Satellite Communications

Reference book:
Satellite Communications, 3rd ed.
Dennis Roddy
McGraw-Hill International Ed.
1.1 Introduction

Features offered by satellite communications

• large areas of the earth are visible from the satellite, thus the
satellite can form the star point of a communications net linking
together many users simultaneously, users who may be widely
separated geographically
•Provide communications links to remote communities
•Remote sensing  detection of pollution, weather conditions,
search and rescue operations.
 1.2 Frequency allocations
• International Telecommunication Union
(ITU)  coordination and planning
• World divided into three regions:
– Region 1: Europe, Africa, formerly Soviet
Union, Mongolia
– Region 2: North and South America,
Greenland
– Region 3: Asia (excluding region 1), Australia,
south west Pacific
• Within regions, frequency bands are allocated to various
satellite services:
•Fixed satellite service (FSS)
•Telephone networks, television signals to cable

•Broadcasting satellite service (BSS)

•Direct broadcast to home  Astro is a subscription-based
direct broadcast satellite (DBS) or direct-to-home satellite
television and radio service in Malaysia and Brunei

•Mobile satellite service

•Land mobile, maritime mobile, aeronautical mobile

•Navigational satellite service

•Global positioning system

•Meteorological satellite service

Frequency band designations in common use for
satellite service
1.3 Intelsat
• International Telecommunications Satellite
•Created in 1964, now has 140 member countries, >40
investing entities
•Geostationary orbit  orbits earth`s equitorial plane.
•Atlantic ocean Region (AOR), Indian Ocean Region
(IOR), Pacific Ocean Region.
•Latest INTELSAT IX satellites  wider range of service
such as internet, Direct to home TV, telemedicine, tele-
education, interactive video and multimedia
Satellite Coverage Maps

Source: http://www.intelsat.com
Coverage maps: Footprints
1.4 U.S DOMSAT (Domestic Satellites)
•Provide various telecommunication service within a
country
•In U.S.A all domsats in geostationary orbit
•Direct-to-home TV service can be classified as high
power, medium power, low power
1.5 Polar orbiting satellites

• Orbit the earth such a way as to cover the north and

south polar regions
• A satellite in a polar orbit passes above or nearly above
both poles of the planet (or other celestial body) on each
revolution. It therefore has an inclination of (or very close
to) 90 degrees to the equator.
• Since the satellite has a fixed orbital plane perpendicular
to the planet's rotation, it will pass over a region with a
different longitude on each of its orbits.
• Polar orbits are often used for earth-mapping-, earth
observation- and reconnaissance satellites, as well as
some weather satellites.
The orbit of a near polar
satellite as viewed from a
point rotating with the Earth.
• in U.S.A, the National Oceanic and Atmospheric
Administration (NOAA) operates a weather satellite
system,
geostationary operational environmental satellites
(GEOS) and
polar operational environmental satellites (PEOS)
2.0 Orbits and Launching Methods
• Johannes Kepler (1571 –1630) derive empirically three
laws describing planetary motion.
• Kepler’s laws apply quite generally to any two bodies in
space which interact through gravitation.
• The more massive of the two bodies is referred to as the
primary, the other, the secondary, or satellite.
2.2 Kepler’s first law
states that the path followed by a satellite around the
primary will be an ellipse. An ellipse has two focal points
shown as F1 and F2

The center of mass of the two-body system, termed the

barycenter, is always centered on one of the foci
• In our specific case, because of the enormous
difference between the masses of the earth and the
satellite, the center of mass coincides with the center of
the earth, which is therefore always at one of the foci.

• The semimajor axis of the ellipse is denoted by a, and

the semiminor axis, by b. The eccentricity e is given by

a2  b2
e
a
For an elliptical orbit, 0 < e < 1. When e = 0, the orbit
becomes circular.
2.3 Kepler`s Second Law
Kepler’s second law states that, for equal time intervals, a
satellite will sweep out equal areas in its orbital plane,
focused at the barycenter.

Thus the farther the satellite from earth, the longer it takes
to travel a given distance
2.4 Kepler’s Third Law

states that the square of the periodic time of orbit

is proportional to the cube of the mean distance between
the two bodies.

The mean distance is equal to the semimajor axis a.

For the artificial satellites orbiting the earth, Kepler’s third

law can be written in the form

a  2
3
… (2.2)
n
a = semimajor axis (meters)
n = mean motion of the satellite (radians per second)
 = earth’s geocentric gravitational constant.
= 3.986005  1014 m3/sec2
Eqt (2.2) applies only to ideal situation  satellite
orbiting a perfectly spherical earth of uniform mass,
with no pertubing forces acting, such as atmospheric
drag.
Section 2.8 will take account of the earth`s oblateness
and atmospheric drag.
With n in radians per second, the orbital period in
seconds is given by

2
P (2.4)
n

This shows that there is a fixed relationship between

period and size
Example 2.1:
Calculate the radius of a circular orbit for which the
period is 1day.

Solution: 2 
The mean motion, in rad/ day, is  n :
1day
5rad
Thus, n  7.272 10 
sec

Also,  : 3.986005 1014  m3  sec 2

1
 3
Kepler`s 3rd Law gives a :  2 
n 
a  42241km
2.5 Definitions of Terms for Earth-Orbiting Satellites
For the particular case of earth-orbiting satellites, certain
terms are used to describe the position of the orbit with
respect to the earth.
• Apogee:
The point farthest from earth. Apogee height is shown
as ha in Fig. 2.3.
• Perigee:
The point of closest approach to earth. The perigee
height is shown as hp in Fig. 2.3.
• Line of apsides:
The line joining the perigee and apogee through the
center of the earth.
• Ascending node:
The point where the orbit crosses the equatorial plane
going from south to north.
• Descending node:
The point where the orbit crosses the equatorial
plane going from north to south.

• Line of nodes
The line joining the ascending and descending
nodes through the center of the earth.

•Inclination
The angle between the orbital plane and the
earth’s equatorial plane. It is measured at the
ascending node from the equator to the orbit,
going from east to north. The inclination is shown
as i in Fig. 2.3.
• Prograde orbit
An orbit in which the satellite moves in the same
direction as the earth’s rotation. Also known as a direct
orbit. The inclination of a prograde orbit always lies
between 0 and 90°

• Retrograde orbit
An orbit in which the satellite moves in a direction
counter to the earth’s rotation. The inclination of a
retrograde orbit always lies between 90 and 180°.

• Argument of perigee
The angle from ascending node to perigee, measured
in the orbital plane at the earth’s center, in the direction
of satellite motion.
Right ascension of the ascending node:
To define completely the position of the orbit in space, the
position of the ascending node is specified.

Mean anomaly:
Mean anomaly M gives an average value of the angular
position of the satellite with reference to the perigee

True anomaly:
The true anomaly is the angle from perigee to the satellite
position, measured at the earth’s center. This gives the
true angular position of the satellite in the orbit as a
function of time.
2.8.1 Effects of a nonspherical earth

For a spherical earth of uniform mass, Kepler’s third

law (Eq. 2.2) gives the nominal mean motion n0.

n0 
a3

The 0 subscript as a reminder that this result applies

for a perfectly spherical earth of uniform mass.

However, earth is not perfectly spherical  an

equatorial bulge and a flattening at the poles (oblate
spheroid)
Taking account of earth`s oblateness, mean motion, n, is
modified to

1  K1 1  1.5 sin 2 i
n  n0 
  … (2.8)
 
a 1 e
2

2 1.5


K1 = 66,063.1704 km2 a (constant).

The earth’s oblateness has negligible effect on the
semimajor axis a.

The orbital period taking into account the earth’s

oblateness is termed the anomalistic period (e. g., from
perigee to perigee). The anomalistic period is
2 … (2.9)
PA  sec
n
With n in radians/second
If n is known, equation 2.8 can be used to solve a, by
solving the root of the following equation

 1  K1 1  1.5 sin 2 i  
n 3  0 … (2.10)
a 1  e 
1.5
a  2 2


Example 2.4, refer text book page 31

Oblateness of the earth produces two rotations on the
orbital plane:
1. Regression of the nodes
2. Rotation of the line of the apsides in the orbital plane

Regression of the nodes, the nodes appear to slide

along the equator. The line of nodes, which is in the
equatorial plane, rotates about the center of the earth.
Thus , the right ascension of the ascending node,
shifts its position.
If the orbit is prograde, the nodes slide westward,
if retrograde, they slide eastward.
For a polar orbit (i = 90°), the regression is zero.
Rotation of the line of apsides in the orbital plane,
the argument of perigee changes with time.

Both effects depend on the mean motion n, the

semimajor axis a, and the eccentricity e. These factors
can be grouped into one factor K given by
nK1
K 3 … (2.11)
a (1  e 2 ) 2

K will have the same units as n.

The rate of regression of the nodes

d
  K (cos i )
dt … (2.12)

Where i is the inclination. The rate of change of the

nodes will have the same units as n.
Rate of change negative  regression westward
If positive  regression is eastward
The rate of rotation of the line of apsides:
d
 K (2  2.5 sin 2 i ) … (2.13)
dt
Units: same as n
Given Epoch time = t0, right ascension of the
ascending node 0 at epoch, argument of the
perigee 0 at epoch, the new values for  and  at
time t is: d
  0  (t  t0 ) … (2.14)
dt

d
  0  (t  t0 ) … (2.15)
dt
2.8.2 Atmospheric Drag

The effects of atmospheric drag are significant for

near-earth satellites,( below about 1000 km).
The drag is greatest at the perigee, which reduce the
velocity at this point, thus the satellite does not reach
the same apogee height on successive revolutions.

The result is that the semimajor axis and the

eccentricity are both reduced. Drag does not
noticeably change the other orbital parameters,
including perigee height. An approximate expression
for the change of major axis is
2
 n0  3
… (2.16)
a  a0   
n 
 0 0 n '
 t  t 
0 
The mean anomaly is also changed. An approximate
expression for the amount by which it changes is
'
n
M  0  t  t0 
2

2 … (2.17)
2.9 Inclined Orbits
A study of the general situation of a satellite in an inclined
elliptical orbit is complicated by the fact that different
parameters are referred to different reference frames.

The orbital elements are known with reference

to the plane of the orbit, the position of which is fixed (or
slowly varying) in space, while the location of the earth
station is usually given in terms of the local geographic
coordinates which rotate with the earth.

Rectangular coordinate systems are generally used in

calculations of satellite position and velocity in space, while
the earth station quantities of interest may be the azimuth
and elevation angles and range.
Transformations between coordinate systems are therefore
required.
 With inclined orbits the satellites are not
geostationary, thus the required look angles and
range will change with time.

Determination of the look angles and range involves

the following quantities and concepts:
1. The orbital elements, as published in the NASA
bulletins and described in Sec. 2.6
2. Various measures of time
3. The perifocal coordinate system, which is based
on the orbital plane
4. The geocentric-equatorial coordinate system,
which is based on the earth’s equatorial plane
5. The topocentric-horizon coordinate system, which
is based on the observer’s horizon plane
The two major coordinate transformations needed are:

• The satellite position measured in the perifocal

system is transformed to the geocentric-horizon system
in which the earth’s rotation is measured, thus enabling
the satellite position and the earth station location to be
coordinated.

• The satellite-to-earth station position vector is

transformed to the topocentric-horizon system, which
enables the look angles and range to be calculated.
2.9.1 Calendars

The mean sun does move at a uniform speed but otherwise

requires the same time as the real sun to complete one orbit of
the earth, this time being the tropical year. A day measured
relative to this mean sun is termed a mean solar day.
Calendar days are mean solar days, and generally they are just
referred to as days.

A tropical year contains 365.2422 days. In order to make the

calendar year, also referred to as the civil year, more easily
usable, it is normally divided into 365 days. The extra 0.2422 of
a day is significant and for example, after 100 years, there
would be a discrepancy of 24 days between the calendar year
and the tropical year.
Julius Caesar made the first attempt to correct for the
discrepancy by introducing the leap year, in which an extra
day is added to February whenever the year number is
divisible by four. This gave the Julian calendar, in which the
civil year was 365.25 days on average, a reasonable
approximation
to the tropical year.

By the year 1582, an appreciable discrepancy once again

existed between the civil and tropical years. Pope Gregory XIII
took matters in hand by abolishing the days October 5 through
October 14, 1582, to bring the civil and tropical years into line
and by placing an additional constraint on the leap year in that
years ending in two zeros must be divisible by 400 to be
reckoned as leap years.
This dodge was used to miss out Gregorian calendar 3 days
every 400 years. The resulting calendar is the, which is the
one in use today.
2.9.2 Universal Time

Universal time coordinated (UTC) is the time used

for all civil timekeeping purposes, and as a standard
for setting clocks.

The fundamental unit for UTC is the mean solar day.

The mean solar day is divided into 24 hours, an

hour into 60 minutes, and a minute into 60 seconds.
Thus there are 86,400 “clock seconds” in a mean
solar day.

Satellite-orbit epoch time is given in terms of UTC. Universal

time coordinated is equivalent to Greenwich mean time (GMT),
as well as Zulu (Z) time
Distinction between system is not critical, the term
universal time (UT) will be used.
Given UT in the normal form of hours, minutes, and
seconds, it is converted to fractional days as

1 minutes seconds
UTday  (hours   ) … (2.18)
24 24 3600

This is converted to degrees as

UT o  360  UTday … (2.19)

2.9.3 Julian Dates
Calendar times are expressed in UT, and although the
time interval between any two events may be
measured as the difference in their calendar times, the
calendar time notation is not suited to computations
where the timing of many events has to be computed.
What is required is a reference time to which all events
can be related in decimal days.
Julian zero time reference is 12 noon (12: 00 UT) on
January 1 in the year 4713B.C.

The Julian day or Julian day number (JDN) is the

(integer) number of days that have elapsed since
Monday, January 1, 4713 BC in the Julian calendar.
That day is counted as Julian day zero.
The Julian Date (JD) is the number of days (with
decimal fraction of the day) that have elapsed since
12 noon Greenwich Mean Time (UT or TT) of that
day. Rounding to the nearest integer gives the Julian
day number.
The Julian day number can be considered a very
simple calendar, where its calendar date is just an
integer.
Ordinary calendar times are easily converted to Julian
dates, measured on a continuous time scale of Julian
days.
First determine the day of the year, keeping in mind
that day zero, denoted as Jan 0, is December 31. For
example, noon on December 31 is denoted as Jan
0.5, and noon on January 1 is denoted as Jan 1.5
The Julian day number can be calculated using the following
formulas:
The months January to December are 1 to 12. Astronomical year
numbering is used, thus 1 BC is 0, 2 BC is −1, and 4713 BC is
−4712. In all divisions (except for JD) the floor function is applied
to the quotient (for dates since 1 March −4800 all quotients are
non-negative, so we can also apply truncation).
Example 2.10:
Find the Julian day for 13h UT on 18 Dec.2000
Solution: JD = 2451897.0417 day
Certain calculations require a time interval measured
in Julian centuries.
Julian century consists of 36,525 mean solar days. The
time interval is reckoned from a reference time of
January 0.5, 1900, which corresponds to 2,415,020
Julian days.
Reference time JDref, the Julian century JC, and the
time in question JD, then the interval in Julian centuries
from the reference time to the time in question is given
by

JD  JDref
T … (2.20)
JC
Example 2.11: Find the time in Julian centuries from the reference
time January 0.5 1900 to 13h UT on 18 December 2000
2.9.4 Sidereal time
Sidereal time is time measured relative to the fixed
stars. It will be seen that one complete rotation of the
earth relative to the fixed stars is not a complete
rotation relative to the sun.
The sidereal day is defined as one complete rotation of
the earth relative to the fixed stars. One sidereal day
has 24 sidereal hours, one sidereal hour has 60
sidereal minutes, and one sidereal minute has 60
sidereal seconds
The relationships between the two systems, given in
Bate et al. (1971), are

1 mean solar day

= 1.0027379093 mean sidereal days
= 24h 3m 56s .55536 sidereal time (2.21)
= 86,636.55536 mean sidereal seconds

1 mean sidereal day

= 0.9972695664 mean solar days
= 23h 56m 04s .09054 mean solar time (2.22)
= 86,164.09054 mean solar seconds
Measurements of longitude on the earth’s surface
require the use of sidereal time (discussed further in
Sec. 2.9.7).
The use of 23 h, 56 min as an approximation for the
mean sidereal day will be used later in determining
the height of the geostationary orbit
2.9.2 The orbital Plane

In the orbital plane, the position vector r and the

velocity vector v specify the motion of the satellite

Figure: Perifocal coordinate system (PQW frame)

Only the magnitude of the position vector is required

r

a 1  e2 … (2.23)
1  e cos v

The true anomaly v is a function of time. The usual

approach to determining v proceeds in two stages.
First, the mean anomaly M at time t is found.

M  n(t  T )
… (2.24)

Here, n is the mean motion, and T is the time of

perigee passage.
The time of perigee passage T can be eliminated from
eq. (2.24) if one is working from the elements specified
by NASA. For the NASA elements:

M 0  n(t0  T )
Therefore,
M0
T  t0  … (2.25)
n
Hence, substituting this in Eq. (2.24) gives
M  M 0  n(t  t0 ) … (2.26)

Consistent units must be used throughout. For example, with n

in degrees/ day, time (t - t0) must be in days and M0 in degrees,
and M will then be in degrees.
Refer example 2.12 page 43
Calculate the time of perigee passage for the NASA elements
given in Table 2.1.

solution
The specified values at epoch are mean motion n = 14.23304826
rev/ day, mean anomaly M0 = 246.6853°, and t0 = 223.79688452
days. In this instance it is only necessary to convert the mean
motion to degrees/ day, which is 360n.
Applying Eq. (2.25) gives

246.6853
T  223.79688452 -
14.23304826  360
 223.79604425 days
Once the mean anomaly M is known, the next step is
to solve an equation known as Kepler’s equation.
Kepler’s equation is formulated in terms of an
intermediate variable E, known as the eccentric
anomaly, and is usually stated as

M  E  e sin E … (2.27)

This equation requires an iterative solution, preferably

by computer to solve for E as the root of the equation

M  ( E  e sin E )  0 … (2.28)
Once E is found, v can be found from an equation known
as Gauss’equation, which is

v 1 e E
tan  tan … (2.29)
2 1 e 2

It may be noted that r, the magnitude of the radius

vector, also can be obtained as a function of E and is
r  a1 e cos E  … (2.30)
For near-circular orbits where the eccentricity is small,
an approximation for v directly in terms of M is
5 2
v  M  2e sin M  e sin 2M … (2.31)
4
Note that the first M term on the right-hand side must be in
radians, and v will be in radians.
Example 2.14, textbook page 45

Magnitude r of position vector r may be calculated by

either equation (2.23) or (2.30).
Perifocal coordinate system:

Orbital plane is the fundamental plane

Origin at the center of the earth (earth orbiting satellite only are
considered)

The positive x axis lies in the orbital plane and passes through the
perigee. Unit vector P points along the positive x axis as shown in
Fig. 2.8.

The positive y axis is rotated 90° from the x axis in

the orbital plane, in the direction of satellite motion, and the unit
vector is shown as Q.

The positive z axis is normal to the orbital plane such that

coordinates xyz form a right-hand set, and the unit vector is shown
as W.
The position vector in this coordinate system, which will be referred
to as the PQW frame, is given by

r = (r cos v)P + (r sin v)Q … (2.32)

The perifocal system is very convenient for

describing the motion of the satellite.
If the earth were uniformly spherical, the perifocal
coordinates would be fixed in space.
However, the equatorial bulge causes rotations of the
perifocal coordinate system, as described in Sec.
2.8.1.
These rotations are taken into account when the
satellite position is transferred from perifocal
coordinates to geocentric-equatorial coordinates
Example 2.15, page 46 of textbook
2.9.6 The geocentric-equatorial coordinate system

The fundamental plane is the earth’s equatorial

plane. Figure 2.9 shows the part of the ellipse above the
equatorial plane and the orbital angles , , and i.

Figure 2.9: Geocentric-equitorial

coordinate system (IJK frame)
It should be kept in mind that  and  may be
slowly varying with time, as shown by Eqs. (2.12)
and (2.13).
The unit vectors in this system are labeled I, J, and
K, and the coordinate system is referred to as the
IJK frame, with positive I pointing along the line of
Aries (reference line being fixed by the stars).
To transformation vector r from the PQW
frame to the IJK frame:

 rI 
r   R~  rp  … (2.33a)
 J r 
rK   Q
 where the transformation matrix R˜ is given by R

(cos  cos   sin  sin  cos i )   cos  sin   sin  cos  cos i  
~ 
R    sin  cos   cos  sin  cos i    sin  sin   cos  cos  cos i  
  sin  sin i  cos  sin i 

This gives the components of the position vector r for

the satellite, in the IJK, or inertial, frame.

The angles  and  take into account the rotations

resulting from the earth’s equatorial bulge, as
described in Sec. 2.8.1.
Example 2.16, page 47, textbook
2.9.7 Earth station referred to the IJK frame

The earth station’s position is given by the

geographic coordinates of latitude E and longitude
E.

North latitudes will be taken as positive numbers

and south latitudes as negative numbers, zero
latitude, is the equator.

Longitudes east of the Greenwich meridian will be

taken as positive numbers, and longitudes west, as
negative numbers.
The position vector of the earth station relative to the IJK frame is R
as shown in Fig. 2.10.

Figure 2.10 Position vector R of the earth relative to the IJK frame
The angle between R and the equatorial plane is denoted by E.

To find R, first find the position of the Greenwich meridian relative to

the I axis as a function of time. The angular distance from the I axis
to the Greenwich meridian is measured directly as Greenwich
sidereal time (GST), also known as the Greenwich hour angle, or
GHA.
The formula for GST in degrees is

GST  99.6910  36,000.7689  T  0.0004  T 2  UT 

… (2.34)
Where,
UTo = universal time in degrees, Eqt. 2.19
T = Time in Julian Centuries, eqt 2.20
Local sidereal time (LST) is found by adding the east longitude of
the station in degrees.

LST = GST + EL

East longitude for the earth station will be denoted as EL.

Longitude was expressed in positive degrees east and negative
degrees west.

For east longitudes, EL=E,

while for west longitudes, EL = 360° + E.

For example, earth station at east longitude 40°

EL = 40°
For an earth station at west longitude 40°,
EL = 360 + (-40) = 320°.
Thus the local sidereal time in degrees is given by
LST = GST + EL
Example 2.17, page 50 textbook

Example 2.18, page 51 textbook

Knowing the LST, the position vector R of the earth

station can located with reference to the IJK frame.
However, when R is resolved into its rectangular
components, account must be taken of the
oblateness of the earth.
The earth may be modeled as an oblate spheroid, in
which the equatorial plane is circular, and any
meridional plane (i. e., any plane containing the
earth’s polar axis) is elliptical, as illustrated in Fig.
2.11.
Figure 2.11 Reference ellipsoid for the earth showing the geocentric
latitude E and the geodetic latitude E
For the reference ellipsoid model, the semimajor axis of
the ellipse is equal to the equatorial radius, the
semiminor axis is equal to the polar radius, and the
surface of the ellipsoid represents the mean sea level.
With semimajor axis aE and semiminor axis bE gives

aE = 6378.1414 km
bE = 6365.755 km
The eccentricity of the earth is
2 2
aE  bE
eE   0.08182 … (2.38)
aE
In Figs. 2.10 and 2.11, what is known as the
geocentric latitude is shown as E. This differs from
what is normally referred to as latitude.
An imaginary plumb line dropped from the earth
station makes an angle E with the equatorial plane,
as shown in Fig. 2.11. This is known as the geodetic
latitude, and for all practical purposes here, this can
be taken as the geographic latitude of the earth
station.
With the height of the earth station above mean sea level
denoted by H, the geocentric coordinates of the earth
station position are given in terms of the geodetic
coordinates by
aE
N … (2.39)
1 eE sin E
2 2

RI   N  H  cos E cos LST  l cos LST … (2.40)

RJ   N  H  cos E sin LST  l sin LST … (2.41)

  
RK  N 1  eE  H sin E  z
2
 … (2.42)
Example 2.19, page 52 textbook
2.9.8 The topocentric-horizon coordinate system

The position of the satellite, as measured from the earth

station, is given in terms of the azimuth and elevation angles
and the range .
These are measured in the topocentric-horizon coordinate
system illustrated in Fig. 2.12 b.
The fundamental plane is the observer’s horizon plane.
The positive x axis is taken as south, the unit vector being
denoted by S.
The positive y axis points east, the unit vector being E.
The positive z axis is “up,” pointing to the observer’s zenith,
the unit vector being Z.
(Note: This is not the same z as that used in Sec. 2.9.7.)
The frame is referred to as the SEZ frame, which rotates with
the earth.
Figure 2.12: Topocentric-horizon
coordinate system (SEZ frame):
(a) Overall view,
(b) Detailed view
As shown in the previous section, the range vector  is
known in the IJK frame, and it is now necessary to
transform this to the SEZ frame.

  S   sin E cos LST sin E sin LST  cos E    I 

      sin LST cos LST 0    J 
 E 
  Z  cos E cos LST cos E sin LST  sin E    K  … (2.44)

Geocentric angle, E is given by

z
 E  arctan … (2.45)
l

The coordinates l and z given in Eqs. (2.40) and (2.42) are known
in terms of the earth station height and latitude.
For zero height, the angle E is known as the geocentric latitude
and is given by

 2

tan E ( H 0 )  1  eE tan E … (2.46)

eE, earth`s eccentricity = 0.08182

Difference between the geodetic and geocentric latitudes
reaches a maximum at a geocentric latitude of 450, when the
geodetic latitude is 45.1920
The magnitude of the range and the antenna look angles are
obtained from
  S  E  Z
2 2 2
… (2.47)

 Z 
  arcsin  … (2.48)
  
We define an angle  as,
E
  arctan
S
Then the azimuth depends on which quadrant  is in and is given
by Table 2.2.

S S Azimuth
degrees
- + 
+ + 180 - 
+ - 180 + 
- - 360 - 

Table 2.2: Azimuth angles

2.9.9 The subsatellite point

The point on the earth vertically under the satellite is

referred to as the subsatellite point. The latitude and
longitude of the subsatellite point and the height of the
satellite above the subsatellite point can be
determined from a knowledge of the radius vector r.

Figure 2.13 shows the meridian plane which cuts the

subsatellite point. The height of the terrain above the
reference ellipsoid at the subsatellite point is denoted
by HSS, and the height of the satellite above this, by
hSS.
Thus the total height of the satellite above the reference
ellipsoid is

h  H SS  hSS … (2.50)

Components for the radius vector r in the IJK frame are

given in equation (2.33)
Figure 2.13 is similar to 2.11, with difference in r
replacing R, height to the point of interest h replacing H,
and subsatellite latitude shown by SS. Thus, from
equation (2.39) to (2.42) are written as,
aE
N … (2.51)
2
1 eE sin SS
2
 rI   N  h  cos SS cos LST … (2.52)

rJ   N  h  cos SS sin LST … (2.53)

  2
 
rK  N 1  eE  h sin SS … (2.54)

Three equations in three unknowns, LST, E, and h.

Referring to Fig. 2.10, the east longitude is obtained from Eq.

(2.35) as

EL = LST - GST

where GST is the Greenwich sidereal time.

3.0 The Geostationary Orbit

A satellite in a geostationary orbit appears to be

stationary with respect to the earth.

Three conditions are required for an orbit to be

geostationary:

1. The satellite must travel eastward at the same

rotational speed as the earth.

• If the satellite is to appear stationary it must rotate at the

same speed as the earth, which is constant
2. The orbit must be circular.
• Constant speed means that equal areas must be swept out
in equal times, and this can only occur with a circular orbit
(see Fig. 2.2).

3. The inclination of the orbit must be zero.

•any inclination would have the satellite moving north and

south, (see Sec. 2.5 and Fig. 2.3), and hence it would not be
geostationary. Movement north and south can be avoided only
with zero inclination
From eqt (2.2) and (2.4), with radius, aGSO
1
 P
2
 3
aGSO   2  … (3.1)
 4 

The period P for the geostationary is 23 h, 56 min, 4

s mean solar time (ordinary clock time).
Substituting this value along with the value for 
given by Eq. (2.3) results in

aGSO  42164km … (3.2)

The equatorial radius of the earth, to the nearest
kilometer, is

aE  6378km … (3.3)

The geostationary height is

hGSO  aGSO  aE

 42,164  6378 … (3.4)

 35,786km

This value is often rounded up to 36,000 km for

approximate calculations.
In practice, a precise geostationary orbit cannot be
attained because of disturbance forces in space and
the effects of the earth’s equatorial bulge.

The gravitational fields of the sun and the moon

produce a shift of about 0.85 °/year in inclination.

The earth’s equatorial ellipticity causes the satellite to

drift eastward along the orbit.

In practice, station-keeping maneuvers have to be

performed periodically to correct for these shifts
3.2 Antenna Look Angles

The look angles for the ground station antenna are the
azimuth and elevation angles required at the antenna so
that it points directly at the satellite.

In Sec. 2.9.8 the look angles were determined in the

general case of an elliptical orbit, and there the angles
had to change in order to track the satellite. With the
geostationary orbit, the situation is much simpler
because the satellite is stationary with respect to the
earth.
The three pieces of information that are needed to
determine the look angles for the geostationary orbit are

1. The earth station latitude, denoted here by E

2. The earth station longitude, denoted here by E

3. The longitude of the subsatellite point, denoted here

by SS (often referred to as the satellite longitude)
Latitudes north will be taken as positive angles, and
latitudes south, as negative angles. Longitudes east of
the Greenwich meridian will be taken as positive
angles, and longitudes west, as negative angles.
eg: latitude 40° S  -40°, longitude 35° W  -35°.

In Chap. 2, when calculating the look angles for lower-

earth-orbit (LEO) satellites, it was necessary to take
into account the variation in earth’s radius. With the
geostationary orbit, this variation has negligible
effect on the look angles, and the average radius of
the earth will be used, R.

R  6371km … (3.5)
Figure 3.1: The geometry used in determining the look angles for a
geostationary satellite
ES = position of the earth station
SS = subsatellite point
S = the satellite
d = range of the earth station to the
satellite
 = angle to be determine
2 types of triangles involved in the geometry of Fig. 3.1:
• spherical triangle (Fig. 3.2a)
• plane triangle (Fig. 3.2b)

6 angles defining the triangles:

a, b, c, A, B, C

Angle a = angle between the

radius to the north pole and the
radius to the subsatellite point,
a = 90°.

Angle b = between the radius to

the earth station and the radius to
the subsatellite point.

Angle c = between the radius to

the earth station and the radius to
the north pole. c = 90° - E.
Other angles A, B, and C are angles between the
planes.

Angle A = angle between the plane containing c and the

plane containing b.

Angle B = angle between the plane containing c and the

plane containing a.

Angle C = angle between the plane containing b and the

plane containing a.
To summarize the spherical triangle
a  900 … (3.6)

c  90  E … (3.7)

B   E  SS … (3.8)

when the earth station is west of the subsatellite point,

B is negative, and when east, B is positive.
When the earth station latitude is north, c is less than
90°, and
when south, c is greater than 90°.
Special rules, known as Napier’s rules, are used to
solve the spherical triangle.
Napier’s rules gives angle b and angle a as

b  arccos cos B cos E  … (3.9)

 sin B 
A  arcsin 
 … (3.10)
 sin b 

Two values will satisfy Eq. (3.10), A and 180° - A,

and these must be determined by inspection. These
are shown in Fig. 3.3
Fig. 3.3 a: angle A is acute (less
than 90°), and the azimuth angle
Az = A.

Fig.3.3 b: angle A is acute, and

the azimuth Az = 360° - A.
Fig. 3.3 c: angle Ac is obtuse and is
given by Ac = 180° - A, where A is the
acute value obtained from Eq. (3.10).
By inspection, Az = Ac = 180°- A

In Fig. 3.3 d, angle Ad is obtuse and is

given by 180°- A, where A is the acute
value obtained from Eq. (3.9). By
inspection, Az = 360° - Ad = 180°+ A.
These conditions are summarized in Table 3.1

Fig 3.3 E B Az degrees

a <0 <0 A
b <0 >0 360 – A
c >0 <0 180 - A
d >0 >0 180 + A

Applying the cosine rule for plane

triangles to the triangle of Fig.3.2
b, d is approximated to:

d  R 2  aGSO  2  R  aGSO  cos(b) … (3.11)

Applying the sine rule for plane
Figure 3.2b triangles to the triangle of Fig. 3.2 b,
the angle of elevation is
 aGSO 
El  arccos sin b  … (3.12)
 d 
 3.3 Polar Mount Antenna

polar mount is a piece of equipment installed into

geostationary satellites to be accessed by swinging the
satellite dish around one axis. This allows one positioner
only to be used to remotely point the antenna at any
satellite.

The dish is mounted on an axis termed the polar axis such

that the antenna boresight is normal to this axis, as shown
in Fig. 3.5 a
 Figure 3.5(a): The polar mount antenna

The angle between the polar mount and the local horizontal
plane is set equal to the earth station latitude, E. Thus the
boresight is parallel to the equatorial plane.
Next, the dish is tilted at an angle  relative to the polar
mount until the boresight is pointing at a satellite
position.

Figure 3.5(b)
Referring to Fig 3.5(b),
The required tilt angle  is:

  90  El0  E … (3.13)

Where El0 = angle of elevation required for the satellite

position due south of the earth station.
For due south situation, angle B in eq. (3.8) equals
zero, thus from eq. (3.9),
b  E
 Figure 3.5(c)

from Eq. (3.12), or Fig 3.5 c

aGSO
cos El0  sin E … (3.14)
d

Combining Eqs. (3.13) and (3.14) gives the required angle

of tilt as:
 aGSO 
  90  arccos

sin E   E … (3.15)
 d 
To calculate d, assume earth station altitude can be
ignored (as eq. 3.11) and R = 6371km.
The value  is sufficiently accurate for initial alignment.
Fine adjustment can be made if necessary.

Example 3.3: Determine the angle of tilt required for a polar

mount used with an earth station at latitude 49 degrees north.
Assume a spherical earth of mean radius 6371 km, and ignore
earth station altitude.

Solution:
Given data E = 49, aGSO= 42164km, R = 6371km

d  R 2  aGSO  2  R  aGSO  cos(b)

Example 3.3: Determine the angle of tilt required for a polar
mount used with an earth station at latitude 49 degrees north.
Assume a spherical earth of mean radius 6371 km, and ignore
earth station altitude.

Solution:
Given data E = 49, aGSO= 42164km, R = 6371km

d  R 2  aGSO  2  R  aGSO  cos(E ) … Eq (3.11) with b = E

 aGSO  … Eq (3.12) with b = E

El0  arccos sin E 
 d 

  90  El0  E  7 deg

3.4 Limits of Visibility

There will be east and west limits on the geostationary

arc visible from any given earth station. The limits will be
set by the geographic coordinates of the earth station
and the antenna elevation. The lowest elevation in
theory is zero, when the antenna is pointing along the
horizontal.

To estimate of the longitudinal limits, consider an earth

station at the equator, with the antenna pointing either
west or east along the horizontal, as shown in Fig. 3.6.
 Figure 3.6: Illustrating the limits of visibility.

The limiting angle is given by:

aE 6378
  arccos  arccos  81.3 … (3.16)
aGSO 42,164
3.5 Near Geostationary Orbits

A number of perturbing forces causes an orbit to depart

from the ideal keplerian orbit.

For the geostationary case, the forces are mainly due to:
 gravitational fields of the moon and the sun
 nonspherical shape of the earth.
 Other significant forces are solar radiation pressure and
reaction of the satellite itself to motor movement within
the satellite.

As a result, station keeping maneuvers must be carried

out to maintain the satellite within set limits of its nominal
geostationary position.
Exact geostationary orbit is not attainable in practice, and orbital
parameters vary with time.
The two-line orbital elements are published at regular intervals, Fig. 3.7
showing typical values.
The period for a geostationary satellite is 23 h, 56 min, 4 s, or 86,164 s.
The reciprocal of this is 1.00273896 rev/ day, which is about the value
tabulated for most of the satellites in Fig. 3.7.

Figure 3.7 Two-line elements for some geostationary satellites

Thus these satellites are geosynchronous, in that
they rotate in synchronism with the rotation of the
earth. However, they are not geostationary.
The term geosynchronous satellite is used in many
cases instead of geostationary to describe these
near-geostationary satellites.

However, in general a geosynchronous satellite does

not have to be near-geostationary. There are a
number of geosynchronous satellites that are in
highly elliptical orbits with comparatively large
inclinations (e. g., the Tundra satellites).
Chapter 4: Radio Wave Propagation

4.1 Introduction

A signal traveling between an earth station and a

satellite must pass through the earth’s atmosphere,
including the ionosphere.
This introduce certain impairments, summarized in
Table 4.1. (Refer text book, page 93)
4.2 Atmospheric Losses

Losses occur in the earth’s atmosphere as a result of

energy absorption by the atmospheric gases. These
losses are treated quite separately from those which
result from adverse weather conditions, which of
course are also atmospheric losses.
To distinguish between these, the weather-related
losses are referred to as atmospheric attenuation and
the absorption losses simply as atmospheric
absorption.
The atmospheric absorption loss varies with
frequency, as shown in Fig. 4.2. (Refer text book,
page 94)
Two absorption peaks will be observed:
1. at a frequency of 22.3 GHz, resulting from
resonance absorption in water vapor (H2O), and
2. at 60 GHz, resulting from resonance absorption
in oxygen (O2).

At other frequencies, the absorption is quite low.

4.3 Ionospheric Effects

Radio waves traveling between satellites and earth

stations must pass through the ionosphere. The
ionosphere has been ionized, mainly by solar radiation.
The free electrons in the ionosphere are not uniformly
distributed but form in layers.
Clouds of electrons may travel through the ionosphere
and give rise to fluctuations in the signal.

The effects include scintillation, absorption, variation in

the direction of arrival, propagation delay, dispersion,
frequency change, and polarization rotation.
All these effects decrease as frequency increases.
Only the polarization rotation and scintillation
effects are of major concern for satellite communications.

Ionospheric scintillations
• are variations in the amplitude, phase, polarization, or
angle of arrival of radio waves.
• Caused by irregularities in the ionosphere which changes
with time.
• Effect of scintillations is fading of the signal.
Severe fades may last up to several minutes.
Polarization rotation:
• porduce rotation of the polarization of a signal (Faraday
rotation)
•When linearly polarized wave traverses in the
ionosphere, free electrons in the ionosphere are sets in
motion  a force is experienced, which shift the
polarization of the wave.
•Inversely proportional to frequency squared.
• not a problem for frequencies above 10 GHz.
4.4 Rain Attenuation

Rain attenuation is a function of rain rate.

Rain rate, Rp = the rate at which rainwater would

accumulate in a rain gauge situated at the ground in
the region of interest (e. g., at an earth station).
The rain rate is measured in millimeters per hour.

Of interest is the percentage of time that specified

values are exceeded. The time percentage is usually
that of a year; for example, a rain rate of 0.001 percent
means that the rain rate would be exceeded for 0.001
percent of a year, or about 5.3 min during any one
year.
The specific attenuation  is

  aR p dB / km
b
… (4.2)

where a and b depend on frequency and polarization.

Values for a and b are available in tabular form in a
number of publications. (eg, Table 4.2, pg 95)
Once the specific attenuation is found, the total
attenuation is determined as:

A  L dB … (4.3)

where,
L = effective path length of the signal through the rain.
Because the rain density is unlikely to be uniform
over the actual path length, an effective path length
must be used rather than the actual (geometric)
length.
Figure 4.3 shows the geometry of the situation.

Figure 4.3: Path length through rain

The geometric, or slant, path length is shown as LS. This
depends on the antenna angle of elevation  and the rain
height hR, which is the height at which freezing occurs.
Figure 4.4 shows curves for hR for different climatic
zones.
Method 1: maritime climates
Method 2: Tropical climates
Method 3: continental climates

Figure 4.4: Rain height as a function of earth station latitude for

different climatic zones
For small angles of elevation (El < 10°), the determination
of LS is complicated by earth curvature.
For El  10°, a flat earth approximation may be used.
From Fig. 4.3 it is seen that
hR  ho
LS  … (4.4)
sin El

The effective path length is given in terms of the slant

length by
L  LS rp … (4.5)

where rp is a reduction factor which is a function of the

percentage time p and LG, the horizontal projection of LS.
Refer Table 4.3, page 97, for values of reduction factors, rp
From Fig. 4.3 the horizontal projection is seen to be

LG  LS cos El … (4.6)

Putting all the factors into one equation, we obtain

the rain attenuation (decibels) as:
b
Ap  aR p LS rp dB … (4.7)
Chapter 12: The Space Link
12.1 Introduction

This chapter describes how the link-power budget

calculations are made. These calculations basically
relate two quantities, the transmit power and the
receive power, and show in detail how the difference
between these two powers is accounted for.

Link budget calculations are usually made using

decibel or decilog quantities. These are explained in
App. G.
12.2 Equivalent Isotropic Radiated Power

A key parameter in link budget calculations is the

equivalent isotropic radiated power, conventionally
denoted as EIRP.
The Maximum power flux density at some distance r
from a transmitting antenna of gain G is

GPS
M  … (12.1)
4r 2
An isotropic radiator with an input power equal to GPS
would produce the same flux density. Hence this
product is referred to as the equivalent 　 isotropic
radiated power, or
EIRP  GPS … (12.2)

EIRP is often expressed in decibels relative to one

watt, or dBW. Let PS be in watts; then
 EIRP   PS    G  dBW … (12.3)

where [PS] is also in dBW and [G] is in dB.

The isotropic gain for a paraboloidal antenna is

G   10.472 fD 
2
… (12.4)

Where,
f is the carrier frequency

D is the reflector diameter in m

 is the aperture efficiency

12.3 Transmission Losses

The [EIRP] is the power input to one end of the

transmission link, and the problem is to find the power
received at the other end.

Losses will occur along the way, some of which are

constant. Other losses can only be estimated from
statistical data, and some of these are dependent on
weather conditions, especially on rainfall.
The first step in the calculations is to determine the
losses for clear weather, or clear-sky, conditions.
These calculations take into account the losses,
including those calculated on a statistical basis, which
do not vary significantly with time. Losses which are
weather-related, and other losses which fluctuate with
time, are then allowed for by introducing appropriate
fade margins into the transmission equation.
 12.3.1 Free Space Transmission

Spreading of the signal is space causes power loss.

The power flux density at the receiving antenna is
EIRP
M  … (12.6)
4r 2
The power delivered to a matched receiver is this
power flux density multiplied by the effective aperture of
the receiving antenna. The received power is therefore:

PR   M Aeff
2
EIRP  GR 2
  
  ( EIRP )(GR ) 
4r 2 4  4r 
… (12.7)
where
r = distance, or range, between the transmit and receive
antennas
GR = isotropic power gain of the receiving antenna. The
subscript R is used to identify the receiving antenna.
In decibel notation, equation (12.7) becomes
2
 4r 
 PR    EIRP   GR   10 log  … (12.8)
  

Free space loss is given by

2
 4r 
 FSL  10 log  … (12.9)
  
 32.4  20 log r  20 log f … (12.10)
Equation (12.8) can then be written as

 PR    EIRP   GR    FSL … (12.8)

The received power [PR] will be in dBW when the

[EIRP] is in dBW and [FSL] in dB.

Equation (12.9) is applicable to both the uplink and

the downlink of a satellite circuit
12.3.2 Feeder Losses

Losses occur in the connection between the

receive antenna and the receiver proper, such as in
the connecting waveguides, filters, and couplers.
These will be denoted by RFL, or [RFL] dB, for
receiver feeder losses.
The [RFL] values are added to [FSL] in Eq. (12.11).

Similar losses occur in the filters, couplers, and

waveguides connecting the transmit antenna to the
high-power amplifier (HPA) output. However,
provided that the EIRP is stated, Eq. (12.11)
can be used without knowing the transmitter feeder
losses. These are 　 needed only when it is desired
to relate EIRP to the HPA output
12.3.3 Antenna misalignment losses
When a satellite link is established, the ideal situation
is to have the earth station and satellite antennas
aligned for maximum gain, as shown in Fig. 12.1 a.

Figure 12.1: (a) aligned for maximum gain, (b) earth-station

antenna misaligned
There are two possible sources of off-axis loss, one
at the satellite and one at the earth station.

The off-axis loss at the earth station is referred to as the

antenna pointing loss, which are usually only a few
tenths of a decibel.

Losses may also result at the antenna from misalignment

of the polarization direction. The polarization
misalignment losses are usually small, and is assumed
that the antenna misalignment losses, denoted by [AML],
include both pointing and polarization losses resulting
from antenna misalignment.
Antenna misalignment losses have to be estimated
from statistical data, based on the errors actually
observed for a large number of earth stations.

Separate antenna misalignment losses for the uplink

and the downlink must be taken into account.
12.3.4 Fixed atmospheric and ionospheric losses

Atmospheric gases result in losses by absorption.

These losses usually amount to a fraction of a
decibel, and decibel value will be denoted by [AA].
Table 12.1 shows values of atmospheric absorption
losses and Satellite pointing loss in Canada.
12.4 The Link-Power Budget Estimation

Losses for clear sky conditions are

 LOSSES    FSL   RFL   AML   AA   PL … (12.12)

The decibel equation for the received power is

 PR    EIRP   GR    LOSSES  … (12.13)

where
[PR] = received power, dBW
[EIRP] = equivalent isotropic radiated power, dBW
[FSL] = free-space spreading loss, dB
[RFL] = receiver feeder loss, dB
[AML] = antenna misalignment loss, dB
[AA] = atmospheric absorption loss, dB
[PL] = polarization mismatch loss, dB
12.5 System Noise

The major source of electrical noise in equipment is from

the random thermal motion of electrons in various
resistive and active devices in the receiver.

Thermal noise is also generated in the lossy components

of antennas, and thermal-like noise is picked up by the
antennas as radiation.

The available noise power from a thermal noise source is

given by
PN  kTN BN … (12.14)
where
TN = equivalent noise temperature (K)
BN = equivalent noise bandwidth (Hz)
k = 1.38x10-23 (Boltzmann’s constant)

For thermal noise, noise power per unit bandwidth,

N0, is constant (a.k.a noise energy)
PN
N0   kTN joules … (12.14)
BN
In addition to thermal noise, intermodulation distortion
in high-power amplifiers result in signal products which
appear as noise, that is intermodulation noise.
12.5.1 Antenna Noise

Antennas operating in the receiving mode introduce noise

into the satellite circuit.

2 groups of antenna noise:

• Noise originating from antenna losses
• Sky noise
Sky noise:
The microwave radiation which is present through out
the universe and which appears to originate from matter
of any form at finite temperatures.
Fig.12.2 shows the equivalent noise temperature of the
sky, as seen by an earth-station antenna.
The lower graph: Antenna pointing directly overhead
The upper graph: Antenna pointing just above the
horizon.
The equivalent noise temperature for satellite antennas
is approx. 290 K
The increased noise in antenna pointing overhead is due to thermal radiation of
the earth. Sets limits of elevation angle to 5 for C band and 10 for Ku band.

Antenna
pointing just
above the
horizon

Peak coincides
with peaks in
atmospheric
absorbtion loss

Antenna
pointing
overhead

Fig 12.2: Irreducible noise temperature of an ideal, ground-based antenna.

12.5.2 Amplifier noise temperature

Consider the noise representation of the antenna

and the low noise amplifier (LNA) in Fig.12.4 below

Fig. 12.4: Circuit used in

finding the equivalent noise
temperature of (a) an amplifier
and (b) two amplifiers in
cascade
The input noise energy from the antenna is:

N 0,ant  kTant

Where,
N0,ant = Input noise energy from the antenna (Joules).
In noise power per unit bandwidth
Tant = Antenna noise temperature
k = Boltzman constant,
The output noise energy, No,out,

N 0,out  Gk  Tant  Te 

Where,
N0,out = Output noise energy
Te = Equivalent noise temperature of amplifier,
referred to the input
Tant = Antenna noise temperature
k = Boltzman constant
G = Amplifier gain
Total noise referred to the input, No,out/G

N 0,in  k  Tant  Te 

12.5.3 Amplifiers in cascade

The cascade connection is shown in Fig 12.4b

For this, the overall gain is

G  G1G2
 Cascaded amplifiers

The noise energy of amplifier

2 referred to its own input
 kTe 2

The noise input to amplifier 2

from the preceding stages is
 G1k  Tant  Te1 

Thus, total noise energy referred to amplifier 2 input is

N 0, 2  G1k  Tant  Te1   kTe 2

This noise energy may be referred to amplifier 1 input by
dividing by the available power gain of amplifier 1

N 0, 2
N 0,1  (12.21)
G1
 Te 2 
 k  Tant  Te1  
 G1 

Define system noise temperature, Ts, where

N 0,1  kTS (12.22)

TS is given by,
Te 2
TS  Tant  Te1  (12.23)
G1
From equation 12.23, the noise temperature of the
second stage is divided by the power gain of the first
stage when referred to the input
To keep the overall noise as low as possible  the first
stage (usually LNA) should have high power gain as
well as low noise temperature
For any number of stages in cascade,
Te 2 Te 3
TS  Tant  Te1    ... (12.24)
G1 G1G2
12.5.4 Noise Factor

Noise Factor, an alternative way of representing amplifier

noise.
Source is taken at room temperature, T0 = 290K.
Input noise is kT0, output noise of the amplifier is,
N 0,out  FGkT0 (12.25)

where,

G = available power gain of the amplifier

F = noise factor
Let Te be the noise temperature of the amplifier, and let
the source be at room temperature, T0, that is T0 = Tant.
With the same noise output, we obtain,

Gk  T0  Te   FGkT0

or
Te   F  1T0 (12.26)

Noise figure is F expressed in decibels.

Noise Figure   F   10 log F (12.27)

12.5.5 Noise Temperature of Absorptive Networks

An absorptive network is one which contains resistive

elements.
It introduces losses by absorbing energy from the
signal and converting it to heat (thermal noise).
Consider an absorptive network with power loss L.
Power loss, L = input power/output power > 1
Referring to Fig.12.5.
Let the network be matched at both ends, to a terminating
resistor RT at one end and an antenna at the other.
Let the system be at ambient temperature Tx.
The noise energy transferred from RT in the the network is
kTx.
Let the network noise be represented at the output
terminals by an equivalent noise temperature TNW0. Then
the noise energy radiated by the antenna is
kT
N rad  x  kTNW ,0 (12.28)
L
Because the antenna is matched to a resistive source at
temperature Tx, the available noise energy which is fed
into the antenna and radiated is Nrad = kTx
The expression for Nrad can be substituted to eq.(12.28) to
give the equivalent noise temperature of the network
referred to the output terminals of the network
 1 (12.29)
TNW , 0  Tx 1  
 L

The equivalent noise temperature of the output can

be transferred to the input on dividing by the
network power gain, 1/L
The equivalent noise temperature of the network referred
to the network input is

TNW ,i  Tx  L  1 (12.30)

Since network is bilateral, eq.(12.29) and (12.30) apply

for signal flow in either direction.
Thus eq.(12.30) gives the equivalent noise temperature
of a lossy network referred to the input at the antenna
when the antenna is used in receiving mode.
If lossy network is at room temperature, Tx = T0.
Comparing eq.(12.26) and (12.30) shows that
FL (12.31)
12.5.6 Overall system noise temperature

Refer to Fig.12.6a of a typical receiving system.

The system noise temperature referred to the input is

TS  Tant  Te1 
 L  1T0 L F  1T0
 (12.32)
G1 G1
Example 12.7
For the system shown in Fig12.6a, the receiver
noise figure is 12dB, the cable loss is 5dB, the LNA
gain is 50dB, and its noise temperature 150K. The
antenna noise temperature is 35K. Calculate the
noise temperature referred to the input.
Solution:
Main receiver, F  10  15.85
1.2

Cable, L  100.5  3.16

LNG, G  105

Hence, noise temperature referred to input

TS  35  150 
 3.16  1  290 3.16  15.85  1  290
  185K
105 105
Example 12.8
Repeat the calculation when the system of Fig.12.6a is
arranged as shown in Fig.12.6b.
Solution:
The cable now precedes the LNA, the equivalent noise
temperature referred to the cable input is
3.16  15.85  1  290
TS  35   3.16  1  290  3.16 150 
105
 1136 K

Ex.12.7 and 12.8 shows why it is important that the LNA

(an amplifier) be placed ahead of the cable.
12.6 Carrier-to-Noise Ratio

A measure of the performance of a satellite link is the

ratio of carrier power to noise power at the receiver
input.

Conventionally, the ratio is denoted by C/ N (or CNR),

which is equivalent to PR/PN.
In terms of decibels,

C 
 N    PR    PN 
… (12.33)
Equations (12.17) and (12.18) may be used for [PR]
and [PN], resulting in

C 
 N    EIRP    GR    LOSSES    k   TS    BN  … (12.33)
 

The G/ T ratio is a key parameter in specifying the

receiving system performance

G 
 T    G R    TS  dBK -1 … (12.35)
Since PN = kTNBN = NoBN, then

C   C 
N   N B 
   0 N

C
     BN 
 N0 
therefore
 C  C 
       BN  … (12.37)
 N0   N 

[C/N] is a true power ratio in units of decibels, and [BN] is

in decibels relative to one hertz, or dBHz.
Thus the units for [C/No] are dBHz.
Substituting Eq. (12.37) for [C/N] gives

C  G 
    EIRP       LOSSES    k  dBHz … (12.38)
 N0  T 
Ex.12.9:
In a link-budget calculation at 12GHz, the free
space loss is 206dB, the antenna pointing loss is
1dB, and the atmospheric absorption is 2dB. The
receiver feeder losses are 1dB. The EIRP is
48dBW. Calculate the carrier-to-noise spectral
density ratio
Solution:
Data are computed in tabular form, with losses
entered as negative numbers. Also, recall that
 k   228.6 decilogs thus   k   228.6 decilogs
The final result, 86.1 dBHz is the algebraic sum of the
quantities given.
12.7 The Uplink
The uplink  earth station is transmitting the signal and
the satellite is receiving it.
Equation (12.38) can be applied to the uplink, but with
subscript U denotes that the uplink is being considered.
C G 
   EIRP U      LOSSES U   k 
  … (12.39)
 N 0 U  T U

Eq (12.39) contains: the earth station EIRP, the

satellite receiver feeder losses, and satellite receiver
G/T. The freespace loss and other losses which are
frequency-dependent are calculated for the uplink
frequency. The resulting carrier-to-noise density ratio
given by Eq. (12.39) is that which appears at the satellite
receiver.
12.7.1 Saturation Flux Density

TWTA
The traveling-wave tube amplifiers (TWTA) are widely
used in transponders to provide the final output power
required to the transmit antenna.

Advantage of TWTA compared to other amplifiers is

that it can provide amplification over a wide bandwidth.

Disadvantage: Causes distortion due to the nonlinear

transfer characteristics of the TWTA
 Fig.7.18: Power transfer
characteristics of a TWT.

At low-input powers: output-input relationship is linear.

At higher power input: the output power saturates
Saturation point  the maximum power output
Intermodulation distortion: A serious effect of the non-
linear transfer characteristic of TWTA
Non linear transfer characteristic is expressed as:
2 3
e0  aei  bei  cei  ... … (7.1)

where,

a,b,c = coefficients which depend on the transfer characteristics

e0 = output voltage

ei = input voltage

The third order term, cei3, give rise to intermodulation

products (most significant contributor)
Suppose multiple carrier are present, separated by
. (Fig.7.20).

Considering frequencies 1 and 2, these will give rise

to 22-1 and 21-2 (As demonstrated in App.E of text
book)

Fig.7.20: Third order

intermodulation
products
Because 22-1 = , we obtain,

2 f 2  f1  f 2  f and 2 f1  f 2  f1  f

which fall on the neighboring carrier frequencies.

To reduce the intermodulation distortion, the

operating point of the TWT must be shifted closer to
the linear portion of the curve.
Reduction of input power  input backoff
 Fig.7.21: Transfer curve for a
single carrier and for one carrier
of a multiple-carrier input.
When multiple carriers are
present, the power output
around saturation, for any one
carrier is less than that achieved
with single-carrier operation.

Input backoff  the difference in decibels between the

carrier input at the operating point and the saturation
input required for single-carrier operation.
Output backoff  corresponding drop in output power
Output backoff is 5dB less than input backoff.
The flux density required at the receiving antenna to
produce saturation of the TWTA  saturation flux
density.
EIRP
From (12.6), flux density, M 
4r 2
In decibel notation
1
 M    EIRP  10 log 2 … (12.40)
4r

Free space loss

2 1
  FSL  10 log  10 log … (12.41)
4 4r 2
Substituting in to Eq.(12.40),
2
 M    EIRP   FSL  10 log … (12.42)
4

2
Let,  A0   10 log … (12.43)
4

 A0    21.45  20 log f  … (12.43)

Combining with (12.42) and rearranging

 EIRP   M    A0    FSL … (12.45)

Considering other propagation losses such as
atmospheric absorption loss, polarization mismatch
loss and antenna misalignment loss, (12.45) becomes

 EIRP   M    A0    FSL   AA   PL   AML … (12.46)

From (12.12) we have,

 LOSSES    FSL   RFL   AML   AA   PL

Thus (12.45) becomes,

 EIRP   M    A0    LOSSES    RFL … (12.47)

(12.47) is for clear-sky conditions, with minimum value
of [EIRP] which the earth station must provide to
produce a given flux density at the satellite.
We denote saturation values with subscript S, thus
(12.47) becomes,
 EIRPS U   S    A0    LOSSES    RFL … (12.48)

Refer Example 12.10

12.7.2 Input backoff

When a number of carriers are present simultaneously in

a TWTA, the operating point must be backed off to a linear
portion of the transfer characteristics to reduce
intermodulation distortion.
Suppose that the saturation flux density for a single-carrier
operation is known. Input BO will be specified for multiple-
carrier operation, referred to the single-carrier saturation
level.
The earth-station EIRP will have to be reduced by the
specified BO, resulting in the uplink value of

 EIRPU   EIRPS U   BO i … (12.49)

From (12.39), (12.48) and (12.49) we get
C  G 
    S    A0    BO  i      k    RFL … (12.50)
 N 0 U  T U

Refer Ex. (12.11)

The total gives the carrier-to-noise density ratio at the satellite receiver as 74.5dBHz
12.7.3 The earth station HPA
Earth station HPA has to supply radiated power plus
transmit feeder losses, TFL (including waveguide, filter,
coupler losses between HPA output and transmit
antenna).
From (12.3), power output of the HPA is given by

 PHPA    EIRP   GT   TFL … (12.51)

EIRP is given by (12.49).

If the earth station transmit multiple carrier, it must be
backed off, denoted by [BO]HPA.
The earth station HPA must be rated for a saturation
power output given by,
 PHPA ,sat    PHPA    BO HPA … (12.52)
12.8 Downlink

The downlink  the satellite is transmitting the signal

and the earth station is receiving it.
Equation (12.38) can be applied to the downlink, but with
subscript D to denote that the downlink is being
considered.

C G 
    EIRP D      LOSSES  D   k  … (12.53)
 N0  D T D

Eq. (12.53) contains: the satellite EIRP, the earth

station receiver feeder losses, and the earth station
receiver G/T. The free-space and other losses are
calculated for the downlink frequency.
The resulting carrier-to-noise density ratio given by Eq.
(12.53) is that which appears at the detector of the earth
station receiver.
Where the carrier-to-noise ratio is the specified quantity
rather than carrier-to-noise density ratio, Eq. (12.38) is
used. On assuming that the signal bandwidth B is equal
to the noise bandwidth BN, we obtain:

C  G 
N    EIRP  D   T    LOSSES  D   k    B  … (12.54)
 D  D
The required EIRP is 38dBW or 6.3kW
12.8.1 Output back-off

Output back-off is 5dB less than input back-off

 BO o   BO i  5dB

If the satellite EIRP for saturation condition is [EIRPS]D,

then
 EIRP D   EIRPS  D   BO o

(12.53) becomes
C G 
    EIRP 
S D   BO  o      LOSSES  D   k  … (12.55)
 N0  D T D
12.8.2 Satellite TWTA output

Satellite power amplifier, TWTA, has to supply the

radiated power plus the transmit feeder losses.
Losses include waveguide, filter and coupler losses
between the TWTA output and satellite’s transmit
antenna.
From (12.3), power output of TWTA is
 PTWTA    EIRP  D   GT  D  TFL D … (12.56)

The saturated power output rating of the TWTA is

 PTWTA  S   PTWTA    BO o … (12.57)
12.10 Combined Uplink and Downlink C/N Ratio

The complete satellite circuit consists of an uplink and a

downlink, as sketched in Fig. 12.9 a.

Figure 12.9
(a) Combined uplink
and downlink
(b) Power flow
diagram for (a)
Noise will be introduced on the uplink at the satellite
receiver input.
PNU = noise power per unit bandwidth
PRU = average carrier at the same point

The carrier-to-noise ratio on the uplink is

(C/ No)U = (PRU/PNU).

Note that power levels, and not decibels, are being

used.
PR = carrier power at the end of the space link
= the received carrier power for the downlink.
=  x the carrier power input at the satellite

where
 = the system power gain from satellite input to
earth station input
This includes the satellite transponder and
transmit antenna gains, the downlink losses, and the
earth station receive antenna gain and feeder losses.
The noise at the satellite input also appears at the earth
station input multiplied by , and in addition, the earth
station introduces its own noise, denoted by PND.
Thus the end-of-link noise is PNU + PND.

The C/No ratio for the downlink alone, not counting the
PNU contribution, is PR/PND, and the combined C/No ratio
at the ground receiver is PR/(PNU + PND).
The power flow diagram is shown in Fig. 12.9 b.

The combined carrier-to-noise ratio can be determined in

terms of the individual link values. To show this, it is
more convenient to work with the noise-to-carrier ratios
rather than the carrier-to-noise ratios, and these must be
expressed as power ratios, not decibels.
Denoting the combined noise-to-carrier ratio value by
No/C, the uplink value by (No/C)U, and the downlink
value by (No/C)D then,

PR
Equation (12.61) shows that to obtain the combined value
of C/N0, the reciprocals of the individual values must be
added to obtain the N0/C ratio and then the reciprocal of
this taken to get C/N0.

The reason for this reciprocal of the sum of the reciprocals

method is that a single signal power is being transferred
through the system, while the various noise powers which
are present are additive.

Similar reasoning applies to the carrier-to-noise ratio, C/ N.