UNIFORM PLANE WAVES (PROPAGATION IN FREE SPACE)€¦ · PROPAGATION IN DIELECTRICS Assuming a simple dielectric, the wave equation is written as: 2E E (Wave2 eqation for E ) ... that

Dr Naser Abu-Zaid Lecture notes in electromagnetic theory 1 Referenced to Engineering electromagnetics by Hayt 8th

edition 2012

Dr Naser Abu-Zaid Page 1 7222012

UNIFORM PLANE WAVES (PROPAGATION IN FREE SPACE) Starting with point form of Maxwells equations for time varying fields in free space

tto

HBE

tto

EDH

0 E

0 H Let

xx zE aE ˆ

Then

y

y

ooyx

t

zH

tz

Ea

HaE ˆˆ

And

xx

oox

y

t

E

tz

Ha

EaH ˆˆ

Collecting results

t

zH

z

E y

ox

t

E

z

Hx

o

y

Good reminder of telegraphist equations

To obtain the wave equations differentiate the first wrt z and the second wrt t and rearranging to get

E for equation wave

ldimensiona One2

2

2

2

t

E

z

E xoo

x

Or reversing differentiations to get

H for equation wave

ldimensiona One2

2

2

2

t

H

z

H y

oo

y

And a general solution is given by

EE

v

ztf

v

ztftzEx 21

From which the velocity of wave propagation may be deduced (by substituting 1f

in the wave equation performing the indicated diffrsquos)


edition 2012


cs

mvoo

81031

TEM waves Transverse ElectroMagnetic waves implies E is perpendicular to H and both lying in a transverse plane (a plane normal to the direction of propagation)

Uniform Plane Waves UPW E and H fields have constant magnitude and phase in the transverse plane (Constant phase and amplitude) For sinusoidal waves

2

1

2

1

coscos

coscos

zktEzktE

v

ztE

v

ztE

tzEtzEtzE

oxooxo

p

xo

p

xo

xxtotalx

timeunit per shift phases

rad

distanceunit per shift phasem

radko

constant phase implies

space) free(in

0constant

constant

1

1

cvkdt

zd

dt

dzkt

dt

d

zkt

p

o

o

o

The wave number in free space is defined as

m

radc

ko

The wavenumber is a property of a wave its spatial frequency that is proportional to

the reciprocal of the wavelength It is also the magnitude of the wave vector (to be seen

later) The wavenumber has dimensions of reciprocal length so its SI unit is m-1

Simply the number of wavelengths per 2π units of distance

Also the wave length is given by


edition 2012


m

f

c

ko

o

2

2πby shifts phase

spatial the over which

distance

Maxwells equations and the wave equations may be written in frequency domain with the help of the transformation

jt

soso jtt

HEHB

E

soso jtt

EHED

H

0 sE

0 sH

3D WAVE EQUATIONS (FREE SPACE) Taking the curl of the first equation namely

sos j HE

Using the identity

FFF2

Substituting

sosss j HEEE 2

Using the rest of Maxwellrsquos equations

sos j HE 2

soos jj EE 2

soos EE 22

) foreqation (Wave E

EE

ationmholtz equVector Hel

sos k 22

With

m

radv

k oo

p

o

Similar approach may be followed to obtain the wave equation for the magnetic field


edition 2012


)for eqation (Wave equation Helmholtz Vector H

HH sos k 22

RELATION BETWEEN AND

For the previous assumption xx zE aE ˆ the wave equation reduces to

zEk

dz

zEdxso

xs 2

2

2

With the frequency domain solution given by

zjk

x

zjk

xoxsoo eEeEzE

0

But from Maxwells equation

sos j HE

zHj

dz

zdEyso

xs

Substituting the solution

zHj

dz

eEeEdyso

zjkx

zjkxo

oo

0

Differentiating and solving for zH ys

zjk

x

o

ozjk

xo

o

oys

oo eEeEzH

0

Identifying

zjk

yoyfoeHzH

zjk

yyboeHzH

0

xo

o

oyo EH

0

x

o

oyo EH

And the intrinsic impedance of free space is

377120

o

o

yo

xoo

H

E

Or


edition 2012


o

o

yo

xoo

H

E

It can be shown that

HaEEa

H

no

o

n or ˆˆ

Or

HEn aaa ˆˆˆ Where

na unit vector in the direction of propagation

Ea unit vector in the direction of

Ha unit vector in the direction of


edition 2012


PROPAGATION IN DIELECTRICS Assuming a simple dielectric the wave equation is written as

)for eqation (Wave22 EEE ss k

Where k is the wave number

Ex 121 Let m

Ae zj

y

o

x

o

s

070ˆ203ˆ402 aaH for a uniform plane

wave traveling in free space Find

1)

2) nstPH x 31321

3) origin the 0 tH


edition 2012


k

Allowing the permittivity to be a complex constant (to be explained later) implies that the wave number may be complex and it is called the complex propagation constant

The propagation constant of an electromagnetic wave is a measure of the change

undergone by the amplitude of the wave as it propagates in a given direction The

propagation constant itself measures change per metre but is otherwise dimensionless

The quantity measured such as voltage or electric field intensity is expressed as a

sinusoidal phasor The phase of the sinusoid varies with distance which results in the

propagation constant being a complex number the imaginary part being caused by the phase change

For a One dimensional problem xxs zE aE ˆ the wave equation reduces to

zEkdz

zEdxs

xs 2

2

2

Define

jjk

So the solution is given by

zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE

Transferring to time domain and considering only the forward part

zteEtzE zxox cos

Define the complex permittivity (dipole oscillations and conduction electrons and holes) as

rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk

Clearly from the time domain expression of tzE x the phase velocity is given

by

s

mv p

And the wave length is (distance required to change the phase by 2 )

m

2

And the magnetic field associated with the forward propagating part is (can be found through the use of Maxwellrsquos equations)

With the intrinsic impedance being a complex quantity given by

je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos

Then xE leads yH by And you may do the same for the backward wave

Lossless medium (Perfect dielectric)

ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro

xE and yH are in phase


edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH

But on the other hand

sssss jj EEEJH

Comparing the two equations

Also from the second equation


edition 2012


jjJ

J

d

c

cJ and dJ are o90 out of time phase and we identify the material as having

large losses or small losses depending on the magnitude of the loss tangent defined by

tantangnet Loss

is the angle by which dJ lead the total current density J

Good Dielectric Approximations

losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j

The above two exact expressions may be approximated using the binomial expansion

1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j

Why Complex Permittivity Loss mechanism occurs in dielectrics even in the absence of free electrons

0 this is due to rotation of the dipoles to align with applied time varying

field or due to the net shift of the electron cloud with respect to the positive

nucleus At high frequencies the polarization P of the material is out of time

phase with applied field This loss mechanism is modeled by a complex permittivity as shown previously even with zero conductivity

j

So again from Ampers law

tyconductividielectricastreated

bemayormechanismLoss

s

mechanismstorageEnergy

sss

d

jjj

EEEH

J

And the loss tangent shown earlier is

tantangnet Loss


edition 2012


Even if the conductivity is nonzero 0 but quite small still we may write

Ampers law as

)(

tyconductividielectricJJ

EEE

EEH

astreatedbemayor

mechanismLossDielectric

s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s

typermittiviEffective

mechanismslossdielectricandConductor

s

sss

j

jj

J

tyconductiviEffective

EE

EEH

And the loss tangent is defined as

tantangnet Loss

Depth of penetration (Skin Depth) Consider a forward traveling wave in a lossy dielectric

zteEtzE zxox cos

At

1z the wave is attenuated by a factor of

amplitudemaximum sit Of

1

1

370

eee z

The quantity

1

is called the skin depth or depth of penetration


edition 2012


Illustration of skin depth


edition 2012


POYNTINGrsquoS THEOREM (POWER THEOREM)

t

DJH

Left dot both sides with E then using the identity

HEEHHE

And with some vector manipulations one can obtain (Follow text book)

Theorem sPoynting the of form alDifferenti

2

1

2

1

HBEDJEHE

tt

Integrating over a volume v enclosed by a surface s

vv v vt

dvt

dvdv HBEDJEHE2

1

2

1

Upon using the divergence theorem for LHS

v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2

1 volumeinto flowingpower Total

s

dsHE volumeofout flowingpower Total


edition 2012


Illustration of power balance for EM fields

And the instantaneous Poynting vector S (or instantaneous power density) is defined as

2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then

zzyyxx StzHtzE aaaHES ˆˆˆ

1) Perfect Dielectric (forward part only)

ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22

The average power density (time averaged Poyntingrsquos vector) is (for time harmonic case)

T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS

The above expression is easily evaluated using phasors by defining

Re2

1ss HES

Doing it for lossy dielectric (Sinusoidal wave)

Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012


Try to fill the rest

Good conductors Approximations (Skin effect)

losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238

Example 125 At frequencies of 1 100 and 3000MHz the dielectric constant

of ice made from pure water has values 0f 415 345 and 32 respectively

while the loss tangent is 012 0035 and 00009 also respectively If a UPW

with amplitude of 100(Vm) z=0 is propagating through the ice fine the time

average power density z=0 and z=10m for each frequency


edition 2012


oj45

Consider a forward traveling wave

zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11

the wave is attenuated by a factor of


1

1

370

eee z

f1

Which is the skin depth again or depth of penetration

For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066

After a few skin depths within the conductor all fields are almost zero

2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012


Attacking the power problem (in good conductors)

zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2

The total average power (loss) crossing the conductor surface at 0z

22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS

What result would be obtained for the power loss if it is assumed that the total current is distributed uniformly in one skin depth

To calculate the total current crossing the surface at 0x rewrite in phasor form

zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo

Assuming this current is distributed uniformly with current density S

IJ uniform

through the cross section bS then

4cos

2

t

J

S

IJ xo

uniform

Then the total instantaneous power dissipated in volume of one skin depth thickness is

4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ

And the time average power loss within this volume is

bLJ

dtPP xo

T

LL ins 4

2

This is exactly the same formula obtained before Conclusion The average power loss in a conductor with skin effect may be calculated assuming that the total current is distributed uniformly in one skin depth Or the resistance of width b and length L of an infinitely thick slab with skin effect is the same as the resistance of a rectangular slab of width b length

L and thickness without skin effect


edition 2012


S

LRdc

b

LRac

For a circular cross-section wire with radius a and at high frequencies

a

L

S

LR

2

More often the surface or skin resistance is defined as the

real part of the intrinsic impedance for a good conductor Thus

fRs

1

since

o

jj 452

111

b

L

With Skin effect Without Skin effect

b

L


edition 2012


WAVE POLARIZATION

Polarization is defined as The locus that the tip of the E field traces

as time varies for a fixed point in space Or the time-varying behavior of E at a given point in space

Consider a wave propagating in the negative z-direction

tztztz yyxx ˆˆ EΕ aa E

Assume each component have a sinusoidal time dependence and Since

tjet Recos then each component maybe written as a real

part of some complex quantity (complex phasor)

phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE

So the time domain representation is obtained as


edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE

Three cases are to be considered Case1 Linear polarization

210 nnxy with

Example Find the polarization (linear circular elliptical) and sense of rotation for the uniform plane wave whose electric field is given by

kztkzttz yx cos5ˆcos10ˆ aaE

Solution

Take 0z

ttt yx cos5ˆcos10ˆ0 aaE

And since tt coscos

ttt yx cos5ˆcos10ˆ0 aa E

2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012


Linearly polarized with an angle of o562650tan 1

Note that y 0x and 0xy So from the beginning we may state

that the polarization is linear but with what angle

Case2 Circular polarization

210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx

If the direction of propagation is in the positive z-direction then the phases for CW and CCW must be reversed Example Find the polarization (linear circular elliptical) and sense of rotation for the uniform plane wave whose electric field is given by

2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos

ttt yx sin10ˆcos10ˆ0 aa E

ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012


Since the wave is propagating in negative z-direction this is a RHCP circular polarization or CW polarization Note that

10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy

So from the beginning we may state that the polarization is CW circular

Case3 Elliptical polarization

210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012


The curve traced is a tilted ellipse

2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor

axismajor oAxial rati

And the tilt angle wrt the y-axis is

cos

2tan

2

1

2 22

1

yx

yx

EE

EE

Example Find the polarization (linear circular elliptical) sense of rotation axial ratio AR and the tilt angle for the uniform plane wave whose electric field is given by

2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E

This is an elliptical polarization with

52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor

axismajor ratio Axial

OB

OAAR

And the tilt angle is

22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012


Since the wave is propagating in negative z-direction this is a RH elliptical polarization or CW elliptical polarization Note that

yx EE 2

y 0x

and

02

2

1

20

2xy

So from the beginning we may state that the polarization is CW elliptical

copyFrom Wikipedia the free encyclopedia

x

y

0t

10

2

t

t

5


edition 2012


THE ELECTROMAGNETIC SPECTRUM

wherec = 299792458 ms is the speed of light in vacuum and

h = 662606896(33)times10minus34 J s = 413566733(10)times10minus15 eV s is Plancks constant[5]



edition 2012


copyFrom Microwave Engineering by David M Pozar copy2005


edition 2012


cs

mvoo

81031

TEM waves Transverse ElectroMagnetic waves implies E is perpendicular to H and both lying in a transverse plane (a plane normal to the direction of propagation)

Uniform Plane Waves UPW E and H fields have constant magnitude and phase in the transverse plane (Constant phase and amplitude) For sinusoidal waves

2

1

2

1

coscos

coscos

zktEzktE

v

ztE

v

ztE

tzEtzEtzE

oxooxo

p

xo

p

xo

xxtotalx

timeunit per shift phases

rad

distanceunit per shift phasem

radko

constant phase implies

space) free(in

0constant

constant

1

1

cvkdt

zd

dt

dzkt

dt

d

zkt

p

o

o

o

The wave number in free space is defined as

m

radc

ko

The wavenumber is a property of a wave its spatial frequency that is proportional to

the reciprocal of the wavelength It is also the magnitude of the wave vector (to be seen

later) The wavenumber has dimensions of reciprocal length so its SI unit is m-1

Simply the number of wavelengths per 2π units of distance

Also the wave length is given by


edition 2012


m

f

c

ko

o

2

2πby shifts phase


distance


jt

soso jtt

HEHB

E

soso jtt

EHED

H

0 sE

0 sH


sos j HE

Using the identity

FFF2

Substituting

sosss j HEEE 2


sos j HE 2

soos jj EE 2

soos EE 22


EE


sos k 22

With

m

radv

k oo

p

o



edition 2012



HH sos k 22



zEk

dz

zEdxso

xs 2

2

2


zjk

x

zjk

xoxsoo eEeEzE

0


sos j HE

zHj

dz

zdEyso

xs


zHj

dz

eEeEdyso

zjkx

zjkxo

oo

0


zjk

x

o

ozjk

xo

o

oys

oo eEeEzH

0

Identifying

zjk

yoyfoeHzH

zjk

yyboeHzH

0

xo

o

oyo EH

0

x

o

oyo EH


377120

o

o

yo

xoo

H

E

Or


edition 2012


o

o

yo

xoo

H

E


HaEEa

H

no

o

n or ˆˆ

Or






edition 2012





Ex 121 Let m

Ae zj

y

o

x

o

s



1)

2) nstPH x 31321

3) origin the 0 tH


edition 2012


k









zEkdz

zEdxs

xs 2

2

2

Define

jjk


zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE


zteEtzE zxox cos


rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


m

f

c

ko

o

2

2πby shifts phase


distance


jt

soso jtt

HEHB

E

soso jtt

EHED

H

0 sE

0 sH


sos j HE

Using the identity

FFF2

Substituting

sosss j HEEE 2


sos j HE 2

soos jj EE 2

soos EE 22


EE


sos k 22

With

m

radv

k oo

p

o



edition 2012



HH sos k 22



zEk

dz

zEdxso

xs 2

2

2


zjk

x

zjk

xoxsoo eEeEzE

0


sos j HE

zHj

dz

zdEyso

xs


zHj

dz

eEeEdyso

zjkx

zjkxo

oo

0


zjk

x

o

ozjk

xo

o

oys

oo eEeEzH

0

Identifying

zjk

yoyfoeHzH

zjk

yyboeHzH

0

xo

o

oyo EH

0

x

o

oyo EH


377120

o

o

yo

xoo

H

E

Or


edition 2012


o

o

yo

xoo

H

E


HaEEa

H

no

o

n or ˆˆ

Or






edition 2012





Ex 121 Let m

Ae zj

y

o

x

o

s



1)

2) nstPH x 31321

3) origin the 0 tH


edition 2012


k









zEkdz

zEdxs

xs 2

2

2

Define

jjk


zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE


zteEtzE zxox cos


rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



HH sos k 22



zEk

dz

zEdxso

xs 2

2

2


zjk

x

zjk

xoxsoo eEeEzE

0


sos j HE

zHj

dz

zdEyso

xs


zHj

dz

eEeEdyso

zjkx

zjkxo

oo

0


zjk

x

o

ozjk

xo

o

oys

oo eEeEzH

0

Identifying

zjk

yoyfoeHzH

zjk

yyboeHzH

0

xo

o

oyo EH

0

x

o

oyo EH


377120

o

o

yo

xoo

H

E

Or


edition 2012


o

o

yo

xoo

H

E


HaEEa

H

no

o

n or ˆˆ

Or






edition 2012





Ex 121 Let m

Ae zj

y

o

x

o

s



1)

2) nstPH x 31321

3) origin the 0 tH


edition 2012


k









zEkdz

zEdxs

xs 2

2

2

Define

jjk


zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE


zteEtzE zxox cos


rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


o

o

yo

xoo

H

E


HaEEa

H

no

o

n or ˆˆ

Or






edition 2012





Ex 121 Let m

Ae zj

y

o

x

o

s



1)

2) nstPH x 31321

3) origin the 0 tH


edition 2012


k









zEkdz

zEdxs

xs 2

2

2

Define

jjk


zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE


zteEtzE zxox cos


rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012





Ex 121 Let m

Ae zj

y

o

x

o

s



1)

2) nstPH x 31321

3) origin the 0 tH


edition 2012


k









zEkdz

zEdxs

xs 2

2

2

Define

jjk


zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE


zteEtzE zxox cos


rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


k









zEkdz

zEdxs

xs 2

2

2

Define

jjk


zjzxo

zjzxo

zjkxo

zjkxo

zxo

zxoxs

eeEeeE

eEeE

eEeEzE


zteEtzE zxox cos


rrororo jjj


edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


2

1

1

1

j

jjk

With

2

1

2

1

1

2

ReRe

jk

2

1

2

1

1

2

ImIm

jk


by

s

mv p


m

2



je

j

j

1

1

jzjzxozjzxo

zjkxozxoys

eeeE

eeE

eE

eE

zH


edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


Since

zteEtzE zxox cos

zteE

tzH zxoy cos



ro 0

01

1

2

ReRe

2

1

2

jk

1

1

2

ImIm

2

1

2

jk

s

mcv

rr

p

1

mf

c

rr

o

rr

22

0amp

1

1

r

ro

j

r

ro

ro

ro

e

j


edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


0amp

r

ro



edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


(Lossy Dielectrics)

ss

sss

j

jjj

EE

EEH


sssss jj EEEJH




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


jjJ

J

d

c



tantangnet Loss



losses Small1

1

or

2

1

1

jjjjk


edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

2

ReRe

jk

8

11ImIm

2

jk

21

j






j




s


sss

d

jjj

EEEH

J


tantangnet Loss


edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



Ampers law as

)(


EEE

EEH

astreatedbemayor


s


s

mechanismloss

Conductor

s

sss

dc

j

jj

So we may write

d


s



s

sss

j

jj

J


EE

EEH


tantangnet Loss


zteEtzE zxox cos

At



1

1

370

eee z

The quantity

1



edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



t

DJH


HEEHHE



2

1

2

1

HBEDJEHE

tt


vv v vt

dvt

dvdv HBEDJEHE2

1

2

1


v v vs

dvt

dvt

dvds HBEDJEHE2

1

2

1

s

v v v

d

dvt

dvt

dv

sHE

HBEDJE2

1

2


s



edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012




2m

WHES

For UPW with

xx tzEtz aE ˆ

yy tzHtz aH ˆ

Then



ztEtzE xox cos

ztE

tzH xoy

cos


edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


ztE

tzS xoz

2

2

cos

2) Lossy Dielectric

zteEtzE zxox cos

zteE

tzH zxoy cos

ztzte

EtzS zxo

z coscos 22


T

zxo

T

dtzteE

T

dtT

cos22cos

2

11

1

22

SS

cos2

11 22

zxo

T

eE

dtT

SS


Re2

1ss HES


Re2

1

jzjzxozjzxo eee

EeeES

cos2

1 22

zxo eE S


edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012




losses High1

1

or

2

1

1

jjjjk

2

1

1

j


1

2

111 2

xx

nnnxx

n

Hence

f

Solution

)(MHz

f

r

S

z=0

S

z=10m

1 415 012 2717 2582

100 345 0035

3000 32 00009 238







edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


oj45


zteEtzE zxox cos

fzteEtzEfz

xox

cos

fzteEEtzJfz

xoxx

cos

At

f

z11



1

1

370

eee z

f1


For copper m

S71085 o

fcopper

0660

f copper

Hz60 mm538

GHz10 mm41066


2

2

pv


edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


o

jj 452

111

If

fzteEtzEfz

xox

cos

Then

o

z

xoy

zteEtzH 45cos

2

yH lags xE by o45


edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



zteEtzE

z

xox cos

o

z

xoy

zteEtzH 45cos

2

zteJ

zteEtzJ

z

xo

z

xox

cos

cos

z

z

xoss eE aHES ˆ4

Re2

12

2


22

0 0 0

2

2

44

ˆˆ4

xoxo

z

b

y

L

x

z

z

z

xo

S

L

JbL

EbL

dxdyeEdP

aasS



zj

xoxs eJzJ

1

So

y

x

z

by

0y

Lx

UPW


edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


j

bJdydzeJdI xo

z

b

y

zj

xo

S

s

10 0

1 sJ

and

4cos

2

t

bJtI xo


IJ uniform


4cos

2

t

J

S

IJ xo

uniform


4cos

2

4cos

2

1

22

0 0 0

2

tbLJ

dxdydztJ

dvP

xo

z

b

y

L

x

xo

v

uniformLinsEJ


bLJ

dtPP xo

T

LL ins 4

2




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


S

LRdc

b

LRac


a

L

S

LR

2



fRs

1

since

o

jj 452

111

b

L


b

L


edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


WAVE POLARIZATION








phasor Complex

with

xjjkzx

jkzxx

tjxx

eeEeEzE

ezEtz

ˆ

ˆReE

phasor Complex

with

yjjkzy

jkzyy

tjyy

eeEeEzE

ezEtz

ˆ

ˆReE



edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


xxkztj

xx kztEeEtz x

cosReE

yy

kztj

yy kztEeEtz y

cosReE


210 nnxy with



Solution

Take 0z


And since tt coscos


2

1tan

cos10

cos5tan0

cos125cos125

cos5cos100

11

2

2222

t

tt

tt

ttt

E

E

x

y

0t 125

2

t

t

125

50tan 1


edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012






210

LHCPor CCW for 22

1

RHCPor CW for 22

1

n

n

n

EE

xy

yx


2

5cos10ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z

2

5cos10ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

5cos


ttt

tt

tt

ttt

tantancos10

sin10tan0

10100sincos100

sin10cos100

11

22

2222

E

E


edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



10 yx EE 2

5 y 0x

and

12

2

1

2

50

2

5xy



210

LHEPor CCW for 0

RHEPor CW for 0

2

Or

LHEPor CCW for 22

1

RHEPor CW for 22

1

n

n

n

n

EE

xy

xy

yx

x

y

0t

10

2

t

t


edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



2cos22

1

2cos22

1

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

OB

OAAR

axisminor



cos

2tan

2

1

2 22

1

yx

yx

EE

EE


2cos50ˆcos10ˆ

kztkzttz yx aaE

Solution Take 0z


edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012


2cos5ˆcos10ˆ0

ttt yx aaE

And since tt

sin2

cos


tt

tt

ttt

tan50tancos10

sin5tan0

sin5cos100

11

2222

E

E


52

2cos51025105102

1

2cos22

1

102

2cos51025105102

1

2cos22

1

224422

224422

224422

224422

yxyxyx

yxyxyx

EEEEEEOB

EEEEEEOA

25

10

axisminor


OB

OAAR


22cos

510

5102tan

2

1

2

cos2

tan2

1

2

22

1

22

1

yx

yx

EE

EE


edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012



yx EE 2

y 0x

and

02

2

1

20

2xy



x

y

0t

10

2

t

t

5


edition 2012







edition 2012




edition 2012







edition 2012




edition 2012



UNIFORM PLANE WAVES (PROPAGATION IN FREE SPACE)€¦ · PROPAGATION IN DIELECTRICS Assuming a simple dielectric, the wave equation is written as: 2E E (Wave2 eqation for E ) ... that

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