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EM350 midterm Problem 2

# EM350 midterm Problem 2 - frequency range The following...

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0 2 4 6 8 10 12 14 16 18 20 1 2 3 4 5 6 7 8 Impedance Vs. Frequency frequency ,MHz Impedance, MRayls Problem 2 a) Here is the function layer_imp.mat that will produce the specific acoustic impedance Z p in MRayls with input of frequency and thickness of the layer. MATLAB CODE function [Zp] = layer_imp(f,l) Cep=2700*10^-3; Zep=3.25; Cw=1480*10^-3; Zw=1.48; T11=cos((2*pi*f/Cep)*l)*Zw; T12=-i*Zep*sin((2*pi*f/Cep)*l); T21=(-i*sin((2*pi*f/Cep)*l)/Zep)*Zw; T22=cos((2*pi*f/Cep)*l); Top= T11 + T12; Bottom= T21 + T22; Zp= Top./Bottom; Using this function made a plot of magnitude of impedance with frequency of 0 to 20 MHz if l= 0.06 mm. MATLAB CODE clc; clear; f=s_space(0,20,1024); l=0.06;

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Zw=1.48; [Zp] = layer_imp(f,l); figure(1); plot(f,Zp) title( 'Impedance Vs. Frequency' ) xlabel( 'frequency ,MHz' ) ylabel( 'Impedance, MRayls' ) b) To simulate the 10 MHz transducer at the Center for NDE the equation given was given to determine the change of the magnitude of acoustic impedance over the

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Unformatted text preview: frequency range. The following code was used to make a plot of the percent acoustic impedance as the thickness layer increases from 0.006 mm to 0.06 mm. MATLAB CODE f2=7; f3=13; l=[0.006:.000005:0.06]; for x=1:length(l) [Zp2(x)] = layer_imp(f2,l(x)); [Zp3(x)] = layer_imp(f3,l(x)); lab_acoustic(x)=(abs(Zp3(x))-abs(Zp2(x)))/Zw*100; end figure(2); plot(l,lab_acoustic) title( 'Lab Impedance Vs. layer thickness' ) xlabel( 'layer thickness, mm' ) 0.01 0.02 0.03 0.04 0.05 0.06 50 100 150 200 250 300 X: 0.008375 Y: 10 Lab Impedance Vs. layer thickness layer thickness, mm percent of acoustic impedance ylabel( 'percent of acoustic impedance' ) From the plot the percent impedance increases rapidly and if the need is that the transducer needs a change less than 10% the thickness layer will have to be below 0.008375 mm....
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EM350 midterm Problem 2 - frequency range The following...

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