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
 
• Propagation velocity

up 



1

– for free space up = 3108 m/s (speed of light)
– for non-magnetic lossless dielectric (r = 1),
up 
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c
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
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From
  j (  j )
and

2
2
 (  j  )
2
Propagation in dielectrics
We can derive
 
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 
  
( 1 
 1)

2
  
2

  
( 1 
 1)

2
  
 
and
2



1
1  j (   )
.
Loss tangent
• A standard measure of lossiness, used to classify a
material as a good dielectric or a good conductor
tan  
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  

'
"

 eff

'
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

up 


1

• wavelength
 
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2

1

f

High loss material or a good
conductor (tan » 1)
• In this case


 1
or > 10, we can approximate
2

  
1 
 1) 


  
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
 f 

1

• propagation velocity

up 
 

• wavelength
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 
2

 2 

1

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 ()
is the distance z that causes E to reduce to E 0 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
R 
L
R 
1 L
t w
 R sh eet 
A


1 L
 wt
Rsheet () L
w
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 x0e
 z
J x   E x0e
 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

I  
; ds  dydz
w
  E x0e
 z
dydz
z0 y0
  w E x 0 e
 



 z
I  w E x 0



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  E x0 L
R 
V
I

E x0 L
w E x 0 

 L
 R skin  
 w
w
1 L
R is called skin resistance
Rskin is called skin-effect resistance
For finite thickness,
t
I  
w
  E x0e
 z
z 0 y 0

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R skin 
dydz  w E x 0 (1  e
1
 (1  e
t /

)
 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
 ( E  H )  d S    J  E dV 
Total power leaving Joule’s law
the surface
for instantaneous
power dissipated
per volume (dissipated by heat)

  E dV 
2
t
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t
1
2
 H dV
2
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
 (E  H )  d S
   J  E dV 

E

t
2
dV 

1
H

t 2
2
dV
Rate of change of energy stored
In the fields = 0
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Example of Poynting theorem in DC
case
J 
From
I
a
az
2
By using Ohm’s law,
E 

J

 J  E dV

I
a 

I
2
 ( a )
 I
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az
2
2
2
1
2
L
  a2
a
2
L
0
0
0
  d   d   dz
 I R
2
Example of Poynting theorem in DC
case
Verify with
 (E  H )  d S
From Ampère’s circuital law,
 H dl  I
2 a H  I
 H 
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I
2 a
a
Example of Poynting theorem in DC
case
S  EH 
Total power
I
az 
a 
2
 S d S 
I
 2
I a
2

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2 a 
2
3
2
2
a 
3
2
I
2 a
a 
2
2 a 
2
3
a

a    d  dz
L
 d   dz 
0
I
0
I L
2
 a
2
 I R
2
W
Uniform plane wave (UPW) power
transmission
• Time-averaged power density
P a vg 
amount of power
for lossless case,
1
2

R e( E  H )
P 
P
avg
d S
P avg 
1
P avg 
1 E x0
2
W/m2
E x0e
 j z
ax 
E x0

2

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2 
az
W/m2
e
j z
ay
Uniform plane wave (UPW) power
transmission

for lossy medium, we can write E ( z , t )  E x 0 e   z cos(  t   z   ) a x
E  E x0e
The phasor is
 z
e
 j z
j
e ax
   e
intrinsic impedance for lossy medium
H 
1

a  E 

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1

a z  E x0e
E x0

e
 z
e
 z
 j z
e
 j z
j
e e
e
 j n
j
ax
ay
j n
Homework
Prob.6.19: In seawater, a propagating electric field is given by

z
6
E ( z , t )  20 e
cos( 2   10 t   z  0 . 5 ) a y V/m
Assuming  = 5 S/m, r’ = 72, r” = 0, find (a)  and β and (b) the instantaneous
form of H(z,t)

z
6
cos(   10 t   z   / 6 ) a x A/m
Prob. 6.26: In air, H ( z , t )  12 e
Determine the power passing through a 1.0 m2 surface that is normal to the
direction of propagation.
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