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.

The center of mass of the two-body system, termed the barycenter, is always centered on one of the foci

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

• 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

a

bae

22

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 orbitis 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

23

na

a = semimajor axis (meters)n = mean motion of the satellite (radians per second) = earth’s geocentric gravitational constant. = 3.986005 1014 m3/sec2

… (2.2)

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

(2.4)n

P2

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: The mean motion, in rad/ day, is

dayn

1

2:

Thus,sec

10272.7 5 radn

Also, 2314 sec10986005.3: m

Kepler`s 3rd Law gives 3

1

2:

na

kma 42241

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 directorbit. 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.

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 (oblatespheroid)

30 an

Taking account of earth`s oblateness, mean motion, n, is modified to

5.122

21

01

sin5.111

ea

iKnn

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

sec2

nPA

With n in radians/second

… (2.8)

… (2.9)

If n is known, equation 2.8 can be used to solve a, by solving the root of the following equation

01

sin5.1115.122

21

3

ea

iK

an

… (2.10)

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

K will have the same units as n.

2231

)1( ea

nKK

… (2.11)

The rate of regression of the nodes

)(cos iKdt

d

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

… (2.12)

)sin5.22( 2 iKdt

d

The rate of rotation of the line of apsides:

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:

)( 00 ttdt

d

)( 00 ttdt

d

… (2.13)

… (2.14)

… (2.15)

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

3

2

0'00

00

ttnn

naa … (2.16)

The mean anomaly is also changed. An approximate expression for the amount by which it changes is

20

'0

2tt

nM … (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 referenceto 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 thereforerequired.

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. 2.62. Various measures of time3. The perifocal coordinate system, which is based

on the orbital plane4. The geocentric-equatorial coordinate system,

which is based on the earth’s equatorial plane5. 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 approximationto 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

)(24

1

3600

seconds

24

minutes hoursUTday … (2.18)

This is converted to degrees as

dayo UTUT 360 … (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 measuredin 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

JC

JDJDT ref

… (2.20)

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

ve

ear

cos1

1 2

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.

Here, n is the mean motion, and T is the time of perigee passage.

)( TtnM

… (2.23)

… (2.24)

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:

Therefore,

Hence, substituting this in Eq. (2.24) gives

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.

)( 00 TtnM

n

MtT 0

0

)( 00 ttnMM

… (2.25)

… (2.26)

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 = 14.23304826rev/ 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

days 25223.796044

360614.2330482

246.6853 - 52223.796884 T

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

EeEM sin

0)sin( EeEM

This equation requires an iterative solution, preferably by computer to solve for E as the root of the equation

… (2.27)

… (2.28)

Once E is found, v can be found from an equation known as Gauss’equation, which is

It may be noted that r, the magnitude of the radius vector, also can be obtained as a function of E and is

For near-circular orbits where the eccentricity is small, an approximation for v directly in terms of M is

Note that the first M term on the right-hand side must be in radians, and v will be in radians.

2tan

1

1

2tan

E

e

ev

… (2.29)

Eear cos1 … (2.30)

MeMeMv 2sin4

5sin2 2 … (2.31)

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 inthe 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 thatcoordinates 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

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

… (2.32)

Example 2.15, page 46 of textbook

2.9.6 The geocentric-equatorial coordinate system

The fundamental plane is the earth’s equatorialplane. 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 PQWframe to the IJK frame:

Q

p

K

J

I

r

r

r

r

r

R~ … (2.33a)

where the transformation matrix R˜ is given by R

i

i

i

i

i

i

sincos

coscoscossinsin

coscossinsincos

sinsin

cossincoscossin

)cossinsincos(cos~

R

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

UTTTGST 20004.07689.000,366910.99

… (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 byLST = 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 kmbE = 6365.755 km

The eccentricity of the earth is

08182.022

E

EEE a

bae … (2.38)

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

EE

E

e

aN

22 sin1

LSTlLSTHNR EI coscoscos

LSTlLSTHNR EJ sinsincos

zHeNR EEK sin1 2

… (2.39)

… (2.40)

… (2.41)

… (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.

K

J

I

EEE

EEE

Z

E

S

LSTLST

LSTLST

LSTLST

sinsincoscoscos

0cossin

cossinsincossin

… (2.44)

Geocentric angle, E is given by

l

zE arctan … (2.45)

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 latitudeand is given by

EEHE e tan1tan 2)0( … (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

222ZES

Zarcsin

… (2.47)

… (2.48)

We define an angle as,

Then the azimuth depends on which quadrant is in and is given by Table 2.2.

S

E

arctan

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

SSSS hHh

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,

SSE

E

e

aN

22 sin1

… (2.50)

… (2.51)

LSThNr SSI coscos

LSThNr SSJ sincos

SSEK heNr sin1 2

… (2.52)

… (2.53)

… (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

31

2

2

4

P

aGSO… (3.1)

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

kmaGSO 42164 … (3.2)

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

kmaE 6378 … (3.3)

The geostationary height is

EGSOGSO aah

… (3.4)

km786,35

6378164,42

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 theearth.

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, andlatitudes 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 negligibleeffect on the look angles, and the average radius of the earth will be used, R.

kmR 6371 … (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 tothe 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.

090a … (3.6)

Ec 90 … (3.7)

SSEB … (3.8)

To summarize the spherical triangle

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

EBb coscosarccos … (3.9)

b

BA

sin

sinarcsin … (3.10)

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:

Applying the sine rule for plane triangles to the triangle of Fig. 3.2 b,the angle of elevation is

Figure 3.2b

)cos(22 baRaRd GSOGSO … (3.11)

b

d

aEl GSO sinarccos … (3.12)

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:

EEl 090 … (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),

Eb

from Eq. (3.12), or Fig 3.5 c

EGSO

d

aEl sincos 0 … (3.14)

Figure 3.5(c)

Combining Eqs. (3.13) and (3.14) gives the required angle of tilt as:

EEGSO

d

a

sinarccos90 … (3.15)

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

)cos(22 baRaRd GSOGSO

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

)cos(22EGSOGSO aRaRd … Eq (3.11) with b = E

E

GSO

d

aEl sinarccos0

… Eq (3.12) with b = E

deg790 0 EEl

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:

3.81164,42

6378arccosarccos

GSO

E

a

a … (3.16)

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), and2. 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 inthe direction of arrival, propagation delay, dispersion, frequency change, and polarization rotation.

All these effects decrease as frequency increases.Only the polarization rotation and scintillationeffects 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 specifiedvalues 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

kmdBaR bp / … (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:

dB LA … (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 climaticzones.

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

El

hhL oR

S sin

… (4.4)

The effective path length is given in terms of the slant length by

pS rLL … (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

ElLL SG cos … (4.6)

Putting all the factors into one equation, we obtain the rain attenuation (decibels) as:

dB rLaRA pSb

pp … (4.7)

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 differencebetween these two powers is accounted for.

Link budget calculations are usually made using decibel or decilog quantities. These are explained in App. G.

Chapter 12: The Space Link

12.1 Introduction

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

24 r

GPSM

… (12.1)

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

SGPEIRP … (12.2)

EIRP is often expressed in decibels relative to one watt, or dBW. Let PS be in watts; then

dBW GPEIRP S … (12.3)

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

The isotropic gain for a paraboloidal antenna is

2472.10 fDG … (12.4)

Where,

f

D

is the carrier frequency

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

24 r

EIRPM

… (12.6)

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:

effMR AP

… (12.7)

44

2

2RG

r

EIRP

2

4))((

rGEIRP R

wherer = 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

4log10

r

GEIRPP RR … (12.8)

Free space loss is given by

2

4log10

r

FSL … (12.9)

fr log20log204.32 … (12.10)

Equation (12.8) can then be written as

FSLGEIRPP RR … (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 andthe 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, oneat 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 adecibel, 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

PLAAAMLRFLFSLLOSSES … (12.12)

The decibel equation for the received power is

LOSSESGEIRPP RR … (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

NNN BkTP … (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)

joules kTB

PN N

N

N 0… (12.14)

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

Antenna pointing overhead

Antenna pointing just above the horizon

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.

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

Peak coincides with peaks in atmospheric absorbtion loss

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

Where,

N0,ant = Input noise energy from the antenna (Joules). In noise power per unit bandwidth

Tant = Antenna noise temperature

k = Boltzman constant,

The input noise energy from the antenna is:

antant kTN ,0

The output noise energy, No,out,

eantout TTGkN ,0

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

eantin TTkN ,0

12.5.3 Amplifiers in cascade

The cascade connection is shown in Fig 12.4b

For this, the overall gain is

21GGG

Cascaded amplifiers

The noise input to amplifier 2 from the preceding stages is

The noise energy of amplifier 2 referred to its own input 2ekT

11 eant TTkG

Thus, total noise energy referred to amplifier 2 input is

2112,0 eeant kTTTkGN

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

1

2,01,0 G

NN

1

21 G

TTTk e

eant

(12.21)

Define system noise temperature, Ts, where

SkTN 1,0

TS is given by,

1

21 G

TTTT e

eantS

(12.22)

(12.23)

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,

(12.24)...21

3

1

21

GG

T

G

TTTT ee

eantS

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,

0,0 FGkTN out (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,

00 FGkTTTGk e

or 01 TFTe (12.26)

Noise figure is F expressed in decibels.

FFFigure Noise log10 (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

0,NWx

rad kTL

kTN (12.28)

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

(12.29)

LTT xNW

110,

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

(12.30) 1, LTT xiNW

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

(12.31)LF

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

(12.32)

1

0

1

01

11

G

TFL

G

TLTTT eantS

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, 85.1510 2.1 F

Cable, 16.310 5.0 L

LNG, 510G

Hence, noise temperature referred to input

KTS 185

10

290185.1516.3

10

290116.315035

55

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

510

290185.1516.315016.3290116.335

ST

K1136

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,

NR PPN

C

… (12.33)

Equations (12.17) and (12.18) may be used for [PR] and [PN], resulting in

NSR BTkLOSSESGEIRPN

C

… (12.33)

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

1-SR dBK TG

T

G

… (12.35)

Since PN = kTNBN = NoBN, then

NBN

C

N

C

0

… (12.37)

NBN

C

0

therefore

NBN

C

N

C

0

[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

… (12.38) dBHz kLOSSEST

GEIRP

N

C

0

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

decilogs 6.228k thus decilogs 6.228 k

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.

… (12.39) kLOSSEST

GEIRP

N

CU

UU

U

0

Eq (12.39) contains: the earth station EIRP, thesatellite 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 satellitereceiver.

12.7.1 Saturation Flux Density

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

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:

...320 iii cebeaee … (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 22-1 and 21-2 (As demonstrated in App.E of text book)

Fig.7.20: Third order intermodulation products

Because 22-1 = , we obtain,

ffff 2122 and ffff 1212

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.

24 r

EIRPM

In decibel notation

24

1log10

rEIRPM

Free space loss

2

2

4

1log10

4log10

rFSL

… (12.40)

… (12.41)

From (12.6), flux density,

Substituting in to Eq.(12.40),

4

log102

FSLEIRPM… (12.42)

Let,

4

log102

0 A … (12.43)

fA log2045.210 … (12.43)

Combining with (12.42) and rearranging

FSLAEIRP M 0… (12.45)

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

AMLPLAAFSLAEIRP M 0 … (12.46)

From (12.12) we have,

PLAAAMLRFLFSLLOSSES

Thus (12.45) becomes,

RFLLOSSESAEIRP M 0 … (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,

RFLLOSSESAEIRP SUS 0 … (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

iUSU BOEIRPEIRP … (12.49)

From (12.39), (12.48) and (12.49) we get

… (12.50) RFLkT

GBOA

N

C

UiS

U

0

0

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

TFLGEIRPP THPA … (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,

HPAHPAsatHPA BOPP , … (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.

… (12.53) kLOSSEST

GEIRP

N

CD

DD

D

0

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

… (12.54) BkLOSSEST

GEIRP

N

CD

DD

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

dBBOBO io 5

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

(12.53) becomes

… (12.55) kLOSSEST

GBOEIRP

N

CD

DoDS

D

0

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

DDTDTWTA TFLGEIRPP … (12.56)

The saturated power output rating of the TWTA is

oTWTASTWTA BOPP … (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 inputThis 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.