Ultrasound Imaging

Spec mapping: OCR H556 Module 6.5 — Medical Imaging (ultrasound as a longitudinal mechanical wave at frequencies above the audible range; the piezoelectric transducer as combined emitter and detector; acoustic impedance $Z = \rho c$Z=ρc; the intensity reflection coefficient $I_r/I_0 = ((Z_2 - Z_1)/(Z_2 + Z_1))^2$Ir​/I0​=((Z2​−Z1​)/(Z2​+Z1​))2 at a boundary between two media; the pulse-echo technique; A-scan and B-scan display modes; the role of impedance-matching coupling gel; non-ionising character as the key safety distinction from X-ray and CT). Refer to the official OCR H556 specification document for exact wording.

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; the risks are small in relation to the diagnostic benefits but 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 routine clinical use since the 1950s, is ultrasound.

Ultrasound imaging uses high-frequency sound waves rather than electromagnetic radiation. Short pulses of ultrasound are emitted by a handheld transducer; 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 uses no ionising radiation and has no known adverse effects at the intensities employed. Ultrasound is the standard imaging modality for obstetric care, for cardiac assessment (echocardiography), for many soft-tissue applications, and (with the Doppler extension treated in the next lesson) for blood-flow measurement.

This lesson covers the physics of ultrasound generation in a piezoelectric transducer, acoustic impedance, reflection coefficients, the pulse-echo technique and A-scan / B-scan display modes — all within Module 6.5 — Medical Imaging of the OCR A-Level Physics A specification (H556).

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What Is Ultrasound?

Ultrasound is simply sound at frequencies above the upper limit of human hearing (20 kHz). It is a mechanical, longitudinal wave — pressure variations propagating through a material at the local speed of sound. This is fundamentally different from X-rays, which are electromagnetic transverse waves propagating in vacuum at the speed of light. The mechanical nature of ultrasound is what makes it non-ionising: a sound wave cannot strip electrons from atoms or break chemical bonds at the intensities used in clinical imaging.

Medical ultrasound typically uses frequencies in the range 2 MHz to 20 MHz (sometimes a little below 2 MHz for very deep abdominal imaging). The choice depends on the clinical task:

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

There is an inevitable trade-off: higher frequencies give higher resolution (the wavelength sets the smallest distinguishable feature) but are more strongly attenuated by tissue and so cannot penetrate as deeply. The clinician chooses the operating frequency to match the diagnostic task, often with a single transducer that can switch electronically among several frequencies.

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The Piezoelectric Transducer

Ultrasound is both 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) or, in the most modern devices, a single-crystal material such as PMN-PT.

Piezoelectricity is the property of certain crystals that they produce a voltage across themselves when mechanically stressed, and conversely they deform mechanically when a voltage is applied across them. The mechanism is the displacement of asymmetric positive and negative ions within the crystal lattice. A piezoelectric crystal acts as a two-way transducer: it converts electrical signals into mechanical vibrations (and hence into sound waves in an adjacent medium), and it converts mechanical vibrations from incoming sound waves back into electrical signals.

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

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How the Transducer Makes a Pulse

The pulse-generation sequence is:

• A brief electrical pulse (typically a few hundred volts, lasting under a microsecond) is applied to the piezoelectric element.
• The crystal mechanically deforms and "rings" briefly at its natural resonant frequency, generating an ultrasound pulse that travels into the tissue.
• A damping backing material (often tungsten-loaded epoxy) absorbs the back-going sound and stops the ringing rapidly — typically after 2–3 cycles. Too much damping reduces sensitivity; too little extends pulse length and degrades axial resolution.
• The crystal then falls electrically silent and acts as a receiver.
• Echoes returning from tissue interfaces deform the crystal, generating small voltages.
• The electronics amplify these voltages and record them as a function of time delay after the original pulse.
• The whole cycle repeats thousands of times per second — typical pulse repetition frequencies are 1–10 kHz.

The resonant frequency of the crystal is set by its thickness: a half-wavelength of ultrasound in the crystal material equals the crystal thickness. For 5 MHz operation in PZT (sound speed $\sim 4000$∼4000 m s$^{-1}$−1), the crystal is about $0.4$0.4 mm thick. Thicker crystals give lower frequencies; thinner crystals give higher.

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Acoustic Impedance

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

The acoustic impedance $Z$Z of a medium is defined as

$Z = \rho c$Z=ρc

where $\rho$ρ is the density of the medium and $c$c is the speed of sound in that medium. The SI units are kg m$^{-2}$−2 s$^{-1}$−1 (also called rayls). Typical values for human tissues and other relevant materials:

Material	Density $\rho$ρ (kg m$^{-3}$−3)	Speed $c$c (m s$^{-1}$−1)	Impedance $Z$Z (kg m$^{-2}$−2 s$^{-1}$−1)
Air	1.3	330	$\sim 430$∼430
Water	1000	1480	$1.48 \times 10^{6}$1.48×106
Fat	920	1450	$1.33 \times 10^{6}$1.33×106
Soft tissue (mean)	1060	1540	$1.63 \times 10^{6}$1.63×106
Muscle	1070	1580	$1.69 \times 10^{6}$1.69×106
Bone (cortical)	1900	4080	$7.75 \times 10^{6}$7.75×106
Lung (gas-filled)	400	650	$\sim 0.26 \times 10^{6}$∼0.26×106

Notice how similar the impedances of different soft tissues are: fat, muscle and "soft tissue" differ by only a few percent. This gives only modest reflections at soft-tissue boundaries — which is exactly what you want for imaging. If most of the wave were reflected at the first interface, nothing would penetrate any deeper into the body.

Bone has a much higher impedance, so reflections at bone boundaries are strong and the bone casts an acoustic shadow behind it: structures lying behind bone are invisible because so little of the beam gets through. This is why ultrasound is poor for imaging through the rib cage and useless for imaging the brain in adults (the skull blocks it).

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

• Ultrasound cannot image through air pockets (bowel gas, lung tissue) — anything beyond an air interface is hidden.
• A coupling gel must be placed between the transducer and the patient's skin, to eliminate the air gap at the entry surface and let the beam pass into the body in the first place.

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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

$\dfrac{I_r}{I_0} = \left(\dfrac{Z_2 - Z_1}{Z_2 + Z_1}\right)^{2}$I0​Ir​​=(Z2​+Z1​Z2​−Z1​​)2

and the fraction transmitted is

$\dfrac{I_t}{I_0} = 1 - \dfrac{I_r}{I_0} = \dfrac{4 Z_1 Z_2}{(Z_2 + Z_1)^{2}}.$I0​It​​=1−I0​Ir​​=(Z2​+Z1​)24Z1​Z2​​.

This formula is the central A-Level ultrasound equation.

• If $Z_1 = Z_2$Z1​=Z2​ (matched impedances), the reflected fraction is zero — the wave passes the boundary as if it were not there.
• If $Z_1 \gg Z_2$Z1​≫Z2​ or $Z_2 \gg Z_1$Z2​≫Z1​ (badly mismatched impedances), nearly all the wave is reflected.
• The formula is symmetric in $Z_1$Z1​ and $Z_2$Z2​: 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 $Z_\text{air} \approx 430$Zair​≈430 rayls; soft tissue $Z_\text{tissue} \approx 1.63 \times 10^{6}$Ztissue​≈1.63×106 rayls. What fraction of ultrasound intensity is reflected at an air–tissue boundary?

Solution.

$$\begin{aligned} \dfrac{I_r}{I_0} &= \left(\dfrac{1.63 \times 10^{6} - 430}{1.63 \times 10^{6} + 430}\right)^{2} \\ &\approx \left(\dfrac{1.63 \times 10^{6}}{1.63 \times 10^{6}}\right)^{2} \\ &\approx 1 - 1.0 \times 10^{-3} \\ &\approx 0.999. \end{aligned}$$I0​Ir​​​=(1.63×106+4301.63×106−430​)2≈(1.63×1061.63×106​)2≈1−1.0×10−3≈0.999.​

Essentially 100% of the ultrasound is reflected at an air–tissue interface — which is why coupling gel is absolutely essential, and why bowel gas and lung tissue cast complete acoustic shadows.

Worked Example 2: Fat–muscle interface

Fat has $Z \approx 1.33 \times 10^{6}$Z≈1.33×106 rayls; muscle $Z \approx 1.69 \times 10^{6}$Z≈1.69×106 rayls. What fraction of intensity is reflected at a fat–muscle boundary?

Solution.

$$\begin{aligned} \dfrac{I_r}{I_0} &= \left(\dfrac{1.69 - 1.33}{1.69 + 1.33}\right)^{2} \\ &= \left(\dfrac{0.36}{3.02}\right)^{2} \\ &\approx 0.0142. \end{aligned}$$I0​Ir​​​=(1.69+1.331.69−1.33​)2=(3.020.36​)2≈0.0142.​

About 1.4% is reflected — a small but easily detectable signal. The other 98.6% continues onward to reach deeper structures and contribute to the rest of the image. This is the sweet spot: enough reflection to detect the boundary, not so much that the deeper tissue is hidden.

Worked Example 3: Soft tissue–bone interface

Soft tissue $Z \approx 1.63 \times 10^{6}$Z≈1.63×106 rayls; cortical bone $Z \approx 7.75 \times 10^{6}$Z≈7.75×106 rayls. What fraction is reflected?

Solution.

$$\begin{aligned} \dfrac{I_r}{I_0} &= \left(\dfrac{7.75 - 1.63}{7.75 + 1.63}\right)^{2} \\ &= \left(\dfrac{6.12}{9.38}\right)^{2} \\ &\approx 0.426. \end{aligned}$$I0​Ir​​​=(7.75+1.637.75−1.63​)2=(9.386.12​)2≈0.426.​

About 43% of the intensity is reflected at a tissue–bone interface. This is why bones appear as very bright features on ultrasound images but also cast strong acoustic shadows behind them — the bulk of the remaining intensity is absorbed by the bone itself, and very little makes it through to deeper tissue.

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The Pulse-Echo Technique

The basic ultrasound imaging technique is pulse-echo. The transducer emits a short pulse and then listens for returning echoes. The time delay between pulse and echo gives the depth of the reflecting boundary:

$d = \tfrac{1}{2} c\,t$d=21​ct

where $c$c is the speed of sound in the medium and $t$t is the round-trip time (there and back, hence the factor of $\tfrac{1}{2}$21​).

For soft tissue, $c \approx 1540$c≈1540 m s$^{-1}$−1 is used as the standard reference value:

$t = \dfrac{2d}{c} = \dfrac{2 \times 0.1}{1540} \approx 1.3 \times 10^{-4}\,\text{s} \quad (\text{for } d = 10\,\text{cm}).$t=c2d​=15402×0.1​≈1.3×10−4s(for d=10cm).

An echo returning 130 $\mu$μs after the pulse therefore came from a boundary 10 cm deep. Pulses are repeated at rates of a few kHz; between pulses, echoes from different depths arrive in sequence, and the transducer/electronics assemble them into a one-dimensional line image. By sweeping the beam across the tissue — either electronically with a phased array of many small transducer elements, or mechanically with a rocking transducer — a full 2-D image is built up.

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A-Scan vs B-Scan

There are two fundamental display modes:

A-scan (amplitude mode). A simple 1-D plot of echo amplitude against time (or equivalently, depth). Each peak represents a tissue boundary; the height of the peak gives the reflection coefficient and the time gives the depth. A-scans are still used in ophthalmic measurements (e.g. to measure the axial length of the eye for cataract surgery) but are rarely seen in general imaging today.

B-scan (brightness mode). A 2-D image formed by sweeping the ultrasound beam across the tissue and displaying each line of echoes as a vertical strip whose brightness corresponds to the echo amplitude. The standard "ultrasound picture" you see in obstetric clinics is a B-scan. Modern scanners display B-scans in real time at frame rates of 30+ per second.

In a B-scan, the column corresponding to a given beam-sweep position is precisely the A-scan from that direction — converted from amplitude-vs-time into brightness-vs-depth. The B-scan is the family of all such columns laid side by side.

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Attenuation of Ultrasound

Like X-rays, ultrasound is attenuated exponentially as it passes through tissue:

$I = I_0\, e^{-\mu x}$I=I0​e−μx

with $\mu$μ the (acoustic) attenuation coefficient — not to be confused with the X-ray $\mu$μ of the previous lessons. Attenuation of ultrasound in soft tissue is roughly proportional to frequency: 10 MHz ultrasound is attenuated about twice as fast as 5 MHz, and so can penetrate only half as deep. This is the fundamental reason for the frequency-resolution trade-off: higher frequency = shorter wavelength = better axial resolution, but higher attenuation = smaller imaging depth.

The mechanisms of ultrasound attenuation are:

• Absorption — conversion of mechanical wave energy into heat as the tissue vibrates against its own viscosity.
• Scattering at small inhomogeneities (cells, small blood vessels) that are below the wavelength scale.
• Beam spreading — the approximately cylindrical or focussed beam of a real transducer diverges with distance, reducing the intensity per unit area even without absorption.

Modern transducers compensate for the depth-dependent attenuation by time-gain compensation (TGC) — applying progressively higher electronic gain to echoes from deeper tissues, so that all depths appear with comparable brightness in the final image.

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Coupling Gel and Impedance Matching

A thin layer of water-based coupling gel is always applied between the transducer and the patient's skin. The gel has acoustic impedance close to that of soft tissue, so the wave passes from transducer into gel into skin with only small reflections at each interface. Without the gel, the air gap between the transducer face and the skin would act as a near-perfect acoustic mirror (reflection coefficient $\sim 99.9\%$∼99.9%, as we calculated in Worked Example 1), and essentially no ultrasound would enter the body. The gel is inert, non-toxic, water-soluble and easily wiped off after the scan — a tiny inconvenience that makes the whole technique workable.