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ENE 428
Microwave Engineering
Lecture 3 Polarization, Reflection
and Transmission at normal
incidence
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Uniform plane wave (UPW) power
transmission
from
Pavg

1
 Re( E  H )
2
1  Ex20 2 z j 
 Re 
e e  az
2 

1 Ex20 2 z

e
cos a z
2 
W/m2
2
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Polarization
• UPW is characterized by its propagation direction and
frequency.
• Its attenuation and phase are determined by medium’s
parameters.
• Polarization determines the orientation of the electric
field in a fixed spatial plane orthogonal to the direction of
the propagation.
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Linear polarization
• Consider E
in free space,
E( z, t )  E0 cos(t   z )a x
• At plane z = 0, a tip of E field traces straight line
segment called “linearly polarized wave”
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4
Linear polarization
• A pair of linearly polarized wave also produces linear
polarization
E ( z , t )  Ex 0 cos(t   z )a x  E y 0 cos(t   z )a y
At z = 0 plane
E (0, t )  Ex 0 cos(t )a x  E y 0 cos(t )a y
At t = 0, both linearly polarized waves
have their maximum values.
E(0,0)  Ex 0 a x  Ex 0 a y
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T
E(0, )  0
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5
More generalized linear polarization
• More generalized of two linearly polarized waves,
E ( z , t )  Ex 0 cos(t   z  x )a x  E y 0 cos(t   z   y )a y
• Linear polarization occurs when two linearly polarized
waves are
in phase
out of phase
y  x  0
y  x  180 .
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Elliptically polarized wave
• Superposition of two linearly polarized waves that
y  x  0 or 180
• If x = 0 and y = 45, we have

E(0, t )  Ex0 cos(t )a x  Ey 0 cos(t  )a y
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Circularly polarized wave
• occurs when Exo and Eyo are equal and
y  x  90
• Right hand circularly polarized (RHCP) wave
y  x  90

E(0, t )  Ex0 cos(t )a x  Ey 0 cos(t  )a y
2
• Left hand circularly polarized (LHCP) wave
y  x  90

E(0, t )  Ex0 cos(t )a x  Ey 0 cos(t  )a y
2
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Circularly polarized wave
• Phasor forms:
from
for RHCP,
E ( z  0)  Ex 0 e jx a x  E y 0e
j y
ay
E ( z  0)  Ex 0 (a x  ja y )
for LHCP,
E ( z  0)  Ex 0 (a x  ja y )
Note: There are also RHEP and LHEP
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Ex1 Given
E( z, t )  8cos(t   z  30 )a x  8cos(t   z  90 )a y
,determine the polarization of this wave
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Ex2 The electric field of a uniform plane wave in
free space is given byE s  100(a z  ja x )e j 50 y
, determine
a) f
b) The magnetic field intensity H s
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c)
S
d) Describe the polarization of the wave
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Reflection and transmission of UPW at
normal incidence
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Incident wave
• Normal incidence – the propagation direction is normal
to the boundary
Assume the medium is lossless, let the incident electric
field to be

E1 ( z, t )  Ex10 cos(t   z )a x

1
or in a phasor form E ( z )  Ex10e j1z a x
since

1
H  a 
E

1
1

1
then we can show that H ( z ) 
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Ex10
1
e j 1z a y
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Transmitted wave
• Transmitted wave
Assume the medium is lossless, let the transmitted
electric field to be

2
E ( z )  Ex20e  j 2 z a x
then we can show that

2
H ( z) 
Ex20
2
e j 2 z a y
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Reflected wave (1)
• From boundary conditions,
Etan1  Etan 2
H tan1  H tan 2
At z = 0, we have
Ex10  Ex20
and
Ex10

1 = 2
1

Ex20
2
are media the same?
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Reflected wave (2)
• There must be a reflected wave

1
E ( z )  Ex10e j 1z a x
and

1
H ( z)  
Ex10
1
e j1z a y
This wave travels in –z direction.
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Reflection and transmission coefficients
(1)
• Boundary conditions (reflected wave is included)
Ex1  Ex 2
from
Ex1  Ex1  Ex2
therefore at z = 0
Ex10  Ex10  Ex20
(1)
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Reflection and transmission coefficients
(2)
• Boundary conditions (reflected wave is included)
H y1  H y 2
from
H y1  H y1  H y2
therefore at z = 0
Ex10
1

Ex10
1

Ex20
2
(2)
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Reflection and transmission coefficients
(3)
• Solve Eqs. (1) and (2) to get
Reflection coefficient
Ex10 2 1
   
  e j
Ex10 2  1
Transmission coefficient
Ex20
22
  
 1     e j1
Ex10 2  1
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Types of boundaries: perfect dielectric
and perfect conductor (1)
j2
2 
0
 2  j 2
From

. 
Ex 20  0
Since 2 = 0

then  = -1 and Ex10+= -Ex10-
Ex1  Ex1  Ex1
Ex1  Ex10e j1z  Ex10e j1z
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Types of boundaries: perfect dielectric
and perfect conductor (2)
Ex1  (e j1z  e j1z )Ex10
  j 2Ex10 sin(1z)
This can be shown in an instantaneous form as
Ex ( z, t )  2Ex10 sin(1z)sin t Standing wave
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Standing waves (1)
When t = m, Ex1 is 0
at
all positions.
and when z = m, Ex1 is 0
at all time.
Null positions occur at
2
1
z  m
m1
z
2
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Standing waves (2)
Since
and
Ex1   H y1
Ex1   H y1
the magnetic field is
or
,
H y1 
Ex10
1
H y1 ( z, t ) 
Hy1 is maximum when Ex1 = 0
(e j1z  e j1z )
2Ex10
.
1
cos 1 z cos t
Poynting vector
S  EH
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24
Power transmission for 2 perfect
dielectrics (1)
Then 1 and 2 are both real positive quantities and 1 = 2 =
0

2  1
 
 real
2  1
Average incident power densities

1
P1i  Re Ex1H y1
2




E
1

x1
 Re  Ex1 * 
2 
1 
1 1  2
 Re  *  Ex10
2 1 
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Ex3 Let medium 1 have 1 = 100  and medium 2
have 2 = 300 , given Ex10+ = 100 V/m. Calculate
average incident, reflected, and transmitted power
densities
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Wave reflection from multiple interfaces (1)
• Wave reflection from materials that are finite in extent
such as interfaces between air, glass, and coating
• At steady state, there will be 5 total waves
1
2
3
Incident
energy
in
-l
0
z
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Wave reflection from multiple interfaces (2)
Assume lossless media, we have
 23 
3   2
,
3   2
then we can show that
Ex20   23 Ex20
H y20 
H

y 20
1
2

Ex20
1
2
E

x 20

1
2
 23 Ex20
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Wave reflection from multiple interfaces (2)
Assume lossless media, we have
 23 
3   2
,
3   2
then we can show that
Ex20   23 Ex20
H y20 
H

y 20
1
2

Ex20
1
2
E

x 20

1
2
 23 Ex20
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Wave impedance w (1)
Ex 2
Ex20e  j 2 z  Ex20e j 2 z
w ( z ) 
   j 2 z
H y 2 H x 20 e
 H x20e j 2 z
e j 2 z   23e j 2 z
 w ( z )  2  j 2 z
.
j 2 z
e
  23e
Use Euler’s identity, we can show that
3 cos  2 z  j2 sin  2 z
 w ( z )  2
2 cos  2 z  j3 sin  2 z
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Wave impedance w (2)
Since from B.C.
at z = -l
Ex1  Ex1  Ex 2
H y1  H y1  H y 2
we may write
Ex10  Ex10  Ex 2
Ex10
1

Ex10
1

Ex20
w
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Input impedance in
solve to get
Ex10
in 1
 

in  1
Ex10
3 cos  2l  j2 sin  2l
in  2
2 cos  2l  j3 sin  2l
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Refractive index
n  r
Under lossless conditions,
   0 0
n
r 
c
0 0


n
r 0
c
vp 
n
v p 0
 
f
n
1
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