homework9sol - ECE 303: ngwwk its? Sfimng (Eva FMW 1L....

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Unformatted text preview: ECE 303: ngwwk its? Sfimng (Eva FMW 1L. ‘9 V‘Lé [mflgi.:mL3,,,:§ 2.1:; iqgl/Tfi «l é-FM“ hm” ' ' ' _ , L" 1' éfis‘mé’a M b) See; qaa‘wa Farm“; mm a». W WWW .. is, n ....~: {1’ "r. n?‘ 3 A 7 7 ._ v I,“ I = ,4 _ . I _ ' V. . a. L! , I (7% 2 ——> Guam: {M1} '2 <9 I. z; r .I I .,No4-T(.L, Awe, MJt/Ocu’éQkLEw. AQMAi. ,o/euv jug 57F VG'A’MEM- ’Ltwafingg .m-a *mwwg 3M Riga; = 052/445‘2- 55*? nm_ _ 0.35 Problem 9.1(b) 0.25 0.05 0.3 ' I 0.4 l I I L 05» 0] wavelength (Hm),- C:\Farhan\e—library\courses\ece 303 2005\prob9wlfib.m Page 1 October 27, 2005 12:15:07 AM % PROBLEM 9.1(b) %constants c = 3e8; muo = 4*pi*1e~7; epsilono = 8.85e—12; I eatao = sqrt(muo/epsilono); %wavelength 1D array lambda = [O.3:.OOl:O.9]*1e—6; %structure n2 = sqrt(3.5); eata2 = eatao/nZ; 12 = O.52e—6/(4*n2); %layer thickness k2 = 2*pi*n2./lambda; %layer wavevector‘lD array n3 = 3.5; eata3 = eatao/n3 7%actual computation using matlabs array processing features Gamma = (eata3/eata2 — l)/(eata3/eata2 + l); . eata = eata2*( l + Gamma*exp(—2*j*k2*12) )./( l — Gamma*exp(—2*j*k2*12) ); Gamma = (eata/eatao — l)./(eata/eatao + l); %plot results ‘ plot(lambda/le—6,abs(Gamma).A2); %labe;s xlabe;{‘wavelength (\mum)'); ylabe;('l\Gammal“{2}'); ‘ E title ‘Problem 9.1(b)‘); g 2 IFI ‘ 0.1 0.25 ' 0.2 0.15 0.05 - Problem 9.1(d) 6 I I 0.4 ._5" .027 wavelength (gm) 1 0.18 0.9 C:\Farhan\e—library\courses\ece 303 2005\prob9_lwd.m' Page 1 October 26, 2005 9:03:24 PM i___m_in_____________mwm______________r__im__________________________________ % PROBLEM 9.1(d) %constants c = 3e8; muo = 4*pi*le—7; epsilono = 8.85e—12; eatao = sqrt(muo/epsilono); %wavelength range lambda = [O.3:.OOl:fi9]*le—6; %lD array omega = 2*pi*c./lambda; %1D array %structure n2 = 1.5; eata2 = eatao/n2; 12 = 0.52e—6/(4*n2); %layer thickness k2 = 2*pi*n2./lambda; %layer wavevector 1D array %NO THIRD LAYER D4 = 2.8; eata4 = eatao/n4; l4 = O.52e—6/(4*n4); %layer thickness k4 = 2*pi*n4./lambda; %layer wavevector 1D array n5 = 3.5; eataS = eatao/nS; %actual computation using matlabs array processing features start from the right side and keep repeating the same code for\all layers % till you reach the left side .- ; o\° Gamma r (eata5/eata4 — l)/(eata5/eata4 + l); L eata = eata4*( l " Gamma*exp(—2*j*k4*l4) )./( l — Gamma*exp(-2*j*k4*l4) ); ' E Gamma = (eata/eata2 — l);/(eata/eata2 + l); « eata = eata2*( 1 ~ Gamma.*exp(—2*j*k2*12) )-/( l — Gamma.*exp(—2*j*k2*l2) )f Gamma = (eata/eatao " l)./(eata/eatao + l); ' %plot results plot(lambda/1e—6,abs(Gamma).AZ); %;abels , V x Xlabel('wavelength (\mum)‘); ‘ : yiabel('i\Gamma]A{2}‘); > title(‘Problem 9.1(d)'); Problem 9.2(b) .N‘=‘8. wavelength» (um) IT‘IZ- Problem 9.2(b) | l N32 N216 I 0.5 0.16,. f " 0.8 * *.9 wa’Velfengthmm) - ‘ C:\Farhan\e—library\courses\ece 303 2005\prob9m2_b.m I r Page 1 October 26, 2005 » 9:24:24 PM W % PROBLEM 9.2(b) %constants c = 3e8; ' muo = 4*pi*le—7; epsilono = 8.85e—12; eatao = sqrt(muo/epsilono); %wavelength range ~ lambda = [O.3:.OOl:.9]*le—6; %1D array omega = 2*pifc./lambda; %lD array %structure N=8; %number of pairs n2 : 2.4; eataZ = eatao/nZ; 127: O.52e—6/(4*n2); %layer thickness k2 = 2*pi*n2./lambda; %layer wavevector 1D array n3 = 1.5; eata3 2 eatao/n3; l3 2 O.52e-6/(4*n3); %layer thickness k3 = 2*pi*n3./lambda; %layer wavevector lD array D4 = 3.5; eata4 = eatao/n4; %actual computation using matlabs array processing features > % start from the right side and keep repeating the same code for all layers till you reach the left side o\0 ea:a = eata4 %starting mpedance for n=l:l:N Gamma = (eata/eata3 — l)./(eata/eata3 « 1); eata = ea:a3*( l +\Gamma.*exp(—2*j*k3*13) )./( 1 — Gamma.*exp(—2*j*k3*l3) ); Gamma = (eata/eataZ * l)./(eata/eata2 n l); - » g eata = ea:a2*( l + Gamma.*exp(—2*j*k2*12) )./( l — Gamma.*exp(-2*j*k2*12) ); F . end Gamma = (eata/eatao a l)./(eata/eatao + 1)} %plot results plot(lambda/le—6,abs(Gamma).A2); %labels x xlabel('Wavelength (\mum)'); ylabel('i\GammalA{2}'); title('Problem 9.2(b)'); 6 Problem 9.3(3) 0.9L 0.8% I I I Angle 91 (degrees) '80 100 C:\Farhan\e—library\courses\ece 303 2005\prob9fl3_a.m Page 1 October 27, 2005 12:08:35 AM % PROBLEM 9.3(a) %constants c = 3e8; muo = 4*pi*1e—7; epsilono = 8.85e—12; eatao = sqrt(muo/epsi10no); %wave1ength lambda = O.52*1e—6; %structure i ln2 = sqrt(3.5); ' ' eata2 = eatao/n2; 12 = 0.52e-6/(4*n2); %1ayer thickness n3 = 3.5; eata3 eatao/n3 %ang1e of incidence'thetaml range theta_1 : [0:.1:89]*pi/180; %1D array %other angles theta~2 = asin((1.0/n2)*sin(theta_1)); %1D array 5 thetagB asin((n2/n3)*sin(theta‘2)); %1D array . I E k22 = 2*pi*n2*cos(thetaa2)/1ambda; %1ayer z-wavevector 1D array % effective impedances , V - : ea:a2_eff = eata2./cos(theta_2); %1D array 1 ea:a1‘eff = eatao./cos(theta_1); %1D array eata3_eff eata3./cos(theta_3); %1D array ll %actua1 computation using matlabs array processing features Gamma = (eata3_eff./eata2_eff — 1)./(eata3_eff;/eata2aeff + 1); eata_eff = ea:a2_eff.*( 1 + Gamma.*exp(e2*j*k22*12) )./( 1 — Gamma.*exp(—2*j*k22*12) ); VGamma = (eatameff./eata1_eff — 1)./(eata_eff./eata1_eff + 1); %p10t results > plot(thetam1*180/pi,abs(Gamma).A2); %;abels x1abe1('Ang1e \theta_{1} (degrees)‘); yLabe1(‘l\Gammal“{2}'); tit1e(’PrOb1em 9.3(a)'); ...
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This note was uploaded on 09/15/2007 for the course ECE 3030 taught by Professor Rana during the Fall '06 term at Cornell University (Engineering School).

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homework9sol - ECE 303: ngwwk its? Sfimng (Eva FMW 1L....

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