19.电阻矩阵+地损耗.doc

X RESISTANCE MATRIX If both dielectric and conducting losses exist in a transmission line the telegraph equation system will be revised as zVCjG dz zId zILjR dz zVd where the resistance matrix accounts for the conducting loss R Let the guided current and voltage waves in z direction in a transmission line be expressed as zIIIIzI T Mc exp 21 zVVVVzV T Mc exp 21 where is the number of conductors c M and are the complex current amplitude vector and complex voltage I V amplitude vector respectively and are the current and voltage at the i th conductor respectively i I i V is the propagation constant of the transmission line j is the attenuation constant and is the phase constant Substitution of these two expressions into the telegraph equations gives zVCjGzI exp exp zILjRzV exp exp or VCjGI ILjRV That will lead to IILjRCjG 2 VVCjGLjR 2 If the transmission line is lossless i e jjGR 0 then 0 2 0 2 IjILCj 0 2 0 2 VjVCLj giving the following eigenvalue equations 0 2 0 IvILC p 0 2 0 VvVCL p where and represent respectively the complex current amplitude 0 I 0 V vector and the complex voltage amplitude vector for the lossless line and these two vectors should be real since the transmission line does not have any losses denotes the phase velocity of the guided wave along the line p v p v Since the transmission line is composed of conductors these eigenvalue c M equations will produce eigenvectors for current and c M c M III 0 2 0 1 0 eigenvectors for voltage c M c M VVV 0 2 0 1 0 T c iiii MIIII 0000 2 1 T c iiii MVVVV 0000 2 1 c Mi 2 1 Since and are real symmetrical the matrices and L C LC CL should be mutually transpose and complex conjugate In fact CLLCLCLC T TTTT where the superscript T symbolizes the transpose and the conjugate It follows that the eigenvectors of and the eigenvectors of LC CL should be mutually orthogonal namely bi orthogonal ij i T ij T ii T jji VIVIIVVI 00000000 where the is Kronecker Delta ij ji ji ij 0 1 It also follows that the eigenvalues of and the eigenvalues of LC CL should be mutually complex conjugate Since the phase velocity is p v real these two matrices and possess the same eigenvalues LC CL cppp Mvvv 222 2 1 Therefore this transmission line system composed of conductors and c M one ground plane possess propagating modes c M In a lossless line system all eigenvectors either and c M III 0 2 0 1 0 should be real c M VVV 0 2 0 1 0 For a lossless system 0RG jj and I 0 I V 0 V the earlier mentioned formulas VCjGI ILjRV are reduced to 00 p IvC V 00 p VvL I If dissipation is involved in a transmission line then the complex power propagated along the longitudinal direction i e axis is z VIIVP T M m mm c 1 and the average power is VIPP T T ReRe The power loss per unit length of the line system is z P P T L z P Re V z I z V I T T Re VVCjGILjRI TT Re VCjGVILjRI T TT Re VGVIRI TT dc PP where and denote respectively the conducting loss per unit length c P d P and the dielectric loss per unit length IRIP T c VGVP T d It will be proven below that the attenuation constant in a dissipated transmission line is given by dc T d T c T L P P P P P P 222 Re where and indicate the attenuation constant for conductors and the c d attenuation constant for dielectrics respectively VI IRI P P T T T c c Re2 2 2 2Re T d d T T VG V P P IV Proof Since zIIIIzI T Mc exp 21 zVVVVzV T Mc exp 21 then 1 exp 2 c M TT mm m PV IIVIVz ReReReexp2 TT T PPIVIVz z P P T L Re2exp22 T T IVzP hence 2 2Re2Re TT L cd TT T IR IVG V P P IVIV To calculate the resistance matrix the attenuation constant for conductors is consider only VI IRI P P T T T c c Re2 2 The numerator in this quotient is the dissipated conducting power per unit length which according to the perturbation theory is approximately c P equal to IRIP T c 2 1 c j M sz j l j RJdl is the number of conductors c M is the contour of cross section of the i th conductor j l is the surface resistance of conductors s R f Rs is the current density of the j th conductor which is approximately z j J equal to 00 1 zF jj J 1 2 c jM is the free charge density on the j th conductor F j If each of conductors is driven by the elements of the i th eigenvector c M here represents 0000 1 2 iiii c VVVVM 0 i Vj1 2 c jM the voltage between the j th conductor and the ground then the free charge density on the j th conductor can be determined in a way given i F j earlier and the surface current density flowing on the j th conductor i z j J can be found in an above mentioned formula 00 1 zF jj J 1 2 c jM The average power transmitted along the transmission line appeared in T P the quotient expression of is approximately equal to c 00000 ReRe TTT TT PIVPIVIV Therefore the attenuation constant for conductors becomes c 2 2Re c c T T T P P IR I IV 00 1 2 2VI dlJR T M j l js c j Since there exist current and voltage modes in the transmission line c M and then there are attenuation c M III 0 2 0 1 0 c M VVV 0 2 0 1 0 c M constants given by c M ccc 21 2 1 00 2 2 c j M i szi l j ij c cT i ii T RJdl P P IV 1 2 c iM To calculate the resistance matrix alone a transmission line is assumed to have conducting loss only viz the telegraph equation for voltage 0 G VVCjGLjR 2 is reduced to VjVCjLjR 2 For a low loss transmission line the above equation becomes 0 VV 0 2 0 VjVCjLjR and the equality of imaginary parts of both sides gives rise to 0 2 0 Im ImVjVCjLjR namely 00 2 VVCR It is inferred from that 00 p IvC V 00 2R IV If the dielectric loss is not considered for the time being then the 0 G attenuation constant for dielectrics becomes zero 0 2Re T d T VG V IV and the attenuation constant is simplified as cdc 00 1 2 2VI dlJR T M j l js c j As a result the equation is transferred as 00 2R IV 00 2 VIR c As mentioned above there are modes for which c M c c M ccc 21 give rise to equations c M c ii c i MiVIR 2 1 2 00 are modal currents given by c M III 0 2 0 1 0 0 2 0 IvILC p are modal voltages given by c M VVV 0 2 0 1 0 0 2 0 VvVCL p are modal attenuation constants given by c M ccc 21 i c 2 1 00 2 c j M i sz l j j T ii RJdl IV 1 2 c iM are modal current densities given by 12 c M zzz JJJ 00 1 ii zF jj J 1 2 c i jM are modal charge densities determined by putting each 12 c M FFF element of modal voltage to the each conductor 0 i V1 2 c iM Each of the following matrix equations c M c ii c i MiVIR 2 1 2 00 contains algebraic equations and therefore there are equations c M 2 c M Let the first matrix equation becomes1i 111 00 2 c R IV where 11121 21222 12 c c cccc M M MMM M RRR RRR R RRR 1111 0000 1 2 T c IIIIM 1111 0000 1 2 T c VVVVM in which the first equation of the first matrix equation is 11111 11 0120100 1221 c Mcc R IR IRIMV Take all the first equations for each of the above matrix 1 2 c iM c M equations are similarly given by 11111 11 0120100 1221 c Mcc R IR IRIMV 22222 11 0120100 1221 c Mcc R IR IRIMV 11 0120100 1221 ccccc c MMMMM Mcc R IR IRIMV The simultaneous equations can be solved for the first row of the resistance matrix R 11121 c M RRR All other rows of can be determined in a similar way and therefore the R resistance matrix is finally found R XI GROUND LOSS If there are two ground planes in a transmission line system then the upper ground is considered to be an additional conductor and its loss computation has been described previously Therefore only one ground plane need to be considered here The z directed narrow strip current of the j th conductor produces a jj Jl magnetic field on the ground plane 1 2 2 jj jj j Jl HHjN The magnetic field has a component tangential to the ground plane cos 2 jjj jtjxjj jj Jly HxHxHx j j j j jj xy zj Jl j H 0 i x x y Ground Loss The current of all conductors will produce a tangential magnetic field at the point on the ground 0 i x 11 2 2 11 1 0 2 NN jjj tijt jj ijj Jl y HxHx xxy where is the number of subsections for all conductor to dielectric 1 N interfaces In accordance with the method of image the total tangential magnetic field at the point due to all conductors will be doubled 0 i x 1 2 2 1 1 02 0 N jjjg titi j ijj Jl y HxHxx xxy It follows from the boundary condition that the current density at is 0 i x 1 2 2 1 1 0 0 N jjjg giti j ijj Jl y JxyHxz xxy In addition to the conducting loss due to all conductors c P 2 1 c j M sj l j RJ dl the following ground loss needs to be added 2 g sg l RJ dl where is the width of ground plane which is usually taken as 5 to 10 times g l of the transverse dimension of the transmission line system After the ground loss is computed the total power dissipated per unit length is given by c P 22 1 c jg M sjsg ll j RJ dlRJ dl and the resistance matrix is calculated in the same way as described above Proposed homework 1 Calculate the resistance per unit length of a parallel wire transmission line and compare the calculated result with that given by 2 2 1 s RD d R d D d where d is the diameter of the wire and D is the