Ultrasound transducer and acoustic impedance

Medical Physics | AQA A-Level Physics

Key Definition

Acoustic impedance: $Z = \rho c$, where $\rho$ is the density of the medium and $c$ is the speed of sound in that medium. Measured in kg m$^{-2}$ s$^{-1}$.

Ultrasound is generated and detected using the piezoelectric effectCertain crystals (such as quartz or lead zirconate titanate) produce a voltage when deformed, and deform when a voltage is applied. This allows the same crystal to act as both transmitter and receiver of ultrasound.. A piezoelectric crystal vibrates when an alternating voltage is applied, producing ultrasound. The same crystal detects returning echoes by converting pressure variations back into a voltage.
The transducer consists of a piezoelectric crystal with electrodes on each face, backed by a damping material that shortens the pulse length and improves resolution.
When ultrasound hits a boundary between two materials with different acoustic impedances, some is reflected and some is transmitted.
The intensity reflection coefficientThe fraction of incident intensity that is reflected at a boundary. $\alpha = \frac{I_r}{I_0} = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}$. When $Z_1 = Z_2$, no reflection occurs. is: $$\alpha = \frac{I_r}{I_0} = \frac{(Z_2 - Z_1)^2}{(Z_2 + Z_1)^2}$$
A coupling mediumA gel applied between the ultrasound transducer and the skin. It has an acoustic impedance close to that of skin ($Z_{\text{gel}} \approx 1.5 \times 10^6$, $Z_{\text{skin}} \approx 1.6 \times 10^6$), preventing the large reflection that would occur at an air-skin boundary ($Z_{\text{air}} = 430$). (gel) is essential because the acoustic impedance of air ($Z \approx 430$ kg m$^{-2}$ s$^{-1}$) is vastly different from skin ($Z \approx 1.6 \times 10^6$), so almost all the ultrasound would be reflected at an air-skin interface. The gel has $Z \approx 1.5 \times 10^6$, closely matching skin.

Worked Example

Calculate the intensity reflection coefficient at a boundary between soft tissue ($Z_1 = 1.63 \times 10^6$) and bone ($Z_2 = 6.40 \times 10^6$).

Show Solution

Substitute into the reflection coefficient equation

$$\alpha = \frac{(6.40 - 1.63)^2}{(6.40 + 1.63)^2} \times 10^{12} / 10^{12}$$

$$= \frac{(4.77)^2}{(8.03)^2} = \frac{22.75}{64.48}$$

Calculate

$$\alpha = 0.35$$

Answer

$\alpha = 0.35$, so 35% of the ultrasound intensity is reflected at a tissue-bone boundary.