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8/2/2019 Lecture 23 Oblique Incidence and Reflection
1/21
EECS 117
Lecture 23: Oblique Incidence and Reflection
Prof. Niknejad
University of California, Berkeley
Universit of California Berkele EECS 117 Lecture 23 . 1/
8/2/2019 Lecture 23 Oblique Incidence and Reflection
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Review of TEM Waves
We found that E(z) = xEi0ejz is a solution to
Maxwells eq. But clearly this wave should propagate inany direction and the physics should not change. We
need a more general formulation.
Consdier the following plane wave
E(x,y,z) = E0e
jxx
jyy
jzz
This function also satisfies Maxwells wave eq. In thetime-harmonic case, this is the Helmholtz eq.
2E + k2E = 0where k =
=
c
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Conditions Imposed by Helmholtz
Each component of the vector must satisfy the scalarHelmholtz eq.
2Ex + k
2Ex = 02
x2+
2
y2+
2
z2
Ex + k
2Ex = 0
Carrying out the simple derivatives
2x 2y 2z + k2 = 0
2x + 2y + 2z = k2
Define k = xx + yy + zz as the propagation vector
Universit of California Berkele EECS 117 Lecture 23 . 3/
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Propagation Vector
The propagation vector can be written as a scalar timesa unit vector
k = kan
The magnitude k is given by k = As well show, the vector direction an defines thedirection of propagation for the plane wave
Using the defined relations, we now have
E(r) = E0ejkr
x = k x = kan xy = k
y = kan
y
z = k z = kan zUniversit of California Berkele EECS 117 Lecture 23 . 4/
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Wavefront
wav
efront
plane
r
k
Recall that a wavefront is a surface of constant phasefor the wave
Then an R = constant defines the surface of constantphase. But this surface does indeed define a planesurface. Thus we have a plane wave. Is it TEM?
Universit of California Berkele EECS 117 Lecture 23 . 5/
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E is a Normal Wave
Since our wave propagations in a source free region, E = 0. Or
E0 e
jkanr = 0ejkanr
=
x
x+ y
y+ z
z
ej(xx+yy+zz)
= j(xx + yy + zz)ej(xx+yy+zz)
So we have
jk(E0 an)ejkanr
= 0This implies that an E0 = 0, or that the wave ispolarized transverse to the direction of propagation
Universit of California Berkele EECS 117 Lecture 23 . 6/
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H is also a Normal Wave
Since H(r) = 1j
E, we can calculate the directionof the H field
Recall that (fF) = f F +f FH(r) =
1
jejkr
E0
H(r) =1
jE0
jkejkr
H(r) = k
an E(r) = 1an E(r)
= k
=
= /Universit of California Berkele EECS 117 Lecture 23 . 7/
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TEM Waves
So we have done it. We proved that the equations
E(r) = E0ejkr
H(r) =1
an 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 E
These are the simplest general wave solutions toMaxwells equations.
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Wave Polarization
Now 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 superposition
of two linearly polarized waves with a 90
time lag
E(z) = xE1(z) + yE2(z)
The first wave is x-polarized and the second wave isy-polarized. The wave propagates in the z direction
In the time-harmonic domain, a phase lag corresponds
to multiplication by jE(z) = xE10e
jz jyE20ejz
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Elliptical Polarization
In time domain, the waveform is described by thefollowing equation
E(z, t) = E(z)ejtE(z, t) = xE10 cos(t z) + yE20 sin(t z)
At a paricular point in space, say z = 0, we have
E(0, t) = xE10 cos(t) + yE20 sin(t)
Thus the wave rotates along an elliptical path in thephase front!
We can thus create waves that rotate in one direction or
the other by simply adding two linearly polarized waveswith the right phase
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Perpendicular Polarization
Let the angle of incidence and relfection we given by iand r. Let the boundary consists of a perfect conductor
Ei = yEi0ejk1r
where k1 = k1ani and ani = x sin i + z cos i
Ei = yEi0e
jk1(x sin i+z cos i)
Hi =1
ian Ei
For the reflected wave, similarly, we haveanr = x sin r z cos r so that
Er = yEr0ejk1(x sin rz cos r)
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Conductive Boundary Condition
The conductor enforces the zero tangential fieldboundary condition. Since all of E is tangential in thiscase, at z = 0 we have
E1(x, 0) = Ei(x, 0) + Er(x, 0) = 0
Substituting the relations we have
yEi0e
jk1x sin i + Er0ejk1x sin r
= 0
For this equation to hold for any value of x and , thefollowing conditions must hold
Er0 = Ei0 i = rUniversit of California Berkele EECS 117 Lecture 23 . 13/
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Snells Law
We have found that the angle of incidence is equal tothe angle of reflection (Snells law)
The total field, therefore, takes on an interesting form.The reflected wave is simply
Er = yEi0ejk1(x sin iz cos i)
Hr =1
1anr Er
Hr = Ei01(x cos i z sin i)ejk1(x sin iz cos i)
The total field is thus
E1 = Ei + Er = yEi0ejk1z cos i ejk1z cos i ejk1x sin i
Universit of California Berkele EECS 117 Lecture 23 . 14/
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The Total Field
Simplifying the expression for the total field
E1 = y2jEi0 sin(k1z cos i) standing waveejk1x sin i
prop. wave
H1 = 2Ei0
1 ( x cos i cos(k1z cos i)ejk1x sin i
+
z j sin i sin(k1z cos i)ejk1x sin i )
Universit of California Berkele EECS 117 Lecture 23 . 15/
I Ob i
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Important Observations
In the z-direction, E1y and H1x maintain standing wave
patterns (no average power propagates since Eand Hare 90 out of phase. This matches our previous
calculation for normal incidenceWaves propagate in the x-direction with velocityvx = /(k1 sin i)
Wave propagation in the x-direction is a non-uniformplane wave since its amplitude varies with z
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TE Waves
Notice that on plane surfaces where E = 0, we are freeto place a conducting plane at that location withoutchanging the fields outside of the region
In particular, notice that E = 0 when
sin(k1z cos i) = 0
k1z cos i =2
1z cos i = m
This holds for m = 1, 2, . . .
So if we place a plane conductor at z = m12cos i , therewill be a guided wave traveling between the twoplanes in the x direction
Since E1x = 0, this wave is a TE wave as H1x = 0Universit of California Berkele EECS 117 Lecture 23 . 17/
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P ll l P l i ti (II)
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Parallel Polarization (II)
Now consider an incident electric field that is in theplane of polarization
Ei = Ei0(x cos i z sin i)ejkr
Hi = yEi01
ejkr
Likewise, the reflected wave is expressed as
Er = Er0(x cos i + z sin i)ejkr
Hr = yEr01
ejkr
Note that ki,r r = x sin i,r z cos i,r
Universit of California Berkele EECS 117 Lecture 23 . 19/
T ti l B d C diti
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Tangential Boundary Conditions
Since the sum of the reflected and incident wave musthave zero tangential component at the interface
Ei0 cos iejk1x sin i
+ Er0 cos rejk1x sin r
= 0
These equations must hold for all . Thus Er0 = Ei0 asbefore
Thus we see that these equations can hold for allvalues of x if and only if i = r
Universit of California Berkele EECS 117 Lecture 23 . 20/
The Total Field (Again)
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The Total Field (Again)
Using similar arguments as before, when we sum thefields to obtain the total field, we can observe astanding wave in the z direction and wave propagation
in the x direction.Note that the magnetic field H = Hy is alwaysperpendicular to the direction of propagation but the
electric field has a component in the x direction.This type of wave is known as a TM wave, ortransverse magnetic wave
Universit of California Berkele EECS 117 Lecture 23 . 21/