MAKERERE UNIVERSITY

DEPARTMENT OF ELECTRICAL ENGINEERNG

ELE3101 ELECTROMAGNETIC FILEDS CLASS NOTES

BY

STEPHEN S. MWANJE

1

Table of Contents

CHAPTER 1: RELATIONSHIP BETWEEN FIELD AND CIRCUIT THEORY.........................3

1.1. INTRODUCTION.................................................................................................3

1.2. CIRCUIT LAWS OBTAINED USING FIELD QUANTITIES..........................................3

1.3. MAXWELL’S EQUATIONS AS GENERALISATIONS OF CIRCUIT EQUATIONS...........5

1.4. BREAK DOWN OF SIMPLE CIRCUIT THEORY IN PROBLEM ANALYSIS..................8

CHAPTER TWO: UNBOUNDED WAVE PROPAGATION............................................9

2.1. THE WAVE EQUATION IN A PERFECT DIELECTRIC..............................................9

2.2. UNIFORM PLANE WAVES.................................................................................11

2.3. FREQUENCY DEPENDENCE OF THE CLASSIFICATION MATERIALS...................13

2.4. WAVE PROPAGATION IN A CONDUCTIVE MEDIUM............................................14

2.5. POWER FLOW IN ELECTROMAGNETIC FIELDS.................................................17

2.6. REFLECTION AND REFRACTION OF UNIFORM PLANE WAVES...........................20

2.7. POLARISATION................................................................................................23

CHAPTER 3: WAVE PROPAGATION IN TRANSMISSION LINES..................................26

3.1. INTRODUCTION...............................................................................................26

3.2. TRANSMISSION LINE EQUATIONS (DISTRIBUTED CIRCUIT ANALYSIS)...............26

3.3. STANDING WAVES ON TRANSMISSION LINES..................................................32

3.4. TRANSMISSION LINES MATCHING CONSIDERATIONS.......................................35

3.5. GRAPHICAL AIDS TO TRANSMISSION LINE CALCULATIONS..............................41

CHAPTER 4: ELECTROMAGNETIC WAVE PROPAGATION IN WAVEGIDES...................49

4.1. THE INFINITE PLANE WAVEGUIDE....................................................................49

4.2. THE RECTANGULAR WAVEGUIDE....................................................................60

4.3. CIRCULAR WAVEGIDES...................................................................................70

CHAPTER 5: WAVE ROPAGATION IN OTHER SYSTEMS..........................................72

6.1. PLASMAS........................................................................................................72

6.2. MICROSTRIP TRANSMISSION LINES.................................................................73

6.3. PROPAGATION IN OPTICAL FIBERS.................................................................................75

REFERENCES:............................................................................................78

APPENDICES..............................................................................................79

APPENDIX A: GRAPHICAL SOLUTION TO DOUBLE STUB MATCHING............................79APPENDIX A: GRAPHICAL SOLUTION TO DOUBLE STUB MATCHING

2

CHAPTER 1: RELATIONSHIP BETWEEN FIELD AND CIRCUIT THEORY

1.1. INTRODUCTION

Conventional circuit theory, where we deal with Voltage, V and Current, I, and field theory, where we use the field vectors E, D, B, H, and J are inter-related. Consideration of circuits from either point of view gives the same results. However, there are certain inherent assumptions in the circuit theory approach, which become invalid as circuit dimensions and the impressed signal wavelength become comparable. This necessitates either the use of field theory, which is the more general approach or a modification of the circuit theory approach.

In this chapter, we shall see how the two are related, and why circuit theory has limitations. It will be shown that the normal expressions can be obtained using field theory, and that Maxwell’s equations, the “four commandments” of electromagnetic field propagation, can be obtained as generalizations of circuit expressions.

1.2. CIRCUIT LAWS OBTAINED USING FIELD QUANTITIES

(1) Ohm’s law:

3

pqAJ,

VCRL

Consider the conducting rod in figure 1.1 with parameters as shown

Figure 1.1: Conducting Rod of Uniform cross-section and current density

If

Ε is the electric field at a point, then

J σΕ = and

q q

p p

J Jl ldl dl J

σ σ σΕ = = = Α

Α∫ ∫g g

1.1

Note that we have assumed a uniform rod with a uniform current density, J.Since:

q

p

dl VΕ =∫ g

(potential difference between p and q)

J IΑ= (Current through the rod)

lR

σ=

Α

(Resistance of the rod)

Equation 1.1 states that V=IR, which is Ohm’s law derived from field theory.

(ii) The series R-L-C circuit

4

01765432

Figure 1.2 shows a simple series R-L-C circuit

Figure 1.2: Simple R-L-C Circuit

Recall Faraday’s law in integral form:

dl dst

∂Ε = − Β∂∫ ∫g g

[Surface not changing] 1.2

Consider the RHS. Since the circuit in figure 1.2 is time invariant, the partial

derivative can be replaced by an ordinary one: furthermore,

dsΒ =Φg, the total

flux (we assume it links all turns). The RHS can therefore be written as:

( )d d dIds LI L

t dt dt dt

φ∂− Β = − = − = −∂ ∫ g

1.3

The right hand side can be broken into five parts:

• The integral from 0 to 1 - V01, is the applied voltage. Note that V01=-V10;

• The integral from 2 to 3;3 3

2 2

Jdl dl IR

σΕ = =∫ ∫g g

1.4

• The integral from 4 to 5, where

dlΕ∫ g

= voltage drop across an element.

Voltage drop across the resistor is not the same as that across the

5

capacitor. Across resistor, energy is actually lost. Across the capacitor,

energy is stored as 5 5

4 4

D Qdl dl

CεΕ = =∫ ∫g g

1.5

Note: D= Q/A, and the integral gives the capacitor plate spacing d

multiplied by Q/A. we then use C =

/ dεΑ

With no charge on the capacitor at t = −∞

, the charge Q will be given by

5

4

1,

t t

Q Idt dl Idtc−∞ −∞

= ∴ Ε =∫ ∫ ∫g

1.6

• The integral from 6 to 7; By virtue of the fact that we assume a perfectly conducting filament, which must have zero tangential electric field; this part of the integral is identically zero.

Combining equation 1.3 to 1.7 then gives us the following result:

10

1 tdIV IR L Idt

dt C −∞

= + + ∫

1.7

Equation 1.7 is the familiar expression for the series R-L-C circuit, but this time derived from field theory. Several assumptions were used:

(A) A filamentary conductor defines the closed path or circuit. This conductor has zero tangential electric field (E) everywhere. For perfect conductor,

tan 0Ε = and

0dlΕ =∫ g

. No voltage drop along conductor.

(B) Maximum circuit dimensions are small compared to the wavelength.

(C) Circuit elements are ideal, i.e., displacement current, magnetic flux and imperfect conductivity are confined to capacitors, inductors and resistors respectively.

The above two examples have demonstrated that ordinary circuits, can be analyzed using field theory.

6

1.1. MAXWELL’S EQUATIONS AS GENERALISATIONS OF CIRCUIT EQUATIONS

Maxwell’s equations can be obtained as generalizations of Ampere’s, Faraday’s, and Gauss’s laws, which are circuit equations.

(i) Ampere’s law:

dl IΗ =∫ gÑ1.8

Note: A Capacitor stores energy predominantly in the electric field while an Inductor stores energy predominantly in the magnetic field.

Stokes theorem coverts the line integral in equation 1.8 around a closed path to an integral over the surface enclosed by the path. Consequently, a more general relation is obtained by substituting for I using the conduction current density, J. An even more general expression is obtained by including the displacement

current density,

D t∂ ∂ to give:

s s s

D Ddl J ds ds J ds

t dt

∂ ∂Η = + = + ∂ ∫ ∫ ∫ ∫g g g gÑ

1.9

This is the loop or mesh form of one of Maxwell’s equations derived from Ampere’s law. Using Stoke’s theorem, LHS of the integral in equation 1.9 can be converted to an open surface integral. We thus get the point form of the equation:

DJ

t

∂∇ΧΗ = +∂

1.10

(ii) Faraday’s Law (for constant flux):

d

dt

φν = −

1.11

Where; V is the induced emf in a circuit and Φ

is the total magnetic flux linking

the circuit.

7

Since voltage is the integral around the circuit of

,dlΕg

and Φ

is the integral of

dsΒg

over the surface enclosed by the circuit, the more general form of equation

1.11 is:

s

dl dst

∂ΒΕ = −∂∫ ∫g gÑ

1.12

The surface may be changing so the time derivative should be inside the integral sign. This is another one of Maxwell’s equations. The point relation is obtained by applying Stokes theorem to get:

t

∂Β∇ΧΕ = −∂

1.13

(iii) Gauss’s law (electric field)

D ds Q=∫ gÑ1.14

Generally, total charge is the integral, over the volume of interest, of the charge density, p. Equation 1.14 becomes:

D ds dvρ=∫ ∫g gÑ1.15

The relation is obtained by applying the divergence theorem (which converts an integral over a closed surface to a volume integral within the volume enclosed) to the LHS of equation 1.15 to give:

D ρ∇ =g

1.16

(iv) Gauss’ law (magnetic field)

0dsΒ =∫ gÑ1.17

The magnetic field does not have source points. Thus, there is no such things as a magnetic charge, implying that magnetic charge = 0 as in equation 1.7. Applying the divergence theorem gives

8

0∇ Β =g

1.18

To summarize these results:

0

J D

D ρ

∇ΧΗ= +

∇ΧΕ=−Β∇ =∇Β=

&

&

g

g

( )

0

dl D J ds

dl ds

D ds dv

ds

ρ

Η = +

Ε =−Β

=

Β =

∫ ∫∫ ∫∫ ∫∫

&g g

&g g

g

g

Ñ

ÑÑÑ

I

II

III

IV

The above field equations have been obtained as generalizations of circuit equations. These four equations contain the continuity equation,

J ρ∇ = −g

or

J ds dvρ= −∫ ∫gÑ1.19

1.1.1. Free space relationships

In free space, and for most practical purposes in air, the conduction current density and the charge density are zero, permitting simplification of Maxwell’s equation:

0

0

D

D

∇ΧΗ=

∇ΧΕ=−Β∇ =∇Β=

&

&

g

g

1.1.2. Harmonic fields

For harmonic time variation of a field, jwteΟΑ = Α

;

jwt

∂Α = Α∂

.In other words,

taking a partial derivative with respect to time for harmonic fields is equivalent to

multiplying the field by

jω. Similarly, a double partial derivative with respect to

9

ComponentHalf a Wavelength

time is equivalent to multiplying by -2w

. For harmonic time variations, Maxwell’s

equations therefore are:

I.

( )jwσ ε∇ΧΗ= + Ε→Circulation of the magnetic field generates

an orthogonal electric field.

II.

JWµ∇ΧΕ = − Η →Circulation of the electric field generates an

orthogonal magnetic field.

III.

, ,D ρερ ε ρ∇ = ∇ Ε = ∇ Ε = →g g g

Source point of an Electric

Field is a charge. Charge enclosed by the surface determines the flux out of the surface.

IV.

0∇ Β = →g

Magnetic field has no source points

Note that the constitutive relations

, ;D ε µ= Ε Β = Η and

J σ= Ε have been

used, and that a homogeneous isotropic medium has been assumed.

1.1. BREAK DOWN OF SIMPLE CIRCUIT THEORY IN PROBLEM ANALYSIS

Simple circuit theory assumes a current (conduction or displacement) which is constant throughout a circuit element, i.e., even if the current is alternating, the same current in the same direction exists at all similarly aligned cross-sections of the circuit element at any instant in time. This is because at low frequencies the wavelength is much greater than the dimensions of the circuit element, so the field strength can be assumed constant. This is illustrated in Figure 1.3A.

10

AwavelengthB

At higher frequencies, wavelength approaches circuit dimensions so that the assumptions of constant electric field and current are no longer valid (Figure. 1.3B). These vary from point to point in circuit element at any instant in time.

Figure .1.3: circuit component relative size at low frequencies.

When simple circuit theory breaks down, it is necessary to use distributed circuit analysis. Circuit quantities (V and I) are permitted to change incrementally along the circuit. Defining relationships are in the form of differential equations. The physical circuit is then described in the form of equivalent impedance, to which simple circuit theory can be applied. This approach will be used when analyzing transmission lines.

1.1.1. Assignment One:

1.1.2. 1.1. Starting with Maxwell’s equations derive the continuity equation

1.2. Show that for harmonic time variation of a field

Α, given as

jwteΟΑ = Α

22

2w

t

∂ Α = − Α∂

1.3. Show that the partial differential equation 2 2

2 2

A A

x tµε∂ ∂=

∂ ∂

has a general

solution of the form:

( ) ( )0 2 0IA f V t f v t= ×− + ×+; with Vo appropriately defined

11

CHAPTER TWO: UNBOUNDED WAVE PROPAGATION

2.

2.1. THE WAVE EQUATION IN A PERFECT DIELECTRIC

Definition: Wave motion:

A group of phenomena constitute a wave if a physical phenomenon occurring at one place at a given time is reproduced at other locations later, the time delay being proportional to the space separation from the first location.

12

Consider, e.g.,

( )1 0f V tΧ− at times

1t

and

2t

(Figure 2.1). At any fixed time (e.g.

t=t1,t=t2 etc) the function only depends on X. Evidently the phenomenon travels in

the positive x direction with a velocity

0V

. Similarly,

( )2 0f X tν+ represents a

phenomenon traveling in the negative x direction.

Figure 2.1: Illustration of a propagating phenomenon.

We shall now develop the equation governing the propagation of fields in a perfect dielectric (no charges, no conduction current), starting with Maxwell’s equations.

( )jwσ ε∇ΧΗ= + Ε I

JWµ∇ΧΕ = − ΗII

D ρ∇ =gIII

0∇ Β =g

IV

We differentiate I w.r.t time and since the curl operation is w.r.t space we can reverse the order of differentiation:

LHS:

( )t

∂ ∇×Η = ∇×Η∂

&

13

RHS:

( )Dt t

ε ε∂ ∂= Ε = Ε∂ ∂

& & &&

Where ε

and µ

have been assumed time- independent.

i.e. 2

2t

εε ∂ Ε∇ × Η = Ε = ∂

&

2.1

Taking the curl of LHS and RHS of II, and use

µΒ= Η& & for time invariant

µ:

µ∇×∇×Ε=− ∇×Η &2.2

Use 2.1:

µε∇×∇×Ε=− Ε &&2.3

Use identity:2∇×∇×Α = ∇∇ Α−∇ Αg

i.e. 2 µε∇∇Ε−∇ Ε=− Ε &&g

2.4

Therefore: 2 µε∇ Ε= Ε&&

2.5

Similarly,2 µε∇ Η= Η&&g

2.6

Equation 2.5 and 2.6 are the wave equations in a perfect dielectric and must be

satisfied by

Ε and

Η for electromagnetic wave propagation. For free space,

0µ µ= and

0ε ε= and, assuming harmonic time dependence, we get Helmholtz

equation (a similar equation can be derived for

Η):

14

2 2 0k E∇ Ε + =

2.7

Where,2 2w µεΚ =

2.8

And

2 2w

w f

v v

π πκ µελ

= = = =

2.9

It can be shown that if E and H are independent of the y and z directions (a common case) 2.5 and 2.6 reduce to

2 2

2 2tµε∂ Ε ∂ Ε=

∂× ∂

2.10

2 2

2 2x tµε∂ Η ∂ Η=

∂ ∂

2.11

Consider equation 2.10 which is equivalent to three scalar equations in

, & .x y zΕ Ε Ε It will be shown later that

0xΕ = for a wave propagating in the x

direction. Taking say the y component (the z component behaves similarly) gives

equation 2.10 as:

2 2

2 2

y y

x tµε

∂ Ε ∂ Ε=

∂ ∂

2.12

This partial differential equation has a general solution of the form (HW 1.3):

( ) ( )0 2 0y If V t f v tΕ = ×− + ×+2.13

With reference to the definition given earlier, it is evident that equation 2.13 describes wave motion.

2.2. UNIFORM PLANE WAVES

15

z

Definition:

A uniform plane wave is an electromagnetic wave in which electric and magnetic fields are orthogonal, both laying in a plane transverse to the direction of propagation, each being uniform in any such plane (Figure.2.2). Note that, the fields in the illustration are functions of x and t only.

16

HExDirection of motion

y

Figure 2.2: UPW propagating in positive x –direction

Writing the wave equation 2.10 in terms of its components;

2 2

2 2x x

x tµε∂ Ε ∂ Ε=

∂ ∂

2.14 (a)

2 2

2 2

y y

x tµε

∂ Ε ∂ Ε=

∂ ∂

2.14(b)

22

2 2ZEE

x tµε ∂∂ =

∂ ∂

2.14(c)

In free space, the divergence of the electric field E is zero, so that:

0yx zE E

x y z

∂∂Ε ∂+ + =∂ ∂ ∂

2.15

The last two terms on the LHS are zero because E is independent of y and z.

Therefore even the first component must be zero. This means that either

ΧΕ is

constant or equal to zero. However, a constant cannot be part of wave motion,

therefore

0ΧΕ =. A similar argument for the magnetic field shows that

0ΧΗ =.

We can therefore conclude that uniform plane waves are transverse.

2.2.1. Intrinsic impedance

For E, H independent of y and z and having no x components, the curl expressions can be written as:

17

yzy za a

x x

∂Ε∂Ε∇ΧΕ=− +∂ ∂

2.16(a)

yzy za a

x x

∂Η∂Η∇ΧΗ = − +∂ ∂

2.16(b)

Substitute into I and II:

y yz zy z y z

E Ea a a a

x x t tε

∂Η ∂ ∂Η ∂− + = + ∂ ∂ ∂ ∂

2.17(a)

y yz zy z y za a a a

x x t tµ

∂Ε ∂Η ∂Ε ∂Η− + = − + ∂ ∂ ∂ ∂

2.17(b)

Equating components in the y and z directions gives:

yz

x tε

∂Ε∂Η− =∂ ∂

2.18(a)

y z

x tε

∂Η ∂Ε=∂ ∂

2.18(b)

yz

x tµ

∂Η∂Ε = −∂ ∂

2.19(a)

y z

x tµ

∂Ε ∂Η= −∂ ∂

2.19(b)

With

( ) 1 2

0v µε −=,

( )1 0y f V tΕ = Χ − for propagation in positive x direction. Then:

( )( ) ( )01

0 1 00

y x v tfv f x v t

t x v t t

∂Ε ∂ −∂= = − −∂ ∂ − ∂

g

; where

( )1

10

ff

x v t

∂=∂ −

18

Using 2.18(a),

0 1, 1z

zv f f dxx

εεµ

∂Η = ⇒ Η =∂ ∫

; But

( )011 1

x v tff f

x x

∂ −∂ = =∂ ∂

So

11z

fx c f c

x

ε εµ µ

∂Η = ∂ + = +∂∫

We can ignore the constant C since it is not part of wave motion, giving:

;y

z yz

EE

ε µµ ε

Η = ∴ =Η

2.20(a)

Similarly,

z

y

µε

Ε = −Η

2.20(b)

Since,

2 2 2 2;y z z yΕ = Ε + Ε Η = Η + Η

µε

Ε =Η

2.21

E is volts/m, and H is in amps/m, so that E/H has dimensions of impedance. This ratio, which depends only on the dielectric, is called the intrinsic impedance of

the medium. In free space the intrinsic impedance is

( )0 0/ 377ν µ ε = ohm

2.3. FREQUENCY DEPENDENCE OF THE CLASSIFICATION MATERIALS

Before obtaining the wave equation in conducting media, it is instructive to establish guidelines by which dielectrics and conductors can be distinguished. Consider equation I

( )jw Eσ ε∇×Η= +

19

We see that the term on the RHS has two components: conduction current

( )σΕ

and a displacement current

( )jwεΕ. While the conduction current is independent

of frequency, the displacement current increases with frequency. This means that as frequency increases, a material can change from a conductor to a dielectric. It therefore makes sense to classify materials depending on the relative magnitudes of conduction and displacement currents:

wε σ> 1

100w

σε

<

Dielectrics

wε σ≈ 1100

100 w

σε

< <

Quasi conductors

wε σ<100

w

σε

<

Conductors

It is therefore possible for the same material to behave as a dielectric, a quasi conductor, or conductor depending on frequency (See example in HW 2.1).

2.4. WAVE PROPAGATION IN A CONDUCTIVE MEDIUM

2.4.1. Propagation Constant for a Conductive medium

Maxwell’s equations for a conductive medium will retain both the conduction and displace current components, but there will be no stored charge. As before, we differentiate I with respect to time; take the curl of II, and carry out the necessary substitutions to get the wave equation for the electric field E. A similar derivation can be used to get the wave equation for the magnetic field H (see equation 2.22)

2 µε µσ∇ Ε= Ε+ Ε&& &2.22(a)

2 µε µσ∇ Η= Η+ Η&& &2.22(b)

For harmonic time dependence, Helmholtz equation for a conducting medium is

20

( )2 2 0w jwµε µσ∇ Ε + − Ε =

2.23(a)

( )2 2 0w jwµε µσ∇ Η + − Η =

2.23(b)

Rearranging

2 2 0γ∇ Ε− Ε =2.24(a)

2 2 0γ∇ Η − Η =2.24(b)

where,

( )2 jw jwγ µ σ ε= +2.25

a jγ β= + is a complex number known as the propagation constant. For a UPW

propagating in the x direction, 2.24 gives:

22

2

E

xγ∂ = Ε

∂

2.26

2.26 has a solution of the form:

( )( )

0

0, Re

x

x jwt

x e

x t e

γ

γξ − +

Ε = Ε

= Ε

( )0Re j wte e βα − ΧΧ = Ε

2.27

Evidently equation 2.27 represents a wave traveling in the positive x direction, attenuating (decaying) according to e- X with β as the phase shift per unit distance. α is therefore called the attenuation constant and β the phase constant of the medium.

Using 2.25 and considering only positive square roots, it can be shown that:

21

2

2 21 1

2w

w

µε σαε

= + −

and

2

2 21 1

2w

w

µε σβε

= + +

2.28

From the definition of

β,

2πβλ

=

so that v =

wfλ

β=

2.4.2. Good dielectric

A good dielectric will always have some losses (as opposed to a perfect

dielectric). However since

( )/ 1wσ ε =, it can then be shown that (HW 2.3):

2a

σ µε

≈

2.29(a)

2

2 21

8w

w

σβ µεε

≈ +

2.29(b)

The wave velocity, v, will be:

12

2 2

11

8

wv

w

σβ εµε

−

= = +

2

0 2 21

8v

w

σε

≈ −

2.30

( ) 12

0v µε=

is the velocity of propagation in the unbounded lossless dielectric. It

can be seen that the effect of small losses is a reduction in the velocity of propagation of the wave.

Good conductor

For a good conductor,

( )/ 1wσ ε ?

This gives:

22

( ) 45jw jw jw w wγ µ σ ε µσ µσ °= + ≈ = <

2.31

2

wa

µσ=

2.32

2

wµσβ =

2.33

2w wv

β µσ= =

2.34

2.4.3. Skin Effect

From 2.32 and 2.33, it is evident that α

and

β will be very large for a good

conductor, especially at high frequencies. This has several consequences:

(i) Velocity of propagation will be very low (see 2.34)

(ii) The wave attenuates very rapidly as it propagates through a conductor.

Consequently, radio frequency waves penetrate only to a small depth in a good conductor before they become negligibly small compared to their surface magnitude. We define the depth of penetration, or skin depth, δ , as the depth at which the wave is 1/e (approximately 37%) of its surface value.

If the electric field strength at the surface is E, then at a depth δ , the field strength Eδ s, is given by:

11s

s

as s se e

eδ

δ− −Ε =Ε = Ε = Ε

Or1sae eδ− −=

=>

1 2s w

δα µσ

= =

2.36

Using the result of equation 2.32 for a good conductor

23

Example: Copper with 7 7

05.8 10 / ; 4 10mhos mσ µ µ π −= × = = ×, the depth at

100Hz, 1 MHz, 1GHz and 100GHz are 6.6m, 6.6x10-2 mm, 2.1x10-3mm

and 2.1x10-4mm respectively.

Surface Impedance:

From the above example, we see that current is confined to a very thin sheet on the surface of a good conductor at high frequencies. It is convenient to define surface impedance,

tans

sJ

ΕΖ =

2.37

Where,

tanΕ is the tangential electric field at the surface and

sJ

is the resulting

linear surface current density (total conduction current per meter width of the surface).

Consider a thick flat plate with a current distribution as shown in figure 2.3:

0yJ J e −ϒ=

2.38

The limit is justified only if the thickness, t>>δs so that

24

Thick conductor

Thickness

tyy

0 0

t

J Jdy Jdy∞

∴ = =∫ ∫

00

0

y JJ e dy

γ

∞− ϒ= =∫

2.39

Figure 2.3: Conduction current distribution in a thick plate

Since,

tan0 tan , sJ J

σσγΕ= Ε =

, then

tans

s

ZJ

γσ

Ε= =

Recall that for a good conductor, 045wγ µσ= <

(equation 2.33)

( ) ( )12s m

jw wZ j

µ µ ησ σ

∴ = = + =

1

s

j

σδ+=

2.40

Surface resistance

1

2ss

wR

µσδ σ

= =

2.41

And Surface reactance

1s

s

Xσδ

=

2.42

We see therefore that a conductor having a thickness >>δs with exponential current distribution has the same resistance as a conductor of thickness δs with the total current as before uniformly distributed throughout its thickness.

Power loss in the conductor is thus 2

effs sJ R=2.43

25

With Jseff

as the effective value of the linear current density

1.1. POWER FLOW IN ELECTROMAGNETIC FIELDS

Consider I:

xH J D J ε∇ = + = + Ε& &

J xH ε= ∇ − Ε&2.44

Dimensions of 2.44 are those of current density (A/2m

). Multiply through by

Ε:

J x εΕ = Ε ∇ Η − Ε Ε&g g g

2.45(a)

Dimensions of 2.44(a) are those of power per unit volume (Amps/m2xVolts/mWatts/m3)

Applying vector identity

xF F x xF∇ Α = ∇ Α − Α ∇g g g

to first term on the right:

x x x∇ Ε Η = Η ∇ Ε − Ε ∇ Ηg g g

Or

x x xΕ ∇ Η = Η ∇ Ε −∇ Ε Ηg g g

Substitute into 2.45(a)

J xE x εΕ = Η ∇ −∇ Ε Η − Ε Ε&g g g g

From II,

µ∇×Ε=−Β=− Η&

And Substituting:

J xµ εΕ =− Η Η− Ε Ε−∇ Ε Η& &g g g g

Since

21

2 t

∂Η Η = Ε∂

&g

and

21

2 t

∂Ε Ε = Ε∂

&g

(see below)

2 2

2 2J x

t t

µ ε∂ ∂Ε = − Η − Ε −∇ Ε Η∂ ∂

g g

2.45(b)

260 ˆjwte aΗ = Η

0 ˆjwtjw e aΗ = Η& 2 20

jwtjw eΗ Η = Η&g

Consider the integral of 2.45(b) over some volume V

( )2 2

2 2v v v

E Jdv dv x dvt

µ ε∂ = − Η + Ε − ∇ Ε Η ∂ ∫ ∫ ∫g g

2.45(c)

Apply divergence theorem to last term:

,v

x dv x ds∇ Ε Η = Ε Η∫ ∫g gÑover S – the Surface enclosing V, gives

2 2

2 2v v s

Jdv dv x dst

µ ε∂ Ε = − Η + Ε − Ε Η ∂ ∫ ∫ ∫g gÑ

2.46

(1) (2) (3) Evidently

1.

JΕg

is power dissipation/ unit volume

v

Jdv⇒ Ε∫ g

is the total power

dissipated in a volume v.2.

21

2εΕ

is Stored electric energy/unit volume and

21

2µΗ

is Stored magnetic

energy/unit volume. Therefore, the volume integral (2) represents total stored energy. The negative time derivative represents the rate of decrease of stored energy.

3. From the law of conservation of energy, the rate of dissipation of energy (1) must equal the rate at which stored energy is decreasing plus the rate at which energy enters the volume V, i.e., (3) must represent the of flow of energy inwards through the surface of V.

s

x dsΕ Η∫ gÑ is the rate of energy flow outwards from the volume V.

s

ds− ΕΧΗ∫ gÑ is the rate of energy flow inwards through surface of V

Poynting’s theorem:

27

P x= Ε Η and called, Poynting’s vector, at any point is a measure of the rate of

flow of energy per unit area at that point. The direction of flow (direction of

Poynting’s vector) is perpendicular to both

&Ε Η. Note that

Ρ is normal to

&Ε Η

Perfect Dielectric (UPW):

Total energy density due to electric and magnetic fields is

( )2 21

2ε µΕ + Η

. Given

that wave velocity is

0v

, the rate of energy flow per unit area

( )2 20

1

2vε µΡ = Ε + Η

0

0

1

2

sin 90

v

v

ε µµ εε µ

°

= ΕΗ + ΕΗ

ΕΗ= = Ε × Η = ΕΗ

1.1.1. Conducting Medium

The normal component of Poynting’s vector at the surface of a conductor accounts for power loss in the conductor. Assuming a flat metal plate with

thickness

sδ?

The tangential components of electric and magnetic fields,

tan tan&Ε Η are related by

tan tansΕ = Ζ Η2.47

28

Where

45s

wµσ

°Ζ = ∠

(see equation. 2.40)

Since

&Ε Η are no longer in time phase we use the complex Poynting’s vector.

1

2x •Ρ = Ε Η

2.48

tan tan

1

2x= Ε Η g

2.49

Then

( )tan tan

1Re

2av x •Ρ = Ε Η

2.50

Note:

tan tan&Ε Η are in space quadrature so that the cross product maintains

both magnitudes. However,

tanΕ leads

tanΗ by 45 in time (see equation 2.47) so

that a factor of cos45 is introduced.

i.e,

0tan tan

1cos45

2avΡ = Ε Η

2.51

2

2 tantan2 2

1 1

2 2s

s

Ε= Ζ Η =Ζ

2.52

Now

sJ

is equal in magnitude to the tangential magnetic field

∴2 2

2

1/

2av s sJ watts mΡ = Ζ

29

( )2s sR J eff=

2.53

i.e., Poynting’s vector can be used to account for power loss in the conductor.

1.2. REFLECTION AND REFRACTION OF UNIFORM PLANE WAVES

We shall consider only normal incidence. (see Jordan & Balmain, “Electromagnetic Waves and radiating systems”, for the case of incidence at

angles 090< )

1.2.1. Perfect conductor

2 0s wδ µσ= =

for σ → ∞

i.e., all energy will be reflected. Let the perfectly

conducting surface be at X=0 (figure 2.4). Then;

Incident wave:j x

ieβ−Ε = Ε

2.54(a)

Reflected wave:j x

reβΕ = Ε

2.54(b)

Fig 2.4 standing

&Ε Η waves near the surface of a perfect conductor

Since the transmitted field is zero, continuity of tangential E field across the boundary requires that:

0,r iΕ +Ε =or

r iΕ = −Ε2.55

At any point –x from the x=0 plane, the total field

TΕ is:

30

( ) j x j xT i rx e eβ β− +Ε =Ε +Ε

( )j ji e eβχ βχ− += Ε −

2 sinij βχ= − Ε

( ) , Re 2 sin jwtT ix t j xeβΕ = − Ε

2 sin sini x wtβ= Ε2.56

Equation 2.56 represents a standing wave of maximum amplitude,

2 iΕ which

varies sinusoidally with distance from the reflecting plane (figure 2.4)

By considering Poynting’s vector (

xΕ Η), it is evident that for a reversal of power

flow, only one of the fields can have a phase reversal (Both reversed ⇒

power

flow direction unchanged). i.e.

( )j x j x

i rT X e eβ β−Η = Η +Η

2 icos xβ= Η

( ) ( ), Re jwtT Tx t eΗ = Η

2 cos cosi x wtβ= Η2.57

Which is also a standing wave. The surface current density 1

s TJ Am −= Η

31

Meanwhile, whereas

&i iΕ Η are in time phase,

&T TΗ Ε and

2π

out of phase,

so that there is no average flow of power.

1.2.2. Perfect Dielectric (Fig. 2.5)

1 X = 0 2

1 1, 1,ε µ η

2 2 2, ,ε µ η

,

,i i

r r

Ε ΗΕ Η

2 2 2, ,

,t t

ε µ ηΕ Η

, , ,i r t → incident, reflected, transmitted respectively. Recall that for a perfect

dielectric.

1 1 2, ;i i r r t tη η ηΕ = Η Ε = − Η Ε = Η Where

η µ ε=

is the intrinsic impedance.

Continuity requirements are that:

i r t

i r t

Η + Η = ΗΕ + Ε = Ε

Given the relationships above, derive equation 2.58- 2.61:

1 22 1

2 1 1 2

r

i

ε εη ηη η ε ε

−Ε −= =Ε + +

2.58

1 22 1

2 1 1 2

r

i

ε εη ηη η ε ε

−Ε −= =Ε + +

2.58

212

2 1 1 2

2t

i

εηη η ε ε

Ε =Ε + +

2.59

32

t r

i i

Η Ε= −Η Ε

2.60

1

2

t t

i i

ηη

Η Ε=Η Ε

2.61

Equations 2.58- 2.61 define the reflection and transmission coefficients for the electric and magnetic fields.

ReAmplitudeofreflectedwave

flectionCoefficientAmplitudeofincidentwave

AmplitudeoftransmittedwaveTransmissionCoefficient

Amplitudeofincidentwave

=

=

The field reflection coefficient,ρ

is given by:

r r

i i

ρ δΕ Ε= = ∠Ε Ε

2.62

In the general case, ρ

is complex, with

1& 180 180 .ρ δ° °≤ − ≤ ≤

Assume that

Ε is in the y Direction i.e.,

( ) ( )

Re

Re

j wt xyi i

j wt xyr r

e

e

β

β δ

−

+ +

Ε = Ε

Ε = Ε

δ →phase difference between

&yr yiΕ Ε at X = 0, which we shall ignore here for

convenience of manipulation because we are only interested in the general nature of the wave.

TY yi yrΕ =Ε +Ε

( ) ( ) Re j wt x j wt x

i re eβ β− += Ε + Ε

33

( ) ( )cos cosi rwt x wt xβ β=Ε − +Ε +2.63

( )cos cos sin sin cos cos sini rwt x wt x wt x xβ β β β=Ε + +Ε −

( ) ( )cos cos sin sini r i rwt x wt xβ β= Ε + Ε Ε − Ε

For

( ) ( )cos & sin ,i r i rx xβ βΑ = Ε + Ε Β = Ε − Ε

It can be shown that (HW 2.4)

( ) ( ) ( )2 22 2cos sin sinTY i r i rx x wt xβ β βΕ = Ε + Ε + Ε − Ε −

2.64

TYΕ is therefore a traveling wave contained in standing (stationary) envelope.

The maximum value at each point, or the shape of the standing wave envelope is

obtained when

( )sin 1wt Xβ− = and is given by :

( ) ( ) 2 22 2cos sinTY i r i rx xβ βΕ = Ε + Ε + Ε − Ε

2.65

Note the oscillation of stored energy in both time and space over

&2 2

x wtπ πβ = =

respectively.

Fig 2.5 standing waves at a dielectric boundary

This envelope in figure 2.5 is a result of the incident and reflected waves reinforcing each other at some points and canceling at other points.

34

Max value:

i rE E+

Min Value:

i rE E−

The standing wave ratio is defined as the ratio of the maximum value to the minimum value of the envelope (normally called VSWR or S).

(max)

(min)

T i r

T i r

E E EVSWR

E E E

+= =−

2.66

1.3. POLARISATION

Polarisation refers to the time-varying behaviour of the electric field vector at a fixed point in space during the duration of at least one full cycle. It refers in the same sense to the behaviour of the electric field radiated by an antenna (e.g., a vertical dipole is said to be vertically polarized, etc).

Knowledge of the polarization of the received signal enables one to align or to set up a suitable antenna system for reception.

(i) General case:

Assume propagation in the Z direction, i.e.,

( ) 0j z

z e β−Ε = Ε

( ) , Re j z jwtz t e eβο

−Ε = Ε2.27

( ),z tε lies in the X- Y plane .

Assume a case where

&y xΕ Ε are present, with different amplitude, with

yΕ

leading

xΕ by

2π

, i.e.

x yjΕ=Ε + Ε2.75

35

( ) ( ) 0, Re jwtx yt j eε = Ε + Ε

2.76

ˆ ˆcos sin cos sinx x x y ywt wt a wt a E wt=Ε −Ε = Ε −

i.e.,

cosx x wtΕ = Ε

&

siny y wtΕ = Ε

So that2

2 21

yx

x y

ΕΕ+ =

Ε Ε

2.77

Evidently the end point of

( )0, tΕ traces out an ellipse and the wave is said to

be elliptically polarized (Figure 2.6).

The ellipticity is defined as the minor to major axis ratio (normally given in dB).

(ii) Linear Polarisation

Let

&y xΕ Ε be in phase,

( )0, cos cosx yt wt wt⇒Ε =Ε +Ε

( ) cosx y wt= Ε + Ε

2.78

The resultant direction, which depends only on the relative magnitudes of the two

fields, is fixed, making an angle arctan

( )y xΕ Ε with the X- axis. The wave is

said to be linearly polarized (Figure 2.6b). This can be considered as elliptical

polarization with an ellipticity of

dB−∞.

(iii) Circular polarization

36

Let

&y xΕ Ε have the same amplitude

aΕ with

yΕ leading

xΕ by

2π

. Then

2.77 gives:

22 2x y aΕ + Ε = Ε

2.79

i.e.,

( )0, tΕ traces out a circle and the wave is said to be circularly polarized (Fig

2.6c). This can be considered as elliptical polarization with an ellipticity of 0 dB

Assignment Two:

2.1. Investigate the behavior of ground with a relative permittivity of 14 and conductivity 0.01 siemens per meter at 100Hz, 1KHZ, 10MHZ, and 100 GHz.

2.2. Using equation 2.25 and considering only positive square roots, shown that:

2

2 21 1

2w

w

µε σαε

= + −

2.28(a)

and

2

2 21 1

2w

w

µε σβε

= + +

2.28(b)

2.3. Show that for a dielectric, the attenuation, phase constants and wave velocity are respectively given as

2a

σ µε

≈

2.29(a)

2

2 21

8w

w

σβ µεε

≈ +

2.29(b)

37

12

2 2

11

8

wv

w

σβ εµε

−

= = +

2

0 2 21

8v

w

σε

≈ −

2.30

2.4. Let

( ) ( )cos & sin ,i r i rx xβ βΑ = Ε + Ε Β = Ε − Ε

It can be shown that (see assignment)

( ) ( ) ( )2 22 2cos sin sinTY i r i rx x wt xβ β βΕ = Ε + Ε + Ε − Ε −

2.64

38

CHAPTER 3: WAVE PROPAGATION IN TRANSMISSION LINES

1.

1.1. INTRODUCTION

In all applications, electromagnetic energy must be guided either for transmission from a point (telephone wires, component interconnections, etc), or for feeding antennas before radiation and consequent unguided (unbounded) transmission can occur.

Figure 3.1: Examples of wave guiding structures

Wave guiding systems are classified into two broad categories:

(i) Transmission Lines:

These are characterized by having at least two conductors, and supporting the TEM mode in normal operation (see examples in fig 3.1).

(ii) Wave guides:

These are guiding systems, which support the transverse electric (TE) or transverse magnetic (TM) modes in normal operation. They are incapable of supporting the TEM mode and are characterized by having a cut- off frequency for each mode below which propagation cannot occur. Examples include rectangular and circular wavegides (Figure 3.1) .

We shall study these guiding systems in their normal mode of operation and derive the important relationships and parameters pertaining to them, starting with Transmission lines in this chapter.

1.1. TRANSMISSION LINE EQUATIONS (DISTRIBUTED CIRCUIT ANALYSIS)

39

dzConductorsHomogenous Isotropic medium

VzI

1.1.1. The Infinite Transmission line

Consider a differential length, δz, taken out of an infinite uniform 2- wire transmission line (fig 3.)

Fig .3.3: Infinite Uniform two – wire transmission line.

Let R be the series resistance per mL be the series inductance per mG be the shunt conductance per mC be the shunt capacitance per m

Notes:

1. The above parameters are uniformly distributed over the whole length of the line.

2. L and C account for the energy storage in the magnetic and electric fields respectively, while R and G account for conductor loss and dielectric loss respectively.

40

LdzRdzCdzGdzVI

Then the differential length δz be represented by the equivalent lumped element circuit shown in fig. 3.4

Fig. 3.4: Equivalent lumped parameter circuit of the differential length, dz.

The input current and voltage are i(z,t) and v(z,t) respectively so that the outputs

are

ii z

z

∂+ ∂∂

and

.v

v zz

∂+ ∂∂

Apply Kirchoff’s voltage and current laws:

v iv v dz iRdz Ldz

z t

∂ ∂ − + = + ∂ ∂

v iiR L

z t

∂ ∂= −∂ ∂

3.1(a)

Similarly,

i vvG C

z t

∂ ∂= − −∂ ∂

3.1(b)

Differentiate 3.1(a) with respect to z: and 3.2(b) with respect to time

2 2

2

v i iR L

z z z t

∂ ∂ ∂= − −∂ ∂ ∂ ∂

3.2(a)

2 2

2

i v vG C

t z t t

∂ ∂ ∂= − −∂ ∂ ∂ ∂

3.2(b)

Substitute into 3.2 (a) using .3.1(b) and 3.2(b):

( )2 2

2 20

v v VRC LG LC RGv

z t t

∂ ∂ ∂− + − − =∂ ∂ ∂

3.3(a)

41

Assignment: Obtain a similar equation for the current I:

( )2 2

2 20

i i iRC LG LC RGi

z t t

∂ ∂ ∂− + − − =∂ ∂ ∂

3.3(b)

For sinusoidal time – variation, we can use phasor notation so that

( )VR jwL I ZI

z

∂ = − + = −∂

3.4(a)

and

( )IG jwC v YV

z

∂ = − + = −∂

3.4(b)

where Z = R + jwL is the series impedance per unit length

Y = G + jwC is the shunt admittance per unit length.

The one dimensional wave equation 3.3 then becomes;

( ) ( )2

22

0V

RG w LC V jw RC LG Vz

∂ − − − + =∂

or2

20

VZYV

z

∂ − =∂

3.5(a)

Similarly,2

20

IZYI

z

∂ − =∂

3.5(b)

Equations 3.5 are the basic differential equations, or wave equations for the

second order with constant coefficients. Let 2ZY γ=, where

γ is some constant,

then we have, 3.5(a) as 2

22

0V

Vz

γ∂ − =∂

with a general solution,

z zV V e V eγ γ+ − −= +3.6

42

Note that zV e γ+ −

, from our earlier consideration, denotes a wave traveling in the

positive Z- direction, while zV eγ−

denotes a wave traveling in the negative Z-

direction, i.e. both waves are present on the transmission line. In the general

case,γ

is given by:

( ) ( )ZY R jwL G jwCγ = = + +

3.7

We also have from 3.4(a):

( )1 1 z zVI V e V e

R jwL z R jwL zγ γ+ − −∂ ∂= − = − +

+ ∂ + ∂

( ) ( )1 z zV e V eR jwL

γ γγ γ+ − −= − − ++

( )z zV e V eR jwL

γ γγ + − − = − +

z z

z z

c c

V e V eI e I e

z z

γ γγ γ

− −+ − −+= − = −

3.8

where

c

R jwL R jwLZ

G jwCγ+ += =

+

3.9

is called the characteristic impedance of the line. It is evident that,

c

V VZ

I I

+ −

+ −= = −

43

Zc

Zc

Zc

cZ

is the impedance seen looking into a uniform infinite transmission line at any

point (figure 3.5).

Fig. 3.5: Characteristic impedence of a transmission line at different points

Lossless line (R = G = 0)

For a lossless line,

jw LC jγ β= =

3.10

i.e.,

LCβ ω=

and

0α =

so that

cLZ C=

3.11

Low loss line

( );R wL G wC<< <<

At very high frequencies (UHF), the condition R<<ωL and G<<ωC is obtained. Using the binomial expansion and neglecting higher order terms for this case;

( ) ( ) 1

2

R Gj R jwL G jwC jw LC LC

L Cγ α β == + = + + ≈ + +

3.12

It can be seen that for a low loss line,

,w LCβ =

which is the same as the lossless case

and

( )1 1

2 2 c c

R GLC RY GZ

L Cα = + = +

cY

is the characteristic admittance

1cZ

=

1.1.1. The Terminated line

44

For an infinite line, we expect that we have only the incident waves,

&V I+ +

with

V andI− −

identically zero. For termination with some impedance

RZ

different

from

,cZ

we shall have both “incident” and “reflected” waves.

45

Direction zIsIR,

z=0Z

R

VR

Length, l

Let a section of line length I, characteristic impedance,

cZ

, and propagation

constant γ

and be terminated in

RZ

as shown in fig 3.6.

Fig. 3.6 Terminated transmission line

RZ

is located at the plane Z = 0 while

,s sV I

are sending end voltage and current

respectively and

&R RV I

are the corresponding receiving end quantities.

In hyperbolic function form, solutions 3.6 and 3.8 are:

1 1cosh sinhV z zγ γ= Α +Β3.13(a)

2 2cosh sinhI z zγ γ= Α +Β3.13(b)

The Boundary conditions are:

RV V= and

RI I= at

0z =

sV V= and

sI I= at

1z z=

So

1 1cosh0 sinh0RV = Α +Β 1 RV⇒ Α =

From 3.13(a); and 3.4(a)

46

( )1 1sinh coshV

z z R jwL IZ

γ γ γ γ∂ = Α + Β = − +∂

At Z = 0,

( )1 1sinh0 cos0 RR jwL Iγ γΑ + Β =− +

Or

( )1 R c R

R jwLI Z I

γ+

Β = − =−

We similarly obtain

2 RIΑ =

and

2R

c

V

ZΒ = −

Equations 3.13 become:

1 1cosh sinhs R c RV V z Z I zγ γ= −

1 1cosh sinhRs R

c

VI I z Z

Zγ γ= −

Now

1l Z=−where

l

is, measured from the receiving end, so that:

cosh sinhs R c RV V l Z I lγ γ= +3.14(a)

cosh sinhRs R

c

VI I l l

Zγ γ= +

3.14(b)

Equations 3.14 relate the voltage and currents at the two ends of the transmission line. The input impedance of the line is given by

( )cosh sinh

cos / sinhs R c R

ins R R c

V V l Z I lZ

I I l V Z l

γ γγ γ

+= =+

3.15

( )cosh sinh

cosh sinhin

R c

R c

Z l Z lZ

l Z Z l

γ γγ γ

+=+

47

where we have set

/ .R R RZ V I=

There are three cases of special interest

i. Short – circuited line

( ) ( )0 0R RZ V= =

tanhin c scZ Z l Zγ= = 3.16

ii. open – circuited line

( )RZ =>∞

cothin c ocZ Z l Zγ= =

3.17

iii. Line terminated in its characteristic impedance

( )R cZ Z=

in cZ Z=3.18

Note that:

a)2

sc oc cZ Z Z=

b) For a line terminated in its characteristic impedance, the input impedance at any point looking towards the load is constant and equal to Zc.

c) For an open circuit or short- circuited line, the input impedance looking towards the load varies from zero to infinity depending on the distance from the load.

Low loss lines

At ultra high frequencies and above, lines designed for these frequencies have

very low losses and we can use the approximations given by 3.12 for

&α β.

Generally, unless we are evaluating attenuation, we can neglect the expression

48

lα in

l l j lγ α β= + in comparison to

lβ at these high frequencies. Equations

3.14 and 3.15 can therefore be written in their lossless form:

cos sins R R cV V l jI Z lβ β= +3.19 (a)

( )cos sins R R CI I l j V Z lβ β= +3.19(b)

( )( )

cos / sin

cos sinc R

s RR c

l j Z Z lZ Z

l j Z Z l

β ββ β

+= +

cos sin

cos sinR c

cc R

Z l jZ lZ

Z l jZ l

β ββ β

+= +

3.20

Note also that

cZ L C≈

is a pure resistance.

1.1. STANDING WAVES ON TRANSMISSION LINES

The voltage and current distributions at any point Z from the termination are obtained by replacing I by Z in equations 3.19. We shall consider the case where

RZ

is real. The case where

RZ

complex can be inferred from these results. We

have equations 3.21, which are familiar standing wave envelopes.

( ) 22 2cos sin ]z R cV V R Rβ β= Ζ + Ζ

3.21 (a)

( ) 22 2cos sinz R cI I z R R zβ β = +

3.21 (b)

,R c cZ R Z L C R = = =

It is normally convenient to consider the standing wave in terms of the voltage standing wave ratio (VSWR) or the current standing wave ratio which are easily

49

measured. These are simply the ratios of the maximum (Vmax, Imax) to the minimum (Vmin, Imin) amplitudes.

1.1.1. Standing wave Patterns

Case 1: R<Rc

The maximum voltage value occurs when

sin 1, 0z cos zβ β= = and the minimum

voltage value occurs when

cos 1,sin 0z zβ β= =

maxc

R

RV V

R=

and

min RV V=

max max

min min

V I

V I∴ =

3.22

Case 2: R>Rc

It can be similarly shown that for this case

max max

min min c

V I R

V I R= =

3.23

Fig 3.7 shows standing wave patterns for R = 0,

, cR R R→ ∞ = and a general

case R

cR≠

50

Fig 3.7: standing wave patterns on a lossless line for various terminations.

Note that the voltage standing wave ratio (VSWR), S, is a measurable quantity. If

we know

cR

- readily calculated from line dimensions, we can measure S and

determine the value of terminating resistance, R, using 3.22 or 3.23. the ambiguity is cleared by determining if it is voltage or current which is a maximum at the termination:

If voltage is maximum, R>Rc

If current is maximum, R<Rc

1.1.2. Reflection Coefficient, Input Impedance and Standing Wave Ratio

We also have, as in the case of fields, the reflection coefficient, ρ

as.

V I

V Iρ

− −

+ += = −

3.24(a)

However,

;cV

ZI

+

+ = ;cV

ZI

−

+ = −

and

( )R c

I IV VZ Z

I I I I

+ −+ −

+ − + −

−+= =+ +

1 / 1

1 / 1R

c

Z I I I I

Z I I I I

ρρ

+ − − +

+ − − +

− − += = =+ + −

51

and

R c

R c

Z Z

Z Zρ −=

+

3.24

Since

1

1 /

V V V VS

V V V V

+ − − +

+ − − +

+ += =− −

it follows that

1

1S

ρρ

+=

−

3.25

and, conversely,

1

1

S

Sρ −=

+

3.26

Also, the input impedance

inZ

at any point is given by:

( )in c

V V VZ Z

+ −+ −

+ − + −

Ι − Ι+= = =Ι Ι + Ι Ι + Ι

1 1

1 1in

inc

ZZ

Z

ρρ

− +

− +

− Ι Ι += = =+ Ι Ι −

3.27(a)

1 1

1inin

YZ

ρρ

−= =+

3.27(b)

At a voltage minimum, V and V+ are π out of phase making the angle of

180 ;V

Vρ

−°

+

= <

or

ρ ρ π= ∠

so that

1 1

1 1inY Sρ π ρρ π ρ

− ∠ += = =

+ ∠ −

3.27(c).

1.1. TRANSMISSION LINES MATCHING CONSIDERATIONS

It is normally necessary to minimize standing waves on transmission lines owing to the following:

52

i. To maximum power carrying capacity: standing waves produce voltage peaks higher than those of the impressed wave form, thus leading to an earlier possibility of dielectric break-down.

ii. To achieve a higher transfer of power to the load. Note that most high frequency lines have a characteristic impedance which is purely resistive, so that maximum power is transferred when load resistance RL = RC the Characteristic impedance.

iii. In communication systems, reflections and re-reflections can cause echoes in the system.

iv. In some systems, like those employing microwaves tubes or high power transmitter tubes, a high level of reflections can lead to destruction of the tube or a drastic shortening of its life – time. Isolators can be used for tube protection but these become unacceptably expensive at high power levels.

In practical systems, steps are therefore always taken to obtain the best match possible. The commonest methods make use of line transformers and /or stub tuning.

1.1.1. Quarter Wave Transformer

Consider a load

L L LZ R jX= + connected through a line of length

4λ and

characteristic impedance

2Z

(fig 3.8). The idea is to match the load

LZ

to the

line with

1cZ Z=.

ZlZc=Z1

λ/4Zin

Fig 3.8: Quarter – Wave transformer network

53

Using equation 3.20 (lossless line) show that the impedance,

inZ

presented to the

main line,

4l λ= is given by:

22

inL

ΖΖ =Ζ

3.28

We require

1inΖ = Ζ

i.e. 22

1,L

Ζ = ΖΖ

or

2 1 LΖ = Ζ Ζ

3.29

In other words, the load ZL is matched to the line characteristic impedance Z1 if

the intermediate quarter wave section has a characteristic impedance

1 LΖ Ζ

.

The

4λ line acts like an ideal transformer of turns ratio

1 LΖ Ζ

. The quarter

wave transformer is normally used for matching lines of different characteristic impedances.

Note that it is a narrow-band device. For broad matching, multi- section

4λ

transformers are used.

For an Example see HW 3.4:

1.1.2. Single Stub-Matching

Stub matching makes use of reactive elements connected in shut or series with the load. Stubs may be open-circuited or short-circuited lengths of transmission line. Their matching ability arises from the fact that the impedance looking into the section as given by equations 3.16 and 3.17 varies with the stub length as the input impedence is the function of the length of the line.

For single stub matching, Fig 3.9 shows a line of normalized characteristic

admittance

1CY = (Normalisation w.r.t the characteristic admittance) terminated

54

in a pure conductive load of normalized admittance

LY G=. We want to obtain

expressions for the length, Io, of a short circuited stub with characteristic

impedance

CY

and its distance d from the load where it is matched to the line.

In general, short-circuit stubs are preferred to open circuit stubs because of their ease of adjustment and better mechanical rigidity.

Yl=G

lo

Yc=1

dYin =1+jB

Yc=1

Fig 3.9: Single-stub matching network.

Principle:

Because of the impedance transforming properties of a transmission line, there will be some point distance “d” from the load at which the normalized input

admittance will be

1inY j= + Β. If we connect a stub with normalized input

susceptance

j− Β at this point, the resultant is

1 1inY j j= + Β− Β =; i.e., the load

will be matched to the line.

We shall consider two approaches to obtaining lo and d:

Approach 1: Obtain d and lo directly

We have:

1 ;1 t

Lin

L

Y jtY j

jY

+= + Β =+

where t=tan βd

With

LY G= (pure, real)

( ) ( )1 1j jGt G jt+ Β + = +

55

To obtain d, equate real and imaginary parts and solve for t to show that:

1cos2 1

Gd

G

λπ

−=+

3.30(a)

or

1 1cos

4 1

Gd

G

λπ

− −=+

3.30(b)

where the alternative solutions 3.30(a) and (b) are obtained according as

we set 2

22

1 costan

cos

dd

d

βββ

−=

or we replace 22cos dβ

by

1 cos2 .dβ+

Note that if

1d

is solution to 3.30,

1 2d nλ± ±

are all solutions.

To obtain lo we have the value of

Β given by:

( ) 11

GG t

G

−Β = − =

3.31

Using equation 3.16 for the input impedance of a short circuited transmission line

and using

tanh tanl j lγ β= for a lossless line, we have;

0cotj j lβ− Β = −or

0

1cot

Gl

Gβ −=

And

10 tan

2 1

Gl

G

λπ

−=−

3.32

The sign of

G

must be chosen to give the correct sign for

:Use GΒ +

for

0 4d λ< <

and

G−

for

.4 2dλ λ< <

56

The above analysis is easy if

LY

is real, but becomes rather involved for

LY

complex. In that case, the second method below is preferred.

Approach 2: Obtain d and lo through dmin

First, Locate the position of a voltage minimum say at a distance

mind

from the

load (fig 3.10). At this point, the reflection coefficient is a negative real quantity and the normalized input admittance is pure real given by (see eqn 3.27(c)):

1

1inY Sρρ

+= =

−

(Standing wave ratio) 3.33

Yin =1+jB

Yl=G

lo

Yc=1

d

Yc=1

dmindo

Fig 3.10: location of stub relative to voltage minimum.

If

0d

is the distance from the voltage minimum to the point where the input

admittance is

1inY j= + Β, we can solve the equation for

0&od I

as before with

S replacing

G

:

10

1cos

4 1

Sd

S

λπ

− −=+

3.34

57

( )1

0 tan2 1

Sl

S

λπ

−=−

3.35

Then

0d d= + (distance dmin of Vmin from the load).

For an example refer to HW 3.5

Series Stubs

It is possible to use a series stub (fig 3.11) for matching, in which case we consider solution in terms of the normalized input impedance. This has been left as an exercise for the Student.

ZlZc=Z1

do Zin=1/S

lo

Vmin

Figure 3.11: Series stub.

1.1.3. Double and Triple Stub Matching

Yl

jB1

b a

jB2

ld

ab

ldd

Figure 3.12: Double Stub Tuner (Left) and Triple Stub Tuner (Right)

Both the tuners in fig 3.12 can be used for matching, the triple stub tuner matching a wider range of loads. We shall consider only the double stub tuner. A

58

common approach to the problem is graphic, but we shall first attempt the analytic approach for completeness.

We can transform the admittance

LY

to plane aa to obtain

L L LY G j= + Β.

Just to the right of the first stub

1,a L LY G j j= + Β + Β

and just to the right of the second stub,

( )1

11L L

b

L L

G j J jtY

jt G j j

+ Β + Β +=+ + Β + Β

3.36

where

tant dβ=.

We have now got the case of the single stub tuner for which we require

1bY j= + Β so that with the second stub having a susceptance

j− Β the load will

be matched to the line. Equating the real part of the RHS of 3.36 to 1 gives:

( ) 22

1

2 2 2

4 111 1

2 (1 )L

L

t t ttG

t t

− Β − Β+ = ± − +

3.37

From 3.37

( )2 2 2

1

1 1 L L

L

t G G t

t

± + −Β = −Β +

3.38

By equating the imaginary part of 3.36 to

jΒ and substituting for

1Β we get:

( )2 2 21L L L

L

G t G t G

G t

± + − −Β =

3.39

59

The upper and lower signs in 3.38 and 3.39 go together. For a match, we then

chose

2j jΒ = − Β.

General comment on stub tuners (comparative)

With a single stub tuner, each load and frequency requires a new position of the stub, which is extremely inconvenient in a practical system. This problem is overcome by using two stubs located at fixed distance from the load. However if

we consider equation 3.37 we see that

LG

must be real, this putting limits on the

expression under the square root sign. A necessary condition is that the value of

the square root term lies between zero and one, i.e., the limits on

LG

are

2

2 2

1 10

sinL

tG

t dβ+≤ ≤ =

This means that some load admittances cannot be matched with a double-stub tuner. This problem is overcome using the triple –stub tuner. If the stubs are

spaced

3 8λ apart and each one can be varied in length over at least half a

wavelength, any admittance can be matched to the line.

Baluns (Balance to unbalance Transformers)

Baluns are used to connect unbalance to balanced transmission line (or balanced loads e.g. many antenna types. This part has been left out to be discussed on study of antennas.

1.2. GRAPHICAL AIDS TO TRANSMISSION LINE CALCULATIONS

The solution of a wide range of transmission line problems is simplified by the use of graphical aids. Most prominent among these graphical aids is the Smith Chart which we shall consider in detail.

1.2.1. The Smith Chart: Development

Recall that the reflection coefficient ρ

is in general given by,

60

R C

R C

ρ Ζ − Ζ=Ζ + Ζ

3.40(a)

We can define the reflection coefficient, ρ(l) at a distance I from the termination:

( ) ( )( )

in C

in C

ll

lρ

Ζ − Ζ=

Ζ + Ζ

where

( )in lΖ is the input impedance at the distance l.

i.e.

1

1C

C

ρ Ζ Ζ −=Ζ Ζ +

3.40(b)

Where ρ

l now refer to any point on the line. Let

u jvρ = + and

( )/ ,C C C n n nR jX r jXΖ Ζ = Ζ + Ζ =Ζ = + where we have assumed that Zc is

real ( a good approximation for most high freguency lines ). 3.40(b) becomes:

1

1n n

n n

r jxu jv

r jx

+ −+ =+ +

X- multiply and equate real and imaginary parts:

( ) ( )( )

1 1 3.419( )

1 3.41( )

n n

n n

r u x v u a

r v x u v b

− − = − + − − − − − − − − − − − − − −

+ − = − − − − − − − − − − − − − − − − − −

Eliminate

nX

( ) ( )2 21 2 1 1n n n nu r ur v r r+ − + + = −

61

Divide through by

( )1nr + and complete the square of the resulting terms

containing 2 & :u u

( )

2

22

1

1 1n

n n

ru v

r r

− + = + +

3.42

If we plot equation 3.42 on rectangular co-ordinates of u and v, we obtain, for any

value of

nr

, a circle on the ρ

- plane with centre

, 01

n

n

ru v

r

= = +

and radius

1

1nr +

Note particularly that

0nr =, the radius is 1, and

nr →∞, the radius is zero. In

other words, for all values of

nr

, the loci (circles ) will lie within the unit circle for

0nr = (Fig.3.12). for the bounding circle

( )0nr =,

2

2

1 11

1 1

n n

n n

jx x

jx xρ

− += = =

+ +

62

Fig. 3.12: Co-ordinate circles for constant normalized resistance

If we now go back to equations 3.41(a) and (b) and eliminate

nr

instead of

nx

,

the locus of any constant value of

nx

on the ρ

-plane is found to be given by:

( )2 2

2 1 11n n

u v x x − + − =

3.43

These loci are again circles of radius

1nx

and center

11,nx

(Fig. 3.13): only

the portions within the bounding circle

( )1ρ =

are plotted.

Fig. 3.13: Co-ordinate circles for constant normalized resistance

A combination of fig. 3.12 and fig.3.13 gives the Smith Chart.

Standing wave data

From a consideration of the equation,

1

1S

ρρ

+=

−

63

it is evident that loci for constant VSWR on the p-plane are also circle with center

(0,0). The circle for S = 1 (matched case,

0ρ =) corresponds to the center of the

chart and the bounding circle for

S =∞ corresponds to the bounding circle

( )1ρ =. For any required value of S, the radial scaling is shown on the scale

next to the Smith chart.

It can also be shown (student to show) that if

mind

is the distance of a voltage

minimum from the point of reflection, then

min 14 2

d nλ φ λ

π = + ±

or

min 1 11

4 2

dn

φλ π

= + ±

3.44

where

λ is the wavelength,

φ is the angle of the reflection co-efficient, and n is

an integer. From equation 3.44, it can be seen that

mindλ

varies from 0.25 to

0.75 as

φ varies from 0 to

2π, so that

mindλ

can be plotted round the

circumference of the chart (fig. 3.14).

64

Fig. 3.14: Radial line loci of constant

mindλ

on the ρ

-plane.

Note that loci of constant

mindλ

are radial lies as shown in fig. 3.14. since the

standing wave pattern is periodic in

2λ (lossless line), the maximum value of

mind λ is 0.5

1.2.2. Some application of the Smith Chart

Exercise.1: Ex. 9.1, page 189 (Chipman):

Determination of reflection co-efficient

A transmission line with characteristic impedance

50 0C jΖ = + is terminated in

an impedance 25- j100ohms. Determine the reflection co-efficient at the load end of the line.

Normalize load impedance:

0.5 2.00.n n nr jx jΖ = + = −

Locate the normalized impedance on the Smith chart (intersection of constant

( )0.5nr = and constant

( )2.00nΧ =−.

65

Draw a radial line through this point from the center of the chart (1,0) to meet

the angle of reflection co-efficient circle 0309ρ∠ =.

Note that constant

ρ loci are concentric circles whose radii relative to the

bounding circle

( )1ρ =

gives the reflection co-efficient. Get the radial

distance

nΖ from the obtain

ρ from the radially scaled chart next to the

Smith chart:

0.82ρ =.

Exercise .2: (pp 191, EX. 9.3 chipman )

Det of S and

mind

.

The value of ρ

is -0.30+ j0.55 at the load end of low-loss transmission line.

Determine S and

mind

.

Express ρ

in polar from: 00.63 118.6ρ = ∠

Establish this point on the chart using the scale for

φ and the constant

ρ circle,

0.63ρ =.

66

Determine S either using the radially scaled chart next to the Smith chart, or, by

moving along the constant ρ

circle to the axis where

0nx =. Here

4.4.nr =

since

0nr R Z= it follows from equation 3.23 that

4.4nS r= =.

mind

is directly obtained by drawing the line of constant

mind λ through the point

of interest to the scale for

mind λ.

Exercise 3: Ex. 2 (pp 193, Ex. 9.4 Chipman):

Determination of load impedance: Slotted line measurements on a coaxial line

operating at 800MHZ with

50c joΖ = + give a VSWR of 2.5 and a voltage

minimum 8.75cm from the termination. Determine the load impedance if the dielectric is air.

Air dielectric

0 & 0.375v C c f mλ⇒ = = =

min / 0.233d λ∴ =

ii Locate the required point on the chart using:

( ) ( )

min 0.233& 2.5 :

0.5; 2.35

2.35 0.5

2.35 0.5 50 0 117 25

n n

n

L

d s

x r

z j

andZ j j j

λ = == − =

∴ = −= − + = − Ω

Impedance Transformation

67

We again assume a lossless line (or a high frequency line which approximates a lossless line when we are not evaluating attenuation). On such a line , the

reflection coefficient magnitude,

ρ is essentially constant everywhere on the

line so that impedance transformation simply consists of moving an appropriate

distance along a constant

ρ circle. Starting at any point, the transformed

impedance at any point can be obtained by moving the right number of wavelengths towards the load (outer scale of the chart ) or towards the generator (inner scale). The normalized transformed impedance is then read off the chart.

Exercise 4: (Ex. 9.6, pp 196-Chipman)

An air dielectric slotted section is connected to an air dielectric transmission line

of the same characteristic impedance

50 0j+ Ω by a reflection less connector.

The transmission line is 3.75m long and is terminated in an antenna. On the slotted section, the VSWR is measured to be 2.25. There are successive voltage minima at 0.180 and 0.630m from the connector. Assuming negligible attenuation on the line and the slotted section, determine the impedance of the antenna and the frequency of he measurements.

Preliminary Information:

Air dielectric for TEM wave 83 10 secv m⇒ = ×

Separation of minima

0.45 2 0.9 mλ⇒ = × =

Frequency

3.33v MHZλ= =

Line length in wavelengths

3.75 / 0.9 4.17wavelengths= =

Normalised impedance at connector: S = 2.25,

min 0.18 / 0.9d λ = = 0.2

(Ref. Ex.3):

1.62 0.86n jΖ = −.

68

Transform impedance by moving 0.17 wavelength towards the load. (Can you see why?. Always subtract an integral number of half wavelengths from

d λ).

.077 0.70,& 50 37.5 35 .ant ant antj z x jΖ = + Ζ = = + Ω

Normalised admittance co-ordinates on the Smith Chart

Recall that a

4λ section of lossless line inverts the normalized impedance

values (ref. equation. 3.28):

22 ,inL

ΖΖ =Ζ

where

2Ζ is the characteristic impedance of the quarter wave

section.

( )( )

1m in

n L

∴ Ζ =Ζ

This implies that any normalized impedance co-ordinate on the Smith Chart can be transformed to the corresponding normalized admittance co-ordinate by transforming it through a quarter wavelength, i.e., rotating through 180.

Exercise. 5: (Ex. 9.9, pp200-Chipman)

A VSWR of 3.25 is observed on a slotted section with a voltage minimum 0.205 wavelengths from the load end of the section. Determine the value of the normalized admittance at the terminal load end.

Wavelengths towards the load: 0.205. Using this and the VSWR of 3.25, we move 0.205 wavelengths towards the load on the constant VSWR circle to

obtain

0.33 0.26.ny j= +

Exercise 6: Stub Matching (problem 9.23 – Chipman)

The VSWR on a lossless transmission line is 3.0.

Where relative to a voltage minimum on the line might stub lines be placed to remove standing waves at the generator side of the stub?

69

Obtain the required short circuit stub length for matching if the characteristic impedance of the stub is the same as that of the transmission line.

Assignment Three B

3.1. Starting from maxwell’s equations, Obtain the one dimension current wave equation in 3.3(b) (below)

( )2 2

2 20

i i iRC LG LC RGi

z t t

∂ ∂ ∂− + − − =∂ ∂ ∂

3.3(b)

3.2. Prove that for a Low loss line at very high frequencies (UHF), the condition

( ) ( ) 1

2

R GR jwL G jwC jw LC LC j

L Cγ α β = + + ≈ + + = +

3.3. Using equation 3.20 (lossless line) show that the input impedance,

inZ

,

presented to a transmission line of length,

4l λ= and characteristic

impedance Z2 by a load of impedance ZL is given by:

22

inL

ΖΖ =Ζ

3.28

3.4. Design a quarter-wave transformer to match an antenna array with an input

impedance of

36 0j+ Ω, operating at 40 MHZ to a generator of output

impedance

500 0j+ Ω located at 30m from the antennas terminals. A

parallel wire transmission line with

500 0C jΖ = + Ω runs from the generator

to the vicinity of the antenna. Assume you will use a parallel wire transmission line on which the phase velocity is 97% of the free space

velocity of light for the

4 txλ.

70

Solution:

134 0 ; 1.82a txj L mΖ = + Ω =

3.5. Example: Design a network to match a

75Ω load to a

50Ω coaxial line

using a ingle stub. Assume an operating frequency of 100MHZ, air dielectric and a lossless line.

Answer:

0

0

d =

=Ι

71

CHAPTER 4: WAVE PROPAGATION IN WAVEGIDES

2.

2.1. THE INFINITE PLANE WAVEGUIDE

We shall consider an electromagnetic wave propagating between two parallel perfectly conducting planes of infinite extent (fig. 3.44).

typically b>>a, infinite in z

Figure 4.1: Parallel infinite conducting planes

We have to solve Maxwell’s equations subject to the boundary conditions

tan 0& 0normalΕ = Η = at the perfectly conducting planes.

Recall the curl equations and the wave equations:

( )

2 2 2 2

jw

jw

and

σ εµ

γ γ

∇ΧΗ = + Ε

∇ΧΕ = − Η∇ Ε = Ε ∇ Η = Η

where

( )jw jwγ µ σ ε= +

In Cartesian co-ordinates, for the non –conducting region where

0σ =, the curl

equations can be written as:

x

z

b

b

72

( )ˆ ˆ ˆ

ˆ ˆ ˆ

x y z

x x y y z z

x y z

a a a

jw a a ax y z

ε∂ ∂ ∂ = Ε + Ε + Ε∂ ∂ ∂Η Η Η

3.45

( )ˆ ˆ ˆ

ˆ ˆ ˆ

x y z

x y y z z

x y z

a a a

jw a a ax y z

µ∂ ∂ ∂ = Η + Η + Η∂ ∂ ∂Ε Ε Ε

3.46

2 2 22

2 2 2w

x y zµε∂ Ε ∂ Ε ∂ Ε+ + =− Ε

∂ ∂ ∂

3.47

2 2 22

2 2 2w

x y zµε∂ Η ∂ Η ∂ Η+ + = − Η

∂ ∂ ∂

3.48

We can reasonably assume that fields are uniform or constant in the y-direction since there are no boundary conditions to be satisfied. The derivatives with respect to y in 3.45 and 3.46 can be put to zero. Recall also that for propagation

in the z-direction,

z y∂ ∂ ≡−. Equations 3.45- 3.48 now become:

y xjwγ εΗ = Ε

y xjwγ µΕ = − Η

zx yjw

xγ ε∂Η− Η − = Ε

∂

zx yjw

xγ µ∂Ε− Ε − = − Η

∂

3.49

yzjw

xε

∂Η= Ε

∂

y jwx

µ∂Ε

= − Η∂

73

2

2 22

22 2

2

wx

wx

γ µε

γ µε

∂ Ε + Ε = − Ε∂∂ Η + Η = − Η∂

3.50

Define h2=γ2+ω2μϵ and rewrite these equations;

Hx=-γh2∂Hz∂x

Hy=jωεExγHy=jωε∂Ez∂x-jωμHy-γ and -γ2Hy=jωε∂Ez∂x+ω2μεHy

-γ2-ω2μεHy=jωε∂Ez∂xHy=-jωεh2∂Ez∂x

-γEx-∂Ez∂x=-jμεHy=-jωμjωεExγ and -γ2-ω2μεEx=γ∂Ez∂xEx=-γh2∂Ez∂xγEy=-jωμHx

γEy=-jωμ∂Hz∂x+jωεEy-γ-γ2Ey-ω2μεEy=-jωμ∂Hz∂x

Ey=jωμh2∂Hz∂x

2.1.1. Field solutions for TE and TM waves

Three categories of guided - wave solutions i. Transverse electric (TE) waves Ez=0 , Hz≠0 ii. Transverse magnetic (TM) waves Ez≠0 , Hz=0 iii. Transverse electromagnetic (TEM) waves Ez=Hz=0

TE Waves: There is always and every where an electric field vector that is transverse to the direction of propagation and Ez=0

We shall use the wave equation to find Eyi.e. ∂2Ey∂x2+γ2Ey=-ω2μεEySince ∂Ey∂y=0, ∂2Ey∂x2=-(γ2+ω2με)Ey=-h2Ey Now write Ey as a product of two functions Eyx,z= Eyox e-γz

Then ∂2Eyo e-γz∂x2=-h2Eyox e-γz

or ∂2Eyo ∂x2=-h2Eyox

The solutions are Eyox=C1sinhx + C2coshx

The boundary conditions are Ey=0 at x=0, aat x=0 Ey=0 requires that C2=0, Eyx,z=C1sinhx e-γzat x=a Ey=0 implies ha=mπ, m=0,1,2,3, …,

74

h the “characteristics value “ or eigenvalue Eyx,z=C1sinmπaxe-γz

To the H fields ∂Ey∂x=-jωμHz, => Hz=-1jωμ∂Ey∂x=-mπjωμcosmπaxe-γzγEy=-jωμHx, => Hx=-1jωμEy=-γjωμC1sinmπaxe-γz

Figure 4.2: Electric and magnetic field distributions of TE1 and TE2 modes in parallel plate waveguides

Transverse magnetic (TM) fields: There is always and every where a magnetic field vector transverse to the direction of propagation and Hz=0

Using the wave equation to find Hy , ∂2Hy∂x2+γ2Hy=-ω2μεHy

Since ∂Hy∂y=0, ∂2Hy∂x2=-(γ2+ω2με)Hy=-h2Hy Writing Hy as a product of two functions Hy= Hyox e-γz

∂2Hyo e-γz∂x2=-h2Hyox e-γz

and ∂2Hyo ∂x2=-h2Hyox

The solutions are Hyox=C3sinhx + C4coshx

The boundary conditions do not directly apply to Hy but can applied to EZ by using Maxwell’s equations

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EZ=1jωε∂Hy∂x = 1jωε∂∂xC3sinhx + C4coshxe-γz

EZ=hjωεC3coshx- C4sinxe-γz

We now apply boundary conditions Ez=0 at x=0,a at x=0 Ez=0 requires that C3=0 at x=a Ez=0 implies ha=mπ, m=0,±1, ±2 ,±3, …

Figure 4.3: Electric and magnetic filed distributions of TM1 and TM2 modes in parallel plate waveguides

1.1.1. Transverse electromagnetic (TEM) waves.

This similar to the previous solutions EXCEPT there are no z fields, i.e Ez=Hz=0

The TE mode vanishes since Hz=0. Also all but the m=0 TM mode vanish when h=0. We remain with only the TMO mode , which is the TEM mode.Hy= C4 e-γz i.e. cosmπax goes to 1 if m=0

Ex=-γjωμC4e-γz

Ez=-jmπωεaC4sinmπaxe-γz=0

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So we have Ex, Hy but Ez =0. Therefore;

Figure 4.4: TEM mode in a parallel plate waveguides. In the top mode only electric fields are showed- magnetic fileds are out of (or into) the page.

1.1.2.

1.1.3. Cutoff frequency, Phase velocity, Wavelength.

TE and TM modes have similar characteristics

i. E and H have sinusoidal standing wave distributions is the x-direction.

ii. X-Y planes are equiphase planes, i.e. surface of constant phase.

iii. The equiphase surface propagate along the waveguide with phase velocity vp=ωβ

Consider Ey for TE waves Ey=C1sinmπaxe-γz

Assume ∝=0 so γ→j β , then write the real wave as

Eyx,z,t=C1sinmπaxcos(ωt-βz),

where sinmπax is the transverse standing wave.

By definition h2=γ2+ω2μϵ . But also h2=mπa2. Solving for γ, we have

γ=h2-ω2μϵ=mπa2-ω2μϵ .

There is the cutoff frequency fcmfor which γ=0. Solving for f we obtain

fcm=m2aμϵ=mvp2a

Propagating Wave: For f>fcm , mπa2-ω2μϵ <0. Using the expressions for fcm

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γ = j βm = jω2μϵ-mπa2 = jω2μϵ-2fcmπvp2 = jβ2-2fcmπvp2

γ=jβm1-2fcmπvpβ2 = jβm1-fcmf2

where β=ωvp and ω=2πf and βm emphasizes that this is for mode m.

Evanescent wave: For f<fcm , mπa2-ω2μϵ >0

γ=αmmπa2-ω2μϵ = βfcmf2-1 for f<fcm

αmmπa2-ω2μϵ is simply recognizing that the square root is negative so this becomes αm. This is known as an evanescent wave where attenuation is NOT due to energy losses but from boundary conditions.

For Propagation, ⋋m = 2πβm*2πβm1-fcmf2 = 2πβ1-fcmf2 = ⋋1-fcmf2

Similarly vpm = ωβm=vp1-fcmf2

We see that ⋋m and vpm vary as a function of the mode frequency.

The intrinsic wave impedance of the mode is obtained by ZTM or ZTE=ExHy=-EyHx

ZTEm=-EyHx=-C1sinmπaxe-jβz-βωμC1sinmπaxe-jβz=βωμ=ωμβ1-fcmf2

ZTEm=η1-fcmf2

For TM modes ZTMm=ExHy=βωμC4cosmπaxe-jβzC4cosmπaxe-jβz=βωμ

ZTMm=η1-fcmf2

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Figure 4.5: variation of impedance ( ZTMmη and ZTEmη ) against frequency ( fcmf )

1.1.1. Dispersion

Let us consider the “simple” case of a uniform plane wave in a medium with zero conductivity. What would happen if we had two waves propagating, each at a slightly different frequency, and a function of ?

– assume the two frequencies are ± with corresponding phase constants ±

If we have two waves propagating, each at a slightly different frequency, where the two frequencies are ± with corresponding phase constants ± then the solution is proportional to

In words, this looks just like a wave at the frequency and associated phase constant , with phase velocity but multiplied with an “amplitude modulation” function

The “velocity” of a phase front for this modulation envelope is

Group velocity

If we have two waves propagating, each at a slightly different frequency, ± , the solution behaves like a wave at the frequency with associated phase constant , traveling at the phase velocity vp = but it is multiplied by an “amplitude modulation” function traveling at the “velocity” . This is called the “group velocity” vgIn the limit of infinitesimal variation we obtain the “group velocity” vg

[ ] ( ) ( ) ( ) ( )j t z j t zj t ze e eδω δβ δω δβω β − − − − = +

[ ] ( ) ( ) [ ] ( ) ( )j t z j t zj t z j t ze e e eδω δβ δω δβω β ω β − − − + − + − − = +

( ) ( ) ( ) ( )j t z t z j t z t ze eω β δω δβ ω β δω δβ − + − − − − + − = +

( ) ( ) ( ) ( )1 2E E

j t z j t ze eω δω β δβ ω δω β δβ

∝ ∝

− − − + − + + =6 4 47 4 48 6 4 47 4 48

[ ] ( ) ( ) ( ) ( )j t z j t zj t ze e eδω δβ δω δβω β − − − − ∝ + [ ] ( )2cos

" "@

j t ze t z

normal amplitudewave modulation

functionfrequency

ω β δω δβ

ω

− = ⋅ ⋅ − ⋅ 142 43 1 4 4 4 2 4 4 4 3

( )cos t zδω δβ⋅ − ⋅

1

g

d dv

d d

ω ββ ω

− = =

79

Dispersion (β-ω) diagrams

Consider the plot of β versus ω fro the TE1/TM1 modes in a parallel plate waveguide

Notes on omega-beta diagrams• plot the frequency vs beta• slope from origin to a point on

the curve is the phase velocity

• slope of tangent is the group velocity

TE1

ω

TE2

ωC1 ωC2

Β=ω/Vp

Unbounded

2

1 cωβ ω µεω

= −

2( )

1

pp p

c

υωυ υ ωβ ω

ω

= = = −

,

1pυ

µε=

1 2

1 cg p

d

d

ωωυ υβ ω

− = = −

1 2 3 4

Eva

nesc

ent

Re

gion

p

p

υυ

g

p

υυ

1

c

f

f

p

p

υυ

Normalized

or p

p

υυ

The velocity of Energy Flow is the group Velocity

2

12E g p

mV V V

a

λ = = −

which is also the component of each mode’s velocity in the z direction.

Figure 4.6: ω-β diagramSpecial relativity says that the velocity of information cannot be greater than c, the “speed of light”. Since “packets” carry information, and group velocity is

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usually (but not always) related to “packet velocity”, vgroup is normally less than c.

1.1.1. Attenuation in parallel plate waveguides

Practical waveguides are made of copper or brass usually coated with silver. Assuming losses very small so that they have negligible effect on the field distribution the attenuation for different modes are (See assignment 4 for derivation of the expression given below)

αcTEM=1ηaωμo2σ

αc,TEm=2m2π2βωηa3ωμo2σ

And

αc,TMm=2Rsηa1-fcmf2Rs

Figure 4.7: Attenuation versus frequency for the parallel plate waveguide.

Observations

• The figure shows the attenuation as a function of frequency for a few modes. Higher order modes have higher losses.

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• TM modes have higher losses than TE modes since they have a tangential J due to tangential Hy i.e. Hy=C4cosmπaxe-jβz

• TE modes have lower losses at higher frequencies since as ω increases surface currents decreases i.e. very strong frequency dependence.

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1.1. THE RECTANGULAR WAVEGUIDE.

Typically b>>a, infinite in z

Figure 4.8: A rectangular waveguide

We assume perfectly conducting waveguide walls which require Etan=0 and Htan=0

Ex, Ez=0 at y=0 and y=b

Hy=0 at y=0 and y=b

Ex, Ez=0 at x=0 and x=a

Hx=0 at y=0 and y=a

We also want the field to vary in the z-direction as e-γz. Aside from the boundary conditions this is no different than the parallel plate waveguide and must satisfy the curl equations ∇XH=jωεE and ∇XE=-jωμH , that we have developed for the parallel plate waveguide using ∂∂z→-γ

For ∇XH=jωεE

∂Hz∂y+γHy=jωεEx

∂Hz∂x+γHx=-jωεEy

∂Hy∂x-∂Hx∂y=jωεEx

For ∇XE=-jωμH

∂Ez∂y+γEy=-jωμHx

∂Ez∂x+γEx=jωμHy

∂Ey∂x-∂Ex∂y=-jωμHx

The wave equations for Ez and Hz reduce to

∂2Ez∂x2+ ∂2Ez∂y2+γ2Ez= -ω2μϵEz

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∂2Hz∂x2+ ∂2Hz∂y2+γ2Hz= -ω2μϵHz

The transverse field components can be written in terms of Ez and Hz.

Hx=-γh2∂Hz∂x+jωεh2∂Ez∂yHy=-γh2∂Hz∂y-jωεh2∂Ez∂x

Ex=-γh2∂Ez∂x-jωμh2∂Hz∂y

Ey=-γh2∂Ez∂y+jωμh2∂Hz∂x

Where h2=γ2+ω2μϵ

Just as for the parallel plate waveguide the field solutions can be classified as

TE where Ez=0

TM where Hz=0

For waveguides, we write the wave equations using a transverse operator ∇tr which can be written as ∇tr=x∂∂x+y∂∂y

And ∇tr2=∂2∂x2+∂2∂y2

The wave equations become ∇tr2Ez+γ2+ω2μϵEz=0

∇tr2Hz+γ2+ω2μϵHz=0

For TM modes, the component equation become

Ex=-γh2∂Ez∂x and Ey=-γh2∂Ez∂y

Etr=xEx+yEy=-xγh2∂Ez∂x-yγh2∂Ez∂y

Similarly Hx=jωεh2∂Ez∂y and Hy=-jωεh2∂Ez∂x

Htr=xHx+yHy=xjωεh2∂Ez∂y-yjωεh2∂Ez∂x

Htr=jωεh2x∂Ez∂y-y∂Ez∂x

Htr=jωεh2x-h2Eyγ-y-h2Exγ

Htr=jωεh2γh2xEx+yEy

Htr=jωεγxyzExEy0001

We can do the component equation for the TE waves in the same way

Htr=xHx+yHy=-γγ2+ω2μϵ∇trHz

Etr=xEx+yEy=jωμγHtrxz

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Where the boundary condition is n.Htr=0 or

∂Hz∂x=0 , ∂Hz∂y=0

1.1.1. Transverse magnetic (TM) modes

We use separation of variables similar to that which we used for parallel plate waveguide.

Ez(x,y,z)=Ezo(x,y)e-γz

Ezo is a function of two variables. So let Ezo(x,y)=f(x)g(y), the wave equation becomes

∇tr2fg+γ2+ω2μϵfg=0

However, ∇tr2fg=g∂2f∂x2+f ∂2g∂y2

Therefore, g∂2f∂x2+f ∂2g∂y2+h2fg=0 where h2=γ2+ω2μϵ

Dividing by fg we obtain 1f∂2f∂x2+1g ∂2g∂y2+h2=0

Rearranging 1f∂2f∂x2+h2=1g ∂2g∂y2

Each side must equal to a constant call it A2 which is determined by the boundary conditions.

1f∂2f∂x2+h2=A2 and 1g ∂2g∂y2=A2

The two equations have similar solutions.

fx=C1cosBx+C2sin(Bx) where B=h2-A2

And gy=C3cosAy+C4sin(Ay)

The complete product solution is Ezo(x,y)=f(x)g(y)

Ezox,y=C1C3cosBxcosAy+C1C4cosBxsinAy+C2C3sin(Bx)cos(Ay)+C2C4sin(Bx)sin(Ay)

The boundary conditions are Ezo=0 at x=0,a; and y=0,b.

At x=0 , Ezo0,y=C1C3cosAy+C1C4sinAy

For Ez o(0,y)=0 we require C1=0 since C3=C4=0 will result into a trivial solution.

Ezox,y=C2C3sin(Bx)cos(Ay)+C2C4sin(Bx)sin(Ay)

For Ez ox,0=0, Ez ox,0=C2C3sinBx. This implies that C2or C3 equals zero.

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We pick C3=0 since picking C2=0 would be a trivial solution. If we let C2C4=C

Ezox,y=CsinBxsinAy.

At x=a, , Ezoa,y=CsinBasinAy=0

=> sinBa=0, and B=mπa where m=1, 2, 3, …

At y=b, , Ezox,b=CsinBxsinAb=0.

This requires that A=nπb where n=1, 2, 3, …

The final expression is Ezox,y=Csinmπaxsinnπby.

Therefore for the propagating modes, ( γ=jβmn), the fields expressions are

Ezx,y,z=Csinmπaxsinnπbye-jβmnz

ξzx,y,z,t=Csinmπaxsinnπbycos(ωt-βmnz)

For evanescent waves, ( γ=αmn)

Ezx,y,z=Csinmπaxsinnπbye-αmnz

And ξzx,y,z,t=Csinmπaxsinnπbycos(ωt)e-αmnz

The other field components can also be calculated using component equations. For TM modes,

Hz=0. So Hx=jωεh2∂Ez∂y , Hy=-jωεh2∂Ez∂x

Ex=-γh2∂Ez∂x , Ey=-γh2∂Ez∂y

So for propagating rectangular TMmn modes,

Ez=Csinmπaxsinnπbye-jβmnz

And Ex=-jβmnCh2mπacosmπaxsinnπbye-jβmnz

Similarly, Ey=-jβmnCh2nπbsinmπaxcosnπbye-jβmnz

Hx=jωεCh2nπbsinmπaxcosnπbye-jβmnz

Hy=-jωεh2mπacosmπaxsinnπbye-jβmnz

Where m=1, 2, 3, … and n=1, 2, 3, …

To find γ=h2-ω2μϵ we note that 1f∂2f∂x2+h2=A2 where fx=C2sin(Bx)

h2=A2+B2=mπa2+nπb2

Knowing h2 γ=mπa2+nπb2-ω2μϵ

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1.1.2. Cut off Frequency in rectangular Waveguides:

We see γ corresponds to a propagating wave only i.e. when γ is imaginary (ω>ωcnm)

At cutoff frequency, mπa2+nπb2-ωcnm2μϵ=0

ωcnm2=1μϵmπa2+nπb2

ωcnm=1μϵmπa2+nπb2

The cutoff frequency is fcnm=ωcnm2π

We can also define the cutoff wave number Kc as

The quantity k=w/c=w/k=ωc=ωεμ is the wave number a uniform plane wave would have in the propagating medium ε, µ

For (ω>ωcnm), γ=jβmn=jω2μϵ-mπa2-nπb2

Where βmn=ω2μϵ-mπa2-nπb2

βmn=β1-fcmf2.

Where β=ωμε and fcm=ωcnm2π

Correspondingly for ⋋cm ⋋cmn=vpfcm=1με1ωcnm2π

⋋cmn=2πμε με mπa2+nπb2=2ma2+nb2

For propagating waves, vpmn=ωβmn=ωβ1-fcmf2=1με 11-fcmf2

⋋mn=2πβ=2πω2μϵ-mπa2-nπb2=⋋1-fcmf2

87

TE11, TM21

TE21,TM21

TE01

TE10

TE20

TE 12,T

M22

TE02

fcmn/

fc10

a/b

1 2 3 4 5

5

4

3

2

1

TE 22

Figure 4.9 :Cut off frequency for various waveguide dimensions

1.1.3. Wave Impedance:

We can also define a wave impedance.

ZTMmn = ExHx = ExoHyo = -jβmnCh2mπacosmπaxsinnπbye-jβmnz-jωεh2mπacosmπaxsinnπbye-jβmnz = βmnωε

ZTMmn=β1-fcmf2ωε=ωμεωε1-fcmf2

ZTMmn=η1-fcmf2

1.1.4. Transverse Electric (TE) modes

Ez=0 and Hz=Ccosmπaxcosnπbye-jβmnz

From which we can derive Hx=jβmnh2sinmπaxcosnπbye-jβmnz

Hy=jβmnCh2nπbcosmπaxsinnπbye-jβmnz

Ex=jωμCh2nπbcosmπaxsinnπbye-jβmnz

Hy=-jωμh2mπasinmπaxcosnπbye-

jβmnzHy=-jωμh2mπasinmπaxcosnπbye-jβmnz

88

The formulae for ωcnm, βmnetc are identical. One different formula is that of impedance which is given by ZTMmn=η1-fcmf2

A very important mode is the TE10 mode (a>b)

Hz=Ccosπaxe-jβmnz and Hx=jβmnCh2πasinπaxe-jβmnz

Hy=0, and Ex=0

Hy=-jωμh2πasinπaxe-jβmnz

Figure 4.10: The TE10 mode in the rectangular waveguide.

β10=ω2μϵ-πa2=2π⋋2-πa2

⋋10=2πβ10=⋋1-⋋2a2

fc10=12aμε

NOTE: If propagation at a specified f is not possible in the TE10 mode, then it is not possible for any mode.

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Figure 4.11: Some selected field patterns in the X-Y plane in rectangular waveguides [2]

1.1.5. Coupling power into waveguides

We have not talked about how to couple power for particular modes into waveguides. The practice is to use a probe (source) that will produce lines of E and H that are roughly parallel to the lines of E and H for that particular mode and that produce the maximum electric field where the field would be maximum for that mode. A single probe will excite the TE10 mode into the waveguide

Figure 4.12: Coupling TE10 and TE20 modes into a rectangular waveguide

To excite the TE20 mode, use two vertical antenna probes while the TE11 mode requires parallel excitation of the electric field at the wall.

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Figure 4.12: Coupling TE11 and TM10 modes into a rectangular waveguide

In practice waveguide dimensions are chosen to allow only one mode to propagate. Square waveguides (where a=b) are undesirable since modes differ only by rotation. In practice pick a≈2b to separate modes and maximize power transmission.

Final Notes on Single mode waveguides:

• Different phase velocities would give different transverse modes and make it difficult to extract energy.

• Chose λ/2 <a< λ to ensure transmission of only the TE10 mode

• Often pick a=.07 λ since values near λ may allow the next mode to propagate and values near λ /2 have large variation of vp and ZTEorTM with f.

1.1.1. Attenuation in Rectangular waveguides:

In rectangular waveguides attenuation occurs due to three mechanisms:

1. Losses due to surface currents flowing in the waveguides walls

2. Dielectric losses due to a dielectric with sigma =/0 or ec=e’-je”

3. Evanescent wave attenuation when f<fc

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1.1. CIRCULAR WAVEGIDES

Detailed analysis on circular waveguides has been left to the student. The following should however be noted:

Figure 4.13: Types of cylindrical waveguides

Figure 4.14: Some of the available modes in rectangular waveguides

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Figure 4.15: Electric and Magnetic fields in Circular waveguides for selected modes

Assignment Four

I.1. Derive the expressions for attenuation of the TEM, TEm and TMn modes in a parallel plate waveguide

I.2. Using clear analysis, quantitatively discuss the losses in circular waveguides.

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CHAPTER 5: WAVE ROPAGATION IN OTHER SYSTEMS

1.

2.

2.1. PLASMAS

2.1.1. Simple Models of Dielectrics, Conductors, and Plasmas

A simple model for the dielectric properties of a material is obtained by considering the motion of a bound electron in the presence of an applied electric field. As the electric field tries to separate the electron from the positively charged nucleus, it creates an electric dipole moment. Averaging this dipole moment over the volume of the material gives rise to a macroscopic dipole moment per unit volume.

A simple model for the dynamics of the displacement x of the bound electron is as follows (withx=dx/dt) mx=eE-kx-mγx 6.1

where we assumed that the electric field is acting in the x-direction and that there is a spring-like restoring force due to the binding of the electron to the nucleus, and a friction-type force proportional to the velocity of the electron.

The spring constant k is related to the resonance frequency of the spring via the relationship ω0=k/m or k=mω0. Therefore, we may rewrite 6.1 as

x+γx+ω02x=emE 6.2

The limit ω0 = 0 corresponds to unbound electrons and describes the case of good conductors. The frictional term γx arises from collisions that tend to slow down the electron. The parameter γ is a measure of the rate of collisions per unit time, and therefore, τ = 1/γ will represent the mean-time between collisions.

The case of a tenuous, collisionless, plasma can be obtained in the limit γ=ω0= 0. Thus, the above simple model can describe the following cases:

a) Dielectrics, γ≠0, ω0≠ ω

b) Conductors, γ=0, ω0≠0

c) Collisionless Plasmas, γ=0, ω0=0

The basic idea of this model is that the applied electric field tends to separate positive from negative charges, thus, creating an electric dipole moment. In this sense, the model contains the basic features of other types of polarization in materials, such as ionic/molecular polarization arising from the separation of positive and negative ions by the applied field, or polar materials that have a permanent dipole moment.

94

The applied electric field E(t) in equation 6.2 can have any time dependence. In particular, if we assume it is sinusoidal with frequency ω, E(t)= Eejωt, then, equation 6.2 will have the solution x(t)= xejωt, where the phasor x must satisfy equation 6.2 re-written in harmonic form as:

-ω2x+jωγx+ω02x=emE 6.3Its solution therefore is: x=emEω02x-ω2+jωγ

6.4

1.1.1. Electromagnetic Waves in Plasmas

To describe a collision less plasma, such as the ionosphere, the simple model above can be specialized by choosing ω0 = γ = 0. Thus, Equation 6.4 becomes:

x=-emEω2 6.5The corresponding electron velocity will also be sinusoidal v(t)= vejωt, where v=x=jωx. Thus, v=jωx=-jωemEω2 6.6Assuming that there are N such elementary dipoles per unit volume, since the individual electric dipole moment is p = ex, then the polarization per unit volume P, will be:

P=Np=Nex=-Ne2mEω2=ε0X(ω)E6.7

The electric flux density will then be: D=ε0E+P=ε01+XωE=ε(ω)E6.8

where the effective permittivity ε(ω) is: ε(ω)=ε0-Ne2mEω2or in a more convenient form, εω=ε0-ε0ωp2ω2=ε0(1-ωp2ω2)

6.9where ωp is the so-called “Plasma Frequency” of the material defined by:

ωp2=Ne2ε0m6.10

The plasma frequency can be calculated from equation 6.10. In the ionosphere the electron density is typically N = 1012, which gives fp = 9 MHzFrom chapter 5, we saw that the propagation wavenumber of an electromagnetic wave propagating in an electric/conducting medium is given in terms of the effective permittivity by:

k=ω√(με((ω))

It follows that for plasma: k=ωμ0ε01-ωp2ω2=1cμε0ω2-ωp2 6.11

where we used c=1μ0ε0

If ω > ωp, the electromagnetic wave propagates without attenuation within the plasma. But if ω < ωp, the wavenumber k becomes imaginary and the wave gets attenuated. At such frequencies, a wave incident (normally) on the ionosphere from the ground cannot penetrate and gets reflected back.

1.2. MICROSTRIP TRANSMISSION LINES

As circuits have been reduced in size with integrated semiconductor electron devices, a transmission structure was required that was compatible with circuit construction techniques to provide guided waves over limited distances. This was realized with a planar form of single wire transmission line over a ground plane,

95

called microstrip. Microstrip employs a flat strip conductor suspended above a ground plane by a low-loss dielectric material. The size of the circuit can be reduced through judicious use of a dielectric constant some 2-10 times that of free space (or air), with a penalty that the existence of two different dielectric constants (below and above the strip) makes the circuit difficult to analyze in closed form (and also introduces a variability of propagation velocity with frequency that can be a limitation on some applications). The solution is to find an effective relative permittivity εreff for the combination.

The advantages of microstrip have been well established, and it is a convenient form of transmission line structure for probe measurements of voltage, current and waves. Microstrip structures are also used in integrated semiconductor form, directly interconnected in microwave integrated circuits.

(a) (b)

Figure 6.1: (a) The Microstrip and (b) The Stripline Conductor

Waves and Impedances in Microstrip

Although the presence of two dielectric regimes in microstrip precludes the strict propagation of TEM waves, the same type of transmission-line characteristics are present, as can be seen from the fact that microstrip can propagate energy down to zero frequency (direct current). Microstrip construction lends itself to small structures that can carry semiconductor devices and surface-mount lumped elements, which can be attached by automatic means.

This extreme usefulness of microstrip makes the lack of an elegant closed-form solution acceptable, and accurate approximations based on the velocity/capacitance method are used to estimate Zo and other parameters. Unwanted modes are dealt with in part by using material with a relatively high dielectric constant, but waveguide modes are present and represent an upper frequency limit. The effects of unwanted waveguide modes can be restricted by choosing dielectric thickness less than λ/4 and strip width w less than λ /2 at the highest frequency of interest. Thus, for a maximum frequency of interest fmax, we chose

The velocity of propagation in microstrip is relatively constant with varying w/h, and Zo can be estimated accurately using a number of methods and software applications. Some downloadable applications include AppCad2, Txline3, Microstrip Calculator4 and Sonnet5.

Some Microstrip Relations

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Note that It's difficult to get more than 200W for Z0 in a microstrip. For the simple closed form solution, the approximations below would be used in the relations above.

Stripline ConductorAlso called shielded microstrip, it uses a different dielectric (different from air) on the upper side of the line. The effective relative permittivity is used in calculations above.

Assuming w≥10h, where er1 = the relative permittivity of the dielectric of thickness h1.

er2 = the relative permittivity of the dielectric of thickness h2.

1.3. Propagation in Optical Fibers

Ray Theory in Dielectric Slab waveguides

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Fig 6.2: (a) Unguided wave since θi<θc and wave refracts out of guide

(b) Guided wave since θi>θc gives total internal reflection. However not any angle can propagate

Fig 6.3

φr is the phase shift from TIR at either B or C. Geometry gives

Assume parpendicular polarization (E out of plane)

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To do it graphically, plot LHS and RHS e.g figure 6.4 for the following parameters: f=30GHz, d=1 cm, εd=2.25ε0 (glass sorrounded by air).

Even m=2,4,6,…; Odd m = 1,3,5,….

Read θis from graphs as

TE1 θi=75.030

TE2 θi=59.470

TE3 θi=43.860

Fig 6.4: Graphical Evaluation of propagation in Optical fibers.

Assignment

6.1. Calculate the plasma frequency in the ionosphere where the electron density is typically N = 1012

6.2. Discuss the advantages and uses of microstrip in today’s world. What are the major challenges to their use and how are they overcome

6.3. Explain why in a collision less plasma, ω0 = γ = 0.

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REFERENCES:

[1] E.C. Jordan and K.G. Balmain, “Electromagnetic Waves And Radiating Systems”, 2nd Edition

[2] Sophocles J. Orfanidis, “Electromagnetic Waves and Antennas”, ECE Department, Rutgers University

[3] C.S. Lee, S. W. Lee and S. L. Chuang, Plot of modal field distribution in rectangular and circular waveguides, IEEE trans. Microwave Theory and Techniques, 33(3). PP 271-274, March 1985

[4] BO THIDÉ, “Electromagnetic Field Theory”, Internet Text Book

[5] Leonard M. Magid, “Electromagnetic Fields, Energy and Waves”, John Wiley & Sons

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APPENDICES

APPENDIX A: GRAPHICAL SOLUTION TO DOUBLE STUB MATCHING

Example of Graphical Solution to transmission line matching problem using two tuning elements:

First step: don't worry about the stub lengths, first find the tuning susceptances:

To get started, recall that to get no reflection at plane B you must be on the "g = 1" circle in plane B'. We can move the "g=1" circle to plane A so that we can ultimately find Y1:

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Now we have to get the load to plane A':

Now we must move on a circle of constant real part of Yload in plane A' to get onto the transformed g=1 circle in plane A; this will actually give us Y1:

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Now for each possible choice of Y1 we need to move back to plane B'; this will gives us Y2:

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