Introduction to Diagnostic Ultrasound Technology PDF

**Introduction to Diagnostic Ultrasound Technology** **Course purpose: **The aim of this course is to give you a good start in understanding the physics of diagnostic ultrasound. It introduces the main concepts and explains why they are important clinically. **Course structure: **The course parall...

**Introduction to Diagnostic Ultrasound Technology** **Course purpose: **The aim of this course is to give you a good start in understanding the physics of diagnostic ultrasound. It introduces the main concepts and explains why they are important clinically. **Course structure: **The course parallels my textbook *The Physics and Technology of Diagnostic Ultrasound: A Practitioner\'s Guide (Second edition)*.\* It\'s important to realise that the textbook provides significantly more detail. So, I strongly recommend that you regard this online course as an **introduction** and use the textbook to fill in the details you need to understand to be an ultrasound expert. **This module:** This module discusses the nature of ultrasound waves, some of their properties and the interaction of ultrasound with human tissues. It parallels Chapter 2 of the textbook. **Chapter 1: Ultrasound Interaction with Tissue** **Lesson 1: Ultrasound Waves and Propagation** Ultrasound is simply a very high-frequency sound. Diagnostic ultrasound frequencies range from 2 MHz to 20 MHz\*. This is around 1000 times higher than the frequency range of audible sound (20 Hz to 20 kHz). As we will see soon, these high frequencies are used to achieve the best possible image resolution\*\*. The ultrasound wave consists of oscillating pressures. It is generated when the ultrasound transducer vibrates against the skin surface (at the ultrasound frequency). When the transducer is moving towards the body it compresses the tissues, causing an increase in pressure (*compression*). When it moves away from the body it decompresses the tissues, causing a decrease in pressure (*rarefaction*). In this way, oscillating changes of pressure are created. These constitute an ultrasound wave that travels through the tissues at a constant speed.  If we could focus on a single point in the tissues, we would see the pressure oscillating as a function of time, increasing and decreasing equally above and below the mean pressure in the tissue.  The amplitude (A) is the maximum change of pressure from its mean value in the tissues. It determines the amount of energy in the ultrasound wave and therefore the level of exposure of the tissues.  The period (T) is the time occupied by one cycle. It is related to the ultrasound frequency (f) (the number of cycles per second) as follows:  *T = 1/f* We can also imagine a situation where we could look at the variation of pressure as a function of depth into the tissue at one instant in time. (This is like taking a photograph of ocean waves at one instant.) We would find a similar pattern of pressure variations with depth.  The wavelength (λ) is the physical length of a single cycle. (This is like the distance from the top of one ocean wave to the top of the next one.) It is very closely related to image resolution; the smaller the wavelength the better the resolution will be. The speed at which the ultrasound wave travels is referred to as the *propagation speed*. It is conventionally represented by the symbol c. The average value of c in normal soft tissue is 1540 m/sec. The equation below describes the relationship between propagation speed, frequency and wavelength.  *c = f ⨉ λ* This can be rearranged to allow the wavelength to be calculated for a given frequency. *λ = c/f*  This table lists a number of typical ultrasound frequencies and wavelengths in soft tissue (assuming c = 1540 m/sec). Notice that for all frequencies the wavelength is less than one millimetre. Also notice that the wavelength becomes smaller as the frequency increases. Since the resolution of an ultrasound image is directly related to the wavelength it is clear that: *The higher the frequency the better the resolution.* Unfortunately, we will see shortly that it is not possible to keep increasing the frequency to achieve ever better resolution. The physics of ultrasound places a limit on the frequency that can be used in a given clinical situation. **Lesson 2: Attenuation** Diagnostic ultrasound gives useful information precisely because it interacts strongly with soft tissue. In this and the following two sections, the major types of interaction will be discussed. These are: - attenuation. - reflection. - scattering. - refraction. Attenuation As an ultrasound wave travels through tissue, it becomes progressively weaker. This is referred to as attenuation. In soft tissue, the primary cause of attenuation is the absorption of some of the wave's energy by the tissue. This happens because tissue is not perfectly elastic and so there is some friction as it moves back and forth in response to the pressure variations in the wave. This friction causes heating of the tissues and so the temperature increases. Since energy cannot be created or destroyed, this process removes energy from the ultrasound wave, making it weaker. Other factors can contribute to attenuation. As discussed in the next section, reflection and scattering from structures in the body cause some of the ultrasound energy to be deflected. This energy is lost from the wave, resulting in weakening (i.e. attenuation) of the ongoing ultrasound. If the ultrasound beam diverges (because of defocusing), the energy in the beam is spread over a greater area and so the intensity decreases. (Think of a torch being defocussed.) A diagram of a red cube with arrows pointing to the sides Description automatically generated Referring to the figure on the left, the amount of attenuation is calculated as the ratio of the initial ultrasound intensity (I~1~) to the final intensity (I~2~). Since it is a ratio, it is usually measured in decibels (dB). So, the definition of attenuation is: *attenuation = 10 ⨉ log (I~1~/I~2~) dB* Using decibels makes it particularly easy to calculate tissue attenuation for a given situation: *attenuation = (α ⨉ L ⨉ f) dB* Here α is the attenuation coefficient of the specific tissue involved (in dB/cm/MHz), L is the total distance travelled by the ultrasound (in cm) and f is the ultrasound frequency (in MHz). For typical soft tissue, the attenuation coefficient α is approximately 0.5 dB/cm/MHz **Example** Consider a situation where the machine is operating at 3 MHz If the transmitted ultrasound travels to a depth of 20 cm in soft tissue, it will be attenuated by 30 dB, as shown below. *attenuation of transmitted ultrasound = (0.5 ⨉ 20 ⨉ 3) = 30 dB* This means the transmitted intensity is reduced by a factor of 30dB = 1,000\*. Similarly, the echo returning from a reflector at this depth will be attenuated by 30 dB as it travels back to the transducer. Thus, the total round path (there and back) attenuation is 60 dB.  This means that the echo coming from 20 cm depth will be 60 dB weaker (i.e. 1,000,000 times lower in intensity) than an echo from a similar reflector at the skin surface. This example highlights how substantial the attenuation of ultrasound is. *\*See Chapter 1 of my textbook for a discussion of decibels.* **A second example** Suppose we now decide to scan using 6 MHz instead of 3 MHz in an attempt to improve image resolution. The round path attenuation will then be 120 dB, corresponding to a total reduction in intensity by a factor of 1,000,000,000,000! Echoes that have been attenuated this much are so small that they cannot be detected by the machine so they will not appear in the image\*.  **Penetration (P)** When the round path attenuation exceeds the maximum that the machine can tolerate, the echoes are too small to detect and they are not displayed. The depth (P) at which this happens (i.e. the depth beyond which the echoes are not detectable) is referred to as the *depth of penetration* (or simply the *penetration*). For a given machine and clinical setting the round path attenuation for echoes coming from the penetration depth is: *attenuation = α ⨉ (2P) ⨉ f = 2α(P ⨉ f)* This is the maximum attenuation the machine can tolerate without the echoes becoming undetectable. For a given machine this value will be constant. The attenuation coefficient of the tissue (α) is also constant for a given clinical situation. Therefore, the bracketed term on the right-hand side of this equation (P × f) must be constant. As the frequency increases the penetration must decrease to keep the product (P × f) constant. *As the frequency increases the penetration decreases.* **Lesson 3: Reflection and Scattering** Reflection and scattering are the two mechanisms that produce echoes.  **Reflection:** the interaction of ultrasound with relatively large and smooth surfaces (think of light reflecting from glass.)  **Scattering:** the interaction of ultrasound with small structures (red blood cells, capillaries, etc.) within the tissues (think of light scattering from the tiny water droplets in a fog.) A diagram of a tissue interface Description automatically generated **Reflection, acoustic impedance** When ultrasound strikes an interface between two tissues, some of the ultrasound energy is reflected and the rest continues into the deeper tissues. The more different the two tissues are the more of the energy is reflected. In this diagram, z~1~ and z~2~ are the *acoustic impedance *values of the first and second tissue respectively. The acoustic impedance of a tissue is  *z = ρ ⨉ c* where ρ is the density of the tissue (its weight per unit volume) and c is the propagation speed. The units are Rayls. **Reflection coefficient** The reflection coefficient R is a measure of the fraction of the ultrasound energy that is reflected. It can be calculated as follows: *R = (z~1~ - z~2~)^2^ / (z~1~ + z~2~)^2^ * A reflection coefficient of 0.01, for example, would mean that 1% of the ultrasound energy is reflected and 99% passes through the interface. It is easy to show that when the two tissues have the same acoustic impedance (z~1~ = z~2~) the reflection coefficient is zero and no energy is reflected. It can also be shown that when z~1~ and z~2~ are very different the value of R is close to one, which means that almost all the energy is reflected and very little passes through the interface.  **Reflection geometry** So far, we have focussed on the special situation where the ultrasound is incident on the interface at right angles (i.e. at an angle of 90° to the interface); this is called *perpendicular incidence*. What about the more general situation?\ When the incident angle is not 90°, the reflected ultrasound does not travel back to the transducer. As a result, the echo from the interface is not detected and so it is not seen in the image. A diagram of a cell Description automatically generated **Scattering** The word scattering describes the interaction of ultrasound with small structures such as red blood cells and capillaries. Scattering differs from reflection in two important ways. 1. Scattered energy is distributed in all directions. Reflected ultrasound goes in a single direction. 2. Scattered energy is generally much weaker than reflected energy. Echoes caused by scattering are generally displayed as low- to mid-level grey tones in the image.  Most of the echo information in a typical ultrasound image comes from scattering from soft tissue, not reflection from interfaces between different tissues. So, an understanding of the character of scattered echoes and the way they look in the image is important. Diagram of a ultrasound beam Description automatically generated **Speckle** As the diagram to the left shows, the echo signal at any moment is the sum of many echoes, each caused by an individual scatterer within the volume of tissue occupied by the ultrasound pulse.  These scatterers are randomly positioned relative to each other, so their echoes add together randomly. This causes random variation in the amplitude of the echo signal received by the transducer. The result is the random granular echo texture in the image that we call *speckle*.  **Example** Notice how the speckle (the echo texture) in this image varies with depth. Close to the transducer the texture is quite fine-grained whereas at greater depths it is coarser. The phantom material is the same at all depths, highlighting the fact that **speckle does not directly reflect a tissue property**. It is important to realise, however, that the **average brightness** of the speckle is meaningful, and it will vary as tissue properties vary. **Lesson 4: Refraction** A triangle with text and green arrow Description automatically generated with medium confidence **Refraction of light** You are probably familiar with refraction of light -- the bending of the light's path as it passes through different materials. Examples include: - a prism -- glass with a triangular cross-section (see diagram); - the bending of light as it passes from water to air. Light is refracted whenever it travels from one medium (e.g. air) into another medium that has a different propagation speed (e.g. glass or water).  **Refraction of ultrasound** Similarly, ultrasound is refracted whenever it passes through an interface between tissues with different ultrasound propagation speeds (for example, from liver tissue to fat). The geometry of refraction is determined by measuring the direction of travel of the ultrasound relative to a line drawn at right angles to the interface, as shown in the diagram. The amount of refraction can then be calculated using Snell's Law: *(sin θ~i~)/c~1~ = (sin θ~t~)/c~2~ * where θ~i~ is the incident angle and θ~t~ is the transmitted angle. This equation explains a number of features of refraction, including: - When two tissues have the same propagation speed (c~1~ = c~2~), θ~i~ = θ~t~ which means there is no refraction, and the beam direction is unaltered. - A special case occurs when the beam is at perpendicular incidence to the interface. In this case, θ~i~ is 0° so sin θ~i~ = 0. According to Snell\'s Law, this means that θ~t~ is also 0°. So there is no change of beam direction (refraction) when the beam is at perpendicular incidence, regardless of the propagation speeds. - When the propagation speed in the second tissue is lower than in the first (c~2~ \< c~1~), θ~t~ is smaller than θ~i~ as shown in the diagram above. - When c~2~ is greater than c~1~, θ~t~ is larger than θ~i~ as shown below. View the four images below. They show that the amount of refraction increases as the incidence angle increases. They also reveal a problem that occurs with large incidence angles when the second tissue has a propagation speed higher than the first.  -- ------------------------  1. 2.  3. 4.  **Critical angle** As shown above, ultrasound cannot penetrate into the second tissue if\ (a) the propagation speed is higher in the second tissue, and\ (b) the incidence angle is equal to or larger than the critical angle. The value of the critical angle (θ~c~) for a given pair of tissues is found by solving Snell\'s Law for the situation where θ~t~ = 90°. This gives us *θ~c~ = sin^-1^ (c~1~/c~2~)* **Summary** When ultrasound passes through an interface between two tissues with different propagation speeds: - the beam path is bent, except for the special case of perpendicular incidence; - the amount of bending increases for large differences in propagation speed and large incident angles; - in the situation where the propagation speed in the second tissue is higher than in the first, a critical angle exists. When the incident angle is larger than this critical angle, total reflection occurs and no energy is transmitted into the second tissue.