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%Méwaﬂ£ 1'? S7” \fd/Uf/O’ﬂ 729. 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 713. 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 Aml 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 Am‘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 rads.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.059ej4y Vnr‘ (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 82.. TV broadcast signai. (a) Since the signal is propagating in air, we have 27rf 3 103 = 9.3 radm”
>< 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 1012)
g _,. y37.7e—j9.3zejar/2 mv_m_1 Therefore, the realntime €04, t) expression can be found from the phasor E(z) expression as so, t) = mammm} 2 —y sin(21r(444 x 10% — 9.3z) mVm"1 83. 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 Am1 1/}.50/60 2: 1207i 2 3779. As a WWWWWWW. 3M
4 .mlmmmmw~mvmmmmmxmwmmww ...
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This note was uploaded on 01/17/2011 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.
 Summer '08
 STAFF

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