The sinusoidal wave is also useful as the building block for more complex waveforms. Any form of repetitive function can be broken up into the sum of a number of sinusoidal waves of different frequencies. This process is referred to as "frequency analysis". Thus, for example, a continuous triangular wave can be broken up into the sum of a number of sinusoidal waves of different frequencies, as shown in Figure 2.4.
Figure 2.4 A non-sinusoidal waveform (like this triangle waveform) with frequency f can be broken up into the sum of sinusoidal waves of different amplitudes and with frequencies f, 2f, 3f, .... etc.
Note that there are frequency components at the repetition frequency of the wave and at integer multiples of it ("harmonics"). We will return to this concept in chapter 12 when we discuss Harmonic Imaging.
Similarly a pulse (i.e. a wave that lasts just a short time) can be broken up into the sum of many different waves with different frequencies. This will be discussed further in chapter 3.
Suggested activities
1 Look at ripples travelling in water and think about the similarities and differences to ultrasound waves.
2 Calculate the wavelength for ultrasound frequencies of 5 MHz and 10 MHz. Check your answers against Table 2.1.
3 Calculate what frequency would give a wavelength of 0.51 mm and make sure you get the same answer as in the table above.
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
We will start with attenuation. As an ultrasound wave travels through tissue it becomes progressively weaker. This is referred to as attenuation (see Figure 2.5).
Figure 2.5 Attenuation of ultrasound as it travels through tissue. The amount of attenuation depends on the type of tissue, the frequency and the total distance travelled (L).
The attenuation is calculated as the ratio of the input ultrasound intensity to the output intensity (I1/I2). Since it is a ratio, it is usually measured in decibels (see Mathematics Review in chapter 1 for further discussion of decibels).
It is calculated as follows:
In soft tissue, the primary mechanism causing attenuation is heating of the tissue due to absorption of some of the wave's energy as it passes through.
Why does this happen? Since tissue is not perfectly elastic there is friction as it moves back and forth in response to the pressure variations in the wave and this means that heat is generated. Energy cannot be created or destroyed and so this process of heating removes energy from the ultrasound wave.
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 in other directions. This energy is therefore lost from the wave, resulting in weakening, i.e. attenuation, of the ongoing ultrasound.
If the ultrasound beam diverges (due to defocussing or other mechanisms) the energy in the beam is spread over a greater area, and so the intensity of the ultrasound decreases. (Think of a torch being focussed and defocussed and how this affects its intensity).
The use of decibels makes it particularly easy to calculate the attenuation for a given situation:
where α is the "attenuation coefficient" for the specific tissue involved (in dB/cm/MHz), f is the ultrasound frequency (in MHz) and L is the total distance travelled by the ultrasound (in cm).
For 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