Interm Electromag Waves
Interm Electromag Waves ECE 6340
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ECE 6340 Intermediate EM Waves Fall 2008 Prof David R Jackson Dept of ECE Notes 26 Equivalence Principle Basic idea We can replace the actual sources in a region by equivalent sources at the boundary z Ei I l E 39 l 39 I ANT I 39 l t I S I l l z Keep original fields E outside S Put zero fields and no sources inside S Equivalence Principle cont Note 0510 and E5 b both satisfy Maxwell s equations Equivalence Principle cent The BC s on Sare violated Use equivalent sources to make BC s valid 71 Z I Equivalence Principle cont Hence Equivalent sources I S I 139 zero s I u 39 fields is l l I l I l l f e zero M s I sources I l I I Outside S these sources radiate the same fields as the original antenna and produce zero fields inside S This is justified by the uniqueness theorem Maxwell39s equations are satisfied along with boundary conditions at the interface Equivalence Principle cont Note about materials If there are zero fieldsthroughout a region it doesn t matter what material is placed there or removed p zero fields I r Scattering by a PEC Scattering by a PEC J source I D Equivalent Sources E I 6 source I 9 9 1 is l 80 Llo e A IMS Example cont Original problem source 1 D ELI Equivalent problem E 0 x I source I II 99 ls I I 80210 I I S Conclusion The conductor can be removed Example cont Integral equation for the unknown current l SOUFCG I Electric Field Integral Equation EFIE This integral equation has to be solved numerically Example Scattering by Dielectric Body Source I E ED incident field Ei E scattered field Exterior Equivalence I I ff b 339 E O quotr 155 I E E source I U Ebg J I 1 a t 7 va 8r 7 Iur I is t isfi S Replace body by free space since material doesn t matter in zerofield region I E E source I l 00 tie 39 807x10 39 e med o f M 2xg 0 E Summary for Exterior Original problem Freespace problem I I e 00 I I lt gt is I r 80 0 6 6 A MS is Z x 39 S Me A E EX Interior Equivalence We fill the outside region with Q 9 dielectric IIquot l e l Eb JS I l l 8 y Me I s i S 8 Original problem sources in free space Integral Equation Note The means calculate the fields if The means calculate outside the surface I Aquot the fields inside the radiated by the U Emur surface assuming an infinite dielectric region Boundary E iM 5 Et ifa f conditions E 1 MT i a JZZMZ 57 PMCHVVT Integral Equation PoggioMierChangHarringtonWuTsai Fields in a Half Space E sources region of interest Zgt0 Equivalent sources 9 9 gt gt IE 393 IR II it X lm Fields in a Half Space cont Put PEC in zerofield region 6 E1 E is PEG 6 Ms Z The electric surface current on the PEG does not radiate E 0 Hence we have 6 PEG Ms Fields in a Half Space cont Now use image theory E H goaluo M S incorrect fields correct fields l Note fields are not correct forZ lt O the image method gives incorrect fields in this region Fields in a Half Space Summary E region of interest Z gt 0 SOUFCGS E Emilio Ms incorrect fields correct fields quot35quotquot Ms 22gtltE Fields in a Half Space cont Alternative better when 11 is known on the interface 6 PMC I is Me does not radiate on PMC and is therefore not included image theory E LD 80 uo correct fields Incorrect fields is 21 22X LR Exampe Radatmn from Waveguide y Ex y 0 3on 0053 Cl IL Z Step 1 Step 2 Example cont M 2xE 69 z PEC A 8 EEO cos j a Ms 2 E0 cos j a Example cont Example cont Apply Duality Z Solve this problem first 7Z39X a Then use ECE 6340 Intermediate EM Waves Fall 2005 Prof David R Jackson ECE Dept Notes 25 Image Theorem Vertical electric dipole VED over an infinite ground plane dipole PEC Boundary Conditions 129 700 N I l 500 V I I h SC I Source condition a single VED Boundary condition E Q everywhere on S PEC Image Picture a r N I N I I orlgmal dipole Image dipole Atz0 EiEZ Hence Er 9 Also EtwQ image Picture cont Hence the image solution for Z gt 0 is the same as forthe original problem For 2 lt 0 the image solution is not valid because the source is not the same as the original source Vertical Magnetic Dipole PEC Image picture EW Z Z 0 E 12 E 2 Hence i9 at 20 All Possible Cases 1 I t PEC PMC The PMC case can be obtained from the PEG case by duality Example Monopole Antenna Gap voltage model Example Monopole Antenna Gap voltage model Magnetic fri model Monopole Antenna cont The image picture is a dipole antenna ZD We 2 2W0 10 10 233 2365 Q h Corner Reflector corner reflector I Image picture 1 I 4 Corner Re ector cont Also we can have this polarization Dipole in a Waveguide H H H H H H H H b H M Ma H H H H H H H ECE 6340 lntermecllate EM Waves Fall 2006 Prof Donald R Wilton ECE Dept Notes 1 Notes adapted from those of Prof David R Jackson Maxwell s Equations y Vm y WmZET Lt Cmz rt Am V x 2 aa Faraday39s Law Vx ig Ampere39s Law V pV Gauss39s Law V O Gauss39s Law magnetic form Continuity Equation From Gauss s Law From Ampere s Law 69 nyf2 i at 5 vVXzvgV at Continuity Equation cont 0 V V Va From last slide a V 6t 6t Hence Continuity Equation Integral Form of Continuity Equation 5L JVde atVdV V S closed Apply divergence theorem 3gt J VZ dV giggals V uxper un vohune net ux outofS Integral Form of Continuity Equation cont A a figEds 4an V Assume Vis stationary 1 611 1 IdeV dQencl 6 dt V dt V Hence ltigt charge conservation Integral Form of Continuity Equation A component of current density crossing dS Generalized Continuity Equation Assume S is moving eg expanding dQ Note Wlt pv XS 1 2015 dt S W difference in charge and surface velocities Integral Forms of Maxwell s Eqs 8 Vx fz 1 at Stokes s Theorem Vx mdS g S circulation per unlt area 0f 5 circulation on boundary of S Integral Forms of Mraxwel s Eqs cont Hence A A a A JVxgdSJ gdSJa 7gds Apply Stokes s theorem A a A g zggQdSJa 7QdS or Ci lis a 7 d5 C S wmfm IE Lugmam mwaagh m 1mni a 5 Integral Forms of Maxwell s Eqs cont Ampere s law Note In statics is is independent of the shape of S since the LHS is Similarly Faraday s law Note In statics the voltage drop around a closed path is always zero the voltage drop is unique Boundary Form of Maxwell s Eqs Evaluate LHS of Ampere39s Law on the path shown 0 limqiwnmL l Maw 6 0C 6 0 C C C6 limqt azi l zw C 6 0 Boundary Form of Maxwell s Eqs cont RHS of Ampere39s Law for the path Boundary Form of Maxwell s Eqs cont Since j is an arbitrary unit tangent vector gx l zzxgsxzgs Similarly from Faraday39s Law Note i always Boundary Form of Maxwell s Eqs cont u I 39H HUM HHH 1M Hlll Evaluate LHS of Gauss39s 0252y2 Law over the surface shown 0 lim Q dSzlimL gym Q dsjg s 5 0S 6 0 S S S6 Boundary Form of Maxwell s Eqs cont quot NH 6 1 M H RHS of Gauss39s Law gig QM p5 d5 16133 Qencl z OS TM Hlll Summary of Maxwel Eq Forms Point Form Integral Form Boundary Form szf g isJid5 ampXJ Zfz S Vx Lg 1 aw ampx 1 29 a S at V M qSQ dS QM Ql ngps S Faraday s Law IfS is stationary I dszij dsz S at sz dt l magnetic flux through 8 Faraday s Law cont lfS is moving Previous form is still valid C S at However A A ds i ijggndszdl Faraday s Law cont Vector identity for a moving path d A 8 A EJQgdS Sja ngS gmx i See appendix for derivation Faraday s Law cont Startwith qgdrJ dS C S Then use ij d52 dS ysx y sz S at C Hence j lZ Cil Ifltc 2sx39 or lt Xsgtlt d W C dt Faraday s Law cont De ne Then Two Forms of Faraday s Law Faraday s law Generalized Faraday s law Two Forms of Ampere s Law Ampere s law Generalized Ampere s law Example 0 cosmt 0tt x Note tie in s 0 is in m a 6 Q C Find the voltage drop vt around the closed path Find the electric field on the path Practical nowthe Find the EMF drop around the closed path magnet39Cf39eld 395 conStam out to some large radius Find the force on a charge q that is on the path and the it decays to zero Example cont Voltage drop 5 A vltgS a 7gds 6 M 8 7502 8t 2 ath sin wt Example cont Electric field 6 2750 ath sin wt 6 wsinwt 2p at s1nat 2 Note There is no 0 orz component from the curl of the magnetic field 7th cos mt Example cont w dw dt 72 2 COS WI m w 7 u m Example cont 3 Example cont Q q dV 9 2it27ztcos mt ath sin wt 27w dt Example cont 3 Example cont Electric field alternative method 3 611 2 6 Q lt 7wgtltzs gtltgt dt 611 2 6 Q VSP Z dt Since vSp 1 27rp a d Z dt dw Z 272p dl dlj Example cont 1 dl I COSat 2 7UE cos wt ZLZIZM cos wt 7rth sin an 7 wt coswt cosmt Ysmwt 6 zagsin mt Vm Voltage drop vt around the closed path vt amtz sin wt V A Electric field on the path 07tsin mt Vm EMF drop around the closed path EMF 27rtcos wt 07H2 sin wt V 27 amt sin wt A 227 cosa Force on a charge q that IS on the path g z N Time Harmonic Representation Assume m is sinusoidal f U A COSW Re A z ej ejm W Denote 71 2 141 6355 Phasor form off Then fzt ReF5eW TimeHarmonic Representation cont f t 2 145 cosat 5 Notation fzt lt gt Fz for scalars Lt lt gt 153 for vectors TimeHarmonic Representation cont Derivative Property R6F5ejm RCF ejm RejaF5eW H4 phasor for the derivative Hence TimeHarmonic Representation cont Consider a differential equation such as Faraday s Law 8 r t V X t t or VxReEzeW Rej 0 lew or ReVgtltEja ejm 0 TimeHarmonic Representation cont Let CCrjCi VxEja x7y7Z Then Recej 0 Does this imply that c O Phasor interpretation choose wt 0 Cr 0 I cejwt Im choose wtzyz2 3120 Re Rope Hence 0 0 00myifco TimeHarmonic Representation cont Therefore VxEja 0 x7y7Z so VXE 0 Q WEwa Hence in the sinusoidal steadystate these two equations are equivalent Maxwell s Equations in Time Harmonic Form Time Averaging of Periodic Quantities 1 T Define d W T CW r Tperiod S Assume a product of sinusoidal waveforms ftACOSdZ0 Q FZAeja 8ZBcosat 9 GZBeJ flgtABcosatacosaz ABECOSa 0052m0 Time Average cont ftgt ABcosa cos2ata The timeaverage of a constant is simply the constant The timeaverage of a sinusoidal wave is zero Hence ltft gm A3005a Time Average cont The phasors are denoted as F 2 A eja G B 615 Consider the following FG A Beja so ReFG ABcosa 8 Hence The same formula extends to vectors as well Example Stored Energy Density Consider 9 9 92W 2 ReijyZeW etc lt gtltgtltgt gt ZR6DXER6DyEReDZE lRe DXE DyE DZE 2 1 or Example Stored Energy Density cont 1 OZEZEQ39 Similarly Example Stored Energy Density cont Example Power Flow 2 2 x Poynting vector ggt x Re xH This formula gives the timeaverage power flow Appendix Proof of Moving Surface Identity Note In the figure St is drawn as planar for simplicity d5hmi j zm d5 gamers S s AHO At 5mm 1 I 1Az dS jtAz dS zAz ds 31 AS StAt Proof of Moving Surface Identity cont Hence d A 1 A A Egggob iigg5 tm ZrgdSitAtgdS SO if dS dsmmi IAt dS sz SO 6 AHOAZAS For the last term Hmy dsz I y db AS AS Proof of Moving Surface Identity cont amp Examine term aws tdt Proof of Moving Surface Identity cont Since xys only has an 2 component we can write A tigtltxs tgtltXs Hence I t dS z txysAt Therefore summing all the dS contributions 1 A i13 i dS ltCl xxs Proof of Moving Surface Identity cont Therefore d A 6 A w dS dS w d dtSL a S at a lt3L 1 Vector identity x ZS ZS gtlt Hence ECE 6340 Intermediate EM Waves Fall 2008 Prof David R Jackson Dept of ECE Notes 3 Types of Current 0 g OV Note the free charge Ji I R J0 density pv will be zero for perfect insulators i u l impressed current C J 0 v V c i conduction ohmic current Linear medium 10 0E Ohm s law Note the electric field is set up in response to the impressed sources Types of Current Ampere s law Vx ijw8 VX I0Ejw8E source conduction displacement Effective Permittivity Vx ng 0Eng 1 I 039ja8 1ijwgijg ja Ijwe 15E a Effective Permittivity cont This quoteffectivequot permittivity accounts for the conductivity Define Ampere s law becomes Vx I wow Ampere s law thus becomes in the same form as for free space Vx zfngo Effective Permittivity cont Note ac is usually called 8 for simplicity However be careful Q8E O39E 2ch Effective Permittivity Principle This allows us to solve a problem involving a homogeneous material as long as we know how to solve the corresponding freespace problem V X E 2 ii a 5390 E freespace problem 1 V X E I it 0 EC E material problem The formulas for the fields remains the same we simply make this simple substitution Example Example A dipole is embedded in an infinite medium of ocean water What is the far field of the dipole First examine problem in free space Example cont Dipole in free space As r gtoo a o I r Eez smgejko koza 080 Example cont ocean k 0sm e 1 10 yogc E6 10 47w 0 80 808r J a Loss Tangent 0 8028 0 Write this as gc 839 jg Note The loss tangent combines losses from atomic and molecular friction together with loss from conductivity 8392R68 Img 8quot2 Imgg tan zw Polarization Current Vx ziingc 1quot 039 E j a gg z39 1 0 Jw80 w8 80 I I I I Source Conduction Freespace Polarization Jc displacement JP 0 0 0V 9quot o o i J Polarization Current cont Model of polarization current Nd dipoles per unit volume x q V 0 0 0 0 9 As the electric field changes we 9 9 imagine that the position x of the 9 0 0 electron changes with the nucleus 0 0 0 being stationary 0 0 O 93 990 X m 61Ndv d dx X N N dt dqdt dqv d9 Hence Xi x dt Polarization Current cont d In general g dt Timeharmonic steady state 1p 2 Jag JowEOXeE Jowgo 8r 1E Joa880E Polarization Current cont ll If magnetic material IS present 7 0 i Vxi Ijwsc H0 Jia jwa 2 0Ejwao jwa 80E 1 Vx 1 0EJwao m5 go Vx 0 polarization current from polarization current from dielectric properties magnetic properties Equivalent Current 80 I eczg jg body a nonmagnetic body Inside body Vx jwecg ngo jw8c80 Define Equivalent Current cont Vx zngo i39ieq Interpretation I l The body is Jl39 I 80 9 0 replaced by its equivalent current in l I t free space Jeq I ECE 6340 Intermediate EM Waves Fall 2008 Prof David R Jackson Dept of ECE Notes 24 Uniqueness Theorem 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 Et or l is specified on S Then ELI is unique inside V Uniqueness Theorem cont Proof Assume different two solutions that have the same sources and tangential field El orbit on S EA LI and E E V gtlt E aw M V gtlt E aw M Subtract VxE E jayHa H Uniqueness Theorem cont Let b E A 3 Zia A In Its Then VxAE jauA Similarly VxA ja8AE These are sourcefree equations Uniqueness Theorem oont Now use the complex Poynting theorem 1 A fwd M 1 H 2 l H 2 2aVIZ1 48 jdV It 1 2 1 2 1 gtxlt 2 39H 39E de EJ M dV WW m NJ gt Uniqueness Theorem cont lt A xA d5 2mVijA 2 EquotIAEIZJQ V 2ja ly39M F lg39A 2jdV 0 V 4 4 Next examine the first term Uniqueness Theorem oont On S Ag 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 1 2 1 2 2ijjZ2 A 73 A jdV 0 Set the real and imaginary parts to zero 1 n 2 1 n 2 VIE2 A 13 AE jdV O 1 2 1 2 ZMAm 4gA jdV 0 V Uniqueness Theorem cont Examine the real part JyquotA 2gquotAE2dVO 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 bmhnm bezem AE Q and Ag Q in V 9 Im E E E Hence Generalization Two regions with an interface El or l is specified on S i 8 L Kquot 2 K 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 L I i 7quot El and ll continuous on Generalization cont Note D and B are automatically continuous at the boundary Z For example Vx z D i aHy 6Hx 6x 6y Z Example Infinitesimal dipole VZAZ szZ u 55 Two possible solutions kr e A 472739 Both satisfy the Helmholtz equation for AZ Example cont We need a 80 at infinity for El or HI Assumption The correct solution goes to zero at infinity Example cont Both solutions tend to zero at infinity if there is no loss we cannot tell which one is the correct one Assume k k39 jk kquot gt 0 We then have A 0 00 as r OO AZ The correct choice is then A Example Current source tangent to and just outside PEC Example cont El 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 Examplte cont Conclusion No fields It An electric current tangent to a PEC body does not radiate 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 ME 0 F gtOO 2 Lim 61 my 0 8r Example e jkr Use WZWJF w r k 6r W jkr jkr wine Ham I I I I le jkr I gt0 Example cont Now use jkr 6 W2 Iquot I jkr jkr W316 kt 2 2115ko Jew 7quot 7quot 7quot 7quot 740 as r gtoo 6w r k a W ECE 6340 ntermedlate 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 ASMAI lgt i At IQQ pv Poynting Theorem TimeDomain Vx z a 8t Vx z ag 8t From these we obtain Vx k 8 t Subtract and use vxe avx wax Poynting Theorem TimeDomain cont toobtain Nowlet Jfa i so V Xgi 0 2 io oo Note 502 Zl l2 Poynting Theorem TimeDomain cont Nowuse t 8t 2 and 86 at at at 6Q 16362 Hence 8 at 2 at 6 1 62 And Similarly u 6t 2 6t