ECE 6341 Spring 2016 Prof. David R. Jackon ECE Dep. Noe 39 1
Finie Source J ( y, ) For a phaed curren hee: p p j( k kyy) J y = J e + (, ) 0 The angenial elecric field ha i produced i: ( p ˆ ˆ) j( k + k y),, = E y z uv z J u e i 0 ( )( p ) j ˆ ( k+ kyy) i 0 + vˆ V z + J v e y Recall ha 1 J J k k dk dk ( 2π ) (, ) p 0 = 2 y y 2
Finie Source (con.) Hence + + 1 E (, y, z) = uv z J u vv z J v ( 2π ) { ˆ ˆ ˆ ˆ} 2 i i e ( y ) dk dk j k + k y y Noe: (, y), (, y) uˆ = uˆ k k vˆ = vˆ k k y (, ) k = k k y k y ( k, ky) ˆv û φ φ k Spaial coordinae Wavenumber plane 3
N Model for Tranform of Field We can alo wrie 1 j( k+ kyy) E, y, z = E k, k, z e dk dk ( 2π ) + + 2 y y Comparing wih he previou reul, we have E ˆ ˆ ˆ ˆ k ky z = uvi z J u vvi z J v,, Similarly, H ˆ ˆ ˆ ˆ k ky z = uii z J v + vii z J u,, Thi moivae he following idenificaion: 4
N Model (con.) Modeling equaion for horizonal elecric urface curren: (,, ) V z = E k k z uˆ y (,, ) V z = E k k z vˆ y (,, ) I z = H k k z uˆ y (,, ) I z = H k k z vˆ y I = J k k uˆ (, ) y I = + J k k vˆ (, ) y 5
N Model (con.) z : I + V I V = E ˆ u I = H ˆ v I = J uˆ 6
N Model (con.) z : I + V I V = E ˆ v I = H ˆ u I = J vˆ 7
Eample J = J ˆ, y Find E( yz,, ) E k ky z u E u v E v (,, ) = ˆ ˆ( ˆ) + ˆ( ˆ) ( uv ˆ ˆ) ( z) ( v ˆ ˆ) V ( z) = + 8
Eample (con.) ˆ uˆ = coφ = ˆ vˆ = inφ = Hence k k k k y ˆv y û φ (, ) k = k k y k k E k ky z V z V z k k y,, = + k k y = V ˆ ˆ i z J u+ Vi z J v k k k ˆ ( ˆ) ˆ = = coφ = ˆ ( ˆ) ˆ ( inφ ) J u J u J J k k J v = J v = J = J k y 9
Eample (con.) Hence k k ky ky E k, ky, z = Vi ( z) J + Vi ( z) J k k k k 1 2 2 = J k Vi ( z) + ky Vi ( z) k 2 or 1 1 2 2 (,, ) = 2 2 i + y i (2 π ) k E y z J k V z k V z e ( y ) dk dk j k + k y y 10
Dyadic Green Funcion G G G G(, y y ; zz, ) G G G G G G y z = y yy yz z zy zz where ij i (,, ) G E yz (, y, z ) (,, ) = ˆδ ( ) δ ( ) δ ( ) = due o he uni-ampliude elecric dipole a J yz j y y z z From uperpoiion: (, ;, ) = (, ;, ) (, ; ) E y z z G y y z z J y z d dy where E Noe: We have ranlaional invariance due o he infinie ubrae. E E = y E z J J J = y J z We aume here ha he curren are locaed on a planar urface z. 11
Dyadic Green Funcion (con.) (, ;, ) = (, ;, ) (, ; ) E y z z G y y z z J y z d dy Thi i recognized a a 2D convoluion: E = G J Taking he 2D Fourier ranform of boh ide, E = G J where (, y;, ) G = Gk k zz 12
Dyadic Green Funcion (con.) E = G J (, y;, ) Gk k zz i called he pecral-domain dyadic Green funcion. I i he Fourier ranform of he paial-domain dyadic Green funcion. Auming we wih he componen of he elecric field due o an -direced curren J (, y ), we have E = G J In order o indenify G, we ue 1 1 E( yzz, ;, ) = J k, k kv zz, + kv zz, 2 2 2 2 y i y i (2 π ) k e ( y ) dk dk j k + k y y 13
Dyadic Green Funcion (con.) 1 1 2 2 E ( yzz, ;, ) =,, 2 J 2 kv i zz kv y i zz (2 π ) k + j( k+ kyy) e dk dk y E k k zz 1 kv zz kv zz J k k 2 2 (, y;, ) =,,, 2 i + y i y k Recall ha E = G J Hence 1 G k k zz kv zz kv zz 2 2, ;, = (, ) + (, ) y 2 i y i k 14
Dyadic Green Funcion (con.) We hen have: G G G G G G G G G G y z = y yy yz z zy zz 1 G k k zz kv zz kv zz 2 2, ;, = (, ) + (, ) y 2 i y i k The oher eigh componen could be found in a imilar way. We could alo find he magneic field componen. We can alo find he field due o a magneic curren. EJ G = G ( he componen of he elecric field due o an elecric durren ) 15
Dyadic Green Funcion (con.) The differen ype of pecral-domain dyadic Green funcion are: EJ G ij EM G ij HJ G ij HM G ij Give elecric field due o elecric curren Give elecric field due o magneic curren Give magneic field due o elecric curren Give magneic field due o magneic curren Noe: There are 36 erm here, hough many are equal by reciprociy or ymmery. There are 20 unique erm (five from each ype of Green funcion). 16
Summary of Reul for All Source Thee reul are derived in Noe 44. V I V I = E u = H v = E = H u v I = J u k V = M + J ωε v z V = M u k I = J + M ωµ v z Definiion of verical planar curren : = J yz,, J y, δ ( z) z z = M yz,, M y, δ ( z) z z 17
Source ued in Modeling z : V i = volage due o 1[A] parallel curren ource I i = curren due o 1[A] parallel curren ource V v = volage due o 1[V] erie volage ource I v = curren due o 1[V] erie volage ource I + V I + V I + - V = = i v = = v V z V z I I z I z I i V z V z V I z I z V 18
Source ued in Modeling (con.) z : V i = volage due o 1[A] parallel curren ource I i = curren due o 1[A] parallel curren ource V v = volage due o 1[V] erie volage ource I v = curren due o 1[V] erie volage ource I + V I + V I + - V = i = = i = V z V z I I z I z I V z V zv v I z I z V v 19
Eample Find HM G Thi i he componen of he pecral-domain dyadic Green funcion ha i ued o obain H from M. Sar wih: H k ky z u H u v H v (,, ) = ˆ ˆ( ˆ) + ˆ( ˆ) ( ˆ uˆ) I ( z) ( ˆ vˆ) I ( z) = + Recall: = (, y, ) (, y, ) V z E k k z uˆ V z = E k k z vˆ (, y, ) (, y, ) I z = H k k z uˆ I z = H k k z vˆ 20
Eample (con.) (,, ) = ( ˆ ˆ) + ( ˆ ˆ) ( ) = ( coφ ) I ( z) + ( inφ )( I ( z) ) = ( coφ ) Iv ( zv ) + ( inφ ) Iv ( zv ) = ( coφ ) Iv ( z)( M u ) + ( inφ ) Iv ( z)( M v ) = ( coφ ) Iv ( z)( M coφ ) + ( inφ ) Iv ( z) M ( inφ ) 2 2 = M ( Iv ( z) co φ + Iv ( z) in φ ) H k k z u I z v I z y 2 k = M Iv z + 2 k I ( z) v k k 2 2 Recall: I = J u k V = M + J ωε v z V = M u k I = J + M ωµ v z 21
Eample (con.) 1 H k k z M ki z ki z 2 2 (,, ) = ( + ) y 2 v y v k Hence, we have 1 G k k zz ki zz ki zz 2 2, ;, = (, ) + (, ) HM y 2 v y v k Noe: The noaion (z,z ) ha been ued in he final reul o emphaize ha he erm depend on boh z and z. 22
Summary of Specral-Domain Recipe Sar wih a given planar curren diribuion. Eample (E from J ) J ( y, ) Take he Fourier ranform of he urface curren. (, ) J k k y Ue he appropriae pecral-domain dyadic Green funcion o find he Fourier ranform of he field of inere. (, ) E = G J k k EJ y Take he invere Fourier ranform (2D inegral in k and k y ) o find he field of inere in he pace domain. 1 EJ jk ( + kyy) = 2 y (2 π ) 23 E G J e dk dk