Prelim 3 Solutions - PRELIM 3 ECE 303 18 November 2004 NAME so/u:9 m SECTION 1(40 points Suppose a satellite is orbiting Jupiter It takes light

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Unformatted text preview: PRELIM 3 ECE 303 18 November 2004 NAME so/u :9 m SECTION 1. (40 points) Suppose a satellite is orbiting Jupiter. It takes light approximately 35 minutes to travel from Jupiter to the Earth during times near closest approach. Assume that the mean interplanetary electron density between the Earth and Jupiter is roughly 4 electrons per cc. (4 x 106 m3). The satellite transmits a series of Gaussian shaped pulses with envelopes ~ exp[(t — to)/21]2, and with carrier frequencies of 100 MHZ and 400 MHZ. Suppose 1: = 1 microsecond (the fill width of the pulse is about 2 us) and the pulses are transmitted simultaneously at the two frequencies. (a) What will be the difference in anival times when the pulses at the two frequencies reach the Earth? (b) How much will the pulses at each frequency be broadened when they arrive? See the cover page for information on plasmas. Also, the formula given in class for the broadening 82 am of a Gaussian shaped pulse was r —> z": 1'[1+(—2 a 2 )2 ]“2 , where z is the distance traveled. r a) ‘ Z. (4) :5»: : J‘J’f‘m’é : ’3 ’4’“? ,wluéh 1’5 44 log! 50 fp /€"“ <“". r ‘1 a. —1 Ill” ~44 _f__oé€ _ 1/;— M/gw L:3§(¢o)cl A15:C([._ {lo/£1) _C(, 2 p41) i" IV;- 3 L G : 5 ’+ mz/zo‘) 50 in: fif, + 2§z(?l){4){lo )] C[ 3“ P ’ _ L 6 J. - -—L- M». At: iwwz- two ~ 2062)“ )[0053‘ It, (1032] 6 : 15*{60)(/(92)/0 P4] = 3,.c7w0'é = 31.?fl5 lolé I ,6 “fin—if“. - é 9(I'QTPL/ ‘JUL .1421 WNW/26- " 55-2. :4 ‘° ref/'2 W) 52$” L I 4 ‘vl 1 31 _ t — COL amafizz{/+Z 9A0) )fi—z—éwfij):‘l°s 3 ‘0 ‘4 cw 1 )l 0H; ’7“ I 3° 3'5: 31?:— = ‘flfrf—r 30% a}; = Meier!” «’4» Z. (I0 ) lo'vz In. (ms/)5 {cumulfo . ? k : —= 3mm s. 7,». apan 217 I0’1 So : //{.s —-) 2”: /.3Z/u$ {bmecpewacj b7 32% af/Wfiflffi) 4-} 400 MHE/ hCW “5’66 ‘9 4.8133 :,ol3§/ 43 ’ -— t ‘ affitwwb’é Mc/ 2’: T/lvooaafi) ~> "97/!‘iwjg/‘f brvnflwww7 PRELIM 3 ECE 303 18 November 2004 \ NAME SECTION 2. (30 points) Suppose a wave traveling in a dielectric with e = 780 is obliquely incident onto a plane interface with another semi—infinite dielectric with e = 350. Suppose the incident wave is right circularly polarized and consider two possible angles of incidence (the angle between the normal to the interface and the direction of propagation of the incident wave), namely 35 and 45 degrees. (a) Give the fraction of the total incident power in the reflected and transmitted waves for each angle of incidence, and (b) Describe in words only (no detailed calculations needed), as completely as you can, the polarization of the reflected and transmitted waves (if they exist), for each angle. 6L; \ - i Aer? Uj‘e'Ful +3 {315‘}— (alt. a : 71v“ $2MF:53'20 76° y® '5 5', 7 #~ 36: 7 (L) I o 94. Ala-f2 ’fi‘f‘f “55° ,5 c/Ofi? 'lv 95 cw} H5 is 7 Q -o ’I ’0 - a, 3} (es-e C059 5 4643 =.€r;‘i 3:9 :f/uq gigsmzsj —6/,Q ‘—”—__" L ) 15 q W c056:.432 8 a "53 z r , yLMé —’7,‘4¢$r “ (q "~8' /('§ - M —. — n flags}?! 4616:, I: + N “278' 3;: 2,3654 2, ’lI/‘fle; ,/q8’_,‘/52' F :W 2,1”— ; '44.} “L h ‘9' H 4- H 2 I 7070 WFMI-AQ’J H'— \ l Tia/(«HZML'V‘V' :7 flp—Hpulej Luau-e will be almost“ Iméarlv palarizeap writ”? E .L film 0*" vw-Z'que-r [le P‘I‘L’M). MON” meosélvi I” P’ltgiSP a9fl0t+ fa)“, ads/:1 ) ifwfll up euaf‘v‘m'w, fe/unzebfl (+14», WM W'ifi'i‘ui‘y I?» “Wu: /f’p+ hwy 39.455 (1—9411. or? Q” W) Tkp +mnsm-jffoJ WW by PILPIJQcJ’ m+a+lvt7 I'M ( SW W Wm be 14ml, "sq-r, 45°“59 H50) 626 => WW wan be (Me/4:75.11 wfilecfr/ f,“ 7‘ J + Wild/11L? / i We" “met 17,45): have“ M m“ H. 3W , gomPr/Q,,gq+,w 'mwvwwer n + H 'H b; le—W' m+€2+!‘;fi é “(Pitta ) mo‘f’f'lfakz) , {ileukby Eva/0‘43 9"“ 541W M (c; lyrethv, SubflOfimU’J-fi; ho +WSM:H‘fl W 3 SW) PRELIM 3 ECE 303 18 November 2004 NAME 50/" 7'1” SECTION 3. (30 points) Suppose light in the visible spectrum (wavelengths from 400 nm to 700 nm) is normally incident from air upon the plane surface of a semi-infinite dielectric with 8 = 880. We wish to maximize the transmission (minimize the reflection coefficient) into this dielectric by using one or more quarter wave matching coatings of other dielectrics. However, we have only three coating materials available to us, with a = 280, 380, and 5.380, none of which will give a perfect “match.” What would you do and Why? Give a reason for your choice and the order and thickness(es) of the coating(s). What fraction of the incident power would be reflected at the center wavelength (550 run) for your arrangement? , I fired, #Saoo film‘r’nesss’s 36,, at, : if”? :7an, cgs‘izzm he) ' “K' d ;.ra7umrqsaé=srseo> ’én abVlo'ws ’9095'zlf’ Change gé 5.3 ' JM Q 7° Ila/E fig [8; .lrrms‘P?” “an” I n.) 2:? ~> / *r P: I‘M3_ : "0293 AU” ’7 " V 7 .3 ‘ £94370 2) ' AW? '4 a , _ a 22»( 3 M Irvzsg'élilf (Very 7w;.¢7?7‘% WSW)!” Ed’flw 1's Gummy? pass, 5:"va . , Z gas 536. 56¢ ’ $3 - 317, @ 60 7 ya g, _, _ ‘5 70 n ‘”2 E5 n/we » L_ l K g z, 3a Z, « Z, / , , 93 fl) Jaw, I l7 1 lo? x/o’g So 7th} artwsemeflf [27; .933 " : ~,635))/z) -7, 1.636 [(30 Which to choose? The first arrangement (1 slab) is a tiny bit better at 550 nm (but both reflect only about 0.1% of the power), but we know from HW and Workshop problems that the 2 slab scheme will work almost as well over a substantially larger wavelength range, i.e., from 400 nm to 700 nm. So the best choice is the 2 slab arrangement shown. Many of you used 3 slabs, giving a 1, 2, 3, 5.3, 8 sequence. This does use small steps, which is good, but the “match” at the center is made considerably worse by adding the 380 layer. Computer calculations in Mathcad and computer plots with Matlab (see later pages) confirm this intuition. The 3 slab choice is OK, the single slab is better, and the 2 slabs (2 and 5.3) is the best. Most of you ignored the question of matching at wavelengths other than 550 nm. Try various arrangements for QB on Prelim 3 1. What about 1, 2, 3, 5.3, 8? Thenn2~isthe-samees-abe¥erlaut-ether—el=ienge- 377 377 377 n 2: 377 n2 2: — n3 := —— n4 2: —— 377 it 550 2 3 5.3 8.5 2 700 where y = [39d as in Wkshp 10 Then generalize some of the above by '79 j 26,, I 3e» ( 57" 63} k 60 2. '. 1. ' Z(x1,x2,y) ;= x1 - 23 ;= Z(n4,n5,y) x1 - cos(y) + 1 - x2 - sm(y) N23,; Z(n3,Z3,y) "a; Z(n2,ZZ, y) 21 — 1 RM ,=<__12 2 21+“ (|r1|) =0.029 (lr1|)2 is 0.012 for 550 nm, 0.029 for 700 nm, and 0.018 for 400 nm Pretty good, even though the "match" at the center frequency is worse than for 2 other arrangements. (but 5e: {its-i"; 3 2. What about just one slab, with a relative dielectric constant of 3 (a 1, 3, 8 arrangement)? Then NZWIV‘: Z(n3,n5,y) (21 — n1) F12: W 21 +n1 (lril)2=0.032 by,“ mus miss / 1/ 1/ Gives .000866 at 550 nm, .032 at 700 nm, .084 at 400 nm 3. Finally, what about 2 slabs, with a 1, 2, 5.3, 8 arrangement, the one I believed would be the best? ’JngAzz Z(n4,n5,y) g3“: Z(n2,ZZ,y) r1 ::(Z1— n1) r. M 21 + HI I (IF1|)2=2.951x 10”3 “M” ‘4' 9”” J 7cm} (WM rm” For this 2 slab setup, (|r1|)2 is 0.0011 at 550 nm, 0.025 at 400 nm and 0.0030 at 700 nm 51>»: p’a'b (1491+ Faye) 46’? WW M2 (We) ECE 303 Prelim 3 #3 v/-’ ./ .‘1; 3—25.3 500 550 600 ...
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This note was uploaded on 09/26/2007 for the course ECE 3030 taught by Professor Rana during the Fall '06 term at Cornell University (Engineering School).

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Prelim 3 Solutions - PRELIM 3 ECE 303 18 November 2004 NAME so/u:9 m SECTION 1(40 points Suppose a satellite is orbiting Jupiter It takes light

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