the scattered energy is generally much weaker than reflected energy and so the echoes due to scattering are generally displayed in the image as low- to mid-level grey tones.
If you look at a typical ultrasound image you will see that the majority of the echo information in the image comes from scattering from within tissue, not reflection from interfaces between different tissues. Thus the nature of the scattered echoes and their appearance in the image are very important.
You will also notice that scattering produces a random granular echo texture in the image. To understand why this happens, consider Figure 2.10.
Figure 2.10 At one instant of time the transmitted pulse will "see" a volume of tissue (shown in light blue); the transducer will receive echoes from any scatterers that are within this volume. In soft tissue there will generally be a large number of scatterers within the volume, and the echo signal seen by the transducer will be the sum of the signals from all these scatterers.
This shows that at each instant the echo signal coming from soft tissue is actually the sum of the echoes from a number of individual scatterers that lie within the ultrasound pulse.
Since these scatterers are randomly positioned relative to each other, their echoes will add together randomly. This causes the echo signal received by the transducer to have a random variation in its amplitude. This phenomenon is termed "speckle" and it gives rise to the "echo texture" that we see in ultrasound images.
Speckle is a random process that is only indirectly related to the distribution of the scatterers. To highlight this, consider Figure 2.11. This is an image of an ultrasound "phantom" – a test object made of a gel material containing scatterers and designed to look like liver tissue when scanned. (The strong white echoes come from "point targets" that are used to check measurement accuracy and other aspects of equipment performance; they will be discussed in chapter 10.)
Figure 2.11 Scan of an ultrasound phantom (test object) showing speckle.
Notice how the echo texture varies with depth. Close to the transducer the texture is quite fine-grained whereas at greater depths it is much coarser. The phantom material, however, is uniform throughout the phantom, highlighting the fact that the speckle does not directly reflect a tissue property.
Suggested activities
1 Calculate the reflection coefficient at a tissue interface for which z1 = 1.5 × 106 and z2 = 1.5 × 104 (i.e. the two impedance values differ by a factor of 100). Note that the reflection coefficient is almost 1 (or 100%) meaning that virtually all the energy is reflected. Repeat the calculation with z1 and z2 interchanged.
2 Calculate the reflection coefficient at a tissue interface for which z1 = 1.5 × 106 and z2 = 1.51 × 106 (i.e. there is a minimal difference between the two impedances). Note that the reflection coefficient is small, which means that very little energy is reflected.
3 Calculate both the reflection coefficient (R) and the transmission coefficient (T) for an interface with z1 = 1.5 × 106 and z2 = 1.8 × 106 (a moderate difference in impedance). Show that (R + T) = 1.
4 Carefully examine scans of different anatomical areas and identify which echoes are caused by reflection and which are caused by scattering. Note the differences in appearance of the two types of echo.
5 Scan a region within a liver from two different directions. Carefully compare the speckle patterns – can you see a difference?
Refraction
You are probably familiar with refraction of light – the bending of the light's path as it passes through different materials (as shown in Figures 2.12 and 2.13).
Figure 2.12 As light passes through a prism its direction of travel changes. This is an example of the refraction of light.
Figure 2.13 Refraction of light coming from an object (darker box) on the bottom of a pool. When a viewer's eye receives the light, the brain processes the information on the assumption that the light has travelled in a straight line (as shown by the broken line), and so it sees the box in the position shown by the lighter box. This is why pools generally look shallower than they really are.
Examples include:
a prism – a piece of glass with a triangular cross-section, often found in a science laboratory (shown in Figure 2.12),
the bending of light as it passes from water to air (see Figure 2.13).
Refraction is also the principle on which optical lenses (such as the one in your camera) are based.
Optical refraction occurs whenever light travels from one medium (e.g. air) into another medium that has a different propagation speed for light (e.g. glass or water). In a similar way, ultrasound is refracted whenever it passes through an interface between tissues with different ultrasound propagation speeds (e.g. from liver tissue to fat).
As with reflection, the geometry is determined by measuring the direction of travel of the ultrasound relative to a line drawn at right angles to the interface.
Figure 2.14 Refraction of ultrasound. The direction of travel of the ultrasound is altered as it passes through the interface between tissues with different ultrasound propagation speeds. In this example the propagation speed is lower in the second tissue (i.e. c2 < c1).
Mathematically the amount of refraction can be determined using Snell's Law:
where θi is the incident angle and θt is the transmitted angle. It is easy to show that if the difference between the two propagation speeds increases then the difference between the two angles will also increase. It can also be shown that if c2 is less than c1 then θt will be less than θi while if c2 is larger than c1 then θt will be larger than θi (see Figures 2.14 and 2.15).
Figure 2.15 Refraction for the situation where c2 is larger than c1. The "bending" of the ultrasound is now in the opposite direction to that shown in Figure 2.14. Note that the beam has been "bent" by just 3.5°.
Figure 2.16 An identical situation to that shown in Figure 2.15 except that the incident angle has increased from 30° to 60°. Note that the effect of refraction is now more pronounced, causing a change of direction of 13°.
Comparing Figures 2.15 and 2.16 shows that the difference between θi and θt also depends on the incident angle θi. As the incident angle increases the bending of the beam increases.
Conversely (see Figure 2.17) when the incident angle is 0° (i.e. when the ultrasound is perpendicular to the interface) the transmitted angle is also 0° and hence no deflection of the beam occurs, regardless of the propagation speeds.
Figure