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ENE 428 Microwave Engineering. Lecture 2 Uniform plane waves. Propagation in lossless-charge free media. Attenuation constant = 0, conductivity = 0 Propagation constant Propagation velocity for free space u p = 310 8 m/s (speed of light) - PowerPoint PPT Presentation
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ENE 428Microwave
Engineering
Lecture 2 Uniform plane waves
Propagation in lossless-charge free media
• Attenuation constant = 0, conductivity = 0
• Propagation constant
• Propagation velocity
– for free space up = 3108 m/s (speed of light)
– for non-magnetic lossless dielectric (r = 1),
1
pu
p
r
cu
Propagation in lossless-charge free media
• intrinsic impedance
• wavelength
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, finda) phase constant
b) wavelength in the polyethelene
c) propagation velocity
d) Intrinsic impedance
e) Amplitude of the magnetic field intensity
Propagation in dielectrics• Cause
– finite conductivity– polarization loss ( = ’-j” )
• Assume homogeneous and isotropic medium
' "( ) 555555555555555555555555555555555555555555H E j j E
" '[( ) ] 5555555555555555555555555555H j E
Propagation in dielectrics
" effDefine
From2 ( ) j j
and2 2( ) j
Propagation in dielectrics
We can derive2
( 1 1)2
2
( 1 1)2
and 1
.1 ( )
j
Loss tangent
• A standard measure of lossiness, used to classify a material as a good dielectric or a good conductor
"
' 'tan
eff
Low loss material or a good dielectric (tan « 1)
• If or < 0.1 , consider the material
‘low loss’ , then
1
2
(1 ).2
jand
Low loss material or a good dielectric (tan « 1)
• propagation velocity
• wavelength
1
pu
2 1
f
High loss material or a good conductor (tan » 1)
• In this case or > 10, we can
approximate
1
2 f
45 .
jje
therefore
2
1 1)
and
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
• propagation velocity
• wavelength
1 1 1m
f
pu
22
Ex2 Given a nonmagnetic material having r = 3.2 and = 1.510-4 S/m,
at f = 3 MHz, find a) loss tangent
b) attenuation constant
c) phase constant
d) intrinsic impedance
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
b) propagation velocity
c) compare these answers with the same wave propagating in a free space
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 to reduce to
z = 1 z = 1/ =
E55555555555555
10E e
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
1L LR
A wt
1 LR
t w Rsheet () Lw
1sheetR
t sheet resistance
from
At high frequency, it will be adapted to skin effect resistance
Currents in conductor
0
0
zx x
zx x
E E e
J E e
Therefore the current that flows through the slab at t is
;xI J dS ds dydz
Currents in conductor
;xI J dS ds dydz
00 0
wz
xz y
I E e dydz
0
0
zxw E e
0 .xI w E A
From
Jx or current density decreases as the slab gets thicker
Currents in conductor
0xV E L
0
0
1xskin
x
E LV L LR R
I w E w w
For distance L in x-direction
For finite thickness,
R is called skin resistanceRskin is called skin-effect resistance
0 00 0
(1 )t w
z tx x
z y
I E e dydz w E e
/
1
(1 )skin tRe
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 = 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
b) The skin resistance
c) The dc resistance
The Poynting theorem and power transmission
2 21 1( )
2 2E H d S J E dV E dV H dV
t t
5555555555555555555555555555555555555555555555555555555555555555555555
Poynting theorem
Total power leavingthe surface
Joule’s lawfor instantaneouspower dissipated per volume (dissi-pated by heat)
Rate of change of energy storedIn the fields
2W/mS E H 555555555555555555555555555555555555555555
Instantaneous poynting vector
Example of Poynting theorem in DC case
2 21 1( )
2 2E H d S J E dV E dV H dV
t t
5555555555555555555555555555555555555555555555555555555555555555555555
Rate of change of energy storedIn the fields = 0
Example of Poynting theorem in DC case
2 z
IJ a
a
55555555555555
By using Ohm’s law,
From
2 z
J IE a
a
5555555555555555555555555555
2 2
2 20 0 0( )
a LId d dz
a
2 22
1 LI I R
a
Example of Poynting theorem in DC case
E H d S555555555555555555555555555555555555555555
From Ampère’s circuital law,
Verify with
H dl I5555555555555555555555555555
2 aH I 55555555555555
2
IH a
a
55555555555555
Example of Poynting theorem in DC case
2
2 32
IS d S a d dz
a
5555555555555555555555555555
2
2 2 32 2z
I I IS E H a a a
aa a
555555555555555555555555555555555555555555
2 222
2 3 20 02
LI a I Ld dz I R
a a
Total power
W
Uniform plane wave (UPW) power transmission• Time-averaged power density
1Re( )
2avgP E H
555555555555555555555555555555555555555555
amount of power avgP P d S5555555555555555555555555555
for lossless case, 00
12
55555555555555j z j zx
avg x yx
EP E e a e a
201
2x
avg zE
P a 55555555555555
W
W/m2
Uniform plane wave (UPW) power transmission
0
z j z jxxE E e e e a
55555555555555
intrinsic impedance for lossy medium nje
0
1 1 z j z jz xxH a E a E e e e a
5555555555555555555555555555
0 njz j z jxy
Ee e e e a
for lossy medium, we can write