Electromagnetic Wave Propagation Lecture 2: Time harmonic ... · I The theory given in this lecture...

Electromagnetic Wave PropagationLecture 2: Time harmonic dependence,

constitutive relations

Daniel Sjoberg

Department of Electrical and Information Technology

August 2016

Outline

1 Harmonic time dependence

2 Constitutive relations, time domain

3 Constitutive relations, frequency domain

4 Examples of material models

5 Bounds on metamaterials

6 Conclusions

2 / 53

Outline






6 Conclusions

3 / 53

Scope

I The theory given in this lecture (and the entire course) isapplicable to the whole electromagnetic spectrum.

I However, different processes are dominant in different bands,making the material models different.

103 106 109 1012 1015 f (Hz)

ε = ε′ − jε′′

+

−

conduction +

−relaxation

+

−

vibrationrotation

+

−transition

ε(0)

ε(∞)

I Today, you learn what restrictions are imposed by therequirements (regardless of frequency range)

1. Linearity2. Causality3. Time translational invariance4. Passivity

4 / 53

Electromagnetic spectrum, c0 = fλ ≈ 3 · 108m/s

Band Frequency WavelengthELF Extremely Low Frequency 30–300Hz 1–10MmVF Voice Frequency 300–3000Hz 100–1000 kmVLF Very Low Frequency 3–30 kHz 10-100 kmLF Low Frequency 30–300 kHz 1–10 kmMF Medium Frequency 300–3000 kHz 100–1000mHF High Frequency 3–30MHz 10–100mVHF Very High Frequency 30–300MHz 1–10mUHF Ultra High Frequency 300–3000MHz 10–100 cmSHF Super High Frequency 3–30GHz 1–10 cmEHF Extremely High Frequency 30–300GHz 1–10mm

Submillimeter 300–3000GHz 100–1000µmInfrared 3–300THz 1–100µmVisible 385–789THz 380–780 nmUltraviolet 750THz–30PHz 10–400 nmX-ray 30PHz–3EHz 10nm–100 pmγ-ray >3EHz <100 pm

5 / 53

Three ways of introducing time harmonic fields

I Fourier transform (finite energy fields, ω = 2πf)

E(r, ω) =

∫ ∞−∞

E(r, t)e−jωt dt

E(r, t) =1

2π

∫ ∞−∞

E(r, ω)ejωt dω

I Laplace transform (causal fields, zero for t < 0, s = α+ jω)

E(r, s) =

∫ ∞0E(r, t)e−st dt

E(r, t) =1

2πj

∫ α+j∞

α−j∞E(r, s)est ds

I Real-value convention (purely harmonic cosωt, preservesunits)

E(r, t) = Re{E(r, ω)ejωt}6 / 53

Some examples

Unit step function: u(t) =

{0 t < 0

1 t > 0

E(r, t) Fourier Laplace Real-value

e−αt2 √

παe−ω

2

4α — —

cos(ω0t) π(δ(ω + ω0) + δ(ω − ω0)) — 1

sin(ω0t) jπ(δ(ω + ω0)− δ(ω − ω0)) — −je−at u(t) 1

jω+a1s+a —

e−at cos(ω0t) u(t)jω+a

(jω+a)2+ω20

s+a(s+a)2+ω2

0—

e−at sin(ω0t) u(t)ω0

(jω+a)2+ω20

ω0

(s+a)2+ω20

—

δ(t) 1 1 —

7 / 53

Different time conventions

Different traditions:

Engineering: Time dependence ejωt, plane wave factor ej(ωt−k·r).

Physics: Time dependence e−iωt, plane wave factor ei(k·r−ωt).

If you use j and i consistently, all results can be translated betweenconventions using the simple rule

j = −iIn this course we follow Orfanidis’ choice ejωt.

8 / 53

Outline






6 Conclusions

9 / 53

The need for material models

Maxwell’s equations are

∇×E(r, t) = −∂B(r, t)

∂t

∇×H(r, t) = J(r, t) +∂D(r, t)

∂t

These are 2× 3 = 6 equations for at least 4× 3 = 12 unknowns.Something more is needed! We choose E and H as ourfundamental fields (partly due to conformity with boundaryconditions), and search for constitutive relations on the form{

DB

}= F

({EH

})This would provide the missing 6 equations.

10 / 53

Examples of models

I Linear, isotropic materials (“standard media”):

D = εE, B = µH

I Linear, anisotropic materials (different in different directions):

D = (εxxx+ εyyy + εzzz) ·E ⇔

Dx

Dy

Dz

=

εx 0 00 εy 00 0 εz

ExEyEz

I Linear, dispersive materials (depend on the history):

Debye material Gyrotropic material

D = ε0E + P B = µ0(H +M)

∂P

∂t= ε0αE − P /τ

∂M

∂t= ωS z × (βM −H)

The models correspond to the physical processes in the material.11 / 53

Basic assumptions

To simplify, we formulate our assumptions for a non-magneticmaterial where D = F (E) and B = µ0H. We require themapping F to satisfy four basic physical principles:

Linearity: For each α, β, E1, and E2 we have

F (αE1 + βE2) = αF (E1) + βF (E2)

Causality: For all fields E such that E(t) = 0 when t < τ , wehave

F (E)(t) = 0 for t < τ

Time translational invariance: If D1 = F (E1), D2 = F (E2), andE2(t) = E1(t− τ), we have

D2(t) =D1(t− τ)

Passivity: The material is not a source of electromagneticenergy, that is, ∇ · 〈P〉 ≤ 0.

12 / 53

The general linear model

The result of the assumptions is that all such materials can bemodeled as

D(t) = ε0

[E(t) +

∫ t

−∞χe(t− t′) ·E(t′) dt′

]+

∫ t

−∞ξ(t− t′) ·H(t′) dt′

B(t) =

∫ t

−∞ζ(t− t′) ·E(t′) dt′

+ µ0

[H(t) +

∫ t

−∞χm(t− t′) ·H(t′) dt′

]The dyadic convolution kernels χe(t), ξ(t), ζ(t), and χm(t) modelthe induced polarization and magnetization.

Linearity, causality, and time translational invariance byconstruction. Passivity will be seen in frequency domain.

13 / 53

Instantaneous response

Some physical processes in the material may be considerably fasterthan the others

10−5 10−4 10−3 10−2 10−1 100−1−0.5

0

0.5

1

t

χ(t)

This means the susceptibility function can be split in two parts χ1

and χ2, operating on very different time scales.

= +

χ(t) = χ1(t) + χ2(t)14 / 53

Instantaneous response, continued

Assume that E(t) does not vary considerably on the time scale ofχ1(t). We then have

D(t)/ε0 = E(t) +

∫ t

−∞[χ1(t− t′) + χ2(t− t′)]E(t′) dt′

= E(t) +

[∫ t

−∞χ1(t− t′) dt′

]E(t) +

∫ t

−∞χ2(t− t′)E(t′) dt′

= E(t) +

[∫ ∞0

χ1(t′′) dt′′

]E(t)︸︷︷︸

=ε∞E(t)

+

∫ t

−∞χ2(t− t′)E(t′) dt′

The quantity ε∞ = 1 +∫∞0 χ1(t

′) dt′ is called the instantaneousresponse (or momentaneous response, or optical response).Thus, there is some freedom of choice how to model thematerial, depending on the time scale of interest!

15 / 53

Instantaneous response, example

0 5 10 15 20 25 30t

0.5

0.0

0.5

1.0

1.5

2.0P(t)/ε0

E(t)

χ(t)

D(t)/ε0

See also the python script materialmodels.py on the home page formore variations of parameters. 16 / 53

Outline






6 Conclusions

17 / 53

Constitutive relations in the frequency domain

Applying a Fourier transform to the convolutions implies

D(ω) = ε0 [ε∞ ·E(ω) + χe(ω) ·E(ω)] + ξ(ω) ·H(ω)

B(ω) = ζ(ω) ·E(ω) + µ0 [µ∞ ·H(ω) + χm(ω) ·H(ω)]

or (D(ω)

B(ω)

)=

(ε(ω) ξ(ω)ζ(ω) µ(ω)

)·(E(ω)

H(ω)

)where we introduced the permittivity and permeability dyadics

ε(ω) = ε0 [ε∞ + χe(ω)] µ(ω) = µ0 [µ∞ + χm(ω)]

This is a fully bianisotropic material model.

18 / 53

Classification of materials

Type ε, µ ξ, ζ

Isotropic Both ∼ I Both 0An-isotropic Some not ∼ I Both 0Bi-isotropic Both ∼ I Both ∼ IBi-an-isotropic All other cases

General media (including nonlinear)

Linear media (bi-an-isotropic)

IsotropicAn-isotropic Bi-isotropic

19 / 53

Modeling arbitriness

Assume a material with both conductive and dielectric properties,

J(ω) = σ(ω)E(ω), D(ω) = ε(ω)E(ω)

The total current in Maxwell’s equations can then be written

J(ω) + jωD(ω) = [σ(ω) + jωε(ω)]︸︷︷︸=σ′(ω)

E(ω) = J ′(ω)

= jω

[σ(ω)

jω+ ε(ω)

]︸︷︷︸

=ε′(ω)

E(ω) = jωD′(ω)

where σ′(ω) and ε′(ω) are both equivalent models for the material.Thus, there is an arbitrariness in how to model dispersivematerials, either by a pure conductivity model (σ′(ω)) or by apure permittivity model (ε′(ω)), or any combination.

20 / 53

When to use what?

Consider the total current as a sum of a conduction current Jc

and a displacement current Jd:

J tot(ω) = σc(ω)E︸︷︷︸Jc(ω)

+ jωεd(ω)E︸︷︷︸Jd(ω)

The ratio can take many different values (using f = 1GHz)

|Jc(ω)||Jd(ω)|

=|σc(ω)||ωεd(ω)|

=

109 copper (σ = 5.8 · 107 S/m and ε = ε0)

1 seawater (σ = 4S/m and ε = 72ε0)

10−9 glass (σ = 10−10 S/m and ε = 2ε0)

18 orders of magnitude in difference! Conductivity model goodwhen |Jc| � |Jd|, permittivity model good when |Jd| � |Jc|.

21 / 53

Poynting’s theorem in the frequency domain

In the time domain we had

∇· (E(t)×H(t))+H(t) · ∂B(t)

∂t+E(t) · ∂D(t)

∂t+E(t) ·J(t) = 0

For time harmonic fields, we consider the time average over oneperiod (where 〈f(t)〉 = 1

T

∫ t+Tt f(t′) dt′):

∇·〈E(t)×H(t)〉+⟨H(t) · ∂B(t)

∂t

⟩+

⟨E(t) · ∂D(t)

∂t

⟩+〈E(t) · J(t)〉 = 0

The time average of a product of two time harmonic signalsf(t) = Re{f(ω)ejωt} and g(t) = Re{g(ω)ejωt} is

〈f(t)g(t)〉 = 1

2Re{f(ω)g(ω)∗}

22 / 53

Poynting’s theorem, continued

The different terms are

〈E(t)×H(t)〉 = 1

2Re{E(ω)×H(ω)∗} = 〈P〉⟨

H(t) · ∂B(t)

∂t

⟩=

1

2Re {−jωH(ω) ·B(ω)∗}⟨

E(t) · ∂D(t)

∂t

⟩=

1

2Re {−jωE(ω) ·D(ω)∗}

〈E(t) · J(t)〉 = 1

2Re {E(ω) · J(ω)∗}

Using the constitutive relation D(ω) = ε(ω) ·E(ω), we rewrite

2Re {−jωE(ω) ·D(ω)∗}= −jωE(ω) · [ε(ω) ·E(ω)]∗ + jωE(ω)∗ · ε(ω) ·E(ω)

= jωE(ω)∗ · [ε(ω)− ε(ω)†] ·E(ω)

where ε(ω)† denotes the hermitian transpose of ε(ω).23 / 53

Poynting’s theorem, final version

Using a permittivity model (J = 0), we find

∇ · 〈P〉 = − jω

4

(E(ω)

H(ω)

)†·(ε(ω)− ε(ω)† ξ(ω)− ζ(ω)†ζ(ω)− ξ(ω)† µ(ω)− µ(ω)†

)·(E(ω)

H(ω)

)Passive material: ∇ · 〈P〉 ≤ 0

Active material: ∇ · 〈P〉 > 0 ⇐= Definitions!

Lossless material: ∇ · 〈P〉 = 0

This boils down to conditions on the material matrix

jω

(ε(ω)− ε(ω)† ξ(ω)− ζ(ω)†ζ(ω)− ξ(ω)† µ(ω)− µ(ω)†

)= 2Re

{jω


)}= −2ω Im

{(ε(ω) ξ(ω)ζ(ω) µ(ω)

)}If this matrix is positive we have a lossy material, if it is zero wehave a lossless material.

24 / 53

Example: “Standard media”

With the material model D = εE, J = σE, B = µH (assumingε, σ, and µ being positive real numbers) we have(

ε(ω) ξ(ω)ζ(ω) µ(ω)

)=

((ε+ σ

jω

)I 0

0 µI

)

and

−ω Im


)}=

(σI 00 0

)This model is lossy with electric fields present, but not with puremagnetic fields. In wave propagation, we always have both E andH fields.

25 / 53

Lossless media

The condition on lossless media,

0 = Im


)}=

(ε(ω)− ε(ω)† ξ(ω)− ζ(ω)†ζ(ω)− ξ(ω)† µ(ω)− µ(ω)†

)can also be written(

ε(ω) ξ(ω)ζ(ω) µ(ω)

)=


)†That is, the matrix should be hermitian symmetric.

26 / 53

Isotropic materials

A bi-isotropic material is described by(ε(ω) ξ(ω)ζ(ω) µ(ω)

)=

(ε(ω)I ξ(ω)Iζ(ω)I µ(ω)I

)The passivity requirement implies that all eigenvalues of the matrix

jω

(ε(ω)− ε(ω)∗ ξ(ω)− ζ(ω)∗ζ(ω)− ξ(ω)∗ µ(ω)− µ(ω)∗

)are positive. If ξ = ζ = 0 it is seen that this requires (using thatε(ω)− ε(ω)∗ = 2j Im ε(ω))

−ω Im ε(ω) > 0, −ω Imµ(ω) > 0

and if ξ and ζ are nonzero we also require (after more algebra)

|ξ(ω)− ζ(ω)∗|2 < 4 Im ε(ω) Imµ(ω)

27 / 53

Kramers-Kronig dispersion relations

The causality requirement implies the Kramers-Kronig dispersionrelations (writing χ(ω) = χr(ω)− jχi(ω) for the real and

imaginary part)

χr(ω) =1

πp.v.

∫ ∞−∞

χi(ω′)

ω′ − ω dω′

χi(ω) = −1

πp.v.

∫ ∞−∞

χr(ω′)

ω′ − ω dω′

where the principal value of a singular integral is defined as

p.v.

∫ ∞−∞

χi(ω′)

ω′ − ω dω′ = limδ→0

[∫ ω−δ

−∞

χi(ω′)

ω′ − ω dω′ +

∫ ∞ω+δ

χi(ω′)

ω′ − ω dω′]

The Kramers-Kronig relations prohibit the existence of a losslessfrequency dependent material (if χi(ω) = 0 for all ω, then the firstequation implies χr(ω) = 0 for all ω).

28 / 53

Some notes

I The Kramers-Kronig relations restrict the possible frequencybehavior of any causal material.

I It requires a model χ(ω) for all frequencies.

I Usually, our models are derived or measured only in a finitefrequency interval, ω1 < ω < ω2.

I We need to extrapolate the models to zero and infinitefrequencies, ω → 0 and ω →∞. Not a trivial task!

29 / 53

Outline






6 Conclusions

30 / 53

Randomly oriented dipoles

Consider a medium consisting of randomly oriented electric dipoles,for instance water. The polarization is defined as

P = lim∆V→0

∑i pi

∆V

A typical situation is depicted below.

E = 0

p

E

p

31 / 53

Physical processes

We now consider two microscopic processes:

1. The molecules strive to align with an imposed electric field, atthe rate ε0αE.

2. Thermal motion tries to disorient the polarization. With τbeing the relaxation time for this process, the rate of changesin P are proportional to −P /τ .

This results in the following differential equation:

∂P (t)

∂t= ε0αE(t)− P (t)

τ

This is an ordinary differential equation with the solution(assuming P = 0 at t = −∞)

P (t) = ε0

∫ t

−∞αe−(t−t

′)/τE(t′) dt′

32 / 53

Physical processes

We now consider two microscopic processes:

1. The molecules strive to align with an imposed electric field, atthe rate ε0αE.

2. Thermal motion tries to disorient the polarization. With τbeing the relaxation time for this process, the rate of changesin P are proportional to −P /τ .

This results in the following differential equation:

∂P (t)

∂t= ε0αE(t)− P (t)

τ

This is an ordinary differential equation with the solution(assuming P = 0 at t = −∞)

P (t) = ε0

∫ t

−∞αe−(t−t

′)/τE(t′) dt′

32 / 53

Dispersion or conductivity model

From the solution we identify the susceptibility function

χ(t) = u(t)αe−t/τ

This is monotonically decaying without oscillations.

Including allthe effects in the D-field results inD(t) = ε0

(E(t) +

∫ t

−∞αe−(t−t

′)/τE(t′) dt′)

J(t) = 0

and shifting it to J results inD(t) = ε0E(t)

J(t) = ε0αE(t)− ε0α

τ

∫ t

−∞e−(t−t

′)/τE(t′) dt′

Both versions have the same total current J(t) + ∂D(t)∂t .

33 / 53




This is monotonically decaying without oscillations. Including allthe effects in the D-field results inD(t) = ε0

(E(t) +

∫ t

−∞αe−(t−t

′)/τE(t′) dt′)

J(t) = 0



τ

∫ t

−∞e−(t−t

′)/τE(t′) dt′


33 / 53




This is monotonically decaying without oscillations. Including allthe effects in the D-field results inD(t) = ε0

(E(t) +

∫ t

−∞αe−(t−t

′)/τE(t′) dt′)

J(t) = 0



τ

∫ t

−∞e−(t−t

′)/τE(t′) dt′


33 / 53

Response to different excitations

Consider two different excitation functions:

I One square pulse E(t) = 1 for 0 < t < t0, and zero elsewhere.

I A damped sine function, E(t) = e−t/t0 sin(ωat)u(t).

The response P (t) can be calculated by numerically performingthe convolution integral

P (t) = ε0

∫ t

−∞χ(t− t′)E(t′) dt′

A python script materialmodels.py is available on the course homepage, with which different material models and excitations can beexplored. It is also available as a windows executablematerialmodels.exe.

34 / 53

Debye model, square pulse excitation

0 5 10 15 20 25 30t

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5P(t)/ε0

E(t)

χ(t)

D(t)/ε0

35 / 53

Debye model, sine excitation

0 5 10 15 20 25 30t

1.0

0.5

0.0

0.5

1.0

1.5P(t)/ε0

E(t)

χ(t)

D(t)/ε0

36 / 53

Debye material in frequency domain

The susceptibility kernel is χ(t) = αe−t/τu(t), with the Fouriertransform

χ(ω) =

∫ ∞0

αe−t/τe−jωt dt =α

jω + 1/τ=

ατ

1 + jωτ

The real and negative imaginary parts have typical behavior asbelow:

0.0 0.5 1.0 1.5 2.0 2.5 3.00.0

0.5

1.0

1.5

2.0

2.5real/neg imag

χ(ω)

and the relative permittivity is εr(ω) = 1 + χ(ω) = ε′ − jε′′.

37 / 53

Harmonic oscillator

The archetypical material behavior is derived from an electronorbiting a positively charged nucleus. The typical forces on theelectron are:

1. An electric force F 1 = qE from the applied electric field.

2. A restoring force proportional to the displacementF 2 = −mω2

0r, where ω0 is the harmonic frequency.

3. A frictional force proportional to the velocity,F 3 = −mν∂r/∂t.

38 / 53

Harmonic oscillator, continued

Newton’s acceleration law now gives

m∂2r

∂t2= F 1 + F 2 + F 3 = qE −mω2

0r −mν∂r

∂t

Introducing the polarization as P = Nqr, where N is the numberof charges per unit volume, we have

∂2P (t)

∂t2+ ν

∂P (t)

∂t+ ω2

0P (t) =Nq2

mE(t)

This is an ordinary differential equation, with the solution(assuming P = 0 for t = −∞)

P (t) = ε0ω2p

ν0

∫ t

−∞e−(t−t

′)ν/2 sin(ν0(t− t′))E(t′) dt′

with ωp = Nq2/(mε0) and ν20 = ω20 − ν2/4.

39 / 53

Interpretation

The susceptibility function of the Lorentz model is

χ(t) = u(t)ω2p

ν0e−tν/2 sin(ν0t)

with typical behavior as below.

0 5 10 15 20 25 30t

0.8

0.6

0.4

0.2

0.0

0.2

0.4

0.6

0.8

1.0χ(t)

40 / 53

Lorentz model, square pulse excitation

0 5 10 15 20 25 30t

1.0

0.5

0.0

0.5

1.0

1.5

2.0P(t)/ε0

E(t)

χ(t)

D(t)/ε0

41 / 53

Lorentz model, sine excitation

0 5 10 15 20 25 30t

1.0

0.5

0.0

0.5

1.0

1.5P(t)/ε0

E(t)

χ(t)

D(t)/ε0

42 / 53

Lorentz model, resonant excitation

0 5 10 15 20 25 30t

6

4

2

0

2

4

6P(t)/ε0

E(t)

χ(t)

D(t)/ε0

43 / 53

Lorentz material in frequency domain

The susceptibility kernel is χ(t) =ω2p

ν0e−νt/2 sin(ν0t) u(t), with the

Fourier transform

χ(ω) =

∫ ∞0

ω2p

ν0e−νt/2 sin(ν0t)e

−jωt dt =ω2p

−ω2 + ω20 + jων

The real and negative imaginary parts have typical behavior asbelow:

0.0 0.5 1.0 1.5 2.0 2.5 3.03

2

1

0

1

2

3

4

5

6real/neg imag

χ(ω)

and the relative permittivity is εr(ω) = 1 + χ(ω) = ε′ − jε′′.44 / 53

Example: permittivity of water

Microwave properties (one Debye model, Kaatze (1989)):

10 20 30 40 50 600

20

40

60

80

f (GHz)

ε = ε′ − jε′′ Water permittivity at 25◦ C

◦ ◦Measured data

––Debye model

Light properties (many Lorentz resonances, Hale & Querry (1973)):

1012 1013 1014 10150

1

2

3

4

5

f (Hz)

ε′

visible

region

10−8

10−4

100

ε′′

45 / 53

Handin 1

In the first handin, you will model and interpret the response of aferromagnetic material, when subjected to a magnetic field.

H

M

The magnetic moment of each atom precesses around the appliedmagnetic field, described by the Landau-Lifshitz-Gilbert equation

∂M

∂t= −γµ0M ×H + α

M

|M | ×∂M

∂t

46 / 53

Outline






6 Conclusions

47 / 53

What is a metamaterial?

I Engineered materials,designed to have unusualproperties

I Periodic structures

I Resonant inclusions

I Negative refractive index

I Negative or near zeropermittivity/permeability

48 / 53

Some examples

49 / 53

Typical frequency behavior of a dielectric

103 106 109 1012 1015 f (Hz)

ε = ε′ − jε′′

+

−

conduction +

−relaxation

+

−

vibrationrotation

+

−transition

ε(0)

ε(∞)

Between the low- and high-frequency asymptotes ε(0) and ε(∞)(green region), there are no bandwidth limitations for metamaterialdesign. Outside the asymptotes there are strong restrictions onbandwidth for metamaterials (note small dip at f ≈ 1015Hz).

50 / 53

Typical frequency behavior of a dielectric

103 106 109 1012 1015 f (Hz)

ε = ε′ − jε′′

+

−

conduction +

−relaxation

+

−

vibrationrotation

+

−transition

ε(0)

ε(∞)

Between the low- and high-frequency asymptotes ε(0) and ε(∞)(green region), there are no bandwidth limitations for metamaterialdesign. Outside the asymptotes there are strong restrictions onbandwidth for metamaterials (note small dip at f ≈ 1015Hz).

50 / 53

Bounds on metamaterials

The requirements of linearity, causality, time translationalinvariance, and passivity, gives bounds on the relative bandwidth B(after some complex analysis involving Kramers-Kronig relations).

maxω∈B|ε(ω)− εm| ≥

B

1 +B/2(ε∞ − εm)

{1/2 lossy case

1 lossless case

51 / 53

Outline






6 Conclusions

52 / 53

Conclusions

I Constitutive relations are necessary in order to fully solveMaxwell’s equations.

I Their form is restricted by physical principles such as linearity,causality, time translational invariance, and passivity.

I A Debye model is suitable for dipoles aligning with animposed field (relaxation model).

I A Lorentz model is suitable for bound charges (resonancemodel).

I There are restrictions on what kind of frequency behavior isphysically possible. If you want extreme behavior (outside low-and high-frequency asymptotes), you get small bandwidth.

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Electromagnetic Wave Propagation Lecture 2: Time harmonic ... · I The theory given in this lecture...

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