Lecture 23 Oblique Incidence and Reflection

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/


2/21

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



3/21

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



4/21

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/


5/21

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?



6/21

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



7/21

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/


8/21

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.



9/21

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



10/21

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



11/21


12/21

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)



13/21

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/


14/21

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



15/21

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 )


I Ob i


16/21

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



17/21

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/


18/21

P ll l P l i ti (II)


19/21

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


T ti l B d C diti


20/21

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


The Total Field (Again)


21/21

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


Lecture 23 Oblique Incidence and Reflection

Documents

Transcript of Lecture 23 Oblique Incidence and Reflection