19
CHAPTER 3: PULSED ULTRASOUND AND IMAGING
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.
.
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.
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.
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 “realtime” (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).
Figure 3.7 Geometry showing the round-path distance
travelled by the ultrasound for an echo coming from a
reflector at depth d in the tissues.
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.
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:
Suggested activities
1. Consider a 5 MHz transmit pulse with a duration of
1 μsec.
a) How many cycles are transmitted?
b) Sketch the spectrum for this pulse.
2. Now consider a 10 MHz pulse with a duration of 0.5
μsec.
a) Sketch the spectrum for this pulse.
b) Compare this to your answer for question 1.
c) Which pulse will give the best image resolution?
3. An L3-7 probe has a bandwidth extending from 3
round path distance
propagation speed
(2 × d )
=
c
t=
Rearranging this equation allows us to calculate the depth
of the reflector from the delay time as follows:
d=
(c × t )
2
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).
20
CHAPTER 3: PULSED ULTRASOUND AND IMAGING
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.
tp =
(2 × P )
c
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:
maximum PRF=
c
(2 × P )
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 will also
be high. 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).
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:
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:
N × FR = PRF
and therefore
FR
PRF
N
The frame rate is directly proportional to the PRF and
inversely proportional to the number of pulses required to
produce each image.
CHAPTER 3: PULSED ULTRASOUND AND IMAGING
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:
FR =
c
(2 × P × N )
SUMMARY
This story has become complex, so let us recapitulate:
t
t
This is more commonly rearranged and written as follows:
( FR × P × N ) =
c
2
t
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:
t
( FR × P × N ) = 77 , 000 cm/sec
t
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.
21
t
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.
(a) Calculate the delay between the transmit pulse
and the last detectable echo.
(b) Suppose that the machine requires 200 transmit
pulses to create an image. Calculate how long it
takes to create an image.
(c) Using the answer to (b), calculate the frame rate
(i.e. how many images will be created each second).
(d) 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.