1 ee 543 theory and principles of remote sensing reflection and refraction from a planar interface

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EE 543Theory and Principles of

Remote Sensing

Reflection and Refraction from a Planar Interface

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Outline

1. Reflection and Transmission at a planar interface

– Boundary conditions– Fresnel reflection coefficients– Special Cases: Total reflection (critical

angle), Total transmission (Brewster angle)

2. Rough Surface3. Scattering from objects

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Reflection and Transmission

• When we consider interaction of em waves with a target, we must account for the effects of boundaries between media.

• These boundary effects give rise to changes in the amplitude, phase and direction of propagation waves.

• These in turn either carry information about the target or cause clutter for the received signal.

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Boundary Conditions

Medium 1

Medium 2

niE, H

* Boundary conditions are a direct consequence of Maxwell’s equations.* They can be derived from the integral form by assuming an infinitesimal closed path, or an infinitesimal volume across the boundary.

t

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dA 2

dA 1

Continuity of the Normal Component

1 2 3

0

ˆˆ ˆ

eS V

S

S dA dA dA L z

ds dv Q

ds

ds nds nds tds

D.

B.

D. D D D

Medium 1

Medium 2 n1D1

D2

z 0

Medium 1

Medium 2 n2

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Continuity of the Tangential Component

Medium 1

Medium 2t1

D1

D2

z 0

Medium 1

Medium 2

t2

t

C S

C S S

dl dst

dl ds dst

E B

H D J

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General Form of Boundary Conditions

1 2 1 2

1 2 1 2

1 2 tan1 tan 2

1 2 tan1 tan 2

ˆ 0

ˆ

ˆ 0

ˆ

n n

n n s

s

n B B B B

n D D D D

n E E E E

n H H H H J

Medium 1

Medium 2

niE, H

Components of E and H (Normal and Tangential):

H

n

Hn

Htan

n

E

En

Etan

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Special Cases – Dielectric Boundary

• For dielectric material or material with finite conductivity, Js = 0; i.e. surface current does not exist. Only volume current exists. Therefore:

1 2 1 2

1 2 1 2

1 2 tan1 tan 2

1 2 tan1 tan 2

ˆ 0

ˆ

ˆ 0

ˆ 0

n n

n n s

n B B B B

n D D D D

n E E E E

n H H H H

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Special Cases – Perfect Conductors• For perfect conductors, there can not be any

voltage difference between any two points on the surface. Thus, Etan2 = 0 on the surface.

• Also, due to the same reasoning there can not be any fields inside a perfect conductor. Therefore:

2 2

1 1

1 1

1 tan1

1 tan1

0

ˆ 0

ˆ

ˆ 0

ˆ

n

n s

s

E H

n B B

n D D

n E E

n H H J

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Example 1

A region contains a perfectly conducting half-space and air. We know that the surface current on the perfect conductor is . What is the tangential H field in air just above the conductor?

ˆ2 A/msJ x

y

x

Js

H

tan

tan tan

1 tan1

1

1 1

1 1

ˆ

ˆ ˆ 2

where in the most general sense:

ˆ

Thus

ˆ ˆ ˆ2 2 A/m

s

y

n H H J

y H x

H H yH

y H x H z

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Plane of Incidence

ni

Defined by unit vectors i and n, where i is the direction of the incident fields, and n is the unit normal vector to the boundary between the two medium.

In this example, x-z plane is the plane of incidence.

x

z

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Plane of Incidence and E, H Fields

EH

i

Incident fields Ei, Hi lie on a plane (P) perpendicular to i.

Therefore, Ei, Hi can be decomposed into two basis (orthogonal) vectors that describe (P).

There are infinitely many possible such pair of components.

For instance, if i=z then x, y or x+y, x-y would be possible orthogonal vectors.

P

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E and H Decomposition

• We observe that E and H can be decomposed into infinitely many pairs of orthogonal vectors that lie on the plane P.

• One choice is to have t = i x n as one component, where i: incidence direction, and n: normal to the boundary.

• Since t is defined as the cross product of i with another vector, t is orthogonal to i, and thus can be a valid component of E or H.

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Polarization• The selection of the orthogonal components

for a reflection and transmission problem is usually done with respect to the plane of incidence.

• These define two orthogonally polarized components of E and H fields.

• These are called:– Perpendicular Polarization (TE, Horizontal)– Parallel Polarization (TM, Vertical)

Defined with respect to the plane of incidence

Defined for E field with respect to the interface

Defined for E field

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Perpendicular (TE) Polarization (Horizontal)

• The electric field is perpendicular (transverse, TE) to the plane of incidence.

ni

E

Plane formed by i and n

REMARK: Since n is on the plane of incidence, this condition (TE) implies that E lies on the interface; i.e. is horizontal.

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Parallel Polarization (TM) (Vertical)

• The electric field does not have a component perpendicular to the plane of incidence; i.e. E lies on the plane of incidence.

• Hence, the magnetic field is perpendicular (transverse, TM) to the plane of incidence.

n

i

H

REMARK: Since n is on the plane of incidence, this condition (TM) implies that E has a component perpendicular to the interface i.e. is vertical.

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Decomposition into Two Polarizations

Therefore, both E and H fields can be represented in the most general sense as a sum of these two orthogonal polarizations; i.e.

E E E

H H H

TE(perpendicular)

TM(parallel)

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Example: Planar Boundary with a Dielectric Interface

x

z

i

E, H

n

n: normal to the surfacei: direction of propagation

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Reflection and Transmission – Perpendicular Polarization (TE)

x

i

Ei

n

z

X

Hi i

o

t

r

r

1, 1

2, 2

E is perpendicular to the plane of incidence. H lies on the plane of incidence.

Plane of incidence: x-z

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Calculation of TE Coefficients

• Due to the law of reflection (from matching the boundary conditions)

• The incident, reflected and transmitted fields can be written in terms of the propagation direction, and reflection and transmission coefficients for TE waves.

1 i r

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TE Geometry

1 1 1

1 1 1

2 2 2

ˆ ˆ

ˆ ˆ

ˆ ˆ

i x z

r x z

x z

k k x k z

k k x k z

k k x k z

1 1

1 1

2 2

ˆ ˆ ˆsin cos

ˆ ˆ ˆsin cos

ˆ ˆ ˆsin cos

i x z

o x z

t x z

Wave vectors:

1 1 1

1 1 1

sin

cosx

z

k k

k k

2 2 2

2 2 2

sin

cosx

z

k k

k k

Let the unit vectors along the direction of propagation be:

incident

reflected

transmitted

1

z

o

1k1i

t

k1r

k2

1 1 1 1 1

2 2 2

i rk k k w

k w

i

t

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Electric Field Expressions, TE

1 11

1 11

2 22

ˆ

ˆ

ˆ

ˆ ˆ

ˆ ˆR R

ˆ ˆT T

x z

x z

x z

j k x k zjk i r

i o o

j k x k zjk o r

r o o

j k x k zjk t r

i o o

E y E e y E e

E y E e y E e

E y E e y E e

Unknowns:

R : Reflection coefficient

T : Transmission coefficient

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Magnetic Field Expressions, TEThe corresponding magnetic fields are obtained from

Maxwell’s equations

1 1

1 1

2 2

1 1

1 1

2 1

1 ˆ ˆ ˆ

1ˆ ˆ ˆR

1 ˆ ˆ ˆT

x z

x z

x z

j k x k zoi i

j k x k zor r

j k x k zot t

EH i E i y e

EH o E o y e

EH t E t y e

1 1

1 1

2 2

ˆ ˆ ˆ ˆcos sin

ˆ ˆ ˆ ˆcos sin

ˆ ˆ ˆ ˆcos sin

i y x z

o y x z

t y x z

where,

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Application of Boundary Conditions

Boundary conditions on the surface (dielectric medium):

1 2

1 2

tan tan0 0

tan tan0 0

z z

z z

E E

H H

x

i

Ei

X

Hi

o

t

1, 1

2, 2

z

XEr

Hr

X

Ht

Et

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BC for Electric Field

1

1 tan tan

2

2 tan

1 2

1 2

tan 0 0

tan 0 0

tan tan0 0

ˆ 1 R

ˆ T

1 R T

x

x

x x

jk x

i r oz z

jk x

t oz z

jk x jk x

z z

E E E y E e

E E y E e

E E e e

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BC for Magnetic Field

1

1 tan tan

2

2 tan

1 2

1 2

tan 10 01

tan 20 02

tan tan 2 1 1 20 0

ˆ cos 1 R

ˆ T cos

cos 1 R cos T

x

x

x x

jk xoi rz z

jk xotz z

jk x jk x

z z

EH H H x e

EH H x e

H H e e

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Equations for TEIn summary, the following has to be satisfied

for all points across the boundary; i.e. for all x values.

1 2

1 2

2 1 1 2

1 R T

cos 1 R cos T

x x

x x

jk x jk x

jk x jk x

e e

e e

This implies that exponential terms have to be equal; i.e.

1 2

1 2 1 1 2 2sin sin

x x

x x

jk x jk xe e

k k k k

Snell’s Law

Phase matching condition

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Solution for TE

2 1 1 2

1 R T

cos 1 R cos T

2 1 1 2

2 1 1 2

2 1

2 1 1 2

cos cosR

cos cos

2 cosT

cos cos

Thus, the boundary conditions result in:

We obtain

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Reflection and Transmission – Parallel Polarization (TM)

x

i

Ei

n

XHi

i

o

t

t

r

1, 1

2, 2

H is perpendicular to the plane of incidence. E lies on the plane of incidence; i.e. is parallel to it.

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Calculation of TM Coefficients

• Due to duality principle, the reflection and transmission coefficients for TM can be obtained from the TE case by letting

-

E H

H E

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Solution for TM

1 1 2 2

1 1 2 2

1 1

1 1 2 2

cos cosR

cos cos

2 cosT

cos cos

HW Problem: Carry out the derivation for TM mode following similar steps as in TE case, and prove the equation above.

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Power Coefficients

• The R, T coefficients describe field (E,H) behavior.

• The power reflection coefficient (reflectivity) and transmission coefficient (transmissivity) are defined with respect to the normal components of the time average Poynting vector.

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Power ConservationSav_z, inc Sav_z, ref

Sav_z, tr

Sav_x, inc

Sav_x, ref

Sav_x, tr

Sav_x, inc

Sav_x, ref

Sav_x, tr

Sav_z, inc Sav_z, trSav_z, ref

= +

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Reflectivity and Transmissivity

ˆ

ˆ

ˆ

ˆ

r

i

t

i

av

av

av

av

z S

z S

z S

z S

*12 Re ( ) ( )avS E r H r

where

Time average Poynting Vector

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Reflectivity and Transmissivity, TE

21

21

21

1

22 2

22 2

21 1

1 1

cosRe

cosRe

cos cosRe Re

cos cosRe Re

o

o

o

o

R ER

E

T ET

E

1 R T Note that the field coefficients satisfy:

While the power coefficients satisfy energy conservation:

1

Due to b.c.

Due to power conservation

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Reflectivity and Transmissivity, TM

2

22 2

1 1

Re cos

Re cos

R

T

Note that cos(2) can be a complex number for certain incidence angles, such as intotal reflection phenomenon.

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Total Reflection (Critical Angle)

• If the wave is incident from a dense to a less dense medium (i.e. k1>k2), it is possible to have no transmission into the second medium.

• Recall Snell’s Law:

1 1 2 2

12 1

2

sin sin

sin sin

k k

k

k

Can be > 1 for certain values of 1

>1 will result in pure imaginary 2

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Transmission Angle for Incidence Beyond Critical Angle

12

12

12

12 1

2

2

2 2

2 2

2 2

2

2

2 2

1 1

2

2

2 2

1 1

2

2

sin sin 1 ?????

cos 1 sin

sin1

sin1

sin1

k

k

k

k

k

k

kj jb

kPure imaginary

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Critical Angle• The smallest incidence angle at which no real

transmission angle exists is called the critical angle.

• Thus all waves that arrive at incidence angles greater than the critical angle suffer total reflection.

21 2

1

21 2

1

2

1

sin real value for

sin com

sin

plex value for

critical angl

e

c

k

k

k

k

k

k

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Total Reflection for Incidence Beyond Critical Angle

2 1 1

2 1 1

1 1 2

1 1 2

cos1

cos

cosR 1

cos

jbR

jb

jb

jb

2 1 1 2

2 1 1 2

1 1 2 2

1 1 2 2

cos cosR

cos cos

cos cosR

cos cos

2

2

1

1

R

R

0

Therefore, for power considerations

and

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Example – Light source under water

For an isotropic (i.e. equal radiation in all directions) light source, only light rays within a cone of c can be transmitted to air. The permittivity of water at optical frequencies is 1.77o. Calculate the value of the critical angle.

1 1 1sin sin 49

1.77oo

c

w

k

k

k1

k2

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Total Transmission (Brewster Angle)

• There exists an angle at which total transmission occurs for TM (parallel = vertically polarized) wave.

2

1

tan B

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Proofo

2 1 o o i

1 1 2 21 2

1 1 2 2

1 22

2 1

11 2 2 2

2

2 222

1

2 212

2

Let ; ;

cos cos0 or

cos cos

cos cos cos

From Snell's Law:

sin sin sin sin

cos cos

sin sin

i i

i

B B

B B

B

B

k w

R

kk k

k

2 22 1

1 2

2 2

2

1 2

2

1

Adding the two: 1 cos sin

Using cos =1-sin

sin

tan

B B

B B

B

B

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Total Transmission

• Note that total transmission phenomenon only occurs for TM (parallel) polarization.

• As a result, randomly polarized waves become polarized on reflection. (TM portion is not reflected back)

• Note that:– for incidence angles smaller that B, R// term is

less than zero.– for incidence angles larger than B, R// term is

greater than zero.

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Brewster Angle Impact on Polarization

Impact on Polarization

o

E E jE

E R jR

Polarization changes direction when R// changes sign.

Consider circular polarization:

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HW Problem

• Plot the magnitude of reflection and transmission coefficients for a plane wave incident on a flat ground for incidence angles [0-90] degrees. Assume a lossless ground, i.e. zero conductivity and r2 = 4.0. Plot both TE and TM cases on the same chart.

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Brewster vs Critical AngleNote that Brewster angle is always less than critical angle.

2

1

2

1

tan

sin

B

c

c

B

2

1

1

2

1 2

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Example

• A perpendicularly polarized em wave impinges from medium 1 to medium 2 as shown below. Calculate:

60o

i

Ei

2, 21, 1

x

z

a) the critical angle

b) kx, kz in terms of w, ,

c) ktz in terms of w, , d) Reflection and transmission coefficients

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About Lossy MediaThe Brewster angle loses its meaning if one of the media is lossy, in which case the Brewster angle θB will be complex-valued.

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Solution (1/3)

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Solution (2/3)

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Solution (3/3)Plane of incidence: x-zTE polarization because E is along y-direction

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Reflection and Transmission for Perfect Conductor Boundary

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Examples

1. Normal incidence of a plane wave on a perfect conductor

2. Oblique incidence of a plane wave on a perfect conductor

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Normal Incidence on a Perfect Conductor

Note that normal incidence is a special case of TE. E lies on the interface, so it is horizontally polarized (TE).

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Total Fields (Perfect Conductor Boundary)

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Oblique Incidence on a Perfect Conductor

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Reflection and Transmission for Lossy Media (Conducting Medium)

• When a plane wave is incident on a finitely conducting medium, the laws of Fresnel and Snell are still valid in a purely formal way.

• Since the medium permittivity is complex, the Fresnel coefficients are now complex.

• This implies that the incident wave will be modified in both amplitude and phase upon reflection.

• The problem of using Snell’s law to find the angle of transmission is more involved.

• The complex value of the angle results in attenuating transmitted waves.

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Plane Wave Parameters for Conducting Material

>>

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Incidence on Lossy Medium

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Lossy medium

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Examples of Conducting Medium

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f=100 MHz = 0.001r = 7

B

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f=100 MHz = 0.02r = 30

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Effect of Conductivity on Brewster Angle

Labels reversed!!!!!!!

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Layered Media

• Examples: soil, vegetation, ionosphere

• The problem of reflection and transmission at a plane boundary can be generalized to a multilayer case by evaluating the fields existing within a layer and then using a matrix technique to sum the effects of all layers.

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Layered Media

Waves in each layer should satisfy:• Maxwell’s equations• Boundary conditions

The wave in each layer is the sum of transmitted and reflected waves from all layers.

z

o

* Will be discussed in depth later.

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Rough Surface

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What is rough?

cos il h

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Rayleigh Criterion

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Scattering from Finite Size Objects

D

Ei, Hi

Es, Hs

E = Einc + Es

H = Hinc + Hs

Use boundary conditions on the surface

S

(1)

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Radiation Condition

ˆˆ( , )ikr

s i

eE F o i E

rr

i or

Generic form of scattered field

(2)

Function of:• object characteristics, • direction of incidence• direction of scattered field

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Object Size

• Depending on object size (D), different approximations can be applied.

1. >> D Low Frequency Approximation (quasi-static, Rayleigh)

2. ~ D resonant region; modal solution

3. << D optical region; ray optics

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Scattering from a Cylinder

• Assume infinitely long and circularly symmetric

zx

y

Ei

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Scattering from a Cylinder

• Approach would be to write Maxwell’s equations.

• The scattered fields and internal fields (i.e. fields transmitted inside the cylinder) should both satisfy Maxwell’s equations.

• Analytic solution is an infinite series summation.• Assumption of infinite length helps reduce the

problem, as no variation with z is expected due to symmetry along that axis.

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Scattering from Cylinders

/ 2 2

, 2

( )( )

( )TM in innz scat o n

n n

J kaE E e e H k

H ka

Outward going wave

( / 2) 2

, 2

( )( )

( )ino n

scat nn n

E J kaH ine H k

iw H ka

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References

• Applied electromagnetism, L. C. Shen, J. A. Kong, PWS

• Microwave Remote Sensing, F. T. Ulaby, et.al. Addison-Wesley