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

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This 33 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 15 views.

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Date Created: 02/06/15
ECE 6340 ntelmedate EM Waves Fall 2006 Prof Donald R Wilton ECE Dept Notes 6 Notes based on those by Prof David R Jackson Power Generated by Moving Charges Work done by a collection of electric charges moving in an electric field AW Aq A5 vaS Ma Ag Power generated per unit volume AW A ASMA1 pVAt IQQ Poynting Theorem TimeDomain Vx z 8t 8 Vx fz 5 8t From these we obtain Subtract and use VX VX Ww Poynting Theorem TimeDomain cont toobtain V x a i Nowlet ggia i 0 692 OM Poynting Theorem TimeDomain cont Nowuse t 8t 2 and 86 at at at 6Q 16362 Hence 8 at 2 at 5 15 26f And similarly y at Poynting Theorem TimeDomain cont Define 2 x Then V5V gi 062 i 21862y2 arz Integrate throughout Vand use divergence theorem A z39 z39 a 1 1 EggQ03 17 zdV a 2dV E g 23wZjdV V Poynting Theorem TimeDomain cont Interpretation 0 6562 dV 2 stored electric energy V 0 2 guy dV 2 stored magnetic energy V 9 039 92 dV 2 dissipated power V 9 X E39 6 4quot gtdV 2 source power V 9 gig at 2 power flowing out of S S Poynting Theorem TimeDomain cont or Poynting Theorem TimeDomain cont 1 Power flow out of surface lt5 m source 1 Poynting Theorem Note on Interpretation 2 i gtlt Does the Poynting vector really represent local power flow vgg 0go2 i218g2lgg2 at 2 2 Consider g V X g M Arbitrary vector function This new Poynting vector is Note that V X V 2 equally valid They both give same TOTAL power flowing out of the volume Note on Interpretation cont Another dilemma A static point charge is sitting next to a bar magnet Z x iQ Is there really power flowing in space Note on Interpretation cont Bottom line we always get the correct result if we assume that the Poynting vector represents local power flow Because In a practical measurement all we can ever measure is the power flowing through a closed surface Complex Poynting Theorem VXEMjw Vx 1ijwsc Hence HVgtltE MquotE jcou 2 EVXEEfjw8 EIZ Subtract and use Complex Poynting Theorem cont Define EgtltH complex Poynting vector 1 Notethat Re Re x 2 ltwgt Then 1 j l 1 gtxlt V EEi M 1wscE2 u 2 Nextuse I 8028 8 gt 80 8j8 39J Complex Poynting Theorem cont i i 1 r r 1 n n V u M 31w8El w H2 w8 lam Iar 139quot i 1 I 1 I 1 II II V Ei M 2aZgE2 Zu 2 Ea8 E2u 2 Integrate over a volume Vand apply the divergence theorem A 1 i i 1 lt dSV EEol M dV2aJZg 1 ng EIZ i39 lzjdV Elz au 2jdV Complex Poynting Theorem cont Final form of complex Poynting theorem j agf M EdVlt SndS V Complex Poynting Theorem cont Interpretation of PS Pszcomplex SOUFCG pOWGF Complex Poynting Theorem cont Interpretation of Pf RePflt lt xgt dS 42 P f 2 complex power flowing outof S Complex Poynting Theorem cont Interpretation of energy terms Complex Poynting Theorem cont Interpretation of dissipation terms iaa39 lz wafiRe l2gt wequot gt lt08 For Simple linear media 03quot a a a 1 H Hence a8 2 Er OW lt22 Complex Poynting Theorem cont Interpretation of dissipation terms cont Similarly im l lz Complex Poynting Theorem cont V wequot 2 au 2jdV 1 2 2ja u39 839E jdV V14 4 Complex Poynting Theorem cont Ps pr ltgbsgtj2wlltgtltgtdV Complex Poynting Theorem cont Rem gtlt91gt 1mm gtlt2wgtgltltgt ltwgtdv I smzfpznvsgce V Denote Pabs complex power absorbed Wabs anbS Calculate Pabs using circuit theory and verify that the result is consistent With the complex Poynting theorem Note W abs O Iossless element Example cont Example cont 1 Qabs 1m Pabs Ea lLIIr 4 1 LI2 2w 111 2 LRe 2 1L 2 2w D 2w 2w Example cont Since there is no stored electric energy in the inductor we can write Hence the circuittheory result is consistent with the complex Poynting theorem Note the inductor absorbs positive VARS Example Antenna Example cont 2R P so Zm Real part 252 Re1 znltgt Re POO Note the farfield Poynting vector is much eaSIer to calculate Hence Imaginary part Example cont 4 g a Example cont XI 2 2392 J1m jd2wltgt ltgtdl in We have that 00 This follows from planewave MB 0 properties in the farfield Hence Xm f i lltgt ltgtl in 4 1 102 0 90 dV m Example cont However it would be very difficult to calculate the input impedance using this formula

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