Slide 1OverviewPatch FieldsPatch Fields (cont.)Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 281Spring 2011Notes 20ECE 6345ECE 6345Prof. David R. JacksonECE Dept.2OverviewOverviewIn this set of notes we apply the SDI method to investigate the fields produced by a patch current.We calculate the field due to a rectangular patch on top of a substrate.  We examine the pole and branch point singularities in the complex plane.  We examine the path of integration in the complex plane.3Patch FieldsPatch FieldsWxyLsxJzxreh10ˆcossxxJ xLp� �=� �� �LFind Ex (x,y,0)4Patch Fields (cont.)Patch Fields (cont.)From Notes 19 we have)()(~22zVkkzVkkGTEityTMitxxxwhere( )( ) ( )( )21, , ,2x yj k x k yx xx x y sx x y x yE G k k z J k k e dk dkp+�+�- +- �- �=��%%( )2 2cos2, sinc2 22 2xsx x y yxLkWJ k k LW kLkpp� �� �� �� �� � � �� �� �=� � � �� �� � � �� � � �-� �� � � �� � � �� �%For the patch current we have5Patch Fields (cont.)Patch Fields (cont.)From the TEN we have( )0 1 1(0) (1) (0)1(0)1cotT Ti inTinT TzV ZYY jY k h===-111000111000zTEzTEzTMzTMkYkYkYkY( )1 22 20 0=/z tk k k-( )1 22 21 1=/z tk k k-( )1 22 2 2=/t x yk k k+0Z1Z1Amphz(TMz or TEz)T denotes TM or TE+-( )0V6Patch Fields (cont.)Patch Fields (cont.)Define the denominator term as ( )( )0 1 10 1 1( ) cot( ) cotTM TM TMt zTE TE TEt zD k Y jY k hD k Y jY k h= -= -( ) (0)T Tt inD k Y=so that7TEtyTMtxxxDkkDkkG11~22( )( )( )( )2 221 1 1, , 02,x yyxxTM TEt tj k x k ysx x y x ykkE x yk D k DJ k k e dk dkp� +�- �- �- +� �� � � �� �= - +� � � �� �� � � �� ����%Patch Fields (cont.)Patch Fields (cont.)We then haveThe final form of the electric field at the interface is then8ddkkdkdkttyx( )( )( )( )( ) ( )( )20 0/20 04 cos cosx y x yj k x k y j k x k yx y t tx y t te dk dk e k dk dk x k y k dk dppff� +� +�- + - +- �- �+�==�� ����K KKPolar CoordinatesPolar CoordinatesUse the following change of variables:xkykftkcosxtkkf =sinytkkf =Advantage: The poles and branch points are located at a fixed position in the complex kt plane.9( )( )( )( )/22 220 01 1 1, , 0 , cos sin( ) ( )cos cosx sx tTM TEt tx y t tE x y J kD k D kk x k y k dk dpff fpf�� �=- +� �� ����%( )/ 20 0,x t tE F k dk dpff�=��Polar Coordinates (cont.)Polar Coordinates (cont.)we have( )( )( )( )2 221 1 1, ,0 ,2x yj k x k yyxx sx x y x yTM TEt tkkE x y J k k e dk dkk D k D%p� +�- +- �- �� �� � � �� �= - +� � � �� �� � � �� ���UsingThis is in the following general form:10PolesPoles0)(0)(tTEtTMkDkD  TEtptTMtptkkkkPoles occur when either of the following conditions are satisfied:TMz:( )0 1 10cot 0TMTM TMzDY jY k h=� - =11Poles (cont.)Poles (cont.)( ) ( )( ) ( )0 00 0V VI I+ -+ -==( )( )00VYI++=r( )( )00VYI--=-sHenceYYThis coincides with the well-known Transverse Resonance Equation (TRE) for determining the characteristic equation of a guided mode.( )0 1 1cot 0TM TMzY jY k h- =(Kirchhoff’s laws)so that1TMZ0TMZTMYsTMYr( )I z+-( )V z0z =( )01 22 20 0=/z TMk kβ-( )01 22 21 1=/z TMk kβ-12Poles (cont.)Poles (cont.)Comparison:( )0 1 1cot 0TM TMzY jY k h- =00111TMzTMzYkYkwewe==( )0 1 1cot 0TM TMzY jY k h- =000111TMzTMzYkYkwewe==( )1 22 20 0=/z tk k k-( )1 22 21 1=/z tk k k-( )01 22 20 0=/z TMk kβ-( )01 22 21 1=/z TMk kβ-TRE (surface-wave mode)Poles in kt plane(A similar comparison holds for the TE case)13Poles (cont.)Poles (cont.)Hence, we have the conclusion thatTMtp TMTEtp TEkkbb==That is, the poles are located at the wavenumbers of the guided modes (the surface-wave modes).14Poles (cont.)Poles (cont.)0TMbRetkImtk0TMb-0k1k0k-1k-The complex plane thus has poles on the real axis at the wavenumbers of the surface waves.( )1/22 2=t x yk k k+15Path of IntegrationPath of IntegrationThe path avoids the poles by going above them.RetkImtk0k1kCRetkImtk0k1kCRhRLlossy caselossy caseThis path can be used for numerical computation.16The path avoids the poles by going above them.lossless caseRetkImtk0k1kCRhRL0= 0 05Rh . k(typical choices)( )1= 1 1RL k .Practical note: If hR is too small, we are too close to the pole. If hR is too large, there is too much round-off error due to exponential growth in the sin and cos functions.Path of Integration (cont.)Path of Integration (cont.)17( ) ( )( )0 1 10110 1cotcotTM TM TMt zzz zD k Y jY k hj k hk kwewe= -� �� �= -� �� �� �� �( )( )1/ 22 20 01/ 22 21 1z tz tk k kk k k= -= -changes0 0( )TMz z tk k D k� -Branch PointsBranch PointsTo explain why we have branch points, consider the TM function:withIf Note: There are no branch cuts needed for kz1 (the function DTM is an even function of kz1).(We need branch cuts for kz0.)18( )( ) ( )( ) ( )( ) ( )( )1/ 22 20 01/ 2 1/ 20 01/ 2 1/ 20 01/ 21/ 20 0z tt tt tt tk k kk k k kj k k k kj k k k k= -= - +=- - +=- - - -Branch Points (cont.)Branch Points (cont.)Note: The representation of the square root of –1 as –j is arbitrary here.RetkImtk0k0k-0tk k-tk( )0tk k- -19( ) ( )( )( )1 21/ 21/ 20 0 0/ 2 / 20 0z t tj jt tk j k k k kj k k k k e eff=- - - -=- - - -Branch Points (cont.)Branch Points (cont.)RetkImtk0k0k-tk1f2fNote: The shape of the branch cuts is arbitrary, but vertical cuts are shown here.Branch cuts are necessary to prevent the angles from changing by 2 :Note: The branch cuts should not cross the real axis when there is loss in the air (the integrand must be continuous).20( ) ( )( )1/ 21/ 20 0 0z t tk j k k k k=- - - -Branch Points (cont.)Branch Points (cont.)We obtain the correct signs for kz0 if we choose the following branches:( )0/ 2 Arg 3 / 2tk kp p- < - <( )( )03 / 2 Arg / 2tk kp p- < - - <RetkImtk0k0k-0zk =+0( )zk j=- +0( )zk j=- +The wave is then either decaying or outgoing in the air region when we are on the real axis.21Branch Points (cont.)Branch Points (cont.)The wavenumber kz0 is then uniquely defined everywhere in the complex plane:RetkImtk0k0k-tk1f2f( ) ( )( )( )1 21/ 21/ 20 0 0/ 2 / 20 0z t tj jt tk j k k k kj k k k k e eff=- - - -=- - - -1`2`/ 4/ 6f