Question

Exercise 4.4. Use parameters from Textbook Page 156. (12 pts total 2 pts each answer.) Absorption and attenuation of ultrasound energy 4.4 Calculate the distance at which the intensity of a 1 MHz and 5 MHz ultrasound beam will be reduced by half traveling through (a) bone, (b) air, and (c) muscle. Parameters: a(dB cm'')-: 4.343a(crn-'). (4.21) A useful rule-of-thumb is that each 3 dB reduction corresponds to a reduction in intensity by a factor of 2. So a 6 dB reduction corresponds to a factor of 4, a 9 dB reduction to a factor of 8 and so on. The frequency dependence of μ for soft tissue is 1 dB cm-MHz-, i.e. at 2 MHz the attenuation coefficient is 2 dB cm For fat the attenuation coefficient is given approximately by 0.7f5 dB. The values of the attenuation coefficient for air and bone are much higher, 45 dB cm MHz and 8.7 dB cm MHz respectively. The intensity of a 3 MHz ultrasound beam entering tissue is 10 mWicm2. Calculate the intensity at a depth of 4 cm. ample 4.3 Solution The attenuation coefficient is 1 dB cm- MHz, and so has a value of 3 dB cm1 at 3 MHz. At a depth of 4 cm, the attenuation is 12 dB, which corresponds to a factor of 16. So the intensity of the beam is 0.625 mW/em 2

Answer 1

From attenuation law, -aft log I log Ie - - log I log I-af log I-af log 2 log Io-log 2og Io-afx log 2 afc log 2 af 2 dB af The change in intensity log 2 2 dB. (a) The distance at which the intensity 1 MHz is reduced to half through bone is 2 dB af 2 dB MHz"(1 МH2) 8.7 dB cm 0.23 cm The distance at which the intensity 5 MHz is reduced to half through bone is, 2 dB X= af 2 dB - MHz(5 MHz 8.7 dB cm 0.046 cm

(b) The distance at which the intensity 1 MHz is reduced to half through air is 2 dB х3 af 2 dB МHz" (1 МHz) 45 dB cm 0.044 cm The distance at which the intensity 5 MHz is reduced to half through air is, 2 dB af 2 dB MIz" (5 МHz) 45 dB cm =0.008 cm (c) The distance at which the intensity 1 MHz is reduced to half through muscle is, 2 dB х3 af 2 dB -1 1 dB cm MHz" (1 МНz) = 2 cm The distance at which the intensity 5 MHz is reduced to half through muscle is 2 dB х3 af 2 dB MHz"(5 МH2) 1 dB cm =0.4 cm