Dr. S. Cruz-Pol, INEL 4152- Electromagnetics

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Dr. S. Cruz-Pol, INEL 4152ElectromagneticsOutlineI. Faraday’s Law & Origin of ElectromagneticsElectromagnetic wavesII. Transformer and Motional EMFIII. Displacement Current & Maxwell EquationsIV. Wave Incidence (normal, oblique)Sandra Cruz-Pol, Ph. D.ECE UPR- Mayagüez, PRI. Lossy materialsII. Multiple layersElectricity MagnetismØ In 1820 Oersted discovered that a Eleven(11) yearssteady current produces a magneticfield while teaching a physics class.This is what Oersteddiscovered accidentally: H dl J dSLVemfVemf NdΨdt E dl N t B dSLsCruz-Pol, Electromagnetics UPRMLen’s Law (-)dΨ dtlater, and at the sametime, (Mike) Faradayin London & (Joe)Henry in New Yorkdiscovered that atime-varying magneticfield would producean electric voltage!sCruz-Pol, Electromagnetics UPRM If N 1 (1 loop)Would magnetism would produceelectricity? The time changeVemf! !! ! " E dl t N B dSLscan refer to B or SElectromagnetics was born! This is Faraday’s Law the principle of motors,hydro-electric generatorsand transformersoperation.!! ! B !Faraday's Law " E dl t dSLs! !! !Ampere's Law "H dl J dSLCruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMsCruz-Pol, Electromagnetics UPRM1

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsMaxwell noticed somethingwas missing Faraday’s Law And added Jd, the For N 1 and B 0Vemfdisplacement currentdΨ Ndt H dl J dS ILenc IS1S1 H dl J dS 0LVemf ! E dl IRLCruz-Pol, Electromagnetics UPRMLS2 H dl JLId dS S2S2ddQD dS Idt S 2dtAt low frequencies J Jd, but at radio frequencies bothterms are comparable in magnitude.Cruz-Pol, Electromagnetics UPRMElectromagnetic SpectrumCruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMMaxwell Equationsin General FormUniform plane em waveapproximationDifferential form Integral Form D ρv D dS ρ dvGauss’s Law for Efield. B dS 0Gauss’s Law for Hfield. Nonexistenceof monopolevs B 0vs B E t H J Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRM D t E dl t B dSLFaraday’s Laws D Ampere’s Circuit H dl J t dS LawLsCruz-Pol, Electromagnetics UPRM2

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsWould magnetism would produceelectricity? Eleven years later,and at the same time,Mike Faraday inLondon and JoeHenry in New Yorkdiscovered that atime-varying magneticfield would producean electric current!VemfdΨ NdtElectromagnetics was born! This is the principle ofmotors, hydro-electricgenerators andtransformers operation.This is what Oersted discoveredaccidentally: E dl L Cruz-Pol, Electromagnetics UPRMLs*Mention some examples of em wavesCruz-Pol, Electromagnetics UPRMSpecial case Consider the case of a lossless mediumσ 0 with no charges, i.e. .ρv 0The wave equation can be derived from Maxwellequations as22c E ω µε E 0What is the solution for this differential equation? The equation of a wave!Phasors for harmonic fieldsWorking with harmonic fields is easier, butrequires knowledge of phasor.Ø The phasor is multiplied by the time factor,ejωt, and taken the real part.φ ωt θRe{re jφ } r cos(ωt φ )Im{re jφ } r sin(ωt φ )Cruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMMaxwell Equationsfor Harmonic fieldsDifferential form*Gauss’s Law for E field. Bµ H0 Gauss’s Law for H field.No monopole Es jωµ H s Then apply the vectorial identity 0Faraday’s Law EE jBωµHt D H JJ jωε E H tD ε EA wave Start taking the curl of Faraday’s law ρv DεE ρ vCruz-Pol,Electromagnetics UPRM* (substituting D H dl J t dS B dS t sandAmpere’s Circuit LawH µB )Electrical Engineering, UPRM A ( A) 2 A And you’re left with ( Es ) 2 Es jωµ (σ jωε ) EsCruz-Pol, Electromagnetics UPRM γ 2 Es3

Dr. S. Cruz-Pol, INEL 4152Electromagnetics 2 E γ 2 E 0To change back to timedomainA Wave From phasorLet’s look at a special case forsimplicitywithout loosing generality: The electric field has only an xcomponent The field travels in z directionThen we haveE ( z, t )E xs ( z ) Eo e γz Eo e z (α jβ ) to time domain E ( z , t ) Eo e αz cos(ωt βz ) xwhose general solution isE(z) Eo e γz Eo' e γzCruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMSeveral Cases of Media1. Free spaceThere are no losses, e.g.(σ 0, ε εo , µ µo )1.Free space2.Lossless dielectric3.Low-loss4.Lossy dielectric5.Good Conductor E ( z, t ) A sin(ωt βz ) x(σ 0, ε εr εo , µ µr µo or σ 0)(σ 0, ε εr εo , µ µr µo or σ ωε )(σ 0, ε εr εo , µ µr µo )Let’s define(σ , ε εo , µ µr µo or σ ωε )Permitivity:εo 8.854 x 10-12[ F/m]Permeability: µo 4π x 10-7 [H/m]Cruz-Pol, Electromagnetics UPRM2The angular frequency Phase constant The phase velocity of the wave The period and wavelength How does it moves? E ( z , t ) Eo e αz cos(ωt βz ) x2γ jωµ (σ jωε )2γ 2 β 2 α 2 ωµ σ 2 ω 2ε 2 From this we obtain22 σ µε σ 1 1 1 and β ω 1 2 ωε2ωε µε So , for a known material and frequency, we can find γ α jβCruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMAny mediumγ α jβ Re γ β α 2 ω 2 µεα ωThe phase of the wave Cruz-Pol, Electromagnetics UPRM3. Lossy Dielectrics(General Case) In general, we had αωSummaryLosslessmedium(σ 0)02 σ 1 1 2 ωε µε Low-loss medium(ε”/ε’ .01)σµε2Good conductor(ε”/ε’ 100)[Np/m]πfµσ[rad/m]βω µεω µεπfµσηjωµσ jωεµεucω/β114πfµεµεµσλUnits2π/β up/f**In free space;Cruz-Pol, Electromagnetics UPRMµε(1 j )upupupfffασ[ohm][m/s][m]εo 8.85 10-12 F/m µo 4π 10-7 H/m ηο 120π Ω4

Dr. S. Cruz-Pol, INEL 4152Electromagneticsεc ε ' jσωεo(Relative) Complex Permittivityk β ω µoεoε rCruz-Pol, Electromagnetics UPRMFor lossless media,The wavenumber, k, is equal toThe phase constant. This isnot so inside waveguides.Cruz-Pol, Electromagnetics UPRMThe attenuation and phase constants can also be expressed as:Intrinsic Impedance, η If we divide E by H, we get units of ohms andthe definition of the intrinsic impedance of amedium at a given frequency.{ ε}µ ε Re { ε }α ω µ εo Imβ ωojωµ η θησ jωεη [Ω] E ( z, t ) Eo e αz cos(ωt βz ) xH ( z, t ) Cruz-Pol, Electromagnetics UPRMEoηe αz cos(ωt βz θη ) yˆCruz-Pol, Electromagnetics UPRMNote E ( z, t ) Eo eH ( z, t ) *Not in-phasefor a lossymediumEoη αz cos(ωt βz ) xe αz cos(ωt βz θη ) yˆLoss Tangent If we divide the conduction current by thedisplacement current E and H are perpendicular to one anotherJ cs Travel is perpendicular to the direction of propagationJ ds The amplitude is related to the impedance σ Esσ tan θ loss tangentjωε Es ωε And so is the phase H lags ECruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM5

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsRelation between tanθ and εcσ H σE jωε E jωε 1 jEωε 2. Lossless dielectric(σ 0, ε ε rεo , µ µ r µo or σ 0) Substituting in the general equations: jωε c Eα 0, β ω µεω12πu λ ββµεThe complex permittivity is"σ %εc ε 1 j ' ε ' jε ''#ωε &ε" σThe loss tangent can be defined also as tan θ ε ' ωεCruz-Pol, Electromagnetics UPRMReview: 1. Free Space Substituting in the general equations:α 0, β ω µε ω / cω12πu cλ ββµ oε oEocos(ωt βz ) yˆ Substituting in the general equations:ωµσω2ωu βµσ2λ ωµη 45 oσ E ( z , t ) Eo cos(ωt βz ) x V / mηo4. Good Conductors(σ , ε ε o , µ µr µo )α β µo o 0 120π Ω 377 ΩεoH ( z, t ) µ o 0εCruz-Pol, Electromagnetics UPRM(σ 0, ε ε o , µ µo )η η A/ mCruz-Pol, Electromagnetics UPRMIs water a goodconductor?2πβ E ( z , t ) Eo e αz cos(ωt βz ) x [V / m]H ( z, t ) Eoωµσ oe αz cos(ωt βz 45 o ) yˆ [ A / m]Cruz-Pol, Electromagnetics UPRMSkin depth, δØ Is defined as thedepth at which theelectric amplitude isdecreased to 37%Skin depthe 1 0.37 (37%)e αz e 1 at z 1 / α δδ 1 / α [m]Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM6

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsExercises: Wave Propagation inLossless materialsShort Cut You can use Maxwell’s or use A wave in a nonmagnetic material is given byFind:! 1!H kˆ Eη!!E η kˆ Hwhere k is the direction of propagation of the wave, i.e.,the direction in which the EM wave is traveling (aunitary vector).Cruz-Pol, Electromagnetics UPRM!H zˆ50 cos(10 9 t 5 y ) [mA/m](a)direction of wave propagation,(b)wavelength in the material(c)phase velocity(d)Relative permittivity of material(e)Electric field phasorAnswer: y, up 2x108 m/s, 1.26m, 2.25,Cruz-Pol, Electromagnetics UPRMExercises: Wave Propagation inLossless materials Power in a waveA wave in a nonmagnetic material is given by A wave carries!H ẑ50e 2 y cos(10 9 t 5y) [mA/m]power andtransmits itwherever it goesFind:(a)direction of wave propagation,(b)wavelength in the material(c)phase velocity(d)Relative permittivity of material(e)Electric field phasorAnswer: y, up 2x108 m/s, 1.26m, 2.25,The power density perarea carried by a waveis given by thePoynting vector.!E xˆ12.57 e j 5 y [V/m]Cruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRM http://www.hitechmv.com/Poynting Vector Derivation ε (E H ) dS t 2 ESTotal power acrosssurface of volume!E xˆ12.57 e j 5 y [V/m]2 vµ2 H 2 dv σE 2 dv vRate of change ofstored energy in E or HOhmic losses due toconduction current Which means that the total power coming out ofPower: Poynting Vector Waves carry energy and information Poynting says that the net power flowing out of agiven volume is to the decrease in time inenergy stored minus the conduction losses. S P E Ha volume is either due to the electric ormagnetic field energy variations or is lost asohmic losses.Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRM[W/m 2 ]Represents theinstantaneouspower vectorassociated to theelectromagneticwave.Cruz-Pol, Electromagnetics UPRM7

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsTime Average PowerTotal Power in W The Poynting vector averaged in time is 1S ave T 1 S dt T0TT (0 1E H dt Re Es H s*2){The total power through a surface S is!Pave Pave dS [W ]}S Note that the units now are in Watts For the general case wave:Es Eo e αz e jβz xˆ [V / m]Hs Eoηe αz e jβz yˆ [ A / m] S ave Eo22ηe 2α z cosθη ẑ[W/m 2 ]For general lossy mediaCruz-Pol, Electromagnetics UPRMExercises: Power Note that the dot product indicates that the surface areaneeds to be perpendicular to the Poynting vector so thatall the power will go thru. (give example of receiverantenna)Cruz-Pol, Electromagnetics UPRMTEM wavexxzzy1. At microwave frequencies, the power density consideredsafe for human exposure is 1 mW/cm2. A radar radiates awave with an electric field amplitude E that decays withdistance as E(R) 3000/R [V/m], where R is the distancein meters. What is the radius of the unsafe region?Transverse ElectroMagnetic plane wave Answer: 34.6 m only perpendicular (transverse).2. A 5GHz wave traveling in!a nonmagnetic medium withεr 9 is characterized by E yˆ 3 cos(ωt βx) zˆ 2 cos(ωt βx)[V/m]Determine the direction of wave travel and the averagepower density carried by the wave If have an electric field Ex(z) Answer: S ave x̂0.05 [W/m 2 ]Cruz-Pol, Electromagnetics UPRM There are no fields parallel to the direction ofpropagation, then must have a corresponding magnetic fieldHx(z)ˆE aˆH aˆk The direction of propagation is aCruz-Pol, Electromagnetics UPRMPolarization:Why do we care? Antenna applications – Antenna can only TX or RX a polarization it is designed to support.Straight wires, square waveguides, and similar rectangular systemssupport linear waves (polarized in one direction, often) Circularwaveguides, helical or flat spiral antennas produce circular orelliptical waves. Remote Sensing and Radar Applications – Many targets will reflect or absorb EM waves differently for differentpolarizations. Using multiple polarizations can give differentinformation and improve results. Rain attenuation effect. Absorption applications – Human body, for instance, will absorb waves with E oriented fromhead to toe better than side-to-side, esp. in grounded cases. Also,the frequency at which maximum absorption occurs is different forthese two polarizations. This has ramifications in safety guidelinesand studies.Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMPolarization of a waveIEEE Definition:The trace of the tip of theE-field vector as a functionof time seen from behind.xxBasic types:z Vertical, ExyyE xs (z) Eox e jβ zxxE x (z) Eox cos(ω t β z) x̂z Horizontal, EyE ys (z) E y Eoy e jβ z δyyE (z) E cos(ω t β z δ ) ŷyoyCruz-Pol, Electromagnetics UPRM8

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsPolarization In general, plane wave has 2 components; in x & yE ( z ) xˆE x yˆE y And y-component might be out of phase wrt to xcomponent, δ is the phase difference between x and y.E x Eox e jβ zEox ax 0 oE y Eoy e jβ z δEoy ay e jδxExxyEyyFront ViewCruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMSeveral Cases Linear polarization: δ δy-δx 0o or 180on Circular polarization: δy-δx 90o and ax ay RHC is -90o Elliptical polarization: δy-δx 90o and Eox Eoy, or δ 0o or 180on even if Eox Eoy Unpolarized- (Natural radiation)Cruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMcos(ξ 90o ) sin (ξ )Linear polarizationCircular polarizationØ Both components haveFront View δ 0same amplitude ax ay,xE x E xo e jβ zE y E yo e jβ zExoØ δ δ y-δ x -90o RightyEyo @z 0 in time domaincircular polarized(RCP)Ø δ 90o LCPE x a x cos(ω t)E x ax cos(ω t)E y ay cos(ω t 90o )in phasor:E(z) ( x̂a ŷae j 90 ) e jkz a ( x̂ jŷ) e jkzE(z, t) ( x̂a cos(ω t kz) ŷa cos(ω t kz 90 o ))E(z, t) ( x̂a cos(ω t kz) ŷasin(ω t kz))xE y ay cos(ω t)Back View:yxyCruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM9

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsElliptical polarizationPolarization exampleAll lightcomes outUnpolarizedradiationentersNothing comesout this time. X and Y components have different amplitudes ax ay, andδ 90oPolarizing glasses Or δ 90o and Eox Eoy,Cruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMPolarization Parameters Rotation angle, ππ ψ 22 Ellipticity angle, tan 2ψ (tan2α o ) cos δsin 2 χ (sin 2α o ) sin δtan χ aηaξ 1Rππ χ 44 tan of Axial ratiotan α o ayaxCruz-Pol, Electromagnetics UPRMPolarization for em wavesCruz-Pol, Electromagnetics UPRMsin(ξ 90 o ) cos(ξ )Examplecos (ξ 90 o ) sin(ξ )sin(ξ 180o ) sin(ξ )cos (ξ 180o ) cos(ξ )Ø Determine the polarization state of a plane wave withelectric field:a.E ( z, t ) xˆ3cos(ωt - βz 30 o ) - ŷ4sin(ωt - βz 45o )b. E ( z, t ) c.d.xˆ3cos(ωt - βz 45o ) ŷ8sin(ωt - βz 45o )E ( z, t ) xˆ 4cos(ωt - βz 45o ) - ŷ4sin(ωt - βz 45o )Es ( y ) 14( xˆ-jzˆ )e -jβya.c.Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRMEllipticb. -90, RHEPLP 135d. -90, RHCP10

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsCell phone & brainDecibel Scale In many applications need comparison of two Computer model forpowers, a power ratio, e.g. reflected power,attenuated power, gain, Cell phone Radiationinside the HumanBrain The decibel (dB) scale is logarithmicG PoutPin V 2 /R P V G[dB ] 10 log out 10 log o2 20 log o Pin Vi Vi /R Note that for voltages, the log is multiplied by 20instead of 10.Cruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMPower RatiosG10x1004210. [dB]10x dB20 dB6 dB3 dB0 dB-3 dB-6 dB-10 dB-30 dBAttenuation rate, A Represents the rate of decrease of the magnitudeof Pave(z) as a function of propagation distance P (z) A 10 log ave 10 log e 2αz Pave( 0 ) 20αz log e - 8.68αz -α dB z [dB]()whereα dB[dB/m] 8.68α [ Np/m]Assigned problems ch 2 1-3,5,7,9,13,16,17,24,26,28, 32,36, 37,40,42, 43Cruz-Pol, Electromagnetics UPRMIncidenceReflection and TransmissionWave incidence Wave arrives at an angle Snell’s Law and Critical angleParallel or PerpendicularBrewster angleCruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM11

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsReflection at NormalIncidencexMedium 1Etε1, σ1 , µ1EiakHiHtIncidentwaveMedium 2 Incident waveakHi!Eis ( z ) Eioe γ 1z xˆzyz 0akrEiε2, µ2, σ2TransmittedwaveErNow in terms of equations HrakIncidentwave!EH is ( z ) H ioe γ 1z yˆ io e γ 1z yˆη1ReflectedwaveReflected waveThe total fieldsakr!Ers ( z ) Ero e γ 1z xˆ At medium 1 and medium 2Er It’s traveling along –z axisHrReflectedwave!EH rs ( z ) H roeγ 1z ( yˆ ) ro eγ 1z yˆη1Normal Incidence Reflection coefficient, ρ''Eη ηρ ro 2 1 ε1 ε 2Eio η2 η1ε1' ε 2' Transmission coefficient, ττ Eto2η 2 Eio η 2 η1!! !E1 Ei Er!!!H1 H i H r!!E2 Et!!H 2 Ht Tangential components must becontinuous at the interface!!!Ei (0) Er (0) Et (0)!!!H i (0) H r (0) H t (0)Normal IncidenceNote: 1 ρ τ Both aredimensionlessand may becomplex 0 ρ 1Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRM12

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsOblique IncidenceØ Normal , anincidenceincidenceEi Eio cos( kix x kiz z ωt )kizØ Plane ofØ Angle ofExpression for fieldsMedium 2 : ε2, µ2Medium 1 : ε1 , µ1kixEr Ero cos( k rx x k rz z ωt )kiθiθrkrzyEt Eto cos( ktx x ktz z ωt )θtz 0θikik i k ix2 k iz2 β1 ω µ1ε 1wherekizktkixk ix β1 sin θ ik iz β1 cosθ iCruz-Pol, Electromagnetics UPRMTangentialSnell LawE must be Continuous!!!Ei ( z 0) Er ( z 0) Et ( z 0)ωi ω r ωt ωkix krx ktx k xkiy kry kty k yFrom this we know thatfrequency is a property ofthe wave. So is color.θi θ rWhen going to adenser medium,the refraction isinward.kix ktxβ1 sin θ i β2 sin θ tn1 c εr1u1So 700nm is not always red!!kix krxβ1 sin θi β1 sin θ r Equating, we getAlso written as,orn1 sin θ i n2 sin θ twhere, the index of refraction of a medium, ni ,is defined as the ratio of the phase velocity infree space (c) to the phase velocity in themedium.Critical angle, θc All is reflected When θt 90o, the refractedwave flows along the surfaceand no energy is transmittedinto medium 2. The value of the angle ofincidence corresponding tothis is called critical angle, θc. If θi θc, the incident wave istotally reflected.Ulaby,Electrical Engineering, UPRMsin θ c n2sin θ t [θ t 90 o ]n1n2ε r2n1ε r1(for µ1 µ 2 )Example; εr1 9; εr 2 4εr 2sin 90oεr14sin 42 o (1) .679sin θ c sin 40 o .64 .67 (sin 73o )sin 50 o .77 .67 (sin?o )13

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsFiber opticsOptical fibers have cylindrical fiber core withindex of refraction nf, surrounded by anothercylinder of lower, nc nf , called a cladding. Light can be guided with total reflectionsthrough thin dielectric rods made of glass ortransparent plastic, known as optical fibers. The only power lost is due to reflections at theinput and output ends and absorption by thefiber material (not perfect dielectric).[Figure from Ulaby, 1999]Use Snell and critical angle toderive: For total reflection:nsin θ 3 sin θ c 2θ 2 θ 3 90 on1Acceptance anglesin θ a (n 2f nc2 )n0Parallel (V) polarizationPerpendicular (H) or Parallel(V) It’s defined as E is to incidence planeEis Eio (cos θ i xˆ sin θ i zˆ )e jβ1 ( x sinθi z cosθi )H is Eioη1e jβ1 ( x sinθ i z cos θ i )Ers Ero (cos θ r xˆ sin θ r zˆ )eH rs Eroη1Erkr jβ1 ( x sinθ r z cos θ r )e jβ1 ( x sinθ r z cosθ r ) yˆEiEts Eto (cos θ t xˆ sin θ t zˆ )e jβ 2 ( x sinθt z cosθt )H ts Equating for continuity, the tangentfieldsWhich components are tangent to theinterface between two surfaces?Etoη2xMedium 1 : ε1 , µ1yˆe jβ 2 ( x sinθt z cosθt ) yˆEtktθrθikiθtyzz 0Medium 2 :ε2, µ2Reflection and TransmissionCoefficients: Parallel (V) Incidence Reflection y and xρ Ero η2 cosθ t η1 cosθ i Eio η2 cosθ t η1 cosθ iτ Eto2η2 cosθ i Eio η2 cosθ t η1 cosθ iAt z 0 (interface):xˆ : Eio (cos θ i )e jβ1 ( x sinθi z cosθi ) Ero (cos θ r )e jβ1 ( x sinθ r z cosθ r ) Eto (cos θ t )e jβ 2 ( x sinθt z cosθt )yˆ :Eioη1e jβ1 ( x sinθi z cosθi ) Eroη1e jβ1 ( x sinθi z cosθi ) Etoη2e jβ 2 ( x sinθt z cosθt )xˆ : Eio cosθ i Ero cosθ r Eto cosθ tyˆ :Eioη1 Eroη1 Etowhereτ (1 ρ )cosθ icosθ tη2Electrical Engineering, UPRM14

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsReflection and TransmissionCoefficients: Perpendicular(H) Incidenceρ Ero η2 cosθ i η1 cosθ t Eio η2 cosθ i η1 cosθ tPropertyNormalIncidenceReflectioncoefficientρ η2 η1η2 η1Γρ 12 Transmissioncoefficientτ 2η 2 iη 2 η1tτ RelationE2η2 cosθ iτ to Eio η2 cosθ i η1 cosθ t1 ρ τ Perpendicularτ 1 ρPowerReflectivityΓ ρPowerTransmissivitysin θ t Cruz-Pol, Electromagnetics UPRMcosθθi1 ηη11 coscosθ t2η cosθ η cosθηη22cosρ12 η 2 2cosθ t 2 η1 1cosθ i 1cosθ 2 η1 cosθ1cosθθi1 ηη11 cosθ t2 Γ ηη2cosηη22cosθ t η1 cosθ i22η 2 cosθ i2η 2 cosθ iη 2 cosθ i η1 cosθ t τ η cosθ η cosθ2t1icosθττ 12 (1( 1 Γ ρ)12 ) cosi θ1coscosθt θ2 1 ρ 12τ 12 1 Γ2T 1 ΓSnell’s Law:Paralleln1sin θ in2Γ H ρH2Γ ρ 2T 1 Γ T 1 Γ where n2 µ r 2ε r 2Brewster angle, θB Is defined as the incidence angle at which the reflectioncoefficient is 0 (total transmission).Γ η 2 cosθ t η1 cosθ B 0η 2 cosθ t η1 cosθ Bη 2 cosθ t η1 cosθ B 0sin θ B 1 (ε 1µ 2 / ε 2 µ1 )1 (ε 1 / ε 2 ) 2* θB isknown ve/Oblique/Oblique-2.htmlCruz-Pol, Electromagnetics UPRM The Brewster angle does not exist for perpendicularCruz-Pol, Electromagnetics UPRMpolarization for nonmagnetic materials.Reflection vs. Incidence angle.Reflection vs.incidence anglefor differenttypes of soil andparallel orperpendicularpolarization.Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM15

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsCruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMDielectric Slab:2 layers Medium 1: Air At the top boundary, ρ12,α1 0 Medium 2: layer of thickness d, low-loss (ice, oil,snow)ε 2 ε 2' jε"2 Medium 3: Lossyγ 2 α 2 jβ2ε3 ε3' jε"3γ 3 α 3 j β3Snell’s Law Phase matching condition at interphase:γ1 sinθ1 γ 2 sinθ 2 γ 3 sinθ3( "( " γ %2 %2 β1 1θ * *1 θ- ' 2- low losscosθcossin θsin2 2 1 1 ' 1if*) #*)β2 # γ 2 & -, & -,Cruz-Pol, Electromagnetics UPRMReflections at interfaces At the bottom boundary, ρ23For H polarization:ρ12 η2 cosθ1 η1 cosθ 2η2 cosθ1 η1 cosθ 2ρ23 η3 cosθ 2 η2 cosθ 3η3 cosθ 2 η2 cosθ 3For V polarization:2( "% γcosθ 3 *1 1 sin θ1 ' *) # γ 3& -,ρ12 η2 cosθ 2 η1 cosθ1η2 cosθ 2 η1 cosθ1ρ23 η3 cosθ 2 η2 cosθ 3η3 cosθ 2 η2 cosθ 3Cruz-Pol, Electromagnetics UPRMMulti-reflection Method Propagation factor:ρ ρ12 τ 21ρ23 L 2τ 12 τ 21ρ21ρ232 L 4τ 12 . ρ12 τ 21τ 12 ρ23 L 2 (1 x x 2 .)L e γ 2d cosθ2τ 12 1 ρ12τ 21 1 ρ21 1 ρ12τ 21ρ23 L 2τ 1211 1 x x 2 .1 xτ 21ρ21ρ232 L 4τ 12ρ12τ 12ρ23 L 2τ 12ρ21ρ23 L 2τ 12ρ212 ρ232L 4τ 122dτ 12 Lρ21ρ23 L 3τ 12ρ21ρ232L 3τ 12ρ23τ 12 L3Cruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM16

Dr. S. Cruz-Pol, INEL 4152ElectromagneticsCont for H Polarizationρ ρ12 τ 21ρ23 L 2τ 12 τ 21ρ21ρ232 L 4τ 12 . ρ12 τ 21τ 12 ρ23 L 2 (1 x x 2 .)τ 12 1 ρ12τ 21 1 ρ21 1 ρ1221 1 x x 2 .1 xandρ21 ρ12Substituting the geometric series:ρ ρ12 And then Substitutingρ (1 ρ12 )(1 ρ12 )ρ23 L1 ρ21ρ23 L 2L e γ 2d cosθ2ρ12 ρ23 e 2 γ 2 d cosθ 21 ρ12 ρ23e 2γ 2d cosθ2Cruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMCruz-Pol, Electromagnetics UPRMAntennasNow let’s review antenna theoryCruz-Pol, Electromagnetics UPRMElectrical Engineering, UPRMCruz-Pol, Electromagnetics UPRM17

Cruz-Pol, Electromagnetics UPRM Cruz-Pol, Electromagnetics UPRM Uniform plane em wave approximation Maxwell Equations in General Form Differential form Integral Form Gauss’s Law for E field. Gauss’s Law for H field. Nonexistence of monopole Faraday’s Law t Ampere’s Circuit Law Cruz-Pol,

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