Zhang K., Li D. Electromagnetic Theory for Microwaves and Optoelectronics

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2.6 Dielectric Layers and Impedance Transducers 111

then the input impedance at the surface of the intermediate medium is equal

to the normal-wave impedance of the input medium and the reﬂection will

be eliminated:

Z(x

) =

Z(x

)

= Z

and the reﬂection is vanished,

Γ (x

) = 0.

The state of Γ (x

) = 0 and ρ = SWR = 1 is know as the state of matching,

i.e., the traveling wave state.

This is known as the quarter-wavelength anti-reﬂection coating and also

as the quarter-wavelength impedance transducer.

Obviously, the single-section quarter-wavelength impedance transducer is

a frequency sensitive or narrow band device.

2.6.2 Multiple Dielectric Layer,

Multi-Section Impedance Transducer

The bandwidth of a single-section transducer is narrow. To broaden the

bandwidth, we may increase the number of the quarter wavelength sections

to form a multiple dielectric layer or a multi-section impedance transducer.

For a N section transducer, the imp edance relation for the neighboring

sections is

= Z

C(i−1)

C(i+1)

The solution of a Multi-section Impedance Transducer is not unique, so

there are a number of designs. The most popular design is the Chebyshev

polynomial design and the binomial design. The former gives a equal ripple

response and the latter gives a ﬂatness response. The design of a multi-

section impedance transducer or a multiple dielectric layer is to be given in

Section 3.7.

2.6.3 A Multi-Layer Coating with Alternating Indices.

The Chebyshev and binomial multi-section transducers are successfully used

in microwave transmission systems. But for microwave or optical coatings,

it may be diﬃcult to ﬁnd a transparent dielectric material with the required

wave impedance or the required index that adheres well to the substrate.

A multi-layer coating with an alternating wave impedance or alternating

indices, as shown in Fig. 2.26, is much easier to make. It may become an

anti-reﬂection (AR) coating as well as a high-reﬂection (HR) coating [38].

The wave impedances of the input and the output media are Z

and

, respectively. There are a number of layer pairs in between the input

and the output media. The wave impedances of the dielectrics in each layer

pair are Z

and Z

and their thickness is λ/4, where λ is the wavelength

112 2. Introduction to Waves

Figure 2.26: A multi-layer coating with an alternating index.

in the medium or waveguide. Thus we have a multi-layer λ/4 coating with

an alternating wave impedance or alternating index.

, Z

Suppose there are m pairs of layers, i = 1 to m, counted from Z

. The

input impedance of the layer with Z

of the ith layer pair is Z

and the

input impedance of the layer with Z

of the ith layer pair is Z

, which is

the input impedance of the ith layer pair. See Fig. 2.26.

For the ﬁrst λ/4 layer, referring to (2.259), we have

which means that the input impedance of the layer with Z

of the ﬁrst layer

pair is given by

Similarly, the input impedance of the layer with Z

of the ﬁrst layer pair is

given by

Substituting the expression for Z

into it yields

= Z

For the ith pair of layers,

= Z

2,i−1

. (2.261)

2.6 Dielectric Layers and Impedance Transducers 113

We come to the conclusion that a pair of λ/4 layers transforms the

impedance by multiplying it by the factor n

. For a coating system of m

pairs of λ/4 layers on the substrate, the input impedance seen at the input

plane Z

with respect to the impedance of the substrate Z

is given by

= Z

. (2.262)

The transformation ratio of the input impedance to the impedance of the

substrate is the 2mth power of the ratio of the indices of the layers. Even

for a ratio n

very close to unity, the multiplier can be made much larger

or much smaller than unity by choosing the relation between n

and n

, and

making a large number of layer pairs m.

1. Anti-reﬂection (AR) coating. For optical elements, n

is usually larger

than n

. The multiple layer pairs act as an anti-reﬂection coating if

> n

and m is large enough. The ratio n

is chosen as

≈ 1, i.e.,

≈ 1,

so that

= Z

≈ Z

, and Γ =

− Z

+ Z

≈ 0.

This is the matching state.

2. High-reﬂection (HR) coating. The multiple layer pairs act as a high-

reﬂection coating if m is large enough, n

< n

, and n

is chosen

¿ 1, i.e.,

¿ 1,

so that

= Z

¿ Z

and Γ =

− Z

+ Z

≈ −1.

This is the high reﬂection state.

In most applications, the low-loss highly reﬂecting mirrors made with

multi-layer dielectric coatings are even better than those made with metal-

lic coatings. Of course, the anti-reﬂection or high-reﬂection property of a

multi-layer dielectric coating is frequency dependent, because the layers are

a quarter-wavelength thick only at a speciﬁc frequency. The band width of

such a multi-layer dielectric coating can b e large enough if the number of layer

pairs is large enough. Usually, the number of layer pairs is up to hundreds

for an optical coating.

114 2. Introduction to Waves

The multi-layer coating with an alternating wave imp edances can also

be analyzed as a DFB structure by means of a periodic system and mode-

coupling theories, refer to Chapter 7.

Problems

2.1 (1) Find the instantaneous Poynting vector of a linearly polarized plane

wave with respect to time and distance.

(2) Prove that the instantaneous Poynting vector of a circularly polar-

ized plane wave is independent of time and distance.

2.2 The expression for an elliptically polarized wave is given as

E = (E

+ jE

j(ωt−kz)

where E

and E

are not necessarily perpendicular to each other.

(1) Find the relations among E

, E

and E

, E

, δ

in (2.71).

(2) Find the angle between E

and E

(3) Find the condition under which E

and E

are perpendicular to

each other.

(4) Find the corresponding magnetic ﬁeld vector H.

2.3 The electromagnetic parameters of earth depend upon the dampness.

For dry earth, ²

≈ 5, σ ≈ 10

−5

S/m, and for wet earth, ²

≈ 10,

σ ≈ 10

−1

S/m.

(1) Find the frequency for σ = ω² for the above two cases.

(2) Find the depth of penetration of a plane wave at 100 MHz in the

above two cases.

2.4 Calculate the attenuation coeﬃcient, phase velocity, and wave

impedance for a plane wave of frequency 10 GHz propagating in glass.

For typical glass at 10 GHz, ²

/²

= 6, ²

/²

= 20.

2.5 Derive the expressions for the phase velocity and group velocity of a

uniform plane wave propagating in a good conductor.

2.6 Prove that the power ﬂow entering a good conductor is equal to the Joule

dissipation in the conductor.

2.7 Derive the expression for the group velocity of a plane wave in a good

conductor.

2.8 Find the group velocity of a plane wave in a conductive medium, If σ, ²

and ²

of the medium are independent of frequency.

Problems 115

2.9 Pure water is a good insulator, its relative permittivity is 81. Find the

power ratio of the reﬂection wave to the incident wave and that of

the transmission wave to the incident wave, for an incident plane wave

passing normally through the plane surface of the water.

2.10 A plane wave is normally incident upon an boundary between two non-

magnetic, lossless media. Find the condition under which the reﬂection

coeﬃcient and the transmission coeﬃcient are equal to each other.

2.11 A plane wave with circular frequency ω is incident normally from vac-

uum upon the plane surface of a nonmagnetic conductive medium with

conductivity σ and permittivity ².

(1) Find the reﬂection coeﬃcient and the transmission coeﬃcient.

(2) Find the reﬂection coeﬃcient and the transmission coeﬃcient for a

low-loss conductive dielectric and a good conductor. Show that the re-

ﬂection coeﬃcient of a good conductor is given by Γ ≈ 1 −

2ω²

/σ =

1 − ωδ/c.

2.12 Give the expressions for instantaneous values of the composed ﬁeld

components for the oblique reﬂection of n wave and p wave at an air –

perfect conductor plane boundary.

2.13 Find the complex and instantaneous Poynting vectors in the direction

parallel and perpendicular to the boundary for the oblique reﬂection of

n wave and p wave at an air – perfect conductor plane boundary.

2.14 A light beam is incident from the air upon a nonmagnetic medium of

index n, show that

(1 + Γ

)(1 − Γ

)

(1 − Γ

)(1 + Γ

)

This is an experimental method for measuring the index of the dielectric

material. In practice, the incident angle is chosen to be 45

◦

. Note that,

both Γ

and Γ

are negative for an incident angle of 45

◦

, and the

measured values are the power reﬂection coeﬃcients, hence the negative

square roots of them must be taken for the calculation.

2.15 Show that Fressnel’s formulas for nonmagnetic dielectric media may be

expressed by the following form, in which only θ

and θ

are included.

= −

sin(θ

− θ

)

sin(θ

+ θ

)

, T

2 cos θ

sin θ

sin(θ

+ θ

)

tan(θ

− θ

)

tan(θ

+ θ

)

, T

2 cos θ

sin θ

sin(θ

+ θ

) cos(θ

− θ

)

116 2. Introduction to Waves

2.16 Prove that for the boundary between magnetic media, i.e., ²

= ²

and

6= µ

, the Brewster angle exists only for the TE wave. Find the

expression for the corresponding Brewster angle.

2.17 Derive the composed ﬁeld expressions for the TM mode in the case of

total reﬂection.

2.18 (1) Calculate the critical angle for a plane wave passing from pure water

into air, for the pure water, ²

/²

≈ 81.

(2) Calculate the critical angle for a plane wave passing from glass into

air, for typical glass, ²

/²

≈ 6.

2.19 (1) Calculate the Brewster angle for a plane wave passing from from

air into pure water and from pure water into air.

(2) Calculate the critical angle for a plane wave passing from air into

glass and from glass into air.

2.20 In the case of total reﬂection,

(1) Prove that the average active Poynting vector along the direction

perpendicular to the b oundary is zero, both in the incident region and

the refraction region.

(2) Find the Poynting vector and the power ﬂow per unit width along

the direction parallel to the boundary.

2.21 Reﬂection mirrors for the HeNe laser consist of alternative coatings of

ZnS and ThF

. The refractive indices of them are 2.5 and 1.6, respec-

tively. The index of the substrate is 1.5 (glass). Find the minimum

number of layer pairs for a p ower reﬂection coeﬃcient greater than

99.5%. The wavelength of the HeNe laser beam is 632.8 nm.

2.22 When a plane wave is reﬂected from the boundary of a dielectric,

(1) under what condition does the circularly polarized incident wave

becomes linearly polarized reﬂected wave.

(2) under what condition does the linearly polarized incident wave be-

comes a circularly polarized reﬂected wave.

Chapter 3

Transmission-Line Theory

and Network Theory for

Electromagnetic Waves

Two powerful tools based on circuit theory, transmission line theory and

network theory, are widely applied to the analysis and simulation of var-

ious electromagnetic wave phenomena. In this chapter, the basic concept

of distributed circuits, the waves propagating along transmission lines, the

transmission-line charts, the elementary network theory and the impedance

transducers are introduced.

3.1 Basic Transmission Line Theory

A transmission line made up of two parallel wires is the earliest system for

the transmission of electromagnetic signals and energies. In general, a trans-

mission line may be made up of any two conductors separated by a dielectric

insulator, for example, parallel wires, parallel plates, or coaxial conductors,

see Figure 3.1. For a transmission line with two conductors, the boundary

conditions can be satisﬁed by TE, TM and TEM modes. Among them, TEM

mode is the dominant mode.

Two diﬀerent approaches are used in the analysis of the TEM wave in

transmission lines, the ﬁeld approach and the circuit approach. In the ﬁeld

approach, we deal with the ﬁeld distribution in a speciﬁc transmission-line

structure, and the result can be used for only this speciﬁc structure. In

the circuit approach, distributed circuit parameters are introduced and the

transmission line can be described as a distributed-parameter electric net-

work. The result of the circuit approach is suitable for transmission lines

of any structure, and the circuit parameters for a speciﬁc structure can be

118 3. Transmission-Line and Network Theory for Electromagnetic Waves

Figure 3.1: Transmission lines.

formulated by means of static ﬁeld theory.

3.1.1 The Telegraph Equations

In the circuit approach, the transmission line can be described as series-

connected inductance and resistance per unit length, denoted by L and R, re-

spectively, and shunt-connected capacitance and conductance per unit length,

denoted by C and G, respectively. These circuit elements are not connected

at discrete points on the line, but are distributed inﬁnitesimally along the

line. The transmission line can then be described as a cascade connected

network chain comp osed of a inﬁnite number of diﬀerential lengths dz, see

Fig. 3.2. The equivalent circuit for these inﬁnitesimal segments of line is

shown in Fig. 3.3, it consists of circuit elements Ldz, Rdz, Cdz and Gdz.

According to Kirchhoﬀ’s law, the circuit equations for the equivalent cir-

cuit of a diﬀerential segment, refer to Fig. 3.3, are written as follows:

U(z + dz) − U(z) =

dU(z)

dz = −(R + jω L )dz I(z), (3.1)

I(z + dz) − I(z) =

dI(z)

dz = −(G + jωC)dz U(z + dz). (3.2)

Let

Z = R + jωL, Y = G + jωC, (3.3)

and neglect the terms including the square of the inﬁnitesimal quantity dz,

gives

dU(z)

= −Z I(z), (3.4)

dI(z)

= −Y U (z). (3.5)

3.1 Basic Transmission Line Theory 119

Figure 3.2: Equivalent network with distributed parameter.

Figure 3.3: Equivalent circuit for a diﬀerential segment.

These are the transmission-line equations. The forms of these equations are

the same as the one-dimensional source-free Maxwell equations for a plane

wave (2.8) and (2.10) as shown in Section 2.1.

Taking the derivative of equation (3.4) then substituting equation (3.5)

to cancel I(z), yields

U(z)

− Y Z U(z) = 0. (3.6)

Similarly, we have

I(z)

− Y Z I(z) = 0. (3.7)

These two diﬀerential equations are known as telegraph equations.

3.1.2 Solution of the Telegraph Equations

We can see that the telegraph equations are just the same as the one-

dimensional scalar Helmholtz’s equations (2.14) and (2.15). The solutions

of (3.6) and (3.7) must be the same as (2.18) and (2.19):

U(z, t) = U(z)e

jωt

−γz

+ U

−

γz

jωt

, (3.8)

I(z, t) = I(z)e

jωt

−γz

+ I

−

γz

jωt

, (3.9)

120 3. Transmission-Line and Network Theory for Electromagnetic Waves

where

γ = α + jβ =

√

Y Z, (3.10)

is the propagation factor of the wave; the real part is the attenuation factor

and the imaginary part is the phase factor.

Substituting (3.8) into (3.4), yields

I(z, t) = I(z)e

jωt

−γz

−

γz

jωt

, (3.11)

where

= −

−

, (3.12)

denotes the characteristic impedance of the transmission line which is the

ratio of the voltage to the current of a traveling wave and is determined by

the conﬁguration of the transmission line [24].

(1) The General Case

In the general case, the propagation factor and the characteristic impedance

are complex,

γ = α + jβ =

(R + jωL)(G + jωC),

where

α =

+ ω

) (G

+ ω

) − (ω

LC − RG)

, (3.13)

β =

+ ω

) (G

+ ω

) + (ω

LC − RG)

, (3.14)

and

R + jωL

G + jωC

. (3.15)

It is just the same as the plane wave in a lossy medium. There are two

attenuated traveling waves propagate on the transmission line with opposite

directions of propagation.

(2) Low Frequency, Large Loss

In the case of relatively low frequency and relatively large loss,

ωL ¿ R, ωC ¿ G.

We have

α ≈

√

RG, β ≈ 0.

There is no wave propagation, only attenuation on the line.