Satellite Unit1

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UNIT-1

Introduction to Satellite Communications



Satellite Communication combines the

missileand microwave technologies

The space era started in 1957 with the

launching of the first artificial satellite

(sputnik)



Satellite Communications

• Satellite-based antenna(e) in stable orbit above earth.

• Two or more (earth) stations communicate via one or

more satellites serving as relay(s) in space.• Uplink: earth->satellite.

• Downlink: satellite->earth.

• Transponder: satellite electronics converting uplink

signal to downlink.



Satellite Communications

• Satellite Transponder is a microwave deviceconsisting of receiver, repeater and regenerator inorbit

• Satellite transmission involves sending signals tosatellite that receive, amplify, and transmit back to

earth



Capabilities of Satellite Transmission

• Point-to-point transmission – To transfer large volume of data

– Voice, data, etc communication

– Video conference• Point-to-multipoint transmission

– Data communication

– Internet – Video conference

• Broadcast services such as television



Satellite Network Configurations





Satellite Services

• FSS Fixed Satellite Services (VSATnetworks,..)

• MSS Mobile Satellite Services (Inmarsat

systems,...)• BSS Broadcasting Satellite Services ( TV,

DVB..)

• RDSS Radiodetermination Satellite Services

(GPS)



Satellite Orbits



Satellite OrbitsGEO

advantages:

- the satellite appears to be fixed (immovable) when viewed from the Earth, notracking required for earth station antennas

- about. 40% of the earth`s surface is in view from the satellite

disadvantages:

- high attenuation level (power loss) (200dB) on the path

- large signal delay (238-284ms)

- polar regions (latitudes > 81 deg.) are not covered

LEO

advantages:

- much smaller attenuation compare GEO satellites

- low signal delay

disadvantages:

- short period satellite visibility (through earth station), many times during theday

- Doppler effect

- many satellites are required for establishing continuous transmission



• Four different types of satellite orbits can be identifieddepending on the shape and diameter of the orbit:

• GEO: geostationary orbit,36000 km above earth

surface• LEO (Low Earth Orbit): 500 - 1500 km

• MEO (Medium Earth Orbit) or ICO (Intermediate

Circular Orbit): ca. 6000 - 20000 km

• HEO (Highly Elliptical Orbit) elliptical orbits

Orbits I



Orbits II

earth

km

35768

10000

1000

LEO

(Globalstar,

Irdium)

HEO

inner and outer Van

Allen belts

MEO (ICO)

GEO (Inmarsat)

Van-Allen-Belts:

ionized particels

2000 - 6000 km and

15000 - 30000 km

above earth surface



Geostationary satellites

• Orbit 35.786 km distance to earth surface, orbit in equatorial plane(inclination 0°)

complete rotation exactly one day, satellite is synchronous to earth rotation

• fix antenna positions, no adjusting necessary

• satellites typically have a large footprint (up to 34% of earth surface!),

therefore difficult to reuse frequencies

• bad elevations in areas with latitude above 60° due to fixed position above

the equator

• high transmit power needed

• high latency due to long distance (ca. 275 ms)

not useful for global coverage for small mobile phones and data

transmission, typically used for radio and TV transmission



LEO systems

• Orbit ca. 500 - 1500 km above earth surface

• visibility of a satellite ca. 10 - 40 minutes

• global radio coverage possible

• latency comparable with terrestrial long distance connections, ca. 5 - 10

ms

• smaller footprints, better frequency reuse• but now handover necessary from one satellite to another

• many satellites necessary for global coverage

• more complex systems due to moving satellites

• Examples:

• Iridium (start 1998, 66 satellites)

• Globalstar (start 1999, 48 satellites)



MEO systems

• Orbit ca. 5000 - 12000 km above earth surface

• comparison with LEO systems:

• slower moving satellites

• less satellites needed

• simpler system design

• for many connections no hand-over needed

• higher latency, ca. 70 - 80 ms

• higher sending power needed

• special antennas for small footprints needed

• Example:

• ICO (Intermediate Circular Orbit, Inmarsat) start ca. 2000



GEO vs LEO

GEO

• advantages:- the satellite appears to be fixed (immovable) when viewed from the Earth, no tracking

required for earth station antennas

- about. 40% of the earth`s surface is in view from the satellite• disadvantages:

- high attenuation level (power loss) (200dB) on the path

- large signal delay (238-284ms)- polar regions (latitudes > 81 deg.) are not covered

LEO• advantages:

- much smaller attenuation compare GEO satellites

- low signal delay• disadvantages:

- short period satellite visibility (through earth station),

- many times during the day

- Doppler effect

- many satellites are required for establishing continuous transmission



Communication Satellites



Overview of LEO/MEO systems

Iridium Globalstar ICO Teledesic

# satellites 66 + 6 48 + 4 10 + 2 288

altitude(km) 780 1414 10390 ca. 700

coverage global ±70° latitude global globalmin.elevation

8° 20° 20° 40°

frequencies[GHz

(circa)]

1.6 MS29.2 ↑

19.5 ↓23.3 ISL

1.6 MS ↑

2.5 MS ↓

5.1 ↑6.9 ↓

2 MS ↑

2.2 MS ↓

5.2 ↑7 ↓

19 ↓

28.8 ↑

62 ISL

accessmethod

FDMA/TDMA CDMA FDMA/TDMA FDMA/TDMA

ISL yes no no yesbit rate 2.4 kbit/s 9.6 kbit/s 4.8 kbit/s 64 Mbit/s ↓

2/64 Mbit/s ↑

# channels 4000 2700 4500 2500Lifetime[years]

5-8 7.5 12 10

costestimation

4.4 B$ 2.9 B$ 4.5 B$ 9 B$



Spectrum Allocation

Frequency Spectrum concepts:

• Frequency: Rate at which an electromagnetic wave reverts its polarity(oscillates) in cycles per second or Hertz (Hz).

• Wavelength: distance between wavefronts in space. Given in meters as:

λ = c/f

Where: c = speed of light (3x108 m/s in vacuum) f = frequency in Hertz

• Frequency band: range of frequencies.

• Bandwidth: Size or “width” (in Hertz) or a frequency band.

• Electromagnetic Spectrum: full extent of all frequencies from zero to infinity.



Radio Frequencies (RF)

• RF Frequencies: Part of the electromagnetic spectrumranging between 300 MHz and 300 GHz. Interesting properties:

– Efficient generation of signal power

– Radiates into free space

– Efficient reception at a different point.

Differences depending on the RF frequency used:

- Signal Bandwidth

- Propagation effects (diffraction, noise, fading)- Antenna Sizes



Microwave Frequencies

• Sub-range of the RF frequencies approximately from

1GHz to 30GHz. Main properties:

- Line of sight propagation (space and atmosphere).

- Blockage by dense media (hills, buildings, rain)

- Wide bandwidths compared to lower frequency bands.

- Compact antennas, directionality possible.

- Reduced efficiency of power amplification as frequency grows:

Radio Frequency Power OUT

Direct Current Power IN

R di F S t



Radio Frequency Spectrum

Commonly Used Bands



Frequency Bands



Frequency allocation

B a n d D o w n l in k B a n d s , G H z U p l in k B a n d s , G H z

U h f M i l i t a r y0 . 25 - 0 . 2 7 (A p p ro x im a t e ly )0 . 29 2 - 0 . 31 2 (A p p r o xi m a t e ly )

S B a n d

L Ba n dC B a n d - C o m m e rc ia l 3 . 7 - 4 .2 5 . 92 5 - 6 . 42 5

X B a n d - M i lit a ry 7 . 25 - 7 . 7 5 7 . 9 - 8 .4

K u B a n d - C o m m e rc ia l 11 .7 - 1 2 . 2 14 .0 - 1 4 . 5

K a B a n d - C o m m e rc ia l 17 .7 - 2 1 . 2 27 .5 - 3 0 . 0

K a B a n d - M il it a ry 20 .2 - 2 1 . 2 43 .5 - 4 5 . 5Q / V G e o s t a t i o n a r y 37 .5 - 4 0 . 5 47 .2 - 5 0 . 2

Q / V N o n - g e o s t a t io n a ry 37 .5 - 3 8 . 5 48 .2 - 4 9 . 2

W B a n d 66 .0 - 6 7 . 0 71 .0 - 7 2 . 0



C-Band Ku-band

Antennas



Satellite Link Performance Factors

• Distance between earth station antenna and satellite

antenna

• For downlink, terrestrial distance between earth station

antenna and “aim point” of satellite – Displayed as a satellite footprint

• Atmospheric attenuation

– Affected by oxygen, water, angle of elevation, and higher

frequencies



Applications

• Traditionally

– weather satellites

– radio and TV broadcast satellites

– military satellites

– satellites for navigation and localization (e.g., GPS)

• Telecommunication

– global telephone connections

– backbone for global networks – connections for communication in remote places or

underdeveloped areas

– global mobile communication

replaced by fiber optics



Current GEO Satellite Applications:

• Broadcasting - mainly TV at present

– DirecTV, PrimeStar, etc.

• Point to Multi-point communications

– VSAT, Video distribution for Cable TV

• Mobile Services

– Motient (former American Mobile Satellite),

INMARSAT, etc.



System Design Considerations

Basic Principles



Signals

• Signals:

– Carried by wires as voltage or current

– Transmitted through space as electromagnetic waves.

– Analog: Voltage or Current proportional to signal. E.g.Telephone.

– Digital: Generated by computers.

Ex. Binary = 1 or 0 corresponding to +1V or –1V.

Current Trends in Satellite



Current Trends in Satellite

Communications

• Bigger, heavier, GEO satellites with multiple roles

• More direct broadcast TV and Radio satellites

• Expansion into Ka, Q, V bands (20/30, 40/50 GHz)

• Massive growth in data services fueled by Internet

• Mobile services:

– May be broadcast services rather than point to point

– Make mobile services a successful business?



Orbital Mechanics

Part 1



Kinematics & Newton’s Law

• s = ut + (1/2)at 2

• v2 = u2 + 2at

• v = u + at

• F = ma

s = Distance traveled in time, t

u = Initial Velocity at t = 0

v = Final Velocity at time = t

a = Acceleration

F = Force acting on the object

Newton’s

Second Law



FORCE ON A SATELLITE

• Force = Mass × Acceleration

• Unit of Force is a Newton

• A Newton is the force required to accelerate

1 kg by 1 m/s2

• Underlying units of a Newton are therefore

(kg) × (m/s2)



ACCELERATION FORMULA

• a = acceleration due to gravity = µ / r 2 k m/s2

• r = radius from center of earth

∀ µ = universal gravitational constant G multipliedby the mass of the earth M E

∀ µ is Kepler’s constant and= 3.9861352 × 105 km3/s2

• G = 6.672 × 10-11 Nm2/kg2 or 6.672 × 10-20 km3/kg s2 in the older units



FORCE ON A SATELLITE : 2

Inward (i.e. centripetal force)

Since Force = Mass × Acceleration

If the Force inwards due to gravity = F IN then

F IN = m × ( µ / r 2

)

= m × (GM E / r 2 )



F 1(Gravitational

Force)

v (velocity)

Why do satellites stay moving and in

orbit?

F 2(Inertial-Centrifugal

Force)



Orbital Velocities and Periods

Satellite Orbital Orbital Orbital

System Height (km) Velocity (km/s) Period

h min s

INTELSAT 35,786.43 3.0747 23 56 4.091

ICO-Global 10,255 4.8954 5 55 48.4

Skybridge 1,469 7.1272 1 55 17.8

Iridium 780 7.4624 1 40 27.0



FORCE ON A SATELLITE



Forces acting on a satellite in a

stable orbit around the earth.

Gravitational force is inversely

proportional to the square of

the distance between the

centers of gravity of the

satellite and the planet the

satellite is orbiting, in this case

the earth. The gravitational

force inward (F IN, the

centripetal force) is directed

toward the center of gravity of the earth. The kinetic energy of

the satellite (F OUT, the

centrifugal force) is directed

diametrically opposite to the

gravitational force. Kinetic

energy is proportional to the

square of the velocity of the

satellite. When these inward

and outward forces are

balanced, the satellite moves

around the earth in a “free fall”

trajectory: the satellite’s orbit.

If F OUT = F IN

the object is in

FREE FALL



FREE FALL???

ORBIT LIMITS



ORBIT LIMITS

Geographical Coordinates



Geographical CoordinatesEarth Centric Coordinate System

The earth is at the center

of the coordinate system

Reference planes coincide

with the equator and the

polar axis



Orbital Plane Coordinates

The earth is at the

center of the coordinate

system but ………

Reference is the plane

of the satellite’s orbit



Balancing the Forces

Inward Force

r

mGM E

F 3

r

−=

−

Equation (2.7)

F −

G = Gravitational constant = 6.672 × 10-11 Nm2/kg2

M E = Mass of the earth (and GM E = µ = Kepler’s constant)

m = mass of satellite

r = satellite orbit radius from center of earth−

r= unit vector in the r direction (positive r is away from earth)



Balancing the Forces

Outward Force F −

2

2

dt

d

m F

−

−

=

r

Equation (2.8)

Equating inward and outward forces we find

2

2

3 dt

d

r

−−

=−

rr

µ

Equation (2.9), or we can write

032

2

=+

−−

µ r dt

d rr Equation (2.10)

Second order differential

equation with six unknowns:

the orbital elements



• We have a second order differential equation

• See text for a way to find a solution• If we re-define our co-ordinate system into polar

coordinates (see Fig.) we can re-write equation

as two second order differential equations in

terms of r 0 and ϕ 0

THE ORBIT



Polar Coordinates

In the plane of theorbit

Polar coordinate system in the plane of the satellite’s orbit. The plane of the orbit

coincides with the plane of the paper. The axis z 0 is straight out of the paper from the

center of the earth, and is normal to the plane of the satellite’s orbit. The satellite’s

position is described in terms of the radius from the center of the earth r 0 and the angle

this radius makes with the x 0 axis, Φo.

O



THE ORBIT

• We have a second order differential equation

• If we re-define our coordinate system into polar coordinates (see

Fig. 2.3) we can re-write equation (2.5) as two second order

differential equations in terms of r 0 and φ 0 .

and

THE ORBIT



THE ORBIT

• Solving the two differential equations leads to six constants

(the orbital constants) which define the orbit, and three laws

of orbits (Kepler’s Laws of Planetary Motion)

• Johaness Kepler (1571 - 1630) a German Astronomer and

Scientist

KEPLER’S THREE LAWS



KEPLER’S THREE LAWS

• Orbit is an ellipse with the larger body (earth) atone focus

• The satellite sweeps out equal arcs in equal time

(NOTE : for an ellipse, this means that the orbitalvelocity varies around the orbit)

• The square of the period of revolution equals a

CONSTANT ´ the THIRD POWER of SEMIMAJOR

AXIS of the ellipse



Review: Ellipse analysis

• Points (-c,0) and (c,0) are the foci.

•Points (-a,0) and (a,0) are the vertices.

• Line between vertices is the major axis.

• a is the length of the semimajor axis.

• Line between (0,b) and (0,-b) is the minor axis.

• b is the length of the semiminor axis.

12

2

2

2

=+b

y

a

x

222cba +=

Standard

Equation:

y

V(-a,0)

P(x,y)

F(c,0)F(-c,0) V(a,0)

(0,b)

x

(0,-b)

ab A π =

Area of ellipse:

KEPLER 1: Elliptical Orbits



The orbit as it appears in the orbital plane, The point O is the center of the earth and the

point C is the center of the ellipse. The two centers do not coincide unless the

eccentricity, e, of the ellipse is zero (i.e., the ellipse becomes a circle and a = b). The

dimensions a and b are the semimajor and semiminor axes of the orbital ellipse,respectively.

KEPLER 1: Elliptical Orbits

e = ellipse’s eccentricity

O = center of the earth (one

focus of the ellipse)

C = center of the ellipse

a = (Apogee + Perigee)/2



KEPLER 1: Elliptical Orbits (cont.)

Equation 2.17 in text:

(describes a conic section,

which is an ellipse if e < 1)

)cos(*11

1φ e

pr +=

e = eccentricity

e<1 ⇒ ellipse

e = 0 ⇒ circle

r 0 = distance of a point in the orbit to the

center of the earth

p = geometrical constant (width of the conic

section at the focus)

p=a(1-e2 )

ϕ 0 = angle between r 0 and the perigee

p



KEPLER 2: Equal Arc-Sweeps

Figure 2.5

Law 2

If t2 - t1 = t4 - t3

then A12 = A34

Velocity of satellite is

SLOWEST at APOGEE; FASTEST at PERIGEE



Kepler’s Laws – 1 & 2



KEPLER 3: Orbital Period

Orbital period and the Ellipse are related by

T 2 = (4 π 2 a3 ) / µ (Equation 2.21)

That is the square of the period of revolution is equal to a

constant × the cube of the semi-major axis.

IMPORTANT: Period of revolution is referenced to inertial space, i.e., to

the galactic background, NOT to an observer on the surface of one of the

bodies (earth).

µ = Kepler’s Constant = GM E



Kepler’s 3rd Law: T² = (4π²a³)/μ

μ μ == 3.986004418 × 103.986004418 × 1055 km/s²km/s²

Numerical Example 1



Numerical Example 1

The Geostationary Orbit:

Sidereal Day = 23 hrs 56 min 4.1 sec

Calculate radius and height of GEO orbit:• T2 = (4 π2 a3) / µ (eq. 2.21)

• Rearrange to a3 = T2µ /(4 π2)

• T = 86,164.1 sec

• a3 = (86,164.1) 2 x 3.986004418 x 105 /(4 π2)

• a = 42,164.172 km = orbit radius

•h = orbit radius – earth radius = 42,164.172 – 6378.14 = 35,786.03 km

S l Sid l D



Solar vs. Sidereal Day

• A sidereal day is the time between consecutive crossings of any

particular longitude on the earth by any star other than the sun.

• A solar say is the time between consecutive crossings of any

particular longitude of the earth by the sun-earth axis.

– Solar day = EXACTLY 24 hrs – Sidereal day = 23 h 56 min. 4.091 s

• Why the difference?

– By the time the Earth completes a full rotation with respect to an

external point (not the sun), it has already moved its center

position with respect to the sun. The extra time it takes to cross

the sun-earth axis, averaged over 4 full years (because every 4

years one has 366 deays) is of about 3.93 minutes per day.

LOCATING THE SATELLITE IN



LOCATING THE SATELLITE IN

ORBIT: 1

Start with Fig. 2.6 in Text ϕ o is the True

Anomaly

See eq. (2.22)

C is the

center of the

orbit ellipse

O is thecenter of the

earth

NOTE: Perigee and Apogee are on opposite sides of the orbit

LOCATING THE SATELLITE



LOCATING THE SATELLITE

IN ORBIT

• Need to develop a procedure that will allow the

average angular velocity to be used

• If the orbit is not circular, the procedure is to use aCircumscribed Circle

• A circumscribed circle is a circle that has a radius equalto the semi-major axis length of the ellipse and also

has the same center

L S lli i O bi



Locate Satellite in Orbit

η = Average angular velocity

E = Eccentric Anomaly

M = Mean Anomaly

M = arc length (in radians) that the

satellite would have traversed sinceperigee passage if it were moving

around the circumscribed circle

with a mean angular velocity η



ORBIT CHARACTERISTICS

Semi-Axis Lengths of the Orbit

2

1 e

pa

−

=where

µ

2h p =

and h is the magnitude of

the angular momentum

See eq. (2.18)

and (2.16)

( ) 2/121 eab −= where µ

C he

2

= See eqn.(2.19)

and e is the eccentricity of the orbit



ORBIT ECCENTRICITY

• If a = semi-major axis,

b = semi-minor axis, and

e = eccentricity of the orbit ellipse,

then

ba

ba

e +

−=

NOTE: For a circular orbit, a = b and e = 0

Time reference



Time reference

• tp Time of Perigee = Time of closest

approach to the earth, at the same time, time

the satellite is crossing the x0

axis, according to

the reference used.

• t- tp = time elapsed since satellite last passedthe perigee.

ORBIT DETERMINATION 1



ORBIT DETERMINATION 1:

Procedure:

Given the time of perigee t p, the eccentricity e

and the length of the semimajor axis a:

∀ η Average Angular Velocity (eqn. 2.25)

• M Mean Anomaly (eqn. 2.30)• E Eccentric Anomaly (solve eqn. 2.30)

• r o Radius from orbit center (eqn. 2.27)

∀ ϕ o True Anomaly (solve eq. 2.22)• x0 and y0 (using eqn. 2.23 and 2.24)

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Satellite Unit1