For pulsed wave transducers, which active-element size creates higher frequency sound pulses?

For pulsed wave transducers, which active-element size creates higher frequency sound pulses?

• A. Thicker
• B. Shorter
• C. Longer
• D. Thinner

Answer: (D)

The frequency of a pulsed-wave transducer is set by the thickness of its active piezoelectric element. In thickness-mode resonance, the fundamental frequency is approximately f = v_crystal / (2t), where v_crystal is the crystal’s speed of sound and t is its thickness. Reducing the thickness (making it thinner) increases this resonant frequency, so you get higher frequency sound pulses. If the element is thicker, the frequency drops. Dimensions like width or length affect beam shape and aperture more than the fundamental frequency. Higher frequency improves resolution but reduces penetration due to greater attenuation.

The frequency of a pulsed-wave transducer is set by the thickness of its active piezoelectric element. In thickness-mode resonance, the fundamental frequency is approximately f = v_crystal / (2t), where v_crystal is the crystal’s speed of sound and t is its thickness. Reducing the thickness (making it thinner) increases this resonant frequency, so you get higher frequency sound pulses. If the element is thicker, the frequency drops. Dimensions like width or length affect beam shape and aperture more than the fundamental frequency. Higher frequency improves resolution but reduces penetration due to greater attenuation.