Radiation and Medical Imaging

Lessons

Senior 6 · Physics

Radiation and Medical Imaging

5 lessons

Lesson 3 of 5

Sound in Medical Imaging: Ultrasound Scanning

25 min

A heartbeat heard through the skin

In an antenatal clinic a midwife spreads cold gel on a pregnant woman's abdomen and presses a small probe against it. A grey, grainy picture appears on the screen, and moments later the room fills with a rapid whooshing beat: the baby's heart, beating more than twice as fast as the mother's. No radiation dose has been given, no needle used, and the examination can be repeated next month. In Unit 1 you met the basics: echoes from boundaries, the acoustic impedance $Z = \rho v$ and the reflected fraction $R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$ . This lesson follows the pulse from the crystal that makes it to the picture on the screen, and asks what limits how deep and how sharp that picture can be.

Making and detecting the pulse

Some crystals and ceramics, such as quartz, barium titanate and lead zirconate titanate (PZT), are piezoelectric: squeezing them produces a voltage across them, and a voltage across them makes them change thickness. An alternating voltage at the crystal's resonant frequency makes it vibrate strongly and send ultrasound into the body. When an echo returns, the same crystal is squeezed by it and produces a small voltage that the machine amplifies.

Definition

The piezoelectric effect is the production of a potential difference across certain crystals when they are mechanically stressed; the reverse piezoelectric effect is the change in their dimensions when a potential difference is applied. A transducer is a device that converts energy from one form to another; an ultrasound transducer converts electrical oscillations to ultrasound and back.

Figure 3.1: Inside an ultrasound transducer.

The crystal resonates when its thickness is half a wavelength of sound in the crystal, $t = \frac{\lambda_{\text{crystal}}}{2}$ , so thinner crystals give higher frequencies.
The backing block damps the vibration after two or three cycles, so the pulse is short. A short pulse separates boundaries that are close together (good axial resolution, about half the pulse length). Between pulses the crystal listens for echoes.
The matching layer and the gel deal with the impedance mismatch at each end of the path, as Worked example 1 shows.

Worked example 1: matching the crystal to the body

Barium titanate has $\rho = 5600\,\mathrm{kg\,m^{-3}}$ and $v = 5500\,\mathrm{m\,s^{-1}}$ ; soft tissue (muscle) has $\rho = 1075\,\mathrm{kg\,m^{-3}}$ and $v = 1590\,\mathrm{m\,s^{-1}}$ . (a) Find both acoustic impedances. (b) What fraction of the intensity would be reflected straight back into a crystal pressed directly on tissue? (c) Find the impedance a matching layer should have so that the reflections at its two faces are equal, and the fraction reflected at each face.

Solution

(a) $Z_c = 5600 \times 5500 = 3.08 \times 10^{7}\,\mathrm{kg\,m^{-2}\,s^{-1}}$ ; $Z_t = 1075 \times 1590 = 1.71 \times 10^{6}\,\mathrm{kg\,m^{-2}\,s^{-1}}$ .

(b) $R = \left(\frac{3.08 \times 10^{7} - 1.709 \times 10^{6}}{3.08 \times 10^{7} + 1.709 \times 10^{6}}\right)^2 = (0.895)^2 = 0.80$ . Four-fifths of the energy would never leave the probe.

(c) Equal reflections need $\frac{Z_c}{Z_m} = \frac{Z_m}{Z_t}$ , so $Z_m = \sqrt{Z_cZ_t} = \sqrt{(3.08 \times 10^{7})(1.709 \times 10^{6})} = 7.26 \times 10^{6}\,\mathrm{kg\,m^{-2}\,s^{-1}}$ . At each face $R = \left(\frac{3.08 - 0.726}{3.08 + 0.726}\right)^2 = 0.38$ .

Why it works: the layer is made a quarter of a wavelength thick. The wave reflected from its back face travels an extra half wavelength, so it cancels the equal wave reflected from its front face, exactly as in the anti-reflection coating on a lens. At the design frequency almost all the energy passes into the body.

From echoes to a picture

The machine times each echo and assumes a speed of $1540\,\mathrm{m\,s^{-1}}$ , the average for soft tissue, to place it at depth $d = \frac{ct}{2}$ . The table recomputes the impedances from density and speed.

Medium	$\rho$ / $\mathrm{kg\,m^{-3}}$	$v$ / $\mathrm{m\,s^{-1}}$	$Z = \rho v$ / $\mathrm{kg\,m^{-2}\,s^{-1}}$
air	1.3	330	429
water (similar to gel)	1000	1500	$1.50 \times 10^{6}$
blood	1060	1570	$1.66 \times 10^{6}$
fat	925	1450	$1.34 \times 10^{6}$
muscle (average)	1075	1590	$1.71 \times 10^{6}$
bone (varies)	1400 to 1900	4080	$5.71 \times 10^{6}$ to $7.75 \times 10^{6}$

Densities and speeds from OpenStax College Physics, Table 17.7.1; impedances recalculated by Ganzaa.

The intensity reflected at a boundary met head-on is $R$ and the intensity transmitted is $1 - R$ . On its way down and back, a pulse also loses energy to absorption and scattering: in soft tissue the attenuation is roughly 0.5 dB per centimetre for each megahertz of frequency, where a loss of $n$ dB means the intensity is multiplied by $10^{-\frac{n}{10}}$ . Deep echoes are therefore far weaker than shallow ones from the same kind of boundary, so the machine applies time-gain compensation (TGC): it amplifies each echo more the later it arrives.

Figure 3.2: The pulse-echo principle and the A-scan, before and after time-gain compensation.

An A-scan (amplitude scan) is a single line: echo voltage against time. It is used to measure distances, such as the length of the eyeball.
A B-scan (brightness scan) shows each echo as a dot whose brightness depends on its strength. An array of many small crystals fires line after line across a fan or rectangle, and the lines build a two-dimensional cross-section. Repeated many times a second, this is the real-time picture in the clinic.
Because only one pulse may be travelling at a time, the deepest echo must return before the next pulse leaves. This limits the pulse repetition frequency, and so the number of lines and frames per second, as Worked example 3 shows.

Worked example 2: how strong is a deep echo?

A 5.0 MHz pulse passes from gel (treat as water) into 2.0 cm of fat, then 3.0 cm of muscle, and is reflected at a muscle–bone boundary (take the bone value $7.75 \times 10^{6}$ ). Using the table, find (a) the fraction of the emitted intensity returning to the probe, ignoring attenuation, and (b) the fraction when attenuation of 0.5 dB per cm per MHz is included.

Solution

Gel to fat: $R = \left(\frac{1.341 - 1.500}{1.341 + 1.500}\right)^2 = 0.0031$ , so $1 - R = 0.9969$ .

Fat to muscle: $R = \left(\frac{1.709 - 1.341}{1.709 + 1.341}\right)^2 = 0.0146$ , so $1 - R = 0.9854$ .

Muscle to bone: $R = \left(\frac{7.752 - 1.709}{7.752 + 1.709}\right)^2 = 0.408$ .

(a) The pulse crosses each soft boundary twice (down and back): $0.9969^2 \times 0.9854^2 \times 0.408 = 0.39$ .

(b) The round trip is 10.0 cm: loss $= 0.5 \times 5.0 \times 10.0 = 25\,\mathrm{dB}$ , a factor $10^{-2.5} = 3.16 \times 10^{-3}$ . Returning fraction $= 0.3937 \times 3.16 \times 10^{-3} = 1.2 \times 10^{-3}$ .

Conclusion: attenuation, not reflection, is what makes deep echoes weak. That is why TGC is essential, and why bone, reflecting about 40% and absorbing strongly, leaves an "acoustic shadow" behind it.

A B-scan of a fetus at 20 weeks, seen in profile. The fan shape comes from a curved array of crystals; strong echoes (bone of the skull and spine) are bright, fluid is black. Along the top the screen records the depth of view (16.3 cm), the frame rate (32 Hz) and the mechanical index (MI 1.1). Photo: Goleisureintl, CC BY 4.0, via Wikimedia Commons

Worked example 3: how many lines in a frame?

The scan in the photo shows tissue to a depth of 16.3 cm at 32 frames per second. Taking $c = 1540\,\mathrm{m\,s^{-1}}$ , find (a) the round-trip time to the deepest point, (b) the maximum pulse repetition frequency, and (c) the maximum number of scan lines in each frame.

Solution

(a) $t = \frac{2d}{c} = \frac{2 \times 0.163}{1540} = 2.12 \times 10^{-4}\,\mathrm{s}$ .

(b) $f_{\text{PRF}} = \frac{1}{t} = \frac{1}{2.117 \times 10^{-4}} = 4.7 \times 10^{3}\,\mathrm{Hz}$ .

Trade-off: looking deeper or refreshing faster leaves fewer lines per frame, so the picture becomes coarser. Depth, frame rate and line density cannot all be increased together.

Choosing the frequency

Higher frequency gives a shorter wavelength, $\lambda = \frac{c}{f}$ , and shorter pulses, so finer detail; but attenuation grows in proportion to frequency, so the pulse cannot go as deep. At 3.5 MHz, $\lambda = 0.44\,\mathrm{mm}$ and an echo from 12 cm deep (24 cm round trip) loses about $0.5 \times 3.5 \times 24 = 42\,\mathrm{dB}$ . At 10 MHz, $\lambda = 0.15\,\mathrm{mm}$ but the same echo would lose 120 dB, a factor of $10^{-12}$ , and be lost in electronic noise. So abdominal and obstetric scans use about 2 MHz to 5 MHz, while the thyroid, tendons, the eye and blood vessels near the surface are scanned at 7 MHz to 15 MHz.

Doppler ultrasound

Echoes from moving red blood cells are shifted in frequency. In Unit 1 you derived $\Delta f \approx \frac{2fv}{c}$ for blood moving straight along the beam. In practice the beam meets the vessel at an angle $\theta$ (Figure 3.3), and only the velocity component along the beam counts:

$\Delta f = \frac{2fv\cos\theta}{c} \qquad (v \ll c)$

Figure 3.3: Doppler geometry.

Colour Doppler shows flow towards the probe in one colour (usually red) and away in another (blue) on top of the B-scan; spectral Doppler plots the speeds against time, beat by beat. Doctors use it to find narrowed arteries, blood clots in leg veins and problems with blood flow to a baby through the umbilical cord. The angle must be kept below about 60°: as $\theta$ approaches 90°, $\cos\theta$ approaches zero and small errors in the angle produce large errors in the speed.

Worked example 4: blood speed, and the danger of a wrong angle

A 5.0 MHz Doppler beam meets an artery at 60° to the flow and the measured shift is 1.30 kHz. (a) Find the blood speed ( $c = 1540\,\mathrm{m\,s^{-1}}$ ). (b) The true angle was 55°. What is the true speed, and what was the percentage error?

Solution

(a) $v = \frac{c\,\Delta f}{2f\cos\theta} = \frac{1540 \times 1300}{2 \times 5.0 \times 10^{6} \times \cos 60^\circ} = \frac{2.002 \times 10^{6}}{5.0 \times 10^{6}} = 0.40\,\mathrm{m\,s^{-1}}$ .

(b) $v = \frac{2.002 \times 10^{6}}{2 \times 5.0 \times 10^{6} \times \cos 55^\circ} = \frac{2.002 \times 10^{6}}{5.736 \times 10^{6}} = 0.35\,\mathrm{m\,s^{-1}}$ . Error $= \frac{0.4004 - 0.3490}{0.3490} \times 100\% = 15\%$ .

A 5° error in the angle gave a 15% error in the speed, and the error grows rapidly at larger angles.

Experiment 3.1: pulse-echo timing in air

The ultrasonic distance sensors sold for school electronics and robotics projects work on exactly the same principle at 40 kHz. Point one at a wall 0.500 m away (measured with a metre rule), trigger a pulse from a microcontroller or a signal generator, and display the transmitted pulse and its echo on an oscilloscope. In one trial the echo arrived 2.92 ms after the pulse (illustrative reading), giving $v = \frac{2d}{t} = \frac{2 \times 0.500}{2.92 \times 10^{-3}} = 342\,\mathrm{m\,s^{-1}}$ , close to the speed of sound in air at about 20 °C. Put a sheet of card halfway to the wall: two echoes appear, and the stronger one comes from whichever surface reflects more. That is an A-scan.

Safety

Ultrasound is not ionising, and no harmful effect has been shown at diagnostic intensities. It does deposit some energy as heat, and at high pressures it can make tiny gas bubbles oscillate (cavitation). Scanners therefore display a thermal index and a mechanical index, and operators keep the output and the scanning time as low as the examination allows: ALARA applies here too. Keepsake scans with no medical purpose are discouraged for this reason.

Warning

Common mistakes. (1) Forgetting the factor 2 in $d = \frac{ct}{2}$ or in the Doppler shift: the sound goes there and back, and the moving blood shifts the frequency twice. (2) Using $\sin\theta$ instead of $\cos\theta$ : $\theta$ is the angle between the beam and the flow. (3) Forgetting that transmission at a boundary is $1 - R$ , and that a pulse crosses each boundary twice. (4) Thinking a higher frequency is always better: it improves detail but costs depth.

Tip

For any ultrasound question, ask three things in order. Where? $d = \frac{ct}{2}$ . How bright? $R$ at the boundary, $(1 - R)$ at every boundary crossed, and the attenuation in dB. How fast? $\Delta f = \frac{2fv\cos\theta}{c}$ . Remember also: high frequency, shallow and sharp; low frequency, deep and coarse.

How to see with sound, by Jacques S. Abramowicz (TED-Ed). Watch how the piezoelectric probe both sends and listens, and why bone and air give strong echoes.

✦

Practice

Explain how one piezoelectric crystal can both produce ultrasound and detect its echoes.
A barium titanate crystal ( $v = 5500\,\mathrm{m\,s^{-1}}$ ) is to resonate at 3.5 MHz. Find its thickness.
Using the table, find the fraction of intensity reflected at a blood–muscle boundary, and explain why the walls of the heart chambers are still visible.
A probe receives an echo 0.13 ms after sending a pulse. How deep is the boundary?
An echo returns from 8.0 cm deep in soft tissue at 3.5 MHz. Find the attenuation in dB over the round trip (0.5 dB per cm per MHz) and the fraction of intensity that survives it.
A 4.0 MHz Doppler beam at 45° to an artery gives a shift of 2.2 kHz. Find the blood speed.
A scan is set to 20 cm depth. Find the maximum pulse repetition frequency and the frame rate possible with 128 lines per frame.
A doctor wants to look at a tendon 1 cm under the skin of the wrist and, later, at a kidney 10 cm deep. Evaluate which frequency to use for each, with reasons.

Show answersHide answers

An alternating voltage makes the crystal vibrate (reverse piezoelectric effect), emitting a pulse. Between pulses, a returning echo squeezes the crystal, which then produces a small voltage (piezoelectric effect) that is amplified and timed.
$\lambda = \frac{5500}{3.5 \times 10^{6}} = 1.57\,\mathrm{mm}$ ; $t = \frac{\lambda}{2} = 0.79\,\mathrm{mm}$ .
$R = \left(\frac{1.709 - 1.664}{1.709 + 1.664}\right)^2 = 1.8 \times 10^{-4}$ . Very weak, but TGC and a sensitive receiver still detect it, and the moving wall stands out against the black (echo-free) blood.
$d = \frac{1540 \times 0.13 \times 10^{-3}}{2} = 0.10\,\mathrm{m} = 10\,\mathrm{cm}$ .
Loss $= 0.5 \times 3.5 \times 16 = 28\,\mathrm{dB}$ ; fraction $= 10^{-2.8} = 1.6 \times 10^{-3}$ .
$v = \frac{1540 \times 2200}{2 \times 4.0 \times 10^{6} \times \cos 45^\circ} = \frac{3.388 \times 10^{6}}{5.657 \times 10^{6}} = 0.60\,\mathrm{m\,s^{-1}}$ .
$t = \frac{2 \times 0.20}{1540} = 2.60 \times 10^{-4}\,\mathrm{s}$ ; PRF $= 3.85 \times 10^{3}\,\mathrm{Hz}$ ; frame rate $= \frac{3850}{128} = 30\,\mathrm{Hz}$ .
Tendon: a high frequency (about 10 MHz to 15 MHz) gives fine detail, and the attenuation over 2 cm is small ( $0.5 \times 12 \times 2 = 12\,\mathrm{dB}$ at 12 MHz). Kidney: a low frequency (about 3 MHz to 5 MHz), because at 12 MHz the 20 cm round trip would lose 120 dB; the image is coarser, but the echoes come back.

Key points

A piezoelectric transducer sends short pulses (backing block) and receives echoes; the matching layer ( $Z_m = \sqrt{Z_1Z_2}$ , a quarter-wave thick) and the gel let the energy into the body.
Echo strength: $R$ at the reflecting boundary, $1 - R$ at every boundary crossed, plus attenuation (about 0.5 dB per cm per MHz); TGC compensates.
A-scan: one line, amplitude against time; B-scan: a 2-D brightness image built line by line; depth, frame rate and line density trade off.
Higher frequency: finer detail, less depth. Doppler: $\Delta f = \frac{2fv\cos\theta}{c}$ , with $\theta$ kept below about 60°. Ultrasound is non-ionising; ALARA still applies.

Sources and credits

17.7: Ultrasound (including the densities and speeds in the table), from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Obstetric ultrasound scan monitor showing 20-week human fetus profile by Goleisureintl, Wikimedia Commons, licensed under CC BY 4.0.
Photo: CarotidDoppler1.jpg by Etan J. Tal, Wikimedia Commons, licensed under CC BY 3.0.
Photo: A medical ultrasound linear array probe, scan head, transducer by Harrison Keely, Wikimedia Commons, licensed under CC BY 4.0.
Video: How to see with sound - Jacques S. Abramowicz, TED-Ed, embedded from YouTube.

The three diagrams, the experiment, the recalculated impedances, the worked examples and the practice exercises are written by Ganzaa.

1A heartbeat heard through the skin

2Making and detecting the pulse

3From echoes to a picture

4Choosing the frequency

5Doppler ultrasound

6Experiment 3.1: pulse-echo timing in air

7Safety

A heartbeat heard through the skin

Making and detecting the pulse

From echoes to a picture

Choosing the frequency

Doppler ultrasound

Experiment 3.1: pulse-echo timing in air

Safety