HW13solution - So/u'I-z'on ECE604 Homework 13 Out Tuesday...

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Unformatted text preview: So/u'I-z'on ECE604 Homework 13 Out: Tuesday, April 19, 2005 Due: Tuesday, April 26, 2005 Notes: £F/l 5— 0 Final: Monday, May 2, 3:20 — 5:20 Room 0 Review Session: Friday, April 29, 4:30 — 5:30 in EE226 1. See attached 2. 3.19c 3. 12.3f 4. 12.3g 5. 12.6a 6. 12.6b Problem #1 Consider a symmetrical optical resonantor (Fig. 1a) formed by two parallel mirrors with reflectivities R = r2, where R is the power reflectivity and r is the field reflectivity. For this problem you will model the resonator as a series of dielectrics (Fig. 1b) with impedances 111,112, and m, respectively. Regions 1 and 3 are assumed to be made of the same material. T11'712 . 711 + T12 The ratio 111an is chosen to be consistent with the field reflectivity r, i.e., we take r = ‘ Z (Z :r -_ Afl%”‘({/¢<fiw .1 / @«J’d d \U": OJ/J ‘ a) For part (a) we'Willassume'?the-resonance conditionis satisfied, i.e., k2! = mn, Where k2 is the propagation constant in region 2. As discussed in class, When kzl = mu, there is no reflected wave in region 1. On resonance we may write the E—field in the three regions as a function of 2 as follows: E1 = E1+C—jklz , Z < 0 E2 = E2+e”jk2Z +E2_e+jk27‘ , 0 < z < e E3 = E3+e—jk1(z_£) , K < Z Similar expressions may be written for H. ii) Use the boundary conditions at z = 0 to obtain expressions for E2+ and E2- in terms of EH. Then use the boundary conditions at z = K to obtain an expression for E3+. Use these results to obtain expressions for the output time-averaged intensity P3+, and the intensities of the forward travelling and backward travelling waves in region 2 (P2+ and P2- , respectively), all in terms of the input time-averaged intensity PH and the power reflectivity R. Also express P2- in terms of P2+. Then evaluate your expressions numerically for R = 99%. iii) Comment on the physical meaning of all the expressions from (ii). (b) Now we no longer assume the resonant condition, i.e., we do not assume kzfl = mu. Instead we wish to calculate the fraction of the power transmitted as a function of frequency. Plot the fraction of the power transmitted assuming R = 99%, fl = 150 um, and refractive index equal to one for optical frequencies in the range 2.975 x 1014 Hz to 3.025 x 1014 Hz. (Hint: there is no need to repeat the calculations of part (a). Rather, this problem can be solved easily by using the impedance transformation formula and writing a simple numerical code). Compare the width of your resonant peaks with the simple expression discussed in class. ‘ Wax-main:fiWu“,ngmwmmnWWMWW-mwfi .. .A.A.‘..w.u...~.,.-,m i~ . x z z r 1‘ vs? 3 a 2 1‘. ii"; - .‘fi . f 5, . 4% I ’1; .H‘ {fig ‘35 :i‘ I: _x > V. 543. “‘5 ‘ E l i .x i‘ 1a '1 I . ;. sh, f is i‘ S.‘ sfi . 1' i‘ , by; i=3 4‘3 .35 .‘2- $13; s'fi‘ “Li? ' i.“‘i in! ‘rgi 3 i -: .xmwasmqumvwvmw‘mm_ K g. E E 3. ;, 3 3 § (lb-067° "HM?“(‘Jsl/O“ 6“- \(‘L‘esammx ‘Gn’WCS Suimm a \ _ S‘SV‘QCI'WQ! A); . lo l 3 NW PM 4» NWN wiwr-t 723m, Lmzzzmm-mamzar an». Wu:- cream: Neva/L‘— (NMENNN mm a map-r. an...“ MW -mmvmmwwmmw... .A'éT-r'GEGE-Z511';:.‘}1:22;:i.7'.1:J: ‘m' ‘ ‘zfiw‘ "“m aw «cg—w, mm. .W -Aé§v6--wr+‘v‘\4vhiwhé—w-«-nim~41~<->e3<~‘r-'w~>,—~“-‘~7v'-IMW-~v-‘—-vri-fu‘ ”5mm; YM—zLTIS—ww‘"' "flu!" NH .;;;_1:‘:~»..“' ":L',.:.-;.n:«:'.:.‘:£;:;;374um“‘“;7:r:rx.m"""“‘:asw‘ _..._ ' 23:34.22:whee-fink-weay—nme—gwdw—vnehu mvmlnflkfifinfiéfiyfi‘wqhmmrwwfit _ Fug-1v u-Aum'I-lrmm mN‘m ”Mutummmm“ www.mmu V—tmwwmmmflv'WA-Hnfl “va H-“ :rm—Www ~' ..::::.: =- < - .E‘1159tersza—3Ei-IzfimizhiéqivsézeMMi :éfléééukwifmflfieufinntmméréflmn—mfi : mz—LL. 1:- - . I“ \-.«mepu-wm.fi-wwmyw.NWW...”—‘—..«Nu—g.”u.,..gwm..u,._...‘ommw.‘mn ié E; a; N 3% a: £4‘ 35 ii a IE E; g l 4 m) :(175+>:L‘i>;+>; my Wflvggm'w“: 'k‘rw m.+J—<.¢Q._ f m f; QIQWESA7QM4 K _. N “’QLM we”? W‘ Péj’kwa/Qég 4 £7” “HMS $91 W-N#I‘C-J .s»+w¢4wl.a . ' _,-<’Pn>zc.~:Q 1,1??? “WA @WQ- 3. 1’ C‘MVM-‘k 4P5? ~. 0". madam) 41$ * W‘NW‘Q“ “‘Q-CQS' Cf-‘W‘d’é 0% ‘ i “ WW1 em gawk M3 mg” 6:ka .,(a»:3%nl¢. ~_>' Maw. n: £9.14; _ «yaw-mmwecm p = we) 1 q : “vii, weaken»)? 6%- £4\(~w5+‘\ “HR- ?wwrM‘w 3 . Fm Wm WM“; wk. quhékg ‘17 ’27? 4° ZP-lKIDO‘ 0M1 /St «rt/341 Clov— ‘QO l/T ! V CNo‘k‘. W \‘5 «0,43%chch M “H“; ‘5 5‘ ”‘1 Chk “(‘0 V?! 0/75“ Aw ”W1 04% vfiM‘ILFE’fi 7404'" ‘7 20" 3, WW”; macaw} fair/2%) b) Plot of problem# 1, part b. Width between each peak is 1e12 Hz (1 T Hz). Power Transmission eoeff. for optical resonator R=99% 0.8 0.5 P3+J’P‘I + [1.4 [12 29925 2.99 2.995 2.99 2.995 3.995 9.91 9.915 3.92 3.925 Frequency: x1914 Hz Zoom in on one of the resonance peak. Power Transmission coeff. for optical resonator R=99°/o [——r'**~r———r———7—‘-r l I l ~l—“ 1 r 1 0.8 — — 0.6 — J FWHM: 3.2GHz P3+/P1+ 0.4 — — 0.2 — i Width at exp(—2): 8.1GHz 0 I l . ' l " 2.9995 2.9996 2.9997 2.9998 2.9999 3 3.0001 3.0002 3.0003 3.0004 3.0005 ' Frequency: x 1e14 Hz B-I‘tc, ' E‘om .egqu-JIOVIS 34‘? (4301661 349(5) - ‘71 ~ gaunt) = % «Ii-7U— v=1f =.,;.? --,o.e v (9%). “P6 23% 1-? VJXh = 0 than: ' ' . __v3§ = ‘£" .IK; #e§§, = ‘F?+/*e v(%) Although the gauge seems simpler, 31: dogs not lead to separated eguflbiebns 'PorA Aomd. §, 0:5 does fine. Lorentz gauge ' w= 21v]; Prr" sinede = «16%; Infin3ede= ~31.(;—:;3) _I____:2.oo it For w= wow I ’ [101:3 Mr] ”rug: =100 6A dipole. 'Provn 59.. mace). w= me = 4:21:1 13’: (-01 so I.“ 445%: = 9-5 3" 1°“ ‘33“ ? ' 1.9» - a.» - V l2~53 For Far-zone, HerteIan diPole, Eo=5¢—-°°4“r e 5 W9 . - he . For- Fug-eone, Loaf (magnetic dipoIe), 5‘ ‘ng—I‘Le' 3 same fines: mag. be made awe: and 150° cut 0? :hase' C ooHdIhon 30¢- umlar ”\WM 3 I = . H1 £103 = 1°15 , '50 4:994 T. (waX'm—IL IZ-M PM fltk/Lxgm 92%“ (xx: 69“» 3711'" e—str eusfimsewvj §D~9 ‘LTF P ’— ‘2, K T. ”In“ Co} (Ff/056.) +’ ‘ , ‘3'“ ”Ts—,7? The plot of the magnitude of E-field with theta for a full wave dipole antenna 180 _- _._ 4—“... as (14°39 ca) —wka .2) $37 8‘ — ——-————b——-—-—-'——~——-—'—-——~4—~——‘————--——‘.—-———3~——~4—-:—:;M—.~fl ...... 19‘ . ' P : V P _) . . , . .. . €17.40. "(”73 _fitem r” K ~ < 3 _ New. 04%. Maw .4‘2-55"7¢€5.").7'§- _ .. é‘rf‘fi- [email protected]~£o DCM 1* [ff]. "*5 .6. x94"; >Cm¢fi 773. -..... “..--.... -a..-fl.~._m.—n-_“-~_- ' “rm m. mu”; .1.- . .../2.7.“.0959..?.,..I1...7./"_,_.(.'?:E)..7'.‘.:..,..99-: 70:323-4? 7"__A:fs 3N.“ , (7f! nub _ 3 ‘ TM ,3 4M, 45¢ Wu.»— Haw/.7 W4” z flu//s_ 32, [flu [050 l . .' .. W 9" “7 Peak/s ocamé/v/ rexMeAré.)....4/T..._¢9§,QI-T7T§9$?).~T TI... 1 ......—.....—...... .. . 'E .1._. I g . i y .. 5 “"595 it} MH‘ 49‘9“” . ._ . ...... .V .. _ .... TA... .. . ... ‘__. .... ...._ ......_..... . .. .. - ...; .. . . ”u... _. ~. .... ”-.....“ .. 6:..."- +~rt‘—~ _ Z. ...... . ... .. .... ...... .-.”.-- . ... . .953 “*3“. “Wu“ 9M..-1\.. 3. .. K~E§1§1€f-3~ 1 ..... . .... :-.‘N 2.85.3 999.422. thwmv) "" _ ._ ..--.. -. _ .. 1| M." '22—»qu Mug.-- -.- - (f [a 134-5 Ii Etta/«7‘1 ”If waOéhmH weasUcQ) .5“??? . .. .. aNL fun—M,” :LL “0%,,“ 1 [61415 f}. MJLJ- #4an 6"”) . _. ...
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