Ultrasound and the piezoelectric effect

In a nutshell

Ultrasounds are used for medical imaging. An ultrasound transducer used the piezoelectric effect to create ultrasounds. A-scans are used to determine the thickness of bones and distances between tissues while B-scans are used to create two-dimensional images of a section through a patient.

Ultrasound scans

Ultrasounds are longitudinal sound waves with a frequency higher than $20\ kHz$ , beyond the human hearing range. They can be used for medical imaging as they are harmless and non-invasive, no surgery is needed. They can be used to identify features in the human body only a few millimeters wide.

The piezoelectric effect

The piezoelectric effect is a phenomenon where some crystals produce an e.m.f. when stretched or compressed. This also happens in reverse where, if a potential difference is applied to opposite ends of a crystal it causes it to stretch or compress as shown below:

Physics; Medical physics; KS5 Year 12; Ultrasound and the piezoelectric effect

1.	Applying voltages on a piezoelectric crystal
2.	No p.d. applied
3.	P.d. applied
4.	P.d. applied with opposite polarity
5.	Unstressed
6.	Crystal compresses
7.	Crystal stretches

This effect is used to generate ultrasounds in ultrasound transducers. An alternating current is applied across two faces of a crystal which will make it continuously compress and expand. The frequency of the current is chosen to be the natural frequency of the crystal such that it resonates and produces ultrasounds.

The transducer can also be used to detect ultrasounds. When an ultrasound reflects back on the crystal it creates vibrations which generate an e.m.f. that gets picked up by a circuit.

A-scans and B-scans

A transducer sends pulses of ultrasounds, each of which gets partially reflected and transmitted at the boundary between two tissues. The reflected pulses are then received by the transducer, the thicker a tissue boundary the more a pulse gets reflected. This makes it possible to scan human bodies.

There are two types of scans:

A-scan; the simplest type of scan. It is used to determine the thickness of bones or distances between tissues. The transducer is placed still and the pulses that are reflected back are displayed on an oscilloscope. These pulses will have been partially reflected back to the transducer, but also partly transmitted through the boundary of tissues. The time interval of the pulse is used with the speed of the ultrasound to calculate the distances between pulses being reflected at different boundaries.

B-scan; the most common type of scan. It used to provide two-dimensional images on a screen. The transducer is moved over the patient's body and the output is sent to a computer. For each pulse the computer creates a row of dots differing in brightness depending on their intensity, which in turn depend on the boundaries between two tissues that the pulses reflect off. This creates a two-dimensional image of a section through the patient. B-scans are essentially multiple A-scans.

The diagram below shows an A-scan of the eye, where the display of an A-scan is shown as voltages. The first voltage pulse, is the pulse that is sending the ultrasound into the eye, whilst pulse two and three are reflections in the eye lense. Pulse four is the pulse coming from the back of the eye. The time between the pulses, can be used to determine the distance the ultrasound has travelled.

The B-scan version, shown at the bottom of the diagram, would display the eye as a series of bright dots.

Physics; Medical physics; KS5 Year 12; Ultrasound and the piezoelectric effect

1.	Transducer
2.	Eye
3.	Path of ultrasound pulse
4.	A-scan display
5.	B-scan display
6.	Bright dot on a B-scan display
7.	Voltage

Below is an image you would expect back on a B-scan, which is the most common type of image from an ultrasound.

Physics; Medical physics; KS5 Year 12; Ultrasound and the piezoelectric effect

Doppler imaging

The doppler effect is a phenomenon which can be observed for a moving object. When an object is moving away, the frequency of waves are decreased, whereas when an object is moving closer, the frequency increases.

Medical physicists have been able to use the doppler effect to monitor blood flow, by using a transducer to send pulses of ultrasound through the skin of a patient. The ultrasound pulse is then reflected back from the moving blood, and the frequency change is measured.

Physics; Medical physics; KS5 Year 12; Ultrasound and the piezoelectric effect

1.	Transducer
2.	Gel
3.	Skin
4.	Blood vessel
5.	Red blood cells

The frequency change, can be used to find the speed of the blood flow, using the following equation:

$\Delta f = \dfrac{2fv \, cos \theta}{c}$

Where $\Delta f$ is the frequency change and $v$ is the speed of the blood flow. The speed of ultrasound in blood is denoted as $c$ (which shouldn't be confused with the speed of light!) and the angle the transducer is held at to the blood vessel is given by $\theta$ . The frequency of the ultrasound pulse leaving the transducer is $f$ .

Some of the uses for doppler imaging include diagnosing blood cots and narrowing vessel walls, as well as measuring blood flow in an umbilical cord during pregnancy.

Physics; Medical physics; KS5 Year 12; Ultrasound and the piezoelectric effect

Example

A transducer sends an ultrasound pulse into a patient at a frequency of $12 \ MHz$ , at an angle of $60 \degree$ .The frequency change observed was $1000 \ Hz$ . The speed of the blood flow in the patient was $13 \ cm \ s^{-1}$ . Calculate the speed of ultrasound in blood.

Firstly, write down your variables:

$\begin{aligned}&f = 12 \ MHz = 12 \times 10^6 \ Hz \\ &\theta = 60 \degree \\ &\Delta f = 1000 \ Hz \\&v = 13 \ cm \ s^{-1} = 0.13 \ ms^{-1} \end{aligned}$ 

Next, write down the equation:

$\Delta f = \dfrac{2fv \, cos \theta}{c}$ 

Rearrange for $c$ :

$c = \dfrac{2fv \, cos \theta}{\Delta f}$ 

Substitute in your values:

$c = \dfrac{2 \times 12 \times 10 ^6 \times 0.13 \, \times cos (60)}{1000} \\$ 

Calculate your final answer to the lowest amount of significant figures given in the question:

$c = 1600 \ ms^{-1}$

The speed of ultrasound in blood is $\underline{1600 \ ms^{-1}}$ .