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Electromagnetic Wave PropagationLecture 3: Plane waves in isotropic and
bianisotropic media
Daniel Sjoberg
Department of Electrical and Information Technology
September 2016
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
2 / 48
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
3 / 48
Plane electromagnetic waves
In this lecture, we dive a bit deeper into the familiar right-handrule, x× y = z.
y
z
x
Typically E = Exx, H = Hyy, and E ×H = ExHyz. This is thebuilding block for most of the course.
4 / 48
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
5 / 48
Propagation in source free, isotropic, non-dispersive media
The electromagnetic field is assumed to depend only on time andone coordinate, z. This implies (ε and µ are scalar constants)
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 ε∂Ez∂z
= 0 ⇒ Ez = const = 0
∇ ·B = 0 µ∂Hz
∂z= 0 ⇒ Hz = const = 0
6 / 48
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
)7 / 48
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
∂
∂z
(E+
E−
)= −1
c
∂
∂t
1
2
(1 11 −1
)(0 11 0
)(1 11 −1
)︸ ︷︷ ︸
=(1 00 −1
)(E+
E−
)
which results in two uncoupled equations
∂E+
∂z= −1
c
∂E+
∂t⇒ E+(z, t) = F (z − ct)
∂E−∂z
= +1
c
∂E−∂t
⇒ E−(z, t) = G(z + ct)8 / 48
Forward and backward waves
The total fields can now be written as a superposition
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 isshown below.
z
E+(z, t)c
E+(z, t+∆t) = E+(z −∆z, t)
∆z = c∆t
z
E−(z, t)c
E−(z, t+∆t) = E−(z +∆z, t)
∆z = c∆t 9 / 48
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) and η =
√µ/ε)
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, we = wm.
10 / 48
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
11 / 48
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
12 / 48
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, thegeneral 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.
13 / 48
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).
z
x
yE
H
λ
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
14 / 48
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!
15 / 48
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 time average 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
17 / 48
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.It depends on which combination of forward and backward wavesare present. This will be used as a means of analysis and design inthe course.
18 / 48
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, the wave number is given by transmission data (phasedelay), and the wave impedance is given by reflection data(impedance mismatch). More on this in future lectures!
19 / 48
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
20 / 48
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.
21 / 48
Complex vectors
The time dependence of an electric field propagating in the +zdirection is
E(z, t) = Re{E0ej(ωt−kz)}
where the complex amplitude can be written
E0 = xA+ + yB+ = xAejφa + yBejφb
where A and B are positive real numbers. We then have
E(z, t) = Re{xAej(ωt−kz+φa) + yBej(ωt−kz+φb)}= xA cos(ωt− kz + φa) + yB cos(ωt− kz + φb)
In the plane defined by kz = φb, this is (where φ = φa − φb)
E(z, t) = xA cos(ωt+ φ) + yB cos(ωt)
22 / 48
IEEE definition of left and right
With your right hand thumb in the propagation direction andfingers in rotation direction: right hand circular.
E(t)
z
x
y
right-polarizedforward moving
E(t)
z
x
y
left-polarizedforward moving
E(t)
−z
x
y
left-polarizedbackward moving
E(t)
−z
x
y
right-polarizedbackward moving
25 / 48
Elliptical polarization
The general polarization state is elliptical
x
y
E(t)
right
lefte
The direction e is parallel to the Poynting vector (the power flow).
26 / 48
Classification of polarization
The complex polarization vector satisfies (with φ = φa − φb)
E0 ×E∗0 = (xAejφa + yBejφb)× (xAe−jφa + yBe−jφb)
= zABej(φa−φb) − zABej(φb−φa) = zAB2j sinφ
The following coordinate-free classification can be given (e is thepropagation direction):
−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) or
E0 = x+ jy (left handed for e = z).
See the literature for more in depth descriptions.27 / 48
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.
An example is given when modelling ferromagnetic materials inhandin 1.
28 / 48
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
29 / 48
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 (available from thecourse home page). The rest of the paper is of course interestingbut not essential to this course.
30 / 48
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
)
31 / 48
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.
32 / 48
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)
33 / 48
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.
34 / 48
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
35 / 48
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
)
36 / 48
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.
37 / 48
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.
38 / 48
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).
39 / 48
Outline
1 Plane waves in lossless isotropic mediaGeneral time dependenceTime harmonic waves
2 Polarization
3 Wave propagation in bianisotropic media
4 Interpretation of the fundamental equation
40 / 48
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.
41 / 48
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
)
42 / 48
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 transverse
wave impedance dyadic Z is defined through the relation
E0t = Z · (H0t × z)
where [E0t,H0t × z] is the corresponding eigenvector.
43 / 48
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
44 / 48
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
45 / 48
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
=1
1/η= η
Ex0Hy
0
,
0Ey0−Hx
, Z =ExHy
=Ey−Hx
=1
−1/η= −η
46 / 48
Eigenvectors
Using that Ht × z = xHy − yHx, the eigenvectors are (top rowcorresponds to β/ω =
√εµ, bottom row to β/ω = −√εµ):
10
1/η0
,
010
1/η
, Z =ExHy
=Ey−Hx
=1
1/η= η
10− 1/η0
,
010− 1/η
, Z =ExHy
=Ey−Hx
=1
−1/η= −η
46 / 48
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.
47 / 48
Conclusions for propagation in bianisotropic media
I We have found a way of computing time harmonic planewaves in any 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(−jω(z2 − z1)W) maps fieldsat 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.
48 / 48
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