EE402 Fall 2004exam1 - [email protected]’ Exam 1 EE 402 jgh Name...

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Unformatted text preview: [email protected]’\ Exam 1 EE 402 jgh 10/22/04 Name Salute”? 1'? €22 two parts; total of 100 points (70 minutes) I have neither given nor received unpermitted aid during this examination. (signature) Part 1 — closed book and notes, no calculator. When finished, turn in to get part 2. l. (9)a. State the symbol used in the text and the SI units for the following: electric field intensity E V/ W permittivity C— F/m electric flux density 3 Q/m‘z permeability [A “/vvx magnetic field intensity,Q H A’m volume current density £3: A/ M“— magnetic flux density? __ M"/ W volume charge density (3,” Chi/1'3 conductivity 0‘ 157m 2.(4)a. For a linear, isotropic media with 11, a, 0, state the differential form of Maxwell’s equations. {3v \/ V E z 0 3%. V E ' ' 51-: ‘1; . E vx E 9. ‘t %1- + G - (3 )b. State the constitutive relations for u, a, c. J; ” (1‘ E §=eé ”E: M 1;» (1)0. State the definti t10 of the instantaneous Poynting vector and give units. E T: ‘kEL w/ m1 3.(4) State the four boundary conditions for two linear, isotropic materials as shown. G e (4% . l ‘ laws—go: Ps ” %%‘%E=O L) Y) \L [a x {‘4 1 ~ \ 4.(4)a. State the time harmonic Maxwell’s equations for a source free linear, isotropic media with 11, 8, o. ,\ Ct 6: 5. (2) State the name, symbol, and SI units of the real and imaginary parts of the propagation constant y. . o; as“? QM e (1 t“. 1.31m 6' 3m. 3%“? Mid/m A . 3- : ol Hfs (5 @iA‘AIiQ 0,7215%: Mex; Wad/m 6. (3) Given a TEM time harmonic field in a linear isotropic homogeneous media given by the phasors Ex(z)ax= (E' eJBZ + Em ejfiz)ax and l—Iy(z)gy— _ (HQ'SJBZ + HnTejBZ)a, define and give units: a. intrinsic wave impedance ~ 1.. ”,1? E m f ‘7 e 17': ‘4 "’ “7?“; t' _ 5W3 Esfi, b. total field impedance at z A r? I}: : 13.1 J? u u U c. reflection coeffic1ent at z 0 A /‘ I‘ "' “ \31 v ‘2 A E G p/ a ._ :M g C wrifi {7‘0} :1 ’ AM \e 1’ " “I; f. ”b Exam 1 BB 402 jgh 10/22/04 Name two parts; total of 100 points (70 minutes) lhave neither given nor received unpermitted aid during this examination. (signature) Part 2 — 8 1/2 x 11 sheet of notes and calculator 1. The electric field intensity of a sinusoidal steady state TEM wave traveling in low-loss linear isotropic material, i.e., G / (co 8) << 1, is given by: _ at} E(z,t) = 20 exp(-0.0094z) cos(127t108t—87tz) ax (uV/m). : EM e (as ( on: - fl 3,) fix Find the following assuming H 2 Ho, and state units: (2)a. frequency f; 2W8”: («J I 037109 7:) V; : 6X1OG : 603M“; (2)b. attenuation constant on; ‘ o< -. 0.00034- (HP/m) (2)c. phase constant B; 18 ‘1‘ 931T 1W1 Ci/m) (2)d. phase velocity vp; ”If ‘ 3:1 __ \ztttote’ : ;,_3‘)(iog im/S) P ' Q ~ $3? (2)e. wavelength 1; . an at: - 2 m - é; - A -: 72 t g” -* 0' {L 3 1%: C—r- e0 ‘ (2)f. the value for a; 32 (43) gr) imr MJgg; ngéfié ‘, fit WW2: 1} 6 7‘ [OI/LL“ I QZW/OQBL141TK‘Q'V) (2)g. intrinsic wave impedance 7]; A.“ )00‘ ’ ‘ 0 +W\i -1055 YMcgwa \ vi: 3).?“ :XW - [430,44 K51) (2)b. estimate for conductivity 0; -n ‘ : 3.540110 (7;) 26‘ ~ 210.99%)“, : 0103211105 64“} JEN him—10% Waite.“ 04:: 47;? =3 0‘: T W (2)1. phasor representation 111(2); V A ~0.0oa+ -38? a god) 2 me 1e 33w (M A F (2)j. phasor representation 11(2)? 016% 1, 05;; 31:1- :33??? Mafia 0 00054? Agngo HMO) 2 1:13.14 : 20 e 933 : OnObQ 9. 9 "' ’1 ‘3 ’QGJSA (2)k. time representation H(z,t); 0‘ 00014—9 PA I A “1C ' 1 at ~ (3“ a __ Ema = 12413133 «93 3— 1 (3.1er (0502:? o g) _«3 k M maxzrr‘log 7. (2)1. time-average Poynting vector Pav(z). , J 120 Q ~0.<l76‘43) a A , q J HER}; __ MM” 5‘ 9! " 1 L7. 13915139‘ ”373' 3 1 (Wk—1M9; ' 251341.414“; 3 Mi) 2. Given the electric field intensity E1 and the magnetic field intensity H1 at the origin for z < 0 at the boundary between the two perfect dielectrics regions defined as shown (note the units): E1: 1807c 21K + 2407: ay + 3007: aZ (mV/m) 6: =0 l H1: 3 21K + 4 any + 5 aZ (mA/m); [h S/Ua showing all work, 5 : 4-60 (10)a. determine the field D2 at the origin for z > o; ’ “M: 91V4¢&§\€\OCH\Q “:13 Pg :0) $5 : O .. MC b‘ = (7?; : 463 0‘60??ij *Z—‘tong? +303'fi‘933 K‘le ”2L Q=2°e V\' ($259.3 : O 1> 323* \ZQfiB—W: O :— \'LOO’\TeQ—;§. Pb“ : «mwée 95 —'L i\ b V) 3“ E)” 9) 3 O 23 "EM? 6:th 196W gx’rz‘mfifl‘g‘, “ EU: a} ‘3‘ Eva 03:3 :2) 51"th30? W €13: 240W A : igofifi} it l4oti‘gx? _ “ ~ " e +2Amroie> ":5 th: ézitl - 160< MON-03', ”a DJ: 1 lgdfiéh 3x + MOWEO 2? (® 3‘ E7. 7‘ btz‘l‘ b I Delgeyfieo g¥>‘l‘0‘240fi\€o€\‘a ‘l- Llagfieogé {914(2) uh; (10)b. determine the field B; at the origin for z > 0. \“(_B;|‘:/’l Er" Mel’béfise 4- A33 Jr 522‘) it} M V‘ X (“1“ H” :' O :3 Hi1: Ht), LSHxaz 3—5 <O\ .\ \ 3 Em "l' 4 3‘3 : V'HZXCZLK —\—- “a? 936 :3 l’lL‘K t. 3 Mw/l “215 2 4- % Bu = )in \i-t't = Arko 3’0 = afloat; 4— 19/40- 3. A TEM wave traveling in a linear isotropic region 1 defined by 2 <0 is incident on another linear isotropic region 2 defined by z > 0, where the properties of each are given as shown. If the sinusoidal steady state phasor representation of the incident electric field is given by E1(z)= 3 exp( -j(161c/3)z) aX (V/m) GT :0 £1 2-1:,360 then determine the following: a): 4' go i if; ; 2}! O (2)21. the intrinsic wave impedance of regionl 1111: (l. : 2L/14:;1Vl__o = i3) 1' bofiC-Ql WD- ’2 § (2)b. the intrinsic wave impedance of region}, 112; 111:. .1 SZAO ~' .; 51A 6L « 260 - no -1107? ( (2)c. the reflection coefficient at the boundary for region 1, 11(0); 41-11. ~ llo'U- (son (60‘? ~ ,L a» A V103) 3 ”2411‘ .. i261"? +5911" ., ‘30 TV ‘ 3 A (2)d. the transmission coefficient at the boundary, T; :11; M; .1 201an : 240R : {1; “ 1121111“ 12m +5011 1901? 3 A» _1_ .L ,_ 1 (3am : MW): 1 i> \‘1 3 ' ‘53") (2)e. the phasor representation of the reflected electric field; A ‘ 4 ~ - ~+ - . , _ Em - T110; by," , (13)“) -1 319; 4 we v @111: EAUL-‘sélgb: e {be t/ml (4)f. the phasor representation of the transmitted magnetic field; ” “*1 A. e a: -— ~1— ém. 141121 N re “W Wt E w 11% “ 2 £ng 5 e “m any“??? 1)“; ”‘45:? “A 4?}? ’3 2%“0 “CS” ‘ 1M: a ' fi .2 %. “19‘3_ ”0“- 213 10.1) QM) Effig‘jlw) _- 1% : {5, (4)g. the total field impedance; 1n region 1 at the boundary, 21(0); , 1+ We \+k‘/2,l _ ,_ ”21(0) : Vla ”‘ __ filo) : K90“) 1' V35 110 H on (4)11. the reflection coefficient at z— — - l in if the frequency IS 400 MHz, F101); “(3‘1 : lite) e “5‘3 fit?) 2 (L2,) e12tlég)t~\) A r» 1‘ "ggx 115+ 1 «4% \mw\ -— L «EL? y‘ L 7. ‘ W11 L I __ w \‘w ‘ 111.2 & fyhh f i \951 207—011 35 ’ $119. mil 1‘11. ...
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