Explain how a piezoelectric transducer generates and detects ultrasound, and relate the principal physical quantities (frequency, acoustic impedance, attenuation, reflection) to medical and industrial applications.
Certain crystals (e.g. quartz, PZT – lead zirconate titanate) show a reversible conversion between mechanical stress and electrical charge.
This wording mirrors the Cambridge syllabus wording: “the crystal changes shape when a p.d. is applied … and generates an e.m.f. when its shape changes.”
\$V(t)=V_0\sin(2\pi ft)\$
The efficiency of the conversion is characterised by the electromechanical coupling factor k (0 < k < 1).
An incoming ultrasound pulse compresses (or stretches) the crystal. By the direct piezoelectric effect this mechanical deformation produces a charge Q(t), which appears as a voltage that can be amplified and recorded.
For a linear transducer the received voltage is proportional to the acoustic pressure:
\$V_{\text{rec}} = S\,p(t)\$
where S is the sensitivity (V Pa⁻¹).
\$\frac{IR}{I0}= \left(\frac{Z1-Z2}{Z1+Z2}\right)^2\$
and the transmission coefficient is 1 – IR/I0.
The intensity of a travelling wave decreases exponentially:
\$I = I_0 e^{-\mu x}\$
Attenuation rises with frequency, which explains why higher‑frequency probes give better resolution but poorer depth of penetration.
| Parameter | Symbol | Typical medical value | Physical significance |
|---|---|---|---|
| Resonant frequency | fr | 2 – 10 MHz | Frequency at which the crystal vibrates most efficiently (maximum k). |
| Bandwidth | Δf | ≈ 30 % of fr | Range of frequencies over which the transducer works effectively; larger Δf → shorter pulse → better axial resolution. |
| Electromechanical coupling factor | k | 0.4 – 0.7 | Fraction of electrical energy converted to mechanical energy (and vice‑versa). |
| Acoustic impedance | Z | ≈ 1.5 MRayl (soft tissue) | Product ρc; determines how well the crystal is matched to the load. |
| Attenuation coefficient | μ | 0.5 – 1 dB cm⁻¹ MHz⁻¹ (soft tissue) | Rate at which intensity falls with distance; larger for higher frequency. |
To maximise transmission into the target medium a thin matching layer is placed between the crystal (impedance Zc) and the tissue (impedance Zt).
Optimal impedance of the matching layer:
\$Zm = \sqrt{Zc\,Z_t}\$
A backing material with high acoustic attenuation absorbs the wave that travels back into the crystal, reducing “ringing” and improving axial resolution.
\$\Delta f = \frac{2 v f_0 \cos\theta}{c}\$
For a transducer with centre frequency f = 5 MHz and bandwidth Δf = 2 MHz:
\$\$\text{Spatial pulse length (SPL)} \approx \frac{c}{\Delta f}
= \frac{1540\ \text{m s}^{-1}}{2\times10^{6}\ \text{Hz}}
\approx 0.77\ \text{mm}\$\$
Axial resolution ≈ ½ SPL ≈ 0.4 mm.
Probe frequency f₀ = 3 MHz, blood speed v = 0.5 m s⁻¹ directly towards the probe (θ = 0°).
\$\$\Delta f = \frac{2 v f_0}{c}
= \frac{2(0.5)(3\times10^{6})}{1540}
\approx 1.95\times10^{3}\ \text{Hz}\$\$
The received echo is shifted by ≈ 2 kHz.
Typical impedances: soft tissue Z₁ ≈ 1.5 MRayl, bone Z₂ ≈ 7.8 MRayl.
\$\$\frac{IR}{I0}= \left(\frac{7.8-1.5}{7.8+1.5}\right)^2
\approx \left(\frac{6.3}{9.3}\right)^2
\approx 0.46\$\$
≈ 46 % of the incident intensity is reflected – a strong echo.
Attenuation coefficient = 0.7 dB cm⁻¹ MHz⁻¹, frequency = 5 MHz, path length = 3 cm.
Total attenuation (dB) = 0.7 × 5 × 3 = 10.5 dB.
Intensity remaining:
\$I = I0 \times 10^{-10.5/10} \approx 0.089\,I0\$
≈ 9 % of the original intensity.
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