PhysicsMedical physics › Ultrasound imaging

Ultrasound imaging

Ultrasound imaging sends a pulse of megahertz sound into the body from a piezoelectric transducer and times the echoes returned from each boundary. Because the reflection depends on the acoustic impedance either side of a boundary, the timings can be assembled into an image, and no ionising radiation is involved.

Builds on Progressive waves and Refraction and total internal reflection.

IN THIS TOPIC

  • Explain how a piezoelectric transducer generates and detects ultrasound pulses.
  • Use the pulse-echo technique and d = ct/2 to locate a boundary in tissue.
  • Say how the duration of the pulse and the wavelength limit the detail a pulse-echo scan can recover.
  • Distinguish an A-scan from a B-scan.
  • Use acoustic impedance Z = ρc and the reflection coefficient to explain what reflects where, and why the gel is needed.
  • Outline the endoscope as total internal reflection put to work.
  • Describe the principle of the MR scanner: protons precessing in a strong field, a radio-frequency pulse tipping them, and a relaxation time that depends on the tissue.
  • Weigh MR against CT and ultrasound for a given diagnostic task.
  • For CIE, use I = I₀e−μx for the attenuation of ultrasound in tissue.
  • For OCR, find the speed of blood from a Doppler shift with Δf/f = 2v cos θ/c.

COMMON MISCONCEPTION

Ultrasound scans photograph the inside of the body with sound.

A crystal that emits and detects

Ultrasound is sound above about 20 kHz, and medical scanners run in megahertz. One component both makes it and detects it, the piezoelectric transducer. That is a crystal which deforms when a pd is applied across it, and which generates a pd when something deforms it. Drive it with an alternating pd at its resonant frequency and it vibrates, launching a sound pulse. Let a returning echo squeeze it and it produces a measurable pd.

So the transducer alternates between two roles. It transmits for a microsecond, then detects the echoes that return. That switching is what lets the whole technique work with one probe held against the skin.

Pulse and echo

Everything else is timing. A pulse leaves the transducer, part of it reflects off each boundary between tissues, and the echoes arrive back after a delay. Sound of speed c that took time t to go there and back has found a boundary at depth

d=ct2d = \frac{ct}{2}NOT ON THE AQA DATA SHEET: LEARN IT
The pulse-echo principle: each boundary returns an echo, and its arrival time gives the depth through d = ct/2transducerboundary 1boundary 2pulseecho 1, time t₁echo 2, laterd = ct/2: the pulse travels there and back
FIG. 1One pulse, two boundaries, two echoes. Each echo's arrival time gives one boundary's depth through d = ct/2, with the factor of two because the pulse travels there and back.

That factor of two is what most exam questions turn on, because the pulse covers the distance twice. An A-scan plots echo strength against time along one line, which suits a single depth measurement such as the length of an eye. A B-scan sweeps the beam and turns each echo into a bright dot on a two-dimensional map, and that is the moving picture of an antenatal scan. Nobody is photographing anything. You are looking at a map of boundaries, assembled from arrival times.

WORKED EXAMPLE

How deep is the organ boundary?

An echo returns 65 μs after the pulse enters soft tissue, where ultrasound travels at 1540 m s−1. Find the depth of the reflecting boundary.

d = ct/2 = (1540 × 65 × 10−6)/2 = 5.0 cm.

Without the factor of two the answer comes out at 10 cm, twice the true depth. Halving the round-trip distance is a required step in the mark scheme.

How much a pulse can tell you

Timing an echo gives a position. It does not give unlimited detail, and the two things that cap the detail are worth knowing because they are the same two on every pulse-echo instrument ever built, from a bat to a ship's sonar to a radar set.

Start with the duration of the pulse. A pulse is not an instant; it lasts some time Δt and therefore occupies a length cΔt of tissue while it travels, and its echo is just as long. Two boundaries a distance Δd apart return echoes separated in time by 2Δd/c, that factor of two arriving for the same reason it arrived in d = ct/2. If that separation is shorter than the pulse itself, the second echo begins before the first has finished, the two run into one blur, and the scanner reports one boundary where there are two. The boundaries are just separable when 2Δd/c equals Δt, so the finest detail the pulse can resolve is about cΔt/2.

The same two close boundaries scanned with a long pulse and with a short pulse: only the short pulse returns two separate echoesthe same two boundaries, half a millimetre aparta long pulse: one blurred echopulse duration, Δttimea short pulse: two echoesΔttimereads as one boundaryreads as two boundaries
FIG. 2One pair of boundaries, scanned twice. Above, a pulse lasting far longer than the gap between the two echoes: the returning blocks overlap and arrive as a single lump, so the pair reads as one boundary. Below, the identical pair with a much shorter pulse: the same two echoes now arrive with clear space between them and both boundaries are seen. Nothing about the patient changed.

So shorter pulses see finer detail, and that pushes you straight into the second limit. A pulse cannot be shorter than about one cycle of the wave carrying it, so the shortest pulse a beam of wavelength λ\lambda can make is roughly λ\lambda long. Detail finer than about a wavelength is out of reach whatever the electronics do, and medical scanners are built around that fact: at 1540 m s−1 a 3 MHz beam has a wavelength near half a millimetre, and half a millimetre is about the resolution such a scan achieves.

Both limits work against penetration depth. Raising the frequency shortens both the wavelength and the pulse, which is the gain, but attenuation climbs steeply with frequency, so the finer beam is attenuated sooner. Shallow structures are scanned at high megahertz and deep organs at low, and the sonographer's choice of probe is that trade-off in practice.

WORKED EXAMPLE

How close is too close?

A scanner sends 3.0 MHz pulses, each three cycles long, into tissue where ultrasound travels at 1540 m s−1. Find the wavelength, the duration of one pulse, and how close two boundaries can lie and still be seen separately.

λ = c/f = 1540/(3.0 × 106) = 5.1 × 10−4 m, so about 0.51 mm.

One cycle lasts 1/f = 3.3 × 10−7 s, so three of them last Δt = 1.0 × 10−6 s.

The echoes must arrive at least Δt apart, and they arrive 2Δd/c apart, so Δd = cΔt/2 = (1540 × 1.0 × 10−6)/2 = 7.7 × 10−4 m, about 0.77 mm.

That is roughly one and a half wavelengths, which is the general result. A pulse-echo instrument resolves down to something near the wavelength of the radiation it uses, and no further, however good the timing.

Acoustic impedance, and the gel

How much of a pulse reflects at a boundary is set by the acoustic impedance of the media either side,

Z=ρcZ = \rho cON THE AQA DATA SHEET

the product of density and sound speed. At every interface between two media the arriving wave splits in two: part of it is reflected straight back and the remainder is transmitted onwards into the second medium, and for the head-on, normal-incidence arrival every exam question assumes, with no absorption at the boundary itself, the two impedances alone set the share. Similar impedances let most of the pulse through. Badly mismatched ones reflect most of it. The fraction of the intensity reflected is

IrI0=(Z2-Z1)2(Z2+Z1)2\frac{I_{r}}{I_{0}} = \frac{(Z_{2} - Z_{1})^{2}}{(Z_{2} + Z_{1})^{2}}ON THE AQA DATA SHEET
Impedance mismatch decides reflection: matched tissues pass the pulse, an air gap reflects nearly all of ittissue to tissuesimilar Z: most passesair to tissuebig mismatch: most reflectsthe coupling gel replaces the air gap
FIG. 3Two boundaries compared: tissue to tissue, similar impedances, most of the pulse passes and a little reflects; air to tissue, badly mismatched, almost everything reflects at the skin.

That equation sends the sonographer reaching for the coupling gel. Air's impedance is thousands of times smaller than tissue's, so an air gap between probe and skin reflects nearly all the ultrasound before it ever gets in. The gel displaces the air with a material whose impedance sits close to tissue, letting the pulse through. This is called impedance matching, and the term is worth using. Inside the body, the useful echoes come from mild mismatches between organs, and the strong ones from tissue meeting bone.

Ultrasound also fades as it travels. Absorption and scattering attenuate the pulse exponentially with depth, just as matter attenuates X-rays, and the attenuation rises steeply with frequency. That sets the sonographer's trade. High frequency resolves finer detail but is attenuated sooner, so shallow work uses high megahertz and deep organs use low. One board asks you to put a number on that fade, and the closing sections take it up.

The other non-ionising imagers

The option puts two more non-ionising instruments alongside ultrasound. The endoscope is total internal reflection put to work. A coherent bundle of optical fibres carries an image out of the body, with each fibre holding its place in the bundle so the picture arrives intact, while a second, incoherent bundle carries light in.

The third instrument is the MR scanner, and it gets a section to itself, because it images without ionising rays and without sound, and gives the best soft-tissue contrast of the methods compared here.

The MR scanner

The patient lies inside a strong magnetic field, a few teslas from a superconducting magnet. The body is mostly water, every water molecule carries two hydrogen nuclei, and each of those protons behaves as a tiny magnet. In the field they do not simply snap into line. They precess about the field direction, the way a leaning spinning top wheels around the vertical, at a frequency set by the strength of the field, and for the fields an MR magnet supplies that frequency sits in the radio range.

The scanner then supplies energy at exactly that frequency. A radio-frequency pulse at the precession frequency is absorbed by the precessing protons and tips them away from the field direction. When the pulse ends they relax back into line, giving the energy up again as a radio signal that coils around the patient pick up.

The relaxation time, the time the protons take to settle back, depends on the tissue their water sits in. Protons in fat relax at a different rate from protons in fluid or in a tumour, so the timing of the returning signal labels the tissue that sent it. That is the contrast mechanism, and it is why MR tells apart soft tissues that X-rays cross almost identically. One board note: AQA states that de-excitation relaxation times will not be examined, so for AQA this paragraph is the why behind the scan, not a required point.

Locating the signal is done with the field itself. The precession frequency depends on the field strength, so gradient coils make the field vary slightly from place to place across the patient, and position along a gradient then maps to frequency. One gradient alone cannot pin down a point, so real scanners switch gradients to encode the remaining directions in the signal's frequency and phase, and a computer reconstructs cross-sectional images in any plane from that encoded signal, with nothing rotating around the patient.

Weigh it as the exam does, both ways at once. In its favour: excellent soft-tissue contrast, the finest of the modalities compared here, sections in any plane, and no ionising radiation, so repeat scans add no radiation dose, though every scan still passes the safety checks on implants, heating and any contrast agent first.

Against it: the superconducting magnet makes it expensive to buy and to run; a scan is slow and loud, with the patient holding still for tens of minutes inside a narrow bore; and the strong field rules out patients with MR-unsafe pacemakers or ferromagnetic implants, which the field would disturb or drag. Many modern implants are MR Conditional and scannable under controlled protocols; the safety label, rather than the presence of a pacemaker, determines whether a scan can proceed.

Attenuation with depth, a CIE equation

CIE alone sets numerical attenuation questions on ultrasound, so read this section only if CIE is your specification. Absorption and scattering strip a fixed fraction of the surviving intensity out of every further millimetre of tissue, which is the recipe for an exponential and gives the law the X-rays lesson already uses:

I=I0e-μxI = I_{0}e^{-\mu x}

with μ the attenuation coefficient of the tissue, in per metre, and x the distance travelled through it. None of the mathematics is new. What is new is a feature of the board's own formula sheet: it prints the radioactive-decay exponential, x = x0e−λt, where x stands for an activity or a number of nuclei, and it does not print this one. An exponential on the page is not the exponential you need, so this equation has to be recalled, for ultrasound and for X-rays alike.

The coefficient is where the sonographer's trade-off appears numerically. μ climbs steeply with frequency, so the higher frequency that resolves finer detail is also attenuated faster. μ depends on the tissue too, and bone attenuates so much more strongly than soft tissue that ultrasound cannot usefully see through the skull or into air-filled lungs.

WORKED EXAMPLE

How much of the pulse comes back?

A pulse enters soft tissue whose attenuation coefficient at this frequency is μ = 23 m−1. A boundary lies 4.0 cm below the skin. Find the fraction of the incident intensity arriving at the boundary, and the fraction returning to the transducer.

Going in, x = 0.040 m, so μx = 23 × 0.040 = 0.92 and I/I0 = e−0.92 = 0.40.

The echo crosses the same tissue again, so the round trip is x = 0.080 m, μx = 1.84, and I/I0 = e−1.84 = 0.16, before anything at all is lost at the boundary itself.

The factor of two is back, and for the same reason it appeared in d = ct/2. Decide which distance the question means before you exponentiate, because an attenuation question about an echo almost always means the round trip.

Doppler ultrasound and the speed of blood, an OCR equation

OCR adds one measurement the other boards leave out, so read this section only if OCR A is your specification. Bounce ultrasound off something that is moving and the echo returns at a shifted frequency, the same Doppler effect that drops the pitch of a siren as it passes. Red blood cells are the moving reflectors, and the shift in the returning echo measures how fast they are travelling.

Two features of the equation are worth understanding rather than memorising. First, the shift is doubled. The cells are moving relative to the transmitter, so they meet an already shifted frequency; they then re-radiate that wave while still moving relative to the receiver, so it shifts a second time. Two shifts, one factor of two. Second, only the part of the velocity along the beam shifts anything, and with θ the angle between the beam and the flow that part is v cos θ. Put the two together:

Δff=2vcosθc\frac{\Delta f}{f} = \frac{2v\,\cos\theta}{c}
Doppler ultrasound: the beam meets the vessel at an angle theta to the flow, and only the component v cos theta shifts the frequencyskinblood vesselblood, speed vtransducerpulse and echoθΔf/f = 2v cos θ/csquare across the vessel, θ = 90°, and there is no shift
FIG. 4The probe is angled deliberately. θ is the angle between the beam and the direction of flow, and only the component v cos θ along the beam produces any shift at all.

Here c is the speed of ultrasound in tissue, about 1540 m s−1, and not the speed of light. The equation is printed on the OCR data sheet, so the work is in choosing the right angle and the right c rather than in recalling it.

The cosine explains how the probe is held. Aim it square across the vessel, θ = 90°, and cos θ is zero: the blood is moving neither towards the probe nor away from it, and the machine reports no flow whatever, however fast the blood is going. Tilt the probe along the flow and the shift grows, which is why scanning angles are chosen deliberately, typically somewhere around 45° to 60°.

WORKED EXAMPLE

How fast is the blood?

A 4.0 MHz beam meets a vessel at 60° to the flow, and the echo returns 780 Hz higher than it left. Taking the speed of ultrasound in tissue as 1540 m s−1, find the speed of the blood.

Rearrange for the unknown first: v = cΔf/(2f cos θ), with cos 60° = 0.50.

v = (1540 × 780)/(2 × 4.0 × 106 × 0.50) = (1.20 × 106)/(4.0 × 106) = 0.30 m s−1.

Notice where that 780 Hz sits: in the audible range. The shift itself is a frequency difference, not a sound, but the machine converts it into a tone through a loudspeaker as well as plotting it, which is why a narrowed artery, which speeds the blood through it, can be heard as well as seen.

ASSESSMENT FOCUS

  • State both halves of the piezoelectric effect. A pd deforms the crystal, which transmits. Deforming the crystal produces a pd, which detects. One sentence in each direction.
  • Every pulse-echo calculation carries the factor of two. Write d = ct/2 before substituting, and check the answer against sense, because organs live a few centimetres down and not tens of centimetres.
  • Asked what limits the detail in a pulse-echo scan, give both limits and tie each to a length. Two boundaries closer than about half the length of the pulse return echoes that overlap, and no pulse is shorter than about one wavelength, so the resolution stops near λ\lambda. Edexcel writes both limits into the statement itself; on the other boards the same argument is what justifies choosing a high frequency, and finishing with the attenuation cost earns the evaluation mark.
  • Z = ρc takes the density and sound speed of the same medium. Mixing media across that product is the common error, and it is easy to make when a question lists four numbers in a table.
  • Explain the gel in impedance language. It excludes the air gap, whose impedance mismatch with tissue would reflect nearly all the intensity straight back at the surface.
  • A-scan against B-scan is a two-mark comparison. One line and a depth trace, against a swept beam and a two-dimensional brightness image.
  • Ultrasound is the non-ionising option, so there is no photon energy argument and no dose to discuss. Comparisons with X-ray imaging score for saying that explicitly.
  • The MR chain scores in order, like the PET chain does. Strong magnetic field, protons precess about it, a radio-frequency pulse at the precession frequency tips them, and they relax and re-emit a radio signal that is detected and processed into the image. The tissue-dependent relaxation time is the contrast mechanism, but AQA states that de-excitation relaxation times will not be examined, so on that board treat it as understanding behind the chain rather than a mark. If asked how the signal is located, the field is made to vary across the body so each position emits at its own frequency.
  • Choosing between MR, CT and ultrasound turns on three axes: dose, soft-tissue detail, and cost. MR gives the finest soft-tissue contrast with no ionising dose but is the most expensive and slowest, and it is barred to patients with MR-unsafe pacemakers or ferromagnetic implants. CT resolves fine detail anywhere, but delivers the largest dose. Ultrasound is cheap, portable, real-time and dose-free, and cannot see through bone or air. Name the axis your reason sits on and the mark follows.
  • CIE only: I = I₀e−μx covers ultrasound as well as X-rays, and CIE prints neither. The decay exponential on the formula sheet is a different equation about a different quantity, so recall the attenuation one, and check whether the question means the depth or the round trip.
  • OCR only: the Doppler shift is doubled because the blood receives a shifted wave and then re-emits one, and cos θ takes the component along the beam. The c in Δf/f = 2v cos θ/c is the speed of ultrasound in tissue, about 1540 m s−1. It is on the data sheet, so the marks are in the angle: at θ = 90° the measured shift is zero.

CHECK YOURSELF

Ultrasound crosses a boundary where Z₁ = 1.6 × 106 kg m−2 s−1 and Z₂ = 1.7 × 106 kg m−2 s−1. Roughly what fraction of the intensity reflects, and why is that useful?

Show a hint

Put the numbers into the reflection coefficient. The answer is small.

Show the answer

Ir/I0 = (0.1/3.3)20.001, so about a tenth of a per cent reflects.

That is useful because imaging needs most of the pulse to carry on to deeper boundaries. Tiny reflections at each interface give faint but detectable echoes from every layer, instead of one blinding echo from the first.

Depth is ct over two, because the echo travelled there and back.

Impedance mismatch sets how much reflects, and the gel exists to remove the worst mismatch of all.

A-scan gives one line of depths. B-scan sweeps that line into a picture.

MR times how quickly tipped protons relax, and the relaxation time depends on the tissue: soft-tissue detail with no ionising dose.

WORKBOOK

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21 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

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CHECK YOUR PROGRESS

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  • Explain how a piezoelectric transducer generates and detects ultrasound pulses.
  • Use the pulse-echo technique and d = ct/2 to locate a boundary in tissue.
  • Say how the duration of the pulse and the wavelength limit the detail a pulse-echo scan can recover.
  • Distinguish an A-scan from a B-scan.
  • Use acoustic impedance Z = ρc and the reflection coefficient to explain what reflects where, and why the gel is needed.
  • Outline the endoscope as total internal reflection put to work.
  • Describe the principle of the MR scanner: protons precessing in a strong field, a radio-frequency pulse tipping them, and a relaxation time that depends on the tissue.
  • Weigh MR against CT and ultrasound for a given diagnostic task.
  • For CIE, use I = I₀e−μx for the attenuation of ultrasound in tissue.
  • For OCR, find the speed of blood from a Doppler shift with Δf/f = 2v cos θ/c.

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