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), diminishing in amplitude until they become undetectable. The time from the transmit pulse to the last detectable echo is tp. The next pulse can only be transmitted after this time.
Now consider a probe which penetrates to a depth P. (Reminder: this means that echoes from structures at depths greater than P are not detectable.) Since we know the relationship between depth and echo arrival time, we can calculate when the last detectable echo will be received.
The time delay between the transmit pulse and the last detectable echo (tp) is given by:
Given the rule cited above, this will be the shortest allowable time between one transmit pulse and the next. It follows therefore that the maximum number of transmit pulses per second will be the reciprocal of this time:
Remember that "Pulse Repetition Frequency" (PRF) means the total number of pulses transmitted each second.
Since the machine generally strives to produce as many images each second as possible, it will make an estimate of the depth of penetration P (taking into account the ultrasound frequency being used) and then it will use the maximum PRF consistent with that depth of penetration.
The above relationship between the maximum allowable PRF and the depth of penetration is an inverse one. Thus, if the depth of penetration is small (e.g. in a carotid artery scan) a high PRF can be used and so the frame rate
