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# Class Note for ECE 6341 at UH

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Date Created: 02/06/15
ECE 6345 Fall 2006 Prof David R Jackson ECE Dept Notes 20 Overview In this set of notes we apply the SDI method to find the field of a finite current sheet Start with the field produced by an infinite phased current sheet derived in Notes 19 Apply superposition to find the field of the finite current sheet Fourier integral Identify a closedform expression for the Fourier transform of the field Identify a TEN model where voltage and current now represent the Fourier transform of the field quantities As an example calculate the field due to a rectangular patch on top of a substrate Finite Source For the infinite current sheet we have 1 x y 10 This produces a field 9 jkxxkyy E x Ewe Note the i subscript on the voltage Where functions denotes the p A A voltage due to a one EIO ZEuo XEvo Amp parallel current source WM 2 VTE 2 Q Vim z 150 12 V1TEZ170 9 Finite Source cont Recall that 150 202 is kwwkxnky Hence p A M 1 N A Em ZVi Z lt2ngt2 1s kxm kyngt39u AkxAky er Z 27 2 is Q AkxAky where Finite Source cont Adding the contributions from all the phased current sheets we have QKTMZZsk k M xm yn 22 m oo n oo xm yn any 1 e jkmquotkwy Akx Aky KTEZZsk k 4 Taking the limit as Finite Source cont From this we can identify From this we can make the following TEN identifications VTMZ2Etkx yjz ISTMz 12j k k s x9 y We Etkxkyz 13132 9110 k x9 y TEN The TEN models are shown below Patch Fields 10 A x x isx 1 cos L Z I isx L 8 Iquot 39l A Patch Fields cont De ne Then Patch Fields cont 11 Patch Fields cont The spectraldomain Green s function is the Fourier transform ofthe spatialdomain Green s function Ex j ijx x x y39 y zz Jsx ltx y gt dx dy 700 700 Gxx JSX From the convolution property of Fourier transforms 12 Patch Fields cont Note more generally TEN o o T TM orTE 13 Patch Fields cont From the TEN we have W 0 1 Z 0 1 K 0 1 YOT leT cotkzlh TM 080 Y0 k 20 TM 081 Y1 k 21 k TE 0 IO Z 010 k TE 1 fl Z 011 k2 k3 k212 14 Patch Fields cont Define the denominator term as DT k Y 0 so that Wag YJM mm cotkzlh DTEUc Y0 1Y1 cotkzlh 15 We then have Patch Fields cont 16 00 l Polar Coordinates Use the following change of variables OO 00 ky dkx dky kt dkt d kt TdkxdkyTTktdktdlt3 I 4 ktdktd3 0055 kt s1n y 17 Polar Coordinates cont K W Using OOOO II CD CD 1 22 k Ex xy0 I t we have Poles Poles occur when the either of the following conditions are satisfied DTM kt 0 kt kt DTEUQ o k 2 kg YOTM jY1TM cotkzlh 0 19 Poles cont This coincides with the wellknown Transverse Resonance Equation TRE for determining the characteristic equation of a guided mode 1 V t V V V 7 Y T gt Kirchhoff s laws I I 1 Hence YTM YTM so that c 20 C Poles cont Comparison poles TRE surfacewave mode YBTM Z 0 7OTM COtkZIh 0 YOTM 050 YTM 080 kZO kzo YTM 051 TM we 1 kZI Y1 kzll 12 12 kzo kg kf kzo 2 5M0 12 2 2 12 kz1 612 kzl k1 TMO 21 A similar comparison holds for the TE case Poles cont Hence we have the conclusion that That is the poles are located at the wavenumbers of the guided modes the surfacewave modes 22 Poles cont The complex plane thus has poles on the real axis at the wavenumbers of the surface waves Imk t t X t I X I gt RCk 23 Path of Integration The path avoids the poles by going above them lossy case lossy case This path may be used for numerical computation 24 Path of Integration cont The path avoids the poles by going above them lossless case hR 005k0 Practical note If hR is too small we are too close to the pole If hR is too large there is too much roundoff error due to exponential growth in the sin and cos functions LR k1 11 typical choices 25 Branch Points To explain why we have branch points consider the TM function DTM t W ijM comm 080 081 t k h kw Jk210021 26 Branch Points cont kzo k5 k 1 12 12 2 k0 kt k0 k Note the representation of 12 12 Jkk0 k k0 thesquarerootof 1as le U2 U2 arbitrary here jkt k0 kt k0 1m kt kt m kt 4g k0 Re kt k0 k0 27 Branch Points cont kzo kt k0 12 kt k0 12 Jgk k0 k k0 em em Branch cuts are necessary to prevent the angles from changing by 27 Note the shape of the branch cuts is arbitrary but vertical cuts are shown here Rek 28 Branch Points cont kzo kt k0 12 kt k0 12 We obtain the correct signs for kzo if we choose the following principal branches 29 Branch Points cont The wavenumber kzo is then uniquely defined everywhere in the complex plane kzo kt k0 12 kt k0 12 Jkt k0 k k0 em eW 30 Riemann Surface The Riemann surface is a pair of complex planes connected by ramps where the branch cuts used to be The angles change continuously over the surface All possible values of the wavenumber are found on the surface Imk t gw z 1 Rek Example top sheet 17r4 27r6 bottom sheet 1 7r 4 27 2 7r6 31 Riemann Surface cont Riemann surface forz 2 j top sheet re y z r o 7Z39lt lt7Z39 bottom sheet X top view 7 lt lt 37 Note a horizontal branch out has been arbitrarily Chosen 32 Riemann Surface cont Riemann surface forz 2 X 3D view top bottom DC x D B side view top view 33 Sommerfeld Branch Cuts Sommerfeld branch cuts are a convenient choice for theoretical purposes discussed more in ECE 6341 1m kzo 0 on branch cut Imk t k 0 k02 k2 2 1m kzo lt 0 top sheet Z t Im kzo gt 0 bottom sheet Rek t k0 k0 34 Complex Plane Ex ff218 d 1aFkta dict top sheet 35

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