Ultrasound Imaging

X-ray and CT imaging are wonderful tools but they share one significant drawback: ionising radiation. Every X-ray exposure carries a small but real risk of radiation-induced cell damage, and although the risks are small in relation to the diagnostic benefits, they are not zero. For many applications — in particular for pregnancy monitoring — doctors need an imaging modality that is completely free of ionising radiation. The answer, in use clinically since the 1950s, is ultrasound.

Ultrasound imaging uses high-frequency sound waves rather than electromagnetic radiation. The sound is emitted in short pulses by a handheld transducer, and the echoes reflected from tissue boundaries inside the body are detected by the same transducer and used to build up an image. The technique is cheap, portable, real-time and safe — it does not use ionising radiation and has no known adverse effects at the intensities employed. It is the standard imaging modality for obstetric care, for cardiac assessment (echocardiography), and for many other soft-tissue applications.

This lesson covers section 6.5.3 of the OCR A-Level Physics A specification (H556), introducing the physics of ultrasound generation, acoustic impedance, reflection coefficients, and the pulse-echo technique.

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What Is Ultrasound?

Ultrasound is simply sound at frequencies above the upper limit of human hearing (20 kHz). Medical ultrasound typically uses frequencies in the range 1 MHz to 20 MHz:

• Abdominal imaging: 2–5 MHz (deeper penetration, lower resolution)
• Cardiac imaging: 2–5 MHz
• Obstetric imaging: 3–7 MHz
• Vascular imaging: 5–10 MHz
• Superficial structures: 7–15 MHz (shallower, higher resolution)
• Intraoperative and endoscopic: up to 20 MHz

There is an inevitable trade-off: higher frequencies give higher resolution (because the wavelength is smaller) but are more strongly attenuated by tissue and so cannot penetrate as deeply. The clinician chooses the frequency to match the diagnostic task.

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The Piezoelectric Transducer

Ultrasound is generated and detected by the same device: a piezoelectric transducer. The active element is a disc of piezoelectric material — classically quartz, nowadays usually a synthetic ceramic called lead zirconate titanate (PZT).

Piezoelectricity is the property of certain crystals that they produce a voltage when mechanically stressed, and conversely they deform mechanically when a voltage is applied across them. A piezoelectric crystal thus acts as a two-way transducer: it can convert electrical signals into mechanical vibrations (and hence into sound waves in an adjacent medium), and it can also convert mechanical vibrations (from incoming sound waves) back into electrical signals.

An ultrasound transducer exploits this two-way behaviour. An electrical pulse applied to the crystal causes it to vibrate briefly at its natural resonant frequency, producing a short burst of ultrasound in the tissue. The same crystal then listens silently for returning echoes, which it converts back to electrical signals for display.

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How the Transducer Makes a Pulse

• A brief electrical pulse is applied to the piezoelectric element.
• The crystal deforms, generating an ultrasound pulse that travels into the tissue.
• The crystal then "rings" briefly at its resonant frequency before damping down (the resonant frequency determines the imaging frequency).
• The crystal falls silent and acts as a receiver.
• Echoes returning from tissue interfaces cause the crystal to vibrate, generating small electrical signals.
• The electronics amplify these signals and display them as a function of the time delay after the original pulse.
• The process repeats, thousands of times per second, building up an image in real time.

The crystal must be efficiently damped (by a backing material behind it) so that it stops ringing quickly after the pulse — otherwise the "dead time" between pulse and echo would be too long to detect close structures. Too much damping, though, lowers the sensitivity. Designing a transducer is a compromise between pulse length (favours damping) and sensitivity (favours resonance).

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Acoustic Impedance

When an ultrasound wave passes from one medium to another, some of it is transmitted and some is reflected. The amount reflected depends on the acoustic impedance mismatch between the two media.

Acoustic impedance Z is defined as:

Z = ρc

where \rho is the density of the medium and c is the speed of sound in that medium. The SI units are kg m⁻² s⁻¹ (also written as Rayls). Typical values for human tissues and some other materials:

Material	Density ρ (kg m⁻³)	Speed c (m s⁻¹)	Impedance Z (kg m⁻² s⁻¹)
Air	1.3	330	\sim 430
Water	1000	1480	1.48 × 10⁶
Fat	920	1450	1.33 × 10⁶
Soft tissue	1060	1540	1.63 × 10⁶
Muscle	1070	1580	1.69 × 10⁶
Bone	1900	4080	7.75 × 10⁶
Lung (gas-filled)	400	650	\sim 0.26 × 10⁶

Notice how similar the impedances of different soft tissues are: fat, muscle, and general "soft tissue" differ by only a few percent. This gives only modest reflections at soft-tissue boundaries, which is actually what you want for imaging (if most of the wave were reflected at the first interface, nothing would penetrate deeper). Bone has a much higher impedance, so strong reflections at bone boundaries tend to cast acoustic "shadows" — which is why ultrasound is poor for imaging beyond bones.

The impedance of air is very different from that of tissue, leading to near-total reflection at any air-tissue interface. This has two important consequences:

• Ultrasound cannot image through air pockets (e.g. bowel gas, lung tissue).
• A coupling gel must be placed between the transducer and the patient's skin to eliminate any air gap. Without the gel, the entire beam would reflect at the skin and nothing would enter the body.

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The Reflection Coefficient

When a sound wave travelling in medium 1 encounters a boundary with medium 2, the fraction of intensity reflected is given by:

I_r / I_0 = ((Z₂ - Z₁) / (Z₂ + Z₁))²

and the fraction transmitted is:

I_t / I_0 = 1 - I_r / I_0

This formula is central to every A-Level ultrasound question.

• If Z_1 = Z_2 (matched impedances), no reflection — the wave passes without any boundary echo.
• If Z_1 \gg Z_2 or Z_2 \gg Z_1 (mismatched impedances), nearly all the wave is reflected.
• The formula is independent of the direction of travel: the reflection coefficient at a given boundary is the same whether the wave goes from 1→2 or from 2→1.

Worked Example 1: Air-tissue interface

Air has impedance Z_{\text{air}} \approx 430 Rayls; soft tissue Z_{\text{tissue}} \approx 1.63 × 10^6 Rayls. What fraction of ultrasound intensity is reflected at an air-tissue boundary?

Solution.

I_r/I_0 = ((1.63 × 10⁶ - 430) / (1.63 × 10⁶ + 430))²
       ≈ (1.63 × 10⁶ / 1.63 × 10⁶)²
       ≈ 1 - 1.0 × 10⁻³
       ≈ 0.999

Essentially 100% of the ultrasound is reflected at an air-tissue interface. This is why coupling gel is absolutely essential.

Worked Example 2: Fat-muscle interface