CHAPTER 18 The use of ultrasound in medicine (pages 356-8)

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1 PHYSICS 2: HSC COURSE 2 nd edition (Andriessen et al) CHAPTER 18 The use of ultrasound in medicine (pages 356-8) 1. Ultrasound is sound above Hz frequency whereas sound for normal hearing in humans is of frequency below Hz Hz is at the limit of human hearing. 2. Ultrasound is transmitted through body material as compression waves. Attenuation of these waves occurs to some extent as some energy is absorbed by the material. At the interface between two different materials, some of the wave is reflected and some is transmitted. 3. The piezoelectric effect is the change in shape of a crystal when subjected to a potential difference. As a result electrical energy is converted to mechanical energy. 4. When subjected to an alternating potential difference the piezoelectric crystal vibrates. This vibration is rapid and produces signals in the ultrasound frequency range. In this way ultrasound waves are produced. When ultrasound strikes a piezoelectric crystal the pressure changes, due to compressions and rarefactions, cause the crystal to vibrate. The vibration generates an alternating potential difference across the crystal. In this way the ultrasound waves are received. 5. Z = ρv where ρ = 1075 v = 1590 Z = 1075 x 1590 = 1.71 x 10 6 The acoustic impedance of muscle is 1.71 x 10 6 kg m -2 s Z = ρv where Z = 1.59 x 10 6 kg m -2 s -1 v = 1570 ms -1 ρ = 1.59 x = 1.01 x 10 3 kg m -3 The density of blood is 1.01 x 10 3 kg m -3 (or 1.01 g cm -3 ) Medical Physics Chapter 18 1

2 7. Z = ρv where ρ = 1.3 kg m -3 Z = 429 kg m -2 s -1 v = = 330 The speed of ultrasound through air is 330 m s (a) (i) Z = ρv where ρ = 1060 kg m -3 v = 1540 m s -1 Z = 1060 x 1540 = 1.63 x 10 6 The acoustic impedance of soft tissue is 1.63 x 10 6 kg m -2 s -1 (ii) Z = ρv where ρ = 1600 kg m -3 v = 4080 m s -1 = 1600 x 4080 = 6.53 x 10 6 The acoustic impedance of bone is 6.53 x 10 6 kg m -2 s -1 (b) (i) I r I 0 = ( Z 2 Z 1 ) 2 (Z 2 + Z 1 ) 2 where Z 1 = 1.63 x 10 6 Z 2 = 6.53 x 10 6 = I transmitted I 0 = I reflected : I transmitted = : = 3:2 9. I r I 0 = ( Z 2 Z 1 ) 2 (Z 2 + Z 1 ) 2 where Z 1 = 1.65 x 10 6 = Z 2 = 1.71 x 10 6 Medical Physics Chapter 18 2

3 10. (a) There is the most reflection at the skin-air interface (as I r / I 0 has the largest value). (b) There is the least reflection at the brain-fat interface (as I r / I 0 has the smallest value). (c) The greatest amount of absorption occurs where there is least reflection. This is at the brain-fat interface. (d) For the fat-bone interface, I r / I 0 = where I 0 = 60 mw cm -2 I r = 1.74 The intensity of the reflected signal is 1.74 mw cm -2 (e) (i) For the fat-muscle interface, I r / I 0 = where I 0 = 80 mw cm -2 I r = 0.88 mw cm -2 Amount of signal travelling into the muscle = mw cm -2 = mw cm -2 (ii) Some of the energy is converted to heat as the ultrasound travels through the muscle and the rest of the energy travels through the muscle to the next interface, where part is reflected and part is transmitted. 11. The ultrasound will travel through the tissue and be reflected off the surface between the lung and the air in the lung, so the ultrasound will not pass through the lung. The difference in acoustic impedance between the surrounding tissue and air is large and hence most of the ultrasound is reflected. 12. If the ultrasound travels through fat to liver, the same percentage of ultrasound will be reflected as when the ultrasound travels through liver to fat. Because of the squared term in the equation to find I r / I 0, it does not matter whether fat or liver is assigned the value Z 1, the ratio will remain the same. 13. A full bladder pushes the lower intestine, which contains a lot of gas, out of the way and lifts the uterus to a good position for taking an image. If the lower intestine were in the way, ultrasound would be reflected from the gas and it would not be possible to obtain an image of the foetus. Medical Physics Chapter 18 3

4 14. I r I 0 = (Z 2 Z 1 ) 2 (Z 2 + Z 1 ) 2 where Z 1 = 1.5 x 10 6 (for aqueous humour) Z 2 = 1.85 x 10 6 (for the lens) = I r = x 15 = 0.16 Intensity of the reflected beam is 0.16 mw cm (a) Most of the ultrasound would be reflected from the interface between air and skin as the difference in acoustic impedance between air and skin is very large. (b) The optimum acoustic impedance of the gel would be the same as that of the skin. Then all the incident ultrasound would be transmitted into the skin. The acoustic impedance of the gel = 1010 x 1540 = 1.56 x 10 6 kg m -2 s -1 (c) Z = ρv where Z = 1.56 x 10 6 kg m -2 s -1 ρ = 1200 kg m -3 v = 1300 The speed of ultrasound in the gel is 1300 m s An ultrasound A-scan is a range measuring system in which an ultrasonic pulse is sent into the body in one line and the time for the pulse to be reflected from an interface in the body is measured. The intensities of the reflected signals are plotted as a function of time and, from this information, the position of features inside the body can be found. For a B-scan, a pulse is sent into the body along one line and the reflected pulses recorded as dots, the brightest corresponding to the signal of greatest intensity. By sending pulses into the body at different angles, a series of B-scans can be used to build up a 2-D picture of a cross-section through the body. Medical Physics Chapter 18 4

5 17. A-scans are used to find the distance of the lens and retina from the front of the eye by measuring the time it takes a pulse of ultrasound to be reflected from these parts of the eye. Hence A-scans are distance-finders or range-finders. 18. (a) The spacings of the pulses indicate how deep the interface is in the body. The first reflection is received after approximately 0.04 ms, the second after 0.2 ms and the spinal column echo from the deepest interface in the body after 0.24 ms. (b) The amplitude of the reflected pulse will be less if the pulse has travelled through material and lost energy and it will be less if the signal is reflected from an interface where the difference in acoustic impedance is small. (c) Time for reflected pulse to be received is approximately 0.24 ms. Time for pulse to reach the spine is 0.12 ms. The pulse travels at 1500 ms -1, hence the distance to the spine is 1500 x 0.12 x 10 3 x 10 2 cm, which is 18 cm. (d) Phase scan 19. (a) speed of sound = 1540 m s -1 distance travelled by pulse = 700 mm = 7 x 10-1 m 0.7 time for pulse to travel this distance = 1540 s = 4.5 x 10-4 s minimum time between pulses = 4.5.x 10-4 s (b) A faster pulse rate would mean that the time between pulses was shorter than 4.5 x 10-4 s. A new pulse would be sent into the body before the reflected pulse was received, leading to interference between the reflected pulse and the incoming pulse and lack of clarity of the position of the image. Medical Physics Chapter 18 5

6 20. Scan type Areas where valuable Areas where not useful A-scans To locate the depth of parts of the eye Where an image of an organ is needed B-scans To obtain a 2-D image of a body organ When the organ is behind bone Sector scans When there is a small window of entry to the When real-time scans are needed body, e.g. through the gap in the skull of a newborn baby Phase scans When real-time scans are needed to monitor foetus movement or heart movement When depth of structures in the body is all that is needed 21. (a) Ultrasound signal sent through heel bone transmitted ultrasound signal received the time of transmission measured the attenuation of the signal recorded process is repeated many times the signal is analysed and compared with standards for normal heel bone (b) This method has limited effectiveness in the diagnosis of osteoporosis as it is not done at the site of possible fractures, i.e. the hip or spine. It can be used as an indicator of the need for further tests using X-rays if an abnormal result is obtained. 22. A continuous signal is directed at the foetal heart and the reflected signal is detected by a separate receiver. The output is electronically filtered so that only the difference in frequency due to the Doppler shift is amplified. This difference is in the audible range. The tone will vary with the speed of movement of the heart and hence with the heartbeats which can therefore be monitored. 23. Real-time imaging is making an image that can be viewed as it is formed. This type of imaging allows movement to be viewed. The images have to be produced at a rate of greater than 16 images per second for movement to be seen. Medical Physics Chapter 18 6

7 24. The video images will probably be colour-coded to show the speed of the blood flow and the direction of blood flow. 25. The images should both show an outline of the organ. These images will be clearly made of a number of spots so they will have a fuzzy appearance. The images may vary in density due to the different amount of reflection from different interfaces in the body. Medical Physics Chapter 18 7

FXA 2008. UNIT G485 Module 4 5.4.3 Ultrasound. Candidates should be able to :

FXA 2008. UNIT G485 Module 4 5.4.3 Ultrasound. Candidates should be able to : 1 Candidates should be able to : ULTRASOUND Describe the properties of ultrasound. ULTRASOUND is any sound wave having a frequency greater than the upper frequency limit of human hearing (20 khz). Describe