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MIT 6 013 - APERTURE ANTENNAS

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L20-1APERTURE ANTENNASDense phased arrays ~apertures:Pairs of radiators cancel (e.g., a,b)Sum of pairs therefore = 0First null at θ = ~λ/DAperture antennas:Derivation of far-field aperture radiation:Parabolic reflectorAperture screenPhased arrayλ/2acancelbθnullArray of 14 elementsDFind surface current Jsthat could produce the same aperture fieldsFind the far fields radiated by an infinitesimal element of JsIntegrate the contributions from the entire apertureSimplify the expressions by using small-angle approximationsNote the Fourier relationship between aperture and far fieldszˆL20-2APERTURE ANTENNASAssume UPW in aperture:-jkyEo-jkyE++ = zˆˆ Eoe ⇒ H = x eηo+jkyEo+jkyE-- = zˆˆ Eoe ⇒ H = -x eηoSatisfies E//continuous at y = 0Equivalent surface current:Js+ = yˆ × [H(y=0 ) - H(y=0-)] = yˆˆ ×η x[Eoo/ + Eoo/η]Jso ]= -zˆ 2E /η [A m-1ozˆI dz = Jsdx dz (Hertzian dipole radiator)Radiated far fields:jη-jkrE(θ,)φ = θˆo sinθ J (x,z) epq∫∫ffAzsdx dz2λrj-jkr = -θˆ sinθ E (x,z) epq∫∫Aozdx dzλrzxyθaperture ALzLxαzαx⎯JsEErpqHrˆpqL20-3FAR-FIELD RADIATIONAssume UPW in aperture:j-jkrE(θ,)φ = - θˆ sinθ E( )pq∫∫Aozx,ze dx dzffλr−+jkr jkrˆ ree−ijkr≅oqFourier relationship:Close to the y axisS(f ) ==∞∞s(t)e−πj2ftt s( ) S(f )e+j2fπt∫∫−∞d t−∞dfj−jkr2πE ( ,ααααˆo+j (x +z )xzffxz) ≅−θ e∫∫E (x,z)eλ dx dzλrozAzxyθaperture ALzLxαzαx⎯JsEErpqHrˆpqzqy0rqrˆrrˆiqrorpqFraunhoferapproximation(else, Fresnel)~Fourier (angle ⇔ space)(frequency ⇔ time)L20-4EXAMPLE: RECTANGULAR APERTUREAssume UPW in aperture:(Observer close to the y axis)j−jkrE(αα ≅−θˆzxo+j2πα παzx+j2L/2λλLx/2ffxz, ) e∫∫zλ−−L/zx2e EL/2oz(x,z)e dx dzrzxyθaperture ALzLxαzαxrEpqrˆqpπαxxLxLLE (x,z)e+j2παλjxxxxxx dx+ πα -jπαsinL/2λλ = E (e - eλ) = E Lλ∫−L/x2oz ozj2παoz xxπαxxLλG(αx,αz) ∝ |EAntenna Gain:2ff| ∝ sinc2(παxLx/λ)sinc2(παzLz/λ)αzNull at αz= λ/LzαxFirst sidelobeNull at αx= λ/LxG(αx,αz)*sinc θ≡(sin θ)/θ [= 1 for θ = 0]πα L ELozsincxxx*λj−jkrπα παE αα ≅−θˆoxxLL(zzffxz, ) Eλoze LrxLz sinc sincλλL20-5RECTANGULAR APERTURE (2)Antenna Gain G(αx,αz):*ηA= “aperture efficiency” ≡ Ae/A ≈ 0.65 typically(non-uniform aperture amplitude and/or phase ⇒ Ae< A)Huygen’s approximation:*True for any uniformly illuminated apertureEˆff ≅θ (j/2λr) (1+cosβ) ∫∫rAE (x,z)e-jkpqoz dx dzPt[W]GoG(αx,αz)I(αx,αz,r) [W m-2]A [m2]βI(ααG(ααxz, ) =xz, ,r )(Pt/ 4πr2)|E |2Ptx= Aoz [W] (A=L L2ηz)o|E2ff|2211222παxL πα LI(ααxz, ,r) = [W/m ] (= ) |Eoz| A sinc (x)sinc2(z z)22ηηooλr λG(ααxz, ) = A(4π/λ22)sinc (παxLx/λ)sinc2(παzLz/λ)4πλ2G=ooA ⇒=A A =2eG πλ4sinθ [ →θ1 as →0]θλL20-6EXAMPLE: SLIT DIFFRACTIONPair of rectangular slots:Uniform illumination*=Spatial domainFourier transformsLD=λ/DAngular domain0λ/Lδ(x,z)δ(x,z-D)z=0DzL20-7Tapered illumination reducessidelobes,broadens beamwidthParabolic reflector⇔FourierθG(θ)E(z)DIFFRACTIONAperture tapering:zFresnel zone plate:Eˆff ≅θ (j/2λr) (1+cosβ) ∫∫rAE (x,z)e-jkpqoz dx dzr = (r21+ r2o)0.5=ro+ λ/2ror1Fresnelzoneblocked forFresnel zone plater = ro+ λreceivertransparent ringsBlocks rings of negative phase, maximizing Huygen’s integralAperture radius = r1maximizes received signal. Additional transparent rings increase it further, yielding lense behaviorAdjacent rings cancel if β≈1r1Negative contributionPositive contributionFresnel zoner0r = r0 + λ/2r = r0 + λCell phoneIncident UPWMIT 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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MIT 6 013 - APERTURE ANTENNAS

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