Satellite Communications System

Satellite Communications System Introduction: Satellite Communication System Satellite Orbits Kepler’s Laws Categories of Satellites Calculations Mari...
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Satellite Communications System Introduction: Satellite Communication System Satellite Orbits Kepler’s Laws Categories of Satellites Calculations Maria Leonora Guico Tcom 126 Lecture 12

Satellite Communication System  Utilizes radio frequencies in the microwave range as

medium and uses satellites to 'bounce' an earth-bound station's uplink signal back down to a receiving earth station. A satellite system consists of:  A transponder (a radio repeater in the sky)  A ground-based station to control this operation  A user network of earth stations (antenna systems on or near earth) that provide the facilities for transmission and reception of communication traffic through the satellite station

Satellite system link  Uplink  Path of the satellite signal from the earth-station

transmitter to the receiver of the satellite.  The freq. signal being transmitted from the earth station to the satellite is called uplink frequency  e.g: uplink freq. for C-band is 6 Ghz  Downlink  Path of the satellite signal from the satellite transmitter to the receiver on the earth  The retransmitted signal from the satellite to the receiving stations is called the down-link  e.g: downlink freq. for C-band is 4 GHz

Satellite communication system

• Uplink and downlink use different carrier frequencies to avoid interference • Frequency translation is done in the transponder

Satellite transponder  Satellite transponder acts like a repeater, consists of a

receiver and a transmitter. The main functions of a satellite transponder are:  To pick up the transmitted signal from the transmission on

the earth  To amplify the signal  To translate the carrier frequency to another frequency  To retransmit the amplified signal to the receiver on the earth

Satellite transponder Band pass filter

Low noise Amplifier (LNA)


Band pass filter (BPF)

Low power Amplifier

Local oscillator Frequency translator Earth station

Earth station

BPF – limits the total noise LNA amplifiers – receive signal and feed it to the frequency translator Freq. translator – convert the high-band uplink frequency to the lowband downlink frequency

Satellite Network Configurations

Ways to Categorize Communications Satellites  Coverage area

 Global, regional, national  Service type  Fixed service satellite (FSS)

 Broadcast service satellite (BSS)  Mobile service satellite (MSS)  General usage  Commercial, military, amateur, experimental

Frequency Bands Available for Satellite Communications

Satellite Orbit

Terms  Apogee – point farthest from earth

in a satellite orbit  Perigee – point closest to earth in a satellite orbit  Elevation angle - the angle from the horizontal to the point on the center of the main beam of the antenna when the antenna is pointed directly at the satellite  Coverage angle - the measure of the portion of the earth's surface visible to the satellite

Classification of Satellite Orbits  Circular or elliptical orbit  Circular with center at earth’s center  Elliptical with one foci at earth’s center  Orbit around earth in different


 Equatorial orbit above earth’s equator  Polar orbit passes over both north and

south poles  Other orbits referred to as inclined orbits

Basic Orbits

Classification of Satellite Orbits (2)  Altitude of satellites  Geostationary orbit (GEO) in which satellite appears to remain stationary at a point above the equator approx 36,000 km  Intermediate Circular Orbits (ICO), or Medium Earth Orbits (MEO)  Low earth orbit (LEO) at altitude less than 1500 km  Highly Elliptical Orbits (HEO) orbits for earth applications were initially exploited by the Russians to provide communications to their northern regions not in coverage by their geo satellite networks. HEOs typically have a perigee at about 500 km above the surface of the earth and an apogee as high as 50,000 km. The orbits are inclined at 63.4 degrees in order to provide communications services to locations at high northern latitudes.

Basic Satellite Orbits Three basic types of orbits are: 1. Polar orbit  Used for navigation, weather satellite, meteorological, etc.  Not used for telecommunication purposes 2. Elliptically inclined orbit  Used for Russian domestic systems, with inclination of 63 degrees and a 12 hour orbit period, but visible for 8 hours only  Three satellites are needed for continuous coverage 3. Circular equatorial orbit  Called geosynchronous orbit  Angular speed is equal to the rotational speed of the earth.  Appears stationary or motionless over a fixed point on the earth’s surface.  The satellite is visible from 1/3 of the earth’s surface, so three satellites are needed for full coverage of the earth

Orbital Calculations • Satellite orbiting the earth must satisfy the equation:


4x10 11 (d + 6400)

where : v = velocity in meters per second d = distance above the earth’s surface in km

• Period of orbit can be found by dividing the circumference by the orbital velocity:

T = c/ v

Example 1 Find the velocity and the orbital period of a satellite in a circular orbit a) 500 km above the earth’s surface b) 36,000 km above the earth’s surface (approx ht. of a geosynchronous satellite)

Answers: a. V = 7.6 km/s b. T = 1.6 hours

Kepler’s Law  Law of planetary motion describes the shape of the orbit, the

velocities of the planet and the distance of the planet (wrt to the sun)  Kepler’s First Law:  A satellite will orbit a primary body following an elliptical path

 Kepler’s Second Law (Law of areas):  for equal intervals of time, a satellite will sweep out equal areas

in the orbital plane, focused at the barycenter (center of mass)

GEO Orbit  Advantages of the the GEO orbit  No problem with frequency changes  Tracking of the satellite is simplified  High coverage area

 Disadvantages of the GEO orbit  Weak signal after traveling over 35,000 km  Polar regions are poorly served  Signal sending delay (about half a second for round trip) is


LEO Satellite Characteristics  Circular/slightly elliptical orbit

 Orbit period ranges from 1.5 to 2 hours  Round-trip signal propagation delay less than 20 ms  Maximum satellite visible time up to 20 min  System must cope with large Doppler shifts

LEO Categories  Little LEOs  Frequencies below 1 GHz  5MHz of bandwidth  Data rates up to 10 kbps  Aimed at paging, tracking, and low-rate messaging  Big LEOs  Frequencies above 1 GHz  Support data rates up to a few megabits per sec  Offer same services as little LEOs in addition to voice and positioning services

MEO Satellite Characteristics  Circular orbit at an altitude in the range of approx 5000 to

12,000 km  Orbit period of 5 – 12 hours  Diameter of coverage is 10,000 to 15,000 km  Round trip signal propagation delay less than 50 ms  Maximum satellite visible time is a few hours

Geostationary Satellites  Orbit above the equator; orbit period is 24 hours

 Antennas in Northern Hemisphere point south and vice

versa  Antenna is aimed at satellite by adjusting its azimuth (horizontal position) and elevation (vertical angle with the ground). Typically requires two motors for automatic positioning to track orbital satellites.  For geostationary satellites, azimuth and elevation adjustment may be done once and then fixed in place.

Declination Angle  Declination is the amount by

which the antenna axis is offset from the earth’s axis  Addition of declination adjustment allows a polar antenna mount to track satellites above the equator  Declination angle is computed by:

R sin L ) θ = arctan ( H  R(1  cos L)

where: R = radius of the earth (6400 km) H = height of the satellite above the earth (36 x 103km) L = earth-station latitude

Example 2 Calculate the angle of declination for an antenna using a polar mount at a latitude of 450.

Answer: Angle of declination = 6.81 degrees

Path Length  Path length from earth to geostationary satellite is 36,00 km

(earth stn at equator, same longitude as satellite)  Increases to 41,700 km for earth station close to either the North or South Pole  Actual path length is found from equation:


(r  h)  (r cos  )  r sin  2


Where: d = distance to satellite in km r = radius of earth (6400 km) h = height of satellite above equator (36x103 km) θ = angle of elevation to and at same longitude as satellite, decreasing to zero at about 800 latitude)

Example 3 Calculate the length of the path to a geostationary satellite from an earth station where the angle of elevation is 300.

Answer: d = 39x103 km

Path Loss Calculations PR (dB)  GT (dBi)  GR(dBi)  (32.44  20log d  20log f ) PT where: PR/PT (dB) = ratio of received to transmitted power GT (dBi) = gain of transmitting antenna wrt isotropic radiator GR (dBi) = gain of receiving antenna wrt isotropic radiator d = distance between transmitter and receiver (km) f = frequency (MHz)

Example 4 A satellite transmitter operates at 4 GHz with a transmitter power of 7 W and an antenna gain of 40 dBi. The receiver has an antenna gain of 30 dBi, and the path length is 40,000 km. Calculate the signal strength at the receiver.

Answer: PR = -88 dBm

Figure of Merit  Received signal levels are very weak for geostationary

satellite systems, hence, require low-noise receivers and high-gain antennas for both satellite and earth stations  Figure of Merit called G/T has evolved to measure the combination of gain and equivalent noise temperature  Gain and noise temperature should be taken at same reference point usually the LNA input

Figure of Merit

G / T (dB)  GR(dBi) 10log(Ta  Teq) where: G/T(dB) = a figure of merit for the receiving system GR (dBi) = gain of antenna system wrt an isotropic radiator Ta = noise temperature of antenna Teq = equivalent noise temperature of receiver

Noise in Antenna  Mainly from extraterrestrial sources (stars, sun) and from the

atmosphere  Sky noise temperature for earth-station receiving antenna is typically 20K or less  Losses in antenna system contribute to noise temperature

Antenna Noise Temperature Ta = (L – 1) 290 + Tsky L Where: Ta = effective noise temp. of antenna and feedline, referenced to receiver antenna input (K) L = loss in feedline and antenna as a ratio of input to output power (not in DB) Tsky = effective sky temp. (K)

Example 5 A receiving antenna with a gain of 40 dBi looks at a sky with a noise temperature of 15K. The loss between the antenna and the LNA input, due to the feedhorn, is 0.4 dB, and the LNA has a noise temperature of 40K. Calculate G/T.

Answer: G/T (dB) = 20.6 dB

Carrier-to-Noise Ratio C (dB)  EIRP(dBW )  FSL(dB)  Lmisc  G / T  k (dBW )  10log B N where: C/N (dB) = carrier-to-noise ratio in decibels EIRP(dBW) = effective isotropic radiated power in dBW = transmitter power (dBW) + antenna gain (dBi) – feedline losses FSL (dB)

= free space loss in dB, w/o compensation for antenna gains


= miscellaneous losses, such as feedline losses, in dB


= figure of merit

K (dBW)

= Boltzmann’s constant expressed in dBW (-228.6 dBW/HzK)


= bandwidth in hertz

 dBW/K-Hz This is a common term used when

analyzing carrier-to-noise (C/N) in a communications link such as a satellite link budget.  Boltzmann's constant: 1.38 x 10-23 watt seconds/K. Since the actual unit of Hz is 1/sec, Boltzmann's constant can also be stated in the linear-form units watt/HzK or in the dB units dBW/HzK. The dB form of Boltzmann's constant is : 10 Log(1.38*10E-23) = -228.6 dBW/HzK

Example 6 The receiving installation whose G/T is 20.6 dB is used as a ground terminal to receive a signal from satellite at a distance of 38,000 km. The satellite has a transmitter power of 50 watts and an antenna gain of 30 dBi. Assume that losses between the satellite transmitter and its antenna are negligible. The frequency is 12 GHz. Calculate the carrier-to-noise ratio at the receiver, for a bandwidth of 1 MHz.

Answer: C/N (dB) = 30.6 dB

Satellite Footprint

Propagation Time  Disadvantage of geostationary satellites over terrestrial

microwave links or fiber optic cable particularly for applications like telephony and full-duplex data communications

Example 7: Telephone communication takes place between two earth stations via a satellite that is 40,000 km from each station. Suppose a person in station 1 asks a question and a person in station 2 answers immediately, as soon as he hears the question. How much time elapses between the end of person 1 and the beginning of person 2’s reply, as heard by person 1? Answer: 0.26s x 2 = 0.52 second

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 (Figure 9.6)

 Atmospheric attenuation  Affected by oxygen, water, angle of elevation, and higher frequencies