Electromagnetic Field Transformations and Modeling Equations Analysis

ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 39 1

Finite Source ( ) , J x y s ( ) ( ) + j k x k y p s p s = , J x y J e x y For a phased current sheet: 0 The tangential electric field that is produced is: ( )( z ) ( ) ( ) + j k x k y p s = TM , , x y z E uV J u e x y t 0 i )( ) ( ( ) ( ) z + j k x k y p s + + TE v v e V J x y 0 i 1 ( ) p s = , J J k k dk dk Recall that ( ) 0 s x y x y 2 2 2

Finite Source (cont.) Hence + + 1 ( ) ( ) z ( ) z = TM TE vV v , , x y z E uV J u J ( ) t i s i s 2 2 ( ) + j k x k y e dk dk x y x y ( ) ( ) = = u u k k v v k k , , , Note: x y x y y yk ( ) ( ) = , k k k , k k t x y x y u v x xk Wavenumber plane Spatial coordinates 3

TEN Model for Transform of Fields We can also write + + 1 ( ) ( ) ( ) + j k x k y = , , x y z , , E E k k z e dk dk x y ( ) t t x y x y 2 2 Comparing with the previous result, we have ( , , t x y E k k z uV = ) ( ) z ( ) z TM TE vV v J u J i s i s Similarly, ( ) ( ) z ( ) z = + TE TM v v I , , H k k z u I J J u t x y i s i s This motivates the following identifications: 4

TEN Model (cont.) Modeling equations for horizontal electric surface current: ( ) ( ) ( ) z = TM , , V z E k k z u t x y ( ) = TE v , , V E k k z t x y ( ( ) ) ( ) ( ) z = TE , , I z H k k z u t x y = TM v , , I H k k z t x y ( ( ) ) = TM s , I J k k u s x y = + TE s v , I J k k s x y 5

TEN Model (cont.) TM : z TM I + TM V = = = TM V E u t TM v I H t TM sI TM I J u s s 6

TEN Model (cont.) TE : z TE I + TE V = = = TE V E v t TE I H u t TE sI TE s v I J s 7

Example ( ) = x J , J x y s sx ( , , ) E x y z Find x ( ) ) ( ) ( ( ) = + x v E v , , E k k z u E u x x y t t ) ( ( ) ( z ) ( ) z = x u V + x v TM TE V 8

Example (cont.) y k k x u = = ( ) x cos = , k k k t x y t k k u y x v = = sin v t x Hence k k k k ( ) ( ) z ( ) z y = + TM TE x , , E k k z V V x x y t t k k k k ( ) z ( ) z y = + TM TE v x V J u V J i s i s t t k k k k ( ) ( ) ( ) y = = = = = = u v v x cos sin J J x u J J J J x J J s sx sx sx s sx sx sx t t 9

Example (cont.) Hence k k k k k k k k ( ) ( ) z ( ) z y y = + TM TE x x , , E k k z V J V J x x y i sx i sx t t t t 1 k ( ) z ( ) z = + 2 x 2 y TM TE J k V k V sx i i 2 t or 1 1 k ( ) z ( ) z = + 2 x 2 y TM TE ( , , ) E x y z J k V k V x sx i i 2 2 t (2 ) ( ) + j k x k y e dk dk x y x y 10

Dyadic Greens Function G G G G G G G G G xx xy xz ( ) x y ; , = , G x y z z yx yy yz zx zy zz where ( ) ( ) = , , G E x y z x y z , , due to the unit-amplitude electric dipole at ( ) ( , , J x y z j x = ij i ) ( ) ( ) x y y z z From superposition: Note: We have translational invariance due to the infinite substrate. ( ) ( ) ( ) x y ; , x y z dx dy = , ; , , , ; E x y z z G x y z z J s E E E J J J x sx We assume here that the currents are located on a planar surface z . = = E J where y s sy z sz 11

Dyadic Greens Function (cont.) ( ) ( ) ( ) x y ; , x y z dx dy = , ; , , , ; E x y z z G x y z z J s This is recognized as a 2D convolution: = G J E s Taking the 2D Fourier transform of both sides, = G J E s ( ) G k k z z = , ; , G where x y 12

Dyadic Greens Function (cont.) ( ) G k k z z , ; , = G J E is called the spectral-domain dyadic Green s function. x y s It is the Fourier transform of the spatial-domain dyadic Green s function. Assuming we wish the x component of the electric field due to an x-directed current Jsx (x , y ), we have = E G J x xx sx In order to indentify , we use G xx 1 1 k ( ) ( ) ( ) = + 2 x 2 y TM TE ( , ; , ) E x y z z , , , J k k k V z z k V z z x sx x y i i 2 2 t (2 ) ( ) + j k x k y e dk dk x y x y 13

Dyadic Greens Function (cont.) 1 1 k ( ) ( ) = + 2 x 2 y TM TE ( , ; , ) E x y z z , , J k V z z k V z z x sx i i 2 2 t (2 ) ( ) + j k x k y e dk dk x y x y 1 k ( ) ( ) ( ) = + 2 x 2 y TM TE ( , ; , ) , , , E k k z z k V z z k V z z J k k x x y i i sx x y 2 t = E G J Recall that x xx sx Hence 1 k ( ) ( ) ( ) = + 2 x 2 y TM TE , ; , , , G k k z z k V z z k V z z xx x y i i 2 t 14

Dyadic Greens Function (cont.) We then have: G G G G G G G G G xx xy xz = G yx yy yz zx zy zz 1 k ( ) ( ) ( ) = + 2 x 2 y TM TE , ; , , , G k k z z k V z z k V z z xx x y i i 2 t The other eight components could be found in a similar way. We could also find the magnetic field components. We can also find the fields due to a magnetic current. = EJ xx ( ) G G the component of the electric field due to an electric durrent xx xx 15

Dyadic Greens Function (cont.) The different types of spectral-domain dyadic Green s functions are: EJ ij G Gives electric field due to electric current EM ij G Gives electric field due to magnetic current HJ ij G Gives magnetic field due to electric current HM ij G Gives magnetic field due to magnetic current Note: There are 36 terms here, though many are equal by reciprocity or symmetry. There are 20 unique terms (five from each type of Green s function). 16

Summary of Results for All Sources These results are derived in Notes 44. = = = = TM V I V E H E H u = TE s = V M TM s I J TM su su v k k TE = + = + TE s TM s t I J M t V M J v sv sz sv sz TE I u Definition of vertical planar currents : ( ( ) ) ( ) = , , x y z , ( ) z J J x y z sz ( ) = , , x y z , ( ) z M M x y z sz 17

Sources used in Modeling Vi = voltage due to 1[A] parallel current source Ii = current due to 1[A] parallel current source Vv = voltage due to 1[V] series voltage source Iv = current due to 1[V] series voltage source TM : z TM I TM I + TM + V TM V + - TM s V TM sI ( ) z ( ) z ( ) z V ( ) z V ( ) z ( ) z ( ) z I ( ) z I = TM TM v TM s V V = TM TM TM s V V i = TM TM v TM s = I I TM TM i TM s I I 18

Sources used in Modeling (cont.) Vi = voltage due to 1[A] parallel current source Ii = current due to 1[A] parallel current source Vv = voltage due to 1[V] series voltage source Iv = current due to 1[V] series voltage source TE : z TE TE I I + TE + TE V V + - TE s V TE sI ( ) z ( ) z ( ) z I ( ) z I ( ) z ( ) z ( ) z V ( ) z V = TE TE TE s = V V TM TE v TE s V V i = TE TE i TE s I I = TE TE v TE s I I 19

Example HM xx G Find This is the xx component of the spectral-domain dyadic Green s function that is used to obtain Hx from Msx. Start with: ( ) ) ( ) ( ) = + x u H u v H v , , H k k z x x y t t ( ) ( ( ) ( z ) ( ) z = x u I + x v TE TM I Recall: ( ) ( ( ) ) ( ) ( ) z ( ) ( ) z = TM u = , , V z E k k z TE u , , I z H k k z t x y t x y ( ) = TE v = , , V E k k z TM v , , I H k k z t x y t x y 20

Example (cont.) ( ) ( ) ( = = ) ( ) ( ( ) ( ( ) ( )( ( )( ( ) z ) ( ) z x u I + x v TE TM , , H k k z z I x x y ( ) ( ) z V ( ( ( ( ) ) ) ) ) ( ) z + TE TM cos sin I z I ( ) = + TE v TE s TM v TM s cos sin I z V I ) ( ( )( ) I ) ) = + TE v TM v cos sin I z M I z M su sv ( ) ) ( ( ) z ( ) ( ) z = + TE v TM v cos cos sin sin I z M M sx sx ( ) = + 2 2 TE v TM v cos sin M I I sx 2 x 2 x k k k k ( ) z ( ) z = + TE v TM v M I I sx 2 t 2 t Recall: = TE s = TM s V M I J su su k k = + = + TM s TE s t V M J t I J M sv sz sv sz 21

Example (cont.) 1 k ( ) ( ) ( ) z ( ) z = + 2 x 2 y TE v TM v , , H k k z M k I k I x x y sx 2 t Hence, we have 1 k ( ) ( ) ( ) = + 2 x 2 y HM xx TE v TM v , ; , , , G k k z z k I z z k I z z x y 2 t Note: The notation (z,z ) has been used in the final result to emphasize that the terms depend on both z and z . 22

Summary of Spectral-Domain Recipe Example (Ex from Jsx) ( ) , J x y Start with a given planar current distribution. sx ( ) , J k k Take the Fourier transform of the surface current. sx x y Use the appropriate spectral-domain dyadic Green s function to find the Fourier transform of the field of interest. ( ) = EJ xx , E G J k k x sx x y Take the inverse Fourier transform (2D integral in kx and ky) to find the field of interest in the space domain. 1 ( ) + j k x k y = EJ xx E G J e dk dk x y x sx x y 2 (2 ) 23