Pulse vs Continuous transducers (Ultrasound)

[Joan B.]

I am a student in a DMS program. Our instructor poised the following question (worth extra credit!) if we can answer it and back it up.

OPERATING FREQUENCY IN PULSED WAVE TRANSDUCERS IS DETERMINED BY:
A. FREQUENCY OF THE VOLTAGE
B. PULSE REPITITION FREQUENCY
C. THE MEDIUM ONLY
D. THE THICKNESS OF THE ELEMENT

I know the correct answer is 'D', but I am having difficulty PROVING it. Can you offer any suggestions as to how I can? (pictures, diagrams, studies I can access?)
The 'assignment/extra credit is due very soon.
Below is what I have so far with references.

Question:
Operating frequency in pulsed wave transducers is determined by:
a. frequency of the voltage
b. pulse repetition frequency
c. the medium only
d. the thickness of the element

The correct answer is: d. the thickness of the element.

Rationale:

Recall that the transducer is a piezoelectric (has the ability to generate an electric charge in response to applied mechanical stress and vice versa. (Reverse-piezoelectric and piezoelectric)

If the transducer is operated in CONTINUOUS mode (applied voltage never turned off) then transducer frequency = voltage frequency (e.g., 5 MHz voltage on a transducer will generate a 5 MHz sound wave).

However, not all transformers are operated in continuous mode. An US transformer acts as source of producing US pulse as well as acting as detection of reflected US pulses. Normally the transducer is operated in pulse mode (or pulse wave). In pulse wave, a short (in terms of time or number of voltage pulses) is applied to the transducer. Then the transducer will "ring" like a bell and produce US waves of frequency directly depending on the thickness of the crystal.

For a transducer with thickness T, the wavelength produced by the ringing transducer is 2 x T (i.e., 0.5 mm transducer will produce US wavelengths of 1 mm or 0.001 m). If the speed of sound in tissue is c and frequency of the US is f, since c = f x wavelength the frequency of the US pulse from the transducer is f = c/(2 x T). If the speed of sound in tissue is assumed to be 1540 m/s then for a 0.5 mm transducer thickness the resulting US frequency is f = 1540 m/s /0.001m = 1.54 MHz.

References:
Ultrasound Physics & Instrumentation Vol. 1 Pegasus Lectures, Inc 2003 chapter 5 pages 31 - 36

Review of Radiologic Physics Walter Huda and Richard Slone 2nd ed chapter 11 (page 177 - 178)

Radiology Review: Radiologic Physics Edward Nickoloff and Naveed Ahmand, 2005Chapter 19p. 181

The Essential Physics of Medical Imaging 2nd edition Bushberg, Seibert, Leidholdt, Boone 2002 chapter 16 p. 484-487.

Wm. Geisler Dept of Physics Marshfield Clinic/Ministry Healthcare Marshfield, WIThank you for your time and efforts.

Joan Buraglio[

 

[Astronuc]

Looks correct.

Manufacturers are a good source of technical and theoretical information.

https://www.olympus-ims.com/en/resources/white-papers/intro-ultrasonic-transducers-ndt-testing/

The performances of the transducer are dependent on the frequency characteristics. The results show that the relative permittivity, piezoelectric strain constant, piezoelectric voltage constant and mechanical quality factor of the piezoelectric transducer increase with increasing thickness of piezoelectric material, and the thickness electromechanical coupling properties and resonant frequency decrease with increasing thickness. In the pulse response test of the transducers, with increasing thickness of the piezoelectric material, the transmitted time and peak-peak value of the ultrasonic wave received by the transducer increase, and the dominant frequency shows the decreasing trend.

https://ieeexplore.ieee.org/document/8340304