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Antennas & Propagation
LECTURE NOTES
VOLUME V
ELECTROMAGNETIC WAVE
PROPAGATION
by Professor David Jenn
(ver1.3)
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Propagation of Electromagnetic Waves
Radiating systems must operate in a complex changing environment that interacts with
propagating electromagnetic waves. Commonly observed propagation effects are depicted
below.
4
SATELLITE
IONOSPHERE
3
1
2
3
4
5
6
5
1
2
TRANSMITTER
6
RECEIVER
DIRECT
REFLECTED
TROPOSCATTER
IONOSPHERIC HOP
SATELLITE RELAY
GROUND WAVE
EARTH
Troposphere: lower regions of the atmosphere (less than 10 km)
Ionosphere: upper regions of the atmosphere (50 km to 1000 km)
Effects on waves: reflection, refraction, diffraction, attenuation, scattering, and
depolarization.
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Survey of Propagation Mechanisms (1)
There are may propagation mechanisms by which signals can travel between the radar
transmitter and receiver. Except for line-of-sight (LOS) paths, the mechanism’s
effectiveness is generally a strong function of the frequency and transmitter-receiver
geometry.
1. direct path or "line of sight" (most radars; SHF links from ground to satellites)
RX
o
TX
o
SURFACE
2. direct plus earth reflections or "multipath" (UHF broadcast; ground-to-air and airto-air communications)
o
TX
RX
o
SURFACE
3. ground wave (AM broadcast; Loran C navigation at short ranges)
TX
RX
o
SURFACE
o
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Survey of Propagation Mechanisms (2)
4. ionospheric hop (MF and HF broadcast and communications)
F-LAYER OF
IONOSPHERE
TX
o RX
o
E-LAYER OF
IONOSPHERE
SURFACE
5. waveguide modes or "ionospheric ducting" (VLF and LF communications)
TX
o RX
o
D-LAYER OF
IONOSPHERE
SURFACE
Note: The distinction between ionospheric hops and waveguide modes is based more on
the mathematical models than on physical processes.
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Survey of Propagation Mechanisms (3)
6. tropospheric paths or "troposcatter" (microwave links; over-the-horizon (OTH)
radar and communications)
TROPOSPHERE
TX
o RX
o
SURFACE
7. terrain diffraction
TX
o RX
o
MOUNTAIN
8. low altitude and surface ducts (radar frequencies)
SURFACE DUCT (HIGH
DIELECTRIC CONSTANT)
TX o
SURFACE
o
RX
9. Other less significant mechanisms: meteor scatter, whistlers
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Illustration of Propagation Phenomena
(From Prof. C. A. Levis, Ohio State University)
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Propagation Mechanisms by Frequency Bands
VLF and LF
(10 to 200 kHz)
LF to MF
(200 kHz to 2 MHz)
HF
(2 MHz to 30 MHz)
VHF
(30 MHz to 100 MHz)
UHF
(80 MHz to 500 MHz)
SHF
(500 MHz to 10 GHz)
Waveguide mode between Earth and D-layer; ground wave at short
distances
Transition between ground wave and mode predominance and sky
wave (ionospheric hops). Sky wave especially pronounced at night.
Ionospheric hops. Very long distance communications with low power
and simple antennas. The “short wave” band.
With low power and small antennas, primarily for local use using direct
or direct-plus-Earth-reflected propagation; ducting can greatly increase
this range. With large antennas and high power, ionospheric scatter
communications.
Direct: early-warning radars, aircraft-to satellite and satellite-to-satellite
communications. Direct-plus-Earth-reflected: air-to-ground
communications, local television. Tropospheric scattering: when large
highly directional antennas and high power are used.
Direct: most radars, satellite communications. Tropospheric refraction
and terrain diffraction become important in microwave links and in
satellite communication, at low altitudes.
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Applications of Propagation Phenomena
Direct
Direct plus Earth
reflections
Ground wave
Most radars; SHF links from ground to satellites
UHF broadcast TV with high antennas; ground-to-air and air-toground communications
Local Standard Broadcast (AM), Loran C navigation at relatively
short ranges
Tropospheric paths Microwave links
Waveguide modes
VLF and LF systems for long-range communication and navigation
(Earth and D-layer form the waveguide)
Ionospheric hops
MF and HF broadcast communications (including most long-distance
(E- and F-layers)
amateur communications)
Tropospheric scatter UHF medium distance communications
Ionospheric scatter
Medium distance communications in the lower VHF portion of the
band
Meteor scatter
VHF long distance low data rate communications
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Multipath From a Flat Ground (1)
When both a transmitter and receiver are operating near the surface of the earth,
multipath (multiple reflections) can cause fading of the signal. We examine a single
reflection from the ground assuming a flat earth.
RECEIVER
d
.
TRANSMITTER
Ro
θ=0
R2
•
ht
ht
C
A
•
θ′ = 0
B ψ R1
•
•
D
hr
ψ
EARTH'S SURFACE
(FLAT)
IMAGE
•
REFLECTION POINT
jφ Γ
ρe
The reflected wave appears to originate from an image.
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Multipath From a Flat Ground (2)
Multipath parameters:
1. Reflection coefficient, Γ = ρ e jφΓ . For low grazing angles, ψ ≈ 0 , the
approximation Γ ≈ −1 is valid for both horizontal and vertical polarizations.
2. Transmit antenna gain: Gt (θ A ) for the direct wave; Gt (θ B ) for the reflected wave.
3. Receive antenna gain: Gr (θC ) for the direct wave; Gr (θ D ) for the reflected wave.
4. Path difference: ∆R = (R1 + R2 ) − Ro
1424
3
{
REFLECTED
DIRECT
Gain is proportional to the square of the electric field intensity. For example, if Gto is the
gain of the transmit antenna in the direction of the maximum (θ = 0 ), then
2
Gt (θ ) = Gto E t norm (θ ) ≡ Gto f t (θ ) 2
where Et norm is the normalized electric field intensity. Similarly for the receive antenna
with its maximum gain in the direction θ ′ = 0
2
Gr (θ ′) = Gro E rnorm (θ ′) ≡ Gro f r (θ ′) 2
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Multipath From a Flat Ground (3)
Total field at the receiver
E tot =
E
ref
{
REFLECTED
+ E
dir
{
DIRECT
F4444448
6444444≡7
− jkR
o
e
1 + Γ f t (θ B ) f r (θ D ) e − jk∆R 
= f t (θ A ) f r (θC )
4π Ro 
f t (θ A ) f r (θ C )


The quantity in the square brackets is the path-gain factor (PGF) or pattern-propagation
factor (PPF). It relates the total field at the receiver to that of free space and takes on
values 0 ≤ F ≤ 2 .
• If F = 0 then the direct and reflected rays cancel (destructive interference)
• If F = 2 the two waves add (constructive interference)
Note that if the transmitter and receiver are at approximately the same heights, close to the
ground, and the antennas are pointed at each other, then d >> ht ,hr and
Gt (θ A ) ≈ Gt (θ B )
Gr (θ C ) ≈ Gr (θ D )
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Multipath From a Flat Ground (4)
An approximate expression for the path difference is obtained from a series expansion:
2
1 ( hr − ht )
Ro = d + ( hr − ht ) ≈ d +
2
d
2
2
1 (ht + hr ) 2
R1 + R2 = d + ( ht + hr ) ≈ d +
2
d
2
2
Therefore,
∆R ≈
and
(
2hr ht
d
| F |= 1 − e − jk 2 h r h t / d = e jkh r h t / d e − jkh r ht
/d
)
− e jkh r h t / d = 2 sin (khr ht / d )
The received power depends on the square of the path gain factor
2
kht hr 

2
2  kht hr 
Pr ∝ | F | = 4 sin 
 ≈ 4

d
d

 

The last approximation is based on h r , ht << d and Γ ≈ -1.
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Multipath From a Flat Ground (5)
Two different forms of the argument are frequently used.
1. Assume that the transmitter is near the ground ht ≈ 0 and use its height as a reference.
The elevation angle is ψ where
h − ht ∆h hr
tanψ = r
≡
≈
d
d
d
Ro
∆h = hr − ht
ψ
d
2. If the transmit antenna is very close to the ground, then the reflection point is very
near to the transmitter and ψ is also the grazing angle:
∆R = b − a = 2 ht sin ψ
ht
a
ψ
ψ
b
ψ
If the antenna is pointed at the horizon (i.e., its maximum is parallel to the ground) then
ψ ≈ θ A.
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Multipath From a Flat Ground (6)
Thus with the given restrictions the PPF can be expressed in terms of ψ
| F |= 2 sin (kht tanψ )
The PPF has minima at:
kht tanψ = nπ ( n = 0, 1, K, ∞ )
2π
ht tanψ = nπ
λ
tanψ = nλ / ht
Maxima occur at:
kht tanψ = mπ / 2 ( m = 1, 3, 5,K , ∞ )
2π
2n + 1
ht tanψ =
π ( n = 0 ,1,K ,∞ )
λ
2
(2 n + 1)λ
tanψ =
4ht
Plots | F | are called a coverage diagram. The horizontal axis is usually distance and
the vertical axis receiver height. (Note that because d >> hr the angle ψ is not directly
measurable from the plot.)
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Multipath From a Flat Ground (7)
Coverage diagram: Contour plots of | F | in dB for variations in hr and d normalized to
a reference range d o . Note that when d = d o then E tot = E dir .
d 
| F |= 2 o  sin (kht tanψ )
 d 
RECEIVER HEIGHT, hr (m)
60
d o = 2000 m
50
ht = 100λ
40
30
20
10
0
1000
2000
3000
4000
5000
RANGE, d (m)
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Multipath From a Flat Ground (8)
Another means of displaying the received field is a height-gain curve. It is a plot of | F |
in dB vs hr at a fixed range.
• The constructive and destructive interference as a function of height can be identified.
• At low frequencies the periodicity of the curve at low heights can be destroyed by the
ground wave.
• Usually there are many reflected wave paths between the transmitter and receiver, in
which case the peaks and nulls are distorted.
• This technique is often used to determine the optimum tower height for a broadcast
radio antenna.
PATH GAIN FACTOR (dB)
10
5
0
-5
-10
-15
-20
0
10
20
30
40
RECEIVER HEIGHT, hr (m)
50
60
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Multipath Example
A radar antenna is mounted on a 5 m mast and tracks a point target at 4 km. The target is 2 m
above the surface and the wavelength is 0.2 m. (a) Find the location of the reflection point on
the x axis and the grazing angle ψ . (b) Write an expression for the one way path gain factor F
when a reflected wave is present. Assume a reflection coefficient of Γ ≈ −1 .
(b) The restrictions on the heights and
distance are satisfied for the following
5m
formula
2m
ψ
ψ
x=0
x=4 km
x
Reflection
Point
(a) Denote the location of the reflection
point by xr and use similar triangles
5
2
tanψ =
=
x r 4000 − x r
xr = 2.86 km
ψ = tan -1 (5 / 2860) = 0.1o
kh h
 2π ( 2)( 5) 
F = 2 sin  t r  = 2 sin 

 d 
 (0.2 )( 4000 
= ( 2 )(0.785) = 0.157
The received power varies as F 2 , thus
( ) = −16.1 dB
10 log F
2
The received power is 16.1 dB below the
free space value
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Field Intensity From the ERP
The product Pt Gt is called the effective radiated power (ERP, or sometimes the effective
isotropic radiated power, EIRP). We can relate the ERP to the electric field intensity as
follows:
• The Poynting vector for a TEM wave:
r
r r*
W =ℜ E ×H =
{
• For the direct path:
}
r
Edir
2
ηo
r
PG
W = t t
4πRo2
• Equate the two expressions: (note that ηo ≈ 120π )
r
Edir
ηo
2
=
Pt Gt
4πRo2
⇒
r
30 Pt G t Eo
Edir =
≡
d
d
where Eo is called the unattenuated field intensity at unit distance.
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Wave Reflection at the Earth’s Surface (1)
Fresnel reflection coefficients hold when:
1. the Earth’s surface is locally flat in the vicinity of the reflection point
2. the surface is smooth (height of irregularities << λ )
Traditional notation:
1. grazing angle, ψ = 90o − θ i , and the grazing angle is usually very small (ψ < 1o )

σ
σ 
 ≡ ε o (ε r − jχ ) ,
2. complex dielectric constant, ε c = ε r ε o − j
= ε o  ε r − j
1424
3
ω
ε
ω

o 
εrc
σ
ωεo
3. horizontal and vertical polarization reference is used
where χ =
Also called
transverse
magnetic
(TM) pol
VERTICAL POL
r
r
E|| = EV
n̂
k̂
i
ψ
SURFACE
HORIZONTAL POL
r
r
E⊥ = E H
ψ
kˆi
n̂
Also called
transverse
electric
(TE) pol
SURFACE
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Wave Reflection at the Earth’s Surface (2)
Reflection coefficients for horizontal and vertical polarizations:
− Γ|| ≡ RV =
(ε r − jχ ) sinψ − (ε r − jχ ) − cos 2 ψ
(ε r − jχ ) sinψ + (ε r − jχ ) − cos 2 ψ
Γ⊥ ≡ RH =
sinψ − (ε r − jχ ) − cos 2 ψ
sinψ + (ε r − jχ ) − cos 2 ψ
For vertical polarization the phenomenon of total reflection can occur. This yields a
surface guided wave called a ground wave. From Snell’s law, assuming µ r = 1 for the
Earth,
sin θ i = sin θ r = (ε r − jχ ) µ r sin θ t
Let θ t be complex, θ t =
π
+ jθ , where θ is real.
2
⇒
µ r =1
sin θ t =
sin θ i
ε r − jχ
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Wave Reflection at the Earth’s Surface (3)
Using θ t =
π
+ jθ :
2
Snell’s law becomes
π

sin θ t = sin  + jθ  = cos( jθ ) = cosh θ
2

cosθ t = − j sin( jθ ) = − j sinh θ
sin θ t = cosh θ =
sin θ i
ε r − jχ
cosθ t = 1 − sin 2 θ t = 1 − cosh 2 θ = sinh θ
Reflection coefficient for vertical polarization:
Γ|| ≡ − RV =
jη sinh θ + ηo cosθ i
jη sinh θ − ηo cosθ i
µo
. Note that Γ|| = 1 and therefore all of the power flow is along the
ε o (ε r − j χ )
surface. The wave decays exponentially with distance into the Earth.
where η =
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Wave Reflection at the Earth’s Surface (4)
Example: surface wave propagating along a perfectly conducting plate
•
•
•
•
•
5λ plate
15 degree grazing angle
TM (vertical) polarization
the total field is plotted (incident plus scattered)
surface waves will follow curved surfaces if the radius of curvature >> λ
INCIDENT WAVE
(75 DEGREES OFF
OF NORMAL)
CONDUCTING
PLATE
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Atmospheric Refraction (1)
Refraction by the lower atmosphere causes waves to be bent back towards the earth’s
surface. The ray trajectory is described by the equation: n Re sinθ = CONSTANT
Two ways of expressing the index of refraction n (= ε r ) in the troposphere:
1. n = 1 + χρ / ρSL + HUMIDITY TERM
Re = 6378 km = earth radius
χ ≈ 0.00029 = Gladstone-Dale
constant
ρ, ρ SL = mass densities at altitude
and sea level
77.6
2. n =
( p + 4,810 e / T )10 −6 − 1
T
θ
REFRACTED
RAY
θ
Re
Re
Re
θ
EARTH'S
SURFACE
p = air pressure (millibars)
T = temperature (K)
e = partial pressure of
water vapor (millibars)
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Atmospheric Refraction (2)
Refraction of a wave can provide a significant level of transmission over the horizon. A
bent refracted ray can be represented by a straight ray if an equivalent earth radius Re′ is
used. For most atmospheric conditions Re′ = 4 Re / 3 = 8500 km
REFRACTED
RAY
TX
TX
RX
LINE OF SIGHT (LOS)
BLOCKED BY
EARTH'S BULGE
ht
RX
hr
ht
hr
EARTH'S
SURFACE
EARTH
RADIUS , Re
REFRACTED RAY
BECOMES A
STRAIGHT LINE
EARTH'S
SURFACE
EQUIVALENT EARTH
RADIUS, Re′
STANDARD
CONDITIONS:
4
Re′ ≈ Re
3
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Atmospheric Refraction (3)
Distance from the transmit antenna to the horizon is Rt = (Re′ + ht ) − (Re′ ) but
Re′ >> ht so that Rt ≈ 2 Re′ ht . Similarly Rr ≈ 2 Re′ hr . The radar horizon is the sum
2
RRH ≈ 2 Re′ht + 2 Re′ hr
Example: A missile is flying 15 m
TX
above the ocean towards a ground
ht
based radar. What is the approximate
range that the missile can be detected
assuming standard atmospheric
EARTH'S
SURFACE
conditions?
2
Rt
Rr
RX
hr
Re′
Re′
Re′
Using ht = 0 and hr = 15 gives a radar
horizon of
R RH ≈ 2 Re′ hr
≈ (2)(8500 × 10 3 )(15)
≈ 16 km
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Atmospheric Refraction (4)
Derivation of the equivalent Earth radius
θ (h )
h
RAY PATH
TANGENT
n (h )
θo
VERTICAL
SURFACE
Break up the atmosphere into thin horizontal layers. Snell’s law must hold at the
boundary between each layer, ε ( h ) sin [θ ( h ) ] = ε o sin θo
h
h3
h2
h1
M
θ (h )
θ2
θ1
n( h3 )
n( h2 )
n (h1 )
THIN LAYER IN WHICH
n ≈ CONSTANT
θo
SURFACE
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Atmospheric Refraction (5)
In terms of the Earth radius,
Re ε o sin θ o = ( Re + h ) ε ( R) sin [θ ( R) ]
3
14
4244
3 1444424444
AT THE
SURFACE
AT RADIUS
R = Re + h
Using the grazing angle, and assuming that ε (h ) varies linearly with h
d


Re ε o cosψ o = ( Re + h )  ε o + h
ε ( h )  cos[ψ ( h )]
dh


Expand and rearrange
d
d


Re ε o {cosψ o − cos[ψ ( h ) ]} =  ε o + Re
ε ( h )  h cos[ψ ( h )] + h 2
ε ( h ) cos[ψ ( h )]
dh
dh


If h << Re then the last term can be dropped, and since ψ is small, cosψ ≈ 1 + ψ 2 / 2


[ψ ( h )]2 ≈ ψ o2 + 2h 1 + Re d ε ( h ) 
Re 
ε o dh

The second term is due to the inhomogenity of the index of refraction with altitude.
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Atmospheric Refraction (6)
Define a constant κ such that
[ψ (h )]2
2h
2h
≈ ψ o2 +
= ψ o2 +
κRe
Re′
where


Re d
κ = 1 +
ε ( h) 
ε
dh
o


−1
Re′ = κRe is the effective (equivalent) Earth radius. If Re′ is used as the Earth radius then
rays can be drawn as straight lines. This is the radius that would produce the same
geometrical relationship between the source of the ray and the receiver near the Earth’s
surface, assuming a constant index of refraction. The restrictions on the model are:
1.
2.
Ray paths are nearly horizontal
ε (h ) versus h is linear over the range of heights considered
Under standard (normal) atmospheric conditions, κ ≈ 4 / 3 . That is, the radius of the Earth
4
is approximately Re′ =   6378 km = 8500 km . This is commonly referred to as “the four 3
thirds Earth approximation.”
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Fresnel Zones (1)
For the direct path phase to differ from the reflected path phase by an integer multiple of
180 o the paths must differ by integer multiples of λ / 2
∆R = nλ / 2 ( n = 0,1,K )
The collection of points at which reflection would produce an excess path length of nλ / 2
is called the nth Fresnel zone. In three dimensions the surfaces are ellipsoids centered on
the direct path between the transmitter and receiver
LOCUS OF REFLECTION
POINTS (SURFACES OF
REVOLUTION)
DIRECT
PATH (LOS)
n=2
RECEIVER
n =1
TRANSMITTER
hr
ht
REFLECTING SURFACE
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Fresnel Zones (2)
A slice of the vertical plane gives the following geometry
d
dr
dt
TX
ht
R1
R2
RX
hr
nth FRESNEL ZONE
REFLECTION POINT
For the reflection coefficient Γ = ρ e jπ = − ρ :
• If n is even the two paths are out of phase and the received signal is a minimum
• If n is odd the two paths are in phase and the received signal is a maximum
Because the LOS is nearly horizontal Ro ≈ d and therefore Ro = d t + d r ≈ d . For the
nth Fresnel zone R1 + R2 = d + nλ / 2 .
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Fresnel Zones (3)
The radius of the nth Fresnel zone is
Fn =
nλ d t d r
d
or, if the distances are in miles, then
Fn = 72.1
nd t d r
(feet)
f GHz d
Transmission path design: the objective is to find transmitter and receiver locations and
heights that give signal maxima. In general:
1. reflection points should not lie on even Fresnel zones
2. the LOS should clear all obstacles by 0.6 F1, which essentially gives free space
transmission
The significance of 0.6 F1 is illustrated by examining two canonical problems:
(1) knife edge diffraction and
(2) smooth sphere diffraction.
Conversions: 0.0254 m = 1 in; 12 in = 1 ft; 3.3 ft = 1 m; 5280 ft = 1 mi; 1 km = 0.62 mi
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Diffraction (1)
Knife edge diffraction
l = CLEARANCE DISTANCE
l>0
l = 0, SHADOW
BOUNDARY
l<0
ht
SHARP
OBSTACLE
hr
d
Smooth sphere diffraction
l = CLEARANCE DISTANCE
BULGE
ht
l = 0, SHADOW
BOUNDARY
l>0
l<0
hr
SMOOTH
CONDUCTOR
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Diffraction (2)
A plot of
E tot
shows that at 0.6 F1 the free space (direct path) value is obtained.
Edir
SHADOW BOUNDARY
r
E
r
Edir
0
FREE SPACE
FIELD VALUE
in dB
-5
-6
l<0
-10
l>0
0
0.6 F1
CLEARANCE DISTANCE, l > 0
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Path Clearance Example
Consider a 30 mile point-to-point communication link over the ocean. The frequency of
operation is 5 GHz and the antennas are at the same height. Find the lowest height that
provides the same field strength as in free space. Assume standard atmospheric conditions.
The geometry is shown below (distorted
scale). The bulge factor (in feet) is given
dd
approximately by b = t r , where d t
1.5κ
and d r are in miles.
0.6F1
TX
ht
d
b
dt
bmax
dr
Re′
RX
hr
The maximum bulge occurs at the midpoint.
d ≈ dt + dr
(15)(15)
bmax =
= 112.5 ft
(1.5)( 4 / 3)
Fn = 72.1
nd t d r
ft
f GHz d
0.6 F1 = 53 ft
Compute the minimum antenna height:
h = bmax + 0.6 F1
= 112.5 + 53 = 165 ft
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Example of Link Design (1)
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Example of Link Design (2)
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Antennas Over a Spherical Earth
When the transmitter to receiver distance becomes too large the flat Earth approximation
is no longer accurate. The curvature of the surface causes:
1. divergence of the power in the reflected wave in the interference region
2. diffracted wave in the shadow region (note that this is not the same as a ground wave)
The distance to the horizon is d t = RRH ≈ 2 Re′ ht or, if ht is in feet, d t ≈ 2ht miles.
The maximum LOS distance between the transmit and receive antennas is
d max = d t + d r ≈ 2 ht + 2hr (miles)
TANGENT RAY
(SHADOW BOUNDARY)
INTERFERENCE
REGION
hr
ht
dt
R ′e
dr
DIFFRACTION
REGION
SMOOTH
CONDUCTOR
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Interference Region Formulas (1)
Interference region formulas
Ro
R2
R1
hr
ψ
ψ
ht
dt
Re′
dr
SMOOTH
CONDUCTOR
The path-gain factor is given by
F = 1 + ρ e jφ Γ e − jk∆R D
where D is the divergence factor (power) and ∆R = R1 + R2 − Ro .
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Interference Region Formulas (2)
Approximate formulas1 for the interference region:
2

F = (1 + Γ D ) − 4 Γ

1
φ − k∆R   2
D sin 2 Γ

2
 

where
−1


2h1h2
h1 + h2
4 S1S 22T
∆R =
J ( S , T ) , tanψ =
K ( S , T ) , D = 1 +
 (power)
2
d
d
 S (1 − S 2 )(1 + T ) 
d1
d2
S1 =
, S2 =
where h1 is the smallest of either ht or hr
2 Re′ h1
2 Re′ h2
S=
d
S T + S2
= 1
, T = h1 / h2 (< 1 since h1 < h2 )
′
′
1
+
T
2 Reh1 + 2 Reh2
J ( S , T ) = (1 −
S12 )(1 −
S 22 ) ,
and K ( S , T ) =
(1 − S12 ) + T 2 (1 − S 22 )
1+ T2
1
D. E. Kerr, Propagation of Short Radio Waves, Radiation Laboratory Series, McGraw-Hill, 1951 (the formulas have been reprinted in many
other books including R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985).
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Interference Region Formulas (3)
The distances can be computed from d = d1 + d 2 and
d
Φ+π 
−1  2 Re′ (h1 − h2 ) d 
d1 = + p cos
,
 , Φ = cos 
3

2
3


p


and
2 
d2
p=
 Re′ ( h1 + h2 ) +

4 
3 
Another form for the phase difference is
k∆R =
1/ 2
2kh1h2
(1 − S12 )(1 − S 22 ) = νζπ
d
where
4 h13 / 2
h13 / 2
ν=
=
,
λ 2 Re′ 1030λ
ζ =
h2 / h1
(1 − S12 )(1 − S 22 ) ,
d / d RH
and d RH = 2 Re′ h1 (distance to the radio horizon).
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Diffraction Region Formulas (1)
DIRECT RAY
TO HORIZON
SHADOW
BOUNDARY
ht
d
hr
DIFFRACTED
RAYS
R ′e
Approximate formulas for the diffraction region (frequencies > 100 MHz):
F = V1 ( X )U1 ( Z1 )U1 ( Z 2 )
where U 1 is available from tables or curves, Z i = hi / H ( i = 1,2 ), X = d / L , and
1
2 3
1
Re′  3
2
 ( R′ )

2/3
V1 ( X ) = 2 π X e − 2.02 X , L = 2 e  = 28.41λ1/ 3 (km), H = 
=
47
.
55
λ
(m)
 4k 
 2k 


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Diffraction Region Formulas (2)
A plot of U1 ( Z )
Fig. 6.29 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985 (axis labels corrected)
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Surface Waves (1)
At low frequencies (1 kHz to about 3 MHz) the interface between air and the ground acts
like an efficient waveguide at low frequencies for vertical polarization. Collectively the
space wave (direct and Earth reflected) and surface wave are called the ground wave.
SURFACE WAVE
ht
hr
d
R′e
The power density at the receiver is the free space value times an attenuation factor
Pr = Pdir 2 As
2
where the factor of 2 is by convention. Most estimates of As are based calculations for a
surface wave along a flat interface. Approximations for a flat surface are good for
d ≤ 50 /( f MHz )1 / 3 miles. Beyond this distance the received signal attenuates more
quickly.
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Surface Waves (2)
Define a two parameters:
kd
p=
(numerical distance)
+ (σ / ωεo )
ε ε ω 
b = tan −1  r o 
 σ 
1.8 × 10 4σ
A convenient formula is σ / ωεo =
. The attenuation factor for the ground wave
f MHz
2 + 0.3 p
− 0.6 p
o
is approximately As =
−
p
/
2
e
sin
b
(
b
≤
90
)
2
2 + p + 0.6 p
2
ε r2
2
Example: A CB link operates at 27 MHz with low gain antennas near the ground. Find
the received power at the maximum flat Earth distance. The following parameters hold:
Pt = 5 W; Gt = Gr = 1; ε r = 12 and σ = 5 × 10 − 3 S/m. The maximum flat Earth range is
d max = 50 /( 27)1 / 3 = 16.5 miles.
p=
πd / λ
 16.5 
= 0.25 d / λ = 0.0225
(1000) ≈ 601 →
2
2
 0.62 
12 + (90 / 27)
d
= 4p
λ
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Surface Waves (3)
Check b to see if formula applies (otherwise use the chart on the next page)
−1  (12 )(8.85 × 10
−12
)( 2π )( 27 × 10 6 ) 
o
=
74
.
5

5 × 10 − 3

b = tan 

Attenuation constant
As =
2 + 0.3 p
2 + p + 0.6 p 2
− p / 2 e −0.6 p sin b ≈ 8.33 × 10 −4
The received power for the ground wave is
Pr = Pdir 2 As
2
=
Pt Gt Aer
2 As
2
=
(
Pt (1) λ2 / 4π
4π d 2
4π d 2
(5)(8.33 × 10 − 4 ) 2
−14
=
=
1
.
52
×
10
W
2
2
(4π )( 4) (601)
) 2 As 2
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Surface Waves (4)
FLAT EARTH
Fig. 6.35 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985
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Ground Waves (5)
SPHERICAL EARTH ( ε r = 15 and σ = 10 − 2 S/m)
Fig. 6.36 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985
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Urban Propagation (1)
Urban propagation is a unique and relatively new area of study. It is important in the
design of cellular and mobile communication systems. A complete theoretical treatment of
propagation in an urban environment is practically intractable. Many combinations of
propagation mechanisms are possible, each with different paths. The details of the
environment change from city to city and from block to block within a city. Statistical
models are very effective in predicting propagation in this situation.
In an urban or suburban environment there is rarely a direct path between the transmitting
and receiving antennas. However there usually are multiple reflection and diffraction
paths between a transmitter and receiver.
• Reflections from objects close to the
BASE
STATION
mobile antenna will cause multiple signals
ANTENNA
to add and cancel as the mobile unit
moves. Almost complete cancellation can
occur resulting in “deep fades.” These
small-scale (on the order of tens of
wavelengths) variations in the signal are
MOBILE
ANTENNA
predicted by Rayleigh statistics.
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Urban Propagation (2)
• On a larger scale (hundreds to thousands of wavelengths) the signal behavior, when
measured in dB, has been found to be normally distributed (hence referred to a
lognormal distribution). The genesis of the lognormal variation is the multiplicative
nature of shadowing and diffraction of signals along rooftops and undulating terrain.
• The Hata model is used most often for predicting path loss in various types of urban
conditions. It is a set of empirically derived formulas that include correction factors for
antenna heights and terrain.
Path loss is the 1 / r 2 spreading loss in signal between two isotropic antennas. From the
Friis equation, with Gt = G r = 4πAe / λ2 = 1
Pr (1)(1) λ2  1 
Ls =
=
=

2
Pt
2
kr
(4π r )  
2
Note that path loss is not a true loss of energy as in the case of attenuation. Path loss as
defined here will occur even if the medium between the antennas is lossless. It arises
because the transmitted signal propagates as a spherical wave and hence power is flowing
in directions other than towards the receiver.
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Urban Propagation (3)
Hata model parameters * : d = transmit/receive distance (1 ≤ d ≤ 20 km)
f = frequency in MHz (100 ≤ f ≤ 1500 MHz)
hb = base antenna height ( 30 ≤ hb ≤ 200 m)
hm = mobile antenna height (1 ≤ hm ≤ 10 m)
The median path loss is
Lmed = 69.55 + 26.16 log( f ) − 13.82 log( hb ) + [44.9 − 6.55 log( hb ) ]log( d ) + a ( hm )
In a medium city: a ( hm ) = [0.7 − 1.1 log( f )]hm + 1.56 log( f ) − 0.8
1.1 − 8.29 log 2 (1.54hm ), f ≤ 200 MHz
In a large city: a ( hm ) = 
2
4.97 − 3.2 log (11.75hm ), f ≥ 400 MHz
− 2 log 2 ( f / 28) − 5.4, suburban areas
Correction factors: Lcor = 
2
− 4.78 log ( f ) + 18.33 log( f ) − 40.94, open areas
The total path loss is: Ls = Lmed − Lcor
* Note: Modified formulas have been derived to extend the range of all parameters.
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Urban Propagation Simulation
Urban propagation modeling using the wireless toolset Urbana (from Demaco/SAIC). Ray
tracing (geometrical optics) is used along with the geometrical theory of diffraction (GTD)
Closeup showing antenna placement
(below)
Carrier to Interference (C/I) ratio (right)
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Measured Data
Two different antenna heights
h = 2.7 m
h = 1.6 m
f = 3.35 GHz
Measured data in
an urban
environment
f = 8.45 GHz
Three different
frequencies
f = 15.75 GHz
From Masui, “Microwave Path Loss
Modeling in Urban LOS Environments,” IEEE Journ. on Selected Areas
in Comms., Vol 20, No. 6, Aug. 2002.
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Attenuation Due to Rain and Gases (1)
Sources of signal attenuation in the atmosphere include rain, fog, water vapor and other
gases. Most loss is due to absorption of energy by the molecules in the atmosphere. Dust,
snow, and rain can also cause a loss in signal by scattering energy out of the beam.
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Attenuation Due to Rain and Gases (2)
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Attenuation Due to Rain and Gases (3)
There is no complete, comprehensive macroscopic theoretical model to predict loss. A
wide range of empirical formulas exist based on measured data. A typical model:
A = aR b , attenuation in dB/km
R is the rain rate in mm/hr
a = Ga f GHz E a
b = Gb f GHz Eb
where the constants are determined from the following table:
Ga = 6.39 × 10 − 5
= 4.21 × 10 − 5
= 4.09 × 10 − 2
E a = 2.03
= 2.42
= 0.699
Gb = 0.851 Eb = 0.158
= 1.41
= −0.0779
= 2.63
= −0.272
f GHz < 2.9
2.9 ≤ f GHz < 54
54 ≤ f GHz < 180
f GHz < 8.5
8.5 ≤ f GHz < 25
25 ≤ f GHz < 164
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Ionospheric Radiowave Propagation (1)
The ionosphere refers to the upper regions of the atmosphere (90 to 1000 km). This region is
highly ionized, that is, it has a high density of free electrons (negative charges) and positively
charged ions. The charges have several important effects on EM propagation:
1. Variations in the electron density ( N e ) cause waves to bend back towards Earth, but
only if specific frequency and angle criteria are satisfied. Some examples are shown
below. Multiple skips are common thereby making global communication possible.
N e max
4
IONOSPHERE
3
2
1
TX
SKIP DISTANCE
EARTH’S SURFACE
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Ionospheric Radiowave Propagation (2)
2. The Earth’s magnetic field causes the ionosphere to behave like an anisotropic medium.
Wave propagation is characterized by two polarizations (“ordinary” and “extraordinary” waves). The propagation constants of the two waves are different. An
arbitrarily polarized wave can be decomposed into these two polarizations upon entering
the ionosphere and recombined on exiting. The recombined wave polarization will be
different that the incident wave polarization. This effect is called Faraday rotation.
The electron density distribution has the general characteristics shown on the next page. The
detailed features vary with
•
•
•
•
location on Earth,
time of day,
time of year, and
sunspot activity.
The regions around peaks in the density are referred to as layers. The F layer often splits
into the F1 and F2 layers.
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Electron Density of the Ionosphere
(Note unit is per cubic centimeter)
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The Earth’s Magnetosphere
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Ionospheric Radiowave Propagation (3)
Relative dielectric constant of an ionized gas (assume electrons only):
ω 2p
εr =1 −
ω (ω − jν )
where: ν = collision frequency (collisions per second)
N ee 2
ωp =
, plasma frequency (radians per second)
mε o
N e = electron density ( / m3 )
e = 1.59 × 10 −19 C, electron charge
m = 9.0 × 10 − 31 kg, electron mass
For the special case of no collions, ν = 0 and the corresponding propagation constant is
k c = ω µoε rε o = ko 1 −
ω 2p
ω2
where ko = ω µoε o .
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Ionospheric Radiowave Propagation (4)
Consider three cases:
1. ω > ω p : k c is real and e − jk c z = e − j k c z is a propagating wave
2. ω < ω p : k c is imaginary and e − jk c z = e − k c z is an evanescent wave
3. ω = ω p : k c = 0 and this value of ω is called the critical frequency, ωc
At the critical frequency the wave is reflected. Note that ωc depends on altitude because
the electron density is a function of altitude. For electrons, the highest frequency at
which a reflection occurs is
ω = ωc ⇒ ε r = 0
IONOSPHERE
REFLECTION
POINT
ωc
≈ 9 N e max
2π
Reflection at normal incidence requires
the greatest N e .
h′
TX
fc =
EARTH’S
SURFACE
1
The critical frequency is where the propagation constant is zero.
Neglecting the Earth’s magnetic field, this occurs at the plasma
frequency, and hence the two terms are often used
interchangeably.
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Ionospheric Radiowave Propagation (5)
At oblique incidence, at a point of the ionosphere where the critical frequency is f c , the
ionosphere can reflect waves of higher frequencies than the critical one. When the wave
is incident from a non-normal direction, the reflection appears to occur at a virtual
reflection point, h ′ , that depends on the frequency and angle of incidence.
VIRTUAL
HEIGHT
IONOSPHERE
h′
EARTH’S
SURFACE
TX
SKIP DISTANCE
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Ionospheric Radiowave Propagation (6)
To predict the bending of the ray we use a layered approximation to the ionosphere just as
we did for the troposphere.
ALTITUDE
M
z3
z2
z1
ψ2
ψ1
ψi
ψ3
ε r ( z3 )
ε r (z2 )
ε r ( z1 )
LAYERED
IONOSPHERE
APPROXIMATION
εr = 1
Snell’s law applies at each layer boundary
sin ψ i = sin (ψ 1 ) ε r ( z1 ) = L
The ray is turned back when ψ ( z ) = π / 2 , or sinψ i = ε r ( z )
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Ionospheric Radiowave Propagation (7)
Note that:
1. For constant ψ i , N e must increase with frequency if the ray is to return to Earth
(because ε r decreases with ω ).
2. Similarly, for a given maximum N e ( N e max ), the maximum value of ψ i that results in
the ray returning to Earth increases with increasing ω .
There is an upper limit on frequency that will result in the wave being returned back to
Earth. Given N e max the required relationship between ψ i and f can be obtained
sin ψ i = ε r ( z)
ω 2p
2
sin ψ i = 1 − 2
ω
81N e max
1 − cos 2 ψ i = 1 −
f2
f 2 cos 2 ψ i
N e max =
⇒
81
f max =
81N e max
cos2 ψ i
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Ionospheric Radiowave Propagation (8)
Examples:
1. ψ i = 45o , N e max = 2 × 1010 / m 3 : f max = (81)(2 × 1010 ) /(0 .707) 2 = 1.8 MHz
2. ψ i = 60o , N e max = 2 × 1010 / m 3 : f max = (81)( 2 × 1010 ) /(0.5) 2 = 2.5 MHz
The value of f that makes ε r = 0 for a given value of N e max is the critical frequency
defined earlier:
f c = 9 N e max
Use the N e max expression from previous page and solve for f
f = 9 N e max secψ i = f c secψ i
This is called the secant law or Martyn’s law. When secψ i has its maximum value, the
frequency is called the maximum usable frequency (MUF). A typical value is less than 40
MHz. It can drop as low as 25 MHz during periods of low solar activity. The optimum
usable frequency (OUF) is 50% to 80% of the MUF.
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Maximum Usable Frequency
The maximum usable frequency (MUF) in wintertime for different skip distances. The MUF
is lower in the summertime.
Fig. 6.43 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985
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Ionospheric Radiowave Propagation (9)
Multiple hops allow for very long range communication links (transcontinental). Using a
simple flat Earth model, the virtual height ( h ′ ), incidence angle (ψ i ), and skip distance (d )
d
are related by tanψ i =
. This implies that the wave is launched well above the horizon.
2h ′
However, if a spherical Earth model is used and the wave is launched on the horizon then
d = 2 2 Re′ h′ .
EFFECTIVE SPECULAR
REFLECTION POINT
IONOSPHERE
IONOSPHERE
h′
ψi
TX
d
Single ionospheric hop
(flat Earth)
EARTH’S
SURFACE
Multiple ionospheric hops
(curved Earth)
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Ionospheric Radiowave Propagation (10)
Approximate virtual heights for layers of the ionosphere
Layer
F2
F1
F
E
Range for h ′ (km)
250 to 400 (day)
200 to 250 (day)
300 (night)
110
Example: Based on geometry, a rule of thumb for the maximum incidence angle on the
ionosphere is about 74 o . The MUF is
MUF = f c sec(74 o ) = 3.6 f c
For N e max = 1012 / m 3 , f c ≈ 9 MHz and the MUF = 32.4 MHz. For reflection from the F2
layer, h ′ ≈ 300 km. The maximum skip distance will be about
d max ≈ 2 2 Re′ h ′ = 2 2(8500 × 10 3 )(300 × 10 3 ) = 4516 km
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Ionospheric Radiowave Propagation (7)
1 + h ′ / Re′ − cosθ
1
=
sin θ
tanψ i
where
d
θ=
2 Re′
For a curved Earth, using the law of sines for a triangle
R/2
ψi
R/2
h′
and the launch angle (antenna
pointing angle above the horizon)
is
∆ = φ − 90 o = 90 o − θ −ψ i
∆
φ
d/2
LAUNCH ANGLE:
o
o
∆ = 90 −θ −ψ i = φ − 90
Re′
θ
The great circle path via the
reflection point is R, which can be
obtained from
R=
2 Re′ sin θ
sin ψ i
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Ionospheric Radiowave Propagation (8)
Example: Ohio to Europe skip (4200 miles = 6760 km). Can it be done in one hop?
To estimate the hop, assume that the antenna is pointed on the horizon. The virtual height
required for the total distance is
d / 2 = Re′ θ → θ = d / (2 Re′ ) = 0.3976 rad = 22.8 degrees
( Re′ + h ′) cos θ = Re′ → h ′ = Re′ /cosθ − Re′ = 720 km
This is above the F layer and therefore two skips must be used. Each skip will be half of
the total distance:. Repeating the calculation for d / 2 = 1690 km gives
θ = d / (2 Re′ ) = 0.1988 rad = 11.39 degrees
h ′ = Re′ /cosθ − Re′ = 171 km
This value lies somewhere in the F layer. We will use 300 km (a more typical value) in
computing the launch angle. That is, still keep d / 2 = 1690 km and θ = 11.39 degrees, but
point the antenna above the horizon to the virtual reflection point at 300 km
300
tanψ i = sin(11.39 o ) 1 +
− cos(11.39 o )
 8500

−1
→ ψ i = 74.4 o
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Ionospheric Radiowave Propagation (9)
The actual launch angle required (the angle that the antenna beam should be pointed above
the horizon) is
launch angle, ∆ = 90 o − θ − ψ i = 90 o − 11.39 o − 74.4 o = 4.21o
The electron density at this height (see chart, p.3) is N e max ≈ 5× 1011 / m 3 which
corresponds to the critical frequency
f c ≈ 9 N e max = 6.36 MHz
and a MUF of
MUF ≈ 6.36 sec 74.4 o = 23.7 MHz
Operation in the international short wave 16-m band would work. This example is
oversimplified in that more detailed knowledge of the state of the ionosphere would be
necessary: time of day, time of year, time within the solar cycle, etc. These data are
available from published charts.
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Ionospheric Radiowave Propagation (10)
Generally, to predict the received signal a modified Friis equation is used:
Pr =
Pt Gt G r
(4πR / λ )
2
Lx Lα
where the losses, in dB, are negative:
L x = Lpol + Lrefl − Giono
Lrefl = reflection loss if there are multiple hops
Lpol = polarization loss due to Faraday rotation and earth reflections
Giono = gain due to focussing by the curvature of the ionosphere
Lα = absorption loss
R = great circle path via the virtual reflection point
Example: For Pt = 30 dBW, f = 10 MHz, Gt = G r = 10 dB, d = 2000 km, h ′ = 300 km,
L x = 9.5 dB and Lα = 30 dB (data obtained from charts).
From geometry compute: ψ i = 70.3o , R = 2117.8 km, and thus Pr = −108.5 dBw
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Ducts and Nonstandard Refraction (1)
Ducts in the atmosphere are caused by index of refraction rates of decrease with height
over short distances that cause rays to bend back towards the surface.
TOP OF DUCT
EARTH’S
SURFACE
TX
• The formation of ducts is due primarily to water vapor, and therefore they tend to occur
over bodies of water (but not land-locked bodies of water)
• They can occur at the surface or up to 5000 ft (elevated ducts)
• Thickness ranges from a meter to several hundred meters
• The trade wind belts have a more or less permanent duct of about 1 to 5 m thickness
• Efficient propagation occurs for UHF frequencies and above if both the transmitter and
receiver are located in the duct
• If the transmitter and receiver are not in the duct, significant loss can occur before
coupling into the duct
72
Naval Postgraduate School
Antennas & Propagation
Distance Learning
Ducts and Nonstandard Refraction (2)
Because variations in the index of refraction are so small, a quantity called the refractivity
is used
N ( h) = [n( h) − 1]10 6
n (h ) = ε r ( h)
In the normal (standard) atmosphere the gradient of the vertical refractive index is linear
with height, dN / dh ≈ −39 N units/km. If dN / dh < −157 then rays will return to the
surface. Rays in the three Earth models are shown below.
True Earth
Re
Equivalent Earth
(Standard Atmosphere)
Re′
Flat Earth
∞
From Radiowave Propagation, Lucien Boithias, McGraw-Hill
73
Naval Postgraduate School
Department of Electrical & Computer Engineering
Monterey, California
Ducts and Nonstandard Refraction (3)
Another quantity used to solve ducting problems is the modified refractivity
M (h ) = N ( h ) + 10 6 (h / Re′ )
In terms of M, the condition for ducting is dM /dh = dN / dh + 157 . Other values of dN / dh
(or dM / dh ) lead to several types of refraction as summarized in the following figure and
table. They are:
1. Super refraction: The index of refraction decrease is more rapid than normal and the ray
curves downward at a greater rate
2. Substandard refraction (subrefraction): The index of refraction decreases less rapidly
than normal and there is less downward curvature than normal
From Radiowave Propagation, Lucien Boithias, McGraw-Hill
74
Naval Postgraduate School
Department of Electrical & Computer Engineering
Monterey, California
Ducts and Nonstandard Refraction (4)
Summary of refractivity and ducting conditions
dN / dh
>0
0
dN
0>
> −39
dh
-39
dN
− 39 >
> − 157
dh
Ray
Curvature
κ
up
none
<1
1
down
>1
4/3
> 4/3
Atmospheric
Refraction
below
normal
normal
Horizontally
Launched Ray
more convex
actual
less
convex
moves
away
from
Earth
above
normal
plane
parallel to
Earth
super-refraction
concave
draws closer
to Earth
-157
< -157
Virtual
Earth
75