Radio Wave Propagation - Exvacuo

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Radio Wave Propagation
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
1
Radio Wave Propagation Terminology
Free space (propagation medium)
u Electromagnetic waves
u Transmitting and receiving
antenna
u
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
2
Transverse Electromagnetic Waves
Dir
e
n
ctio
of
P
ag
rop
atio
n
z
y
Magnetic Field
Electric Field
x
u
The waves propagate as transverse electromagnetic waves (TEM) - i.e.
the electric field, the magnetic field, and the direction of travel of the waves
are all mutually perpendicular.
3
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
Speed & Wavelength of em Waves
u
The speed of propagation (ν) and the wavelength (λ) of an
electromagnetic wave are given, respectively, by:
v=
c
v
and λ =
f
∈r
where c = 3x108 m/s, ∈ r
=
medium’s relative permittivity
or dielectric constant, and f = frequency of wave in Hz.
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
4
Characteristic Impedance
u
u
The characteristic impedance of a medium is the ratio of
the electric field intensity and the magnetic field intensity,
i.e., η = E/H.
For free space, η = 120π = 377 Ω. For other media:
(η
η is also called, the intrinsic impedance)
η=
µ
377
or
∈
∈r
where µ = medium’s permeability, in H/m
and ∈ = m e d i u m ’s permittivity in F/m
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Electromagnetic spectrum
Radio waves, infrared,
visible light,
ultraviolet, X rays,
and gamma rays
are all different
forms of
electromagnetic
radiation.
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Relation between power density and
transmitted power
u
u
Let P t be the transmitted power into
free space.
The power density at any point at
distance R is
SR =
Pt
4πR 2
Watts / m 2
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
7
Example 1:
u
The power density from a point source in free space at 20
km is 200 micro-watts/m 2. Find the power density at 25 km
away from the source
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
8
Relation between power density and
electric field
u
u
u
For an uniform plan wave the E and H
are related by: η = E R/HR
ER is the rms value of the electric field
The average poynting vector at distance R
from the transmitter
SR =
Hence
ER =
Pt
E 2R
E 2R
=
E
xH
=
=
R
R
4πR 2
η 120π
30Pt
R
Volts / m
9
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
The Effective Isotropic Radiated
Power
Since a transmitting antenna focuses energy in a specific
way, it has “gain” over an isotropic radiator in a particular
direction. One can speak of the effective isotropic radiated
power:
EIRP = G t Pt
where Pt = total transmitter power,
and G t = gain of transmitter antenna
If a transmitting antenna has a gain Gt , then
ER =
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
30Pt G t
R
10
Transmission between two antennas
11
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
Friis transmission formula
u
The average power density at the receiving
antenna is
Sr = G t
Whereby
u
Pt
ξDP ξAP
= t t 2t = t 2 t 2 t
2
4 πR
4 πR
λR
Dt =
4π
λ2
At
Has been used
The power intercepted by the receiving antenna
with effective area A r is
Pint = S r A r =
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
ξ t Pt A t A r
λ2 R 2
12
Friis transmission formula
u
The receiving power Pr delivered to the receiver is
ξA PA
 λ 
Pr = ξ r t 2t t2 r = G t G r Pt 

λR
 4πR 
2
Or
 λ 
Pr = G t G r Pt 

 4πR 
2
Friis transmission formula
13
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
Path Loss
u
As the signal propagates there will be loss due to the
spreading of the wave outward from the source
Pr =
G t G r Pt
 4 πR 


 λ 
2
P
 4πR  1
= t


 λ  G t G r Pr
2
hence
P 
c
 4 πdf 
Path Loss = 10 Log  t  = 10 log
 − G t (dBi ) − G r(dBi ) ; R = d and λ =
f
 c 
 Pr 
2
Or
Path Loss = Pt ( dB) − Pr (dB)
= 33 .44 + 20 log( d km ) + 20 log( f MHz ) − G t ( dBi) − G r ( dBi)
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Example 2:
In a satellite communication system, the free-space conditions may be
assumed. The satellite is at a height of 36,000 km above the earth,
the frequency used is 4000 MHz, the transmitting antenna gain is 15
dB, and the receiving antenna gain is 45 dB.
Calculate
(a) the free space transmission loss
(b) the received power when the transmitted power is 200W.
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
15
Radio Propagation Mechanisms:
Effects of environment
During the propagation of electromagnetic waves the
waves experience:
Different environmental characteristics: land, sea, space, etc.
Reflection
u Refraction
u Diffraction
u Interference
u
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Reflection
Radio waves behave like light waves:
u These are reflected by a surface where the
angle of incidence, θi = the angle of reflection,
θr.
Incident
Ray
Normal
θi
θr
Reflected
Ray
Conductor
The reflection coefficient is defined as:. ρ = E reflected
E incident
ρ is unity for a perfect conductor and will be less than 1 in case of practical situations
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Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
Refraction
u
Radio waves will be bend or refracted when they travel from
one medium with refractive index, n 1 to another with
refractive index, n 2.
The angles involved are given by :
n1
θi
n1<n2
sinθ t v 2
ε
=
= 1
sinθ i v1
ε2
ε1and ε 2
n2
θt
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
are dielectric constants of medium
1 and 2 respectively.
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Diffraction
u
Diffraction is the phenomenon which results in radio waves that normally
travel in a straight line to bend around an obstacle.
Direction of wave propagation
T
R
Obstacle
Obstruction
u
Diffraction generates unwanted side lobes and prevents the desired
narrow pencil beam of radiation.
19
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
Interference
u
When two em-waves that left the source, travel different length paths
and arrive at a point at the same instant of time,
l
there will be interference due to the phase shift of the two waves..
T
R
ht
hr
Phase shift!
u
When the antenna is located near ground the direct signal
and ground reflection will reach the receiving antenna.
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Interference
Lobes
Null
Null
Ground surface
u
If the path difference corresponds to half wavelength then
there will be phase opposition between the two, resulting in
cancellation at the receiver.
u
If the path difference is one wavelength, the two waves
reinforce each other resulting in lobe and null antenna
pattern.
If the target lies in null direction, then it cannot be detected.
u
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Radio Wave Propagation Modes
u
depends on several factors such as
u
u
u
frequency of operation,
distance between transmitting and receiving antennas
Radio wave propagation can be :
u
u
u
Ground wave or surface wave propagation
Sky wave or ionospheric propagation
Space wave propagation
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Ground Wave or Surface Wave
Propagation
u
In ground wave propagation the electromagnetic
wave is guided along the surface of the earth.
u
The ground wave is of practical importance at
broadcast and lower frequencies i.e. for medium
waves, long waves and very long waves.
l
l
It is used in local broadcasting.
All the broadcast signals received during daytime are due to
ground wave propagation.
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Ground Wave or Surface Wave
Propagation
u
The ground wave, therefore, suffers attenuation while
propagating along the curvature of the earth,
depending on:
l
u
frequency, surface irregularities, permittivity and conductivity.
Earth's attenuation increases as the frequency
increases and hence the mode of propagation is
suitable for low and medium frequencies up to 2 MHz
only.
25
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
surface wave attenuation due to wave
diffraction and tilting
u
As the wave progress over the curvature of the earth, the wave
fronts starts gradually tilting more and more.
u
The tilt of wave causes short circuit of the electric field
component and hence the field strength goes on reducing.
Wave tilting
Earth surface
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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surface wave attenuation due to wave
diffraction and tilting
The surface wave dies because of the losses mentioned above. It
may be noted that maximum range of surface wave propagation
depends
u
u
not only on the frequency but on the power as well.
The range of transmission can be increased by increasing the
power of the transmitter in the VLF band but this method can not
be effective at the MF band where the tilting due to diffraction is
more effective.
u
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Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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The field strength of ground wave at a distance “d” from
transmitter is given by
E=
u
120πh t I
λd
The voltage induced in receiving antenna placed at this
point is given by
V = Exhr =
120πh thr I
λd
where ht and hr are the effective heights of transmitting
and receiving antennas respectively.
I = antenna current, λ = wavelength,
d = distance from transmitter.
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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Example 3:
A 150 m antenna transmitting at 1.2 M Hz has an
antenna current of 8A. Find the voltage received
by a receiving antenna of height 2 m and 40 km
away from transmitting antenna
Prepared by Dr. Abbou Fouad Mohammed, Multimedia University,
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