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# Class Note for ECE 6340 with Professor Jackson at UH

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COURSE
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TYPE
Class Notes
PAGES
45
WORDS
KARMA
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This 45 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 13 views.

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Date Created: 02/06/15
Wave Impedance Assume TMZ wave traveling in the 2 direction From previous notes So x j ZW gtlt2gtltH l or 2gtltEl Z7M H l Wave Impedance cont ijZ jkzz 2 wave 2 gt kZ because 6 gt e sign Zwave sign Zwave Wave Impedance cont TEZ wave 5 Transverse Equivalent Network TEN Start With Where VZ Ve jkzz V ejkzz Note VZ behaves as a voltage function Also write Emmy mxy Macy assume TMzwavegt Note We can assume that both the unit vector and the transverse amplitude function are both real because of the real theorem and its corollary 6 Transverse Equivalent Network cont For the magnetic field we have fi 1 sz gt H 1 AfoEXEI t TM Z Now use Elxy z 2 gm xy many VZ VZ Vejkzz V e ikzz Hence 1 A A 39kz it ZTM gngMWtx9y V e Jz 1 ZTM jkzz gemxx V e 7 Transverse Equivalent Network cont Define and Note 2 behaves as a current function We then have 8 Transverse Equivalent Network cont Summary 9 Equivalent cont v Waveguide I Z D C ZOZTMorZTE kz kz of waveguide TL VZ Z Transverse Equivalent Network cont Note on Power PTEN 2 1V1 2 WG 1 x l X Sz 2E H 25 Ht mxmwxnwmz x2TMltxygtwtltxygtIltzgtz 1 quot A A 3V ll tXy2 ETM XhTM 1 QM XETM39Z TMXZX TM39 5 5Wquot thltxygt2 11 Transverse Equivalent Network cont The total complex power is then PWG j SZWGdS V1Jltx y2 dS S S or If we Choose J ll txy2dS 1 5 12 TEN Junction 39 80 8r JUHCtIOHI I W Z At Z 0 W400 memo imam Hence V0 Var 10 10 10 IO gt gt These conditions are also TM 7 TM true ataTLjunction Zoo vkzo V0 V0 Zm vkzl 13 Example Rectangular Waveguide TE10 mode incident I 7 i 8 quot L 9 Z 0 I r I I I TEN Z k 207191 007 20 gtAgtT 4 14 Example Rectangular Waveguide cont am am 25 k0 0 2 20 7139 k02 j Similarly a wyOko 770 ZTE 771 01 770 376730313461771 Q 15 Example Numerical Results Xband waveguide f 10 GHz Choose 8r22 te on fc 6749 GHZ air lled guide 255 51025 Q 255 28518 Q The resu39ts are 16 Example cont Pf 100 M2 100 0282952 800 3 100 800 920 Note 13 q 100 T2 This is because the impedances of the two guides are different 17 Example cont Alternative calculation 2 Vt TE TE 132100 22012 100 2231 V Ill 22 22on 18 Matching Elements A couple of commonly used matching elements 1 y TEN 0 l I l ZOTE x Inductive post i y TEN C TE Z0 x Capacitive diaphragm 19 Diseointiinuty Rectangular waveguide with a post a radius of post top View higher order mode region 20 Discontinuity cont top view Z higherorder mode region 0 The TL discontinuity is chosen to give the same TEN 3 ZOTE L ZOTE reflected and transmitted TE1O waves as in the 0 actual waveguide Note the discontinuity is a approximately a shunt load because the tangential electric field is approximately continuous while the tangential magnetic field is not Discontinuity cont Flat strip model top view Narrow strip model for post w 4a Assume center of post is at x x0 In this model the equivalent circuit is exactly a shunt inductor 22 top view Discontinuity cont X Einc sin je jkioz y E 2 field radiated by strip current that is the field scattered by the strip 23 Discontinuity cont Field radiated by the post current noy variation sca 00 m x 39 moz Ey xyZ 2A SIDK 6 sz By symmetry the m1 a scattered field should 00 m x have no y variation mOZ cha xy z 2A s1n je kz m1 a 2 12 kmO k2 sca 00 M x jkmOZ z 0 Cl Hx xyzZ TE sm 6 z quot11 Zm0 a 1 r A r A m7rx moz i ZXE Him xyZZZ Esm a 6sz t 2 t quot11 MO The field has been represented in terms of TEmO waveguide modes therefore there is only ay component of the electric field Note that TMm0 modes do not exist 24 Discontinuity cont top view From boundary conditions EyltzogtEyltzo Hence Ejm z 0 Eym z 2 0 25 Discontinuity cont Esca Z 2 0 2 cha Z Z 0 y EA sinpmrx 2 SA sinpmrx m1 m1 a a Equate terms of the Fourier series Al A1 A1 Eca10 0 Eca10 0 E3 0 E3 0 Modeling equa on 26 Discontinuity cont This establishes that the circuit model for the strip must be a parallel element 27 Discontinuity cont Magnetic field Hxz0 Hxz 0 sz x From the Fourier series for the magnetic field we thus have Z i A Asinm xjJWltxgt Represent the strip current as sz x ijm sinmxj jm 2sz x sinm xjdx m0 28 Discontinuity cont Therefore Hence In order to solve forjm we need to enforce the condition that Ey 0 on the strip 29 Discontinuity cont Assume that the strip is narrow so that a Maxwell function describes accurately the shape of the current on the strip 10 is the unknown 30 where Discontinuity cont Hence Please see the Appendix 31 Discontinuity cont To solve for enforce the electric field integral equation EFIE 7239x 10 E3 xy z 2 sm je jkz Z unitamplitude incident mode a mirx mo Eyscai xyz 2AM s1n a jerz Z Hence 32 sin 7Z39X a Discontinuity cont J m7rx a ZT mirx w w 21m s1n x0 ltxltx0 m1 2 a 2 2 2 ZZE0 mm W w Zcm10 s1n x0 ltxltx0 m1 2 a 2 2 33 Discontinuity cont ZTE sm jzlo quot 0 smm7zxj xO Kltxltx0 a 61 ml 2 a 2 2 This is in the form fxIOgx x0 ltxltx0g To solve for the unknown I we can use the idea of a testing function We multiply both sides by a testing function and then integrate over the strip xO w2 xO w2 i Txfxdx10 i Tltxgtgltxgtdx xO w2 xO w2 34 Discontinuity cont smimidx 00 xO wZ a quot11 xO wZ a Galerkin s method The testing function is the same as the basis function Discontinuity cont Discontinuity cont Hence 37 Discontinuity cont Summary ac1 0 00 mirw m7er ZZTE 62 Cm 2 J0 sm m0 m 2a a m1 38 Discontinuity cont Reflection Coefficient A A F ilncz leszl A1 1 Also Zin ZIZE c o From these equations X may be found 39 Result Discontinuijty39 cont 40 Appendix In this appendix we evaluate the cm coefficients Appendix cont 1 Sm m7rx39x0 x a Jd mizx39 m7239X0 s1n cos 1 a 2 mmc39 m7239X0 39 cos s1n a a This term integrates to zero odd function dx Appendix cont 43 Appendix cont W Next use the transformation x39 sjn9 2 dx39 geos d 44 Appendix cont 7r2 cm 2 Esmfmrxoj J cosm7zw sin jd 7 a 0 2a 1 or cm smm7m0 cosm7zw sm 6 d6 7 a 0 2a Next use the following integral identify for the Bessel function 1 J z J cosz s1n6 m9 616 7r 0 so that JO 2 cos 2 sin 6 d6 0 1 7r 45 Hence Appendix cont Cm Sin m7zx0 jJO mzw a 2a ECE 6340 Intermediate EM Waves Fall 2008 Prof David R Jackson Dept of ECE Notes 12

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