What determines the dimensions of this spectrum? The centre frequency is simply the frequency of oscillation of the pulse (4 MHz in this example); the bandwidth (B) is related to the pulse duration (τ) by a very simple equation:
In this case the pulse duration is 1 μsec and so the bandwidth is 1 MHz. The shape of the spectrum is determined by the overall shape of the transmit pulse (i.e. the way its amplitude increases and decreases).
Notice that the relationship between pulse duration and bandwidth is an inverse one. This means that if the pulse gets shorter then the bandwidth becomes larger. In other words, a bigger range of frequencies must be combined together if you want to create a very short pulse. For example, Figure 3.4 shows the spectrum for a 4 MHz pulse just two cycles long. Notice that the centre frequency is still 4 MHz but the bandwidth has doubled because the pulse duration has halved.
Figure 3.4 Spectrum showing the effect of halving the transmit pulse duration compared with Figure 3.3.
Conversely, if the pulse becomes very long, the bandwidth will become small. At the extreme limit, the bandwidth becomes zero for a continuous wave which lasts for ever (i.e. there is no range of frequencies, just a single frequency, as shown in Figure 3.5).
Figure 3.5 Spectrum for a continuous wave signal.
Why does the bandwidth matter? An ultrasound transducer (the active element in the probe) inherently has a limited bandwidth, i.e. it can only process electrical and ultrasound signals within a defined range of frequencies. Since it is difficult for the manufacturers to increase transducer bandwidth, they must ensure that the transmit pulse is designed so that the full range of frequencies that it contains fall within the range that the transducer can process. (If the bandwidth of the transducer is too small, it will force the transmit pulse to become longer.)
Pulse repetition frequency (PRF)
When the ultrasound machine sends a pulse into the body, it travels along a defined pathway (referred to as the ultrasound "beam") and the echoes return to the probe along the same beam, as shown in Figure 3.6. The echoes therefore provide information to the machine about the tissues and structures that are within the beam.
Figure 3.6 The transmit pulse travels into the tissue along a well-defined path or "beam" and the echoes return to the probe along the same path. Thus the echoes provide information about the tissues lying along this path.
In order to build up an image, the machine must repeat the process of transmitting a pulse and receiving the echoes from the body numerous times, moving the beam so that it passes through a different area of tissue each time. (The movement of the beam is referred to as "scanning"; it will be discussed in more detail in chapter 4.) Thus the machine must produce a series of transmit pulses – usually 100 or more – to produce a single image.
Furthermore, the ultrasound machine must create each image in a fraction of a second so that a rapid sequence of images can be created and displayed in the form of a "real-time" (i.e. movie-like) image.
Clearly this means that the machine needs to send out transmit pulses as rapidly as possible. For example, if each image takes 100 transmit pulses to produce and we require an imaging rate of 20 frames per second (i.e. 20 images per second), it is easy to see that the machine must transmit a total of 2000 pulses each second.
The term used to describe the number of transmit pulses each second is the "Pulse Repetition Frequency" (abbreviated PRF).
For the machine's electronics, transmitting 2000 pulses per second is no problem. However, we will see in the next section that there is an important consideration (namely the depth of penetration of the ultrasound) that limits the number of pulses that can be transmitted each second.
Suggested activities
1 Consider a 5 MHz transmit pulse with a duration of 1 μsec.How many cycles are transmitted?Sketch the spectrum for this pulse.
2 Now consider a 10 MHz pulse with a duration of 0.5 μsec.Sketch the spectrum for this pulse.Compare this to your answer for question 1.Which pulse will give the best image resolution?
3 An L3-7 probe has a bandwidth extending from 3 MHz to 7 MHz (i.e. it can process frequencies in this range). What is the shortest possible pulse duration that can be transmitted using this probe?
Pulse echo principle
We now come to the fundamental concept that underlies diagnostic ultrasound – the pulse echo principle.
Simply put, by carefully measuring the time between the transmission of the transmit pulse and the reception of a given echo, the ultrasound machine can calculate the distance between the probe and the structure that caused that echo. Consider the situation shown in Figure 3.7.
Figure 3.7 Geometry showing the round-path distance travelled by the ultrasound for an echo coming from a reflector at a depth d in the tissues.
The total "round-path" distance travelled by the ultrasound (from the probe to the reflector and then from the reflector back to the probe) is simply (2 × d). The time taken to travel this distance (and therefore the time delay between the transmit pulse and the echo) is calculated as:
Rearranging this equation allows us to calculate the depth of the reflector from the delay time as follows:
Thus we see that there is a simple proportional relationship between the arrival time of an echo (t) and the depth of the structure causing the echo (d).
For the ultrasound machine to use this relationship, it must assume a value for c. It assumes the average propagation speed in soft tissue (1540 m/sec).
As an example, consider a reflector at a depth of 1 cm. Using the above equation gives a delay time of 13 μsec. This is a useful number to remember – for every centimetre of depth the echo delay is 13 μsec. For a reflector at 15 cm depth, for instance, the delay time will be (15 × 13 μsec) = 195 μsec.
Pulse repetition frequency limitations
We saw at the end of the previous section that real-time scanning requires the ultrasound machine to create images as quickly as possible. We will now see how the time taken for ultrasound to travel in the tissues limits the rate of imaging. This occurs because of a fundamental rule that the ultrasound machine must obey (see Figure 3.8):
The machine must not transmit again until all detectable echoes caused by the previous transmit pulse have been received.
If the machine violates this rule, echoes from the new transmit pulse will overlap with the echoes from the previous pulse and an artifact known as "range ambiguity" will occur. We will consider this in more detail in chapter 8.
Figure 3.8 The voltage seen on the transducer as a function of time. The first transmit pulse is followed by a series of echoes (only three are shown here for simplicity),