Conversely, we can see that having a large depth of penetration (as generally required in an abdominal scan, for example) will force the machine to use a relatively low PRF and so decrease the rate of imaging.
Frame rate limitations
Now that we know how to calculate the maximum allowable PRF that the machine can use, it is relatively easy to calculate the maximum possible imaging rate. Note that the number of images created each second is termed the "Frame Rate" (abbreviated FR).
If the machine requires N transmit pulses to create each image, it is easy to see that the total number of pulses required each second is simply (N × FR). Since the total number of transmit pulses available each second is by definition the PRF, we have:
and therefore
The frame rate is directly proportional to the PRF and inversely proportional to the number of pulses required to produce each image.
Now we can introduce the previously derived equation relating the maximum allowable PRF and the depth of penetration (P). This gives us the following relationship between the maximum possible frame rate, the depth of penetration and the number of transmit pulses required to create each image:
This is more commonly rearranged and written as follows:
Assuming that the frame rate (FR) is expressed in frames/ sec and the penetration (P) is measured in cm, then we can substitute the usual value for the propagation speed in soft tissue, converting it to cm/sec for consistency (i.e. c = 1540 m/sec = 154,000 cm/sec). This gives:
The product of the three quantities on the left of this equation is thus limited by a physical constant which we cannot alter (the propagation speed divided by two). Therefore, if we want to increase any one of these three quantities (e.g. frame rate), one or both of the other two must be decreased to compensate.
For example, consider an abdominal scan where we require a depth of penetration of 25 cm. Suppose that the ultrasound machine needs a total of 100 transmit pulses to create each image. Then the above equation tells us that the maximum possible frame rate is 30 frames per second.
While this is quite adequate for most applications, there are some clinical areas (such as echocardiography) where we may want to increase the frame rate above this value. Suppose we wanted a frame rate of 60 frames per second, but still required the 25 cm depth of penetration. Then the equation above tells us that the number of pulses for each image would need to drop to 50.
In echocardiography it is often possible to achieve this by narrowing the field of view of the transducer so that less pulses are needed to create each image.
We will revisit the issue of frame rate under the heading Temporal Resolution in chapter 10. In particular, we will take account of the fact that many machines are able to produce several beams at once. This "multiple beamforming" capability speeds up the acquisition of echo information, increasing the frame rate significantly.
SUMMARY
This story has become complex, so let us recapitulate:
To create an image, the machine transmits ultrasound pulses and collects the echoes that come back from structures within the tissues.
The time taken for each pulse to travel into the tissue and for the echoes to return depends on the depth of penetration (P) of the ultrasound; the greater the depth the longer it takes for the echoes to return.
The machine must transmit a number of pulses (N, typically 100 or so) to create each image.
Thus the overall time taken to produce one image will be determined by both P and N.
The number of images that can be produced each second (i.e. the frame rate FR) is therefore also a function of both P and N.
High frame rates can be achieved if the depth of penetration is small and/or if a relatively small number of transmit pulses are needed to create each image.
Suggested activities
1 Consider a 10 MHz probe used for breast scanning. Suppose that it has a depth of penetration of 8 cm.Calculate the delay between the transmit pulse and the last detectable echo.Suppose that the machine requires 200 transmit pulses to create an image. Calculate how long it takes to create an image.Using the answer to (b), calculate the frame rate (i.e. how many images will be created each second).Check that (FR × P × N) = 77,000. Don't be concerned if you don't get exactly 77,000 – this will be due to rounding errors.
2 Note the actual frame rate at which a clinical ultrasound machine operates and whether the frame rate changes when you switch from a low frequency probe to a high frequency probe. Consider whether the differences that you observe are consistent with the above discussion.
Principles of image formation
The previous sections have described the basis of ultrasound imaging:
The machine's probe transmits a short pulse of ultrasound into the patient's body.
This pulse travels along a defined pathway referred to as the ultrasound "beam".
Structures within the beam cause some of the transmitted energy to be reflected and scattered.
Some of the reflected and scattered energy travels back along the beam to the probe, where it is detected and converted into an electrical signal.
The signal due to an individual structure is referred to as an "echo". As we have seen, by measuring the exact arrival time of an echo the machine can calculate the distance from the probe to the structure that caused it. The machine also knows the position of the beam relative to the probe, and so it is able to define relatively accurately where the structure is within the ultrasound image. This allows it to place a point in the image to represent the structure, as shown in Figure 3.9.
Figure 3.9 (Left) Reflected ultrasound returning to the probe will give rise to an echo signal. (Right) On the machine's display the echo will be displayed as a dot. Note that it is placed in the image (the black rectangle) at the correct depth and on a "line of sight" that corresponds to the centre-line of the beam.
The brightness of the dot is determined by the intensity of the echo. A weak echo will be shown as a dark grey, a moderate intensity echo as a mid-grey and a strong echo as white. Black is shown when no echoes are detected.
Three standard
