The Physics and Technology of Diagnostic Ultrasound (Second Edition)

Study Guide

Robert Gill

• Publisher: High Frequency Publishing
• Year: 2020
• ISBN:
  9780648869610
• Topic: Medicina
• Subtopic: Tecnología y suministros médicos

This Study Guide is a companion to the popular ultrasound physics textbook "The Physics and Technology of Diagnostic Ultrasound: A Practitioner's Guide". It contains over 120 short questions and provides model answers for each. It has been designed for both students and teachers. Students will find it valuable as a learning aid and as a resource to test their knowledge and understanding. Teachers, supervisors and tutors will find it a useful teaching asset and an excellent starting point for writing quiz and exam questions.

C: Model answers

The following answers list the main points that should be mentioned in your answer. They are not exhaustive and the order in which points are listed is not necessarily significant. Where the question asks for one or more diagrams, refer to the textbook The Physics and Technology of Diagnostic Ultrasound: A Practitioner’s Guide (Second Edition) by Robert Gill for relevant diagrams.

Chapter 2: Ultrasound interaction with tissue

1. Energy = work (measured in Joules).
   Power = energy per second (measured in Watts).
   Intensity = power per unit area (measured in Watts/cm²).
   Intensity is the most relevant parameter when you are assessing patient exposure since it takes into account whether the ultrasound energy is spread out through a significant volume of tissue or whether it is concentrated in a small volume (e.g. due to focussing or when the beam is stationary as in pulsed Doppler).
2. Wavelength = physical distance occupied by one cycle.
   Period = time taken for one cycle.
   Pulse duration = time taken for one transmit pulse.
3. If f = 4.5 MHz then

   

   
   If f = 12.5 MHz then

   

4. Amplitude = amount by which the pressure in the patient’s tissues is increased and decreased (relative to normal pressure) due to the presence of the ultrasound wave.
   Intensity is proportional to (amplitude²).
5. Absorption of ultrasound energy by the tissues.
   Reflection of ultrasound energy.
   Scattering of ultrasound energy.
   Defocussing of the ultrasound beam, as occurs in edge shadowing.
   Absorption is the dominant factor in normal tissue.
6. Attenuation:

   

   (note: this means the intensity is reduced by a factor of 10^3.5 = 3,200)

7. 

8. Depth of penetration = maximum depth from which echoes can be detected and displayed. Echoes from greater depths are too weak to display. We can show that

   

   So depth of penetration is inversely proportional to frequency.
9. Using the above equation

   

   therefore:

   

   and so:

   

10. Reflection is the term used when some or all of the energy incident on a relatively smooth surface does not pass through the surface but instead travels back into the first tissue again. The direction in which it travels is easily determined, since the incident angle is equal to the reflection angle. (Note these angles are measured between the direction of travel of the ultrasound and a line perpendicular to the interface.)
    When the incident angle is 90° the reflected ultrasound (i.e. the echo from the interface) travels directly back to the probe and so it will be detected and displayed. When the incident angle is not 90° the reflected ultrasound travels away from the probe and it will not be detected. The interface will then be seen in the image only if it also scatters sufficient ultrasound to be visible.
11. The fraction of the ultrasound power that is reflected is defined as the reflection coefficient (R). Since the factor of 10⁶ is in the top and bottom of the equation it will cancel out, so R can be calculated as follows:

    

12. Energy cannot be created or destroyed, so the sum of the reflected and transmitted power must be equal to the power incident on the interface. It follows that the sum of the reflection and transmission coefficients must be 1. Therefore the fraction of the power that is transmitted through the interface is:

    

13. Reflection is described in the answer to question 10 above.
    Scattering is the term used when ultrasound interacts with a small object. A fraction of the ultrasound energy is ‘scattered’, with the remaining energy continuing on unchanged. The scattered energy is distributed in all directions. This is the major difference between reflection and scattering.
    Reflection gives rise to linear echoes that represent the reflecting tissue interface. Often these echoes are quite strong, especially when the ultrasound is at perpendicular incidence to the tissue interface.
    The echoes produced by scattering are generally lower in level. In addition, the interaction between multiple scattered echoes produces the characteristic ‘speckle’ appearance seen in ultrasound images of soft tissue.
14. See the answer to question 13 above.
    Frame averaging (persistence) and compound imaging both tend to reduce speckle.
15. Refraction is the term used when the direction of the ultrasound beam changes as a result of passing through an interface between two tissues that have different ultrasound propagation speeds. The change of angle is calculated using Snell’s Law:

    

    where θ_i and θ_t are the incident angle and transmitted angle respectively and c₁ and c₂ are the propagation speeds in the first and second tissues respectively.
16. Using Snell’s Law (see the answer to question 15 above):

    

    and so

    

    Yes, this interface does have a critical angle (defined as the incident angle for which θ_t = 90°) since c₂ > c₁.
17. As mentioned above, the critical angle is the value of the incident angle for which the transmitted angle is 90°. This can only occur when the propagation speed in the second tissue is greater than the propagation speed in the first tissue.
    When the incident angle exceeds the critical angle, the ultrasound is reflected by the interface and does not pass through it. This causes artifacts such as edge shadowing and mirror image, and it prevents the second tissue from being imaged.

Chapter 3: Pulsed ultrasound and imaging

1. (a) Assume the echo comes from the centre-line of the ultrasound beam.
   (b) Depth = (c × t)/2 where c is the propagation speed and t is the echo arrival time (i.e. the difference between when the transmit pulse was transmitted and when the echo was received).
2. Using the equation in the answer to question 1:

   

3. Pulse duration.
   Two objects will be ‘resolved’ (i.e. seen in the image as two separate objects) if their echoes do not overlap in time. This will be the case as long as the arrival times of the two echoes differ by at least an amount equal to the pulse duration (τ). Using the relationship between echo arrival time and depth cited in the answer to question 1 above, this means that the difference in depth between the two objects must be greater than (c × τ)/2 which is therefore the formula for the axial resolution.
4. PRF = number of transmit pulses per second.
   Machine must wait until all detectable echoes have been received before it can transmit again. The deeper it is penetrating the longer it must wait and so the lower the PRF will be.
   The last detectable echo comes from the penetration depth P. It will be received at a time 2P/c after the last transmit pulse. If the machine transmits again at this time (since this will maximise the PRF) the period of the PRF will be 2P/c and so the PRF will be c/2P. Thus the relationship between penetration and PRF is a reciprocal one.
5. PRF:

   

   and

   

6. Continuing from the answer to question 4 above, let us assume it requires N transmit pulses to create each image. Then the number of images per second is simply the PRF divided by N. Using the result above:

   

   This is often written as:

   

7. Frame rate:

   

8. With both the linear and curved probes, a group of transducer elements form the aperture that is used to create the beam. The beam is moved along the probe by incrementing the aperture, generally by dropping off one element and replacing it with another at the other end of the aperture. For example, elements 1 - 128 are used for the first beam, 2 - 129 for the second etc. The beam is normally perpendicular to the transducer face and so a linear array scans a rectangular region and a curved array scans a region that spreads out with depth.
   The phased array uses all the transducer elements at once. It keeps the starting point of the beam fixed in the centre of the transducer and steers the beam in different directions to produce a sector scan.
9. See description in the answer to question 8.
10. Linear array: peripheral vascular, neck, breast.
    Curved array: abdomen.
    Phased array: heart.
11. B-mode: standard two-dimensional image where the brightness at a point indicates the echo strength.
    M-mode: the beam is kept in a fixed position and it produces an image which is a single line of echoes along this beam; this image is swept slowly across the screen to create a time-motion display of tissue movement.
    A-mode: Again the beam is kept in a fixed position; in this mode the amplitude is displayed as vertical deflection on a graph, with time (i.e. depth) on the horizontal axis.

Chapter 4: Transducers and focussing

1. The term piezoelectric is used to describe materials that produce an electrical voltage when they are deformed (e.g. squeezed or expanded). The inverse piezoelectric effect is where the dimensions of the material change in response to an applied electrical voltage.
2. The transducer is made of piezoelectric material (see the answer to question 1 above). An electrical voltage that oscillates at the ultrasound frequency is applied between the top and bottom faces of the transducer. The transducer oscillates in thickness as a result, creating pressure variations in the tissues that become an ultrasound wave.
3. When an echo reaches the transducer it produces oscillations in pressure that cause the transducer thickness to oscillate. This causes an oscillating electrical voltage to be produced between the f...