Maxwell’s Equations Wave Equations Plane Waves - Fermilab
Lecture 1- Review Maxwell’s equations Wave equations Plane Waves Boundary conditionsA. Nassiri - ANLMassachusetts Institute of TechnologyRF Cavity and Components for AcceleratorsUSPAS 20101
Maxwell’s EquationsThe general form of the time-varying Maxwell’sequations can be written in differential formas: D H J t B E t D ρ B 0Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators2
A few other fundamental relationshipsJ σE ρ J tD ε E B µ H " Ohm' s law"" continuity equation"" constitutive relationships"here ε ε 0ε r (permittivity) and µ µ 0 µ r (permeability)with ε 0 8.854 10 12 F/m, µ 0 4π 10 7 H/mMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators3
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators4
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators5
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators6
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators7
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators8
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators9
A few other fundamental relationshipsMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators10
Integral form of the equations D ρ B E t B 0 D H J tMassachusetts Institute of Technology D ds QS B E d t dsCS B ds 0S D H d J t ds CS RF Cavity and Components for Accelerators11
Wave EquationsIn any problem with unknown E, D, B, H we have 12unknowns. To solve for these we need 12 scalar equations.Maxwell’s equations provide 3 each for the two curlequations. and 3 each for both constitutive relations (difficulttask).Instead we anticipate that electromagnetic fields propagate aswaves. Thus if we can find a wave equation, we could solve itto find out the fields directly.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators12
Wave equationsTake the curl of the first Maxwell: H J (ε E ) t ( E) J ε t H J ε µ t t 2 H J µε t 2Now use H ( H ) 2 H on the LHS0Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators13
Wave EquationsThe result is: H2 H µε J2 t2Similarly, the same process for the second Maxwell produces 2 E Jρ2 E µε 2 µ t tεNote how in both case we have a wave equation (2nd order PDE)for both E and H with fields to the left of the sign and sourcesto the right. These two wave equations are completely equivalentto the Maxwell equations.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators14
Solutions to the wave equationsConsider a region of free space (σ 0) where there are no sources(J 0). The wave equations become homogeneous: 2 E E µε 2 0 t22 H 2 H µε 02 tActually there are 6 equations; we will only consider onecomponent:e.g. Ex(z,t) 2 Ex 1 2 Ex1220wherevc z 2 v 2 t 2µ 0ε 0Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators15
Solutions to the wave equationTry a solution of the form f(z-vt) e.g. sin[β(z-vt)]. By differentiatingtwice and substituting back into the scalar wave equation, we findthat it satisfies!f(z)t 0zf(z-vt1)t t1f(z-vt2)Massachusetts Institute of Technologyt t2RF Cavity and Components for Acceleratorszz16
Plane Waves First treat plane waves in free space. Then interaction of plane waves with media. We assume time harmonic case, and source free situation.We require solutions for E and H (which are solutions to thefollowing PDE) in free space 2 E k0 E 02No potentials here!(no sources)Note that this is actually three equations: 2 Ei 2 Ei 2 Ei2 k0 Ei 0222 y z xMassachusetts Institute of TechnologyRF Cavity and Components for Acceleratorsi x, y , z17
How do we find a solution?Usual procedure is to use Separation of Variables (SOV).Take one component for example Ex.E x f (x )g ( y )h(z )ghf ′′ fhg ′′ fgh′′ k02 fgh 0f ′′ g ′′ h′′Functions of a single k02 0variable sum constant -k02fg hf ′′ k x2 ;fg ′′h′′2 k y ; k z2ghand sok x2 k y2 k z2 k02Massachusetts Institute of Technologywith k0 RF Cavity and Components for Accelerators2πλ ωc18
Mathematical SolutionWe note we have 3 ODEs now.d2 f2k x f 02dxd 2g2k yg 02dyd 2h2k zh 02dzEx A eMassachusetts Institute of Technologysolution isf e jk x xsolution is g e jk y ysolution is h e jk z z( j kx x k y y kz zRF Cavity and Components for Accelerators)19
But, what does it mean physically?Ex A e( j kx x k y y kz z)This represents the x-component of the travelling wave E-field(like on a transmission line) which is travelling in the directionof the propagation vector, with Amplitude A. The direction ofpropagation is given by k k x xˆ k y yˆ k z zˆMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators20
Physical interpretationThe solution represents a wave travelling in the z direction withvelocity c. Similarly, f(z vt) is a solution as well. This lattersolution represents a wave travelling in the -z direction.So generally,E x (z , t ) f [(x vt )( y vt )(z vt )]In practice, we solve for either E or H and then obtain theother field using the appropriate curl equation.What about when sources are present? Looks difficult!Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators21
Generalize for all componentsIf we define the normal 3D position vector as: r xxˆ yyˆ zzˆ then k r k x x k y y k z zsoGeneral expressionfor a plane waveE x Aesimilarly E y BeE z Ce jk r jk r sign droppedhere jk r jk r E E0 ewhere E0 Axˆ Byˆ CzˆMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators22
Properties of plane wavesFor source free propagation we must have ·E 0. If we satisfythis requirement we must have k·E0 0. This means that E0is perpendicular to k.The corresponding expression for H can be found bysubstitution of the solution for E into the E equation. Theresult is: H k0ωµ 0nˆ EWhere n is a unit vector in the k direction.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators23
Transverse Electromagnetic (TEM) waveNote that H is also perpendicular to k and also perpendicular toE. This can be established from the expression for H.EE and H lie on theplane of constantphase (k·r const)Massachusetts Institute of TechnologyHRF Cavity and Components for AcceleratorsNote that:Eˆ Hˆ nˆ or kˆk,nDirection of propagation24
Plane waves at interfaces (normal incidence)Consider a linearly polarized (in x-direction) wave travelling inthe z direction with magnitude Eiµ1ε1σ1 zMassachusetts Institute of TechnologyRF Cavity and Components for AcceleratorsArbitraryorientation!25
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Metallic BoundaryMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators27
Metallic BoundaryDielectricMetalEHSkin depthMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators28
Boundary conditionsMaxwell’s equations in differential form require known boundaryvalues in order to have a complete and unique solution. Theso called boundary conditions (B/C) can be derived by consideringthe integral form of Maxwell’s equations.We deal with a general dielectric interface and two specialcases. First the general case. For convenience we considerthe boundary to be planar.ε1µ1σ1nε2µ2σ2Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators29
General 2Tangential E continuousHt2n x (H1-H2) JsEquivalentε1µ1σ1Bn1 nε2µ2σ2Bn2Normal B continuousMassachusetts Institute of TechnologyRF Cavity and Components for Acceleratorsε1µ1σ1D1nnD2nε2µ2σ2n·(D1-D2) ρs30
Special case (a) Lossless dielectricε1µ1σ1 0nε2µ2σ2 0Et1 Et2tangential E fields continuous)Ht1 Ht2tangential H fields continuous (no current)Dn1 Dn2normal D fields continuous (no charge)Bn1 Bn2normal B fields continuousMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators31
Special case (b) Perfect Conductorε1µ1σ1 0σ2 nPerfect Electric Conductor Et2 Ht2 0Et1 0Tangential Electric field on conductor is zero.n H1 JsH field is discontinuous by the surface currentn . D1 ρNormal D(E) field is discontinuous by surface chargeBn1 0Normal B(H) field is zero on conductor.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators32
Boundary conditionsContinuity at the boundary for the tangential fields requires:Now define :Ei Er Et(1)Hi Hr Ht(2)Ei Z1HiEr Z1HrFix signs whendefining impedance!Et Z2HtSubstituting into (1) and (2) and eliminating Er givesTransmission coefficientMassachusetts Institute of TechnologyEt2Z 2τ Ei Z1 Z 2RF Cavity and Components for Accelerators33
Plane Wave in Dispersive Media Recall the Maxwell’s equations: E jω B B jω D J B 0 D ρvMassachusetts Institute of Technology E ( x , y , x; t ) E ( x , y , z )e jωt B E t Bj ωt E ( x , y , z )e t 1 ( E ) B E jωBjωRF Cavity and Components for Accelerators34
Plane Wave in Dispersive Mediafar, for lossless media, we considered J 0, and ρv 0 but,there are actually two types of current and one of them shouldnot be ignored. So Totalcurrent is a sum of the Source current and Conductioncurrent. J Jc Jo set J c σE H j ωD J σ H jωεE σE J o jω ε j E J oω Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators35
Plane Wave in Dispersive MediaDefining complex permittivityσε ε jωMaxwell’s equations in a conducting media (source free) can bewritten as E jωµH H jωεE H 0 E 0Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators36
Plane Wave in Dispersive MediaWe have considered so far:Plane Wavesin Free spacePlane Wavesin IsotropicDielectricPlane Wavesin anisotropicDielectricPlane Wavesin DissipativeMedia E jωµ 0 H H jωε 0 E H 0 E 0 E jωµH H jωε 0 E E 0 H 0 E jωµH H jωεE D 0 B 0 E jωµH H jωεE E 0 H 0Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators37
Plane Wave in Dispersive MediaWave equation for dissipative media becomes: ( E ) jωµ H 2 ( E ) E jωµ( jωεE ) 2 2 E ω µεE 2 2 H ω µεHThe set of plane-wave solutions are: E x̂E0 e jκz E0 jκzH ŷ e η Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators38
Plane Wave in Dispersive Media 2 2 22Substituting into E ω µεE and H ω µεHyields the dispersion relationκ 2 ω2µεandµη εIs the complex intrinsic impedance of the isotropic media.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators39
Plane Wave in Dispersive MediaDenoting the complex values:κ κ R jκ Iη η e jφthen, j ( κ R jκ I ) z jκz x̂E0 e x̂E0 e jκ R z e κ I z x̂ E xE x̂E0 e E 0 j ( κ R jκ I ) z jφ E 0 j ( κ R jκ I ) z ŷ eH ŷ eeη η Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators40
Plane Wave in Dispersive MediaLoss tangent is defined fromσ κ κ R jκ I ω µ ε ω µ ε j ω σ ω µε 1 j ωε σis defined as loss tangentωεσ σ ε ε j ε 1 j ε′ jε′′ωε ω ε′′tan δ ε′Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators41
Plane Wave in Dispersive MediaSlightly lossy case:σ 1ωεσ σ κ ω µε 1 j ω µε 1 j ωε 2ωε κ R ω µεσ µσ κ I ω µε2ωε 2 ε2 εdp σ µMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators42
Plane Wave in Dispersive MediaHighly lossy case:σ 1ωε σ σ κ ω µε 1 j ω µε j ωε 2ωε σ ωµ (1 j )22dp δωµσMassachusetts Institute of TechnologySkin depthRF Cavity and Components for Accelerators43
Plane Wave in Dispersive MediaMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators44
Reflection & TransmissionSimilarly, substituting into (1) and (2) and eliminating EtReflection coefficientρ E r Z 2 Z1 Ei Z 2 Z1Not 1-ρWe note that τ 1 ρ, and that the values of the reflectionand transmission are the same as occur in a transmission linediscontinuity.Z1Z2ρMassachusetts Institute of TechnologyRF Cavity and Components for Acceleratorsτ45
Special case (1)(1) Medium 1: air; Medium 2: conductorZ1 377Ω Z 2 Z m 1 jσδ2Z 2So Et τ Ei EiZ1Et2then use H t H t Ei 2 H iZ2Z1This says that the transmitted magnetic field is almost doubledat the boundary before it decays according to the skin depth.On the reflection side Hi Hr implying that almost all theH-field is reflected forming a standing wave.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators46
Special case (2)(2) Medium 1: conductor; Medium 2: airReversing the situation, now where the wave is incidentfrom the conducting side, we can show that the wave isalmost totally reflected within the conductor, but that thestanding wave is attenuated due to the conductivity.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators47
Special Case (3)(2) Medium1: dielectric; Medium2: dielectricµ0Z1 ,ε1µ0Z2 ε2 ε1 1ε2ρ ε1 1ε2This result says that the reflection can be controlled by varyingthe ratio of the dielectric constants. The transmission analogycan thus be used for a quarter-wave matching device.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators48
λ/4 Matching PlateAir: εr 1Z0Plate εr' ?Dielectric εr 4ZpZ2λ/4Transmission line theory tells us that for a matchZ p Z0Z2We will see TL lectures laterZ0376.7Z 0 376.7Ω, Z 2 188Ω2εrSoMassachusetts Institute of TechnologyZ p 266Ωand ε r' RF Cavity and Components for AcceleratorsZ0 2Z249
ApplicationsThe principle of λ/4 matching is not only confined to transmissionline problems! In fact, the same principle is used to eliminatereflections in many optical devices using a λ/4 coating layer onlenses & prisms to improve light transmission efficiency.Similarly, a half-wave section can be used as a dielectric window.Ie. Full transparency. (Why?). In this case Z2 Z0 and thematching section is λ/2. Such devices are used to protect antennasfrom weather, ice snow, etc and are called radomes.Note that both applications are frequency sensitive and that thematching section is only λ/4 or λ/2 at one frequency.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators50
Oblique IncidenceThe transmission line analogy only works for normal incidence.When we have oblique incidence of plane waves on a dielectricinterface the reflection and transmission characteristics becomepolarization and angle of incidence dependent.We need to distinguish between the two different polarizations.We do this by first, explaining what a plane of incidence is, thenwe will point out the distinguishing features of each polarization.We are aiming for expressions for reflection coefficients.We note again that we are only dealing with plane wavesMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators51
Plane of IncidenceSurfacenormalyPlane of incidence containsboth direction of propagationvector and normal vector.Direction ofpropagationxDielectricinterface inx-z planezMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators52
Parallel & Perpendicular IncidencePlane of incidence is the x-y planeyyEHHExxE is Parallel to theplane of incidenceMassachusetts Institute of TechnologyE is Perpendicular to theplane of incidenceRF Cavity and Components for Accelerators53
Perpendicular incidenceyHrErθrθiHiε1µ1ε2µ2xHtMassachusetts Institute of TechnologyEiθtEtRF Cavity and Components for Accelerators54
Write math expression for fields!Ei zˆE0 exp[ jβ1 (x sin θ i y cosθ i )]E0H i ( xˆ cosθ i yˆ sin θ i ) exp[ jβ1 ( x sin θ i y cosθ i )]Z1Er zˆρ E0 exp[ jβ1 ( x sin θ r y cosθ r )]H r ( xˆ cosθ r yˆ sin θ r )ρ E0Z1exp[ jβ1 ( x sin θ r y cosθ r )]Et zˆτ E0 exp[ jβ 2 ( x sin θ t y cosθ t )]H t ( xˆ cosθ t yˆ sin θ t )Massachusetts Institute of Technologyτ E0Z2exp[ jβ 2 ( x sin θ t y cosθ t )]RF Cavity and Components for Accelerators55
How did you get that?Within the exponential: This tells the direction of propagationOf the wave. E.g. for both the incident Ei and HiPropagatingIn medium 1jβ1 ( x sin θ i y cosθ i )A component in the – x directionAnother component in the –y directionOutside the exponential tells what vector components of the fieldAre present. E.g. for Hr(xˆ cosθ r yˆ sin θ r ) x and y components of HrMassachusetts Institute of Technologyρ E0Z1RF Cavity and Components for AcceleratorsPerpendicular reflectioncoefficientE0/Z1 converts E to H56
Apply boundary conditionsTangential E fields (Ez) matches at y 0Tangential H fields (Hx) matches at y 0exp( jβ1 x sin θ i ) ρ exp( jβ1 x sin θ r ) τ exp( jβ 2 x sin θ t )We know that τ 1 ρ, so then the arguments of theexponents must be equal. Sometimes called Phase matchingin optical context. It is the same as applying the boundaryconditions.jβ1 sin θ i jβ1 sin θ r jβ 2 sin θ tMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators57
Snell’s laws and Fresnel coefficientsθ r θiThe first equation givesand from the second using β 2πλsin θ t µ1ε1sin θ iµ 2ε 2By matching the Hx components and utilizing Snell, we canobtain the Fresnel reflection coefficient for perpendicularincidence.Z 2 cosθ i Z1 cosθ tρ Z 2 cosθ i Z1 cosθ tMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators58
Alternative formAlternatively, we can use Snell to remove the θt and write it interms of the incidence angle, at the same time assumingnon-magnetic media (µ µ0 for both media).ε2 sin 2 θ iε1ρ εcosθ i 2 sin 2 θ iε1cosθ i Note how both formsreduce to the transmissionline form when θi 0This latter form is the one that is most often quoted in texts,the previous version is more generalMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators59
Some interesting observations If ε2 ε1 If ε1 ε2Then the square root is positive, ρ Is reali.e. the wave is incident from more dense toless denseANDε2sin θ i ε12Then ρ is complex and ρ 1This implies that the incident wave is totallyinternally reflected (TIR) into the more densemediumMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators60
Critical angleWhen the equality is satisfied we have the so-called criticalangle. In other words, if the incident angle is greater than orequal to the critical angle AND the incidence is from moredense to less dense, we have TIR.θ ic sin 1ε2ε1For θi θic Then ρ 1 as noted previously.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators61
Strange resultsNowε1sin θ t sin θ i so since ε1 ε 2 sin θ t 1 !ε2cosθ t 1 sin 2 θ t jAwhere A cosθ t is imaginary!ε1sin 2 θ i 1ε2What is the physical interpretation of these results? To seewhat is happening we go back to the expression for thetransmitted field and substitute the above results.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators62
Transmitted fieldpreviouslyEt zˆτ E0 exp[ jβ 2 (x sin θ t y cosθ t )] zˆτ E0 exp[ jβ 2 x sin θ t ]exp[ αy ]where α β 2 A ω µ 2ε 2cos θt jAε1sin 2 θ i 1ε2Physically, it is apparent that the transmitted field propagatesalong the surface (-x direction) but attenuates in the y directionThis type of wave is a surface wave fieldMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators63
Exampley2 air1 waterxHiAssume:εr 81σ 0µr 1EiLet θi 45 evaluate θ ic sin 1Massachusetts Institute of Technology1 6.38 81so θ i θ ic TIRRF Cavity and Components for Accelerators64
Example (ctd)Using Snell sin θ t 81sin 45 6.381cosθ t j 81sin 2 45 1 j 6.28α β2 A This means that ifthe field strength onthe surface is1Vm-1,thenλ06.28 39.5λ0Nep / mτ 1 ρ Et τ Ei 1.42Vm -1Massachusetts Institute of Technology2πChoose signto allow forattenuationin y direction1 0.50.707 81 1 1 0.50.707 81 1.42 44.6 RF Cavity and Components for Accelerators65
Evaluate the field just above the surfaceLets evaluate the transmitted E field at λ/4 above the surface. 39.49 λ0 1Et 1.42 exp Vm732.µ 4λ0 73.2 10 6 85.8dB 20 log 1.42 This means that the surface wave is very tightly bound to thesurface and the power flow in the direction normal to thesurface is zero.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators66
What about the factork0?ωµ02πf1 ωµ 0 λ0ωµ 0 cωµ 0 cµ 0k02πε0µ0This term has the dimensions of admittance, in fact11Y0 Z 0 η0ε0µ0where Z0 impedance of free space 377Ω 1 H nˆ EAnd nowη0Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators67
Propagation in conducting mediaWe have considered propagation in free space (perfect dielectricwith σ 0). Now consider propagation in conducting media whereσ can vary from a finite value to . 2 E JρStart with E µε 2 µ t tεAssuming no free chargeand the time harmonicform, gives 2 2 E ω 2 µεE jωµσE 22 E γ E 0whereMassachusetts Institute of Technologyγ 2 jωµσ µεω 2RF Cavity and Components for AcceleratorsComplex propagationcoefficient due tofinite conductivity68
Conduction current and displacement currentIn metals, the conduction current (σE) is much larger than thedisplacement current (jωε0E). Only as frequencies increase tothe optical region do the two become comparable.E.g.σ 5.8x107 for copperωε0 2πx1010x 8.854x10-12 0.556So retain only the jωµσ term when considering highly conductivematerial at frequencies below light. The PDE becomes: E jωµ 0σE 02Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators69
Plane wave incident on a conductorConsider a plane wave entering a conductive medium at normalincidence.Free spaceConducting mediumExHyMostly reflectedSome transmittedxMassachusetts Institute of TechnologyRF Cavity and Components for Acceleratorsz70
Mathematical solution 2 Ex jωµ 0σE x 02 zThe equation for this is:The solution is: jωµ 0σ zjωµ 0σ (1 j )ωµ 0σE x E0 eWe can simplify the exponent:γ 2So now γ has equal realωµ 0σ α z β zEEee αβwithand imaginary parts.x02Alternatively writeMassachusetts Institute of TechnologyRF Cavity and Components for AcceleratorsE x E0 e zδe jzδ71
Skin DepthThe last equationE x E0 e zδegives us the notion of skin depth: jzδ On the surface at z 0 we have Ex E0at one skin depth z δ we have Ex E0/eMassachusetts Institute of TechnologyRF Cavity and Components for Acceleratorsδ2ωµ 0σ 1α 1βfield has decayed to 1/eor 36.8% of value on thesurface.72
Plot of field into conductorE0E0/ezδMassachusetts Institute of Technology2δ .RF Cavity and Components for Accelerators73
Examples of skin depth6.61 10 2 δ ωµ 0σf2Copperatatat60Hz1MHz30GHzSeawaterat1 kHzMassachusetts Institute of Technologyσ 5.8x107 S/mδ 8.5x10-3 mδ 6.6x10-5 mδ 3.8x10-7 m2.52 10 2δ fσ 4 S/mδ 7.96mRF Cavity and Components for Accelerators74
Characteristic or Intrinsic Impedance ZmDefine this via the materialas before:Zm µ0µ0 σεcε jωBut again, the conduction current predominates, which meansthe second term in the denominator is large. With thisapproximation we can arrive at:Z m (1 j )ωµ 0 1 j 2σσδFor copper at 10GHz Zm 0.026(1 j) ΩMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators75
Reflection from a metal surfaceSo a reflection coefficient at metal-air interface isZm Z0 1 since Z m Z 0ρ Zm Z0We also note that as σ , Zm 0 and that ρ -1 for the caseof the perfect conductor. Thus the boundary condition for a PECis satisfied in the limit.The transmission coefficient into the metal is given by τ 1 ρMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators76
Conductors and dielectricsMaterials can behave as either a dielectric or a conductordepending on the frequency. H σE jωεErecallDisplacement current densityConduction current density3 choicesωε σ displacement current conductor current dielectricωε σ displacement current conductor current quasi conductorωε σ displacement current conductor current conductorMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators77
A rule for determining whether dielectric or conductorDielectricsQuasi ConductorsMConductorsconductorσωε 10Mσ1 ωε 100σ1 100100 ωεσ100 ωε210-1-2copperquasi conductordielectricgroundseawaterN8Massachusetts Institute of Technology9RF Cavity and Components for Accelerators10Freq 10N1178
General case: (both conduction & displacement currents) σ γ jωµσ µεω ω µε 1 ωεj 222If we now let γ α jβ, square it and equate real and imaginaryparts and then solve simultaneously for α and β. We obtain:2 1 σ 1 α ω µε 1 ωε 2 122 1 σ 1 β ω µε 1 ωε 2 Massachusetts Institute of TechnologyRF Cavity and Components for AcceleratorsNp/m12rad/m79
ApproximationsBy taking a binomial expansion of the term under the radicaland simplifying, we can obtain:Good dielectricασ2µεβω µεZwµεMassachusetts Institute of TechnologyGood conductorRF Cavity and Components for Acceleratorsωµσ2ωµσ2ωµ(1 j )2σ80
Example Problem 1:An FM radio broadcats signal traveling in the y-dirrection in airhas a magnetic field given by the phasorH (y ) 2.92 10 3 e j 0.68πy ( xˆ zˆ j )A m 1(a) Determine the frequency (in MHZ) and wavelengt h (in m).(b) Find the corresponding E (y ).(a) we haveβ ω µ o ε o 0.68π rad m 1from whichω 102MHz2π H z H x H xˆ zˆ jωε o E y yf E (y ) 1.1e j 0.68πy ( xˆ j zˆ )V m 1Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators81
Example Problem 2:A uniform plane wave of frequency 10 GHz propagates in a sufficientlylarge sample of gallium arsenide (GaAs, εr 12.9,µr 1, tanδc 5x105),which is a commonly substrate material for high-speed solid-statedevices. Find (a) the attenuation constant α in np-m-1,(b) phase velocityνpin m-s-1,and (c) intrinsic impedance ηc in Ω.Since tan δ c 5 10 4 1, we can use the approx for a good dielectric.(a) We haveσ µ ωε tan δ c µ 2π 1010 5 10 4 µ α ε22 ε2ε2π 1010 5 10 4 µ r ε r µ 0 ε 0 2 2π 1010 5 10 42 3 108Massachusetts Institute of Technology12.9 0.188 np m 1RF Cavity and Components for Accelerators82
Example Problem 2:ω(b) Since phase velocity ν p βwhere β ω µε , we have13 108νp 8.35 10 7 m s 1 .Note that the12.9µεphase velocity is 3.59 times slower that in the air.377µ 105Ω.(c) The intrinsic impedance ηc ε12.9Note that the intrinsic impedance is 3.59 times smallerthat that in air.Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators83
Example Problem3:A recent survey conducted in USA indicates that 50% of the populationis exposed to average power densities of approximately 0.005 µW-(cm)2due to VHF and UHF broadcast radiation. Find the correspondingamplitude of the electric and magnetic fields.Consider the uniform plane wave propagatin g in a lossless medium :Εx E 0 cos(ωt βz )1ηH y E 0 cos(ωt βz )where β ω µε and η µ . The Poynting vector for this wave is given byεE 02 E0 2[1 cos 2(ωt βz )]Ρ Ε H zˆE 0 cos (ωt βz ) zˆ2η η Massachusetts Institute of TechnologyRF Cavity and Components for Accelerators84
Example Problem3:S av2E1 Tp1 Tp0 [1 cos 2(ωt β z )]dt() Ρ ztdtz,ˆ 0TpTp 02η S av2E0 zˆ2ηE 02S av zˆ 0.005µW (cm ) 22ηso E 0 2 377 5 10 9 10 4 194 mV m 1E 0 194 mV m 1 515µA m 1H0 377ΩηMassachusetts Institute of TechnologyRF Cavity and Components for Accelerators85
Massachusetts Institute of Technology RF Cavity and Components for Accelerators 12 Wave Equations In any problem with unknown E, D, B, H we have 12 unknowns. To solve for these we need 12 scalar equations. Maxwell’s equations provide 3 each for the two curl equations. and 3 each for both constitutive relations (difficult .
Motive Wave. It is a five wave trend but unlike a five wave impulse trend, the Wave 4 overlaps with the Wave 1. Ending Diagonals are the last section ("ending") of a trend or counter trend. The most common is a Wave 5 Ending Diagonal. It is a higher time frame Wave 5 trend wave that reaches new extremes and the Wave 3:5 is beyond the .
The converses are easily proved, rr S(x) 0, and rr V(x) 0 There are two very important consequences for the Maxwell equations. (a)From the source free Maxwell equations (eqs. three and four) one nds that B r A (1.12) E 1 c @ tAr (1.13) (b)Current conservation follows by manipulating the sourced maxwell equations (eqs. one and two) @ tˆ .
Maxwell’s theory combined in one stroke electricity, magnetism, and light into an eternal braid1. The Maxwell equations predict the existence of light 1J. R. Pierce famously wrote “To anyone who is motivated by anything beyond the most narrowly practical, it is worthwhile to understand Maxwell’s equations simply for the good of his soul .
Wave a and Wave c are constructed of five waves as Elliott originally proposed. As opposed to the five wave impulse move in Elliott’s original version that could form either a Wave 1, Wave 3, Wave 5, Wave A or Wave C the harmonic version can only f
So, the wave 1, wave 3 and wave 5 are parts of impulsive wave in upward direction. [6] Though Elliott waves follow many rules but three basic rules are followed by each wave to interpret Elliott wave. These guidelines are unbreakable. These rules are as follow: Rule 1: Wave 2 is not retracted more than 100% of wave 1.
So, the wave 1, wave 3 and wave 5 are parts of impulsive wave in upward direction. [2] Though Elliott waves follow many rules but three basic rules are followed by each wave to interpret Elliott wave. These guidelines are unbreakable. These rules are as follow: Rule 1: Wave 2 is not retracted more than 100% of wave 1.
ANSYS Maxwell V16 Training Manual . 2013 10 Release 14.5 D. ANSYS Maxwell About ANSYS Maxwell –ANSYS Maxwell is a high-performance interactive software package that uses finite element .
instruction method where learners of different levels form small groups and work together towards a specific objective. Learners take the responsibility of their own learning and of those in the group so the success of one member is a success of all members. Piaget (1932, in Webb, 2009: 3) argues that cognitive conflict leads to higher levels of reasoning and learning. When a student notices a .
