typical soft tissue the attenuation coefficient α is approximately 0.5 dB/cm/MHz.
Thus, for example, transmitted ultrasound with a frequency of 3 MHz travelling to a depth of 20 cm in soft tissue will be attenuated by
(0.5 × 20 × 3) = 30 dB
This means the transmitted intensity is reduced by a factor of 1,000 by the time it reaches a depth of 20 cm.
The echo returning from a reflector at this depth will be similarly attenuated by 30 dB as it travels back to the transducer. The total round-path (i.e. there and back) attenuation is therefore 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.
More generally (see Figure 2.6), the round path attenuation for a depth d (cm) is:
round path attenuation (in dB) = α × (2 × d) × f
since (2 × d) is the total round path distance travelled by the ultrasound.
Figure 2.6 As the transmitted pulse travels into the body it is attenuated. The echo from a given structure is then attenuated by an equal amount as it returns to the skin surface and the transducer. The sum of these two attenuation amounts is termed the round-path attenuation.
Returning to the example above, suppose we now decide to use a 6 MHz probe instead, 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 will be so small that they will be undetectable and will not appear in the image.
Attenuation is therefore a fundamental limitation of ultrasound.
When the round-path attenuation exceeds the maximum that the machine can tolerate, the echoes will be too small to detect and they will not be displayed. The depth (dmax) 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 penetration).
For a given machine and clinical situation, the penetration can be calculated as follows. Since dmax is the depth of penetration, we can write:
This will be constant for a given machine and set of operating conditions, depending on factors such as the transmitted power and receiver sensitivity. The attenuation coefficient of the tissue (α) is also constant for a given clinical situation. Thus the bracketed term on the right-hand side of this equation (dmax × f) must be constant for a given machine and tissue type.
This last observation (i.e. that the depth of penetration multiplied by the frequency is constant) has profound implications. It tells us that:
The depth of penetration is inversely related to the frequency.
Returning to the example above, we can see that when we double the probe frequency from 3 MHz to 6 MHz we must expect the depth of penetration to halve.
More generally, this relationship means that relatively low frequencies (e.g. 3 to 5 MHz) must be used to scan deep regions (e.g. in abdominal and obstetric ultrasound) to ensure adequate penetration. For superficial regions (e.g. thyroid, breast, peripheral vascular) considerably higher frequencies can be used (e.g. 6 to 10 MHz) since less penetration is required.
Note that the use of higher frequencies for scanning superficial areas will yield better image resolution than can be obtained when deeper regions are scanned using lower frequencies. Since we generally want the best possible image resolution, we will therefore:
Use the highest frequency compatible with the depth of penetration required.
Suggested activities
1 Observe which probes are used for different types of examination in your workplace. Notice whether there is a trend to use higher frequencies for more superficial areas.
2 Suppose a given ultrasound machine has a maximum depth of penetration of 20 cm when operating at 4 MHz. Calculate the penetration you would expect at frequencies of 2.5, 3, 5, 7.5, 10 and 15 MHz. Put the results into a table and think about typical clinical applications where you could use each of these frequencies.
Reflection and scattering
Reflection
Reflection and scattering are the two mechanisms that produce echoes and so create the information shown in the ultrasound display.
As with light, the word "reflection" is used to describe the interaction of ultrasound with relatively large and smooth surfaces. (Think of light reflecting from glass.) "Scattering" refers to 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.)
Before we can discuss reflection, we must first introduce the concept of the acoustic impedance (sometimes called the characteristic impedance) of a tissue. It is defined as:
where z is the acoustic impedance, ρ is the density of the tissue (i.e. the weight per unit volume) and c is the ultrasound propagation speed. The units for acoustic impedance are called Rayls.
Acoustic impedance is a useful concept because it is a measure of how the tissue "appears" to the ultrasound. Two tissues with similar acoustic impedance values will appear similar to the ultrasound, while tissues with very different impedances will look very different. If two tissues happened to have the same value of acoustic impedance they would look identical to the ultrasound.
It is important to realise that tissues may look very similar on ultrasound and yet be totally different histologically. As an example, pus can look very similar in an ultrasound image to soft tissue.
How is this related to reflection?
Consider the three situations shown in Figure 2.7 (a), (b) and (c). It is clear that the amount of energy reflected is determined by the degree to which the acoustic impedances in the two tissues are different.
Figure 2.7 (a) Total reflection of the ultrasound energy at an interface between two tissues with a very large acoustic impedance difference (e.g. a soft-tissue - air interface). All of the energy is reflected and none is transmitted into the second tissue. The interface will be seen in the image as a strong linear structure.
Figure 2.7 (b) Total transmission of the ultrasound energy when the two tissues have identical acoustic impedance (i.e. z1 = z2). No energy is reflected and so there will be no echo – the interface will not be seen in the ultrasound image.
