Transcript ENE 429 Antenna and Transmission Lines

ENE 325
Electromagnetic
Fields and Waves
Lecture 12 Plane Waves in Conductor,
Poynting Theorem, and Power Transmission
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Review (1)

Wave equations
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
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 )
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Review (2)
where
  j(  j ).
This  term is called propagation constant or we can write
 = +j
where  = attenuation constant (Np/m)
 = phase constant (rad/m)
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Review (3)

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
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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, 
is the distance z that causes E to reduce to E0e1
z = -1

z = -1/  = -.
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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.
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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
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Currents in conductor
E x  E x 0 e  z
J x   E x 0 e  z
Therefore the current that flows through the slab at t   is
I   J x dS
; ds  dydz
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Currents in conductor
From
I   J x dS

; ds  dydz
w
I     Ex 0 e z dydz
z 0 y 0
  w Ex 0e



 z
 I  w Ex0



0
A.
Jx or current density decreases as the slab gets thicker.
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Currents in conductor
For distance L in x-direction
V  Ex0 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
 (1  et /  )

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Currents in conductor
Current is confined within a skin depth of the coaxial cable.
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Ex1 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)
skin depth
b)
skin resistance
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c) dc resistance
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The Poynting theorem and power
transmission
Poynting theorem
 1 2
 1 2
 (E  H ) dS   J E dV  t  2  E dV  t  2  H dV
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/m2
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Example of Poynting theorem in DC case
 1 2
 1 2
 (E  H ) dS   J E dV  t  2  E dV  t  2  H dV
Rate of change of energy stored
In the fields = 0
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Example of Poynting theorem in DC case
From
I
J  2 az
a
By using Ohm’s law,
J
I
E   2 az
 a 
a
2
L
I2

 d   d  dz
2 2 
 ( a ) 0
0
0
1 L
2
 I


I
R
2
 a
2
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Example of Poynting theorem in DC case
Verify with


 EH dS
From Ampère’s circuital law,
 H dl  I
2 aH  I
H
I
2 a
a
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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 
Total power
I 2
 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
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Uniform plane wave (UPW) power
transmission

Time-averaged power density
Pavg

1
2
W/m
 Re( E  H )
2
P   Pavg d S
Ex 0  j  z
1
 j z
for lossless case,
P avg  Ex 0e
ax 
e
ay
2

1 Ex20
 P avg 
az W
2 
amount of power
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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    e
H
1

a  E 

1

jn
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
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Uniform plane wave (UPW) power
transmission
from
Pavg

1
 Re( E  H )
2
1  Ex20 2 z j 
 Re 
e e  az
2 

1 Ex20 2 z

e
cos a z
2 
W/m2
Question: Have you ever wondered why aluminum foil is not allowed in
the microwave oven?
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