161_1_Lecture_15

161_1_Lecture_15 - EE161 Electromagnetic Waves Fall 2010...

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EE161 Electromagnetic Waves Fall, 2010 Instructor: Dr. Shenheng Xu Electrical Engineering Dept., UCLA © Prof. Y. Ethan Wang
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Lecture 15 • Large Aperture Antenna • Rectangular Aperture with Uniform Field istribution Distribution • Effective Area of an Antenna • Friis Power Transmission Formula Antenna Arrays rrays with Uniform Excitation Arrays with Uniform Excitation
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adiation by Large Aperture Antennas Radiation by Large Aperture Antennas Wire antenna Aperture antenna Current element Radiation field Aperture field distribution Radiation field irchhoff’s diffraction theorem: Kirchhoff s diffraction theorem: The new wave front can be considered as the combination of all the waves rriving at the observation point radiated from the previous wave front arriving at the observation point radiated from the previous wave front Conditions: d>> for Kirchhoff’s diffraction theorem to hold dimension of antenna R e jkR Physical optics approximation Far field conditions: R>2d 2 / ) , ( a a a y x E Aperture field
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adiation by Large Aperture Antennas Radiation by Large Aperture Antennas herefore the radiation far field is ~ Therefore the radiation far field is, jkR y x e j n s n p ) , ( f    a a a a a a a dy dx y x jk y x E R R E ) sin cos ( sin exp ) , ( ) , , ( lement factor Aperture distribution factor Element factor 2D Fourier Transform F.T. Polarization irection same as that of the perture field. ) , ( a a y x ) , ( y x k k cos sin k k x direction aperture field. 2 2 ) , ( ~ ) , , ( f R E The power sin sin k k y 2 2 0 0 2 2 ) , , ( R R S density
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Rectangular Aperture – Uniform Distribution gp Uniform field distribution ) , ( 0 E y x E a a a 2 2 and 2 2 for y a y x a x l y l l x l 0 otherwise For plane (  we have  2 2 0 ) sin exp( ) ( ~ y y x x l l l l a a a dy dx jkx E f 2 2
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Rectangular Aperture – Uniform Distribution sin 2 sin k u Define The field pattern of the rectangular aperture is 2 2 0 ) ( ~ y y x x a l l a l l a jux dy dx e E f y julx jul l ju e e E x 2 / 2 / 0 j e e u l E julx jul y x 2 2 2 / 2 / 0 2 2 (F.T. ) u x a ) 2 / sin( x ul ) 2 / ( sinc x ul x x sin sinc Therefore, ) 2 / sin( 2 ) ( ~ 0 x y ul u l E f 2 / ) 2 / sin( 0 x x y x ul ul l l E ) 2 / ( sinc 0 x y x ul l l E x ) / sin ( sinc ) ( ~ 0 x y x l l l E f Finally, the field pattern is As the physical area of the aperture is y x p l l A then one has n nc ~ ) / sin ( sinc ) ( 0 x p l A E f Thus the power density should be, n nc 2 2 2 0 0 p A E S ) / sin ( sinc ) , ( 0 x l S R S
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161_1_Lecture_15 - EE161 Electromagnetic Waves Fall 2010...

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