What are Lambda and Z Parameters in Ultrasound B-mode Imaging?

AI Thread Summary
Lambda in ultrasound B-mode imaging is defined as 513 x 10^-6 m, with ZR values of 1.2 cm and 1 cm, leading to different calculations for LR based on whether Z is less than or greater than ZR. When Z is less than ZR, LR equals D, which is 0.5 cm; when Z exceeds ZR, LR is calculated using the formula involving the tangent function. The discussion also touches on the time calculation t = Z/c, resulting in 77.9 x 10^-6 s. Participants emphasize the importance of clarifying the variables used in the equations for better understanding. Overall, the conversation highlights the mathematical relationships in ultrasound imaging parameters.
nao113
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Homework Statement
1. The transducer array of a linear-array real-time imaging instrument has 32 unfocused elements. Each element is 0.5 cm wide, and there is a nonradiating gap of 0.5 cm between neighboring elements. Only one element at a time is excited.
a) At a frequency of 3 MHz, find the lateral resolution of this instrument at a depth of 1 cm and a depth of 8 cm. The lateral resolution of a linear- array imager is given by the beam size or the line spacing, whichever is larger.
Use relationship:
c = 1540 m/s, c =frequency x l
b) If echo information out to a depth of 12 cm is desired, calculate the minimum time required to scan the entire array.
Relevant Equations
Z < ZR => LR = D
Z > ZR => LR = 2 tan(sin-1 lambda/D) x Z
Screen Shot 2022-05-26 at 19.16.11.png


a. lambda = 513 X 10^-6 m
ZR = 1, 2 cm
Z = 1 cm => Z < ZR => LR = D = 0.5 cm
Z = 8 cm => Z > ZR => LR = 2 tan(sin^-1 (lambda/D)) x Z
= 2 tan(sin^-1 (513 x 10^-6 m/0.005 m)) x 0.08
= 1.65 cm

b. t = Z/c => 0.12/1540 = 77.9 x 10^-6 s
 
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It won't hurt to actually explain what all these letters mean in your equations. This in case you have some question, eventually. So far I don't see the purpose of your post.
 
nasu said:
It won't hurt to actually explain what all these letters mean in your equations. This in case you have some question, eventually. So far I don't see the purpose of your post.
Hello, thank you for the advice, I already changed it, I wonder whether I answered them correctly or not.
 
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