1 | clear all; close all; clc |
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2 | %% Double Cross filter |
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3 | n1=1; |
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4 | n2=1; |
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5 | n=1;%ref. indx of media between filters |
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6 | Thetp=90; |
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7 | |
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8 | a1=1e-4; |
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9 | A1=0.53; |
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10 | |
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11 | c=299792458; |
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12 | Thet=pi/180*0; |
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13 | SPlam=c/50e+9; |
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14 | aar=[0.01 0.1 1 2 5]*1e-3; |
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15 | color=colormap(jet(length(aar))); |
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16 | |
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17 | figure |
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18 | hold on |
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19 | for cc=1:length(aar) |
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20 | |
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21 | b=aar(cc); |
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22 | a=1e-4; |
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23 | g=(SPlam+4*a+2*b)/2.27; |
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24 | disp(['g=' num2str(round(g*1e+4)/10) ' a=' num2str(a*1e+3) ' b=' num2str(b*1e+3) ' Freq=' num2str(round(c/SPlam*1e-9)) 'GHz' ]) |
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25 | |
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26 | d=g/2; |
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27 | loss=0; |
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28 | c=3e+8; |
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29 | FreqArr=1e+9:1e+8:2e+11; |
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30 | for cnt=1:length(FreqArr) |
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31 | |
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32 | freq=FreqArr(cnt); |
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33 | |
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34 | % Stot= S(freq, Thet, g, a, b, n1, n2)*T(freq,loss,d,n)* S(freq, Thet, g, a, b, n1, n2); |
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35 | Stot= S(freq, Thet, g, a, b, n1, n2); |
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36 | |
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37 | TTE(cnt)=abs(1./Stot(2,2))^2; |
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38 | TTM(cnt)=abs(1./Stot(4,4))^2; |
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39 | end |
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40 | plot(FreqArr*1e-9,TTE,'color',color(cc,:),'linew',2) |
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41 | legendInfo{cc} = ['b= ' num2str(aar(cc)*1e+3) ' mm']; % or whatever is appropriate |
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42 | |
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43 | end |
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44 | % Ttot=sin(Thetp).^2.*TTE+cos(Thetp).^2.*TTM; |
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45 | grid on |
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46 | set(gca,'fontsize',16) |
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47 | % set(gca,'xscale','log') |
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48 | xlabel('Frequency, [GHz]') |
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49 | ylabel('Transmissivity, [1]') |
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50 | title('Single cross filter 50GHz') |
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51 | legend(legendInfo,'location','northeast') |
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