L18-1( )SINUSOIDAL RADIATION BY ANTENNASμJtq− r=oπ∫pqcApdV4rqpqVqHB = ∇× A , E = −∇Φ − jωA = −∇×jωεk =ω μ ε =2πooλo()( )μω−AeoJjkrω=pq d4rπ∫qVqpq()( )1ρω−jkrΦω= e dpqV4rπε∫qopqVqAlgorithm for general problem:{ } { }ρ⇒,J Φ,A ⇒{E,H}()ρ−1qptrqcΦ=pqdV4rπε∫opqVqRetarded potentials:Radiation ≅ static solution if krpq= << 1, or rpq<< Finding fields from potentials:2πrpqλλ2πzrpyxrpqVpVq2ρ∇ Φ=−εPoisson 2∇ A =−μ ∇×oJ (B A)equations:VL18-2()()μωoJ−jkrAepω=∫qpq dV4rπqVqpqμωI( )dˆ−jkr ≅θ(rˆcos −θsinθ) o epq4rπpqddWhere: Jzqq=ˆJ for - <<z22And provided that d << r(so r can come outside the integral) +q(t)I(t)yx-q(t)VAc[m2]zdzyφxθI(t)rrˆφˆθˆHERTZIAN DIPOLE RADIATORzˆrrˆθˆˆr sinθφ∇×A1Hd= = et∂∂r ∂∂θ ∂∂φμ2oμθorsinArrAθrsinθAφL18-3∇×A1jkIdHeˆ− jkr⎡⎤==φ1+sinθμπo4r⎢⎥⎣⎦jkr() ()∇×HE =jωεokIdηo− jkr⎧⎪⎡⎤11ˆ⎡11⎤⎫=+j e2rˆcosθ+θ 1++⎪4rπ⎨⎢⎥jkr2⎢jkr⎥⎬⎪⎩⎣⎦jkr⎣jkr⎦⎪Far field: r >> λNear field: r << 2πλ2πjkIdHeˆ−jkrff≅θφsin4rπjkIdηEeˆo−jkrff≅θθ sin4rπHeˆId− jkrnf≅θφsin4rπ2SOLVING FOR HERZIAN DIPOLE FIELDS(yields Biot-Savart law)IdηEj≅ −θoe− jkrnf3(rˆ2cos+θˆsin)4kπ rφθˆJsin (r −r') J×(r −r')H( r, t) =φ∫∫∫dv. Since ˆJ sin θ=J ×, H( r,t) =dvV'4π−|rr'|2|r− r'|∫∫∫V'4π−|rr'|32sinθ⎭θL18-4Far field: r >> (radiation)λNear field: r << (quasistatic)2πλ2πjkIdHeˆ−jkrff≅φ sinθ4rπjkIdηEˆojkrff≅θθ es−in4rπjIdηEeo− jkrˆnf≅ −θ(rˆ2cos+θsin )4kπ r3HˆIdjkrnf≅ φθe−sin4rπ2zEθ∝ L ∝ sinθθLxNEAR AND FAR FIELDS OF HERTZIAN DIPOLEE=ffηoH×ffrˆ (UPW)μ(ηΩoo = = 377 )εoθL18-5POWER RADIATED BY HERTZIAN DIPOLEG(θ) for isotropic radiatorzθx0.751.0Half-power pointθB1.5S(θ) ∝ |E|2∝ sin2θ11S(ee{E×H*21t) = R {S}=R } = rˆ|Eo|22 2ηoη2okId = rˆsin22θ [W/m ]24πr3-dB antenna beamwidth θB2PdS2Rr=φ∫∫ππrsinθ dθ002kIdπη2Id=πη sin θ dθ = kId ≅ 395 []oW41∫3oππ02λ2L18-6SHORT DIPOLE ANTENNASjkI(z)δηE'ˆojkr 'ff≅ θ'e−sinθ'4rπ '()()()Wire ≅ TEM line open-circuited, so current is a truncated sinusoid ⇒ triangle.∫∫d2'jkηd2EE=θr','dz = oˆsin θ′jkr 'ffθ'Ize−ffdz−−d24rπd2'jkηd2jk ≅≅θˆˆsin θoeI−jkr∫zdz θsin θ I d ejkrπ4roe 4rff−d2Has same far fields as does a Hertzian dipole, except Id → IodeffzzId/20(z)‐d/2⎯rθIIooI(z)deff≅ d/2⎯r'^r •⎯rzδ⎯rzθ’o−ηπL18-7[]2η2PooIdefT=≅kIf12πλodeff395 W2122PT2dηπ = I R ⇒ Radiation resistance R = oeffohms2orr2I()λojXRadiation resistance Rr+V-IoFor a triangular current distribution,center fed, deff≅ d/2.Short dipole antenna example:AM radio at 1 MHz has λ = 300 m. Car antenna 2m long has:2ii377 3.14 12⎛⎞R0r=≅⎜⎟.02 ohms3⎝⎠300jX can similarly be found by measuring or computing SANTENNA RADIATION RESISTANCEinto an antenna; it results from net electric or magnetic energy in the near field.3=L18-8ANTENNA GAIN OVER ISOTROPIC()()()()()jkη|sinθη Iod |2Prθφ θφ2ηoeff/2,, |Er,, | /2oθφ ≡o4rπ3G, = = = sin2θη 2P4AAππr22P4ro2kI d / 4 r212oeffππExample: What maximum⎯E [V/m] is produced at 10 km by a short dipole radiating PA= 1 kW?Solution: |(Er,θφ,)|2=θPr(, ,φ)2ηo(1.5sin2θ)P=A4πr2⇒φ |E(r,θ,)23| =(1.5sin θ)10 ηo/2π108=30 mv/mG(θ) for isotropic radiatorzθx0.751.0Half-power pointθB1.5S(θ) ∝ |E|2∝ sin2θGo= 1.5 (Max. gain)L18-9ANTENNA DIRECTIVITYAND RADIATION EFFICIENCY( )()()Gain over isotropicPr,θ,φ( )G,θφ ≡()P42Aπr()Antenna directivityPr,θ,φD,θφ ≡P4Tπr2∫∫D()θ,φsin θθφ d d =4π since P(r,θ,φ)r2sin θθφ d d =PT44ππG(θφ, ) P==TPower transmitted η n effiD(r "radiatio ciency" ≤1θφ, ) PAPower availableMIT OpenCourseWare http://ocw.mit.edu 6.013 Electromagnetics and Applications Spring 2009 For information about citing these materials or our Terms of Use, visit:
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