Review of TEM WavesConditions Imposed by HelmholtzPropagation VectorWavefrontE is a ``Normal'' WaveH is also a ``Normal'' WaveTEM WavesWave PolarizationElliptical PolarizationOblique Inc. on a Cond. BoundaryPerpendicular PolarizationConductive Boundary ConditionSnell's LawThe Total FieldImportant ObservationsTE WavesParallel Polarization (I)Parallel Polarization (II)Tangential Boundary ConditionsThe Total Field (Again)EECS 117Lecture 23: Oblique Incidence and ReflectionProf. NiknejadUniversity of California, BerkeleyUniversity of California, Berkeley EECS 117 Lecture 23 – p. 1/21Review of TEM WavesWe found that E(z) = ˆxEi0e−jβzis a solution toMaxwell’s eq. But clearly this wave should propagate inany direction and the physics should not change. Weneed a more general formulation.Consdier the following “plane wave”E(x, y, z) = E0e−jβxx−jβyy−jβzzThis function also satisfies Maxwell’s wave eq. In thetime-harmonic case, this is the Helmholtz eq.∇2E + k2E = 0where k = ω√µǫ =ωcUniversity of California, Berkeley EECS 117 Lecture 23 – p. 2/21Conditions Imposed by HelmholtzEach component of the vector must satisfy the scalarHelmholtz eq.∇2Ex+ k2Ex= 0∂2∂x2+∂2∂y2+∂2∂z2Ex+ k2Ex= 0Carrying out the simple derivatives−β2x− β2y− β2z+ k2= 0β2x+ β2y+ β2z= k2Define k = ˆxβx+ ˆyβy+ ˆzβzas the propagation vectorUniversity of California, Berkeley EECS 117 Lecture 23 – p. 3/21Propagation VectorThe propagation vector can be written as a scalar timesa unit vectork = kˆanThe magnitude k is given by k = ω√µǫAs we’ll show, the vector direction ˆandefines thedirection of propagation for the plane waveUsing the defined relations, we now haveE(r) = E0e−jk·rβx= k · ˆx = kˆan· ˆxβy= k · ˆy = kˆan· ˆyβz= k ·ˆz = kˆan·ˆzUniversity of California, Berkeley EECS 117 Lecture 23 – p. 4/21Wavefrontwavefront planerkRecall that a wavefront is a surface of constant phasefor the waveThen ˆan· R = constant defines the surface of constantphase. But this surface does indeed define a planesurface. Thus we have a plane wave. Is it TEM?University of California, Berkeley EECS 117 Lecture 23 – p. 5/21E is a “Normal” WaveSince our wave propagations in a source free region,∇ · E = 0. OrE0· ∇e−jkˆan·r= 0∇e−jkˆan·r=ˆx∂∂x+ ˆy∂∂y+ ˆz∂∂ze−j(βxx+βyy+βzz)= −j(βxˆx + βyˆy + βzˆz)e−j(βxx+βyy+βzz)So we have−jk(E0· ˆan)e−jkˆan·r= 0This implies that ˆan· E0= 0, or that the wave ispolarized transverse to the direction of propagationUniversity of California, Berkeley EECS 117 Lecture 23 – p. 6/21H is also a “Normal” WaveSince H(r) =1−jωµ∇ × E, we can calculate the directionof the H fieldRecall that ∇ × (fF) = f∇ × F + ∇f × FH(r) =1jωµ∇e−jk·r× E0H(r) =1jωµE0×−jke−jk·rH(r) =kωµˆan× E(r) =1ηˆan× E(r)η =µωk=µωω√ǫµ=pµ/ǫUniversity of California, Berkeley EECS 117 Lecture 23 – p. 7/21TEM WavesSo we have done it. We proved that the equationsE(r) = E0e−jk·rH(r) =1ηˆan× E(r)describe plane waves where E is perpendicular to thedirection of propagation and the vector H isperpendicular to both the direction of propagation andthe vector EThese are the simplest general wave solutions toMaxwell’s equations.University of California, Berkeley EECS 117 Lecture 23 – p. 8/21Wave PolarizationNow we can be more explicit when we say that a waveis linearly polarized. We simply mean that the vector Elies along a line. But what if we take the superpositionof two linearly polarized waves with a 90◦time lagE(z) = ˆxE1(z) + ˆyE2(z)The first wave is ˆx-polarized and the second wave isˆy-polarized. The wave propagates in the ˆz directionIn the time-harmonic domain, a phase lag correspondsto multiplication by −jE(z) = ˆxE10e−jβz− jˆyE20e−jβzUniversity of California, Berkeley EECS 117 Lecture 23 – p. 9/21Elliptical PolarizationIn time domain, the waveform is described by thefollowing equationE(z, t) = ℜE(z)ejωtE(z, t) = ˆxE10cos(ωt − βz) + ˆyE20sin(ωt − βz)At a paricular point in space, say z = 0, we haveE(0, t) = ˆxE10cos(ωt) + ˆyE20sin(ωt)Thus the wave rotates along an elliptical path in thephase front!We can thus create waves that rotate in one direction orthe other by simply adding two linearly polarized waveswith the right phaseUniversity of California, Berkeley EECS 117 Lecture 23 – p. 10/21Oblique Inc. on a Cond. BoundaryLet the ˆx-ˆy plane definethe plane of incidence.Consider the polarizationof a wave impingingobliquely on the boundary.We can identify twopolarizations,perpendicular to the planeand parallel to the plane ofincidence. Let’s solvethese problems separately.Any other polarized wavecan always be decom-posed into these two casesEiHiErHrψiψrkikrψ = ∞A perpendicularly polarizedwave.University of California, Berkeley EECS 117 Lecture 23 – p. 11/21Perpendicular PolarizationLet the angle of incidence and relfection we given by θiand θr. Let the boundary consists of a perfect conductorEi= ˆyEi0e−jk1·rwhere k1= k1ˆaniand ˆani= ˆx sin θi+ ˆz cos θiEi= ˆyEi0e−jk1(x sin θi+z cos θi)Hi=1ηian× EiFor the reflected wave, similarly, we haveˆanr= ˆx sin θr−ˆz cos θrso thatEr= ˆyEr0e−jk1(x sin θr−z cos θr)University of California, Berkeley EECS 117 Lecture 23 – p. 12/21Conductive Boundary ConditionThe conductor enforces the zero tangential fieldboundary condition. Since all of E is tangential in thiscase, at z = 0 we haveE1(x, 0) = Ei(x, 0) + Er(x, 0) = 0Substituting the relations we haveˆyEi0e−jk1x sin θi+ Er0e−jk1x sin θr= 0For this equation to hold for any value of x and θ, thefollowing conditions must holdEr0= −Ei0θi= θrUniversity of California, Berkeley EECS 117 Lecture 23 – p. 13/21Snell’s LawWe have found that the angle of incidence is equal tothe angle of reflection (Snell’s law)The total field, therefore, takes on an interesting form.The reflected wave is simplyEr= −ˆyEi0e−jk1(x sin θi−z cos θi)Hr=1η1ˆanr× ErHr=Ei0η1(−ˆx cos θi−ˆz sin θi)e−jk1(x sin θi−z cos θi)The total field is thusE1= Ei+ Er= ˆyEi0e−jk1z cos θi− ejk1z cos θie−jk1x sin θiUniversity of California, Berkeley EECS 117 Lecture 23 – p. 14/21The Total FieldSimplifying the expression for the total fieldE1= −ˆy2jEi0sin(k1z cos θi)| {z }standing wavee−jk1x sin θi| {z }prop.