ENE 428
Microwave Engineering
Lecture 2 Uniform plane waves
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Propagation in lossless-charge free
media
• Attenuation constant  = 0, conductivity  = 0
• Propagation constant
   

1
• Propagation velocity u p  


– for free space up = 3108 m/s (speed of light)
– for non-magnetic lossless dielectric (r = 1),
c
up 
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r
Propagation in lossless-charge free
media
• intrinsic impedance



• wavelength

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2

Ex1 A 9.375 GHz uniform plane wave is
propagating in polyethelene (r = 2.26). If the
amplitude of the electric field intensity is 500 V/m
and the material is assumed to be lossless, find
a) phase constant
 = 295 rad/m
b) wavelength in the polyethelene
 = 2.13 cm
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c) propagation velocity
v = 2x108 m/s
d) Intrinsic impedance
 = 250.77 
e) Amplitude of the magnetic field intensity
H = 1.99 A/m
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Propagation in dielectrics
• Cause
– finite conductivity
– polarization loss ( = ’-j” )
• Assume homogeneous and isotropic medium
 H   E  j( '  j " )E
 H  [(   " )  j ' ]E
Define
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 eff     "
Propagation in dielectrics
From
and
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 2  j (  j )
 2  (  j )2
Propagation in dielectrics
We can derive
 
 
and
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
 
( 1 
 1)

2
  

2
 
( 1 
 1)

2
  
2

1

.
 1  j (  )
Loss tangent
• A standard measure of lossiness, used to classify a
material as a good dielectric or a good conductor
   "  eff
tan  

'

 '
RS
Low loss material or a good
dielectric (tan « 1)

• If
1 or < 0.1 , consider the material ‘low

loss’ , then


2


   
and
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


(1  j
).

2
Low loss material or a good dielectric
(tan « 1)
• propagation velocity

1
up  


• wavelength

RS
2


1
f 
High loss material or a good
conductor (tan » 1)

• In this case

1 or > 10, we can approximate

 
1 
 1) 


  
2
therefore
  
and
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
j


2
  f 
 j 45

e .

High loss material or a good
conductor (tan » 1)
• depth of penetration or skin depth,  is a distance where
the field decreases to e-1 or 0.368 times of the initial field
1
1 1

 
 f   
• propagation velocity
• wavelength
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
u p   


2

 2
m
Ex2 Given a nonmagnetic material having r
= 3.2 and  = 1.510-4 S/m, at f = 30 MHz,
find
a) loss tangent 
tan = 0.03
b) attenuation constant 
 = 0.016 Np/m
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c) phase constant 
 = 1.12 rad/m
d) intrinsic impedance
 = 210.74(1+j0.015) 
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Ex3 Calculate the followings for the wave
with the frequency f = 60 Hz propagating in
a copper with the conductivity,  = 5.8107
S/m:
a) wavelength
 = 117.21 rad/m
 = 5.36 cm
b) propagation velocity
v = 3.22 m/s
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c) compare these answers with the same wave
propagating in a free space
 = 1.26x10-6 rad/m
= 5000 km
v = 3x108 m/s
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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,  E = E0e1
is the distance z that causes
to reduce to
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
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; ds  dydz
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 
RS
1
 (1  et /  )

Currents in conductor
Current is confined within a skin depth of the
coaxial cable
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Ex4 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
 = 7.66x10-4 m
b)
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The skin resistance
c) The 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
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W/m2
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 d S
From Ampère’s circuital law,
 H dl  I
2 aH  I
H
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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
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W
Uniform plane wave (UPW) power
transmission
• Time-averaged power density
Pavg

1
 Re( E  H ) W/m2
2
P   Pavg d S
for lossless case, P avg  1 Ex 0 e  j  z a x  Ex 0 e j  z a y
2

1 Ex20
 P avg 
a z W/m2
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
H
1

a  E 

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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