Transcript ENE 429 Antenna and Transmission Lines

DATE: 25-27/06/07
ENE 429
Antenna and
Transmission lines
Theory
Lecture 2 Uniform plane waves
Review

Wave equations
2

E

E
2
 E  
  2
t
t
2

H

H
2
 H  
  2
t
t

Time-Harmonics equations
2 E s   2 E s  0
2 H s   2 H s  0
where
  j (  j )
Time-harmonic wave equations
or
where
2 H s   2 H s  0
  j (  j )
This  term is called propagation constant or we can write
 = +j
where  = attenuation constant (Np/m)
 = phase constant (rad/m)
Transverse ElectroMagnetic wave
(TEM)
http://www.edumedia.fr/a185_l2-transverseelectromagnetic-wave.html
Solutions of Helmholtz equations

The instantaneous forms of the solutions
E  E0 e z cos(t   z )a x  E0e z cos(t   z )a x
H  H 0 e z cos(t   z )a y  H 0e z cos(t   z )a y

The phasor forms of the solutions
E s  E0 e z e j z a x  E0e z e j z a x
incident wave
reflected wave
H s  H 0 e z e j z a y  H 0e z e j z a y
Attenuation constant 

Attenuation constant determines the penetration of the
wave into a medium

Attenuation constant are different for different
applications

The penetration depth or skin depth, 
 1
E
E
is the distance z that causes
to reduce to 0 e
z = 1

z = 1/  = 
Good conductor
1
1
 

 f 
 At high operation
frequency, skin depth
decreases
 A magnetic material is not
suitable for signal carrier
 A high conductivity
material has low skin depth
Currents in conductor

To understand a concept of sheet resistance
from
L
1 L
R

A  wt
1 L
R
 Rsheet () L
t w
w
 Rsheet
1

t
sheet resistance
At high frequency, it will be adapted to skin effect resistance
Currents in conductor
Ex  Ex 0e z
J x   Ex 0e z
Therefore the current that flows through the slab at t   is
I   J x dS
; ds  dydz
Currents in conductor
From
I   J x dS

; ds  dydz
w
I     Ex 0e  z dydz
z 0 y 0
  w Ex 0e



 z
 I  w Ex 0



0
A.
Jx or current density decreases as the slab
gets thicker
Currents in conductor
For distance L in x-direction
V  Ex 0 L
R
Ex 0 L
V
1 L
L


 Rskin  
I w Ex 0  w
 w
R is called skin resistance
Rskin is called skin-effect resistance
For finite thickness,
t
w
I     Ex 0e z dydz  w Ex 0 (1  e t )
z 0 y 0
 Rskin 
1

t / 
 (1  e )
Currents in conductor
Current is confined within a skin depth of the
coaxial cable
Ex A steel pipe is constructed of a material for
which r = 180 and  = 4106 S/m. The two radii
are 5 and 7 mm, and the length is 75 m. If the total
current I(t) carried by the pipe is 8cost A, where
 = 1200 rad/s, find:
a)
The skin depth
b)
The skin resistance
c) The dc resistance
The Poynting theorem and power
transmission
Poynting theorem
 1 2
 1
2
(
E

H
)
d
S


J
E
dV


E
dV


H
dV




t 2
t 2
Total power leaving Joule’s law
the surface
for instantaneous
power dissipated
per volume (dissipated by heat)
Rate of change of energy stored
In the fields
Instantaneous poynting vector
S  EH
W/m 2
Example of Poynting theorem in DC
case
 1 2
 1
2
(
E

H
)
d
S


J
E
dV


E
dV


H
dV




t 2
t 2
Rate of change of energy stored
In the fields = 0
Example of Poynting theorem in DC
case
From
I
J
az
2
a
By using Ohm’s law,
J
I
E 
az
2
 a 
a
2
L
I2

 d   d  dz
2 2 
 ( a ) 0
0
0
1 L
2
 I


I
R
2
 a
2
Example of Poynting theorem in DC
case


Verify with  E  H d S
From Ampère’s circuital law,
 H dl  I
2 aH  I
H
I
2 a
a
Example of Poynting theorem in DC
case
I
I
I 2
S  E  H  2 az 
a  2 3 a 
2 a
a 
2 a 
2

I
Total power  S d S   2 3 a   d dz
2 a 
 I 2 a 2 L
I 2 L
2
 2 3  d  dz 


I
R
2
2 a  0
 a
0
W
Uniform plane wave (UPW) power
transmission

Time-averaged power density
Pavg

1
 Re( E  H ) W/m2
2
P   P avg d S
for lossless case, P avg  1 E e j z a x  Ex 0 e j  z a y
2 x0

1 Ex20
 P avg 
az W
2 
amount of power
Uniform plane wave (UPW) power
transmission
for lossy medium, we can write
E  Ex 0e z e j z e j a x
intrinsic impedance for lossy medium
H
1

a  E 

1

   e j
a z  Ex 0e z e j z e j a x
Ex 0

e  z e  j  z e j e  jn a y
n
Uniform plane wave (UPW) power
transmission
from
Pavg

1
 Re( E  H )
2
1  Ex20 2 z j 
 Re 
e e  az
2 

2
E
1 x 0 2 z

e
cos  a z
2 
W/m2
Question: Have you ever wondered why aluminum foil is not allowed in
the microwave oven?
Polarization

UPW is characterized by its propagation
direction and frequency.

Its attenuation and phase are determined by
medium’s parameters.

Polarization determines the orientation of the
electric field in a fixed spatial plane orthogonal to
the direction of the propagation.
Linear polarization

Consider E in free space,
E ( z, t )  E0 cos(t   z )a x

At plane z = 0, a tip of E field traces straight
line segment called “linearly polarized wave”
Linear polarization

A pair of linearly polarized wave also produces
linear polarization
E ( z, t )  Ex 0 cos(t   z )a x  E y 0 cos(t   z )a y
At z = 0 plane
E (0, t )  Ex 0 cos(t )a x  E y 0 cos(t )a y
At t = 0, both linearly polarized waves
Have their maximum values
E (0, 0)  Ex 0 a x  Ex 0 a y
t
E (0, )  0
4
More generalized linear polarization

More generalized of two linearly poloraized
waves,
E ( z, t )  Ex 0 cos(t   z  x )a x  E y 0 cos(t   z   y )a y

Linear polarization occurs when two linearly
polarized waves are
in phase
out of phase
 y  x  0
y  x  180
Elliptically polarized wave

Super position of two linearly polarized waves
that
y  x  0 or  180

If x = 0 and y = 45, we have

E (0, t )  Ex 0 cos(t )a x  E y 0 cos(t  )a y
4
Circularly polarized wave

occurs when Exo and Eyo are equal and
y  x  90

Right hand circularly polarized (RHCP) wave
y  x  90


E (0, t )  Ex 0 cos(t )a x  E y 0 cos(t  )a y
2
Left hand circularly polarized (LHCP) wave
y  x  90

E (0, t )  Ex 0 cos(t )a x  E y 0 cos(t  )a y
2
Circularly polarized wave

Phasor forms:
from
E ( z  0)  Ex 0e jx a x  E y 0e
j y
for RHCP,
E ( z  0)  Ex 0 (a x  ja y )
for LHCP,
E ( z  0)  Ex 0 (a x  ja y )
ay
Note: There are also RHEP and LHEP
Ex Given
E( z, t )  3cos(t   z  30 )a x  8cos(t   z  90 )a y
,determine the polarization of this wave
Ex The electric field of a uniform plane wave in
free space is given by E s  100(a z  ja x )e j 50 y
, determine
a) f
b) The magnetic field intensity H s
c)
S
d) Describe the polarization of the wave