner converts this time with 1540 m s⁻¹: displayed depth = (1540 × 41.4 × 10⁻⁶)/2 (1) = 3.19 cm (1). The displayed depth is 1.9 mm too deep (1).

Question 24

A B-scan is used to produce an image of a foetus. Describe fully how the image is produced, from the generation of each ultrasound pulse to the formation of the two-dimensional image.

Accepted Answer

An alternating pd applied across the piezoelectric crystal makes it vibrate and emit a short ultrasound pulse (1). Coupling gel excludes air between the probe and the skin, so the pulse is transmitted into the body instead of reflecting at the surface (1). At each boundary between tissues of different acoustic impedance, part of the pulse reflects back as an echo (1). The returning echoes deform the same crystal, generating pds that the scanner records (1). The depth of each boundary follows from d = ct/2, the factor of two because the pulse travels there and back (1). Each echo is displayed as a spot whose brightness follows the echo strength, and sweeping the beam across the body builds these spots into a two-dimensional image of the boundaries (1).

Question 25

For each of the following patients, choose the most suitable imaging method from MR, CT and ultrasound, giving a reason for each choice: (a) routine monitoring of a foetus; (b) a detailed image of suspected soft-tissue damage inside a knee; (c) urgent imaging of the brain of a patient fitted with an older pacemaker labelled MR Unsafe.

Accepted Answer

(a) Ultrasound: it is non-ionising, so there is no dose to the foetus, and it images in real time (1). (b) MR: relaxation times differ between soft tissues, so it gives the finest soft-tissue contrast, again with no ionising dose (1). (c) CT, because MR is ruled out: the device is MR Unsafe, so the scanner's strong magnetic field could disturb or damage it. (A modern MR Conditional pacemaker can be scanned under controlled protocols, which is why the safety labelling, not the pacemaker itself, decides.) (1) CT still resolves the brain's soft tissue well enough to diagnose, and it is fast, at the price of an X-ray dose (1).

Question 26

An endoscope carries two separate bundles of optical fibres. Describe how total internal reflection keeps light travelling along a single fibre however the endoscope bends, distinguishing between the roles of the two bundles, and explain why each fibre needs a cladding of lower refractive index than its core.

Accepted Answer

A ray inside the core that meets the core-to-cladding boundary beyond the critical angle undergoes total internal reflection, and since this happens at every reflection the light stays trapped in the core all the way along the fibre, however it bends (1). The coherent bundle carries the image: every fibre occupies the same relative position in the bundle at both ends, so light picked up at one point of the tissue leaves at the corresponding point of the picture and the image arrives intact (1). The second, incoherent bundle only carries light in to illuminate the tissue, so it does not matter which fibre ends up where, and it is left unordered, which is cheaper to make (1). The cladding, of lower refractive index than the core, provides the boundary total internal reflection needs; without it, light could leak from a fibre into a neighbouring one where they touch, scrambling or losing part of the image (1).

Question 27

Describe the sequence of events in an MR scanner that produces an image, from the patient entering the magnet to the signal being located within the body.

Accepted Answer

The patient lies in a strong, steady magnetic field of a few tesla, and the protons of the body's water precess about the field direction at a frequency fixed by its flux density (1). A radio-frequency pulse at that precession frequency is absorbed by the protons and tips them away from the field direction (1). When the pulse ends the protons relax back into line and re-emit the energy as a radio signal, which coils around the patient detect (1). The relaxation time depends on the tissue the water sits in, so two tissues return signals of different size at the moment the scanner reads them, and that difference becomes the contrast (1). Gradient coils make the flux density vary from place to place across the patient, so each position precesses and emits at its own frequency, and a computer reconstructs a cross-sectional image from the frequencies and phases received (1). (Board note: AQA states that de-excitation relaxation times will not be examined, so on that board the relaxation-time step is the reason behind the contrast rather than a required point.)

Question 28

An MR scanner tips the protons with radio waves of frequency 64 MHz, and about 10 eV is needed to ionise a molecule of tissue. Calculate the energy of one 64 MHz photon in joules and in electronvolts, and explain what the result shows about the dose an MR scan delivers. Take h = 6.63 × 10⁻³⁴ J s and e = 1.60 × 10⁻¹⁹ C.

Accepted Answer

E = hf = 6.63 × 10⁻³⁴ × 64 × 10⁶ (1)
E = 4.24 × 10⁻²⁶ J (1)
In electronvolts, E = 4.24 × 10⁻²⁶/(1.60 × 10⁻¹⁹) = 2.7 × 10⁻⁷ eV (1)
That is smaller than the 10 eV needed to ionise a molecule by a factor of about 3.8 × 10⁷, so no number of these photons can ionise anything and the scan gives no ionising dose (1)

Question 29

A gradient coil makes the magnetic flux density inside an MR scanner increase steadily along the patient at 10 mT per metre. The protons precess at 42.6 MHz for each tesla of flux density. Calculate the difference in flux density between two points 12 cm apart along the patient, and the difference between their precession frequencies. Explain how this lets the scanner place the signal it receives.

Accepted Answer

ΔB = 10 × 10⁻³ × 0.12 (1) = 1.2 mT (1)
Δf = 42.6 × 10⁶ × 1.2 × 10⁻³ = 51.1 kHz (1)
Each position along the gradient precesses, and so emits, at its own frequency, so the frequency of a received signal states where along the patient it came from (1)