# HW 8 Sol - fag 303’” “%Méwaﬂ£ 1...

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Unformatted text preview: fag 303’” “ %Méwaﬂ£ 1'? S7” \fd/Uf/O’ﬂ 7-29. AM radio waves. Substituting a = mg, t) = 1230 c0s(7.5 x 105:: — [32) and a = 995,,(2, t) = §’(E0/n)cos(7.5 x 106: — ﬂz) into [7.18a] yields 6% am aaey VXCg=——t' —> m» (—ﬁ)E0[—- sin(7.5 x 106: — 52)] : —p0%(7.5 x 105)[— sin(7.5 x 10% — [m] resulting in #005 x 106) —> [3 = “~— 77 Similarly, substituting % and % into [7.180], we have _m 6g 85g 836 6% V 36 z: -—-— = —— “ii—my 2 A x x 6t 6" 6t —’ az m at a —(-ﬁ)%3[— sin(7.5 x 10% -— 52)] = eoEo(7.5 x 106)[-_ sin(7.5 x 10% — ago] resulting in y) g = 600.5 x 105) Solving these two equations simultaneously, the values for [3 and 17 can be calculated as . 1 6 62 a (7.5 x 105mm) —> 13 = m = 0.025 rad—m4 3 x 108 772 = E -—-> a? m @- : 3779 50 EC Note that the g and 3—6 expressions also satisfy {7.1813} and [7.18d] as _— Base) 8:5 =0 phasor into the phasor form of [7.181;], we ﬁrst ﬁnd the corresponding electric field phasor E(y) as E: 1 V x H: “L (2611749)) jWEO jweo 6y _ ‘ ‘ _4 ‘ :29 3491.83 X 10_4e_34y = 4&4 X 1.83 x 10 euﬂy V_m_1 Jwéo {UGO 71.39. Maxwell’s equations. (a) We follow an approach similar to Example 7-13. Substituting the H091) i Next, we substitute E in the phasor form of [7.18a] to ﬁnd H as Ham -_1_V X E=__'_1_ (483431)) jwtto Jwﬂo 6?] , («3292.1 1.83x10~4e“i4y JWMD JWEO A =2 A :z 2 w #060 1.83 x 10—43—949 A-m-l Note that for this expression to be the same as the H(y) expression given as H(y) = 21.83 x 10*43‘3'431 A-m‘1 the following condition must hold: 16 :1 = 24x3><103=12 109 -—1 wzﬂoeo —} w #060 x rad s With this value of w, H(y) and its counterpart E(y) satisfy both [7.18a} E(y) satisﬁes [7.1%] (since E(y) = frEﬂy) and therefore V - D [7.18d] (since H(y) = 2Hz(y) and so V ‘ B = ,ro - H = 0). (b) Substituting w = 1.2 x 109 rad-s.-l and [7.180]. In addition, = 60V - E = 0) and H(y) satisﬁes into the E(y) expression found in part (a) yields E(y) : —xo.059e-j4y V-nr‘ (c) The electric ﬂux density D and the di3p1acement—current density Jd in phasor form can be found as D = 50E : —RO.6Ile_j4y pC—rn‘2 Jd =jweoE : —s(1.2 x 109x835 x 10—12)(o.611 x 10"]2e_j4y)(ejW/2) 2 — 27333—ﬂ4y—‘T/2) nA—m"2 8-2.. TV broadcast signai. (a) Since the signal is propagating in air, we have 27rf 3 103 = 9.3 rad-m” >< ﬁ=w horseman/c: from which we ﬁnd to 1: 2.79 x 109 rad—5‘1 and f = w/(Zn') : 444 MHz. (b) Using [7.18c] V X H = jweoE where H(z) = \$13-\$62) = 30.18—3'9Jze—jn/2 mA‘an and performing the partial differentiation yields E( )_ l 98Hm(z)) N§Y O_1(mj9.3)e~j9.3ze—j7r/2 2 7W 6‘2 " j(2.79 x 109)(8.85 x 10-12) g _,. y37.7e—j9.3ze-jar/2 mv_m_1 Therefore, the realntime €04, t) expression can be found from the phasor E(z) expression as so, t) = mam-mm} 2 —y sin(21r(444 x 10% — 9.3z) mV-m"1 8-3. Uniform plane wave. (a) For free space, we have 0.) 27rf 271' 5*wW—z"?-r Since f n 18 GHz and c z 3 x 108 m—s‘”, we calculate the phase constant as 21r(13 x 109 2 2120 - ‘1 ﬂ 3 X 108 1r rad m and the wavelength as 27r 27r A = ——— 2 —--— L: . 2 ﬂ 12 W 00167m 167cm (b) Using [7.18a] my) = — TLV x E(y) = __1_ (_23Em(y)) me jams By ' 5m; 22% _—.._ ﬂsejﬁy z zéejﬁy Jwiuo V #0 n where n is the intrinsic impedance of free Space which is n = result, the magnetic ﬁeld phase: H(y) can be found as H(y) m 2(247rrle3'120’w A-m-1 1/}.50/60 2: 1207i 2 3779. As a WWWWWWW. 3M 4 .mlmmmmw~mvmmmmmxmwmmww ...
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