L4-1REVIEW OF UPW BASICSEx=ωˆEocos(t−kz)EHy=ωˆocos(t−kz)ηoExample : xˆ-polarized UPW traveling in +z direction1c3==x108m/sμεooE(z) = xˆE−jkzoeEH(z) = yˆoe−jkzηoE×H: zˆ direction of propagationzyHz(,0)xEz(,0)2πλ=kwavelengthdirection of propagationcλ =f2πωμω=(rads/s) 2πf k(rads/m) == η=oλ coε0ˆL4-2HOW DO WAVES CONVEY POWER, ENERGY?But E ⊥ HRecall:This is Poynting’s TheoremWhat does it mean?E [V/m] i J [A/m23] = Pd [W/m ]Manipulate Ampere’s law to get E • J()()∂BD∂HE⋅∇× −E⋅∇×H=−H⋅()−E⋅(J+⋅ )∂tt∂Vector Identity()∂BD∂3∇⋅EH× = H−⋅ E−⋅J E−⋅ (W/m)∂tt∂For symmetry, compute∂BH(⋅∇×E=− )∂t∂DE(⋅∇×H=J+ )∂tPOYNTING THEOREML4-3()∂∂BDPoynting’s Theorem:∇ ⋅×EH = H−⋅ E−⋅ E−⋅J∂∂ttB =μH D =εE()d1⎛⎞22d1⎛⎞∇⋅ EH× = −⎜⎟μH −⎜εE⎟ E3−()⋅J (W/m)dt⎝⎠2 dt⎝2⎠Stored magneticStored electricPower densityenergy density, energy density, dissipated/m3, WmWeWdPoynting vector, S [W/m2]Energy is conserved!⎛volts amps watts⎜⋅=⎝mm m2Poynting VectorSE=×H (W/m2)EHSd1⎛⎞ 2d|H| dBNote:⎜⎟μ |H| =μ |H| =H idt⎝⎠2 dt dt⎞⎟⎠L4-4INTEGRAL POYNTING THEOREMGauss’s Theorem (not Gauss’s Law)Use:∫∫A ⋅=nˆ da ∇⋅A dvSV()∫∫E×⋅H nˆ da = ∇⋅×SV(E H) dvndaˆdvdvdvdvdvTherefore:The Poynting vector SE= ×Hgives both the magnitudeof the power density (intensity) and the direction of its flow.()()⎡d1⎛⎞d1⎛⎞=−∫22 ⎢⎜⎟μH −⎜εE⎟ −E⋅J dv V⎣dt⎝⎠2 dt⎝2⎠d1⎛⎞221∫∫E×⋅H nˆ da = −⎜⎟εE +μH dv − SV⎝⎠∫E⋅J dvdt 2 2VPower emerging = released stored energy - dissipation [W]⎤⎥⎦UNIFORM PLANE WAVE EXAMPLEL4-5Ex=ωEocos()t−kz⎛⎞EHy=ωo⎜⎟cos()t−kz⎝⎠ηo1WEeo=ε22ocos()ωt−kz21μWE=ωo22mo()−η2costkz2oThe time average is “intensity” [W/m2]〈S(r,θφ, 〉⎛⎞E2S(t) =×E H =zˆ⎜⎟ocos22()ωt −kz (W/m )⎜⎟η⎝⎠o1E2⇒〈 Sz〉=ˆo=2 ηozE2Sz=ωˆocos2()t−kzI ()2θφ,,r [w/m]η0Sz( at t= 0)0L4-6COMPLEX NOTATION – POYNTING VECTORDefining a meaningful and relating it to is not obvious.Let’s work backwards to find the time average and thenSSThus, we can defineand〈S〉1〈〉SR= e(E×H∗2)SE= ×H∗SRecall:E =+E =+jtω= ω+ ωrijE H HrjHi e cos t jsin tS(t) =×E H =Re⎡E⋅ejtω⎤⎡×Re H⋅ejtω⎤⎣ ⎦⎣ ⎦=ω[E cos( t) −riE sin(ωt)]×[H ω−ωrcos( t) Hisin( t)]()()()1⇒〈 S(t)〉=⎡⎤E × + ×2⎣⎦rrH EiHi11 = R E ×∗2eH [R=+{(EjE)×(H−er i rjHi)}]2S(by definitionUPW REFLECTED BY PERFECT CONDUCTORL4-7E=ωxˆˆE+−cos(t−kz)+xEcos(tω+kz) = 0 at z = 0 (perfect conductor) ⇒=EE−+−⇒= E xˆ2E+sinωt ⋅sinkz α +β α−β (recall: cosα−cosβ=−2sin isin )22We[J/m3] = 2 εE2 sin2ωt sin2kzPerfectE = 0 every half cycle (ωt = 0, π, etc.)Conductor(Where does the energy go?)ωt = πω/2t = -π/2ωt = 0zz = 0xEStanding waves, oscillate without movingForward plus reflected wave(Solving for unknown reflection)Never any WehereL4-8STANDING WAVE EXAMPLE - CONTINUEDEEH =ωyˆ [+−cos( t −kz) −ycos(ωt +kz)]ηηooE2=⋅zsˆ+in2kz⋅sin2ωtηo2EH =ωyˆ+cos t ⋅coskzηo(It’s in the H field!)E2S =×E H =zˆ 4+cosωtsinωt ⋅coskzsinkzηoE =ωxˆ [E+−cos( t −kz) +E cos(ωt +kz)]EE−+=− ⇒1⇒〈S0〉= = Re()E×H∗2Note= 0 when ωt = π/2, 3π/2, etc.)ωt = 0z = 0ωt = πSyHzωt = π/2W[32⎛⎞2E21moJ/m] = μ+222⎜⎟cos ωt cos kz = 2εE c⎝⎠+osωt cos kz2 ηo2MIT OpenCourseWare http://ocw.mit.edu 6.013 Electromagnetics and Applications Spring 2009 For information about citing these materials or our Terms of Use, visit:
View Full Document
Unlocking...