EECS 117. Lecture 23: Oblique Incidence and Reflection. Prof. Niknejad. University of California, Berkeley

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1 University of California, Berkeley EECS 117 Lecture 23 p. 1/2 EECS 117 Lecture 23: Oblique Incidence and Reflection Prof. Niknejad University of California, Berkeley

2 University of California, Berkeley EECS 117 Lecture 23 p. 2/2 Review of TEM Waves We found that E(z) = ˆxE i0 e jβz is a solution to Maxwell s eq. But clearly this wave should propagate in any direction and the physics should not change. We need a more general formulation. Consdier the following plane wave E(x,y,z) = E 0 e jβ xx jβ y y jβ z z This function also satisfies Maxwell s wave eq. In the time-harmonic case, this is the Helmholtz eq. where k = ω µǫ = ω c 2 E + k 2 E = 0

3 University of California, Berkeley EECS 117 Lecture 23 p. 3/2 Conditions Imposed by Helmholtz Each component of the vector must satisfy the scalar Helmholtz eq. 2 E x + k 2 E x = 0 ( 2 x y z 2 Carrying out the simple derivatives ) E x + k 2 E x = 0 β 2 x β2 y β2 z + k2 = 0 β 2 x + β 2 y + β 2 z = k 2 Define k = ˆxβ x + ŷβ y + ẑβ z as the propagation vector

4 University of California, Berkeley EECS 117 Lecture 23 p. 4/2 Propagation Vector The propagation vector can be written as a scalar times a unit vector k = kâ n The magnitude k is given by k = ω µǫ As we ll show, the vector direction â n defines the direction of propagation for the plane wave Using the defined relations, we now have E(r) = E 0 e jk r β x = k ˆx = kâ n ˆx β y = k ŷ = kâ n ŷ β z = k ẑ = kâ n ẑ

5 University of California, Berkeley EECS 117 Lecture 23 p. 5/2 Wavefront r k wavefront plane Recall that a wavefront is a surface of constant phase for the wave Then â n R = constant defines the surface of constant phase. But this surface does indeed define a plane surface. Thus we have a plane wave. Is it TEM?

6 University of California, Berkeley EECS 117 Lecture 23 p. 6/2 E is a Normal Wave Since our wave propagations in a source free region, E = 0. Or ( ) E 0 e jkâ n r = 0 ( e jkâ n r ) = So we have ( ˆx x + ŷ y + ẑ ) e j(β xx+β y y+β z z) z = j(β xˆx + β y ŷ + β z ẑ)e j(β xx+β y y+β z z) jk(e 0 â n )e jkâ n r = 0 This implies that â n E 0 = 0, or that the wave is polarized transverse to the direction of propagation

7 University of California, Berkeley EECS 117 Lecture 23 p. 7/2 H is also a Normal Wave Since H(r) = jωµ 1 E, we can calculate the direction of the H field Recall that (ff) = f F + f F H(r) = 1 jωµ ( e jk r) E 0 H(r) = 1 jωµ E 0 H(r) = k ( jke jk r) ωµân E(r) = 1 ηân E(r) η = µω k = µω ω ǫµ = µ/ǫ

8 University of California, Berkeley EECS 117 Lecture 23 p. 8/2 TEM Waves So we have done it. We proved that the equations E(r) = E 0 e jk r H(r) = 1 ηân E(r) describe plane waves where E is perpendicular to the direction of propagation and the vector H is perpendicular to both the direction of propagation and the vector E These are the simplest general wave solutions to Maxwell s equations.

9 University of California, Berkeley EECS 117 Lecture 23 p. 9/2 Wave Polarization Now we can be more explicit when we say that a wave is linearly polarized. We simply mean that the vector E lies along a line. But what if we take the superposition of two linearly polarized waves with a 90 time lag E(z) = ˆxE 1 (z) + ŷe 2 (z) The first wave is ˆx-polarized and the second wave is ŷ-polarized. The wave propagates in the ẑ direction In the time-harmonic domain, a phase lag corresponds to multiplication by j E(z) = ˆxE 10 e jβz jŷe 20 e jβz

10 University of California, Berkeley EECS 117 Lecture 23 p. 10/2 Elliptical Polarization In time domain, the waveform is described by the following equation E(z,t) = R ( E(z)e jωt) E(z,t) = ˆxE 10 cos(ωt βz) + ŷe 20 sin(ωt βz) At a paricular point in space, say z = 0, we have E(0,t) = ˆxE 10 cos(ωt) + ŷe 20 sin(ωt) Thus the wave rotates along an elliptical path in the phase front! We can thus create waves that rotate in one direction or the other by simply adding two linearly polarized waves with the right phase

11 University of California, Berkeley EECS 117 Lecture 23 p. 11/2 Oblique Inc. on a Cond. Boundary Let the ˆx-ŷ plane define the plane of incidence. Consider the polarization of a wave impinging obliquely on the boundary. We can identify two polarizations, perpendicular to the plane and parallel to the plane of incidence. Let s solve these problems separately. Any other polarized wave can always be decomposed into these two cases H i k r k i E i E r H r A perpendicularly polarized wave. ψ r ψ i ψ =

12 University of California, Berkeley EECS 117 Lecture 23 p. 12/2 Perpendicular Polarization Let the angle of incidence and relfection we given by θ i and θ r. Let the boundary consists of a perfect conductor E i = ŷe i0 e jk 1 r where k 1 = k 1 â ni and â ni = ˆx sin θ i + ẑ cosθ i E i = ŷe i0 e jk 1(x sin θ i +z cos θ i ) H i = 1 η i a n E i For the reflected wave, similarly, we have â nr = ˆx sin θ r ẑ cosθ r so that E r = ŷe r0 e jk 1(x sin θ r z cos θ r )

13 University of California, Berkeley EECS 117 Lecture 23 p. 13/2 Conductive Boundary Condition The conductor enforces the zero tangential field boundary condition. Since all of E is tangential in this case, at z = 0 we have E 1 (x, 0) = E i (x, 0) + E r (x, 0) = 0 Substituting the relations we have ) ŷ (E i0 e jk 1x sin θ i + E r0 e jk 1x sin θ r = 0 For this equation to hold for any value of x and θ, the following conditions must hold E r0 = E i0 θ i = θ r

14 University of California, Berkeley EECS 117 Lecture 23 p. 14/2 Snell s Law We have found that the angle of incidence is equal to the angle of reflection (Snell s law) The total field, therefore, takes on an interesting form. The reflected wave is simply E r = ŷe i0 e jk 1(x sin θ i z cos θ i ) H r = 1 η 1 â nr E r H r = E i0 η 1 ( ˆx cosθ i ẑ sin θ i )e jk 1(x sin θ i z cos θ i ) The total field is thus E 1 = E i + E r = ŷe i0 ( e jk 1z cosθ i e jk 1z cos θ i ) e jk 1x sin θ i

15 University of California, Berkeley EECS 117 Lecture 23 p. 15/2 The Total Field Simplifying the expression for the total field E 1 = ŷ2je i0 sin(k 1 z cos θ i ) }{{} standing wave e jk 1x sin θ i }{{} prop. wave H 1 = 2E i0 η 1 ( ˆx cosθ i cos(k 1 z cosθ i )e jk 1x sin θ i + ẑ j sinθ i sin(k 1 z cos θ i )e jk 1x sin θ i )

16 University of California, Berkeley EECS 117 Lecture 23 p. 16/2 Important Observations In the ẑ-direction, E 1y and H 1x maintain standing wave patterns (no average power propagates since E and H are 90 out of phase. This matches our previous calculation for normal incidence Waves propagate in the ˆx-direction with velocity v x = ω/(k 1 sin θ i ) Wave propagation in the ˆx-direction is a non-uniform plane wave since its amplitude varies with z

17 University of California, Berkeley EECS 117 Lecture 23 p. 17/2 TE Waves Notice that on plane surfaces where E = 0, we are free to place a conducting plane at that location without changing the fields outside of the region In particular, notice that E = 0 when This holds for m = 1, 2,... sin(k 1 z cosθ i ) = 0 k 1 z cosθ i = 2π λ 1 z cosθ i = mπ So if we place a plane conductor at z = mλ 1 2cos θ i, there will be a guided wave traveling between the two planes in the ˆx direction Since E 1x = 0, this wave is a TE wave as H 1x 0

18 University of California, Berkeley EECS 117 Lecture 23 p. 18/2 Parallel Polarization (I) k r E r H r Now the wave is polarized in the plane of incidence. E i k i H i ψ r ψ i ψ = The approach is similar to before but the tangential component of the electric field depends on the angle of incidence

19 University of California, Berkeley EECS 117 Lecture 23 p. 19/2 Parallel Polarization (II) Now consider an incident electric field that is in the plane of polarization E i = E i0 (ˆx cosθ i ẑ sinθ i )e jk r H i = y E i0 η 1 e jk r Likewise, the reflected wave is expressed as E r = E r0 (ˆx cosθ i + ẑ sin θ i )e jk r H r = y E r0 η 1 e jk r Note that k i,r r = x sin θ i,r ± z cosθ i,r

20 University of California, Berkeley EECS 117 Lecture 23 p. 20/2 Tangential Boundary Conditions Since the sum of the reflected and incident wave must have zero tangential component at the interface E i0 cosθ i e jk 1x sin θ i + E r0 cos θ r e jk 1x sin θ r = 0 These equations must hold for all θ. Thus E r0 = E i0 as before Thus we see that these equations can hold for all values of x if and only if θ i = θ r

21 University of California, Berkeley EECS 117 Lecture 23 p. 21/2 The Total Field (Again) Using similar arguments as before, when we sum the fields to obtain the total field, we can observe a standing wave in the ẑ direction and wave propagation in the ˆx direction. Note that the magnetic field H = Hŷ is always perpendicular to the direction of propagation but the electric field has a component in the ˆx direction. This type of wave is known as a TM wave, or transverse magnetic wave