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Class Note for ECE 6340 with Professor Jackson at UH

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This 22 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 19 views.

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Date Created: 02/06/15
ECE 6340 Intermediate EM Waves Fall 2008 Prof David R Jackson Dept of ECE Notes 24 Uniqueness eorem Shows what BC s are necessary to uniquely determine fields Justifies image theory and the equivalence principle Theorem Assume 1 Sources 1quot M are specified in V 2 El or l is specified on S Then ELI is unique inside V Uniqueness Theorem oont Proof Assume different two solutions that have the same sources and tangential field El orbit on S E LI and E E V gtlt E aw M V gtlt E 4qu M Subtract VxE E jayHa H Uniqueness Theorem cont Let a D In IE II In In1 Eb a b Then VxAE jauA Similarly VxA ja8AE These are sourcefree equations Uniqueness Theorem cont Now use the complex Poynting theorem 4ltEgtltEgt ds Uniqueness Theorem cont lt A xA d5 2mVIGyquotA 2 EquotIAEIZJQ V 2ja ly39M F lg39A 2jdV 0 V 4 4 Next examine the first term Uniqueness Theorem oont On S A2 x Aw amp A5 X Am a O This follows since AEIO or A t0 on S Uniqueness Theorem cont Hence 1 2 l H 2 20JZy A 48 A jdV 2ja GMAEF lg39A 2jdV 0 V 4 4 Set the real and imaginary parts to zero 1 n 2 1 n 2 VIE1 A 13 AE jdV O 1 2 1 2 ZMAm 4gA jdV 0 V Uniqueness Theorem cont Examine the real part jyquot A 2 5quotAE2dV 0 V Assume 8quotgt0 in V or Iuquotgt0 in V Then AEZO or AEZQ in V Uniqueness Theorem cont In either case from Maxwell s equations we have that both must be zero AEQ and A zg in V Hence Im Ea 61 II In 11 Generalization Two regions with an interface El or ll is specified on S 2 i in i 82 El and t continuous on 1 V quot The Interface sources Note If there are surface currents on the interface then we require the appropriate boundary conditions on the interface to be satisfied H RHAV 12 Generalization cont Note D and B are automatically continuous at the boundary INgt For example Vx z D i aHy 6Hx 6x 6y Z 13 Example Infinitesimal dipole VZAZ szZ u 55 X Two possible solutions jkr jkr e e A IL A Ll Z 472739 Z 472739 Both satisfy the Helmholtz equation for AZ 14 Example cont We need a 80 at infinity for El or HI Assumption The correct solution goes to zero at infinity 15 Example cont Both solutions tend to zero at infinity if there is no loss we cannot tell which one is the correct one Assume kk39 jkquot kquotgt0 We then have A 0 Z A gt 00 as r gt oo 6 J r The correct ch0ce IS then AZ 1 4727quot 16 Current source tangent to and just outside PEC Surround Sc by SW 1 Example 17 Example cont Er 9 on SSCSOO Also the sources inside Vare specified no sources in V Hence E are unique inside V One solution that satisfies the source condition and BC is E 9 9 Hence this must be the unique solution 18 Example cont Conclusion No fields M An electric current tangent to a PEC body does not radiate 19 Example cont Similarly No fields A magnetic current tangent to a PMC body does not radiate Sommerfeld Radiation Condition This is a more quotpowerfulquot boundary condition at infinity that does not require the medium to be lossy Let wExEyEZ Assume that Vzw kzw S Then t is unique if 1 Lim M5 0 F gtOO 2 Lim 61 my 0 r gtoo 6739 Example e jkr Use 7y gy I jkw jkr jkr wine Ham I I I I le jkr I 1quot Example cont Now use ejkr 7y 7y I ag 2 1 jkl jkl I 1 I jk j e jk e 2jke1kr ekr I I I I 740 as r gtoo

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