Production and use of ultrasound

1. Overview

Ultrasound imaging is a sophisticated diagnostic tool that utilizes high-frequency longitudinal waves (above 20 kHz) to probe the internal structures of the human body. The fundamental physics of ultrasound involves the piezoelectric effect for wave generation and detection, the interaction of sound with matter through reflection, refraction, and attenuation, and the quantitative analysis of acoustic impedance. By measuring the time delay and intensity of reflected pulses, clinicians can reconstruct detailed images of soft tissues and organs without the ionizing radiation risks associated with X-rays. The technique relies on the principle that different tissues have different acoustic impedances, leading to varying degrees of reflection at tissue boundaries. These reflections are then processed to create a visual representation of the internal anatomy. The resolution of the image is directly related to the frequency of the ultrasound waves used; higher frequencies provide better resolution but penetrate less deeply into the body. Therefore, the choice of frequency is a trade-off between resolution and penetration depth, depending on the specific application.

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Key Definitions

• Ultrasound: Longitudinal sound waves with a frequency greater than 20 kHz. In medical diagnostics, frequencies typically range from 1 MHz to 15 MHz. Higher frequencies offer better resolution but have reduced penetration depth.
• Piezoelectric Effect: The generation of an e.m.f. across the faces of a crystal when it is mechanically deformed (compressed or stretched). This effect is used to detect returning ultrasound waves.
• Converse Piezoelectric Effect: The change in shape of a piezoelectric crystal when a potential difference (p.d.) is applied across its faces. This effect is used to generate ultrasound waves.
• Transducer: A device that converts energy from one form to another. An ultrasound transducer converts electrical energy into sound energy (transmitter) and sound energy into electrical energy (receiver). It contains the piezoelectric crystal and associated circuitry.
• Specific Acoustic Impedance ($Z$): The product of the density ($\rho$) of a medium and the speed of sound ($c$) in that medium. It represents the resistance of the medium to the transmission of sound. Mathematically, $Z = \rho c$.
• Intensity Reflection Coefficient ($\alpha$): The ratio of the reflected intensity ($I_R$) to the incident intensity ($I_0$) at a boundary between two media. It depends on the acoustic impedances of the two media ($Z_1$ and $Z_2$) and is given by $\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$.
• Attenuation: The exponential decrease in the intensity of a wave as it travels through a medium, caused by absorption (conversion to heat) and scattering. Attenuation limits the depth of penetration of ultrasound.
• Linear Absorption Coefficient ($\mu$): A constant for a specific medium and frequency that determines the rate at which ultrasound intensity is attenuated per unit length. The intensity $I$ at a depth $x$ is given by $I = I_0 e^{-\mu x}$, where $I_0$ is the initial intensity.

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Content

3.1 The Piezoelectric Crystal

The active element in an ultrasound transducer is a piezoelectric crystal, such as quartz or synthetic ceramics like Lead Zirconate Titanate (PZT). These materials possess a non-symmetrical molecular structure.

• Molecular Mechanism: When a p.d. is applied, the ions within the crystal lattice shift, causing the crystal to expand or contract. If an alternating p.d. is applied, the crystal vibrates at the same frequency as the electrical signal.
• Resonance: To maximize the amplitude of the ultrasound waves, the frequency of the applied alternating p.d. is set equal to the natural frequency of the crystal. This is achieved by cutting the crystal to a thickness equal to half the wavelength ($\lambda/2$) of the desired ultrasound.

3.2 Acoustic Impedance and Reflection

The reflection of ultrasound at the boundary between two tissues is determined by the difference in their acoustic impedances. A large difference in acoustic impedance leads to a strong reflection, while a small difference results in a weak reflection. This principle is crucial for creating contrast in ultrasound images.

• Calculating Reflection Coefficient: The intensity reflection coefficient ($\alpha$) quantifies the fraction of the incident ultrasound intensity that is reflected at an interface. As defined above, it is given by $\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$, where $Z_1$ and $Z_2$ are the acoustic impedances of the two media.

• Worked Example: Consider an ultrasound wave traveling from soft tissue ($Z_1 = 1.63 \times 10^6 \, \text{kg m}^{-2} \text{s}^{-1}$) to bone ($Z_2 = 6.36 \times 10^6 \, \text{kg m}^{-2} \text{s}^{-1}$). The intensity reflection coefficient is:

  $\alpha = \left(\frac{6.36 \times 10^6 - 1.63 \times 10^6}{6.36 \times 10^6 + 1.63 \times 10^6}\right)^2 = \left(\frac{4.73 \times 10^6}{7.99 \times 10^6}\right)^2 \approx (0.592)^2 \approx 0.35$.

  This means that approximately 35% of the ultrasound intensity is reflected at the soft tissue-bone interface.

3.3 Attenuation of Ultrasound

As ultrasound travels through tissue, its intensity decreases due to absorption and scattering. This attenuation limits the depth to which ultrasound can effectively penetrate.

• Attenuation Coefficient: The linear absorption coefficient ($\mu$) is a measure of how quickly the ultrasound intensity decreases with distance. The intensity $I$ at a depth $x$ is given by $I = I_0 e^{-\mu x}$, where $I_0$ is the initial intensity. The units of $\mu$ are typically $\text{cm}^{-1}$.

• Worked Example: Suppose ultrasound with an initial intensity of $I_0 = 100 \, \text{mW/cm}^2$ is used to image tissue with a linear absorption coefficient of $\mu = 0.5 \, \text{cm}^{-1}$. We want to find the intensity of the ultrasound at a depth of $x = 4 \, \text{cm}$.

  Using the formula $I = I_0 e^{-\mu x}$, we have:

  $I = 100 \, \text{mW/cm}^2 \cdot e^{-0.5 \, \text{cm}^{-1} \cdot 4 \, \text{cm}} = 100 \, \text{mW/cm}^2 \cdot e^{-2} \approx 100 \, \text{mW/cm}^2 \cdot 0.135 \approx 13.5 \, \text{mW/cm}^2$.

  This shows that the intensity of the ultrasound has significantly decreased after traveling 4 cm through the tissue.

3.4 Applications of Ultrasound

Ultrasound has a wide range of applications in medical diagnostics and therapy.

• Diagnostic Imaging: Ultrasound is used to visualize organs, tissues, and blood flow. Common applications include:

  • Obstetrics: Monitoring fetal development during pregnancy.
  • Cardiology: Assessing heart function and detecting heart disease.
  • Abdominal Imaging: Examining the liver, kidneys, gallbladder, and pancreas.
  • Musculoskeletal Imaging: Evaluating muscles, tendons, and ligaments.

• Therapeutic Ultrasound: High-intensity focused ultrasound (HIFU) can be used to:

  • Ablate Tumors: Destroy cancerous tissue by heating it.
  • Break up Kidney Stones: Fragment kidney stones into smaller pieces that can be passed more easily.
  • Drug Delivery: Enhance the delivery of drugs to specific tissues.

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Key Equations

Equation	What it gives
${Z = \rho c}$	Specific acoustic impedance of a medium
${\dfrac{I_R}{I_0} = \dfrac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}}$	Fraction of the incident intensity reflected at a boundary, called the intensity reflection coefficient $\alpha$
${\dfrac{I_T}{I_0} = 1 - \dfrac{I_R}{I_0}}$	Fraction transmitted into the second medium (from conservation of energy at the boundary)
${I = I_0 e^{-\mu x}}$	Intensity after travelling a distance $x$ through a medium
${2d = ct}$	Depth $d$ of a boundary from the echo time $t$: the pulse travels there and back

Symbols and units:

• $\rho$ = density ($\text{kg m}^{-3}$) and $c$ = speed of sound in the medium ($\text{m s}^{-1}$), so $Z$ is in $\text{kg m}^{-2}\,\text{s}^{-1}$.
• $Z_1$ and $Z_2$ = acoustic impedances of the two media. The ratios $\frac{I_R}{I_0}$ and $\frac{I_T}{I_0}$ have no unit.
• $\mu$ = linear absorption (attenuation) coefficient, in $\text{m}^{-1}$ or $\text{cm}^{-1}$, with $x$ in the matching unit.

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Common Mistakes to Avoid

• ❌ Wrong: Using the whole echo time to find the depth, as $d = ct$.
• ✓ Right: The pulse travels to the boundary and back, so $2d = ct$. An echo received $65\ \mu\text{s}$ after the pulse in soft tissue ($c = 1540\ \text{m s}^{-1}$) comes from a depth $d = \frac{1540 \times 65 \times 10^{-6}}{2} = 0.050\ \text{m}$, which is 5.0 cm.
• ❌ Wrong: Forgetting to square in the reflection coefficient, or putting $Z_1 + Z_2$ on top.
• ✓ Right: $\frac{I_R}{I_0} = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}$: the difference goes on top and the whole fraction is squared, so the answer is always between 0 and 1. Because of the square, the order of $Z_1$ and $Z_2$ does not matter.
• ❌ Wrong: Saying the coupling gel is only there to make the probe slide easily.
• ✓ Right: Air has a tiny acoustic impedance (about $400\ \text{kg m}^{-2}\,\text{s}^{-1}$) compared with skin (about $1.6 \times 10^6\ \text{kg m}^{-2}\,\text{s}^{-1}$), so about 99.9% of the ultrasound would be reflected at an air gap. The gel has an impedance close to that of skin, so it removes the air and lets most of the ultrasound into the body.
• ❌ Wrong: Mixing units in $I = I_0 e^{-\mu x}$, for example $\mu$ in $\text{cm}^{-1}$ with $x$ in metres.
• ✓ Right: $\mu x$ must have no unit. With $\mu = 0.5\ \text{cm}^{-1}$ and $x = 4\ \text{cm}$, $\mu x = 2$, not $0.5 \times 0.04 = 0.02$.
• ❌ Wrong: Describing the piezoelectric effect the wrong way round.
• ✓ Right: An alternating p.d. across the crystal makes it vibrate and send out ultrasound. Returning ultrasound squashes and stretches the crystal, which generates an e.m.f. that is detected. The same crystal does both jobs: it sends a short pulse, then listens for the echoes.

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Exam Tips

1. Know these by heart: be ready to write down that specific acoustic impedance is density multiplied by the speed of sound in the material, $Z = \rho c$, and to quote $I = I_0 e^{-\mu x}$ from memory. Practise using $\frac{I_R}{I_0} = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}$ until it is routine.
2. Explaining how the image is built up: Use this chain. A short pulse is sent into the body. Part of it is reflected at each boundary between tissues. The transducer detects the echoes. The time delay of each echo gives the depth of the boundary ($2d = ct$), and the intensity of the echo shows how different the impedances are on the two sides of the boundary.
3. Transmitted fraction: If a question asks how much ultrasound enters the second medium, work out $\frac{I_R}{I_0}$ first, then use $1 - \frac{I_R}{I_0}$. At a soft tissue–bone boundary $\frac{I_R}{I_0} \approx 0.35$, so about 65% is transmitted.
4. Follow the pulse in order: In multi-step questions, apply attenuation on the way in, reflection at the boundary, then attenuation on the way back. With $\mu x = 0.20$ each way and $\frac{I_R}{I_0} = 0.35$, the detected echo is $I_0 \times e^{-0.20} \times 0.35 \times e^{-0.20} \approx 0.23\, I_0$.
5. Units of $Z$: $\text{kg m}^{-3} \times \text{m s}^{-1} = \text{kg m}^{-2}\,\text{s}^{-1}$. Quote this unit with any impedance you calculate.