Electromagnetic Wave Propagation Lecture 4: … · Electromagnetic Wave Propagation Lecture 4:...

transcript

Electromagnetic Wave PropagationLecture 4: Propagation in lossy media,

complex waves

Daniel Sjoberg

Department of Electrical and Information Technology

September 13, 2012

Outline

1 Propagation in lossy media

2 Oblique propagation and complex waves

3 Paraxial approximation: beams (not in Orfanidis)

4 Doppler effect and negative index media

5 Conclusions

2 / 46

Outline

5 Conclusions

3 / 46

Lossy media

We study lossy isotropic media, where

D = εdE, J = σE, B = µH

The conductivity is incorporated in the permittivity,

J tot = J + jωD = (σ + jωεd)E = jω

(εd +

which implies a complex permittivity

εc = εd − jσ

Often, the dielectric permittivity εd is itself complex, εd = ε′d − jε′′d,due to molecular interactions.

4 / 46

Examples of lossy media

I Metals (high conductivity)

I Liquid solutions (ionic conductivity)

I Resonant media

I Just about anything!

5 / 46

Characterization of lossy media

In a previous lecture, we have shown that a passive material ischaracterized by

(ε ξζ µ

)}= −ω Im

{(ε ξζ µ

)}≥ 0

For isotropic media with ε = εcI, ξ = ζ = 0 and µ = µcI, thisboils down to

εc = ε′c − jε′′cµc = µ′c − jµ′′c

⇒ ε′′c ≥ 0

µ′′c ≥ 0

6 / 46

Maxwell’s equations in lossy media

Assuming dependence only on z we obtain{∇×E = −jωµcH∇×H = jωεcE

∂zz ×E = −jωµcH

∂zz ×H = jωεcE

Nothing really changes compared to the lossless case, for instanceit is seen that the fields do not have a z-component. This can bewritten as a system

ηcH × z

(0 −jkc−jkc 0

ηcH × z

)where the complex wave number kc and the complex waveimpedance ηc are

kc = ω√εcµc, and ηc =

√µcεc

7 / 46

The parameters in the complex plane

For passive media, the parameters εc, µc, and kc = ω√εcµc take

their values in the complex lower half plane, whereas ηc =√µc/εc

is restricted to the right half plane.

��

Equivalently, all parameters (jωεc, jωµc, jkc, ηc) take their values inthe right half plane.

8 / 46

Solutions

The solution to the system

ηcH × z

(0 −jkc−jkc 0

ηcH × z

)can be written (no z-components in the amplitudes E+ and E−)

E(z) = E+e−jkcz +E−ejkcz

H(z) =1

ηcz ×

−jkcz −E−ejkcz)

Thus, the solutions are the same as in the lossless case, as long aswe “complexify” the coefficients.

9 / 46

Exponential attenuation

The dominating effect of wave propagation in lossy media isexponential decrease of the amplitude of the wave:

kc = β − jα ⇒ e−jkcz = e−jβze−αz

Thus, α = − Im(kc) represents the attenuation of the wave,whereas β = Re(kc) represents the oscillations.

The exponential is sometimes written in terms of γ = jkc = α+ jβas

e−γz = e−jβze−αz

where γ can be seen as a spatial Laplace transform variable, in thesame way that the temporal Laplace variable is s = ν + jω.

10 / 46

Power flow

The power flow is given by the Poynting vector

P(z) =1

−jβz−αz ×(

ηcz ×E0e

−jβz−αz)∗}

η∗c

)|E0|2e−2αz = P(0)e−2αz

11 / 46

Characterization of attenuation

The power is damped by a factor e−2αz. The attenuation is oftenexpressed in logarithmic scale, decibel (dB).

A = e−2αz ⇒ AdB = −10 log10(A) = 20 log10(e)αz = 8.686αz

Thus, the attenuation coefficient α can be expressed in dB permeter as

αdB = 8.686α

Instead of the attenuation coefficient, often the skin depth (alsocalled penetration depth)

δ = 1/α

is used. When the wave propagates the distance δ, its power isattenuated a factor e2 ≈ 7.4, or 8.686 dB ≈ 9 dB.

12 / 46

Characterization of losses

A common way to characterize losses is by the loss tangent(sometimes denoted tan δ)

tan θ =ε′′cε′c

=ε′′d + σ/ω

ε′d

which usually depends on frequency. In spite of this, it is oftenseen that the loss tangent is given for only one frequency. This isacceptable if the material properties vary only little with frequency.

13 / 46

Example of material properties

From D. M. Pozar, Microwave Engineering:

Material Frequency ε′r tan θ

Beeswax 10GHz 2.35 0.005Fused quartz 10GHz 6.4 0.0003Gallium arsenide 10GHz 13. 0.006Glass (pyrex) 3GHz 4.82 0.0054Plexiglass 3GHz 2.60 0.0057Silicon 10GHz 11.9 0.004Styrofoam 3GHz 1.03 0.0001Water (distilled) 3GHz 76.7 0.157

The imaginary part of the relative permittivity is given byε′′r = ε′r tan θ.

14 / 46

Approximations for weak losses

In weakly lossy dielectrics, the material parameters are (whereε′′c � ε′c)

εc = ε′c − jε′′c = ε′c(1− j tan θ)

µc = µ0

The wave parameters can then be approximated as

kc = ω√εcµc ≈ ω

√ε′cµ0

(1− j

2tan θ

)ηc =

√µcεc≈√µ0ε′c

(1 + j

2tan θ

)If the losses are caused mainly by a small conductivity, we haveε′′c = σ/ω, tan θ = σ/(ωε′c), and the attenuation constant

α = − Im(kc) =1

2ω√ε′cµ0

ωε′c=σ

√µ0ε′c

is proportional to conductivity and independent of frequency.15 / 46

Example: propagation in sea water

A simple model of the dielectric properties of sea water is

εc = ε0

(81− j

)that is, it has a relative permittivity of 81 and a conductivity ofσ = 4S/m. The imaginary part is much smaller than the real partfor frequencies

f � 4 S/m

81 · 2πε0= 888MHz

for which we have α = 728 dB/m. For lower frequencies, the exactcalculations give

f = 50Hz α = 0.028 dB/m δ = 35.6m

f = 1kHz α = 1.09 dB/m δ = 7.96m

f = 1MHz α = 34.49 dB/m δ = 25.18 cm

f = 1GHz α = 672.69 dB/m δ = 1.29 cm

16 / 46

Approximations for good conductors

In good conductors, the material parameters are (where σ � ωε)

εc = ε− jσ/ω = ε(1− j

)µc = µ

The wave parameters can then be approximated as

kc = ω√εcµc ≈ ω

√−jσωµ =

√ωµσ

2(1− j)

√µcεc≈√

−jσ/ω =

√ωµ

2σ(1 + j)

This demonstrates that the wave number is proportional to√ω

rather than ω in a good conductor, and that the real andimaginary part have equal amplitude.

17 / 46

Skin depth

The skin depth of a good conductor is

ωµσ=

1√πfµσ

For copper, we have σ = 5.8 · 107 S/m. This implies

f = 50Hz δ = 9.35mm

f = 1kHz δ = 2.09mm

f = 1MHz δ = 0.07mm

f = 1GHz δ = 2.09µm

This effectively confines all fields in a metal to a thin region nearthe surface.

18 / 46

Surface impedance

Integrating the currents near the surface z = 0 implies (withγ = α+ jβ)

∫ ∞0J(z) dz =

∫ ∞0

σE0e−γz dz =

Thus, the surface current can be expressed as

J s =1

airmetalJ(z) = σE0e

−γz

where the surface impedance is

Zs =γ

σ=α+ jβ

σ(1 + j) =

σδ(1 + j) =

√ωµ

2σ(1 + j) = ηc

19 / 46

Outline

5 Conclusions

20 / 46

Generalized propagation factor

For a wave propagating in an arbitrary direction, the propagationfactor is generalized as

e−jkz → e−jk·r

Assuming this as the only spatial dependence, the nabla operatorcan be replaced by −jk since

∇(e−jk·r) = −jk(e−jk·r)

Writing the fields as E(r) = E0e−jk·r, Maxwell’s equations for

isotropic media can then be written{−jk ×E0 = −jωµH0

−jk ×H0 = jωεE0⇒

{k ×E0 = ωµH0

k ×H0 = −ωεE0

21 / 46

Properties of the solutions

Eliminating the magnetic field, we find

k × (k ×E0) = −ω2εµE0

This shows that E0 does not have any components parallel to k,and the BAC-CAB rule implies k× (k×E0) = −E0(k · k). Thus,the total wave number is given by

k2 = k · k = ω2εµ

It is further clear that E0, H0 and k constitute a right-handedtriple since k ×E0 = ωµH0, or

k×E0 =

ηk ×E0

22 / 46

Preferred direction

What happens when k is not along the z-direction (which could bethe normal to a plane surface)?

I There are then two preferred directions, k and z.

I These span a plane, the plane of incidence.

I It is natural to specify the polarizations with respect to thatplane.

I When the H-vector is orthogonal to the plane of incidence,we have transverse magnetic polarization (TM).

I When the E-vector is orthogonal to the plane of incidence, wehave transverse electric polarization (TE).

23 / 46

TM and TE polarization

From these figures it is clear that the transverse impedance is

ηTM =ExHy

=A cos θ

= η cos θ

ηTE =Ey−Hx

1ηB cos θ

cos θ

24 / 46

Transverse wave impedance

The transverse wave vector kt = kxx corresponds to the angle ofincidence θ as

kx = k sin θ

The transverse impedance is

Et = Zt · (Ht × z), Zt = η cos θxx+η

cos θyy︸︷︷︸

isotropic case

The transverse wave impedance can be generalized to bianisotropicmaterials by solving the eigenvalue problem from last lecture

Ht × z

(0 −z × II 0

)·[(εtt ξttζtt µtt

)−A(kt)

]·(I 00 z × I

Ht × z

and studying the eigenvectors [Et,Ht × z]. The eigenvaluekz/ω = n/c0 corresponds to the refractive index.

25 / 46

Complex waves

When the material parameters are complexified, we still have

k2c = k · k = ω2εcµc

with a complex wave vector

k = β − jα ⇒ e−jk·r = e−jβ·re−α·r

The real vectors α and β do not need to be parallel.

26 / 46

Outline

5 Conclusions

27 / 46

The plane wave monster

So far we have treated plane waves, which have a seriousdrawback:

I Due to the infinite extent of e−jkzz in the xy-plane, the planewave has infinite energy.

However, the plane wave is a useful object with which we can buildother, more physically reasonable, solutions.

28 / 46

Finite extent in the xy-plane

We can represent a field distribution with finite extent in thexy-plane using a Fourier transform (where kt = kxx+ kyy):

Et(x, y; z) =1

(2π)2

∞∫∫−∞

Et(kx, ky; z)e−jkt·r dkx dky

Et(kx, ky; z) =

∞∫∫−∞

Et(x, y; z)ejkt·r dx dy

The z dependence in Et(kx, ky; z) corresponds to a plane wave

Et(kx, ky; z)e−jkt·r = Et(kx, ky; 0)e

−jkt·re−jkzz

The total wavenumber for each kt is given by k2 = ω2εµ and

k2 = |kt|2 + k2z = k2x + k2y + k2z ⇒ kz(kt) = (k2 − |kt|2)1/2

29 / 46

Initial distribution

Assume a Gaussian distribution in the plane z = 0

Et(x, y; 0) = Ae−(x2+y2)/(2b2)

The transform is itself a Gaussian

Et(kx, ky; 0) = A2πb2e−(k2x+k

30 / 46

Paraxial approximation

The field in z ≥ 0 is then

Et(x, y; z) =1

2πAb2

∞∫∫−∞

e−(k2x+k

2/2−j(kxx+kyy)−jkz(kt)z dkx dky

The exponential makes the main contribution to come from aregion close to kt ≈ 0. This justifies the paraxial approximation

kz(kt) = (k2 − |kt|2)1/2 = k(1− |kt|2/k2)1/2

(1− 1

|kt|2k2

+O(|kt|4/k4))

= k − |kt|2

2k+ · · ·

31 / 46

Computing the field

Inserting the paraxial approximation in the Fourier integral implies

Et(x, y; z) ≈1

2πAb2

∞∫∫−∞

e−(k2x+k

2−j z

2k)−j(kxx+kyy)−jkz dkx dky

F 2e−(x

2+y2)/(2F 2)e−jkz

where F 2 = b2 − jz/k = 1jk (z + jkb2) = q(z)/(jk).

I q(z) = z + jz0 is known as the q-parameter of the beam.

I z0 = kb2 is known as the Rayleigh range.

The final expression for the beam distribution is then

Et(x, y; z) ≈A

1− jz/z0e− x2+y2

2b2(1−jz/z0) e−jkz

32 / 46

Beam width

The power density of the beam is proportional to

e−x2+y2

2b2Re 1

1−jz/z0

and the beam width is then

B(z) =b√

Re 11−jz/z0

Re 1+jz/z01+(z/z0)2

= b√1 + (z/z0)2

where z0 = kb2. For large z, the beam width is

B(z)→ bz/z0 =z

kb, z →∞

33 / 46

Beam width

The beam angle θb is characterized by

tan θb =B(z)

kbSmall initial width compared to wavelength implies large beamangle.

34 / 46

How can beams be used?

Beams can be an efficient representation of fields, determined bythree parameters:

I Propagation direction zI Polarization A(ω)I Initial beam width b(ω)

High frequency propagation in officespaces (Timchenko, Heyman, Boag,EMTS Berlin 2010).

Raytracing in optics, FRED

35 / 46

Outline

5 Conclusions

36 / 46

The Doppler effect

Classical formulas:

λb= fa

c− va

fb =c− vbλa

= fac− vbc

The relativistically correct formula is

fb = fa

√c− vc+ v

≈ fa(1− v

)where v =

vb − va1− vavb/c20

Lots more on relativistic Doppler effect in Orfanidis. Do not divetoo deep into this, it is not central material in the course.

37 / 46

Negative material parameters

Passivity requires the material parameters εc and µc to be in thelower half plane,

��

This means we could very well have εc/ε0 ≈ −1 and µc/µ0 ≈ −1for some frequency. What is then the appropriate value fork = ω

√εcµc = k0

√(εc/ε0)(µc/µ0) = k0

√(−1)(−1),

k = +k0 or k = −k0?38 / 46

Negative refractive index

Simple solution: consider all parameters in the right half plane andapproach the negative axis from inside the half plane, using thestandard square root (with branch cut along negative real axis):

jkc =√(jωεc)(jωµc)

Reµǫ

ImImRe

ImRejωǫ · jωµ

jωǫjωµ jk =

√jωǫ · jωµ

The refractive index is then

n =jkcjk0

= −j√(jεc/ε0)(jµc/µ0) = −1

39 / 46

Consequences of negative refractive index

With a negative refractive index, the exponential factor

ej(ωt−kz) = ej(ωt+|k|z)

represents a phase traveling in the negative z-direction, eventhough the Poynting vector 1

2 Re{E ×H∗} = z 12 Re(

1η∗c)|E0|2 is

still pointing in the positive z-direction.

I The power flow is in the opposite direction of the phasevelocity!

I Snel’s law has to be “inverted”, the rays are refracted in thewrong direction.

First investigated by Veselago in 1967. Enormous scientific interestsince about a decade, since the materials can now (to someextent) be fabricated.

40 / 46

Negative refraction

41 / 46

Realization of negative refractive index

I Artificial materials,“metamaterials”

I Periodic structures

I Resonant inclusions

I Small lossesrequired

I Theoretical andpractical challenges

I Very hot topic sinceabout 10 years

To describe the structure as a material usually requiresmicrostructure a� λ, which is not easily achieved.

42 / 46

Band limitations

If the negative properties are realized with passive, causal materials,they must satisfy Kramers-Kronig’s relations (ε∞ = lim

ω→∞ε(ω))

ε′(ω)− ε∞ =1

πp.v.

∫ ∞−∞

ε′′(ω′)ω′ − ω dω′

ε′′(ω) = − 1

πp.v.

∫ ∞−∞

ε′(ω′)− ε∞ω′ − ω dω′

These relations represent restriction on the possible frequencybehavior, and can be used to derive bounds on the bandwidthwhere the material parameters can be negative.

43 / 46

Example: two Lorentz models

0.1 1 10

j²(!)-² jm

ReIm !

j²(!)+1jj²(!)-1.5jj²(!)+1j

0.1 1 10

² =-1m

² =1.5m

An εm between εs and ε∞ is easily realized for a large bandwidth,whereas an εm < ε∞ is not. With fractional bandwidth B:

maxω∈B|ε(ω)− εm| ≥

1 +B/2(ε∞ − εm)

{1/2 lossy case

1 lossless case

44 / 46

Outline

5 Conclusions

45 / 46

Conclusions

I Lossy media leads to complex material parameters, but planewave formalism is the same as in lossless media.

I At oblique propagation, the transverse fields are mostimportant.

I The paraxial approximation can be used to describe beams.The beam angle depends on the original beam width in termsof wavelengths.

I The Doppler effect can be used to detect motion.

I Negative refractive index is possible, but only for very narrowfrequency band.

46 / 46

Electromagnetic Wave Propagation Lecture 4: … · Electromagnetic Wave Propagation Lecture 4:...

Documents