Review of Electromagnetic Waves & Waveguides1.pdf

7/29/2019 Review of Electromagnetic Waves & Waveguides1.pdf

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Electromagnetic Fields

Amanogawa, 2006 Digital Maestro Series 1

Review: TimeDependent Maxwells Equations

( ) ( )( ) ( )D t E tB t H tG G

G G

= =

( )( ) 0D tB tG

G = =

( ) ( )

( ) ( )B t

E tt

D tH t Jt

G

G

G

G G

= = +


2/144



Electromagnetic quantities:

Vectorquantitiesin space

Electric Field

Magnetic Field

Electric Flux (Displacement) DensityMagnetic Flux (Induction) Density

Current Density

Displacement Curren

E

H

D

B

J

D

t

G

G

G

G

G

G tCharge Density

Dielectric Permittivity

Magnetic Permeability


3/144



In free space:

In a material medium:

If the medium is anisotropic, the relative quantities are tensors:

[ ] [ ][ ] [ ]12

0

70

8.854 10 As/Vm or F/m

4 10 Vs/Am or Henry/m

= = = =

0 0;

relative permittivity (dielectric constant)

relative permeability

r r

r

r

= = = =

;

xx xy xz xx xy xz

r yx yy yz r yx yy yz

x zy zz zx zy zz

= =


4/144



In engineering it is very important to considertime-harmonic fields

with a sinusoidal time-variation. If we assume a steady-statesituation (after all transients have died out) most physical situationsmay be investigated by considering one single frequency at a time.

This assumption leads to great simplifications in the algebra. It is

also realistic, because in practical electromagnetics applicationswe often have a dominant frequency (carrier) to consider.

The time-harmonic fields have the form

We can use the complex phasor representation

( ) ( ) ( ) ( )0 0cos cosE HE t E t H t H tG G G G= + = +

( ) { } ( ) { }0 0Re ReE Hj jj t j tE t E e e H t H e eG G G G = =


5/144



We define

Maxwells equations can be rewritten for phasors, with the time-derivatives transformed into linear terms

( )( )

0

0

E phasor of

H phasor of

E

H

j

j

E e E t

H e H t

G G G

G G G

= == =

( )( )222

E phasor of

E phasor of

E tj

t

E t

t

G

G

G

G

= =


6/144



In phasor form, Maxwells equations become

where all electromagnetic quantities are phasors and functions onlyofspace coordinates.

E H

H J E

D

B 0

D E

B H

j

j

G G

G G G

G

G

G G

G G

= = + = == =

( )F E Bq vG G G

G= +


7/144



Lets consider first vacuum as a medium. The wave equations for

phasors become Helmholtz equations

The general solutions for these differential equations are wavesmoving in 3-D space. Note, once again, that the two equations areuncoupled.

This means that each equation contains all the necessary

information for the total electromagnetic field and one only needs tosolve the equation forone field to completely specify the problem.The other field is obtained with a curl operation by invoking one ofthe original Maxwell equations.

2 20 0

2 2 0 0

E E 0

H H 0

G G

G G

+ = + =


8/144



At this stage we assume that a wave exists, and we do not yet

concern ourselves with the way the wave is generated. So, for thesake of understanding wave behavior, we can restrict the Helmhlotzequations to a simple case:

We assume that the wave solution has an electric field which is

uniform on the {x,y}-plane and has a reference positiveorientation along the x-direction. Then, we verify that this is areasonable choice corresponding to an actual solution of theHelmholtz wave equations. We recall that the Laplacian of a

scalaris a scalar

and that the Laplacian of a vectoris a vector

2 2 22

2 2 2

f f ff

x y z

= + + 2 2 2 2 E E E Ex x y y z zi i i

G = + +


9/144



The Helmholtz equation becomes:

Only thex-component of the electric field exists (due to the chosen

orientation) and only the z-derivative exists, because the field is

uniform on the {x,y}-plane.We have now a one-dimensional wave propagation problemdescribed by the scalardifferential equation

( )22 2 20 0 0 02E E E E 0x x x xi iz

G G + = + =

2 20 02

EE 0x x

z

+ =


10/144



This equation has a well known general solution

where the propagation constant is

The wave that we have assumed is a plane wave and we haveverified that it is a solution of Helmholtz equation. The generalsolution above has two possible components

For the simple wave orientation chosen here, the problem ismathematically identical to the one solved earlier for voltagepropagation in a homogeneous transmission line.

( ) ( )exp expj z B j z +

0 0c = =

( )exp zA j z Forward wave, moving along positive( )exp zB j z Backward wave, moving along negative


11/144



If a specific electromagnetic wave is established in an infinite

homogeneous medium, moving for instance along the positivedirection, only the forward wave should be considered.

A reflected wave exists when a discontinuity takes place along thepath of the forward wave (that is, the material medium changes

properties, either abruprtly or gradually).

We can also assume that the amplitude of the forward plane wavesolution is given and that it is in general a complex constant fixed

by the conditions that generated the wave

We can write at last the phasor electric field describing a simpleforward plane wave solution of Helmholtz equation as:

0j

E e=

0E ( ) j j zx xz E e e i

G =


12/144



The corresponding time-dependent field is obtained by applying the

inverse phasor transformation

The phasor magnetic field is obtained directly from the Maxwellequation for the electric field curl

( ) ( ) } { }( )

0

0

, Re E Re

cos

j t j j z j tx x x x

x

E z t z e i E e e e i

E t z i

G = == +

( )( )0 0

0

0

E H

H

j j zx

j j zx

E e e i j

E e e i

j

G G

G

= = =


13/144



We then develop the curl as

( ) ( )( ) ( )

0

0 0

0

det

E 0 0

x y z

j j zx

x

j jj z j z

y z

j j zy

i i i

E e e i x y z

z

E e e E e ei i

z y

j E e e i

= = = =

= = 0

( )Ex z


14/144



The final result for the phasormagnetic field is

We define

( )

( )

0

0 00

0

0 000 0

H

E

j j z

y y

j j zy

j j z y x y

j E e ez i

j

E e e i

E e e i z i

G

= = = = = =

00

0

377 Intrinsic impedance of vacuum = =


15/144



We have found that the fields of the electromagnetic wave are

perpendicular to each other, and that they are also perpendicular(ortransverse) to the direction of propagation.

x

z

y

E

H


16/144



Electromagnetic power flows with the wave along the direction of

propagation and it is also constant on the phase-planes. Thepower density is described by the time-dependent Poynting vector

The Poynting vector is perpendicular to both field components, andis parallel to the direction of wave propagation.

When the wave propagates on a general direction, which does notcoincide with one of the cartesian axes, the propagation constantmust be considered to be a vector with amplitude

and direction parallel to the Poynting vector.

( ) ( ) ( )P t E t H tG G G=

| | G


17/144



The condition of mutual orthogonality between the field

components and the Poynting vector is general and it applies toany plane wave with arbitrary direction of propagation. The mutualorientation chosen for the reference directions of the fields followsthe right hand rule.

( , , )E x y zG

( , , )H x y zG

, PG

G

x

y

z


18/144



So far, we have just verified that electromagnetic plane waves arepossible solutions of the Maxwell equations for time-varying fields.One may wonder at this point if plane waves have practical physicalrelevance.

First of all, we should notice that plane waves are mathematicallyanalogous to the exponential basis functions used in Fourieranalysis. This means that a general wave, with more than onefrequency component, can always be decomposed in terms ofplane waves.

Forperiodic signals, we have a discrete set of waves which areharmonics of the fundamental frequency (analogy with Fourierseries).

For general signals, we must consider a continuum offrequencies in order to decompose in terms of elementary planewaves (analogy with Fourier transform).


19/144



From a physical point of view, however, the properties of a plane

wave may be somewhat puzzling.

Assume that a steady-state plane wave is established in an idealinfinite homogeneous medium. On any plane perpendicular to thedirection of propagation (phase-planes), the electric and magnetic

fields have uniform magnitude and phase.

The electromagnetic power, flowing with a phase-plane of the wave,is obtained by integrating the Poynting vector, which is alsouniform on each phase-plane. For a plane where the Poyntingvector is non-zero, the total power carried by the wave is infinite

In many practical cases, we approximate an actual wave with aplane wave on a limited region of space, thus considering anappropriate finite power.

( ) ( ) ( )plane plane

P t E t H t G G G


20/144



Review of Boundary Conditions

Consider an electromagnetic field at the boundary between twomaterials with different properties. The tangent and the normalcomponent of the fields must me examined separately, in order tounderstand the effects of the boundary.

Medium 11 ; 1

Medium 22 ; 2

boundary1tH

G

2tHG

2nHG

1nHG 1H

G

2HG


21/144



Tangential Magnetic Field

Ampres law for the boundary region in the figure can be written as

Medium 11 ; 1

Medium 22 ; 2

boundary

1HtG

2HtG

3Hn

G

a4Hn

G

b

.x

y

H HH Ey x

zJ jx y

= + G


22/144



In terms of finite differences approximation for the derivatives

If one lets the boundary region shrink, with a going to zero fasterthan b,

4 3 1 2H H H H En n t t zJ jb a

= +

t

t

z

t

t sa

J a Jfor perfect conducto

for materials wi

rs

(sur

th finite co

face cur

nducti

ren

v ty

)

i

t2 10

2 1

H H lim

( )

H 0

H

= = = Tangential components are conserved

3 42 1

0

H HH H lim ( E )n nt t z z

a

J a j a a

b

= + +


23/144



For a general boundary geometry

In the case of a perfect conductor, the electromagnetic fields goimmediately to zero inside the material, because the conductivity is

infinite and attenuates instantly the fields. The surface current isconfined to an infinitesimally thin skin, and it accounts for thediscontinuity of the tangential magnetic field, which becomesimmediately zero inside the perfect conductor.

For a real medium, with finite conductivity, the fields can penetrateover a certain distance, and there is a current distributed on a thin,but not infinitesimal, skin layer. The tangential field components onthe two sides of the interface are the same. Nonetheless, theperfect conductor is often a good approximation for a real metal.

t t sn J1 2 (H H ) =G Gn unit vector normal to the su e rfac


24/144



Tangential Electric Field

Faradays law for the same boundary region can be written as

Medium 11 ; 1

Medium 22 ; 2

boundary

1EtG

2EtG

3En

G

a4En

G

b

.x

y

E EE H

y x jx y

= G


25/144



In terms of finite differences approximation for the derivatives

If one lets the boundary region shrink, with a going to zero fasterthan b,

For a general boundary geometry

4 3 1 2E E E E Hn n t t jb a

=

t t2 1 E E 0 = Tangential components are conserved3 4

2 1

0

E EE E lim ( H )n nt t z

a

j a a

b

= +

t tn 1 2 (E E ) 0 =G


26/144



Normal components

Consider a small box that encloses a certain area of the interface

with

Medium 11 ; 1

Medium 22 ; 2

boundary

1 1D Bn nG G

2 2D Bn nG G

w

Area

.x

y

+ + + + + + s

s interface charge density=

El t ti Fi ld


27/144



Integrate the divergence of the fields over the volume of the box:

VoluVolume

Surface

me

dd

ds

rr

Divergence theorem

Flux of D out of the box

D

D n

=

=

G

G G

G G

G

w

Volume

Surface

dr

ds

Divergence theorem

Flux of B out of the box

B 0

B n

= =

G

G G

G Gw

El t ti Fi ld


28/144



If the thickness of the box tends to zero and the charge density is

assumed to be uniform over the area, we have the following fluxes

The resulting boundary conditions are

The discontinuity in the normal component of the displacementfield D is equal to the density of surface charge.

The normal components of the magnetic induction field B arecontinuous across the interface.

n n

s

n n

Area

Area

Area

G

G

1 2

1 2

= (D D )

= T

D-Flux out of box

B-Flux out of bo

otal interface charge =

= (B ) 0x B

= =

n n s n n1 2 1 2D D B B 0= =



29/144



For isotropic and uniform values ofandin the two media

Even when the interface charge is zero, the normal components of

the electric field are discontinuous at the interface, if there is achange of dielectric constant .

The normal components of the magnetic field have a similardiscontinuity at the interface due to the change in the magnetic

permeability. In many practical situations, the two media may havethe same permeability as vacuum, 0, and in such cases the normalcomponent of the magnetic field is conserved across the interface.

n n n n s

n n n n

1 2 1 1 2 2

1 2 1 1 2 2

D D E E

B B H H 0

= = = =

G G G GG G G G



30/144



SUMMARYIf medium 2 is

perfect conductor

t

G

1H

t

G

2

H

t

G

1E

t

G

2E

n

G

1H

n

G

2H

n

G

1E

n

G

2E

1, 1

1, 1

1, 1

1, 12, 2

2, 2

2, 2

2, 2 t t t s

t

n1 2 1

2

H J

H H

H 0

==

= G GG

G G

t t t

t

1 2 1

2

E 0

E E

E 0

= ==

GG

G G

n n n

n

11

2

2 H 0

H H1 2

H 0

==

G G G

1 12

2

1 E

E E1 2

E 0

s

+ n

n

n n s ==

GG

G G



31/144



Examples:

An infinite current sheet generates a plane wave (free space onboth sides)

The E.M. field is transmitted on both sides of the infinitesimally thinsheet of current.

x

y

+ z- z

Js

H

s

( ) cos( )

Phasor J

s so x

so x

J t J t i

J i

=

G

G



32/144



BOUNDARY CONDITIONS

1 2 (H H ) Jt t sn =G G

1 2

1 2

1 0 1

1 2

1 2

H H

E E

E H

Symmetry H H

H H2 2

t t so x

t t

t t

t t

so so

J i

J J

=== == =

G GG GG G

G G



33/144

g


A semi-infinite perfect conductor medium in contact with free space

has uniform surface current and generates a plane wave

The E.M. field is zero inside the perfect conductor. The wave is onlytransmitted into free space.

x

y

+ z- z

Js

H

J cos( )s so x J t iG

Perfect

Conductor

Free Space



34/144


BOUNDARY CONDITIONS

1 2 (H H ) Jt t sn =G G

1 2 1

2

1 2

1 2

H H H 0

E 0

Asymmetry H H

H H 0

t t t so x

t

t t

t so t

J i

J

= ==

= =

G G GG

G G



35/144


Electromagnetic Waves in Material Media

In a material medium free charges may be present, which generatea current under the influence of the wave electric field. The current

Jc is related to the electric field E through the conductivity as

The material may also have specific relative values of dielectricpermittivity and magnetic permeability

J Ec =

r o r o= =



36/144


Maxwells equations become

In phasor notation, it is as if the material conductivity introduces an

imaginary part for the dielectric constant . The wave equation forthe phasor electric field is given by

We have assumed that the net charge density is zero, even if aconductivity is present, so that the electric field divergence is zero.

E H

H E E ( )E

j

j j j

= = + =

2

c

2

E E E H

(J E)

E ( )E

j

j j

j j

= = = + = +



37/144


In 1-D the wave equation is simply

with general solution

These resemble the voltage and current solutions in lossytransmission lines.

22

2

E( )E Ex x xj j

z

= + =

( )

( )

E ( ) exp( ) exp( )

1( ) exp( ) exp( )

1exp( ) exp( )

x

xy

z A z B z

E jH z A z B zj z j

A z B z

= + + = =

=



38/144


The intrinsic impedance of the medium is defined as

For the propagation constant, one can obtain the real and imaginaryparts as

j je

j

= = +

1 / 22

1 / 22

( )

1 12

1 12

j j j= + = + = + = + +



39/144


Phase velocity and wavelength are now functions of frequency

The intrinsic impedance of the medium is complex as long as theconductivity is not zero. The phase angle of the intrinsicimpedance indicates that electric field and magnetic field are out ofphase. Considering only the forward wave solutions

( )( )

1 / 22

1 / 22

21 1

2 21 1

pv

f

= = + + = = + +

E ( ) exp( ) exp( ) exp( )

1 1H ( ) exp( ) exp( ) exp( )

x

y

z A z A z j z

A z j A z j z j

= = = =



40/144


In time-dependent form

where the integration constant has been assumed to be in general acomplex quantity as

{ }

{ }1

( , ) Re exp( )exp( )exp( )

exp( )cos( )

1( , ) Re exp( )exp( )exp( )

exp( )

exp( )

exp( )cos( )

x

y

E z t z j z j t

z t z

H z t A z j z j j t

A j

A

j

A z t z

= +

=

+

=

exp( )A j



41/144


Classification of materials

Perfect dielectrics - For these materials = 0Propagation constant

0

r o r o= =Medium Impedance

= r o

r o

j

j

=

Phase velocity

1p

r o r o

v= =

Wavelength

2 1p

r o r o

v

f f

= = =



42/144


Imperfect dielectrics For these materials 0 but (/)


43/144


If (/)


44/144


Good conductors For these materials 0 but (/)>>1( )

1 1exp( ) (1 )

4 2 2

4 2 4

exp( )4

1 1

2 2(1 )

p

j j j j

j j f j

f

v f

j jj

f

fjj

f

j

= + = = = + = +

= = =

= + = + =

+



45/144


The simple rule of thumb is that approximations for good conductorcan be applied when

Note that for a good conductor the attenuation constant and thepropagation constant are approximately equal.The medium impedance has nearly equal real and imaginaryparts, therefore its phase angle is approximately 45.This means that in a good conductor the electric and magnetic

fields have always a phase difference = 45 = /4.

10



46/144


Also, in a good conductor the fields attenuate very rapidly. The

distance over which fields are attenuated by a factor exp(1.0) is

A typical good conductor is copper, which has the following

parameters:

1 1Skin depth

f= = =

75.80 10 [S/m]

o

o

=



47/144


Copper remains a good conductor at extremely high frequencies.Another good conductor example is sea water at relatively lowfrequencies

At a frequency of25 kHz

4.0 [S/m]80 o

o

36,000



48/144


Perfect conductor - For this ideal material For this material, the attenuation is also infinite and the skin depthgoes to zero. This means that the electromagnetic field must go tozero below the perfect conductor surface.

General medium - When a material is not covered by one of the limitcases, the complete formulation must be used. We can classify a

material for which the conditions (/)10 areinvalid as a general medium.

The simple rule of thumb for general medium is

10 0.1> >



49/144


Power Flow in Electromagnetic Waves

The time-dependent power flow density of an electromagnetic waveis given by the instantaneous Poynting vector

Fortime-varying fields it is important to consider the time-average

power flow density

where Tis the period of observation.

( ) ( ) ( )P t E t H tG G G=

0 0

1 1( ) ( ) ( ) ( )

T TP t P t dt E t H t dt

T T

G G G G= =



50/144


Considertime-harmonic fields represented in terms of their phasors

The time-dependent Poynting vectorcan be expressed as the sumof the cross-products of the components

(Note that: 1cos sin sin 22

t t t = )( )

2

2

( ) ( ) Re{E} Re{H} cos

Im{E} Im{H} sin

Re{E} Im{H} Im{E} Re{H} cos sin

E t H t t

t

t t

G G G G

G G

G G G G

= + +

{ }{ }( ) Re E exp( ) Re{E} cos Im{E} sin

( ) Re H exp( ) Re{H} cos Im{H} sin

E t j t t t

H t j t t t

= = = = G G G G

G G G G



51/144


The time-average power flow density can be obtained by integrating

the previous result over a period of oscillation T . The pre-factorscontaining field phasors do not depend on time, therefore we haveto solve for the following integrals:

2

00

20

0

2

00

1 1 sin 2cos2 4

1 1 sin 2sin2 4

1 1 sincos sin2

1

12

0

2

TT

T

T

T

T

t tt dtT T

t tt dtT T

tt t dt T T

= + = = =

= =


Th fi l lt f th ti fl d it i i b


52/144


The final result for the time-average power flow density is given by

Now, consider the following cross product ofphasor vectors

( )0

1( ) ( ) ( )

1

Re{E} Re{H} Im{E} Im{H}2

TP t E t H t dt

T

G G G

G G G G

= = +

( )*

E H Re{E} Re{H} Im{E} Im{H}

Im{E} Re{H} Re{E} Im{H}j

G G G G G G

G G G G = +

+


B bi i th i lt bt i th f ll i


53/144


By combining the previous results, one can obtain the followingtime average rule

We also call complex Poynting vectorthe quantity

NOTE: the complex Poynting vector is not the phasor of the time-dependent powernorthat of the time-average power density!

Phasor notation cannot be applied to the product of two time-

harmonic functions (e.g.,P( t)), even if they have same frequency.

{ }*01 1( ) ( ) ( ) Re E H2TP t E t H t dtTG G G G G= = *1

P E H2

G G G=

{ } { }( ) Re P ( ) Re P exp( )P t P t j tdon't t( )ryG G G G= =


Consider a 1 D electro magnetic wave moving along the z direction


54/144


Consider a 1-D electro-magnetic wave moving along the z-direction,

with a specified electric field amplitudeEo

The time-average power flow density is

Power in a lossy medium decays as exp(-2 z)!

E ( ) exp( )exp( )

H ( ) exp( )exp( )exp( )

x o

o

y

z E z j z

Ez z j z j

= =

{ }{ }

**

2 22 2

1 1( ) Re E H Re2 2

1 1

Re cos2 2

j z z j z joo

z zj

o o

EP t E e e e e e

e e

E e E

G G G

= = = =


Consider the same wave with a specified amplitude for the


55/144


Consider the same wave, with a specified amplitude for themagnetic field

The time-average power flow density is expressed as

If is the attenuation constant for the electromagnetic fields 2 is the attenuation constant for power flow.

H ( ) exp( )exp( )

E ( ) exp( )exp( )exp( )

y o

x o

z H z j z

z H z j z j

= =

{ }*2 2

1( ) Re

21

cos2

j z z j z j

o o

zo

P t H e e H e e e

H e

G

= =


If the wave is generated by an infinitesimally thin sheet of uniform


56/144


If the wave is generated by an infinitesimally thin sheet of uniform

currentJso (embedded in an infinite material with conductivity )we have for propagation along the positive z-direction (normal tothe plane of the current sheet):I

For this ideal case, an identical wave exists, propagating along the

negative z-direction and carrying the same amount of power.

22

2 2

( ) cos

8

so soo o

zso

J JH E

JP t e

G = =

=


Poynting Theorem


57/144


Poynting Theorem

Consider the divergence of the time-dependent power flow density

The curls can be expressed by using Maxwells equations

This is the differential form ofPoynting Theorem.

( )( ) ( ) ( ) ( ) ( ) ( ) ( )P t E t H t H t E t E t H tG G G G G G G = =

2 2 2

( ) ( ) ( ) ( ) ( )

1 1( ) ( ) ( )

2 2

H EP t H t E t E t E t

t t

E t E t H tt t

G GG G G G G =

= Density of

dissipatedpower

Rate of change

of stored electricenergy density

Rate of change

of stored magneticenergy density


Now integrate the divergence of the time-dependent power over a


58/144


Now, integrate the divergence of the time-dependent power over a

specified volume V to obtain the integral form ofPoynting theorem

2 2 2

Power Flux through S( ) ( )

1 1( ) ( ) ( )

2 2

V S

V V V

P t dV P t ds

E t dV E t dV H t dVt t

G G

w = = =

Power dissipated

in volumeRate of change

of electric energystored in volume

Rate of changeof magnetic energy

stored in volume


Typical applications


59/144


Typical applications

L

inP t( )G

outP t( )G

= ?

1 m2

2

Watts( ) ( ) exp( 2 )

m

( )1 Nepersln2 ( ) m

out

in

out inP t P t

P tL P t

L

G

G G

G = =



60/144


Example:

Pay attention to the logarithms:

2 2

Watts Watts( ) 30 ; ( ) 5 ; 20 m

m m

Nepers= 0.0448

m

in out P t P t LG G = = =

( ) ( )ln ln( ) ( )

out inin out

P t P t

P t P t

G G

G G

=


SURFACE A SURFACE B


61/144


Area = Area(A) = Area(B)

Power IN ( ) ( ) Area

Power OUT ( ) ( ) Area

( ) ( ) exp( 2 )

= Power IN Power OUPower dissipated T

A A

A

B B

B

B A

P t dS P t

P t dS P t

P t P t L

G G

G G

G G

= = = =

=

L

out

B

P t( ) Power OUT GinA

P t( ) Power IN GPower dissipated

between A and B?


Example


62/144


p

2Area = 5 m

2

8.2244637 General Lossy medium

130.88 0.725rad 130.88 41.5

; 1.0 cm; 1.0 GHz; 10 V/m

; ; 0.45755 S/m

34

40.0 Ne/m; ( ) 0.286 W/m ;

( ) ( ) exp( 2 )

in

out i

o

nB A

o o

P t

P t P t L

L f E

D

G

G G

= = =

= = = = = =

= == = 2Power IN Area ( )

Power OUT Area ( )

= Power IN PowerPower dissipat Te OUd

0.12845 W/m ;

1.43 W

0.6423 W

0.7876 W

in

B

P t

P t

G

G

= =

=

==


Incidence on Perfect Conductor


63/144


Consider first normal incidence at an interface between a dielectricand a perfect conductor. Total reflection occurs, as in a short-circuited transmission line.

Medium 11 = r1 o1 = r1oMedium 2Perfect

Conductor

2Incident wave

Reflected wave

z0

x

y

Interface{x,y}-plane

0

E

H

0

==G

G

E 1.0=


Because ofinterference between incident and reflected wave, there


64/144


is a standing wave in medium 1.

Medium 11 = r1 o1 = r1oMedium 2

Perfect

Conductor

2

z0

x

y

0

E

H

0

==G

G

EG

HG

oE2

o

E2

/ 2 /


Consider now incidence at an angle. We choose an electric field


65/144


perpendicular to the plane of incidence.

Medium 11 = r1 o1 = r1o

Medium 2

Perfect

Conductor

2

z0

x

y

0

E

H

0==G

G

E

G

HG

Gx

EG

HG


Only the normal component, corresponding to zis reflected.


66/144


Note: zz >

Medium 11 = r1 o1 = r1oMedium 2Perfect

Conductor

2

z0

x

y

0

E

H

0

==G

G

EG

HG

oE2

oE2

z/ / 2 =2 / z z =


2 2 2


67/144


max

max

max

m

max

in

First minimum

First maximum

4 4 cos

45 0.35

15 0.259

2 2 co

2 2

s

2

; cos

Exa

0 0.25

mples:

z

z

z

z

z

z

z

z

z

D

D

D

=

= = =

= =

=

= = =

Medium 2

Perfect

Conductor

2

z0

x

y

0

E

H

0

==G

G

EG

H

G

G

x

EG

HG


If we place a second perfect conductor interface, parallel to thei th i id d l th di ti b fl ti


68/144


previous one, the wave is guided along the x-direction by reflection.

z0

x

y

0EH

0==G

G

E

G

H

G

G

EG

HG

Perfect

Conductor

2

Perfect

Conductor

2

0EH

0==G

G


Parallel Plate Waveguide


69/144


Assume uniform waves along they-direction ( )y

0 =

Assume no fringing effects w a>> Propagation along thez-direction

0

a

x

y

z

w

a


Maxwells equations


70/144


Maxwell s equations

E E H

det E

(1)

(2E H

E E EE (E

)

E

3)

H

H

z y xx y z

x z y

x y z y x z

ji i i

j

y z

jx y z z x

jx y

= = =

=



71/144


H (4)

(5

H E

det H H )

(6)

E

E

E

H

H H HH H

z y xx y z

x z y

x y zy x z

ji i i y z

j

x y z z x

jy

j

x

= = =

=


From (1) & (2) & (5)


72/144


From (1) & (2) & (5)

Wave equation forTransverse Electric (TE) modes

2

2

2

2

2 2 22 2

E H

E H

E E E

(1)

(3)

(5)

H

From

H

E

y x

y z

y y yx z

y

jz zz

jx xx

jz x z x

j

= =

+ = =


From (4) & (6) & (2)


73/144


From (4) & (6) & (2)

Wave equation forTransverse Magnetic (TM) modes

2

2

2

2

2 2 22 2

E E

H E

H E

H H H

Fro

(4)

(6)

(2)m H

y x

y z

y xy z y

y

jz zz

jx

z x

j

xx

jz x

= = + = =


Transverse Electric (TE) modes


74/144


This solution satisfies the boundary conditions:

E

H

E

H

y

x

x a

0E 0

== =Boundary Conditions

( ) ( )E sin 2 zx xj z j zj x j xoy o x EE x e j e e e = = x

z

H

E


We have


75/144


We have

and from boundary conditions at the conductor plates

22 2 2 2

2

4x z

= = + =

( )0

sin 0

0

1,

)

2, 3

) y

x x

x E

a a m

m

x a

= = =

===

1 / 22 22

cos

sin 12

x

z

m

a

m m

a a

= = = = =


For each possible index m we have a mode of propagation. Modes


76/144


For each possible index m we have a mode of propagation. Modesare labeled TE10 , TE20 , TE30 , .

The first index gives the periodicity (number of half sinusoidaloscillations) between the plates, along the x-direction. The second

index is zero to indicate uniform solution along the y-direction.

Note that the solution m = 0 (or mode TE00) is not acceptable,because it would require a field configuration with uniform electricfield tangent to the metal plates. This is an unphysical boundary

condition, which is possible only for the case of trivial solution ofzero field everywhere.

E

H

x z

m00TE 0

0

Unphysical !!!

& ==


A mode can propagate only if the frequency is sufficiently high, so


77/144


p p g y q y y g ,

that z > 0.We have the cut-off condition when

Exactly at cut-off the wave would bounce between the plates,without propagation along the wave guide axis.

12 2 2

2

2 2

1 02

2

x cc

z

pc

c

m a

a m

m m

a a

f mv m

a

Cut - off frequency for mode

= = = = = = =

= =


When the frequency is below the cut-off value


78/144


The mode attenuates entering the guide as an evanescent wave.

12

22

12 2

2

( )

1

1

2

12

2

j j

z

z

c

z

caf

j e

fm

m

a

mj

a

m

a

e

>

= =

= =

< > =

=


Transverse Magnetic (TM) modes


79/144


The magnetic field can be tangent to the conductor plates. In fact, it

is maximum at the plates, since the reflection coefficient is H= 1.The solution is of the form:

( ) ( )H cos2

z zx xj z j zj x j xoy o x

HH x e e e e

= = +x

z

E

H

E

H

E

H


At the metal plates


80/144


Modes are labeled TM00 , TM10 , TM20 , TM30 ,

Note that the solution m = 0 (or mode TM00) is acceptable, becausethe magnetic field can be uniform and tangent to the metal plates.

( )H Hcos 1

0))

0, 1, 2, 3

y o

x x

xx a a a m

m

= = = = =

=

E

H

x

m00TM 0

0

Physical !!!

& =


The TM00 mode is like a portion of a uniform plane wave slidingbetween the plates of the waveguide.


81/144


Both the electric and the magnetic field are transverse (normal tothe guide axis) therefore this mode is usually known as TransverseElectro Magnetic mode (TEM). For this mode we have

The TEM mode is the fundamental mode. It can propagate at anyfrequency.

All other TM modes have the same cut-off frequency condition asthe TE modes with identical indices.

2 20

0

z x cc

pc

c

v

f Cut - off frequency for TEM mode

= = = = = =


The apparent wavelength along the guide axis is also called theguide wavelength


82/144


( ) ( )

2

2 2

2 2

sin

2cos

cos sin

1 /

1

2

1 /

g zz

xc c

cc

g

c c

c

f f

ma

f

f

Since :

= = = = = = =

= = =

=

=


There is a corresponding apparent velocity along the guide axis, orguide phase velocity


83/144


The expressions for guide wavelength and guide velocity are alsoidentical for TE and TM modes.

( ) ( )2 2s

1 / /

n

1

i

p

p

p

z

pz

c c

z

v v

v

v

f f

= =

= =


Consider a TE wave with electric field amplitude Eo. The totalamplitude of the magnetic field is


84/144


The magnetic field has two components with amplitude

oo

EH =

2

sin sin

H sin sin

H cos cos

2

cos

o ox o

g

o oz

g

o

c

c

E E

m

E EH

a

H

since :

since :

= = =

= = = = =

= =


Consider a TM wave with magnetic field amplitude Ho. The totalamplitude of the electric field is


85/144


The electric field has two components with amplitude

o oE H

0

2

sin sin

E sin sin

E co

2

s c

os

o

c

s

x o og

z o o o

g

c

c

E H H

E

a

H H

m

since :

since :

= = =

= = =

= =

= =


The xcomponent of the magnetic field for the TE wave isassociated with the wave moving along the zdirection (axis of the


86/144


waveguide). The guide impedance for the TE modes is defined as

The xcomponent of the electric field for the TM wave is associatedwith the wave moving along the zdirection (axis of the waveguide).The guide impedance for the TM modes is defined as

( ) ( )2 2g 1 / 1 /c cTMg

f f

= = =

( ) ( )g

2 2

1 1

1 / 1 /

g

TE

c cf f

= = =


If there is a discontinuity along the guide axis (e.g., a change indielectric medium), one can use transmission line theory to analyze

th d b h i i di id ll i t f t i i d


87/144


the mode behavior individually in terms of transmission andreflection. Sections of the guide can be replaced by a transmissionline, with the guide impedance as the characteristic impedance.

Note that the guide impedance is a function of frequency for all

modes, except for the fundamental TEM mode

The reflection coefficient at a discontinuity is of the usual form

The power reflection coefficient is again ||2 and the powertransmission coefficient is 1||2.

( )2g 1 0( ) 0 /TEMcf TEM f

= =

2 1

2 1

g g

g g

= +


The phasor fields for TE modes are summarized as follows

El t i Fi ld i l t t


88/144


Electric Field: a single transverse component

Magnetic Field: two components, obtained from Faradays law:

( ) E sin sinz zj z j zo x y o ymE x e i E x e ia

= =

E E

H si

E ( )

n

cos

z

z

y x x zy x y z

j zo x

j

z

zo

xz

j Hi i

z xm

E x e ia

mjE x e

i H i

ia

= +

+ = =

+


The following relationships are useful to introduce the mediumimpedance in the TE field expressions above


89/144


Note once again that there is no allowed solution form = 0 in the

case of TE modes. The first allowed TE mode is the TE10.

g

sin

cos

1

x

cc

z

TEg g

= = = = = =

=


The phasor fields for TM modes are summarized as follows

Magnetic Field: a single transverse component


90/144


Magnetic Field: a single transverse component

Electric Field: two components, obtained from Amperes law:

( ) H cos cosz zj z j zy o x y o ymH x e i H x e ia

= =

H H

E c

H ( )

os

sin

z

z

y x x zy x y z

j zo x

x

z

j zo z

z

i i

z xm

H x e ia

mjH x

j E E

e ia

i i

+ = =

+

= +


The following relationships are useful to introduce the mediumimpedance in the TM field expressions above


91/144


The field expressions simplified for the TEM mode resemble a

uniform plane wave propagating along the axis of the guide

Remember, the TM00 or TEM mode is the fundamental mode.

gsin

cos

z

Tg

c

Mg

x

c

= = = = = =

=

H

E

z

z z

j zy o y

j z j zx o x o x

H e i

H e i E e i

== =


Wave Dispersion

A plane wave by itself does not carry information For transmissionf i f ti it i t h f t f fi it


92/144


A plane wave by itself does not carry information. For transmissionof information it is necessary to have a frequency spectrum of finitesize, as obtained by modulation of a wave, for instance.

Information does not travel at the guide phase velocity, but itpropagates according to the group velocity

To illustrate the nature of the group velocity, consider the simple

case of an amplitude modulated signal (assume >> )

pz

g

v

dvd

group ve

guide phase veloc

lo

ity

city

= =

( )) ( )( ) 1 cos cosy o oE t E m t t+


This signal has three components

( ) cos cos cosE t E t tm E t


93/144


( ) ( ) ( )( )

( )( )

cos

co

( )

cos2

cos cos cos

s

2

o

o

y o

o

o o o

o

o o

E t

mE

mE t t

t t

E t t

E

m

t

E t

= +

+

=

++

o o+o


The line at angular frequency o is the carrier. The modulationinformation is contained in the two side frequency lines at o

d +


94/144


and o+.Now, consider an amplitude modulated wave propagating in a

parallel plate wave guide. The zcomponents of the propagationfactor depend on frequency and are different for the two sidefrequencies. In general, we have

21 z mm z z z z= = +

1 m 2

oo+m


The dispersion relation () is approximately linear when


95/144


Under this assumption, we can write

( )( ) ( )( ) ( )cos

2

cos

(

2

, cos)

o o z

o o z z

zy o

z

o

mE t

E z t E t

z

z

mE t z

+ +

+

+

1

zm oz( ) z2

o +mo

z


( )( , ) scoy o o zE z tm

zE t=


96/144


The modulation envelope travels at the group velocity

( ) ( )

( ) ( )( )( ) ( )

( ) ( )cos

cos cos

1 cos cos

cos

2

2

coso o

o o z

o o

o o

z z

o z o z

z z

z z

E t z

mE

mE

mE t

t z t z

E m t z

t z t z

z

z

t

tz

modulated amplitude

++

= +

= +

+

/g zv =


15 0 v

MODULATION ENVELOPE


97/144


4.000.00 0.50 1.50 2.00 3.00

-10.0

-5.0

0.0

5.0

15.0

-15.0

10.0

gz

v =

pzz

v=

CARRIER


For the parallel plate wave guide

1 / 2 1 / 22 2f


98/144


( ) ( )( ) ( )

2

2 2

2 2

2 2

1 1

1 / 1 /

1 / 1 /

cz

c

p ppz

zc c

g p c p cz

pz g p

pz p g p

f

f

v vv

f f

d

v v

v v

v f fd

v

v v v vSince

= = = = = = = =

=


Information travels at the group velocity, which is always less thanthe corresponding phase velocity in the given medium.

The group and phase velocities for each mode propagating in the


99/144


The group and phase velocities for each mode propagating in the

wave guide are frequencydependent. This means that frequencycomponents of a broadband signal travel at different speed andchange their phase relationship as they propagate along the wave

guide. The group and phase velocities of the modes are alsomodedependent. This means that if a signal is distributed over anumber of different modes, the components spread out over timeduring propagation.

This phenomenon is called dispersion. Wave guides are in generaldispersive media.

Note: For the fundamental TEM mode in parallel plate wave guide

0c pz p g f v v v no dispersion = =


gv

Slope


100/144


Dispersion diagram

z1 2

1c

2

g

pz

v

v

0 at cutoff

pv1

=Slope

pzvSlope

g


The power flow follows the Poynting vector, with the same directionas the propagation vector. The group velocity accounts for the

effective motion of the power flow in the direction parallel to theaxis of the wave guide


101/144


axis of the wave guide.

P

gL v t2 sin = L L

pL v t2 =

22 si

si

n

npp gg

L

vt vv

L

v= = =


The guide phase velocity corresponds to the apparent motionillustrated by the following diagrams

L v t/ sin / 2


102/144


P

Lp

L v t/ 2

P

L

pzL v t/ sin / 2=

pzL v t/ sin / 2=

pL v t/ 2


Therefore, we obtain for the guide phase velocity

22 pvLL


103/144


From the results above, we have again

2

s n

2

siin

p

ppz

pz

vL

vt

v

Lv= = =

2

sinsin

p

p

p

p p

pz

g

pz g

vv

v v

vv

v

v v

= =


Rectangular Wave Guide


104/144


Assume perfectly conducting walls and perfect dielectric filling thewave guide.

a wiCon derventio is always the side of the wave gun : ide.

a

b

x z

y


It is useful to consider the parallel plate wave guide as a startingpoint. The rectangular wave guide has the same TE modes

corresponding to the two parallel plate wave guides obtained byconsidering opposite metal walls


105/144


considering opposite metal walls

TEm0

E

TE0n

E

a

b


The TE modes of a parallel plate wave guide are preserved ifperfectly conducting wallsare added perpendicularly to the electric

field.


106/144


On the other hand, TM modes of a parallel plate wave guidedisappear if perfectly conducting walls are added perpendicularly tothe magnetic field.

EThe added metal plate doesnot disturb normal electricfield and tangent magneticfield.H

HThe magnetic field cannotbe normal and the electric

field cannot be tangent to aperfectly conducting plate.

E



107/144


The remaining modes are TE and TM modes bouncing off each wall,all with non-zero indices.

TEmn

TMmn


We have the following propagation vector components for themodes in a rectangular waveguide

x y z2 2 2 2 2 = = + +


108/144


At cut-offwe have

x y

x yz g

m n

a b

m n

a b

222 2 2 2

2 22

;

2 2

= = = = = =

( )z c m nfa b

2 222 0 2

= =


The cut-off frequencies for all modes are

c m nfa b

2 2

12

= +


109/144


with cut-off wavelengths

with indices

a b2

TmTE M

m

mn

nn

0, 1, 2, 3, 1, 2, 3,0, 1, 2, 3, 1, 2, 3,

(but

m mode

not allowed)

o s

0

sde

== == ==

c

m n

a b

2 2

2 = +


The guide wavelengths and guide phase velocities are

g zz 2 2

2 2 = = = =


110/144


pz

z c

c

v

ff

2 2

1 1 1 1

11

= = =

z

c

c

m n

a b

f

f

2

2 2

11

= =


The fundamental mode is the TE10with cut-off frequency

( )c mf TE a10 2 =


111/144


The TE10 electric field has only the y-component. From Ampereslaw

z

x y z

x y z

yi i i

x y

j

z

E

det

E = 0 E E = 0

E H

=

y x

x

j

z

z

E H

E

=

zxE

y

y x

j

x y

H 0

E E

= zj H


The complete field components for the TE10mode are then

E sin z zy o jxEa

e =


112/144


with

H sin

H cos

E1E

E1

z

z

j zz

x o

j zz o

y z

y

x

xj

jE e

a

j xE

z j

a

aj ze

a

= = =

=

=

22

za =


The time-average power density is given by the Poynting vector

{ } ( )1 1

Re Re2 2

*( ) E H { sin z yj z

oP t ixE e

= =


113/144


{ } ( )

( ) ( )

( ) ( ) ( )

*

* *

2 2

E

H

2 21 2Re2

sin co( )}

sin s ni c

s

os

z z

y

xj z zj

o z o

z

z

zo o

x j xE e E e

a ai i

E Ex x xi ja a a

a

a

a

=

( )

22

sin2

x

o z

z

i

E x

ia

=


The resulting time-average power density flow is space-dependent

on the cross-section (varying along x, uniform along y)

22( ) sin

oE xzP t i =


114/144


The total transmitted power for the TE10 mode is obtained byintegrating over the cross-section of the rectangular wave guide

( ) sin2

P t iza =

) ( )2 2

2

2 2( ) sin0 0 02 2

1 1

sin 22 2 4

sino o

b

o

E Ex aa b z zP t btot a

E abzb u u

dx dy u du

=

= = =

=

2

2 2

0area

average 1

|E( , )|

1

4 2 2

TE

o o z

x y

E Ezab ab

= =


The rectangular waveguide has a high-pass behavior, since signalscan propagate only if they have frequency higher than the cut-off

for the TE10 mode.

For mono mode (or single mode) operation only the fundamental


115/144


For mono-mode (or single-mode) operation, only the fundamentalTE10 mode should be propagating over the frequency band ofinterest.

The mono-mode bandwith depends on the cut-off frequency of thesecond propagating mode. We have two possible modes toconsider, TE01and TE20

( )

( ) ( )

01

20 10

1

2

1 2

c

c c

f TEb

f TE f TEa

=

= =


( ) ( ) ( )c c cf TE f TEa

ab f TE01 20 102

2

1

= = =If

Mono-mode bandwidth


116/144


0 ( )cf TE10 ( )cf TE20( )cf TE01f

( ) ( ) ( )c c cf TE f TE fab TEa 10 01 202

< IfMono-mode bandwidth

0 ( )cf TE10 ( )cf TE20)cf TE01 f


Mono-mode bandwidth

( ) ( )c cab f TE f TE 20 012

<


117/144


In practice, a safety margin of about 20% is considered, so that theuseful bandwidth is less than the maximum mono-mode bandwidth.This is necessary to make sure that the first mode (TE10) is well

above cut-off, and the second mode (TE01 or TE20) is stronglyevanescent.

0 ( )cf TE10 ( )cf TE01)cf TE20

Useful bandwidth

0

f

( )cf TE10 ( )cf TE01)cf TE20

Safet mar in


( ) ( )10 01c cf TE f TEa b =(square wave guide)If


118/144


In the case of perfectly square wave guide, TEm0 and TE0n modes

with m=n are degenerate with the same cut-off frequency.

Except for orthogonal field orientation, all other properties ofdegenerate modes are the same.

0 ( )cf TE10 ( )cf TE20( )cf TE01

f

( )02cf TE


Example - Design an air-filled rectangular waveguide for thefollowing operation conditions:

a) 10 GHz is the middle of the frequency band (single-modeoperation)

b) b = a/2


119/144


b) b = a/2

The fundamental mode is the TE10 with cut-off frequency

For b=a/2, TE01 and TE20 have the same cut-off frequency.

co o

cf TEa aa

810

1 3 10( ) Hz2 22 = =

co o

co o

c c cf TE

b a a ab

cf TE

a aa

8

01

820

1 2 3 10( ) Hz

2 22

1 3 10( ) Hz

= = = = = =


The operation frequency can be expressed in terms of the cut-offfrequencies

01 10( ) ( )( ) c cf TE f TE

f f TE


120/144


10

10 01

8 89

2 2

( )2

( ) ( ) 10.02

1 3 10 3 1010.

2.25 10 1.125 1

0

02

10

2 2

c

c c

a

a

f f TE

f TE f TE GHz

m b

a a

m

= ++= =

= +

= = =


Maxwells equations forTE modes

Since the electric field must be transverse to the direction ofpropagation for a TE mode, we assume

E 0=


121/144


In addition, we assume that the wave has the following behavioralong the direction of propagation

In the general case of TEmn modes it is more convenient to startfrom an assumed intensity of the z-component of the magnetic field

zj ze

( ) ( )H cos cos

cos cos

z

z

j zz o x y

j zo

H x y e

m nH x y e

a b

=

=

E 0z =


Faradays law for a TE mode, under the previous assumptions, is

E Hj =


122/144


E E H

det E E H

E E

(1)

(2)

0E E H (3

E H

)

y z y xx y z

x z x y

x yy x z

j ji i i z

j jx y z z

jx y

j

= =

= =

=


Amperes law for a TE mode, under the previous assumptions, is

H Ej =


123/144


(4)

(

H H E

det H H E

H H HH H E 0

5)

(6)

z z y xx y z

z x z y

x y zy x z

j ji i i y

j jx y z x

jx y

+ =

=

= =


From (1) and (2) we obtain the characteristic wave impedance for

the TE modesEE yx

TE

= = =


124/144


At cut-off

H HTE

y x z

= = =

2 2

2 2

2

0 2

1

c

pc

c c

z

c

m nf

a bv

f

m n

a b

= +

= = =

+

=


In general,

( )

2 2 222 2

2 41

2z

m n

a b

= =


125/144


and we obtain an alternative expression for the characteristic waveimpedance ofTE modes as

( )

22 1

2

zc

ca b

=

1 22

1TE oz c

= =


From (4) and(5) we obtain

H H E H

1 H 1 HH

z z y x TE y

z z

j j jy

+ = =


126/144


2

2 2

2

2 2

H HH

2

H HH

H

2

H

H H E

z zy

TE z

zz

z x z y T

cz z zy z

z

E x

cz z zx zz

jj

j j y yj j

j j

y y

jx

jx

jx

= =

= =

= =

= =


We have used

2

2 2 2 2 2 2

1 1 1

2cz x y m n

b

= = = + +


127/144


The final expressions for the magnetic field components of TEmodes in rectangular waveguide are

2

2

H sin cos2

H cos sin2

H cos cos

z

z

z

j zc

x z o

j zcy z o

j zz o

m m nj H x y e

a a b

n m nj H x y e

b a b

m nH x y ea b

=

=

=

a b


The final electric field components for TE modes in rectangularwave guide are

E H


128/144


2

2

E H

cos sin2

E H

sin cos2

E 0

z

z

x TE y

j zcTE z o

y TE x

j zcTE z o

z

n m nj H x y e

b a b

m m nj H x y e

a a b

=

=

=

=

=


Maxwells equations for TM modes

Since the magnetic field must be transverse to the direction ofpropagation for a TM mode, we assume

H 0z =


129/144


In addition, we assume that the wave has the following behavioralong the direction of propagation

In the general case of TMmn modes it is more convenient to startfrom an assumed intensity of the z-component of the electric field

zj ze

( ) )E cos cos

cos cos

z

z

j zz o x y

j zo

E x y e

m nE x y e

a b

=

=

z


Faradays law for a TM mode, under the previous assumptions, is

E Hj =


130/144


E E H

det E E H

E E

(1)

(2

EE H

)

(3)E

z z y xx y z

z x z y

x y zy x z

j ji i i y

j jx y z x

jx y

+ =

=

=


Amperes law for a TMmode, under the previous assumptions, is

H Ej

=


131/144


(4)

(5)

(6

H E

det H E

H H 0 H H E )

x y zz y x

z x y

x y y x z

i i i j j

j jx y z

jx y

= =

=


From (4) and (5) we obtain the characteristic wave impedance forthe TM modes

EE

H H

yx zTM

y x

= = =


132/144


We can finally express the characteristic wave impedancealternatively as

Note once again that the same cut-off conditions, found earlier forTE modes, also apply forTM modes.

y

2

1zTM oc

= =


From (1)and(2) we obtain

EE E Hy

z z y xTM

j j jy

+ = =


133/144


2

2 2

2

2 2

1 E 1 EE

/

EE

E E

E 2

E E

E H

E2

z z

y TM zz

z

xz x z y

cz z z

y zz

T

c

M

z z zx z

z

j

j

j j y yj j

j j

y y

jj

x

jx

x

= =

= =

= =

= =


The final expressions for the electric field components ofTM modesin rectangular waveguide are

2


134/144


2

2

E cos sin

2

E sin cos2

E sin sin

z

z

z

j zcx z o

j zcy z o

j zz o

m m nj E x y e

a a b

n m nj E x y e

b a b

m nE x y ea b

=

=

=


The final magnetic field components forTM modes in rectangularwave guide are

H E /x y TM=


135/144


Note: all the TM field components are zero if either x=0 or y=0.This proves that TMmo or TMon modes cannot exist in therectangular wave guide.

2

2

sin cos2

H E /

cos sin2

H 0

z

z

j zczo

TM

y x TM

j zcz oTM

z

n m n

j E x y eb a b

m m nj E x y ea a b

= =

= =


Field patterns for the TE10mode in rectangular wave guide

x

zSide view


136/144


Cross-section

y

x

E

H

y

x

zTo view

E

H


The simple arrangement below can be used to excite the TE10 in a

rectangular waveguide.


137/144


The inner conductorof the coaxial cable behaves like an antennaand it creates a maximum electric field in the middle of the cross-section.

Closed end

TE10


Waveguide Cavity Resonator

d

xm

a

=


138/144


The cavity resonatoris obtained from a section of rectangular waveguide, closed by two additional metal plates. We assume againperfectly conducting walls and loss-less dielectric.

x

y

z

a

b

y

z

a

nb

l

d

==


The addition of a new set of plates introduces a condition for

standing waves in the zdirection which leads to the definition ofoscillation frequencies

2 2 21

2c

m n lf

a b d

= + +


139/144


The high-pass behavior of the rectangular wave guide is modified

into a very narrow pass-band behavior, since cutoff frequencies ofthe wave guide are transformed into oscillation frequencies of theresonator.

2 a b d

In the wave guide, each mode isassociated with a band of frequencieslarger than the cut-off frequency.

In the resonator, resonant modes canonly exist in correspondence ofdiscrete resonance frequencies.

0 0f f1cf 2cf 1rf 2rf


The cavity resonator will have modes indicated as

The value of the index corresponds to periodicity (number of halfsine or cosine waves) in the three directions. Using z-direction as

m lmnl nTMTE


140/144


sine or cosine waves) in the three directions. Using z direction asthe reference for the definition of transverse electric or magnetic

fields, the allowed indices are

The mode with lowest resonance frequency is called dominant

mode. In the case ad> b the dominant mode is the TE101.

0, 1, 2, 3 1, 2, 3

0, 1, 2, 3 1, 2, 3

1, 2, 3 0, 1, 2, 3

m n

m m

n n

l

MT T

l

E

= === =

=

with only one zero indexor allowed


Note that for a TM cavity mode, with magnetic field transverse tothe z-direction, it is possible to have the third index equal to zero.This is because the magnetic field is going to be parallel to the thirdset of plates, and it can therefore be uniform in the third direction,with no periodicity.

Th l t i fi ld t ill h th f ll i f th t


141/144


The electric field components will have the following form that

satisfies the boundary conditions for perfectly conducting walls

E cos sin sin

E sin cos sin

E sin sin cos

x ox

y oy

z oz

m n lE x y z

a b d

m n lE x y z

a b d

m n lE x y za b d

=

=

=


The magnetic field intensities are obtained from Amperes law

H sin cos cosz y y zx E E m n lx y zj a b d

=


142/144


Similar considerations for modes and indices can be made if the

other axes are used as reference for transverse fields, leading toanalogous resonant field configurations.

H cos sin cos

H cos cos sin

x z z x

y

y x x yz

E E m n l

x y zj a b d

E E m n lx y z

j a b d

=

=


A cavity resonatorcan be coupled to a wave guide through a smallopening. When the input frequency resonates with the cavity,

electromagnetic radiation enters the resonator and a lowering in theoutput is detected. With carefully tuned cavities, this scheme canbe used forfrequency measurements.


143/144


OUTPUT

INPUT

Movable piston changesthe resonance frequencies


Examples of resonant cavity excited by using coaxial cables.


144/144


The termination of the inner conductor of the cable acts like an

elementary dipole (left) or an elementary loop (right) antenna.

E

H

E

H

Documents

Review of Electromagnetic Waves & Waveguides1.pdf