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# Class Note for ECE 6340 with Professor Jackson at UH 2

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## About this Document

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Date Created: 02/06/15
ECE 6340 mntermediate EM Waves Fall 2005 Prof Donald R Wilton ECE Dept Notes 9 Notes based on those of D R Jackson Fields of 3 Guided Wave Assume Em 2 E0 e b yaz 2 0 e z Then Fields of 3 Guided Wave cont Proof for Ey Now solve for Hx Fields of a Guided Wave cont H 1 8E25Ey jam 8y 82 8E2 yE Jaw 8y y Substituting this into the equation for Ey yields the result Multiply by jayjagc k2 or Fields of 3 Guided Wave cont 8H Z k2 Ey jam 8x The other components may be found similarly TEM Wave Wavenumber property To avoid having a completely zero field 72k20 Use 7 so kZ2 k2 O N yaJ ote k2 ja TEM Wave cont Lossless TL kzzcmLCzkza us The phase velocity is equal to the speed of light in the dielectric TEM Wave cont Static property E06 y 2 E006 y 6 l0 x9 6 72 06 y 2 006 y 6 7 jkz jk Em x y and E x y are 2D static field functions TEM Wave cont Proof VXEIOZ x y Z Ex0 Ey0 0 Therefore only a Z component of the curl exists We next prove that this must be zero TEM Wave cont Now use VXE VXltE10 6 6 V x Em We X Em Z 7Z e VXElO e ZXEZO 2vx gte 2ltVx ogt Also 2W x g 2 jw jw z 0 TEM Wave cont Hence 2VXEZO 0 TEM Wave cont Also V E Z O No charge density in the timeharmooic steady state for a homogeneous medium Therefore V E 9 72 2 0 t0 V lo 5 We 0 VEmwzEro97e 0 TEM Wave cont transmission line TEM Wave cont VZCDOC y 0 I constant on A or B E0 0 inside conductors The potential function is therefore unique and is the same as the static potential function TEM Wave cont Similarly V x lo 9 V ro 0 so are VltIgtmltxygt V2 m VZCIDm 0 acpm 0 6n TEM Mode Magnetic Field Vx ngc so EO7H 7 TEM Magnetic Field cont Also 7 jkz Z jk wJAt sc 277 j 80 ngc 08c 6c mac ja so Ex 2 77Hy E 2 77H y x This can be written as TEM Mode Charge Density TEM mode TEM Charge Density cont Hence Note 8 72 SC Example Microstrip Line Ignore substrate and ground plane 1 ip x y 877 39x W Example cont Line in free space with a static charge density 1 7T This was first derived p0 x Q by Maxwell using S UV2y x2 conformal mapping Hence Jxw U IO Kw22 x2 In this result 10 is the total current Amps on the strip Example Coaxial Cable Find E 51 2 V C13p 0 c1361 V0 c1309 0 1a 81 0 0 pap 5p gt Ibzclln0c2 Example cont Boundary conditions C1 In a C2 V0 c1 lnbc2 0 SO 6111 161 11 1bV0 Hence c1 V0 c2 c11nb V01nb In g In E b 7 Therefore 1 V0 1np Volnb 2 V0 mp 112 112 112 b b b b Example cont A 81 EtOltx9y VDx9 a 0 Example cont 2XE zxpEp 77 77 A E a 4 1 77 E V0 6 ij 0 1n This result is a valid at any frequency Example cont V VO e J I 27zaH a 27m V0 ij 77a 1n a TEM Mode Telegrapher s Eqs Telegrapher s Eqs cont Note v is path independent in C the xy plane I a S O Telegrapher s Eqs cont Use VX 81 agg ag 0 ay 62 a 0x 62 So Telegrapher s Eqs cont Ef cdy me d A A Note L is the mag netostatic DC value Telegrapher s Eqs cont If we add R into the equation This is justifiable if the mode is approximately a TEM mode small conductor loss Telegrapher s Eqs cont Ampere s law i r dx Z 39y c so lt xdx ydy c Telegrapher s Eqs cont Now use Telegrapher s Eqs cont Hence a 6i 69 69 533 dx dyltjydx xdy a a a lt dex Qxdylt dx dy Cl C Telegrapher s Eqs cont clx WHH397Q dy I A i n l Cl Cigdx 501 4Q39 qd1 2 QA Q C C 401 gcdy C Z ampCidl ileak Ci c QCv lleak Gv Telegrapher s Eqs cont Hence 2 CV GV 82 t or Telegrapher s Eqs alternate derivation 0 gS dgzvzAzt vztR1R2Azi Telegrapher s Eqs alternate derivation

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