Slide 1OverviewCavity ModelCavity Model (cont.)Slide 5Eigenfunction ExpansionSlide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Maxwell CurrentUniform CurrentUniform ModelUniform Model (cont.)Probe InductanceRLC ModelRLC Model (cont.)Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 291Spring 2011Notes 26ECE 6345ECE 6345Prof. David R. JacksonECE Dept.2OverviewOverviewIn this set of notes we introduce the cavity model (method of eigenfunction expansion) to solve for the input impedance of the rectangular patch antenna.3Cavity ModelCavity ModelLeWexy0 0( , )e ex yPMC0 00 0eex x Ly y W= +D= +D0effe rck k e=Let( )1effrc r effj le e�= -1 1 1 1effd c sp swlQ Q Q Q= + + +Assume no z variation (the probe current is constant in the z direction.)Note: L is from Hammerstad’s formula W is from Wheeler’s formula4Cavity Model (cont.)Cavity Model (cont.)Next, we derive the Helmholtz equation for Ez.Substituting Faradays law into Ampere’s law, we haveicH J j EE j HwewmѴ = +Ѵ =-( )( )( )22 22 21icieieieE J j EjE j J k EE E j J k EE k E j Jwewmwmwmwm- Ѵ Ѵ = +Ѵ Ѵ =- +� �� - � =- +� + =5Cavity Model (cont.)Cavity Model (cont.)HenceDenote2 2 iz e z zE k E j Jwm� + =2 2( , ) ( , )( , )zex y E x yk f x yyy y=� + =where( , ) ( , )izf x y j J x ywm=Then6Eigenfunction ExpansionEigenfunction ExpansionIntroduce “eigenfunctions”For rectangular patch we have, from separation of variables,( , )mnx yy2 2( , ) ( , )mn mn mnx y x yy l y� =-0mnCny�=�2 22( , ) cos cosmne emne em x n yx yL Wm nL Wp pyp pl� � � �=� � � �� � � �� �� � � �� �= +� � � �� �� � � �� �7Cavity Model (cont.)Cavity Model (cont.)Assume an “eigenfunction expansion”Hence,( , ) ( , )mn mnm nx y A x yy y=�2 2( , )ek f x yy y� + =2 2, ,( , )mn mn e mn mnm n m nA k A f x yy y� + =� �Using the properties of the eigenfunctions, we have( )2 2,( , ) ( , )mn e mn mnm nA k x y f x yl y- =�This must satisfy8Cavity Model (cont.)Cavity Model (cont.)Multiply by and integrate.Note that the eigenfunctions are orthogonal, so thatDenote' '( , )m nx yy' '( , ) ( , ) 0 ( , ) ( ', ')mn m nSx y x y dS m n m ny y = ��2, ( , )mn mn mnSx y dSy y y< >=�( )2 2, ,mn e mn mn mn mnA k fl y y y- < >=< >We then have9Cavity Model (cont.)Cavity Model (cont.)Hence, we haveFor the patch we then have2 2,1,mnmnmn mn e mnfAkyy y l� �< >=� �< > -� �2 2,1,iz mnmnmn mn e mnJA jkywmy y l� �� �< >=� �� �< > -� �� �The field inside the patch cavity is then given by,( , ) ( , )z mn mnm nE x y A x yy=�10Cavity Model (cont.)Cavity Model (cont.)***,1( , )21( , )212iin z zViz zSimn mn zm nSP E x y J dVh E x y J dSh A J dSy=-=-=-����To calculate the input impedance, we first calculate the complex power going into the patch as11Cavity Model (cont.)Cavity Model (cont.)*,*2 2,22 2,1,2,1 1,2 ,,1 12 ,iin mn mn zm niimn zmn zm nmn mn e mnimn zm nmn mn e mnP h A JJh j JkJh jkyywm yy y lywmy y l=- < >� �� �< >=- < >� �� �< > -� �� �� �< >� �� �=-� �� �< > -� �� ����orAlso,212in in inP Z I=so22inininPZI=12Cavity Model (cont.)Cavity Model (cont.)Hence we havewhere222 2,,1 1,imn zinm nmn mn e mninJZ j hkIywmy y l� �< >� �� �=-� �� �< > -� �� ��, 0 0m n m n� �= ==� ��13Cavity Model (cont.)Cavity Model (cont.)Rectangular patch:where2 220cos cosmne emne eeffe rcm x n yL Wm nL Wk kp pyp ple� � � �=� � � �� � � �� � � �= +� � � �� � � �=( )1effrc r effj le e�= -2 20 0, cos cose eL Wmn mne em x n ydx dyL Wp py y� � � �=� � � �� � � �� �14Cavity Model (cont.)Cavity Model (cont.)soTo calculate , assume a strip model as shown below.( ) ( )0 0, 1 12 2e emn mn m nW Ly y d d� �� �= + +� �� �� �� �01, 00, 0mmmd=�=���,imn zJypa0 0( , )e ex y0 0( , )e ex ypW15Maxwell CurrentMaxwell Current( )0 0220, ,2 224p pe einszpep pW WIJ y y yWy yW ap� �= � - +� �� �� �- -� �� �=For a “Maxwell” strip current assumption, we haveNote: The total probe current is Iin.16Uniform CurrentUniform Current0 032, ,2 24.482p pe einszpp p pW WIJ y y yWW a e a� �= � - +� �� �= BFor a uniform strip current assumption, we haveNote: The total probe current is Iin.17Uniform ModelUniform Model( ) ( )00202200220 0 021, cos coscos cos ' 'cos cos cos ' sin sin ' 'pepeppppWyeimn z inWe e pyWeeinWp e eWe e einWp e e einm xn yJ I dyL W WI m xny y dyW L WI m x n y n yy y dyW L W WIWppyppp p p+-+-+-� � � �=� � � �� � � �� � � �� �= +� � � �� �� � � �� � � � � �= -� � � � � �� � � � � �=���0 0cos cos sinc2e eppp e e en Wm x n yWL W Wpp p� �� � � � � �� �� � � � � �� � � � � �� �Assume uniform strip current model:0ey y y�= +UseIntegrates to zero18Uniform Model (cont.)Uniform Model (cont.)Hence0 0, cos cos sinc2e epimn z ine e en Wm x n yJ IL W Wpp py� � � � � �=� � � � � �� � � � � �sinc2pen WWp� �� �� �Note: It is the term that causes the series for Zin to converge.We cannot assume a prove of zero radius, or else the series will not converge – the input reactance will be infinite.19Probe InductanceProbe InductanceHenceNote that(1,0) = term that corresponds to dominant patch mode current (impedance of RLC circuit).222 2( , )(1,0),1 1,imn zpm nmn mn e mninJjX j hkIywmy y l�� �� �� �=-� �� �-� �� �� ��222 2,,1 1,imn zinm nmn mn e mninJZ j hkIywmy y l� �< >� �� �=-� �� �< > -� �� ��20RLC ModelRLC Modelwhere,,m nin inm nZ Z=�22,1,imn zmnmn mninJP hIymy y=(These coefficients are not a function of frequency.)We can write,2 2m nmnine mnPZ jkwl� �=-� �-� �21RLC Model (cont.)RLC Model (cont.)Assume a hypothetical lossless substrate:Comparing, we have the conclusion that2 20mn mn mny l y� + =0real number:mnrc r mn rk k ke e e� �= = = =mn mnkl =2 20mn mn mnky y� + =Helmholtz equation:Eigenvalue equation:( , )mnkm n=wavenumber of resonant patch mode for a lossless substrate22RLC Model (cont.)RLC Model (cont.)( )( )( ),2 22 22 2 22 2 21m nmnine mnmneff mnmnmn effmneff mnPZ jk kPjk …