Lecture 2 - web page for staff
Download
Report
Transcript Lecture 2 - web page for staff
ENE 428
Microwave Engineering
Lecture 2 Uniform plane waves
RS
Propagation in lossless-charge free
media
• Attenuation constant = 0, conductivity = 0
• Propagation constant
• Propagation velocity
up
1
– for free space up = 3108 m/s (speed of light)
– for non-magnetic lossless dielectric (r = 1),
up
RS
c
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
RS
c) propagation velocity
v = 2x108 m/s
d) Intrinsic impedance
= 250.77
e) Amplitude of the magnetic field intensity
H = 1.99 A/m
RS
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
RS
From
j ( j )
and
2
2
( j )
2
Propagation in dielectrics
We can derive
RS
( 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
RS
'
"
eff
'
Low loss material or a good
dielectric (tan « 1)
• If
1
or < 0.1 , consider the material ‘low
loss’ , then
2
and
RS
(1 j
).
2
Low loss material or a good dielectric
(tan « 1)
• propagation velocity
up
1
• wavelength
RS
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
RS
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
RS
2
2
1
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
RS
c) phase constant
= 1.12 rad/m
d) intrinsic impedance
= 210.74(1+j0.015)
RS
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
RS
c) compare these answers with the same wave
propagating in a free space
= 1.26x10-6 rad/m
= 5000 km
v = 3x108 m/s
RS
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/ =
RS
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
RS
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
RS
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
RS
; ds dydz
Currents in conductor
From
I J x dS
I
; ds dydz
w
E x0e
z
dydz
z0 y0
w E x 0 e
z
I w E x 0
0
A.
Jx or current density decreases as the slab
gets thicker
RS
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
RS
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
RS
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)
RS
The skin resistance
c) The dc resistance
RS
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
RS
t
1
2
H dV
2
Rate of change of energy stored
In the fields
Instantaneous poynting vector
S EH
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
RS
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
RS
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
RS
I
2 a
a
Example of Poynting theorem in DC
case
S EH
Total power
I
az
a
2
S d S
I
2
I a
2
RS
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
RS
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
RS
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.
RS