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# hw10 - Problem 7.1 The magnetic ﬁeld of a wave...

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Unformatted text preview: Problem 7.1 The magnetic ﬁeld of a wave propagating through a certain nonmagnetic material is given by H = 250cos(109r —~ 5y) (mA/m). Find (a) the direction of ane propagation, (b) the phase velocity, (c) the wavelength in the material, (d) the relative permittivity of the material, and (e) the electric ﬁeld phasor. Solution: (3) Positive y-direction. (b)to=109 radls,k=5 radlrn. (c) 7L = 21t/k = 21t/5 = 1.26 m. c 2 3x108 2 (d)3r=(u—) =(2x108) =2.25. (c) From Eqi.) (7.39b), ' ES = —nt‘c x it, n: g=£g=%=251.33 (52), k = y, and ii = ism-f5)? x 10'3 (Aim). Hence, if = 4251.333: x 250e~f5r x 10*3 = —i‘r12.57e“j5y (Wm), and E(y,t) = my“) = —i]2.57cos(109t — 5y) (Wm). Problem 7.3 The electric ﬁeld phasor of a uniform plane wave is given by E = 5* main-22 (V/m). If the phase velocity of the wave is 1.5 x 108 mfs and the relative permeability of the medium is ,u, = 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: ~ (3) From E = 5710315934 (Vim), we deduce that k = 0.2 radfm. Hence, . Zn 221: k-Tc- — 03-— 101t—31.42m. (b) 8 _ ﬂ _ 1.53 10 z 6 __ f- A _ 31.42 4.77x10 Hz_4.77MHz. (c) From 2 2 c l C I 3 = = — — = ——- —— = 1:6 . up \ﬂ‘rE’r, & :“r (“p) 2-4 (1-5) 7 (d) _. 1” .ur 2.4 n—‘ﬂzlzm —=120n1/___= E 8, L67 451.94 (9), ~ 1 , ~ 1 . H = h—(—z) x E = ﬁ(4.) x SrlOeJO'ZZ = £22.13e1'0-22 (mA/m), H(z, t) = £22.13cos (cut +0.2z) (mA/m), with a.) = 21tf : 9.541: x 106 rad/s. Problem 7.8 For a wave characterized by the electric ﬁeld E(z,t) = ﬁaxcmmr — kz) + iaycos(o)t — kz + 5), identify the polarization state, determine the polarization angles (7,1), and sketch the locus of ‘E(0, t) for each of the following cases: (a) a; = 3 Vim, ay = 4 Vim, and 5: 0, (b) a; = 3 VIII}, a), = 4 Vim. and 8 =180°, (c) a,t = 3 Vim, ay = 3 V/m, and 5 = 45", (d) ax = 3 Wm, at), = 4 Vim, and 5 = —135°. Solution: we = whey/ax), [Ecl- (7.60)], tan27= (tan2wo)cosﬁ [Eq. (7.5%)], sinZX = (sin21p'o) sinﬁ [Eq. (7.59b)]. Linear Linea: Left elliptical Right elliptical (a) E(z, t) = i3cos(mt - kz) + i4cos(o)t — kz). (b) E(z, t) = i3cos(cot -—kz) - 37400561): —— kz). (c) E(z,t) = i3 cos(cot — kz) + 373cos(tot —— kz+45°). (d) E(z,t) = i3cos(tnz —- kz) +§I4cos(mt -— kz—135°). Figure P18: Plots of the locus of E(0, t). Problem 7.10 A linearly polarized plane wave of the form Ii: = ﬂag—1'": can be expressed as the sum of an RHC polarized wave with magnitude an and an LHC polarized wave with magnitude aL. Prove this statement by ﬁnding expressions for (IR and (IL in terms of (1,. Solution: E = ﬂare—1"“, RHC wave: ER = a]; (ft+ Ste-j"/2)e_m 2 an (1‘: — ji)e_ﬂ‘z, LHC wave: EL = 0L (1? + irejx/2)e‘jkz = aL(i + ji)e_ﬂ‘z, E = in + EL, ﬂax = 0120? —ﬁ) +0111? + 1'?)- By equating real and imaginary parts, :21 2 cm +aL, 0 = —aR +aL, or aL = (Ix/2, Q]; = (II / 2. Problem 8.1 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 tr:r = 25. Determine: (a) the reﬂection and transmission coeﬁicients, (b) the standing-wave ratio in the air medium, and (c) the average power densities of the incident, reﬂected, and transmitted waves. Solution: (4!) = 3:5 = 241: (a). se Th =Tio =1201r (Q): 112 From Eqs. (8.83.) and (8.9), 112—111 _ 241:-— 1201: _ —96 r: _——-—_~—=—0.67, n2+m 241t+120n 144 1:: 1+1“: l—O.67=0.33. (b) S: I+|FJ=1+0.67 :5. 1—111 1—-0.67 (c) According to qu. {8.19) and (8.20). _ i 2 s‘ = 1—9!— = — = .13 w 2 a“ 2m, 2x1201t 0 1m ’ 5;, = [rlzsiﬂ = (0.67)2 x 0.13 = 0.06 Win12, ZIEtillz 27h ' 2 1207‘ 2 x —- = -— = , — _ =0. _ SQV IT] 2112 [1| 1125;“, (0 33) x 24“ x0 13 07 Wlm Problem 8.2 A plane wave traveling in medium 1 with a” = 2.25 is normally incident upon medium 2 with 8,2 = 4. Both media are made of nonmagnetic, non- conducting materials. If the electric ﬁeld of the incident wave is given by E" = 94cos(6n x 109: — 30m) (Vim), (a) obtain time-domain expressions for the electric and magnetic ﬁelds in each of the two media, and (b) determine the average power densities of the incident, reﬂected and transmitted waves. Solution: (5!) . Ei = §4cos(6n x 109: — 30m) (Wm), 110 no no 377 = -—— = = — = _— = 5 "I an [—235 1.5 1.5 ‘2 110 no 377 = "2 J— 81'2 VIZ 2 112-111 1/2— 1/1-5 1": =———=—o.143, 112+m 1/2+1f1.5 r=1+F=1—0.143=0.857, E’ = I‘Ei = -—0.57j‘rcos(61t x 109: + some (Wm). Note that the coefﬁcient of x is positive, denoting the fact that Er belongs to a wave traveling in —x-direction. E1 = Ei+151r = y[4cos(61r x 109: — 30m) —O.57cos(6n x 109: + 301m] (Aim), Hi = £1 003(67: x 109: — 30m) = 2 15.91 cos(61t x 109: — 30m) (mAfm), TI: 11' = cos(6r: x 109: + 30m) = 22.27cos(6it x 109! + 30m) (mA/m), 1 H} =Hi+Hf : 2[15.91 cos (6:: x 109: — 30m) + 2.27cos(61r x 109: + 30mg] (mAlm). Since k] = Elk/#81 and k2 = (D #8 I k2—1/alk1—‘Iz—2530R—4OK (radlm), E2 = E‘ = 5r4tcos(6n x 109! — 40m) = y3.43cos(6n x 109: — 40m) (Wm), H; =Ht =2-4—Tcos(61: x109t—40mx)= 218.19cos(61:x1091‘—401Lx) (mA/m). TI2 (b) 42 16 =————=“ . w 2 2n, 2x251.33 x318 ("1 lm)’ S; = 41125; = -i (0-143)2 x 0.032 = —i0.65 (mW/mz), = 15612 2112 2 2 _ ..___=~ _ 2‘ 2112 x2X188-5 x3117 (mW/m) Within calculation error, Siav + SEW = SEW. Problem 8.9 The three regions shown in Fig. 8-32 (P89) contain perfect dielectrics. For a wave in medium 1 incident normally upon the boundary at z = -d, what combination of 81-2 and d produce no reﬂection? Express your answers in terms of an , 8,3 and the oscillation frequency of the wave, f. I-—-d--—-i Figure P83: Three dielectric regions. Solution: By analogy with the transmission-line case, there will be no reﬂection at z = —d if medium 2 acts as a quarter-wave transformer, which requires that 12 d = -—- 4 and 112 = vnms. The second condition may be rewritten as 1/2 T10 110 110 = ———-— , or a —---‘/e 8, var: [Val] Vera] U n U A, = — = ——= 2 «a: we: Hansel/4 and d: C . 4f(8n8r3)‘/4 Problem 8.10 For the conﬁguration shown in Fig. 8-32 (P83), use transmission- line equations (or the Smith chart) to calculate the input impedance at z = —d for 8,1 = 1, 5,2 = 9, 5,3 = 4, d = 1.2 m, and f = 50 MHz. Also determine the fraction of the incident average power density reﬂected by the structure. Assume all media are lossless and nonmagnetic. Solution: In medium 2, 1.0 c 3 x 108 A - - = = 2 m J2}; 12/9;r2 5 x107 x 3 Hence, 21: [32 = l—z = 1: rad/m, [32d 2 1.21: rad. At 2 = —d, the input impedance of a transmission line with load impedance ZL is given by Eq. (2.63) as ZL+jZotanB2d)_ lid—‘1) =2” (20+jzttan132d In the present case, Zn = T]; = Tao/11‘s” ——— 110/3 and ZL = T]3 = 110/, {81-3 = 110/2, where 110 = 1201: (\$2). Hence, n3+jn2tanﬁgd) no %+j(%)tanl.21t _ . _d = -—-——_—.———-..— = —— ———-——-— = 0.35- 0.14 . Z!“ ) “2 (“2+Jn3tanﬂgd 3 %+j(%)tan1.2it 1M J ) Atz= -—d, Zr. —21 _ 110(0-35—10-14)—110 = _ __—_________- = -j162.l4°. 21.. +21 110(0-35-1'0-14) +T|o "3 I“ Fraction of incident power reﬂected by the structure is [1"]2 = [0.49[2 = 0.24. ...
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