exam2_W10_key

# exam2_W10_key - N S...

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Unformatted text preview: N S |J|l|ll|||||Illllllllllﬂlllllllllll||||l|l|||ll| *30193291* ECEN 360 Exam #2 0"? Mar. 25-30, Winter 2010 Prof. Stephen M. Schultz — 422—1693 Name: 145 Y Instructions — Please Read Graphing Calculator This exam consists of ll problems 0 Questions 1-10 are short answer and are worth 7 points each. Be sure to include your work on the exam in case partial credit is awarded. 0 Questions 11 is a longer question worth 30 points. 10 points will be given for a clear description of your solution. 20 points will be given for getting the correct answer. 2. BOA- min“, What is the propagation direction of a plane wave if the electric ﬁeld is given by : + 25») E0 8'17“? and the magnetic ﬁeld is given by H=(2i'—?+2)Eae’ik”’ _) ':‘ \é _/\ ;’,I“I 2‘. S L‘HJ _’ 7C JV?“ ,3, a, >2 A 4 ‘- A A 4' V 2 3: 91%) Ml 0 ~rl +2 r’ l me) i q; A 8. “pl ’6 "I r I Oi "Y ‘3‘ “HM/PH JD [3,“: U," I,( ‘ 711 yr! ;iII.‘ ‘ A A A K ' 0,3é5x-0J83y -Ofrgi' ' -jk!: A plane wave given by f’ x nge is normally incident on a dielectric boundary which lies in the x-y plane. If the reﬂected wave is E' = JEEgeﬂ“: and - - w A — ta.- the transmitted wave is E ' = xE‘e J 7 G , then which of the following represent the correct boundary conditions used to determine the unknowns E; and 5;? 1 1 a. E;+E;=E; and —(E;+Eg)=——E; 7.71 772 @ E;+E;=E; and i(E;VE:)=-l—E; 771 772 c. (E; + a; )9‘” : age-1*:- and —(E;, + a; = —E;e r71 772 __ __ 1 . ._ 1 d. (E; +E;)e<”-=-~ = agent? and —(E; _ Egg-n- z Efﬁe”; 771 77: .' , ( ~ A rOllrglilt5r)§ L‘f fr? I-p 'L‘O‘f : r r \M ?‘ ll“ :7 .52 r 5—” : Eat 3. A plane wave with a frequency of 1 (31-12 is propagating in a nonmagnetic medium. The resulting electric ﬁeld phasor for the wave is given by u A _ JO, aim—+5} A ﬁ0_ "farm—5] _ ‘ E = x238 ’ ' e 4 +y256" ' e 4 . What 15 the equatlon for the magnetic ﬁeld phasor of this wave? '2 : 90 3 h ‘ ‘m \ ‘ It I I ...s m‘ .(‘g 4 A j‘IVlf A ’ M? at“ r 2' .r a. 30. x + as e r A J A ’ “to ‘ i2 '— C’ H - 4? K N: 1M, 3m? 1.90? J n .1 ,l '302 A ‘3 '3”? A ) Q Ma! ’h i {: ’émgHO 0L1 2 700 47 ZLJOD : — cuUI {a a: —i.Li¥l'0'U-0‘H.g‘/ Tm" H 25 “Wig? (Inﬂux! 9 ‘70- Ubié‘lf E \"leo"' “.39er ’ ‘ my \ , , an F», __ If, ' -30 “3f”? 4 b Q <(H’e y ) L t, 2) -. H :J GAE/30.0%) a a Y ' (01370.0? 4. A plane wave with a frequency of 1 GHz is propagating in a nonmagnetic medium. The resulting electric ﬁeld phasor for the wave is given by _ A 101 —J‘l4o:+£l A a _ ﬁlm—El . I E = x25 6" ‘ e \ ‘U + y 25 {’0’ e " 1”. Which ofthe followmg is the polarization state of the wave? a. Linear I Favjpr O“), (a ramp“ 1km», b. Right hand c1rcular @Leﬁ hand circular .5 ,gO—t mm '3' M, A W2:l d. Right hand elliptical I; 3 25 e e. G _ x +9 9 j e. Left hand elliptical E x” (D '- -. " d .-., . E‘ Ed irf‘k/Lwﬂr {US-“‘1' "”I‘:‘l ii _I " ‘l 5’” L: / C) l 0 AV hf" Wt H W3 2 i O ,— 3. A plane wave with a frequency of 1 GHZ is propagating in a nonmagnetic medium. The resulting electric ﬁeld phasor for the wave is given by f r:\ i” ,. _ —f! 40:T— 7 -ji 40:——I E 2 9258'“: e ‘ 4’ + )725 6—30: 6 ‘ U. What is the conductivity, 6, ofthe material that the wave is traveling in? Gr; o’x-fjﬁ ; awg'ia . . .;= _ . ' r .e‘ l ‘ - "" mm) =, 700% 9W) [ “‘lef‘r '6 WW) 6 mt,“ f O“ : (2 V00! Q «[00 Cum .7 Ar» : 0'3 0‘1 ‘75?" 6. What is the correct expression for the incident electric ﬁeld phasor for the conﬁguration shown in Figure 1 if the frequency of the plane wave is 1GH2, region 1 is air (€51), region 2 is Teﬂon (852.1), the incident angle is 8=30 degrees, and the electricﬁeld amplitude is 2 Wm, and the electric ﬁeld vector lies in the xz plane (parallel to the plane of incidence)? X region} regionIZ ____ _.. "—4.2: h E Figurel E"? “on 'i / ’S'lnegqﬂ‘: ( {if I i‘O-‘r \ 71.7).) i '- 1 a ‘35”(5'r156c N ’r “N r 1.7m ﬂ 2) a F , A _3(,o,§x +9.31) , , (L73; r2) 6 Wm _\' (3) (h) 1,0 30,8 (3'.=60()nm) “hum 550.6 -: g 55‘” 3 “e: 0 . . . . . <———-—? 0 IO 20 30 40 SD 60 70 80 9t) rotation stage angle, 6 (degrees) Figure 2 7. Figure 2a shows an experiment used to measure the index of refraction of a piece of glass. In this experiment the transmitted power is measure as a function of rotation angle. If Figure. 2b shows the normalized transmitted power, which of the following is closest to the index of refraction of the glass? The incident region is air. _ — c: _ 51- I'1sanrtple‘1-3 A l 7 § 1 :7 ’00: b- nsample:2-O i r t 0- nsamplezz5 l“; ’“if'w‘k ’i‘\{‘ l":\ f I )7 d. n :"_O a ‘ sample 3 : ﬁlm" I {Wiltth ﬁﬂ/ 0) r‘ i in 1 me m at r x [yon-ml ,nrfiylgﬂrf 7., a7 1'. t MEJA ('1‘, 070 . ?\ 4 {116059, f {it (6894 ,0 F. V . t . l \J’ I l" i d 5f Fiﬁ r r - r} l'm'? “"11”, - "055% , r11 rose)... “0.6% — tat-1,41, f4) 1 7 Sam ’- ‘m-u?‘ (L55 l“5trl7<z7¢ I 1 i H n" s , “NHL 5 any; _ﬂ furs-‘0'5l'lgr‘t F II “ﬂat J ’ '1 ? d w. lb or 5mm (7 r" V)" L595” {Lalo/3M1 7 N’Ccszm t (11 4 3"! w '5? " v/ r)? r l,::?_ ._ (\kl , (if: Ji 1 |"‘“..r F ’ ,_ F 3’kl7- rf‘arzr I“ (“him , J— , a h ' F55?“ T‘c-s‘l‘u ' . L andmlgl 4 reﬂection coefﬁcient C) , :5 [\J ‘ A‘Vﬂ‘ﬁﬁ“ ‘r‘ ._‘ E l _ ...'."_l ____ _.L __..._. _. _1 | ‘ l . l i O 5 10 i5 20 25 30 35 40 incident angle Figure 3. Reﬂection coefﬁcients for a wave incident from glass (with a refractive index of n=1.56) into air. 8. An electromagnetic wave with a frequency of lGHz is incident from glass to air at an angle of 8:3 5“. Ifthe electric ﬁeld of the incident wave is given by E = cos 35“ + j» — 25in35”]e‘-‘ What is the polarization state of the reflected wave? Use Figure 3 for the reflection coefﬁcients. i , G r f J? It é? Linear " " ' " b. Circular CH : ,m g 5 a _ if ,3 g :o‘. c. Elliptical with major axis in the y-direction d. Elliptical with major axis somewhere in the x2 plane ‘ 1 ,1) f e No reﬂected wave ' __.~ um r f Not a valid plane wave tr. m T l 9. A left hand circular polarized wave is incident from glass to air at an angle of 8:35”. If the power of the incident wave is 1W, which of the following is closest to the power of the transmitted wave? a. 0.12 b. 0.56 5—“ 0'5 C. 0.77 E/Iffn‘ls e. 0.99 f1 : 6.5 r7” - -0, _ .7 J’ {ﬁghgl TU [(0,1)2 h H @,7 - 4' 5 I“:0177 10. A center-fed Hertzian dipole is excited by a current ID=20A. If the dipole is 2.50 in length, determine the maximum radiated power density at a distance of 1km. «— A h - In {‘2 ‘ \ -L. «- Liw- If 2 {5-77) _ _. TI: 5,.) .. fa p 0 Hi gkk {Hi I, MT 5L 50) 71/15) Q [2 _.. é-Iérz A, 20 (“’5‘ J 31:52 A ‘ r )7 _ f If) ‘3‘ a: plug ’— N Figure 4 11. A linearly polarized laser beam with a wavelength of ?-.=600nm and time average power density of l rnW/in2 is incident from a glass substrate (with a refractive index of [F] .56) into air at an angle of 9:33 degrees measured from the surface normal (see Figure 4). If the laser beam is oriented such that electric ﬁeld vector always lies within the plane of incidence, what is the time varying electric ﬁeld of the transmitted beam? It should be in the form E=£E1c05(a)r—kxx—kyy—kz z—gb,)+j2l‘32 cos(a)t-—er—kyy—kz z—¢2)+ 2E3 cos(a)f—kxx-kyy—kz z—ngg), where you need to calculate the unknowns E1, E2, E3, (9, kx, ky, kg, 431 4);, and (133. The interface between the air and water is located at the 2:0 plane. (See Figure 4 for the deﬁnition of the coordinate system.) Use Figure 3 for the calculation of the reﬂection coefﬁcients. (H NggA HUI iranSmli'i’Cd rtﬂgie Swat : use 3m "2%" {9L 1 5751!?” CaiCL/iCI-i'e “\‘Inrﬁmrs‘nan ropfél-(lnrf t . r I; I. (exp! ’ 74:) A; :r‘A‘o‘.ﬁ;lr-Ir._/ “u (ydéu‘e WGW 'ng /,r nut-,4 "in : J f’l; we 0; 7”_ i" m -. 2 “wt *4), r0504- I l 3 FM C604. 0. reset t”: (0559.. r .r‘. 4' n_. .zl-vr '92=":a\ u; i l .r1i7oi 4 '1‘ {G rad Sgt} Tc L. .33 p. |,J : ’. 5'75 ,- r. /\ 1-“— Li! (-— 3“th +rg59t w)_zﬂg 7m 35:?- ._ L. 0 , JZ °Io No x 455M); 24 (Lo (“gum we Man FIr‘lA «Pom POM( dgﬂﬂqf L31 :ﬁléum ,g £11 41 “3'3 ’ Quill \er) 3 0-46” Wm E40 2': CoﬁOt )1 +3‘Ilhet 2A 0* .- ’r..\£:,l;cos9¢ i‘ +sin9¢ i) : (\.51<.)(o.bqe) ( 0.697 u? + 0.?5 {W (‘65 (OMEN! 50‘ Wm“ “PM Pow : , ' c. e, b t o A 505”?er +5.51% 2 {2 :58 X 40.43 a ) V/m ’SE '1‘ 3'97‘0‘9" 2557-:0‘2) + 0 ,‘ ' Q .L 00132.(03( a}! “ﬁe +5’AWOL‘A #357110 .2) a? r ' .IHF/D' ' — gl7055 L éfr¢z {d3 C "a Kira.qu ,1, 5330.43 I“; :0 a Appendix Time Harmonic iVlaxwell’s Eguations VxE=—jmy§ VxH=j+ijE v-B=0 V-D2pr D=5E B=ﬂH 802835612 F/m “(5411* 1e-7 H/rn Boundary Conditions Tangential E: E“ 2 152: \$3431“ 52): 0 Normal D: Din - 193,l = A }%'(D] — D2 ) 2 10.: TangentialH: Hl,—H3_,=JS ’AX(—I_HE)='J75 Normal B: BI” - 33,, i 0 ’A'lél _ E? l: 0 General propagation constant: ,1’2 = -a)2#66 = —602#[595r — fig] 0) 2’ = 0' + jﬁ’ General plane wave equations: E 2 EC 8—}: or E = Ea ‘3 1;: E = —]7 JEX g H = —1' )7; X E 7? Lossless propagation constant: k = ark/ya . 27f Lossless propagation constant: k t a) ya 2 —‘ /. “ Wavelength: ,3 : _ f . 1 C Phase Velocrty: v = _ = W {as Jug? - . t? hitrins1c Impedance: 77 : I'— \ g Skin depth: 6 = i a Poynting Vector: § = E x if] . . 1 — — i Tune average power densny: Sm : ERe{E x H } Snell’s Law: 21, sin(19I ) = nI \$11091) . : : # Index of refraction: H V‘s" 5’ J m 5 Reﬂection Coefﬁcients: F = = 2M * E. :32 cosé?I + 771 cosH, E _ 77: cos 6! — m cos 6, E 772 cos E91 + 771 cos 6, Transmission coefﬁcients a = g = ‘ E, 772 cos 9, + 77] cos 6’, T _ a _ 2272 cosﬂ ' E: 772 cos61 +7}?i c036?r I" = 1 + I: T. = )COSgi cos (9, . . . P 2 Reﬂectivity (power reﬂection) R; = ?’ =]1‘_[ antiiEL—K’i . . . — it u e I Antenna Radiation Equation: A (R) = —— [J _ _ dv 47: iii — R'l Far-Field Hei‘tzian Dipole equations: sin (6) _ A Jrglrkng e—_{!:R\ E=81 47 [R ...
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## This note was uploaded on 09/02/2010 for the course ELECTRICAL 360 taught by Professor Schultz during the Spring '10 term at BYU.

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exam2_W10_key - N S...

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