Ultrasound Imaging Consider an acoustic pulse that travels from a transducer thr
ID: 2076180 • Letter: U
Question
Ultrasound Imaging Consider an acoustic pulse that travels from a transducer through the skin, bone, brain, and is reflected back from the second brain/bone interface to return to the transducer (see figure below). Consider the pulse being normally incident onto the interfaces. We also consider the transducer is in close contact with the skin and we neglect absorption. The acoustic impedences of the structures are Z_skin = 1.6 times 10^6 Pa s m^-1 = 7.8 times 10^6 Pa s m^-1 and Z_brain = 1.58 times 10^6 Pa s m^-1 a. What is the portion of the signal coming back from the second brain bone interface to the transducer? b. In practice, why is gel spread between the transducer and the skin?Explanation / Answer
a) we know that acoustic impedance Z=density of the part*velocity of sound i that part
now reflectance or the ratio of incident to reflected wave determines the amount of reflection
for skin to bone reflection the reflection coefficient R (S to B) =R1= (Zskin-Zbone)^2/(Zskin+Zbone)^2=.435
then transmitance T(S to B) =T1=1- R (S to B)=1-.435=.565
again we calculate R and T for bone and brain
then R (Bone to brain)=R2= (Zbrain-Zbone)^2/(Zbrain+Zbone)^2 =.439
T(Bone to Brain)= T2=1- R (Bone to brain)= 1-.439=.561
Using I=I0. T1.T2.R2.T2.T1=I0.(.565*.561*.439*.561*.565)=.044.I0
that means 44% signal is coming back from second surface to the transducer.
b) If there is nothing between transducer and the skin as medium, then we find air filling the vaccum in there. The problem with air is it acts like a wall to the sound and reflects approximately 99.5% back to the first medium. But if gel of medium viscosity is used as the interface between trasducer and skin then the gel will pass the ultrasound ideally just like the velocity of the sound 1518m/s with minimal refraction.
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