Hmwk12 Solutions

Hmwk12 Solutions - ® ECE 307 ‘ t 1 my agi‘ 5mm The...

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Unformatted text preview: ® ECE 307. ‘ t 1 my agi‘ 5mm The magnetic field of a ivave propagating through a certain nonmag- netic material is given by H = 250cos(109t — 5y) (mA/m). Find (a) the direction of wave propagation, (b) the phase velocity, (c) the wavelength in the material. (d) the relative permittivity of the material, and (e) the electric field. Solution: (a) Positive y-direction. (b)w=109rad/s,k=5rad/m. a) 109 up=I:?=2X108m/s. (c) A: Zn/k = Zn/S = 1.26 ml 2 g 2 c 3X10 (dyer—(ID) —(2Xms) _2.25. (c) From Eq. (7.3%), ,E(y,t) = mafia“) = —i12.57cos(1092—5y) (‘V/m). ‘ Ii . i! = —nl} x f1, - . i 120;: 1201: - n=\/E=—Tr=fi=251.33 (n), 'jf ' ii = y, and ii = iSOe’jsy x 10-3 (A/m). Hence, ~ 1 . E = 451.339 x 250.3"st x 10‘3 = —:212.57e-f5y (Vim), and i l Write general expressions for the electric and magnetic fields of a l-GHz sinusoidal plane wave traveling in the 4-y-direction in a lossless nonmagnetic medium with relative permittivity e, : 9. The electric field is polarized along the x—direction, its peak value is 3 Wm and its intensity is 2 Wm at t = 0 and y = 2 cm. Solution: For f = lGHz, u, = l, and a, = 9. (o: 21tf= 211'.><109 rad/s, 211: 21:, 21rf 21tx109 _T_K‘/E—’_T‘/£—'_ 3x103 E(y,t) = i3cos(21t x 109: — 201cy+¢0) (V/m). k \/§ = 201: rad/m, l aAtt=0andy:2cm,E=2V/m: 2 = 3cos(-201t x 2 x10‘2+¢0)= 3cos(—O.41t+¢o). Hence. 2 (lo—0.41: = cos“ (E) = 0.84 rad, which gives . $0 = 2.1 rad =120.l_9° N. ,. and E(y,t) =i3cos(21r>< 109t—201ty+ 120.19°) (V/m). The electric field phasor of a uniform plane wave is given by i) = 5' lOejO'zz (Wm). If the phase velocity of the wave is 1.5 x 108 mjs and the relative permeability of the medium is ii, = 2.4. find (a) the wavelength. (b) the frequency f of the wave, (c) the relative permittivity of the medium, and (d) the magnetic field H(z,t). Solution: (a) From 32 = meal” (Vim), we deduce that k = 0.2 rad/m. Hence, 2n_ 21: A:?_E=10n=3l.42m. (b) 8 up 1.5x10 6 =—=—————-=4.77 10H=4.77MH. f 7L 31.42 X z z (c)From 2 2 c l c 1 3 : , £:-—- — =— —— “P ‘F—M, ‘ map) 2.4(15) (d) _ E~ M- E. 11— 8-1201t 8r—l201t l67—45194 (Q), ii = 111(4) x f; = %(—i)x§r10ej°‘2‘ = £22.13ej0‘21 (mA/m), H(z,t) = i22.13cos(mt + 0.22) (mA/m), with (n = 21tf = 9.541: ><106 rad/s. The electric field of a plane wave propagating in a. nonmagnetic material is given by ' E = [93 sin(21t x 107: — 0.413) + i4cos(21t x 107: — 0.4m] (V/m). Determine (a) the wavelength, (b) er, and (c) H. Solution: (a) Since k = 0.41:. 21: 21! A, 2 — = — : k 0.41: 5 m (b) 7 _ a) _ 21: x 10 _ 7 up— k _ 0.41: _5><10 m/s. But u _ L P «5 Hence, 2 2 c 3 x l08 8“ (a?) ‘ (5x107) ‘36' (c) 1 A 1 H = T—‘k x E = an x [y3 sin(21c x 107: — 0.4m) + i4cos(21t x107r— 0.4m] 4 = 2% sin(21t x 1071— 0.410;) - 91—} cos(21t x 107: - 0.4m) (Aim), with 201: __19_~1 _ _ n_\/E_r_ 6 —201t—62.83 (Q). Wave Polarization An REC-polarized wave with a modulus of 2 space in the negative z—direction. Write down the ex field vector, given that the wavelength is 6 cm. (V/m) is traveling in free pression for the wave’s electric “ ‘ution: For an RHC wave traveling in —i, let us try the following: E = iacos(cnt + kz) + iasin(u)t + k:). Modulus [E] = Val-i-a2 = ax/i = 2 (Wm). Hence, 2 (12—,_,=\/§. V2 Next, we need to check the sign of the y-component relative to that of the i-component. We do this by examining the locus of E versus t at z = 0: Since , \ the wave is traveling along -2, when the thumb of the right hand is along —2 (into ‘ the page), the other four fingers point in the direction shown (clockwise as seen from ! above). Hence, we should reverse the sign of the i-component: E = in/Ecosfint + kz) —5'\/§sin(tut +kz) (Wm) with 2 2 7t 1: k —- T — W —- and i 27‘ s m w=kc=Tx3x10 =1tX10 (rad/s), ; @ 7,_, “Tillie-electric field of a uniform nronagating in free space is given by E = (i + jy)2oe-J‘7=zl6 (V/m). Specify the modulus and direction of the j electric field intensity at the z = 0 plane at: = 0, 5 and 10 ns. - Solution: E(z,t') = minim} = new: + find-WWW] =‘ me“: + f’ejn/z)20e'j“z/6ej""] = i20cos(mt —nz/6)+5'20cos(a)t—rtz/6+1t/2) = i20cos(mt — 1tz/6) — 3'20 sin(mt — 112/6) (V/m), I 5 2 2 ll2 |E| = [Ex +Ey] = 20 (V/m), I v w = tan"l = —(a)t — 1cz/6). . x Ex !' From . _C_kc_1t/6X3X108_ -, f—X—E—T—2.5XIO HZ, m = 21tf= 51: ><107 rad/s. At 2 = O, 0 I att = 0, w = —0x = —51: x107t= —0.251c = —45° at: = 5ns, _ -. _0,51: = _90° att = 10ns. Therefore, the wave is LHC polarized. Locus of E versus time. W @ Compare the polarization states of each of the following pairs of plane waves: “ (a) wave 1: E1 = 32C08(OJI — icz)1— 3‘72 sinlicnt - k3), wave 2: E2 = fi2cos(mt + k:)—1—'}"2 sin(cot + k:), . (b) wave 1: E1 = :22cos(mt- — kz) — i23in(cot — kz), wave 2: E2 = £2 COS(Cx)t +.kz) -— 5'2 sinlmt + k2). Solution: (:1) E; = iZcos(mt—kz) +5’28'1n(0)t—kz) '= i2cos(mt — k2) +5'ZCOS((Dt- kz - 7r/2), E1 = i2e‘jkz-l-5'2e‘jkze‘jn/2, _ / W0=tan-1=tan-11=45o, ax 5 = —TE/2 Hence, wave 1 is RHC. Similarly, ~ E2 = sizei"z + 5’2ejkze'j“/2. i Wave 2 has the same magnitude and phases as wave 1 except that its direction is ; along -2 instead of +2. Hence, the locus of rotation of E will match the left hand instead of the right hand. Thus, wave 2 is LHC. (b) a ' E1: i2cos(a)t —kz) —§'25in((ot—kz), E = iZe‘jkz+i23"jkzej“/2. Wave 1 is LHC. ~ E2 = 32eka +i2eik=efW/2. Reversal of direction of propagation (relative to wave 1) makes wave 2 RHC. In a medium characteriZed by e, = 9, u, = 1, and o = 0.1 S/m, determine the phase angle by which the magnetic field leads the electric field at 100 MHz. ' Solution: The phase angle by which the magnetic field leads the electric field is —9n where 61‘ is the phase angle of nc. o 0.1 x 361: me = 2nx103x10-9x9 =2' Hence. quasi-conductor. u .e” "/2 1201: _ a - -|/2 TIC: §(1—1F) =— 1—]— : 125.67(1— j2)“/2 = 71.49+j44.18 = 84.04z3__I-72°. Therefore 911 = 31.72°. 1 Since H = (1 /nc)l‘c x E, H leads E by —e,‘, or by —31.72°. In other words, H lags : E by 31.72°. © , Ignoring reflection at the air-soil boundary, if the amplitude of :1 2-61-12 1nc1dent wave is 10 V/m at the surface of a wet soil medium, at what depth will it be down to l mV/m’? We: soil is characterized by ur = 1, a, = 16, and 6 = 5 x 10-4 S/m. Solution: E(z) = soc-“z = 1013-“, 0' 5 x 10-4 x 361: -4 l we '2nx2x109x1o-9x16 "8X10 ' r . . ' . . Hence, medium 18 a low-loss d1electnc. ct_o* u o 1201c_5x10'4><1201t 2 s z 5' V5 ‘ me 10-3 ___ lee—0.0242, 1,1104 = -0.024z, z = 383.76 m. = 0.024 (Np/m), @ . Based on wave attenuation and reflection measurements conducted i at 1 MHz, it was determined that the intrinsic impedance of a certain medium is 28.1&(Q) and the skin depth is 5 m. Determine (a) the conductivity of the materiaL ; ' (b) the wavelength in. the'medium, and (c)*the phase velocity. ' - ' " 1‘ Solution:‘(a) Since the phase angle of 11c is 45°, the material is a good conductor. 'I Hence. nc = (1 +j) = 23.1e145° = 23.1 cos45° + j28.1 sin45°, QIR 01' 0|? = 28.1cos45° = 19.87. Sincea =1/8, = 1/5 = 0.2 NP/m, _ a _ 0.2 ' 19.37" 19.87 (b) Since (1 = [3 for a good conductor, and a = 0.2, it follows that B = 0.2. ; Therefore, 0' = 0.01 S/m. 21: Zn x=—=-—= = . . _ B 0.2 1011: 314m (c) up = f). :106 x 31.4 = 3.14 ><107 m/s. ® The electric field of a plane wave propagating in a nonmagnetic medium is given by E = i25e‘30‘cos(21t x 109: — 40x) (V/m). Obtain the corresponding expression for H. Solution: From the given expression for E, a) = 2:: x 109 (rad/s), a: 30 (Np/m), B: 40 (rad/m). 2 2 I 032 I or—fi = ~03 #8 = -m# Eo€,=-:€n «)2 2 _ 2aB=leEIl_—_CTZ_E;I I Using the above values for a), a, and [3, we obtain the following: 8,], = 1.6, 8;, = 5.47. _ n “1/~ ‘1/2 I ' ' = % (1 _ = Li; (1- =157.9e’36'85 (£2). 84 ' ' ~ 1~ ~ 1 . —_ _ r “ A -40 A — x — _' . ° H— ckXE-\157'96j36355xx225e ‘6 J x- y0.16e 306 40xe 13685, H = mafia/W} = -yo.16 e‘30‘cos(21t x 109: — 40x— 36.85°) (A/rn). _ copper block is 30 cm in height (along z). to a wave incident upon the block from above, a current is induced in the block in the 9 positive x~direction. Determine the ratio of the a—c resistance of the block to its‘ d-c resistance at l ' l _. __———_. d-c resistance Rdc = l . ,_ a o-sow’ ‘ . l . a—c re51stance R3,; = — crw6s ' R 0.3 - F“ = 8— : oat/mic .—. 0.3[1c x103 x 4:: x10'7 x 5.3 ><107]1/2 = 143.55. dc ' S 7K © The inner and outer ccndlictors of a coaxial cable have radii of 0.5 cm and 1 cm, respectively. The conductors are made of copper with e, = 1, ,u, = 1 and G = 5.8 X 107 S/m, and the outer conductor is 0.1 cm thick. At 10 MHz: (2) Are the conductors thick enough to be considered infinitely thick so far as the flow of current through them is concerned? 5)) Determine the surface resistance Rs. (c) Determine the a-c resistance per unit length of the cable. §9¥u§i9n= -- a.) 85 = [mm-V2 = [n ><107 x 4:: x10'7 x 5.8 x 1071-”2 = 0.021 mm. ' Hence, 0.1 cm ' 0.021mm ~ 50' d as Eerie? 999919994? Plenty thick- 5) Rs=i 1' =s.2x10-4n. 06; = 5.8x 107 x2.1x10‘5 Rs 1 1 l8.2><10‘4 1 1 w \. =___=__.__. __=.039 C) RI ((#1)) 21: (5x10-3+10-2) O ( m) ‘ ® EM PowerDensity " M i " _ The magnetic field of a plane wave traveling in air is given by H = 1225 sin(21t x 107! — ky) (mA/m). Determine the average power density carried by the wave. ’ Solution: ’1' H = 1225 sin(21t x 107: — k'y) (mA/m), E = may x H = 211025 sin(21t x 107: — ky) (mV/m), , 2 .r . 2 no?) x1o*6=y%03(25)2x10-6=90.12 (W/mz). g Say = (2X 1?) C39) {5’ I t, __ _ me electric-field phasor of a uniform plane wave traveiing down- ward in water is given by it = aloe—0115i“: (Vim), where i is the downward direction and z = 0 is the water surface. If G = 4 S/m. (a) obtain an expression for the average power density, (b) determine the attenuation rate, and (c) determine the depth at which the power density has been reduced by 40 dB. Solution: (a) Since 0. = B = 0.2, the medium is a good conductor. a , 0.2 , . . m:(1+j);=(1+1)T=(1+1)0.05=o.o7e145 (9). From Eq. (7.109), _ A IEOF 40: _ A 100 Sm, — zzmcle c039“ _ 22 X 0.07 (b) A = —8.680Lz '= -8.68 x 0.2: = —l.74z (dB). (c) 40 dB is equivalent to 10". Hence, e‘O-tzcos45° = time-“42 (W/mz). io~4 = 9-2“: = [0-42, [11(10-4) = —o.4z, or z = 23.03 m. . ._.-.... ‘54.. . ml.ll.“ ® , A wave traveling in a lossless. nonmagnetic medium has an electric field amplitude of 24.56 V/m and an average power density Of 4 Wlmz. Determine ' the phase‘velocity of the wave. - solution: E 2 2 .SaV=Li9 ' "=1E0l a. 2n , '- 253V or n_ (24.56)2_7549 _ 2x4 ' ' ‘ But 377 2 n=——=—v = _— = . «a? «a: . 5' (75.4) 25 c 3x108 uP:—-—: =6x107m/s. @ At microwave frequencies, the power density considered safe for human exposure is 1 (mW/cmz). A radar radiates a wave with an electric field amplitude E that decays with distance as E(R)' = (3,000/ R ) (V/m), where R is-the distance in meters. What is the radius of the unsafe region? . Solution: 1 (mW/cmz) = 10-3 W/cmz = 10 W/mz, R X2x1201r_ RZ" (1.2x10“ [/2 10: (3x103)2 1 __1.2x104 —— = 34.64 . 10 m' «EC/E 30 Z. Hw sag SVQfiLMS @ Reflection and Transmission at Normal incidence A plane wave in air with an electric field amplitude of 10 V/m is incident normally upon the surface of a lossless, nonmagnetic medium with a, = 25. Determine: (a) the reflection and transmission coefficients, (b) the standing—wave ratio in the air medium, and (c) the average power densities of the incident, reflected, and transmitted waves. Solution: (a) T10 1207! = Q,‘ =—- —-=247E flTli Tlo V Hz «a 5 ( ) r: 112-m_ 241:-1201: —96 um. m=m=‘°-67" flr=l+l‘=1—0.67=0.33. ' S=l+[1"[:l+0.67 =5. _ p _ 1— r] 1—0.67 .. ~ 5‘12 100 C S' = L o = ‘ 2 () W 2% 2X120n—OJ3W/m, ; 5;" = 11125;" = (0.67)2 x 0.13 F 0.06 W/mz, Sév = ITIZiE‘biz 2T“ ‘7. T _ ' = l 2 a __ 2:. 2112 l l m 3,, (033) x 247: x0.13_o.o7W/m .p __ H I. A ZOO-MHZ left-hand circularly polarized plane wave with an electric field modulus of 10 V/m is normally incident in air upon a dielectric medium with e, = 4 and occupying the region defined by z 2 0. ' (3) Write an expression for the electric field phasor of the incident wave, 3 the field is a positive maximum at z = O and t = 0. (1)) Calculate the reflection and transmission coefficients. (c) Wn'te expressions for the electric field phasors of the reflected wave, the transmitted wave, and the total field in the region 1 g 0. iven that the the incident ayerage power reflected by the boundary and transmitted into the second medium. Solution: (a) a) 21: x 2 x 103 41: kl-z-W-r’ad’mr u) a) 41: 81c = __ = _ = — 4 = "- rad/ - k2 up2 c E" 3 3 .m LHC wave: Ei =‘_a0(i+ yein/2)e-jkz = 0002 + 1-5,)e—J'Izz7 Ei(z,t) = itaocos(o)t — kz) — yaosin((nt — kz), IE‘I = [aécoszm — kz) +a3sin2(mt — kz)]V2 = a0 = 10 (V/m). Hence, .0 _ E' = 10(i+,j3’)e""u/3 (V/m).' :4, Cb) n1=no=1201c (:2), 112:111 r_n2—m 601c—1201: —6O 1 ‘7 "mm, ‘601:+12011= 180 =‘3’ “HIE? . .1 - 10 A - La) E‘:101‘(x+}y)el"‘z=-?(x+j3')e’4m/3 (V/m), ~ 1 .. _~ 2 ~ E‘: 101(x+1y)e fizz: 39(i+j§1)e’-’8"2/3 (Wm), 17:. = fi‘ +E‘ = 10(i+ j?) [KM/3 — gym/3] (Wm). {43) %ofrefiected power: 100x II"!2 = % = 11.11%, 2 1 % oftransmitted power = 100x mzm = 100 X X [20" = 88_89%. Repeat Problem 19' after replacing the dielectric medium with a poor conductor characten'zed by a, = 2.25, p, = 1, and o = 10"4 S/m. Solution: (:1) Medium 1: c0 21c><2x108 41: “*:-—$fiE——? mm” 5‘ ll :1 O u g D :4 Medium 2: oz ' 10“ x 361: ——= :4 10—3. 0122 211x2x103x2.25x10-9 x Hence, medium 2 is a low-loss dielectn'c. <12 1201: 02 1201: 10-4 1201: _2 _ X =—x——=1.26><10 NP/m), 2 «E, 2 «2.25 2 1-5 ( 131 = «We = “if: = 2:: (rad/m), u; joz, 1201c , __3 1201: = — 1 —— = —— 1 2 — = 801: Q . m (+ ) v§0+flxo) 15. 1) = com Hue—1’“, IE'I = do = 10 (V/m), E' = 1001+ 1'9)e-J4*z/3 (V/m). b) 1-:112-711 =801c—1201c__ _’ _ a 112+Th 801t+1201¢_ 0'21 T—l+l“—l—0.2_O.8. (c) : ~ E' = 1or(11+ j?)锑" = —2(12+jy)ei"'=/3 (V/m), E: = lot(i+jy)e—azze—jflzz = 8(i+jy)e—l.26x10'zze-i2nz (V/m), E. = E‘ + E' = 10(12 + jy)[e'1'"‘=/3 — clam/3] (Vlm). (d) % of reflected power = 100m2 = 100(0.2)2 = 4%, % of transmitted power = 100|112$ = 100(0.8)2 x w = 96%. g 112 801! ' A 50-MH2 plane wave with electric field amplitude of 30 V/m is normally incident in air onto a semi-infinite, perfect dielectric medium with e, = 36. Determine (a) F, (b) the average power densities of the incident and reflected waves. and (c) the distance in the air medium from the boundary to the nearest minimum of the electric field intensity, Soiutinn:(a) p2 1201: 1201: = :1207: .Q, = _=—:—=201r Q, m 110 ( ) 112 £2 «32 6 ( ) nz—n. 20n—1201t r: =—__—o.71. ‘ nz+m 201t+1201t Hence, [F1 = 0.71 and 6,, = 180°. S, _|_§;‘£_ (30)2 a- _ 2n. 2 x 1201: 5‘" = 11125;, = (0.71)2 x 1.19 = 0.60 (w1m2). (c).In medium 1 (air), = 1.19 (W/mz), _3x10‘ C 1"? 5x107 =6m. ef—M—RXG-ISm 41c _ 41c _' ’ 1m— ). _ ' (m = 1m- 7‘ = Ls— 1.5 =0m(at the boundary). 1 ' A parallel polarized plane wave is incident from air onto a dielectric medium with e, = 9 at the Brewster angle. What is the refraction angle? ' i l i 81=Og=tan“ 3=tan"3=7157° Bm , sine. sine] sin7 1 .57° Sine; = = = ————— = 0.32, «an 3 3 or 92 = 18.44°. (22> conductor with E, (C\) mania.“ 1n = no 21201: = 377 (9), X1: Medium 2: I 0382_ 21tx5x107x10—9 —' Repeat Problem after replacing the dielectric medium with a :1: #:3113940? .2-78 X 10‘3 Sim- 62 2.78 x 10-3 x 361: Hence, Medium 2 is a quasi—conductor. I -1/2 —1/2 145) = 120::(1—1'3) _ N32 112— 82( 52 = 1201t(1-j1)_1/2 =1201t(\/§)‘1/2em'5°=(292.88+ 1121.31) (9). j 112 —m (292.88+ j121.31) —377 _ a — = =— . 9 0.12:0.22/1Ji. F 112+n1 (292.88+j121.31)+377 00 +1 ' " '53'2 302 2‘ i V‘ c 3x103 ' -— = —— =1.19 / , i c, _ _ _ (b) St“ 2111 2x1201c mm) A-1—f—5X107 |S§v|=|1"|2S;v=(0.22)2(1.l9)=0.06 (W/mz). For 9, = 114.5° = 2 rad, < t parallel polarized beam of light with an electric 9,)»l + = 2&6) max—— — 41: 2 4 +0=3m, 20 (V/m) is incident in air on polystyrene with p1- : I and a = 2.6. If the incidence angle at the air—polystyrene planar boundary is 50°. determine (a) the reflectivity and transmissivity. and (b) the power carried by the incident. reflected. and transmitted beams if the .spot on the boundary illuminated by the incident beam is l in2 in area. " («5 (b) - Ru Pf: ' 2 law 2“! -(82/s|)cos0i + El \— (52/81)9089i + v (82/E|)—Sin29i Acosei = -—2.6cos50°+\/2.6-sin250° _ 0 08 ascossm + \/2.6—sin250° ' ' g If"? = (0.08)2 = 6.4 x 10-3, ; (20V 2 x 120: x cos50° = 0.34 W, ' Pl" = Rm = (6.4 x IO") x 0.34 = 2.2 x10'3 w, P" = Turf = 0.9936 x 0.34 = 0.338 w. field amplitude?“ : =6m. l j l 1‘ ! r l I A perpendicularly polarized wave in air is obliquely incident upon a planar glass-air interface at an incidence angle of30°. 'I‘ne wave frequency is 600 THZ (l THz = 10'2 Hz), which corresponds to green light, and the index of refraction of the glass is 1.6. If the electric field amplitude of the incident wave is 50 Wm, determine (a) the reflection and transmission coefficients, and (b) the instantaneous expressions for E and H in the glass medium. >__$gigltior_r:m(_a) For nonmagnetic materials, (tag/E.) = (Hz/n1 )2.> r cosei — ‘/(n2/n.)zesin29i cos30°~ i/(l.6)2—sin230° .L = = 6089; + V (nz/nl)2 — sinzfli cos30° + V (1.6)2 — sin2 30° IL: 1+1“; = 1—027 =0.73. = —0.27, (b) In the glass medium, . u . a sine! : smeI = srn3 : 0-31, n2 [.6 or 6. = 1821". _ B _ E _ fl _ - ' n2 - £2 _. n2 _ L6 _751r—235.62 (9), k a) 21cf 23f» anémx10‘2x1.6 2 2—: up c/n c 3x103 = 6.41: x 10‘ rad/m, it D Egg-jkfixsinefizéosei), ' ii; = (—icose. +2sin6‘) %e-i"zlflinfi+z°°sm, 2 and the corresponding instantaneous expressions are: E1(x,z,t) = i36.5cos(wt — kzxsinfli — kzzcoset) (V/m), H‘J_(x,z,t) = (—icosG. — icoset)0.16cos((nt — kzxsine. -— kzzcose.) (MRI), with (0 = 21: x 1015 rad/s and k2 = 6.41: ><106 rad/m. @ a) Show that the reflection coefficient 1" L can be written in the form _ sin(6‘ —- 9i) FL — sin(9.+65) ' r _'—néeos0i—n,coset _ (nz/n.)cosei—coset 1’ _ flzcosei‘l-Tll C059: —(1l2/TlI)C059i+COSGiI Using Snell's law for refraction: £:_Nsin 6‘ J n| _ sinei ’ we have I. _ sin6.cos 6i — coselsinei __ sin(9. - 6;) ‘L - sin O‘cos 0i + case. sinOi - sin(9. + 6i) ' ® 2 r . ,. . , . .. @y Show that for nonmagnetic metisz the reflection coefnc1ent 1 H can be written in the form WK 9t — 6i) [I = tamed-9i) . Solution: From Eq. (8.66a), 1"“ is given by nzcosepmcosei _ (nz/kaSGI—cosei nzcosefimcosei — (nz/m)coset+cosei' 1"u = For nonmagnetic media, #1 = p2 = yo and 1‘3 _ ii _ "_1 m 82 ’12 —--~ , _.. Snell’s law of refraction is . ' sm 8, n1 sin 9i — n2 ' Hence, sin 6t cos 9 — cos 6- . . F sin 9i ‘ ‘ , sm 6, cos 6, - sm 9, cos 6, ” — sin 9: _ sin 9, cos 9; + sin 6, cos 9, ' , cos 9, + cos 6i sm 6, To show that the expression for 1"” is the same as = — “ tan(et+ei)’ we shall proceed with the latter and show that it is equal to the former. tan(9t — 91) _ sin(0, — 9i)cos(6t + 9i) tan(et + — (303(91— Sin(9t + I Using the identities I 28inxcos y = sin(x+ y) + sin(x — y), and if we let x = 9: — 6i and y = 0‘ + 6, in the numerator, while letting x = 6; + 6, and' i y = 9, — 9, in the denominator, then tan(6t - 9i) _ sin(291)+ sin(—26i) {811(9: + 9i) — sin(291) + sin(29i) ' But sin29 = 25in6cose, and sin(—6) = — sin 6, hence, tan(6t — 6;) _ sin 9, cos 6‘ — sin 6, cos 6, tan(6t + 6i) — sine‘cos Q: + sin 6, cos 6, ’ ...
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This note was uploaded on 12/06/2010 for the course ECE 302 taught by Professor F during the Spring '10 term at Cal Poly.

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Hmwk12 Solutions - ® ECE 307 ‘ t 1 my agi‘ 5mm The...

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