When the propagation speed is greater in the second medium than in the first (i.e. c2 > c1, see Figures 2.15 and 2.16) it can be shown that there is a particular value of the incident angle for which the transmitted angle (θt) will be 90° (see Figure 2.18).
Figure 2.18 When the incident angle is equal to the "critical angle" (65° in this case) the transmitted ultrasound is refracted so that the transmitted angle is 90°.
This value of the incident angle is called the critical angle.
What is the significance of the critical angle? Since the ultrasound has a transmitted angle of 90° it is barely entering the second tissue – it is just running along the interface between the two tissues.
What happens when the incident angle is greater than the critical angle? In this case ultrasound is not transmitted into the second tissue at all, and instead all the energy is reflected, as shown in Figure 2.19.
Figure 2.19 When the incident angle exceeds the critical angle, all of the ultrasound energy is reflected from the interface. Notice that the geometry of reflection is then the same as discussed in the previous section, i.e. the reflected angle is equal to the incident angle.
To repeat this point:
If the propagation speed is higher in the second tissue, then a critical angle exists; for incident angles greater than the critical angle, total reflection occurs.
Thus we have seen that there are two separate mechanisms capable of causing total reflection of ultrasound from an interface between two tissues:
when there is a very large difference in the acoustic impedance of the two tissues, or
when the propagation speed in the second tissue is higher and the incident angle exceeds the critical angle.
How do we calculate the critical angle?
Recognising that θt = 90° and so sin θt = 1.0, Snell's Law can be written:
As an example, consider the situation shown in Figures 2.15, 2.16 and 2.18 where c1 = 1450 m/sec (a typical propagation speed for fat) and c2 = 1600 m/sec (a typical value for muscle). Using the equation above you will find that the critical angle for this tissue interface is 65°, as shown in Figure 2.18.
Summing up, if ultrasound passes through an interface between two tissues with different propagation speeds, the beam path will be bent except for the special case of perpendicular incidence.
The amount of bending increases when the difference in propagation speeds increases and it also increases for 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.
Suggested activities
1 Look for examples of refraction of light in everyday life.
2 Calculate the critical angle for the example above (where c1 = 1450 m/sec and c2 = 1600 m/sec) and make sure you get the same answer (i.e. 65°). (Hint: calculate the ratio of the two propagation speeds, then calculate the angle that has this as its sin by using your calculator's inverse sin function (sin-1); this angle will be θi.)
Chapter 3: Pulsed ultrasound and imaging
Pulsed ultrasound
Pulse duration and bandwidth
As mentioned in chapter 1, ultrasound imaging makes use of short "pulses" of ultrasound which are transmitted into the body. What do we mean by a pulse of ultrasound, and why do we need to use pulses?
The term pulse refers to ultrasound energy that starts and then stops again shortly afterwards, i.e. a short "burst" of ultrasound energy. Figure 3.1 shows a typical ultrasound pulse. (Remember that the opposite of a pulse is "continuous wave" ultrasound which continues indefinitely.)
Figure 3.1 A typical transmit pulse, in this case lasting a total of four cycles. Note that the amplitude builds to a maximum and then decreases again. The overall time is referred to as the "pulse duration".
Generally the pulse used for ultrasound imaging is 3 - 5 cycles in length, and so the pulse duration will be 3 - 5 times the period of the ultrasound wave. (Remember, the period is the duration of a single cycle.) For example, suppose the pulse shown in Figure 3.1 has a frequency of 4 MHz. Since the period corresponding to a frequency of 4 MHz is 0.25 μsec and since the pulse is 4 cycles long, the pulse duration is simply (4 × 0.25 μsec) = 1.0 μsec.
When ultrasound is reflected or scattered by structures in the body, echoes will return to the probe and they will produce electrical signals that the machine processes to create the ultrasound image. Since the transmit pulse has the form shown in Figure 3.1, each echo will have exactly the same duration and shape (it too will last just a few cycles).
Why is it necessary to use such short pulses of ultrasound? We will see in later chapters that this produces images with good resolution.
The shorter the transmit pulse the better the image resolution.
As we have just seen, a short transmit pulse can be achieved by (a) transmitting only a few cycles and (b) using as high an ultrasound frequency as possible, since the higher the frequency the shorter the period and so the shorter the total pulse duration will be. This is consistent with the statement in chapter 2 that the higher the frequency the better the resolution.
What are the implications of making the transmit pulse short? To answer this, we need to consider the concept of ultrasound frequency again.
As we saw in chapter 2, a sinusoidal continuous wave is the simplest form of ultrasound wave, having a single well-defined frequency, amplitude etc. It is also a useful building block to describe more complex waveforms. Thus we saw that a distorted continuous wave could be broken up into the sum of a number of sinusoidal waves with specific frequencies (f, 2f, 3f, ... etc).
The situation with a pulse is somewhat different. Frequency analysis shows that an ultrasound pulse can be broken up into the sum of an infinite number of sinusoidal waves with frequencies spanning a defined frequency range (see Figures 3.2 and 3.3). While it may seem remarkable that adding an infinite number of continuous waves together can give a pulse that lasts just a short period of time, it is true.
Figure 3.2A short pulse of 4 MHz ultrasound can be shown mathematically to be the sum of an infinite number of continuous sinusoidal waves with frequencies ranging from 3.5 MHz to 4.5 MHz. Just five of the continuous waves are shown in this diagram for simplicity.
A more convenient way of showing the frequency range is the "frequency spectrum" – a diagram showing the range of frequencies that go to make up the pulse and the power at each frequency – as shown in Figure 3.3.
Figure 3.3 The "spectrum" of the pulse in Figure 3.2. Note the definitions of the centre frequency and