Propagation in lossless-charge free media
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Transcript Propagation in lossless-charge free media
ENE 428
Microwave Engineering
Lecture 2 Uniform plane waves
RS
Propagation in lossless-charge free
media
• Attenuation constant = 0, conductivity = 0
• Propagation constant
1
• Propagation velocity u p
– for free space up = 3108 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
RS
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
RS
eff "
Propagation in dielectrics
From
and
RS
2 j ( j )
2 ( j )2
Propagation in dielectrics
We can derive
and
RS
( 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
'
'
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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
RS
u p
2
2
m
Ex2 Given a nonmagnetic material having r
= 3.2 and = 1.510-4 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, = 5.8107
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 = E0e1
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
RS
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
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1
(1 et / )
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 = 4106 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 8cost 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 EH
RS
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
RS
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 jn a y
n