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# hw10solution - 5 l ‘ o n ECE604 Homework 10 D u Out...

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Unformatted text preview: 5 l + ‘ o n ECE604 Homework 10 D u Out: Tuesday, March 29, 2005 Due: Tuesday, April 5, 2005 Problems given by number are from the text by Ramo, Whinnery and Van Duzer (3rd edition, 1994). 1. Write a computer program (e. g., in Matlab, or equivalent) to calculate the frequency- dependent reﬂectivity R in terms of power (R = [1:42) for a multi-layer mirror composed of y. the following layer sequence: (air)(I-lL HL HL) (H) (substrate) \__.__..___V________J I.__v_J 9 pairs one extra "H" layer This means that the wave is incident from air, through a series of 19 dielectric layers, then onto the substrate. Take the substrate to be silica glass (index of refraction n a: 1.45 ). The high-index (H) and low index (L) layers are made from ZnS (nH = 2.3) and Mng (nL =1.38) , respectively. The thicknesses of H and L layers are each chosen to be quarter—wave at a free—space wavelength of lum (frequency = 3x1014 Hz). That is, the thicknesses dH and dL are chosen such that anH = 2"- and nLdL = 4 , for A, = lum. 4 Using your program, plot the power reﬂectivity [9'2 as a function of frequency over the range 2-1014 Hz to 4-1014 Hz. Repeat assuming 11H = 1.8 and nL = 1.38 and plot your results. What is the impact of having a smaller index contrast between the two materials? 2. 6.6d 3. 6.70 4. 6.116 Hint: Just look at the z-components of the time-average Poynting vector on both sides of the boundary 5. 6.11f 6. 6.12e 7. 6.14e %S stands for the index of substrate %L stands for the index of 10W“n layers %H stands for the index of high—n layers S=l.45; .3; 1.8- 8 I I m H I'hbd? II II "E Awl-‘NN .3 :0.0l:4;%freq=2el4~4el4Hz —l)AO.5;%start with substrate + first H layer Z=H*(S*cos(pi*f/6)+i*H*sin(pi*f/6))./(H*cos(pi*f/6)+i*S*sin(pi*f/6)); %for the term (pi*f/6), is intepreting k*l as a function of freq., through %some easy steps replacing Lambda with f in the k*l term. ' for n=l:l:9 %layers 1 through 9 . H Z=L*(Z.*cos(pi*f/6)+i*L*sin(pi*f/6))./(L*cos(pi*f/6)+i*Z.*sin(pi*f/6)); Z=H*(Z.*cos(pi*f/6)+i*H*sin(pi*f/6))./(H*cos(pi*f/6)+i*Z.*sin(pi*f/6)); end R=abs((Z—l)./(Z+l)); R=R.*R;%reflectivity The results are plotted below: .For the ﬁrst ﬁgure, 1’1H=2.3, 111:1.38 ’ For the second ﬁgure, IlH=1.8, 111:1.38 The impact of having smaller index contrast decreases the bandwidth of high reﬂectivity band as well as the maximum reﬂectivity. 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 2.2 2.2 2.4 2.4 2.6 2.6 2.8 2.8 Xe14 Hz Xe14 Hz 3.2 3.2 3.4 3.4 3.6 3.6 3.8 3.8 =2.3,m=1.38 (mcpmM 5““) Max. power reﬂectivity 99.99% 0.9995 0.999 0.9985 0.998 0.9975 0. 997 0.9965 0.996 0. 9955 0.995 ' 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2 X914 HZ 11348. nL=1.38 (J2 ~99 «midi s mi c. Max. power reﬂectivity 98.51% 1 0.98 0.96 0.94 0.92 0.9 0.88 0.86 0.84 0. 82 2.5 2.6 2.7 2.8 2.9 3 3.1 3.2 Xe14 Hz 3.3 3.3 3.4 3.4 3.5 (3.66 Since. ~Fi|m '66 Brain (assumed. than competed with skiar‘a depth), ﬁelds penetrate, and regaon 1:: right as analogous to a. _ ’cpnﬁnu‘mg transmassam line. as shown. with oonduatance °L° G=—L '1.- ‘m shunt distance d lawoﬂ {‘rom short Admat‘t'ancc OF shorted lane: IS Ys=o3z° 539‘ ”"dYoeerd 'Y ‘ ‘(_ d_-_>I Thus, admittance “be \eH: a? eonduetance is L ‘ . Y._ = G -3Yocct Pd: Yo - dYo act ed [Ram‘sheetnrej'asmnee] 5 "1° ‘l ~YL _. '. Reﬂection coca: ﬂ = Y: +1.. _ a. 5:: dcgﬁgﬁ E" 61¢ PoNcr loss In conduejns suer, ghen b3 5%. 5 a8 (5) 35 Na. ' 42* IJ'SIZRS lJ-s‘ =lHt‘ : 2-HH~ = 2E4]? 7 --NL=J‘(4—E'YZ; )ES 115.. .2 331 _ 4-25 9%.: = ’Q‘ s . En} ‘ Q WW0"! Onsl'ces w'aUn 5% (,1 (:0) 9-H: Fm" E15. 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