Post on 28-Dec-2015
1
ECE 480
Wireless Systems
Lecture 3
Propagation and Modulation of RF Waves
2
Circular Polarization
Magnitudes of the x – and y – components of are equal
Phase difference is
E z
2
2
Left – Hand Circular (LHC) Polarization
2
Right – Hand Circular (RHC) Polarization
3
Left – Hand Circular (LHC) Polarization
x ya a a 2
j j k z
j k z
j
ˆ ˆE z xa y ae e
ˆ ˆa x j y e
e j
2
2
4
j tE z,t Re E z e
ˆ ˆx a cos t k z y a cos t k z
ˆ ˆx a cos t k z y a sin t k z
2
Convert to polar form
x yE z,t E z,t E z,t
a cos t k z a sin t k z a
12 2 2
12 2 2 2 2
y
x
E z,tz,t tan
E z,t
a sin t k ztan t k z
a cos t k z
1
1
5
Linear polarization
= f (z, t)
f (z, t)
Circular polarization
z,t t k z
f (z, t)
= f (z, t)
E z,t a
x yE z,t a cos t k z a cos t k z
12 2 2
y
x
atan constant
a
1
6
Back to LHC Polarization
Consider the LHC wave at z = 0
t
Inclination angle decreases with time
7
Right – Hand Circular (RHC) Polarization
x ya a a 2
z,t t k z
E z,t a
8
The direction of polarization is defined in terms of the rotation of as a function of time in a fixed plane orthogonal to the direction of propagation
E
9
Example RHC Polarized Wave
An RHC polarized plane wave with electric field modulus of 3 mV/m is traveling in the + y direction in a dielectric medium with
0 04 0
f = 100 MHz
Obtain expressions for E y,t and H y,t
10
Solution
The wave is traveling in the + y direction.
Therefore, the field components are in the x and z directions.
ˆ ˆ ˆz x y
E H
direction of propagation
11
Assign a phase angle of 0 o to the z component of E y
(arbitrary)
E yThe x component of will have a phase shift
2
Both components have a magnitude of a = 3
x z
j j k y j k y
j k y
ˆ ˆE y x E z E
ˆ ˆx ae e z ae
mVˆ ˆx j z e
m
2
3
12
j k y
j k y
ˆH y y E y
ˆ ˆy x j z e
mAˆ ˆz j x e
m
1
13
3
radf
s 82 2 10
r radk
c m
8
8
2 10 4 4
3 10 3
0 12060
4
13
Converting back to the time domain
j t
j k y j t
j t
j k y j t
E Re E y e
ˆ ˆRe x j z e e
mVˆ ˆx sin t k y z cos t k y
m
H Re H y e
ˆ ˆRe z j x e e
mAˆ ˆx cos t k y z sin t k y
m
3
3
3
1
20
14
Elliptical Polarization
Most general case
x ya a 0 0 0
EThe tip of traces an ellipse in the x – y plane
Can be left – handed or right - handed
Major axis:
Minor axis:
Rotation Angle
Ellipticity Angle
15
Rotation angle
xacosa
1
2 2
The shape and rotation are defined by the ellipticity angle
atan
a R
1
4 4
aR
a
axial ratio R = 1 Circular
R = Linear
16
y
x
o
atan
a
0
00 90
tan tan cos
tan sin sin
0
0
2 22 2
2 24 4
o
o
45
0
Circular
Linear
17
Polarization States for Various Combinations of and
18
Positive values of corresponding to sin > 0 define left – handed rotation
Negative values of corresponding to sin < 0 define right – handed rotation
How is the type of polarization determined?
a x and a y are, by definition, > 0
y
x
awill vary from
a 0
o 00 90
Two possible values of in this range and
if cos
if cos
0 0
0 0
if sin
if sin
0 0
0 0
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Example: Polarization State
Determine the polarization of a plane wave with an electric field given by
o o mVˆ ˆE z, t x cos t k z y sin t k z
m 3 30 3 45
Solution
Convert the sin term to a cos term by subtracting 90 o
o o mVˆ ˆE z, t x cos t k z y cos t k z
m 3 30 3 45
Convert the – cos term to a + cos term by adding 180 o
o o mVˆ ˆE z, t x cos t k z y cos t k z
m 3 30 3 135
20
o oj k z j j k z jˆ ˆE z x e e y e e 30 1353 4Convert to phasor form
o o oy x 135 30 105
y o
x
atan tan .
a
1 10
453 1
3
o o
tan tan cos
tan . cos
. .
.
02 2
106 2 105
3 442 0 2588
0 8900
21
o o o
oo
oo
tan . . , . .
..
..
1
1
2
2 0 89 41 7 221 7 138 3
41 720 85
2138 3
69 152
There are two possible solutions for since the tan function is positive in both the first and third quadrants
Which is correct?
o
cos
.
0
69 15
22
o o
sin sin sin
sin . sin
. .
.
02 2
106 2 105
0 9603 0 9659
0 9276
By a similar analysis,
o. 34 0
The wave is elliptically polarized and the rotation is left - handed
23
Plane – Wave Propagation in Lossy Media
E E 2 2 0
C ' j "
' "
2 2 2
can be written as
j
= attenuation constant
= phase constant
24
j j
' j "
2 2 2
2 2
2
Equate the real and imaginary parts
'
"
2 2 2
22
Solve for and
' "
'
' "
'
1
22
1
22
1 12
1 12
25
For a uniform plane wave traveling in the + z direction with an electric field
xx
d E zE z
d z
22
20
xˆE x E z
the wave equation becomes
The solution is
z z j zx x xˆ ˆ ˆE z x E z x E e x E e e 0 0
x z j
c c
ˆ Ek EˆH y e e
0
cc
"j
' '
1
21
26
The magnitude of is xE z
z j z zx x xE z E e e E e 0 0
Decreases exponentially with e - z
xy
c
EH
also decreases exponentially with e - z
Define: Skin Depth, s
s 1
Distance that a wave must travel before it is attenuated by
.e
10 3679
27
In a perfect dielectrics, , 0 0
In a perfect conductors, , 0
28
Expressions are valid for any linear, isotropic, homogeneous medium
"
'"
'"
'
2
2 2
2
10
10 10
10
Low – Loss Dielectric
Quasi – Conductor
(Semiconductor)
Good Conductor
29
Low – Loss Dielectric
Consider x1
21
xx , x
1
21 1 12
For
" "j ' j j ' j
' '
1
21 1
2
Divide into real and imaginary parts
" Np
' m
'
2 2
- Same as for lossless medium
30
c
" "j j
' ' '
1 1
2 2
xx , x
1
21 1 12
cc
"j
' '
1
21
c
Same as for the lossless case
31
Good Conductor
c
"f
f
fj j j"
2 2
1 1
jj
1
2
32
Semiconductors – Must use exact solution
33
Example – Plane Wave in Seawater
A uniform plane wave is traveling downward in the + z direction in seawater, with the x – y plane denoting the sea surface and z = 0 denoting a point just below the surface.
The constitutive parameters of seawater are:
r r
S, ,
m 1 80 4
The magnetic field intensity at z = 0 is given by
o mAˆH ,t y cos t
m 30 100 2 10 15
a. Determine expressions for
b. The depth at which the amplitude of E is 1% of its value at z = 0
E z,t and H z,t
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Solution
a. The general expressions for the phasor fields are
z j zx
x z j z
c
ˆE z x E e e
EˆH z y e e
0
0
r
"
' 3 90
5
480
2 10 1036
9 10
Seawater is a good conductor at 1 KHz
35
j j
c
Npf .
m.
.j e . e
3 7
4 4
10 4 10 4 0 126
0 126
0 1261 2 0 044
4
The general expression for E x0 is
jx xE E e 0
0 0
j z j z j tx
. zx
ˆE z,t Re x E e e e e
Vx E e cos t . z
m
0
0
0 126 30 02 10 0 126
36
jx z j z j t
j
. z ox
E eˆH z, t Re y e e e
. e
Ay . E e cos t . z
m
0
0
4
0 126 30 0
0 044
22 5 2 10 0 126 45
at z = 0:
ox
AˆH , t y . E cos t
m 3
0 00 22 5 2 10 45
Compare with original expression
o mAˆH ,t y cos t
m 30 100 2 10 15
x
x
. E
mVE .
m
30
0
22 5 100 10
4 44
o o
o
0
0
45 15
60
37
. z o mVˆE z,t x . e cos t . z
m 0 126 34 44 2 10 0 126 60
. z o mAˆH z,t y e cos t . z
m 0 126 3100 2 10 0 126 15
Note that they are no longer in phase. The electric field always leads the magnetic field by 45 o.
b. Set the amplitude to 0.01
. z. e
ln .z m
.
0 1260 01
0 0136
0 126
38
Electromagnetic Power Density
Define: Poynting Vector
WS E H
m
2
Direction of S is in the direction of propagation, k
AˆP S ndA S A cos
Power through a surface, A
n unit vector normal to the surface
39
Plane Wave in a Lossless Medium
j k zx
x j k z
ˆE z x E e
EˆH z y e
0
0
Consider a plane wave traveling in the + z direction
Want to find the power density vector, S
40
Time – Domain Approach
x
x
ˆE z ,t x E cos t k z
EˆH z ,t y cos t k z
0
0
x
S z ,t E z ,t H z ,t
Ez cos t k z
2
0 2
41
T
av
x
x
S S z ,t d t TT f
Ez cos t k z d t
Ez
0
2 2
0 2
0
2
0
1 1 2
2
2
�
Time average of S
42
Phasor – Domain Approach
av
j k z j k zxx
x
S Re E H *
E *ˆ ˆRe x E e y e
Ez
00
2
0
1
2
2
avS Re E H * is valid for any media
43
Plane Wave in a Lossy Medium
x y
z j zx y
z j zx y
c
ˆ ˆE z x E z y E z
ˆ ˆx E y E e e
ˆ ˆH z x E y E e e
0 0
0 0
1
av
x yz
c
S Re E H *
z E Ee Re
*
2 2
0 02 1
2
jc c e
44
jc c e
zav
c
EˆS z z e cos
2
0 2
2
x yE E E 1
2 2 20 0 0
Note that the average power decays withze 2
zav
c
EˆS z z e cos
2
0 2
2
45
Homework
The electric field of a plane wave is given by
x yˆ ˆE z ,t x a cos t k z y a cos t k z
Identify the polarization state, determine the polarization angles (, ), and sketch the locus of E (0, t) for each of the following cases
ox y
ox y
ox y
ox y
V Va , a ,
m mV V
a , a ,m mV V
a , a ,m mV V
a , a ,m m
3 4 0
3 4 180
3 3 45
3 4 135
46
Homework
In a medium characterized by
r r
S, , .
m 9 1 0 1
Determine the phase angle by which the magnetic field leads the electric field
47
Radiation and Antennas
• An antenna may be considered as a transducer that converts a guided EM wave to a transmitted wave or an incident wave to a guided EM wave
• Antenna dimensions are generally referred to in wavelength units
48
49
Reciprocity
• Antenna radiation pattern: The directional function that characterizes the distribution pattern radiated by an antenna
• Isotropic antenna: A hypothetical antenna that radiates equally in all directions
• Used as a reference radiator to compare antennas
• Reciprocal antennas: Antennas that have the same radiation patterns for transmission as for reception
50
Two aspects of antenna performance
1. Radiation Properties
• Direction of the radiation pattern
• Polarization state of the radiated wave in the TX mode (Antenna Polarization)
• In the RX mode, the antenna can extract only that component of the wave whose E – field is parallel to that of the antennas polarization direction
2. Antenna Impedance
• Pertains to the impedance match between the antenna and the generator
51
Radiation Sources
Two basic types
1. Current sources
• Dipole and loop antennas
2. Aperture fields
• Horn antennas
52
Far – Field Region
The far – field region is at a distance R where the wave may be considered to be a plane wave
ff
DR
22 D = Maximum effective size of the antenna
= Wavelength of the signal
53
Example: Far – Field Distance of an Antenna
A parabolic reflector antenna is 18" in diameter operates at 12.4 GHz. Find the operating wavelength and the far – field distance of this antenna.
Solution
ff
c. m
f .
m" x . m
. "
.DR . m
.
8
9
22
3 100 0242
12 4 10
118 0 457
39 37
2 0 457217 3
0 0242
54
Antenna Arrays
• Can control the phase and magnitude of each antenna individually
• Can steer the direction of the beam electronically
55
Retarded Potentials
Consider a charge distribution as shown
The electric potential V (R) at a point in space specified by the position vector R is given by
v i
'
RV R d '
R'
1
4
i
v
i
R
'
R' R R
= position vector of an elemental volume
= elemental volume
= charge density inside the volume
= distance between the volume and the point
56
If the charge density is time – varying, the obvious solution is
Problem: Does not account for reaction time
Any change in the charge distribution will require a finite amount of time to change the potential
v i
'
R , tV R ,t d '
R'
1
4
57
Retarded Vector Potential
i
'
J R , t t 'A R ,t d '
R'
4
v i
'
R , t t 'V R ,t d '
R'
1
4
p
R't '
u
Retarded Scalar Potential
Delay Time
Valid under both static and dynamic conditions
58
Time – Harmonic Potentials
In a linear system, the parameters all have the same functional dependence on time
Consider a sinusoidal time – varying charge distribution
j tv i v i
v i
R ,t Re R e
R
= phasor representation of v iR ,t
59
j k R'v i
'
R eV R d ' V
R'
1
4
j k R'i
'
J R eA R d '
R'
4
p
ku
60
H A
1In terms of A
H j E
E Hj
1
E j H
H Ej
1
61
The Short (Hertzian) Dipole
Approach: Develop the radiation properties of a differential antenna and use that model to predict other configurations
Characteristics of a Short Dipole
• Current is uniform over the length
50
l
j ti t cos t Re e A
0 0
0
I I
I I
62
At the point Q
j k R'
'
J eA R d '
R'
0
4
ˆk J zc s
02 I
s = cross – sectional area of dipole d ' s d z
limits of integration z 2 2
l l
Assume R' ~ R
63
j k R 2
2
j k R
eˆA z d z
R
ez
R
00
00
4
4
l
l
l
I
I
j k Re
R
Spherical propagation factor
Considers both the magnitude and phase change wrt R
Change to spherical coordinates
64
R Range
Zenith angle
Azimuth angle
65
j k R
R
ˆ ˆz R cos sin
eˆ ˆA R cos sinR
ˆ ˆ ˆR A A A
0 0
4
lI
j k R
R
j k R
eA cos
R
eA sin
R
A
0 0
0 0
4
4
0
l
l
I
I
66
H A
1
E Hj
1
j k R
j k RR
j k R
R
k jH e sin
k R kR
2 k jE e cos
k R kR
k j jE e sin
k R kR kR
H H E
20
2
20
0 2 3
20
0 2 3
1
4
1
4
1
4
0
l
l
l
I
I
I
67
Electric Field lines
68
Far – Field Approximation
R
RkR
, negligiblek R kR
2 3
21
1 1
j k R
R
k eE sin
R
E
EH
0
0
4
0
lj I
Independent of
Proportional to sin
69
Power Density
avS Re E H *
For the short dipole:
av
2
ˆS R S R ,
kS R , sin
32 R
WS sin
m
2 2 20 0 2
2
20 2
lI
70
Define: Normalized Radiation Intensity
max
S R , ,F ,
S
Radiation is maximum when 2
(azimuth plane)
max 2
kS S sin
32 R
R
2 2 20 0 2
0 2
220
2
15
l
l
I
I
71
F , F sin 2
No energy is radiated by the dipole along the direction of the dipole axis and maximum radiation (F = 1) occurs in the broadside direction.