Electromagnetic Wave Propagation Lecture 3: Plane waves in ......Electromagnetic Wave Propagation Lecture 3: Plane waves in isotropic and bianisotropic media Daniel Sj oberg Department

Electromagnetic Wave PropagationLecture 3: Plane waves in isotropic and

bianisotropic media

Daniel Sjoberg

Department of Electrical and Information Technology

September 6, 2011

Outline

1 Plane waves in lossless mediaGeneral time dependenceTime harmonic waves

2 Polarization

3 Wave propagation in bianisotropic media

4 Interpretation of the fundamental equation

Daniel Sjoberg, Department of Electrical and Information Technology

Outline


2 Polarization




Plane electromagnetic waves

In this lecture, we dive a bit deeper into the familiar right-handrule, x = y × z.

y

z

x

This is the building block for most of the course.


Outline


2 Polarization




Propagation in source free, isotropic, non-dispersive media

The electromagnetic field is assumed to depend only on onecoordinate, z. This implies

E(x, y, z, t) = E(z, t)

D(x, y, z, t) = εE(z, t)

H(x, y, z, t) =H(z, t)

B(x, y, z, t) = µH(z, t)

and by using ∇ = x ∂∂x + y ∂

∂y + z∂∂z → z ∂

∂z we have

∇×E = −∂B∂t

z × ∂E

∂z= −µ∂H

∂t

∇×H =∂D

∂t=⇒ z × ∂H

∂z= ε

∂E

∂t

∇ ·D = 0∂Ez∂z

= 0 ⇒ Ez = 0

∇ ·B = 0∂Hz

∂z= 0 ⇒ Hz = 0


Rewriting the equations

The electromagnetic field has only x and y components, and satisfy

z × ∂E

∂z= −µ∂H

∂t

z × ∂H

∂z= ε

∂E

∂t

Using (z × F )× z = F for any vector F orthogonal to z, and

c = 1/√εµ and η =

√µ/ε , we can write this as

∂

∂zE = −1

c

∂

∂t(ηH × z)

∂

∂z(ηH × z) = −1

c

∂

∂tE

which is a symmetric hyperbolic system (note E and ηH have thesame units)

∂

∂z

(E

ηH × z

)= −1

c

∂

∂t

(0 11 0

)(E

ηH × z

)Daniel Sjoberg, Department of Electrical and Information Technology

Wave splitting

The change of variables(E+

E−

)=

1

2

(E + ηH × zE − ηH × z

)=

1

2

(1 11 −1

)(E

ηH × z

)with inverse(

E

ηH × z

)=

(E+ +E−E+ −E−

)=

(1 11 −1

)(E+

E−

)is called a wave splitting and diagonalizes the system to

∂E+

∂z= −1

c

∂E+

∂t⇒ E+(z, t) = F (z − ct)

∂E−∂z

= +1

c

∂E−∂t

⇒ E−(z, t) = G(z + ct)


Forward and backward waves

The total fields can now be written

E(z, t) = F (z − ct) +G(z + ct)

H(z, t) =1

ηz ×

(F (z − ct)−G(z + ct)

)A graphical interpretation of the forward and backward waves is

(Fig. 2.1.1 in Orfanidis)


Energy density in one single wave

A wave propagating in positive z direction satisfies H = 1η z ×E

y

z

x

x: Electric fieldy: Magnetic fieldz: Propagation direction

The Poynting vector and energy densities are (using the BAC-CABrule A× (B ×C) = B(A ·C)−C(A ·B))

P = E ×H = E ×(1

ηz ×E

)=

1

ηz|E|2

we =1

2ε|E|2

wm =1

2µ|H|2 = 1

2µ1

η2|z ×E|2 = 1

2ε|E|2 = we

Thus, the wave carries equal amounts of electric and magneticenergy.


Energy density in forward and backward wave

If the wave propagates in negative z direction, we haveH = 1

η (−z)×E and

P = E ×H =1

ηE × (−z ×E) = −1

ηz|E|2

Thus, when the wave consists of one forward wave F (z − ct) andone backward wave G(z + ct), we have

P = E ×H =1

ηz(|F |2 − |G|2

)w =

1

2ε|E|2 + 1

2µ|H|2 = ε|F |2 + ε|G|2


Outline


2 Polarization




Time harmonic waves

We assume harmonic time dependence

E(x, y, z, t) = E(z)ejωt

H(x, y, z, t) =H(z)ejωt

Using the same wave splitting as before implies

∂E±∂z

= ∓1

c

∂E±∂t

= ∓ jω

cE± ⇒ E±(z) = E0±e

∓jkz

where k = ω/c is the wave number in the medium. Thus, the

general solution for time harmonic waves is

E(z) = E0+e−jkz +E0−e

jkz

H(z) =H0+e−jkz +H0−e

jkz =1

ηz ×

(E0+e

−jkz −E0−ejkz)

The triples {E0+,H0+, z} and {E0−,H0−,−z} are right-handedsystems.


Wavelength

A time harmonic wave propagating in the forward z direction hasthe space-time dependence E(z, t) = E0+e

j(ωt−kz) andH(z, t) = 1

η z ×E(z, t).


The wavelength corresponds to the spatial periodicity according toe−jk(z+λ) = e−jkz, meaning kλ = 2π or

λ =2π

k=

2πc

ω=c

f


Refractive index

The wavelength is often compared to the correspondingwavelength in vacuum

λ0 =c0f

The refractive index is

n =λ0λ

=k

k0=

c0c

=

√εµ

ε0µ0

Important special case: non-magnetic media, where µ = µ0 andn =

√ε/ε0.

c =c0n, η =

η0n,︸︷︷︸

only for µ = µ0!

λ =λ0n, k = nk0

Note: We use c0 for the speed of light in vacuum and c for thespeed of light in a medium, even though c is the standard for thespeed of light in vacuum!


EMANIM program


Energy density and power flow

For the general time harmonic solution

E(z) = E0+e−jkz +E0−e

jkz

H(z) =1

ηz ×

(E0+e

−jkz −E0−ejkz)

the Poynting vector and energy density are

P =1

2Re[E(z)×H∗(z)

]= z

(1

2η|E0+|2 −

1

2η|E0−|2

)w =

1

4Re[εE(z) ·E∗(z) + µH(z) ·H∗(z)

]=

1

2ε|E0+|2 +

1

2ε|E0−|2


Wave impedance

For forward and backward waves, the impedance η =√µ/ε relates

the electric and magnetic field strengths to each other.

E± = ±ηH± × z

When both forward and backward waves are present this isgeneralized as (in component form)

Zx(z) =[E(z)]x

[H(z)× z]x=Ex(z)

Hy(z)= η

E0+xe−jkz + E0−xe

jkz

E0+xe−jkz − E0−xe−jkz

Zy(z) =[E(z)]y

[H(z)× z]y=

Ey(z)

−Hx(z)= η

E0+ye−jkz + E0−ye

jkz

E0+ye−jkz − E0−ye−jkz

Thus, the wave impedance is in general a non-trivial function of z.This will be used as a means of analysis and design in the course.


Material and wave parameters

We note that we have two material parameters

Permittivity ε, defined by D = εE.

Permeability µ, defined by B = µH.

But the waves are described by the wave parameters

Wave number k, defined by k = ω√εµ.

Wave impedance η, defined by η =√µ/ε.

This means that in a scattering experiment, where we measurewave effects, we primarily get information on k and η, not ε and µ.In order to get material data, a theoretical material model must beapplied.

Often, wave number is given by transmission data (phase delay),and wave impedance is given by reflection data (impedancemismatch). More on this in future lectures!


Outline


2 Polarization




Why care about different polarizations?

I Different materials react differently to different polarizations.

I Linear polarization is sometimes not the most natural.

I For propagation through the ionosphere (to satellites), orthrough magnetized media, often circular polarization isnatural.


Complex vectors

The time dependence of the electric field is

E(t) = Re{E0ejωt}

where the complex amplitude can be written

E0 = xE0x + yE0y + zE0z = x|E0x|ejα + y|E0y|ejβ + z|E0z|ejγ

= E0r + jE0i

where E0r and E0i are real-valued vectors. We then have

E(t) = Re{(E0r + jE0i)ejωt} = E0r cos(ωt)−E0i sin(ωt)

This lies in a plane with normal

n = ± E0r ×E0i

|E0r ×E0i|


Linear polarization

The simplest case is given by E0i = 0, implying E(t) = E0r cosωt.

x

y

E0r


Circular polarization

Assume E0r = x and E0i = −y. The total fieldE(t) = x cosωt+ y sinωt then rotates in the xy-plane:

x

y

E(t)

The rotation direction needs to be related to the propagationdirection to determine left or right.


IEEE definition of left and right

With your right hand thumb in the propagation direction andfingers in rotation direction: right hand circular.



Elliptical polarization

The general polarization state is elliptical

The direction e is parallel to the Poynting vector (the power flow).


Classification of polarization

The following classification can be given (e is the propagationdirection):

−je · (E0 ×E∗0) Polarization

= 0 Linear polarization> 0 Right handed elliptic polarization< 0 Left handed elliptic polarization

Further, circular polarization is characterized by E0 ·E0 = 0.Typical examples:

Linear: E0 = x or E0 = y.

Circular: E0 = x− jy (right handed for e = z) orE0 = x+ jy (left handed for e = z).

See the literature for more in depth descriptions.


Alternative bases in the plane

To describe an arbitrary vector in the xy-plane, the unit vectors

x and y

are usually used. However, we could just as well use the RCP andLCP vectors

x− jy and x+ jy

Sometimes the linear basis is preferrable, sometimes the circular.


Outline


2 Polarization




Observations

Some fundamental properties are observed from the isotropic case:

I The wave speed c = 1/√εµ and wave impedance η =

√µ/ε

depend on the material properties.

I Waves can propagate in the positive or negative z-direction.

I For each propagation direction, there are two possiblepolarizations (x and y, or RCP and LCP etc).

We will generalize this to bianisotropic materials, where(D(ω)B(ω)

)=

(ε(ω) ξ(ω)ζ(ω) µ(ω)

)(E(ω)H(ω)

)See sections 1 and 2 in the book chapter “Circuit analogs for wavepropagation in stratified structures” by Sjoberg. The rest of thepaper is interesting but not essential to this course.


Wave propagation in general media

We now generalize to arbitrary materials in the frequency domain.Our plan is the following:

1. Write up the full Maxwell’s equations in the frequency domain.

2. Assume the dependence on x and y appear at most through afactor e−j(kxx+kyy).

3. Separate the transverse components (x and y) from the zcomponents of the fields.

4. Eliminate the z components.

5. Identify the resulting differential equation as a dynamicsystem for the transverse components,

∂

∂z

(Et

Ht × z

)= −jω

(W11 W12

W21 W22

)·(

Et

Ht × z

)


1) Maxwell’s equations in the frequency domain

Maxwell’s equations in the frequency domain are

∇×H = jωD = jω[ε(ω) ·E + ξ(ω) ·H

]∇×E = −jωB = −jω

[ζ(ω) ·E + µ(ω) ·H

]The bianisotropic material is described by the dyadics ε(ω), ξ(ω),ζ(ω), and µ(ω). The frequency dependence is suppressed in thefollowing.


2) Transverse behavior

Assume that the fields depend on x and y only through a factore−j(kxx+kyy) (corresponding to a Fourier transform in x and y)

E(x, y, z) = E(z)e−jkt·r where kt = kxx+ kyy

Since ∂∂xe

−jkxx = −jkxe−jkxx, the action of the curl operator isthen

∇×E(x, y, z) = e−jkt·r(−jkt +

∂

∂zz

)×E(z)


3) Separate the components, curls

Split the fields according to

E = Et + zEz

where Et is in the xy-plane, and zEz is the z-component. Byexpanding the curls, we find(−jkt +

∂

∂zz

)×E(z) = − jkt ×Et︸︷︷︸

parallel to z

− jkt × zEz︸︷︷︸orthogonal to z

+∂

∂zz ×Et︸︷︷︸

orthogonal to z

There is no term with ∂Ez∂z , since z × z = 0.


3) Separate the components, fluxes

Split the material dyadics as(Dt

zDz

)=

(εtt εtzzεz εzzzz

)·(Et

zEz

)+

(ξtt ξtzzξz ξzzzz

)·(Ht

zHz

)εtt can be represented as a 2× 2 matrix operating on xycomponents, εt and εz are vectors in the xy-plane, and εzz is ascalar.

The transverse components of the electric flux are (vectorequation)

Dt = εtt ·Et + εtEz + ξtt ·Ht + ξtHz

and the z component is (scalar equation)

Dz = εz ·Et + εzzEz + ξz ·Ht + ξzzHz


4) Eliminate the z components

We first write Maxwell’s equations using matrices (all components)

∂

∂z

(0 −z × I

z × I 0

)·(EH

)=

(0 −jkt × I

jkt × I 0

)·(EH

)− jω

(ε ξζ µ

)·(EH

)

The z components of these equations are (take scalar product withz and use z · (kt × zEz) = 0 and z · (kt ×Et) = (z × kt) ·Et)(00

)=

(0 −jz × kt

jz × kt 0

)·(Et

Ht

)− jω

(εz ξzζz µz

)·(Et

Ht

)− jω

(εzz ξzzζzz µzz

)(EzHz

)from which we solve for the z components of the fields:

(EzHz

)=

(εzz ξzzζzz µzz

)−1 [(0 −ω−1z × kt

ω−1z × kt 0

)−(εz ξzζz µz

)]·(Et

Ht

)


4) Insert the z components

The transverse part of Maxwell’s equations are

∂

∂z

(0 −z × I

z × I 0

)·(Et

Ht

)=

(0 −jkt × z

jkt × z 0

)(EzHz

)− jω

(εtt ξttζtt µtt

)·(Et

Ht

)− jω

(εt ξtζt µt

)(EzHz

)Inserting the expression for [Ez, Hz] previously derived implies

∂

∂z

(0 −z × I

z × I 0

)·(Et

Ht

)= −jω

(εtt ξttζtt µtt

)·(Et

Ht

)+ jωA ·

(Et

Ht

)where the matrix A is due to the z components.


4) The A matrix

The matrix A is a dyadic product

A =

[(0 −ω−1kt × z

ω−1kt × z 0

)−(εt ξtζt µt

)](εzz ξzzζzz µzz

)−1 [(0 −ω−1z × kt

ω−1z × kt 0

)−(εz ξzζz µz

)]In particular, when kt = 0 we have

A =

(εt ξtζt µt

)(εzz ξzzζzz µzz

)−1(εz ξzζz µz

)and for a uniaxial material with the axis of symmetry along the zdirection, where εt = ξt = ζt = µt = 0 andεz = ξz = ζz = µz = 0, we have A = 0.


5) Dynamical system

Using −z × (z ×Et) = Et for any transverse vector Et, it is seenthat (

0 −z × II 0

)·(

0 −z × Iz × I 0

)·(Et

Ht

)=

(Et

Ht × z

)The final form of Maxwell’s equations is then

∂

∂z

(Et

Ht × z

)=

(0 −z × II 0

)·[−jω

(εtt ξttζtt µtt

)+ jωA

]·(I 00 z × I

)·(

Et

Ht × z

)or

∂

∂z

(Et

Ht × z

)= −jωW ·

(Et

Ht × z

)Due to the formal similarity with classical transmission lineformulas, the equivalent voltage and current vectors(

VI

)=

(Et

Ht × z

)are often introduced in electrical engineering (just a relabeling).


Outline


2 Polarization




Propagator

If the material parameters are constant, the dynamical system

∂

∂z

(Et

Ht × z

)= −jωW ·

(Et

Ht × z

)has the formal solution(

Et(z2)

Ht(z2)× z

)= exp

(− jω(z2 − z1)W

)︸︷︷︸

P(z1,z2)

·(

Et(z1)

Ht(z1)× z

)

The matrix P(z1, z2) is called a propagator. It maps the transversefields at the plane z1 to the plane z2. Due to the linearity of theproblem, the propagator exists even if W does depend on z, but itcannot be represented with an exponential matrix.


Eigenvalue problem

For media with infinite extension in z, it is natural to look forsolutions on the form(

Et(z)

Ht(z)× z

)=

(E0t

H0t × z

)e−jβz

This implies ∂∂z [Et(z),Ht(z)× z] = −jβ[E0t,H0t × z]e−jβz. We

then get the algebraic eigenvalue problem

β

ω

(E0t

H0t × z

)= W ·

(E0t

H0t × z

)=

(0 −z × II 0

)·[(εtt ξttζtt µtt

)−A

]·(I 00 z × I

)·(

E0t

H0t × z

)


Interpretation of the eigenvalue problem

Since the wave is multiplied by the exponential factorej(ωt−βz) = ejω(t−zβ/ω) = ejω(t−z/c), we identify the eigenvalue asthe inverse of the phase velocity

β

ω=

1

c=

n

c0

which defines the refractive index n of the wave. The transversewave impedance dyadic Z is defined through the relation

E0t = Z · (H0t × z)

where [E0t,H0t × z] is the corresponding eigenvector.


Example: “standard” media

An isotropic medium is described by(ε ξζ µ

)=

(εI 00 µI

)For kt = 0, we have A = 0 and

W =

(0 −z × II 0

)·[(εtt ξttζtt µtt

)−A

]·(I 00 z × I

)

=

(0 µIεI 0

)=

0 0 µ 00 0 0 µε 0 0 00 ε 0 0


Eigenvalues

The eigenvalues are found from the characteristic equationdet(λI−W) = 0

0 =

∣∣∣∣∣∣∣∣λ 0 −µ 00 λ 0 −µ−ε 0 λ 00 −ε 0 λ

∣∣∣∣∣∣∣∣ = −∣∣∣∣∣∣∣∣λ −µ 0 00 0 λ −µ−ε λ 0 00 0 −ε λ

∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣λ −µ 0 0−ε λ 0 00 0 λ −µ0 0 −ε λ

∣∣∣∣∣∣∣∣ = (λ2 − εµ)2

Thus, the eigenvalues are (with double multiplicity)

β

ω= λ = ±√εµ


Eigenvectors

Using that Ht × z = xHy − yHx, the eigenvectors are (top rowcorresponds to β/ω =

√εµ, bottom row to β/ω = −√εµ):

Ex0Hy

0

,

0Ey0−Hx

, Z =ExHy

=Ey−Hx

=

√µ

ε= η

Ex0Hy

0

,

0Ey0−Hx

, Z =ExHy

=Ey−Hx

= −√µ

ε= −η


Propagator

The propagator matrix for isotropic media can be represented as(after some calculations)(

Et(z2)Ht(z2)× z

)=

(cos(β`)I j sin(β`)Z

j sin(β`)Z−1 cos(β`)I

)·(

Et(z1)Ht(z1)× z

)For those familiar with transmission line theory, this is the ABCDmatrix for a homogeneous transmission line.


Conclusions for propagation in bianisotropic media

I We have found a way of computing time harmonic waves inany bianisotropic material.

I The transverse components (x and y) are crucial.

I There are four fundamental waves: two propagation directionsand two polarizations.

I Each wave is described byI The wave number β (eigenvalue of W)I The polarization and wave impedance (eigenvector of W)

I The propagator P(z1, z2) = exp(−jk0(z2 − z1)W) mapsfields at z1 to fields at z2.

I A simple computer program is available on the course web forcalculation of W, given material matrices ε, ξ, ζ, and µ.


Documents

Electromagnetic Wave Propagation Lecture 3: Plane waves in ......Electromagnetic Wave Propagation Lecture 3: Plane waves in isotropic and bianisotropic media Daniel Sj oberg Department