Satellite Communication Principles
Abdullah Al-Ahmadi
College of Engineering
Majmaah University∗
∗
URL: http://faculty.mu.edu.sa/aalahmadi; Electronic address: [email protected]
1
Part I
Introduction
A.
Key Dates in Satellite Communication History
1957 USSR Sputnik-I, the first man-made satellite, a battery-driven radio beacon.
1958 US Explorer-I, the first US satellite.
1958 US Score satellite launched and broadcast the President Eisenhower’s State of the
Union address.
1962 Bell Labs launched Telestar I and II, which are considered the fist “classic” telecommunications satellite.
• They had 6389 MHz uplinks and 4169 MHz downlinks and solar batteries (longer
lifetime ... early satellites like Sputnik and Explorer had purely battery-driven
systems and died less than a year after launch).
Mid-1980s GPS came online publicly. Operated by the US Air Force, it was the first
globally available positioning system.
• Developed in the 70s with the initial impetus to guide ICBMs.
2000 LEO telephony systems (Iridium, Globalstar, Orbcomm).
B.
Types of Earth Orbits
• There are 4 types of earth orbital altitudes:
• Low Earth Orbit (LEO) is any orbit with an altitude less than 1500 km.
– These are cheapest to launch, but zip across the sky very quickly. Due to low
launch and replacement costs, many communications satellites use networks of
LEO satellites to trunk data and mobile telephony from one point on the earth
to another.
2
∗ This orbit is also used for surveying and weather satellites.
– Medium Earth Orbit (MEO) is any satellite orbit above 1500 km and less than
36,000 km.
∗ These orbits are less populated because they are more expensive to launch
into.
∗ Earth coverage is better here, however, so some satellite systems that require
multiple simultaneous satellites for operation (GPS) operate here.
– Geosynchronous Earth Orbit (GEO) is the designation for any satellite that is
36,000 km.
∗ This orbit is again highly populated because of the unique functionality a
stationary satellite provides for broadcasting information to fixed dishes on
earth.
– High Earth Orbit (HEO) is any orbit above the GEO 36,000 km altitude.
∗ Extremely rare for anything other than scientific spacecraft or de-orbited
GEO satellites.
3
Part II
Orbital Mechanics
I.
CIRCULAR ORBITS
• To achieve a stable orbit, a spacecraft must be first beyond the earth’s atmosphere
(400, 000 ft ≈ 122 km) .
• Newton’s laws of motion are:
1
at2
s = ut +
2
v 2 = u2 + 2at
(1)
(2)
v = u + at
(3)
F = ma
(4)
where
– s is the distance traveled from time t = 0.
– u is the initial velocity of the object at time t = 0.
– v is the final velocity of the object at time t.
– a is the acceleration of the object.
– m is the mass of the object.
– F is the force acting on the object.
• Equations 4 states that the force acting on a body is equal to the mass of the body
multiplied by the resulting acceleration of the body.
• Alternatively, the resulting acceleration is the ratio of the force acting on the body to
the mass of the body.
a=
• The force has a unit of Newton N :
4
F
m
(5)
Figure 2.1 (p. 18)
Forces acting on a satellite in a stable
orbit around the earth (from Fig. 3.4 of
reference 1). Gravitational force is
inversely proportional to the square of
the distance between the centers of
gravity of the satellite and the planet the
satellite is orbiting, in this case the earth.
The gravitational force inward (FIN, the
centripetal force) is directed toward the
center of gravity of the earth. The kinetic
energy of the satellite (FOUT, the
centrifugal force) is directed diametrically
opposite to the gravitational force. Kinetic
energy is proportional to the square of
the velocity of the satellite. When these
inward and outward forces are balanced,
the satellite moves around the earth in a
free fall trajectory: the satellite s orbit.
For a description of the units, please see
the text.
Figure 1: Forces
acting
on a satellite
in Pratt,
a stable
the earth.
Satellite
Communications,
2/E by Timothy
Charles point
Bostian, &around
Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
– A Newton is the force required to accelerate a mass of 1 kg with an acceleration
of 1 m/s2 .
• Thus, the lighter the mass of the body, the higher the acceleration.
• In a stable orbit, two main forces are acting on the satellite as shown in Figure 1:
1. A centrifugal force FOUT :
– Due to the kinetic energy of the satellite.
– Tries to push the satellite to a higher orbit.
2. A centripetal force FIN :
– Due to the the gravitational attraction of the planet about which the satellite
is orbiting.
– Attempts to pull the satellite toward the earth.
• If these two forces are equal, the satellite will remain in a stable orbit.
• The acceleration a, due to centripetal force at distance r from the center of the earth
is:
a = µ/r2 km/s2
5
(6)
• The constant µ is the product of the universal gravitational constant G and the mass
of the earth ME .
µ = GME = 3.986004418 × 105 km3 /s2
(7)
• The constant µ is called Kepler’s constant.
– G = 6.672 × 10−11 Nm2 /kg2
– Me = 5.98 × 1024 kg
• The mean earth radius is 6378.137 km.
• From Equation 4, the centripetal force acting on the satellite, FIN is given by:
FIN = m × µ/r2
(8)
= m × GME /r
2
• The acceleration a, due to centrifugal force is given by:
a = v 2 /r
• Hence, the centrifugal force FOUT is:
FOUT = m × v 2 /r
(9)
• If the forces on the satellite are balanced, FIN = FOUT , then using Equations 8 and 9:
m × µ/r2 = m × v 2 /r
• Hence, the velocity v of a satellite in a circular orbit is given by:
v = (µ/r)1/2
• In circular orbits, the distance traveled by a satellite in one orbit around a planet is
equal to the circumference of that orbit, 2πr.
6
• Since time is equal distance divided by velocity, then the period of the satellite’s orbit,
T , is:
2πr
v
h
i
= (2πr) / (µ/r)1/2
∴ T = 2πr3/2 / µ1/2
T =
Example: Calculate the orbital period and velocity for a satellite in the GEO at
35, 786.03 km above the surface of the earth.
Solution:
• Orbital velocity:
v = (µ/r)1/2
1/2
3.986004418 × 105
=
35, 786.03 + 6378.137
= 3.0747 km/s
• Orbital period:
T = 2πr3/2 / µ1/2
=
2π × [35, 786.03 + 6378.137]3/2
[3.986004418 × 105 ]1/2
= 86164.08246 seconds
= 23 hours 56 minutes 4.08 seconds
II.
KEPLER’S LAWS OF PLANETARY MOTION
• Jahannes Kepler (1571-1630) developed three laws of planetary motion. Kepler’s three
laws are:
1. The orbit of any smaller body about a larger body is always an ellipse, with the
center of mass of the larger body as one of the foci.
2. The orbit of the smaller body sweeps out equal areas in equal time. Figure 2
7
Figure 2.5 (p. 22)
Illustration of Kepler s second law of planetary motion. A satellite is in orbit about the planet earth, E. The orbit is an ellipse
with a relatively high eccentricity, that is, it is far from being circular. The figure shows two shaded portions of the elliptical
plane in which the orbit moves, one is close to the earth and encloses the perigee while the other is far from the earth and
encloses the apogee. The perigee is the point of closest approach to the earth while the apogee is the point in the orbit that
is furthest from the earth. While close to perigee, the satellite moves in the orbit between times t1 and t2 and sweeps out an
area denoted by A12. While close to apogee, the satellite moves in the orbit between times t3 and sweeps out an area
denoted by A34. If t1 – t2 = t3 – t4, then A12 = A34.
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
Figure 2: Illustration of Kepler’s second law of planetary motion
3. The square of the period of revolution of the smaller body about the larger body
equals a constant multiplied by the third power of the semimajor axis of the
orbital ellipse.
– That is:
T2 =
4π 2 a3
µ
where T is the orbital period, a is the semimajor axis of the orbital ellipse
and µ is Kepler’s constant.
– If the orbit is circular, then a becomes distance r.
III.
DESCRIBING THE ORBIT OS A SATELLITE
• Referring to Figure 3:
– The semimajor axis of the ellipse is denoted by a, and the semiminor axis, by b.
– The relationship between the semimajor axes a and the semiminor axis b is as
follows:
p
b = a (1 − e2 )
8
Figure 2.6 (p. 24)
The orbit as it appears in the orbital plane, The point O is the center of the earth and the point C is the center of the ellipse.
The two centers do not coincide unless the eccentricity, e, of the ellipse is zero (i.e., the ellipse becomes a circle and a = b).
The dimensions a and b are the semimajor and semiminor axes of the orbital ellipse, respectively.
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Figure 3: Orbit of aCopyright
satellite© 2003 John Wiley & Sons. Inc. All rights reserved.
– The eccentricity e is given by:
√
e=
a2 − b 2
a
– where e is the eccentricity and it has values between 0 and 1
0≤e≤1
– The Perigee is the distance between the earth and the spacecraft at the closest
distance to the planet and it is equal to rb = (1 − e) a.
– The Apogee is the distance between the earth and the spacecraft at the furthest
distance to the planet and it is equal to ra = (1 + e) a.
Example: A satellite is in an elliptical orbit with a perigee of 1000 km and
apogee of 4000 km. Using a mean earth radius of 6378.14 km, find the period
of the orbit in hours, minutes, and seconds.
Solution:
9
For a semimajor axis length a, earth radius re , perigee height hp , and apogee ha :
2a = 2re + hp + ha
= 2 (6378.14) + 1000 + 4000
= 17756.28 km
∴ a = 8878.14 km
Using Kepler’s third law:
T
2
4π 2 a3
=
µ
4π 2 (8878.14)3
=
3.986004418 × 105
= 6.9308728 × 107 s2
∴ T = 8325.1864 s
= 2 hours 18 minutes 45.19 seconds
IV.
LOOK ANGLE DETERMINATION
• Earth surface was divided into a grid-like structure of orthogonal lines: Latitude and
Longitude.
– Latitude is the angular distance measured in degrees north or south the equator.
– Longitude is the angular distance measured in degrees from a given reference
longitudinal line.
• There are 360◦ of longitude measured from 0◦ at the Greenwich.
• There are 90◦ of latitude.
– Latitude +90◦ is the north pole.
– Latitude −90◦ is the south pole.
• When GEO satellites are registered in Geneva, their subsatellite location is given in
degrees east to avoid confusion.
10
subsatellite point would see the satellite at zenith (i.e., at an elevation angle of 90°). The pointing direction from the satellite
to the subsatellite point is the nadir direction from the satellite. If the beam from the satellite antenna is to be pointed at a
location on the earth that is not at the subsatellite point, the pointing direction is defined by the angle away from nadir. In
general, two off-nadir angles are given: the number of degrees north (or south) from nadir. In general, two off-nadir angles
are given: the number of degrees north (or south) from nadir; and the number of degrees east (or west) from nadir. East,
west, north, and south directions are those defined by the geography of the earth.
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright
© 2003 John Wiley & Sons. Inc. All rights reserved.
Figure 4: The subsatellite
point
– Thus, INTELSAT primary location is at 335.5◦ E (not 24.5◦ W).
– Since all U.S satellites are located west of Greenwich, it has become accepted to
be described in terms of degrees W.
• The coordinates to which an earth station antenna must be pointed to communicate
with a satellite are called the Look Angles.
• In order to determine the look angle, we first have to calculate the subsatellite point.
• The subsatellite point is the location on the surface of the earth that lies directly
between the satellite and the center of the earth. Figure 4
• It is the Nadir pointing direction from the satellite.
• To an observer at the subsatellite point:
– The satellite will appear to be directly overhead in the Zenith direction.
A.
Elevation Angle Calculation.
• From Figure 5:
11
– rs is the vector from the center of the earth to the satellite.
– re is the vector from the center of the earth to the earth station.
– d is the vector from the earth station to the satellite.
– The central angle γ measured between re and rs is the angle between the earth
station and the satellite.
– ψ is the angle measured from re to d.
– The central angle γ is given by:
cos (γ) = cos (Le ) cos (ls − le )
– For a satellite to be visible from the earth station, then γ should be:
re
−1
γ ≤ cos
r
s
6378.137
−1
γ ≤ cos
42164.17
◦
γ ≤ 81.3
– This called a visibility test.
– Then the elevation angle El can be calculated by:
rs
−1
El = tan
− cos γ / sin γ − γ
re
= tan−1 [(6.6107345 − cos γ) / sin γ] − γ
B.
Azimuth Angle Calculation.
• First we have to calculate an intermediate angle α by:
−1 tan |(ls − le )|
α = tan
sin (Le )
• Having the intermediate angle α, the azimuth look angle Az can be found from:
– Case 1: Earth station in the Northern Hemisphere with:
1. Satellite to the SE of the earth station: Az = 180◦ − α
12
Figure 2.12 (p. 33)
The geometry of elevation angle
calculation. The plane of the paper is the
plane defined by the center of the earth,
the satellite, and the earth station. The
central angle is γ. The elevation angle EI is
measured upward from the local horizontal
at the earth station.
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
Figure 5: The geometry of elevation
2. Satellite to the SW of the earth station: Az = 180◦ + α
– Case 2: Earth station in the Southern Hemisphere with:
1. Satellite to the NE of the earth station: Az = α
2. Satellite to the NW of the earth station: Az = 360◦ − α
Example: An earth station is situated at Majmaah University (25.89◦
N,45.35◦ E), needs to calculate the look angle to ARABSAT 5B (26◦ E).
Solution:
1. Find the central angle γ
cos γ = cos (Le ) cos (ls − le )
= cos (26.89) cos (26 − 25.89)
= 0.8988
γ = 26
The central angle γ is less than 81.3◦ , so the satellite is visible from the earth
station.
13
2. Find the elevation angle El
El = tan−1 [(6.6107345 − cos γ) / sin γ] − γ
= tan−1 [(6.6107345 − cos (26)) / sin (26)] − 26
= 59.611◦
3. Find the intermediate angle α
tan |(ls − le )|
α = tan
sin (Le )
−1 tan |(26 − 45.35)|
= tan
sin (25.89)
◦
= 38.81
−1
4. Find the azimuth angle
• Since the satellite is SW of the earth station then:
Az = 180 + α
= 180 + 38.81
= 218.81◦
14
Part III
Orbital Perturbation
• Sources of Orbital Perturbations:
1. Other celestial bodies.
(a) Moon: because of its closeness to the earth.
(b) Sun: because of its size.
2. Longitudinal variations:
(a) Earth oblateness.
(b) Mass concentration (Mascons):
Areas with higher and lower density.
(252◦ E, 75◦ ) and (162◦ , 348◦ )
• Orbital Forcing:
– Cycles of orbital perturbation on Earth’s orbit around the sun.
– Largely due to Jupiter, although other planets contribute slightly.
Figure 2.14 (p. 40)
– Influences long-term climate (Milankovitch cycles).
– Main results of orbital forcing:Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
15
1. Eccentricity variations.
∗ Eccentricity will vary slightly over time due to the influence of other
planets.
∗ The variation is between e=0.005 and 0.058, which occurs over 413,000
year cycles. (Currently at e=0.012).
∗ The planet receives more insolation (level of sunlight and energy from
the sun) at perihelion than at aphelion.
∗ As eccentricity increases, the perihelion portion of the orbit becomes
warmer and the aphelion becomes cooler.
2. Precession variations.
∗ The earth’s off-axis tilt of 23◦ itself spins around slowly over time.
∗ The earth experiences precession in 26,000 year cycles.
3. Obliquity variations.
∗ Earth rotation axis wobbles between 21.5◦ and 24.5◦ (Currently at
23.44◦ ).
∗ Occurs in 41000 year cycles.
∗ As obliquity increases:
· Seasonal variations become more severe across the planet as higher
latitudes receive more insolation in the summer and less in the winter.
4. Inclination variations.
∗ Earth’s orbital inclination wobbles in and out of the sun’s equatorial
plane.
∗ The earth passes through the plane of the solar system twice a year.
∗ During this phase, observed meteor activity always increases.
16
Part IV
Launches and Launch Vehicles
• A satellite cannot be placed into a stable orbit unless two parameters are correct:
1. The velocity vector.
2. The orbital height.
• A GEO satellite must be in an orbit at:
– A height of 35786.03 km above the surface of the earth (42164.17 km from the
center of the earth).
– Inclination of zero degrees.
– An ellipticity of zero.
– A velocity of 3074.7 m/s tangential to the earth in the plane of the orbit.
I.
PLACING SATELLITES INTO GEOSTATIONARY ORBIT
A.
Geostationary Transfer Orbit and AKM
• Place the spacecraft with the final rocket stage still attached into low earth orbit.
• After a couple of orbits, the final stage is reignited and the spacecraft is launched into
GTO (Geostationary Transfer Orbit).
• The GTO has a perigee that is the original LEO altitude and an apogee that is the
GEO altitude. Figure 6
• Since the rocket motors fires at apogee, it is commonly refereed to as the apogee kick
motor (AKM).
• The AKM is used in:
1. Circularize the orbit at GEO.
2. Remove any inclination error.
17
Figure 2.19 (p. 48)
ustration of the GTO/AKM approach to
eostationary orbit (not to scale). The
ombined spacecraft and final rocket
age are placed into low earth orbit (LEO)
ound the earth. After careful orbit
etermination measurements, the final
age is ignited in LEO and the spacecraft
serted into a transfer orbit that lies
etween the LEO and the geostationary
bit altitude: the so-called geostationary
ansfer orbit or GTO. Again, after more
areful orbit determination, the apogee
ck motor (AKM) is fired on the satellite
nd the orbit is both circularized at
eostationary altitude and the inclination
duced to close to zero. The satellite is
en in GEO.
Figure 6: GTO/AKM approach
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
B.
Geostationary Transfer Orbit with Slow Orbit Raising.
• The thrusters are used to raise the orbit from GTO to GEO over a number of burns.
Figure 7.
• The satellite has two power level of thrusters.
– One for more powerful orbit raising maneuvers.
– One for on-orbit (low thrust) maneuvers.
C.
Direct Insertion to GEO
• The final stages of the rocket are used to place the satellite directly into GEO rather
than using the satellite own propulsion.
18
ure 2.20 (p. 49)
tion of slow orbit raising to geostationary
ot to scale). The combined spacecraft
al rocket stage are placed into low earth
EO) around the earth. As before (see
2.19), the spacecraft is injected into GTO
this case, once the satellite is ejected
e final rocket stage, it deploys many of
ments that it will later use in GEO (solar
, etc.) and stabilizes its attitude using
rs and momentum wheels, rather than
spin-stabilized. The higher power
rs are then used around the apogee to
he perigee of the orbit until the orbit is
r at the GEO altitude. At the same time
orbit is being raised, the thruster firings
designed gradually to reduce the
ion to close to zero.
Figure 7: Slow Orbit Raising
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
19
Part V
Orbital Effects in Communications
Systems Performance
I.
DOPPLER SHIFT
• The frequency of a moving radio transmitter varies with the transmitter’s velocity.
• The true transmitter frequency fT is the frequency that the transmitter would send
when at rest.
• The received frequency fR is higher than fT when the transmitter is moving toward
the receiver.
– And lower than fT when the transmitter is moving away from the receiver.
• The relationship between the transmitted and received frequencies is:
fR − fT
∆f
VT
=
=
fT
fT
vp
or
∆f = VT fT /c = VT /λ
where
– VT is the component of the transmitter velocity directed toward the receiver.
– vp = c the phase velocity of light (2.9979 × 108 ≈ 3 × 108 m/s).
– λ is the wavelength of the transmitted signal.
• If the transmitter is moving away from the receiver, then VT is negative.
• This change in frequency is called the Doppler Shift, The Doppler effect or Doppler.
20
• For GEO satellites, the effect is negligible.
Example: A LEO satellite is in a circular orbit with an altitude, h, of 1000
km. A transmitter on the satellite has a frequency of 2.65 GHz. Find:
1. The velocity of the satellite in orbit.
2. The component velocity toward an observer at an earth station as the
satellite appears over the horizon, for an observer who is in the plane
of the satellite orbit.
3. The Doppler shift of the received signal at the earth station.
4. The Doppler shift for this signal when it is received by the same observer
if it carries a Ku-band transmitter at 20 GHz
Answer:
1. The period of the satellite is:
T 2 = 4π 2 a3 /µ
= 4π 2 (6378 + 1000)3 /3.986004418 × 105
= 3.977754 × 107 s2
T = 6306.94 s
The circumference of the orbit is 2πr = 46357.3 km so the velocity of the satellite
in the orbit is:
vs = 46357.3/6306.94 = 7.35 km/s
2. The component of velocity toward an observer in the plane of the orbit as the
satellite appears over the horizon is given by vr = vs cos θ, where θ is the angle
between the satellite velocity vector and the direction of the observer.
cos θ = re / (re + h)
= 6378/7378
= 0.8645
21
hence
vr = vs cos θ
= 6.354 km/s = 6354 m/s
3. The Doppler shift in the received signal is:
∆f = VT /λ
= 6354/0.1132
= 56130 Hz = 56.130 kHz
4. The Doppler shift at the receiver is:
∆f = VT /λ
= 6354/0.015
= 423.6 kHz
II.
RANGE VARIATION
• The position of a satellite with respect to the earth exhibits a cyclic daily variations.
• The variation in position will lead to a variation in range between the satellite and the
receiver.
• If time division multiple access (TDMA) is being used;
– careful attention must be paid to the timing of the frames within the TDMA
busts.
∗ so that the individual user frames arrive at the satellite in the correct sequence at the correct time.
22
Figure 2.21 (p. 52)
Eclipse geometry. (Source: Spilker, Digital Communications by Satellite, Prentice Hall, 1977.) During the equinox
periods around the March 21 and September 3, the geostationary plane is in the shadow of the earth on the far side of
the earth from the sun. As the satellite moves around the geostationary orbit, it will pass through the shadow and
undergo an eclipse period. The length of the eclipse period will vary from a few minutes to over an hour (see Figure
2.22), depending on how close the plane of the geostationary orbit is with respect to the center of the shadow thrown by
the earth.
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
Figure 8: Solar eclipse
III.
SOLAR ECLIPSE
• A satellite is eclipse when the earth prevents sunlight from reaching it;
– that is, when the satellite is in the shadow of the earth. Figure 8.
• For GEO satellites, eclipses occur during two periods that being 23 days before the
equinoxes (about March 21 and about September 23) and ends 23 days after the
equinoxes periods. Figure 9
• During full eclipse, a satellite receives no power from its solar array and must operate
entirely from its batteries.
IV.
SUN TRANSIT OUTAGE
• During the equinox periods, the orbit of the satellite will also pass directly in front of
the sun on the sunlit side of the earth. Figure 10
• The sun is a hot microwave source with an equivalent temperature of 6000 to 10000
K at the frequencies used by communication satellites (4 to 50 GHz).
• The earth station antenna will therefore receive not only the signal from the satellite
but also the noise temperature transmitted by the sun.
23
Figure 2.22 (p. 52)
Dates and duration of eclipses. (Source: Martin, Communications Satellite Systems, Prentice Hall 1978.)
Satellite Communications, 2/E by Timothy Pratt, Charles Bostian, & Jeremy Allnutt
Figure 9: Dates
and duration
of eclipses.
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
• The added noise temperature will cause the fade margin of the receiver to be exceeded
and an outage will occur.
24
gure 2.23 (p. 53)
ematic of sun outage conditions. During
equinox periods, not only does the earth s
dow cause eclipse periods to occur for
stationary satellites, during the sunlit
on of the orbit, there will be periods when
sun appears to be directly behind the
lite. At the frequencies used by
munications satellites (4 to 50 GHz), the
appears as a hot noise source. The
ctive temperature of the sun at these
uencies is on the order of 10,000 K. The
ise temperature observed by the earth
on antenna will depend on whether the
mwidth partially, or completely, encloses
sun.
Figure
Sun outage.
Satellite Communications,
2/E 10:
by Timothy
Pratt, Charles Bostian, & Jeremy Allnutt
Copyright © 2003 John Wiley & Sons. Inc. All rights reserved.
25